- @ et of 7 a ee a ig tig Pots uit neue nt Aft aT . eee Bie t i os ME tect state ; i af i 21 3 DEN + i a ‘ ete fy, hi \j ea iS see ba } " iy mB Pete fis i “ » ¥ MES Steet al U) Boox_/_ 215 ‘ AccEssIon_SO0.3/2. woe z + a < Pa ¥ mt CONCRETE PLAIN AND REINFORCED CONCRETE PLAIN AND REINFORCED VoL. IL. THEORY AND DESIGN OF CONCRETE AND REINFORCED STRUCTURES BY Tue Late FREDERICK W. TAYLOR, SANFORD E. THOMPSON, 5.B. AND EDWARD SMULSKI, C.E. With a Chapter by HENRY C. ROBBINS FOURTH EDITION TOTAL iSSUE, THIRTY THOUSAND NEW YORK TOHN WILEY’ & SONS, Ino. Lonpon: CHAPMAN & HALL, Limrrep 1925 CopyRIGHT, 1925 By E. W. CLARK 38rd AND SANFORD E. THOMPSON Copyrighted in Great Britain Printed a Lies mAs PRESS OF Et 1/28 BRAUNWORTH & CO BOOK MANUFACTURERS BROOKLYN, N. Y, PREFACE TO FOURTH EDITION Tuis treatise is adapted for use both by the practical engineer and as a text-book for students of engineering. As a text-book it provides the student with material that is an essential part of his equipment for any line of professional engineering work. The Fourth Edition presents an entirely rewritten book in three volumes. The first volume treats of reinforced concrete design and construction; the second covers concrete materials and con- struction; and the third, special structures in reinforced concrete. Prefaces to Volumes II and III will be found at the beginning of these volumes. The authors, in preparing the work, have drawn upon their long and varied experience and practice as engineers. Because of this practical experience, they can speak with authority and can make specific recommendations with confidence. Reference to the Table of Contents indicates the breadth and scope of the treatment. The numerous illustrations are selected, not as pictures, but each with the definite purpose of conveying information on a particular point. The diagrams indicate the action of stresses. The line drawings show construction details based on the experience of the authors, to indicate the best way of attaining a required result. The photographs have been selected to show typical structures or structural details. The treatment of flat slabs, which the engineer—even one having broad experience in other types of construction—finds so difficult to design properly, is particularly comprehensive and practical. Not only are the theory and methods of the main design presented in simple and usable form, but all of the supplementary features, such as column heads, wall panels, openings, and beams, are taken up in practical fashion and with numerous illustrations. The treatment of column design includes the various types and adaptation to special uses and Vv Sf — vi PREFACE various City Building Codes. The portions dealing with founda- tions, footings, and piles cover all conditions. The treatment of building construction is particularly compre- hensive, covering not merely general structural features but also intimate consideration of various types of design, with special chap- ters on Wall Bearing Construction, Basement Walls, Roofs, Stair- ways, Fire Exits and Elevator Shafts, Steel Window Sash, Struc- tural Plans, Different Types of Buildings, and Chimneys. Two particularly interesting and novel chapters are those on the construc- tion in concrete of Theaters and Auditoriums and the Architectural Treatment of Exterior and Interior of Reinforced Concrete Build- , ings, the latter chapter prepared by Mr. Henry C. Robbins. SANFORD EH. THOMPSON, EDWARD SMULSKI. New York, June, 1925. CONTENTS CHAPTER I MATERIALS AND Mrtruops ror MAKING CONCRETE PAGE ET ei vy wilh hw OER se ee re we wa pe eee 1 (EP ge toy y 2 =o a FE on = ae 2 re ah ohn isk ris ein ap win ace as wa doef ale vide ee 2 Coarse Aggregates............ PI WRI. Wer Pat Nee honk home Soe 3 I re es eee en oS hes Hee ee ak EEL 4 Tt ee eG re ea ee i ee a ee 4 SEO CP ere BTACLCTISUICS «5 o5 erin ac ks ye cae ee ee CIN 5 CHAPTER II REINFORCEMENT Metre remiorcne Miceli... ae Ey) s ks | tadib ee fee nile p yee tenn ee geen a Selection of Grade of Steel for Reinforcement.......... EET ONG OH ee 8 Os MOSS nSG) oe 2) 001) fly oe rae rete eerta arr 10 CHAPTER III Tests oF REINFORCED CONCRETE eT ieee ce cee Cle en ch ees ee A ee Le eS 35 Tests of Beams to Determine Effect of Diagonal Tension................. 38 behavior of Beam Failing by Diagonal Tension...............5.......... 46 Pee without near Reinforcement... i... Gi. cna gene Sade a eee le 46 Beams Reinforced for Tension and Compression...........-.--++-+++++-- 48 fests ot Bond between Concrete and Steel. ......-0. 6.206262 e cece eee eds 51 SEAS eae Lng. a ANS bh AP. TRAY Seley eee e Peres 51 og. eM keyosrce orot | 2 S79 7 a ee ee i ra a gre rere rorare rier 56 Splices of Tensile Reinforcement at Points of Maximum Stress............ 61 Ree Tee ee enh eo oe eR OED, PS Oe 61 Peer Otis GCAMS) 6c cce. cies da ee Coe wow ee pede oietwaal Pe 64 vill CONTENTS PAGE Distribution of Concentrated Load on Wide Slabs.......----.+++++++s505 69 Tests to Determine Distribution of Load by Slab to Joists............-++- 74 Tests of. Plain Concrete Columns... .. 0. 2omn- 3-5 ++ < 06 =n eee a 76 Tests of Columns Reinforced with Vertical Steel. ........-..------+-+05 78 Tests of Spiral Columns... 2. . )26). 20 22 ey 21 ©) e seein ares ener 82 Tests of Square Columns with Rectangular Bands... cds Se aAeee aegis coe 88 Tests of Columns with Structural Steel Reinforcement.........--.+.+---: 88 Tests of Long Columns... . 05.0. 062 pir >: +: + ee 93 Torsional Resistance of Concrete and Reinforced Concrete te, sree eee 94 Tests of Flat Slab Construction. ....0. 00... -5 + >>> > | =e 98 Test of Flab Slab—Two-way System. .. 2.00... .)2 54+ +2) 5 ie seats 99 Test of Flat Slab-—Smulski System. .....+..--/.-+++ 500% >= 73 ea 104 Tests of Reinforced Concrete Buildings under Load . a. . us boeken 113 Tests of Octagonal Cantilever Flat Slabs.......-.-++-++++2srerssrsseces 116 CHAPTER IV THEORY OF REINFORCED CONCRETE General Principles of Reinforced Concrete Beams. ... +15 > 3: eens 123 ‘Assumptions.......--++2:e0s0ree sce t ett en ieee 126 Analysis of Rectangular Beams. ......-----++++++ssrrttcrss ster stseees 126 Notation: ....seseee + cece ewe tere emised ons ¢ scleen cyt: in 128 Formulas for T-Beams.....4- 2-6) 00-4 29RE TEU «>: -e 133 Reinforced Concrete Beams with Steel in Top and Bottom. sea 137 Wedge-shaped Beams........... 22000 272.005 5s « 5 » vip sinter 140 T-Beams with Unsymmetrical Flange..........-.+++++ssssesrtressctss 142 Shearing Stresses in a Beam or Slab.......-+-----+sssr erst uses sees 143 Diagonal Tension... 02s ety Sed re eee = «00 3 vee ea 147 Column Formulas: :.... $$ 02542 foe ee ob te fos 8 + oe eS ee 158 Members Subjected to Direct Tension......---+--+++ssserssr ests .. 162 Notation... c.cccue ces ce ce ee oe memo ce esis sl 9) chan 167 Plain Concrete Section under Direct Stress and Bending: /)5. eee 169 Reinforced Concrete Section under Direct Stress and Flexuredss-2),. so 173 Rectangular Section with Symmetrical Reinforcement.........------+++-: 174 How to Combine Thrust with Bending Moment........------+++:s+++00: 187 Members Subjected to Direct Tension and Bending. .:.; ..... 0:6 in een 189 Flat Slab Theory... .....:0+> ++ sen nin eo: tenet ot) 90 eee a ee 194 CHAPTER V REINFORCED CONCRETE DESIGN Ratio of Moduli of Elasticity. ......----+- sees e er ereceeeecees ss aan te oe 201 Formulas for Design of Rectangular Beams.........+++++ererreresrcrres 203 Design of Slabs... 20...) ccc.eee ce temer tess 00: + =) ai alee! 208 Design of T-Beams.... 25.0004 0-525 + heroes 0's > =)s Ap tiie 215 CONTENTS Ix PAGE eon Compression Reinforcement. ov iliac. vid Deda de Oe cee 230 Rectangular Beam with Compression Reinforcement...............++.55. 232 Examples of Beams with Steel in Top and Bottom...................... 240 (ANSE TS ee ee 241 Meemalamior Diagonal Tension... . 20... ee esi neem tees 247 Semi-graphical Method of Spacing Stirrups............ 6... s seer eee 252 Diagonal Tension for Moving Loads.............. 000. e eee eee eee 258 NI BE ECCT OLE La Saleen souk ka i eo Rk te ed Le te alors 260 Anchorage and Splicing of Reinforcement.........-... 600-5 seer eee ee. 265 MET SCS CS cee oe eo de eee ele le ea BEGG Pea GG ea BAS 270 Protective Covering for Reinforcement.............---2 see ee eee eee ees EE Clear Distance between Parallel Bars in a Beam..............----. 0005: 213 Bending Moments for Use in Design of Beams...........----++.0++ +005. 275 ee saa aa oe 277 Formulas for Single Span Beams...... 6... -- sees eee eee teens 278 Continuous Beams of Equal Spans...........-. ee eee eee teen ines 278 Continuous Beams of Equal Spans Running into Columns................ 279 Continuous Beams of Unequal Spans or with Non-uniform Loading........ 280 Effect of Varying Moments of Inertia on Bending Moment.............-- 281 ee are ON tIN0Us DSSTIS owe. eg ee ee eee 281 Process of Designing Continuous Rene ee Ce ree eee 285 Points at which Horizontal Reinforcement Should be Bent..............-. 290 Reinforcement for Temperature and Shrinkage Stresses. .............---- 298 CHAP IU VI ‘Design or Fiat SLAB STRUCTURES Columns in Flat Slab Construction........ "ao eaairtd glieaie Ne Scobey Ae hg eeu py 305 Interior Columns for Symmetrical Arrangement of Pandan, Oved ts Tatts 307 ss SESH TIE ieee an hl en al il ie ae A ae al2 eli LOAGS,.,...-.5+-+-+> Fee ti oe Dee eet Geter eoretee te Perea 319 se ak fie lek hp Ry CMe ee ck oe we 322 eee. e. s eb iies eoiee de ge er eee ee B20 Bending Moments in Flat Slabs...........-:.-- 0s esse cree eects 328 Bending Moments in Interior Panels in Flat Sai ewe stir eatae eet pros 331 ey So ee ne ena area eo Cee a 303 Vormuias for Slab Thickness... .....0525 00. (Gia Ca ae le ne aaa 336 Thickness of Slab as Determined by Positive Bending Moment........... 340 Compression Stresses in Concrete in Fat Sla bee ic ccs ler Am eae 341 Shearing Stresses in Flat Slabs..........- +. eee cere eee eee renter ees 346 Meetarcoment in Plat Slabs... . 6 col... ne ee NE, BS EE Ine 351 NN ERR icc. ie ee ove wiog ee en ue RAMAN EL ek AES 358 MEE ee caine nw 2 oe, Pans Shoe We eRe NER E RI oe Oh 358 at er ire es bs pe ratansontem bh ace An MeN Be ESO A 362 Circumferential or Smulski (S. M. I.) System.......------+--5- ptt oie IOOE x CONTENTS PAGE Explanation of Action of the Smulski System....................5+-505: 369 Three-way. System . 1 oishssSeye ee Pa ener eer 374 Openings in Flat Slab... 1.0% 3.50 SERRE. a IS Sk ee 375 Design of Beams in Flat Slab Construction.....................+.++-+: 377 Wall Beams. & £5 siwkeiin ebietecere sone ale ene lesecenpre= fo-/e GARG RUSRET Rs oeaiete enn 379 General Remarks.:). ¢.cs- eo Se 385 Problems in Designing Flat Slab Construction...............-.+-.5+555 389 Example of Flat Slab Design............. PE 390 Chicago and New York City Flat Slab Regulations............. Seok os 396 Bending Moment Coefficients... 0.02.20. 00-52 See eee eens eet ees aas 398 CHAPTER VII CONCRETE AND REINFORCED CONCRETE COLUMNS Plain Concrete Columns and Piers...2 ey oo. oe de 403 Reinforced Concrete Colummns. . 454. oi 0. oe ae cin ees ote se 404 Columns with Longitudinal Bars... 25.0220 4... «4; ©5040), eee 405 Spiral Columns. o. 2... 0.0 000s cece ee ee ees oo» eel eo 419 Octagonal Spiral Columns... . 6.000. oo ce ep «ou 0 arte 429 Square Spiral Columns. ..¢% 2.6500. vee. ee oe oe 429 Oblong Spiral Columns. - 2... 05. 0c net ee 2 2 eee 430 Details of Spiral Columns’...... 062. 0.0. oc soe: oar 431 Long Columms.:.. 2.2.5. Wi tes eens cone he ee 434 Structural Steel Columns Imbedded in Concrete..................-4+-5-- 435 Rules for Structural Steel and Cast-iron Columns................-.++.-- 435 Rules for Structural Steel of Various Building Codes................-.-.- 437 Details of Design of Steel Section. .............-.++- 5 ae= saya teste ony 440 Economies in Column Design... 0... 2.5 -. «+ 0» +» aceite eae nee oon 445 Reduction of: Live Loads in Buildings... ....... . . ... « . scjp one ene 452 Design of Columns in a Building Several Stories High..............-+--. 456 Columns Subjected to Bending... 03 6005500550 ce + | ween eee eee 458 CHAPTER VIII FOUNDATIONS AND FOOTINGS Determination of Carrying Capacity of Soil................-.0 esse eens 469 Bearing Power of Soils. 20 00.00 200000. 470 General Rules of Design... ... 2.60 005. 0.0 ae. 472 Plain Conerete Footings... 044 ...0c0-- 00h as 19 ee 480 Reinforced Concrete Footings:. ..........-. 6605+. evs ea 481 Independent Column Footings. «...............5-5 0. 42) i 485 Design of Independent. Footings...................-5. 5+. =iaeninn ners 491 Simple Slab Footing. ... 2.0... 200 cee eee tee ee es eer 503 Sloped and Stepped Footings. ....27.0.........). 0 Joie 506 CONTENTS xl PAGE Rectangular Footings... .......... cece eee e eee e eee e nee e tenet eeeteees 510 Pe epePe tO OIE OOUITIES 2. oe oe ce ee ne ee tener reese wenaseers 514 Independent Footings with Piles. ........ 1... eee seen eect eee e eee eee ees 516 Continuous Wall Column Footings......... ay Pee JO concnton eee ee 518 Pore POGUINES.. 6. hi ee eee eter e resent ester eeee 522 Strap Beams to Connect Footings. ..........seeeeeeeeeeeeeereeeeeeeees 533 Sg ee ee ee ene tee ee ecece Sa cee re ee ee pees) CHAPTER IX PILES Wood Piles corn t ens, facia elo so wa ee: he 6 88 0 0 8 0 eee eevee ee eeeoeeeneoe §42 vw So a ee eee eo oe eee Bae ee ras ie «okcle OU CHAPTER. & Bumpinc ConsTRUCTION Relative Cost of Buildings of Different Materials.............. Sr eiie ie OU Actual Cost of Reinforced Concrete Buildings..............++eeees B eteare 567 a eh. nll Pel ene eee wae tees ewes ee’ 569 Meme ra DCSOTIUIOW es sic. sc oe ee te ete treo eneeeees 572 Beam and Girder Design of Floors. ..........-- eee c eee eee e eee eens 575 Example of Beam and Girder Design. ..........-- ++ esse tere eeees cere 578 Tint elab Floor Construction. . 6.2.22. 6 eee ee ee ee tee eee eens 587 Light-weight Floor Construction. ........+ ++ esses er ret eeceeeerete te 588 Reinforced Concrete Hollow Tile Floor Construction............+++e++++: 589 Weumple of Hollow Tile Floor. ..-........-.-+2 see ree cece een tne rene: 598 SI aD cary Se e te e es ee ee ee ee we eu se 602 Combination of Structural Steel and Concrete. .......--. seer cere cece 611 Basement and Ground Floor Slabs... 2.0.0... 66sec teeter eee eee enees 618 Pg ecb nse se enue see Met te bee ee tases 620 CHAPTER XI WALL-BEARING CONSTRUCTION Brickwork.5.......»- Ce CURR CUR FO he Soa it gine nine feck we 629 “Coppa: WCU eae e Bee Ae os Ct er che ened ee 634 CHAPTER XII BASEMENT WALLS BMPR OT Construction... fs. c ee ee es toes oce mond tame eines 637 auch TE SREY el ai a a ee ee a a) or 638 Xl CONTENTS ; PAGE Wall Supported at Top and Bottom.............. 2 es 638 Wall Supported by Colunttiges 2.2) on oeeeete eee pee oe yt a ee 643 Example of Basement Wall Supported by Column....................... 645 CHAPTER XIII Roor CONSTRUCTION LORIN Goss 2 degas e heee. - Seewh ne so in: ep va guednte 9 2 9 ck re ee 648 Drainage... 0.0.0 cers we bo a eee ue ote nine eae + see nen 649 Insulation... 5 jahge ee ee ne RR 2 2 650 Roof Coverings. oui 2s. «+ os. 60 ep ee Ge Se oe as oe ce 654 Roof Désign’ (te 0 08 og seco ae a ' 659 Long Span Rosf Construction.; . 20%. es. See ee 661 Concrete Arches for Long-span Roofs... <2. $05 52 2a ei ee 668 Long-span Roof Construction. .. 4. deeply gee ss a eg ec 762 eI Ny Fy nk ce ky wos cee kk wh Rage eg eee Ae 763 ee ER RIS op a a 765 (FRE N25, oe RM aS a a 772 em et 1 OS EVUG IOI G6 ig whe beck es hehe peck eka ve eae wee et es 778 Spe ey SGU M me AIGIN a) yc hn ph eile ends os Ge Wks ee en ve ewe 778 CHAPTER XIX REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS Warehouse Construction....... at Oe ee REPRE Se Xr AAI Stee 785 (UE ALG Te UTP ae a ON ay aa Sr eae 792 I eeCNIOR ro oes 5 oc ht eee eeews wee rw aa es) ee 792 Memriotaidimes and Garages... 00. Se ee bp eee ees 795 ET: remem er gs) 8a Ye pall vie oes Be 6 a AOU ee ee es 803 Perelen srervicnt Houses and Hospitals... 22. 6... de ns bee cee cece 805 NEG yy eee ve ee ey tea ta Cries Me ONE ota 808 CHAPTER XX REINFORCED CONCRETE CHIMNEYS UTI ee RASS Soo. Tt se dae De vedas Seomw Golvgepie Ce anne) Ames 812 Theory and Formulas........... Pa Ee eee ere ENS PU eral: 813 meer rioow Circular Beams, .... 2.650 ce ete ewe eed ey ww ee 822 ieee: meintorcea Concrete Chimneys... /...-........ 536.0 0e gree an nee 822 eer erate amtie vol esien eG. ks SUS We ieee ee we dels 829 xiv CONTENTS CHAPTER XXI RETAINING WALLS PAGE Weight of Barth... 2.242. -- pbk be ce ered's bbe bt eet tulad an 833 Earth Pressure... .. . . ... <5 cc fame Bee 6e oh eee 0 ptt a ete anata ae 835 Design of Retaining Walls of Gravity Section............+...++sesseeeee 839 Cantilever Walls... ss. 5. co ub bb oR te teed be pe 849 Example of T-shaped Retaining Wall................++...45. ee ree 856 Wall with Counterforts: ..... 0b. ..554.065%9 028. 0.) Sa 863 Special Designs of Retaining Walls.............-. 5-6. s esses eres tee aee 869 CHAPTER XXII TABLES AND DIAGRAMS Creneral. oc; cw ec 0 ot bejask enue.» ot lcel aleene steels Bienen ante he nn 879 Constants for Rectangular Beams and Slabs.............---++++++s05-:- 880 Safe Loading for Rectangular Beams............... 2. esses eeeere ee eees 884 Safe Loading for Slabs............ 0002500002051 00 +00 >= 9 gO ee 886 Design of T-Beams..... 00... 000.0 0 Rev ee oe 2 + > =o 5 ose 894 Design of Stirrups........0.-.00+-205 +++ e se ++ + gee 899 Beams with Steel in Top and Bottom............... 2. se sees eee eee eee 904 Flat-Slab Design... 25.0000 0. 85022 oe cee ny oetnt ne 911 Column Design... .....-.60. 08 s2 > seh ai ep + 2 oe ee 915 Members Subjected to Direct Loading and Bending..............------- 934 General Data... occ. ... 0 cue Sone eyeqei en eae + one +e 2 eer 942 CONCRETE PLAIN AND REINFORCED CHAPTER I MATERIALS AND METHODS FOR MAKING CONCRETE ___In this chapter are treated the materials used in concrete and ERRATUM Page v (Preface).—Lines 7 and 8, beginning with the semicolon on line 7, should read as follows: the second volume will cover Theory and Design Formulas for Continuous Beams, Frames, Building Frames, and Arches; while the third will treat of Con- crete Materials and Construction. ESSENTIALS OF GOOD CONCRETE Concrete consists of hard particles of inorganic material, called aggregates, such as sand, gravel, crushed stone, slag, or cinders, cemented together with Portland cement and water. When all the ageregate is fine, like sand, the mixture is termed mortar. In making durable concrete, the prime essentials are as follows: Hard, durable aggregates; Aggregates graded, or stepped, in size, from fine to coarse; Aggregates clean, that is, entirely free from organic impurities; Cement sufficient in quantity to give the required strength or water-tightness ; Water clean and free from organic matter or deleterious minerals; Quantity of water to produce the consistency needed for the work, always avoiding a wet, sloppy mix; XIV CONTENTS CHAPTER XXI RETAINING WALLS Weight of Earth........... PH CRE ete 2, « Earth Pressure... «+ o:+: {vic AOC te set Sh erent, rae ere Design of Retaining Walls of Gravity Section............... Gantilever Walls: . .<.¢.sc et te ie ie te Example of T-shaped Retaining Wall................-+.++. Wall with Counterforts <2 5.44 520: 0.50. 520% se 2 2 Special Designs of Retaining Walls...........--..--+++++++ CHAPTER XXII S, ¢) ohm \w) Bid. 0 6),8 Pe o © @ 8 « 6) © 6, @ @ ate CONCRETE PLAIN AND REINFORCED CHAPTER I MATERIALS AND METHODS FOR MAKING CONCRETE In this chapter are treated the materials used in concrete and certain fundamental principles relating to concrete. The properties of concrete are fully discussed in Volume II of this treatise, which covers selection and testing of materials, proportioning, strength, water-tightness, resistance to sea water, fire resistance, effect of impurities, and the various other characteristics, as well as dis- cussion and advice on mixing and placing concrete, form construc- tion, architectural treatment of surfaces, and the laying of floors, roads, and pavements. In this chapter, only such facts and con- clusions are given as are necessary for the understanding of the subject matter of this volume. ESSENTIALS OF GOOD CONCRETE Concrete consists of hard particles of inorganic material, called aggregates, such as sand, gravel, crushed stone, slag, or cinders, cemented together with Portland cement and water. When all the aggregate is fine, like sand, the mixture is termed mortar. In making durable concrete, the prime essentials are as follows: Hard, durable aggregates; Aggregates graded, or stepped, in size, from fine to coarse; Aggregates clean, that is, entirely free from organic impurities; Cement sufficient in quantity to give the required strength or water-tightness; Water clean and free from organic matter or deleterious minerals; Quantity of water to produce the consistency needed for the work, always avoiding a wet, sloppy mix; 2 MATERIALS AND METHODS FOR MAKING CONCRETE Mixing of the mass thoroughly to insure homogeneity and also properly to work the cement; Transporting and placing in such a way as to avoid separation of aggregates; Ramming or puddling so that the concrete will ail every part of the forms; Temperature of concrete maintained appreciably above freezing until it is thoroughly hard, to avoid retarded hardening. CEMENT Portland cement is used universally for reinforced concrete and also for plain concrete, except where abnormal conditions necessitate a special cement. Cement should be tested, except fon very small and unimportant jobs, and should pass the requirements of the American Society for Testing Materials for tensile strength in 1:3 mortar; fineness; setting time; and soundness. Samples of cement, except for small jobs, are preferably taken at the mill before shipment. If this is impracticable, about 20 bags out of a carload of, say, 200 barrels (800 bags) are opened, and a small quantity taken from each and mixed together to make a single sample of about 2 quarts. Cement that is unsound, that is, cement that checks or warps when steaming, should not be used. Sometimes the unsoundness disappears if the cement is held for a period. A particularly dangerous characteristic in cement is a “ flash set,” which is evidenced in practice by the concrete setting so quickly as to harden in the mixer or in ordinary transportation to place. Con- crete thus prematurely hardened is practically worthless and must not be reworked. FINE AGGREGATES Poor sand is the most common cause of defective concrete. The presence of organic matter, such as vegetable loam, is the most objectionable characteristic, as such matter, even when present in very small quantities, prevents the concrete from hardening satis- factorily. Concrete containing organic matter may remain soft and friable for months and never attain full strength or resistance to abrasion. AGGREGATES 3 Sand, Therefore, Should Always be Tested.—It is more neces- sary to test the sand than the cement, as the quality of sand cannot be determined by inspection. With the exception of the presence of organic matter, the quality of a sand that is most detrimental to concrete is excessive fineness. Not only must an excess of fine particles of dust be avoided, but the average size of the grains should be coarse. Sand, for most uses, is preferably graded from fine to coarse, with not more than 30 per cent passing through a 50-mesh sieve and not less than 85 per cent passing through a No. 4 sieve. The most important test of sand is for organic matter. The Abrams-Harder test is described in Volume II of this treatise. Crushed stone screenings may be used as a substitute for sand or may be mixed with sand, provided they do not contain an excess of dust, which tends to come to the surface of the concrete and reduce its durability. Limestone screenings, for this reason, make a very poor material for roads and sidewalks. COARSE AGGREGATES Coarse aggregate should consist of crushed stone, gravel, or other approved inert materials with similar characteristics, or combina- tions thereof, having clean, hard, strong, durable, uncoated particles free from injurious amounts of soft, friable, thin, elongated or lami- nated pieces, alkali, organic or other deleterious matter. Coarse aggregate should range in size from fine to coarse within the limits given in table below. | The following table indicates desirable gradings for coarse aggre- gates of certain nominal maximum sizes: | Nominal Percentage by Weight Passing through Percentage Passing, Maximum Standard Sieves with Square Openings not More than Size of Aggregate, in. 3in. | 2in. |13in.| lin. | Zin. | }in. |No. 4 Sieve|No. 8 Sieve 3 ND Elbe. oe BU (Sieh eel ip cas eat 10 5 JEP SOT RA ONS Te asaic, UY (9 Da aig 1 POE Fee 10 5 rg ae ae oa ee ee 40-75} 8.4 ss 10 5 Lo SR con i emer Ree we ee oe ee 10 5 o,f game ce Slee gee IR ONE BO ae 10 5 Re ate Li A oe oust » Sif aals o's 95 10 5 4 MATERIALS AND METHODS FOR MAKING CONCRETE The test for size and grading of aggregate should be made in accordance with the Standard Method of Test for Sieve Analysis of Aggregates for Concrete. (A.S. T. M. C 21-24). PROPORTIONS OF CONCRETE The amount of cement to use in concrete, that is, its proportion to the aggregates, depends upon tke nature of the structure and the strength or water-tightness required. The proportions are usually expressed in terms of cement, fine aggregate, and coarse aggregate; proportions 1 : 2 : 4, for example, meaning one part cement by volume to two parts of sand or other fine aggregate, to four parts of gravel or broken stone or other coarse aggregate. A bag (94 lb.) of cement is always assumed to be 1 cubic foot. A 1:2:4 mix, therefore, means 1 bag cement, 2 cubic feet fine aggregate, and 4 cubic feet coarse aggregate. Proportions in practice are apt to run from 1:1:2, often used in columns, to 1 : 4 : 7 for mass concrete employing specially graded aggregates. The usual proportions for reinforced concrete are 1li'22 A For concrete in large masses, such as dams, a great deal of money can be saved by making extensive tests and special measurements of materials when laying, so as to Regs the maximum density and the minimum of cement. MIXING AND PLACING Concrete mixers, driven by power, are used almost universally for all kinds of concreting. ‘The size and type of machine must be adapted to the structure and to the quantity of concrete to be placed: ina day. For ordinary work, a revolving drum with interior blades is used. The time of mixing is an important element. The authors advise mixing a batch of concrete not less than { minute after the dry mate- rials and water are in the mixer. After the proportions have been fixed, the most important ele- ment in good mixing is the amount of water to use. The quantity of water required for a batch, with the materials as they generally come, should be determined by trial with the aid of the slump test. The water for every batch should be measured. When the standard quantity has been fixed, the volume may be changed slightly to STRENGTH AND OTHER CHARACTERISTICS 5 allow for difference in moisture of sand or for the working of the concrete in the mass, since as the depth of the concrete increases, the upper part of the mass becomes wetter. Concrete is often seriously injured by an excess of water in mixing. The strength is reduced by excess water, the wearing qualities are lessened, danger of checking or crazing of the surface is increased and laitance, a whitish, nearly inert substance, rises to the surface. For reinforced concrete, the best consistency is that which will flow very sluggishly without flattening out and without separa- tion of the fine and coarse material. If the concrete is deposited through a sloping chute, the slope should not be flatter than one vertical to three horizontal. For many mixtures, a one to two slope should be maintained. The criterion in fixing the maximum amount of water in concrete is (1) to avoid separation of cement and fine aggregate from the coarse aggregate, and (2) to prevent excess water or thin mortar from rising to the top of the mass as it is being placed. 7 While the greatest danger in mixing concrete is that of having the mix too wet, excessive dryness also must be avoided. In rein- forced concrete, the mix must be soft enough to thoroughly imbed the steel and fill all parts of the forms. In mass concrete, a drier mixture can be used than in reinforced concrete. Enough water must be added to make the mortar flush slightly to the surface and to fill all voids. When, for small pieces of work, concrete is mixed by hand, the same precautions must be taken as in machine-mixed concrete. Materials must be carefully measured, a bag of cement being assumed to be one cubic foot; the materials must be spread out in uniform lay- ers; they must be thoroughly mixed dry by turning two or three times with a square-pointed shovel; water must be added to obtain the required consistency; the mass must be turned about four times wet, and the mixture taken to place without delay. In important construction, specimens of the concrete should be taken at intervals, molded in 6 in. by 12-in. cylinders to be crushed at the age of 7, 14, or 28 days. STRENGTH AND OTHER CHARACTERISTICS For reinforced concrete structures, a compressive strength of the concrete of 2000 Ib. per sq. in. at the age of 28 days is a common 6 MATERIALS AND METHODS FOR MAKING CONCRETE requirement. The strength is determined by testing cylinders 6 in. in diameter by 12 in. high. For concrete in large masses under low stress, and for such struc- tures as cellar walls, a strength of 1 500 lb. per sq. in. at 28 days is satisfactory; while for columns, where small dimensions are pre- ferred, 3.000 lb. per sq. in. may be specified. The variations in strength to accord with any requirements are obtained by proper proportioning of the cement and grading of the aggregates. For water-tight work, concrete 2 000 lb. per sq. in. at 28 days is usually required. In work of this kind, the size of the voids, as well as the density of the concrete, is important; therefore, a finer sand may sometimes be advantageously used to make the concrete flow smoothly and to avoid stone pockets due to separation of materials. Concrete laid in sea water, or in alkali waters, requires special precautions, as discussed in Volume II. For fire resistance, reinforced concrete is the best all-round build-- ing material. Aggregates like trap are better than gravel, which has a large coefficient of expansion. or limestone, which is liable to be | affected chemically by the heat. For the purpose of design, the following strengths of 6 by 12 cylinders at 28 days are assumed for various proportions of concrete: Strength of Cylinder at 28 days. Lb. per sq. in. Mix ye es Ba 17:13:38 1:24 2F25 Strength at 28 days, fc’. 3 000 2 500 2 000 1 600 n 10 12 15 15 eee n=ratio of moduli of elasticity. CHAPTER II REINFORCEMENT Reinforcement for concrete may consist of any kind of ductile metal, the prime factors being strength, reliability, and economy. In the United States, steel is used practically exclusively, while in Europe wrought iron is used to a great extent. QUALITY OF REINFORCING STEEL The Specifications of the American Society for Testing Materials, endorsed by the authors, require the following properties for rein- forcement. Chemical Properties.—The steel shall conform to the following requirements as to chemical composition: Been havtis [ Bessemer not over 0.10 per cent. . | Open-hearth not over 0.05 per cent. Physical Properties.—The bars shall conform to the require- ments as to tensile properties given in the table below. Tensile Properties of Concrete Reinforcement Bars Plain Bars Deformed Bars : Rg et eer 5 ena as a aR eee Properties wal d Considered Structural] Inter- Structural] Inter- ; Hard ; Hard Bars steel mediate den steel mediate eae grade grade z grade grade £ Tensile strength,} 55 000 70 000 80 000 55 000 70 000 80 000 | Recorded Ib. per sq. in.... to to min. to to min. only 70000 | 85000 70 000 85 000 Yield point, min., Ib. per sq. in....| 33 000 40 000 50 000 33 000 40 000 50 000 55 000 Elongationin 8in., min., per cent. .}1 400 000*|1 300 000*|1 200 000*|1 250 000*/1 125 000*| 1 000 000* 5 it ae SS) |S ae al ieee acereeaee (haar eae ama Tens. str. | Tens. str. | Tens. str. | Tens. str. | Tens. str. Tens. str. * Deduct 1 per cent for each increase of $-inch above 3-inch diameter, or for each decrease of zs-inch below ;4-inch diameter. 8 REINFORCENEMT Bend-test Requirements.—The test specimen shall bend cold around a pin without cracking on the outside of the bent portion, as follows: Bend-Test Requirements Se Ee Plain Bars Deformed Bars Thickness or Cold- ere Structural| Inter- Structural| Inter- wwisiee of Bar : ; Hard ; Hard Bars steel mediate eae steel mediate ete grade grade & grade grade = Under 7 in....:.- 180 deg. | 180 deg. | 180 deg. | 180 deg. | 180 deg. | 180 deg. | 180 deg. =f Gh he = ob a=t @ = ror ad =At a = 2t $ in. or over....- 180 deg. | 90 deg. 90 deg. 90 deg. 90 deg. 90 deg. | 180 deg. d— ese ak Greet O—so) de—war d = 3t Exprianatory Note: d = the diameter of pin about which the specimen is bent; t = the thickness or diameter of the specimen. The bending test is the most important test in the specification, and no steel which fails in this test should be used under any cir- cumstances. Steel with high elastic limit, whether due to high carbon or to manipulation in manufacture, should be purchased with these reservations even if the working stress is to be no higher than is used with structural grade steel, say, 16 000 lb. per sq. in., because it is liable to be brittle. In case a lot of steel has been delivered without previous test by the purchaser, one bar selected at random in every 100 should be subjected to this test; and if it fails to pass, the portion from which it is taken should be rejected. On small jobs, where it is impossible to get proper tests of steel, it is of impor- tance to purchase material from a reliable dealer having direct connection with reliable mills. Otherwise there is a likelihood that material rejected on some other job may be furnished. SELECTION OF GRADE OF STEEL FOR REINFORCEMENT Steel of Structural-steel Grade has been universally accepted in the past as the most satisfactory material for reinforcement. It ‘5 manufactured under such standard conditions that, for unim- portant structures, it may be used without other test than the bend- ing test given above. The allowable stress for this grade should not exceed 16 000 Ib. per sq. in. This gives a factor of safety of a little over 2, for the live and dead load. The factor of safety for SELECTION OF GRADE OF STEEL FOR REINFORCEMENT 9 the live load alone will depend upon the ratio of the live to dead load. For structures subjected to severe weather conditions or to locomotive smoke; it may be advisable to reduce the stresses so as to prevent formation of open cracks. Intermediate Grade Steel has the advantage over structural- steel grade in its higher elastic limit. It is possible to use a stress in steel of 18000 lb. and still have a somewhat higher factor of safety than for the structural-steel grade. Therefore, if inter- mediate grade steel can be bought at a unit price not exceeding that of structural-steel grade by more than 8 to 10 per cent, it may be used with economy. Since actually there is no appreciable differ- ence in cost between the two grades, the intermediate grade steel is gradually replacing the structural-steel grade. In some cases, intermediate grade is specified even if the allowable stress in steel is only 16 000 lb. per sq.in. The factor of safety is thereby increased considerably by the additional strength of intermediate steel, with- out any appreciable extra cost. Intermediate grade steel should be properly tested. The only disadvantage in using intermediate grade steel with 18 000-Ib. stress is that the cracking of the concrete at working loads is larger, because the stretch in the steel, being proportional to the stress, is larger. Under ordinary conditions for ordinary building construction work, this objection is not important. Tests made in Europe (see Vol. IJ) have proved conclusively that the cement protects the steel from ordinary corrosive action until the elastic limit of steel is nearly reached. Where it is desirable, how- ever, for any reason, to avoid cracking, the stress in steel should be kept as low as for structural grade steel. For temperature reinforcement, a steel of high elastic limit is particularly valuable. Hard Grade Steel.—If properly tested as required by the pre- viously given specifications, hard grade steel may be safely used for reinforced concrete. Although hard grade steel has a higher elastic limit than the intermediate grade, the allowable stress should not be larger than 18 000 lb. per sq. in. because an accidental increase in loading would produce excessive cracking. With a stress of 18 000 Ib. per sq. in., the factor of safety for hard grade steel is larger than for either of the two previously described grades. What has been said as to the economy of using intermediate grade steel applies also the use of hard grade steel. 10 REINFORCEMENT Many engineers do not approve of the use of hard grade steel because of its brittleness when of poor quality, and because it is pro- hibited in ordinary structural steel work. Brittleness in reinforcing steel, however, is less dangerous in reinforced concrete than in many classes of structural steel work because the concrete protects the steel from shock, and also because smaller sections of steel are used in concrete beams than in steel beams. The large and irregular shapes in structural steel render it much more sensitive to irregular cooling during the process of manufacture than is the case with reinforcing bars. Syapes USED FOR REINFORCEMENT Small Angles.—Small angles are sometimes used as reinforce- ment, mostly in columns or in Melan arches. Reinforcing Bars.—The most common shapes used for rein- forcement are round and square bars, and to a much smaller extent, flat bars. ‘The bars may be either plain or deformed. The com- mon types of deformed bars are shown in Fig. 1, p. 11. Plain vs. Deformed Bars.—Steel bars used in reinforced concrete may be either plain or deformed. Plain bars are round, square, or flat bars passed through regular rolls. . Deformed bars are bars provided with irregularities of the surface. formed either in the process of rolling or by twisting after rolling. When the area of cross section of deformed bars is not constant, its effective area is equal to the minimum cross section. The purpose of the deformations of the surface is to increase the bond between the bars and the concrete, by adding to the sur- face bond of the plain bars the mechanical bond of the projecting . deformations. ‘Tests show that deformed bars of proper design have a considerably larger bond resistance than plain bars, so that larger unit bond stresses are allowable. (See p. 54.) Either plain or deformed bars may be used with safety, provided the unit bond stresses are kept within the limits set for the two types of bars. In ordinary beams and slabs, the bond stresses are likely to be low, so that there is no difficulty in keeping the bond stresses within working limits of plain bars. In such cases, there is no special advantage in using deformed bars. In footings and short beams carrying heavy loads, on the other hand, it is often impos- TYPES OF REINFORCING BARS Cold Twisted Square Bar Herringbone Bar Monotype Bar Fic. 1.—Types of Deformed Reinforcing Bars. (See p. 10.) 11 “ 12 REINFORCEMENT sible to utilize the full working value of the steel in tension, owing to the difficulty of providing sufficient bond. The use of deformed bars in such members is therefore advantageous and economical. They are also useful for temperature reinforcement, as the distri- bution of cracks depends upon the bond between the bars and the concrete. In road construction, deformed bars are useful for the same reason. As a general proposition, where deformed bars are easily obtain- able and their cost is not appreciably higher than that of plain bars, they are recommended in preference to plain bars because the factor of safety against actual collapse is larger for deformed bars than for plain bars. The use of deformed bars is particularly recommended for structures subjected to dynamic loads. Properties of Bars Used as Reinforcement.—The table below gives the sizes of plain bars used in practice, their areas, weights, and perimeters. Round and Square Bars Areas, Weight and Perimeters Se ee es Round Bars | Square Bars Site, 4) ot ae pees ake 2 ee inch Area, |Weight per | Perimeter, Area, |Weight per | Perimeter, sq. in. foot, lb. in. sq. In. foot, lb. in. 1/4 0.049 0.17 0.78 0.062 0.21 1.00 5/16 0.077 0.26 0.98 0.098 0.33 1.25 3/8 0.110 0.38 1.18 0.141 0.48 1.50 7/16 0.150 0.51 76 0.191 0.65 1.75 1/2 0.196 0.67 1.57 0.250" 0.85 2.00 9/16 0.248 0.84 Eger 0.316 1.08 2.25 5/8 0.307 1.04 1.96 0.391 1.33 2.50 11/16 0.371 1.26 2.16 0.473 1.61 2.75 3/4 0.442 1.50 2.36 0.563 1.91 3.00 13/16 0.518 1.76 2.55 0.660 2.24 3.25 7/8 0.601 2.04 2.75 0.766 2.60 3.50 15/16 0.690 2.35 2.95 0.879 2.99 3.75 1 0.785 2.67 3.19 1.000 * 3.40 4.00 11/8 0.994 3.38 3.53 1.266 4.30 4.25 5 TYPES OF REINFORCING BARS 13 The standard sizes in the table are set in bold type. The other sizes can be obtained with difficulty, especially when deformed bars are desired. Therefore, the designer is cautioned against using them. Cold Twisted Bars have the same dimensions and weight as the corresponding square bars. Corrugated Bars, round and square, have the same dimensions as the corresponding round or square plain bars. Their weight is somewhat greater, however, on account of the projections, which are not included in the effective area. Kahn Trussed Bars have special dimensions as given in table below. Dimensions, in. Weight per Tome oce bigs i | SCLineal Poot, Area, sq. in. D B lb. plist es a wis as 3 13 1.4 0.41 Ay eee OS ark: a 2,3, yews 0.79 I te SS oe ado ti Ss i; 24 4.8 1.41 PEE Ee ss the ie 23 6.8 2.00 By Wes Settee oe 66! ona: 2g os 2 34 10.2 3.00 If trussed bars are used as beam reinforcement, the wings are sheared near the ends and bent up to serve as diagonal tension reinforcement. Kahn New Rib Bars.—Same remarks as for corrugated bars. Havemeyer Bars.—Round and square Havemeyer bars have the same properties as the corresponding round and square plain bars. Since the projections form a part of the effective area, the weight of these bars is the same as that of plain bars. Havemeyer flats are of the dimensions given below. mize, in. | 1X ‘AX S/d X 8/12 X geld X Bld X 2/18 X BLE X ILE X 9 eI|-H Area, sq. in.|0.2500!0.3750|0. 4690! 0.4688 |0. 5625/0. 7500/0.6563| 0.7656 [0.8750 Weight per | : foot, UN es 1.280 1.590 | 1.590 {1.913 |2.550 |2.230 | 2.600 |2.980 14 REINFORCENEMT Elcannes Bars.—These are rolled in sizes from 2? in. to 14 in. with a variation of } in. The bars correspond in area and weight to square bars. Kahn Bars with Sheared Diagonals Fig. 2.—Sections of Kahn Trussed Bars. (See p. 13.) Herringbone Bar.—The sizes and weights of the Herringbone bars are as follows: Size, inches 13 14 1 z 3 5 Weight per foot, Ib...) 5.13 | 3.62 2.38 1.72 1.28 0.91 Monotype Bars——These bars are rolled in one type only. The size, however, is arranged so that it is possible to get bars equivalent to standard sizes of round and square bars. The weight of bars is slightly higher than for plain bars because of the additional weight of lugs. Wire Used as Reinforcement.—The wire used for spirals in spiral columns, is cold-drawn from rods which are hot-rolled from billets. The cold drawing causes the wire to attain tensile strength from 80000 to 100000 Ib. per sq. in. It is usually designated by SPECIAL REINFORCEMENT 15 gage numbers. The sizes of wire used in practice are given below. Gage | Equivalent | Area Gage Equivalent Area Number | Diameter, in.| Sq. in. || Number | Diameter, in. SG. ii: 7-0 0.4900 0.189 5 0.2070 0. 034 6-0 0.4615 0.167 6 0.1920 0.029 5-0 0.4305 0.146 7 0.1770 0.025 4-0 0.3938 0.122 8 0.1620 0.021 3-0 0.3625 0.103 9 0.1483 0.017 2-0 0.3310 0.086 | 10 0.1350 0.014 0 0.3065 0.074" 4 11 0.1205 0.011 1 0.2830 0.063 12 0.1055 0.009 2 0.2625 0.054 13 0.0915 0.007 3 0.2437 0.047 14 0. 0800 0.005 4 O22535.,. 4 0.040 SPECIAL REINFORCEMENT Special reinforcement consists of expanded metal and wire mesh of various sorts. Expanded Metal, shown in Fig. 3, p. 15, is produced by special machines from soft open-hearth steel plates. are slit and expanded by a cold drawing process in such a way that ao Fig. 3.—Expanded Metal. (See p. 15.) These plates in the final state the strands form diamond-shaped meshes. The cold drawing raises the elastic limit of the steel to around 55 000 lb. per sq. in. The standard sizes of expanded metal are given in the table on page 16. 16 REINFORCEMENT Standard Sizes of Expanded Metal Adaptable to Concrete Reinforcing Designation *3-13-075 *3-13-10 *3-13-125 73-09-15 *3-9-175 *3-9-20 *3—-9-25 *3-9-30 *3-9-35 3-6-40 3-6-45 3-6-50 3-6-55 3-6-60 *3-1-10 *3-1-100 Sectional Area, sq. in. per ft. Width 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. Li; | 75 075 10 125 15 175 20 25 30 35 Weight, Ib. per sq. it. 0.27 0.37 0.46 0.55 0.64 0.73 0.92 1.10 1.28 1.46 1.65 1.83 2.01 2.19 2.74 3.63 Width of Standard Sheet 6/ QO” 6/ QO” 5/ pues af! 0” 6/ QO” 5/ UN At Oe VE OQ” 6/ 0” 7 0!’ 6’ Bet iy g”’ ye 3! Ae Q’’ iy OM 4’ ae Number of Sheets in a Standard Bundle 10 7 i 5 5 5 5 2 2 2 2 2 2 2 1 1 eS All of the above have a diamond-shaped opening, 3 by 8 in. All items listed above are furnished in 8, 12 and 16-ft. lengths. Items marked thus * are also furnished in 10-ft. lengths. * 3-14-75 and 3-1-100 are manufactured to order only. Fig. 4.—Welded Wire Fabric. ‘ ‘, Ary Welded Transverse Wires Ly No “; Longitudinal Wires (See p. 17.) SPECIAL REINFORCEMENT 17 Welded Wire Fabric consists of two sets of parallel cold- drawn wires placed at right angles to each other and electrically welded at the intersections. The wires are galvanized before welding. Welded Wire Fabric | Transverse Sue ; Area in Longitudinal Wires Only, Wires | pe eainal Wy Ires per Foot of Width, sq. in. ae Brees: at Spacing of Longitudinal Wires Gage aes teed of wire, | of wire, |—— ) ae Seo inal aan es me ein: 6 OK 0 0.307 0.074 | 0.443) 0.295) 0.221} 0.177; 0.148 6 12 i 0.283 0.063 0.377) 0.252} 0.189} 0.151] 0.126 8 Ly 2 0.263 0.054 | 0.325} 0.217) 0.162) 0.130) 0.108 8 12 3 0.244 0.047 0.280; 0.187; 0.140} 0.112) 0.093 9 12 4 0.225 0.040 | 0.239) 0.160) 0.120) 0.096) 0.080 9 12 ‘i 0.207 0.034 | 0.202} 0.135) 0.101) 0.081} 0.067 10 12 6 0.192 0.029 0.174] 0.116} 0.087} 0.069) 0.058 igen cae, 12 7 0.177 0.025 0.148} 0.098} 0.074) 0.059) 0.049 10 12 8 0.162 0.021 0.124, 0.082; 0.062} 0.049} 0.041 12 Ml 12 9 0.148 0.017 0.1041 0.069} 0.052) 0.041} 0.025 12 12 10 02135 0.014 0 0.043) 0.034) 0.029 | 0.057 Triangle Mesh.—Triangle Mesh woven wire reinforcement for concrete is made with either solid or stranded longitudinal members, properly spaced by means of diagonal cross wires, so arranged as to form a series of triangles between the longi- tudinal or tension members; the longitudinal members are spaced 4 in., the cross wires either 2 or 4 in. as desired, providing either a 2- or 4-in. mesh. ‘The sizes of both longi- tudinals and cross wires are varied in order to provide the cross-sectional areas of. steel required to meet the conditions. Fig. 5.—Triangle Mesh Wire Fabric. i (See p. 17.) 18 REINFORCEMENT Information about triangle mesh may be taken from table below: Triangle-mesh Wire Fabric Number and Gage of Wires, Areas per Foot Width and Weights per 100 Square Feet. Longitudinals Spaced 4 Inches. Cross Wires Number 14 Gage Spaced 4 Inches. ee ee ee EE St | Total Effective : Number and Gage ee tr ae Longitudinal | A ee Ni Style of Wires, each ace gitu oe S | Sectional Area Weight lbs. Number Longitudinal Square ches | square inches Ree per foot width per foot width 100 square feet 032 1—No. 12 gage .026 .032 22 .040 | ES .034 .040 20 049 fo OS .043 .049 28 058 Pe 3 ee .052 .058 32 068 te ees .062 .068 30 080 pene Uf tae 074 .080 40 093 jo ee 087 .093 45 107 be ra oe 101 .107 50 126 1p ee Pal 20 oot 57 146 Bt og .140 . 146 65 153 1— + inch .147 . 153 68 168 1—No. 2 gage . 162 .168 74 180 Qt aioe 174° .180 78 208 De ete ee 202 .208 89 245 Dad. nek See 239 .245 103 267 See Gan .261 . 267 111 287 Se Og 281 287 119 309 a tT ae 303 . 309 128 336 Se aa aes 330 .336 138 365 See a .359 .305 149 395 pee Be 389 395 | 160 Length of Rolls: 150-foot, 200-foot and 300-foot. Widths: Approximately 16-in., 20-in., 24-in., 28-in., 32-in., 36-in., 40-in., 44-in., 48-in., 52-in. and 56-in. Norz.—Material may be furnished either plain or galvanized. Unless other- wise specified, shipments will be made of material not galvanized. SPECIAL REINFORCEMENT 19 Expanded Metal Lath.—Expanded metal may be used as lath for plaster or stucco. . Two kinds of metal lath are obtainable, one painted black, and the other cut from galvanized sheet. The weights given in the table below are for painted lath. Galvanized lath weighs 0.4 lb. per sq. yd. more than painted lath. Expanded Metal Lath ‘ é pecan BO Size of pues Sq. Yd. in | Weight per Designation | sq. yd. in Sheet, in a Bundle Bundle, Ib Bundle, Ib. ’ "| Bundle ; Ge 22-P 4.37 24 x 96 10 if cor detoaad aaa els 24-F 3.40 24 x 96 15 26.66 90.67 25K 3.00 24 x 96 15 26.66 80.00 26-F 2.55 24 xX 96 15 26.66 68.00 gpk 2.33 24 x 96 1b 26.66 62.22 24-H 2.90 28 x 96 14 29.00 84.10 26-H 2.20 28 x 96 14 29 00 63.80 | | Rib Lath.—There are several types of expanded metal lath on the market which are provided with stiffening ribs running longitudinally with the lath. Rib lath is used for suspended ceilings, for partitions, and for other purposes. The function of the ribs is to stiffen the lath, so that the fur- ring strips or other sup- ports may be spaced farther apart. Figure 6, p. 19, shows two types of @S8RARRARAABRRAREESAARES Rib Lath, which are inch Hy oad Lath marketed under the trade name of Hy-rib. The ribs of the lath are 2 in. and ? in. deep respectively. The 2-in. Rib Lath is used where greater rigidity and strength are required, Fic. 6.—Special Type of Rib Lath. (See p. 19.) CHAPTER III TESTS OF REINFORCED CONCRETE* The selected tests presented in this chapter were originally carried out to determine the principles of the theory and design of reinforced concrete. They are given here to illustrate the principles and con- clusions presented in the following chapters. The following data are obtained by tests: Stress at which first cracks appear and corresponding stretch in concrete; | Location of neutral axis; Relation between ultimate compressive fiber stress and strength of concrete in compression ; Distribution of stresses over a section; Relation between bending moment and moment of resistance based on stresses in steel; Effect of the percentage of reinforcement; Effect of mix of concrete and age. The results and conclusions are given on the following pages. Formulas for design based on the tests may be found in Chapter V, pp. 201 to 302. Behavior of Rectangular Beams During Loading.—When a reinforced concrete beam is subjected to a loading test its behavior is not the same throughout the progress of loading. Three stages can be distinguished: first stage, before the appearance of the first crack; second stage, after first crack is developed, but before either of the materials passes its elastic limit; third stage, after the elastic limit of steel or concrete has been passed. *To avoid breaking the continuity of treatment of theory and design this chapter on Tests is placed to precede the Theory instead of following it in regular sequence. When using the book as text book this Tests chapter may be taken up subsequently to the chapter on Theory preferably using it merely as supple- mentary material. Notation used in this chapter is explained in Chapter IV. 20° BEHAVIOR OF RECTANGULAR BEAMS 21 First Stage—Before the appearance of the first crack, a rein- forced concrete beam behaves like a homogeneous beam. Com- pression is resisted by concrete, and tension is resisted by concrete and steel in proportion to their moduli of elasticity. The position of the neutral axis nearly coincides with the center of gravity of a section obtained by replacing the steel by an area of concrete equal to the area of steel multiplied by the ratio of moduli of elasticity. This is called the first stage. It lasts until, at an elongation or stretch of concrete equal to the ultimate stretch of plain concrete, the first cracks appear in the beam. Second Stage.—This stage begins when the concrete cracks. At first, the cracks are not visible to the naked eye and do not extend up to the reinforcement. At increased loads, the number of cracks increases. They widen and move up toward the center of the beam. The neutral axis moves up. The larger the amount of reinforcement, the larger the number of cracks and the smaller their width, as illus- trated in Fig. 7, p. 23. In this stage, tension is resisted by steel and by the portion of concrete above the crack and below the neutral axis. The cracks never extend all the way up to the neutral axis (which rises as the cracks develop). The stretch is zero at the neutral axis and increases in a straight line to its maximum at the level of the steel. There must be, therefore, a portion of concrete below the axis where the deformation is smaller than the ultimate stretch for the concrete, and there must be a fiber which is just at the point of breaking. Above it, the concrete is intact and is carrying tensile stresses. The ratio of the tensile stresses carried by the concrete to the total amount of tension decreases with the increase in the load. For equal intensity of loading, the amount of stresses carried by concrete is larger for smaller percentages of steel. For large loads and per- centages of steel, commonly used in design, the amount of the stresses resisted by concrete is negligible. The assumption made in the design formulas, given in the chapters on Theory and Design, that steel resists all tension stresses and concrete in tension is disregarded, is, therefore, accurate enough for practical purposes. This assump- tion is still further justified by the fact that the tensile strength of concrete may be totally destroyed by shrinkage or temperature con- traction cracks and by construction joints. In analyzing tests, how- ever, it is necessary to take the tension in concrete into account, since its existence explains why the moment of resistance, based on the 22 TESTS OF REINFORCED CONCRETE stress in steel and figured by formula M = A,fjd, is smaller than the bending moment, and why the actual stresses in steel obtained from deformations are smaller than the theoretical stresses. (See Fig. 9, p. 29.) The amount of tension carried by the concrete may be estimated by comparing the moment of resistance, based on the actual stresses in steel, with the bending moment. Third Stage. Beams Failing by Tension in Steel.—When the steel reaches its elastic limit, one or two of the cracks, which have been small up to this point, begin to widen and extend toward the top. This is shown by the loads underlined in Fig. 7, p. 23. The deflec- tion increases appreciably as the eracks widen and extend toward the top (the neutral axis rising), the compressive area becomes smaller, and finally the beam fails by the total destruction of the compressive area. This condition is brought about by a small addition to the load at which the steel passes the elastic limit. Thus, the passing of the elastic limit of steel marks the failure of the beam. Ultimate strength of steel 1s never reached, and is, therefore, of no consequence in reinforced concrete design. Beams Failing by Compression.—For beams failing by crushing of concrete, the third stage is marked by cracks in the top of the beam, which appear after the clastic limit of the concrete in com- pression has been reached. At increased load, wedge-shaped pieces of concrete spall off and the beam fails. Appearance of First Crack and Corresponding Stretch in Con- crete.—Numerous tests prove that the appearance of the first cracks in reinforced concrete corresponds to about the same stretch as in plain concrete. This stretch may be taken approximately as 0.00012 of its length (corresponding. to 3 600 lb. per sq. in. in the steel) for 1: 2:4 stone concrete, and 0.00018 (corresponding to 5400 lb. per sq. in. in the steel) for cinder concrete.’ At this stretch, however, as discussed below, the cracks are very minute and not visible to the naked eye. | The early conclusion of Mr. Considére in France, as the result of his tests, that the stretch of concrete when reinforced was 0.002 of its _ length, or about twenty times the stretch of concrete without rein- forcement, has been disproved by further experiments. Professor Turneaure,2 in testing moist beams, observed, at about the same stretch at which the first cracks developed in plain concrete beams, 1 Technologic Paper No. 2, U.58. Bureau of Standards, 1912, p. 39. 2 Proceedings, American Society for Testing Materials, 1904, p. 498. 4 SJI0M INO T ) a O Vv ("Ge “a 29g) ‘QJ9LNUOAZ) OPTUBIDH “YUSTTBDIOJUIOY JO osvUBOIEg UI UOTPETIG ee \ ones ee PSE. xe Boos Ce - oe = seen & . gow Ci nace Ss oi ee # GOS! A ‘e8sor'T “H{ SInoy puv Aorqduinyy “] preyory ‘sassoyy Aq sysoy, “26d ‘ZI6I ‘Z ‘ON Jodeg o1dojouyoay, ‘sprepuvyg Jo nvoing “gx Y}IM syovsiy JO UOT}soOg UI UOIVUVA—) ‘DI 24 TESTS OF REINFORCED CONCRETE dark marks which he called watermarks. Part of these watermarks developed later into actual cracks. Professor Bach * investigated the subject further and came to the conclusion that watermarks are places where adhesion between particles of concrete becomes weak- ened just previous to the formation of cracks. In plain concrete, each watermark develops into a crack. In reinforced concrete, on the other hand, only a part of the watermarks actually open, be- cause the steel strengthens these weakened spots and either retards the appearance of actual cracks or prevents their formation altogether. Professor Bach’s Tests.—Professor Bach’s tests ? in Stuttgart, summarized below, give the relation between actual and computed tensile stresses in concrete and steel just before the first crack, in beams with different mixes of concrete, different percentages of steel, and various conditions of storage. All of the values are high, as Bach evidently obtained a stronger concrete than the same propor- tions give ordinarily. It must be noted further, as has been empha- sized elsewhere, that the actual stresses in steel at this stage are | much smaller than the computed stresses. This does not affect the accuracy of the ordinary formulas for practical design, as it does not increase the factor of safety nor the load at elastic limit. The Influence of Mix of Concrete——The table on p. 25 gives the tensile stresses in concrete at first crack, the actual stresses in steel and the stresses computed by formula fs = iG As would be expected, the richer concretes resist larger stresses before cracking than the leaner concretes. The difference between computed and actual stresses in steel is also larger for rich concretes than for lean concrete. It is interesting, however, to note that the ratio between the computed stresses in steel and the actual stresses equals 35 for all mixes of concrete. Influence of Storage-—The tensile stress in concrete, fi, was smaller for beams stored dry than for beams kept wet, the difference amounting on an average to about 20 per cent. Concrete stored dry tends to shrink, causing initial tensile stresses in concrete because free movement of concrete is prevented by the adhesion of concrete to steel. Concrete kept wet, on the other hand, tends to expand, which, prevented by the steel, causes initial compressive stresses in concrete. When loaded, the initial tensile stresses increase tension 3 Bach. Spannungen unmittelbar vor der Rissbildung. Deutscher Ausschuss, Heft 24, 1913. PROFESSOR BACH’S TESTS 25 on the section while initial compressive stresses decrease it.. To concrete in building construction the values for dry storage are applicable because, even if the concrete is kept wet during construc- tion, in course of time it will dry out and the ultimate amount of shrinkage will be substantially the same as if it were held in dry storage. Actual and Computed Stresses at First Crack for Different Proportions of Concrete. (See p. 24) Age of beams at test, 45 days; aggregates, Rhine sand and gravel; ratio of steel, p = 0.0056. Wet storage. Compiled from tests by C. Bacu. Strength of Plain Concrete Tensile Stresses at First Crack Lb. per sq. in. Lb. per sq. in. In concrete In steel Proportions Compressive| Tensile Computed Actual by formula ft stresses M fs fs = het 123:: 4 2 100 198 290 3 900 13 400 REE be 3 750 270 380 5 150 17 200 CR Gast 4 400 330 485 6 G00 23 000 Tensile stresses, f;, and actual stresses, fs, are figured by formulas on page 207, in which the tensile stresses in concrete are taken into account. The stresses computed by formula, fs = , on the other hand, are figured neglecting the M Agia tensile value of concrete. Influence of Percentage of Steel.—Professor Bach’s tests* show that in concrete beams of the same proportions the actual unit stresses in concrete and steel at first crack are constant, irrespective of the per- centage of steel in the beam. ‘The theoretical stresses in steel at the first crack, however, figured by the ordinary formulas neglecting the tensile resistance of concrete, vary with the percentage of steel. Similar results, as shown in the table below, were obtained in the tests by the Bureau of Standards carried on by Mr. Richard L. Humphrey and Mr. Louis H. Losse.® 4Bach. Spannungen unmittelbar vor der Rissbildung. Deutscher Aus- schuss, Heft 24, 1913. > Technologic Paper No. 2, U. 8. Bureau of Standards, p. 39, 26 TESTS OF REINFORCED CONCRETE Actual and Computed Stresses with Different Percentages of Steel. (See p. 25). ee | Tensile Stresses in Steel at First Crack Lb. per sq. in. Proportions Experimenters of = P Days|Actual}| Computed by ordinary formula Concrete stress lb. per sq.in.| p = .005 | p =.01 |p =.02 Bachsuicinnncet eee 1522.5: 2) 45.1°6 600 |, 25 000 13 000 8 000 Bureau of Standards.| 1:2: 4 28 | 4200} 25000 13 000 8 000 Influence of Consistency.—A wet consistency reduces the strength. At an age of forty-five days for 1 : 2:3 concrete, p = 0.0056, and for percentages of water by weight, varying between 6.8 and 10.0 per cent, the tensile stress, fi, at first crack ranged from 395 lb. to 310 lb. per sq. in., while the compressive strength of the same con- crete ranged from 3800 Ib. to 2360 lb. per sq. in., and the tensile strength in direct pull, from 485 Ib. to 245 Ib. per sq. in. Influence of Age.—Increase in strength with age, as determined by Bach, is shown in the table below. Actual and Computed Stresses at Different Ages. (See p. 26) Proportions of concrete, 1 : 2:3. Ratio of steel, p = 0.0056 Compiled from tests by C. Bacu * an a a ee er va | Stresses at First Crack Lb. per sq. in. Age Age Age Age 28 days 1 45 days | 6 months | 1 year f, Actual stresses in concrete... .| 360 380 466 495 f, Actual stresses in steel....... | 4900 5 100 6 300 6 700 fs Computed stresses in steel....} 16 400 17 500 21 700 23 000 ee ee eee The computed stresses, fs, are the stresses in steel figured by Formula (7), p. 207. * Bach. Spannungen unmittelbar vor der Rissbildung. Deutscher Ausschuss, Heft 24, 1913. POSITION OF NEUTRAL AXIS 27 Position of Neutral Axis.—The position of the neutral axis in rein- forced concrete beams varies with the percentage of steel and the strength of concrete and also with the intensity of the loading. For beams with large percentages of steel, the initial position of the neutral axis is lower than for beams with smaller percentages of steel. With the same steel and stronger concrete, the neutral axis is higher than with a weaker concrete. In any beam, the neutral axis at the beginning of the loading nearly coincides with the center of gravity of a section in which the steel is considered as replaced by an area of concrete equal to the area of steel times the ratio of the moduli of elasticity. With the progress of the loading, it moves upward. Figure 8. p. 27, illus- arn oe Moment | Computed = f ‘A : tae 7 ' y 20 Initial = Moment S44 center Line ' Computed ~ = = al tA = Center Line ; H | Moment.7y44 Center Line 4 Initial lean O75) ae i] ' “4 ‘ ' Initial S ' ' Crack, & : ‘ f 3 “gon i gt ' " ye Bar | oy Bar: | 1 Bar - ———— - - Fic. 8.—Change in Position of Neutral Axis During Loading for Different Percentages of Steel.° (See p. 27.) trates the typical movement of the neutral axis during loading for beams with different percentages of steel. As is evident from the figure, the position of the neutral axis for different loadings was deter- mined by plotting at proper levels the deformation of the upper concrete fiber and the deformation of steel. The intersection of the line obtained by connecting the two points and the vertical section of the beam gives the position of the neutral axis. For usual per- centages of steel, the distance from the compressive side of the beam under working loads is three-tenths to four-tenths of the depth. The formula for location of neutral axis is given on p. 129. Stresses in Steel for Varying Intensity of Load.—Figure 9, p. 29, gives the typical deformation of steel and of the upper fiber of concrete in inches per inch of length in beams with different per- centages of steel, based on the tests of Messrs. Humphrey and Losse.’ 6 Bach ‘“‘Biegeversuche mit Eisenbetonbalken,” Berlin, 1907, pages 7 and 8. 7 Technologic Paper No. 2, U. 8. Bureau of Standards, 1912, p. 39. 28 TESTS OF REINFORCED CONCRETE The deformation curve for steel is not a straight line, but a com- posite curve, the shape of which varies with the percentage of rein- forcement. The deformation and, therefore, the stresses in steel at the first stage of the loading, that is, before the first crack, are com- paratively small and proportional to the load, so that the deformation curve for this stage is almost a straight line. At deformation equal to the ultimate deformation in plain concrete beams, cracks appear in the concrete, and the tensile stresses borne by it are transferred to the steel, causing an abrupt change in the steel deformation curve. As is evident from the change in deformation, as shown in the dia- gram, the change in the direction of the curve is much larger for smaller percentages of steel, because the amount of tensile stress, constant for beams of equal cross sections, which is transferred from the concrete to the steel, is distributed over a smaller amount of steel, and the increase in the unit stress in steel is therefore larger. On the deformation diagram, the load at first crack is marked by the change in deformation from a straight line to a curve. After the first crack, a large proportion of the total tensile stresses is carried by the steel. The concrete, however, still carries a small proportion of tensile stresses, dependent in amount upon the per- centage of reinforcement in the beam. Because of these stresses carried by the concrete, the deformation in steel at different intensi- ties of loading does not vary proportionally to the load. | It is absolutely necessary that this be taken into account in analyzing results from tests not carried to the breaking point, for instance, in tests of completed buildings. The actual stresses in steel corresponding to the deformation are smaller, for reasons indicated above, than the stresses computed by formula f; = SEN for the same load. In beams with small per- centages of steel, concrete carries a considerable portion of the stresses up to the breaking point of the beam. (This is shown by the deformation curve for beam with 0.49 per cent of reinforcement.) For larger percentages of steel, the dash line on the diagrams, which indicates the theoretical deformation of the steel obtained from Formula (9), p. 131, strikes the actual deformation curve at the deformation corresponding to a stress of 39 000 Ib. to 48 000 Ib. per sq. in. This indicates that, near the elastic limit, the actual stresses agree very well with the theoretical stresses. The fact that at design load the actual stresses in steel are much STRESSES IN STEEL FOR VARYING INTENSITY OF LOAD 29 lower than those obtained from formulas may create a wrong impres- sion as to the actual capacity of the beam. To get the right idea, ie S eee ae : oe : : : : ; : Deformation Per Unit of Length Deformatien Per Unit of Length Beam With 0,49 PerCent Reinforcement Beam With 0,98 PerCent Reinforcement 4 lal hy V/ 3 I~} é a 8 os} 8 = asl edhe ies c Peerarey (key bsliea te co ek meal aig ha bait ie Bilis ia ic bash Weight -+ isl had afta lies of Beam © Defatmation Per Unit of Length a3 Detbrmek Per Unit ire Beam With 1.30 PerCent Reinforcement Beam With 1.72 Per€ent Reinforcement Fic. 9.—Deformations in Steel and Concrete Due to Loading.* (See p. 27.) it must be remembered that the conditions at the elastic limit, and not those at the intermediate loads, govern the strength. It is * Technologic Paper No. 2, U. S. Bureau of Standards, 1912, p. 39. 30 TESTS OF REINFORCED CONCRETE desirable to get a certain factor of safety. This means that it is required to build the beam so that it will not fail under a smaller load than the design load multiplied by the factor of safety. When the distribution of stresses in a beam varies, i.e., when the distribu- tion at the design load is different from the distribution at the elastic limit, two methods of procedure are possible. The first method is to multiply the design load by the factor of safety and then use the dis- tribution and magnitude of stresses at the elastic limit. The second method is to use the design load in computations with unit stresses equal to the elastic limit divided by the factor of safety. However, the distribution of stresses used in this method is the same as exists at the elastic limit. ‘The second method is commonly used, with the understanding that the stresses used in design indicate only the factor of safety and not the actual stress. The formulas on p. 131, therefore, although they do not repre- sent the actual conditions of stresses at the design load, give the required factor of safety and are recommended for use in design. The behavior of a beam reinforced with steel in two layers is substantially the same as that of a beam with one layer. The only difference is that the steel is not as effective as if all the bars were placed in one layer. The bars in the second layer are nearer the neutral axis. Their moment arm is, therefore, smaller than that of the bars in the first layer. Also, their unit stress is smaller because it varies with the distance from the neutral axis. The method used in design where the depth of the beam is taken as the distance from the center of gravity of the bars to the top, is not exact, because it does not take into account the difference in stresses between the two layers. It is, however, exact enough for practical purposes. Tests of Beams Failing by Compression.—From tests of beams failing by compression, the following conclusions may be drawn: The computed compression stresses in the extreme fiber in con- crete, at the load causing failure, are much larger than the strength of concrete in compression obtained from control cylinders of stand- ard dimensions (8 X 16 in.) made of the same material as the beams, cured in the same way, and tested at the same time. As would be expected, the difference is much greater for the stresses obtained from formulas based on straight-line distribution of stresses than from formulas based on parabolic distribution. The difference between the fiber at failure and the strength of TESTS OF BEAMS FAILING BY COMPRESSION dl control cylinders may be explained as follows: In a beam, only the extreme fibers are exposed to the maximum stresses. The adjoining fibers, being nearer the neutral axis, are subjected to smaller stresses. If any weakness should develop in the extreme fibers in any part of the beam, causing them to reach the point of crushing, actual crush- ing would not take place, because the less heavily stressed adjoining fibers would come to the assistance of the heavily stressed fibers and would assume a part of the excess stress without exceeding their ultimate strength. “2- ee Round Bars “1-1 x Round Bar yl-l z" Round Bar PEE ” Round Bar 45° 2. a Round Bars --~ ver a Round Bar -3 » ie "Round Bars 3 -4 "Round Bars’ Fig. 10.—Effect of Diagonal Tension. Design and Loading of Test Beams. Selected Beams from Group I. (See p. 38.) 42 TESTS OF REINFORCED CONCRETE Beam 25, although the area of bars bent at 45° was 1.8 sq. in. against 1.98 sq. in. bent at 18° in Beam 25. No marked difference was found in strength for beams with bars bent at 30°, 40°, and 45° respectively. Tests of Beams to Determine the Efficiency of Web Reinforcement. p. 38.) Compiled from Tests by C. Bacu * a On Sn enn Area of Cross : Stresses at Section of B : Maximum Increase in ars j , Load Maximum , Load by 3 , Maxi- Ib. per sq. in. |Hooking of ® | Hori- ; > faa! mum Horizontal ., |zontal| Bent Bo (total) | ta & i iF | Ts v >, sq. in. |Sq. in. | | . Per cent nn ee Beams Loaded at Two Points | (See Deflection Cause of XI- eae Failure maxi- UM) Paes (See load eal note) in. | in. 0.09 | 0.03 D 0.19 | 0.06 D 0°35) O41 D- 0.28 | 0.09 | : D 0.45 | 0.14 D 0.38. 40218 D 0.47 '|. 0.15 T 0.41 | 0.14 D 0.42 | 0.15 D 0.47 | 0.16 ft, 0.41 | 0.14 D 0:36 4-081 D 11°38: OLS eee 35 900} 990) 14630) 185 | ........ TO. G1 Baa 54 300} 1 580} 22 580) 286 iy 164 Sr9l Bee 88 000] 2 430] 36 1380) 451 | ........ 25 | 3.83 | 1.92] 75 900) 2 210} 32 980) 407 22 27 | 3.83 | 1.92] 98 600] 2 860) 42 760) 528 | 20 29 | 3.91 | 1.78} 92 400) 2 680) 39 500) 501 19 30 | 3.91 | 1.78] 106 900) 3 090) 45 480 567 | ......-- 32 | 3.91 | 3.57} 98 600) 2 940} 42 350) 529 | ........ 36 | 3.91 | 2.64] 101 200} 2 940) 42 960} 535 12 38 | 3.91 | 2.73] 108 900] 3 160| 45 830) 573 11 AQ | 3.95 | 2.71) 100 100} 2 880) 41 690) 529 5 50 | 3.87 | 1.77| 81800] 2 440) 35 340 488 | ........ Sor anes TI one nv Nore: D = diagonal tension failure B = bond failure bond failure tension failure es) I Jia * Deutscher Auschuss fiir Eisenbeton, Hefte X, XII, XX, 1908 and 1912. + Areas of bars are converted directly from the metric dimensions. 10 are approximate to nearest sixteenth inch. Diameters in Fig. TESTS OF REINFORCED CONCRETE BEAMS 43 Position of Loads i) ‘ag -] 2" Round Bars 52 \Q- 12° Round Bars 53 \.9-1;2" Round Bars 2-1" Round Bars ~~--. 9.11" oy" aed a5 Round Bars aay Round 45° 45 60 ‘ 3 4 : nn u“ ” -2-7"-7- 14 54 sr y 8 Round Bars tt Round. W, é ‘ any yas. DD hy gy 19" Sy TI" Ny 9 8 Eat 16 a 16 " “ Round Bars -t Round -4 Ny SSM! 4 Sg", ey Ny -2"Round Bars 30° 35 et 745° , \ \ Fig. 11.—Effect. of Diagonal Tension. Design and Loading of Test Beam, Selected Beams from Group II. (See p. 38.) 44 TESTS OF REINFORCED CONCRETE (10) Bent bars, as well as stirrups, are effective reinforcement for diagonal tension. Compare Beams 30 and 38, p. 42, both of which failed by tension in steel. (11) The strength of beams with bars having sharp bends was smaller than for beams with a circular bend with a radius equal to about 12 diameters. (12) It is evident from comparison of the stresses at maximum loads with the elastic limit of steel that almost all the beams with bent bars failed by tension in longitudinal steel. Tests of Beams to Determine the Efficiency of Web Reinforcement. (See p. 38.) Compiled from Tests by C. Bacu * Area of (roast : Section of Stresses a Deflection Bars t Maximum Increase at Load Maximum Maxi- Load by Cause 5 one | mum lb. per sq. in. Hooking of Mat Half of aia Load Horizontal maxi- |Failure ., |zontaliBent mum S | (total) Bars eee mum | (See g Ib. load | note) S| te ae fe fs v eae 3 ag in. |sq. in. Per cent in. in. ! : : l ! Beams Loaded at Eight Points nL nner Ble SOs ake 46 900} 1°31.0)'22 420):252 | 222 eee 0.22 | 0.10 | DB 2) sooo) ae ee 67 300} 1 870} 32 020) 360 | ...2.505 0.37.1 0.15 |; DB S| 33 a0) ol eee 51 300] 1 420} 24 390} 273 | .......: 0.24 | 0.09 D G4. yoeO7 Ree eee 93 900} 2 552| 43 250} 492 | ........ 0.73 | 0.22 55 | 3.81 | 1.91] 73 300] 2 150) 36 890) 404 | ........ 0.50 | 0.20 8) 56 | 3.80 | 1.91} 100 300] 2 920] 50 300) 549 | ........ 1.01 | 0.27 ik 58 | 3.86 | 2.32| 95 300) 2 750) 46 690; 516 6 0.74 | 0.25 ie 60 | 3.91 | 2.37] 95 300) 2 760) 46 220) 518 4d 0.64 | 0.24 TL 62 | 3.94 | 2.74} 99 400) 2 860] 48 010; 539 17 0.88 | 0.26 - 64 | 3.89 | 1.70} 106 200) 2 950} 50 450; 566 18 0.88 | 0.26 T 66 | 3.91 | 2.73) 101 900) 2 900] 49 180) 552 | ........ 0.75 | 0.26 T | ee eee Note: D = diagonal tension failure B= bond failure T = tension failure DB = diagonal tension and bond failure * Deutscher Ausschuss fiir Eisenbeton, Hefte X, XII, XX, 1908 and 1912. + Areas of bars are converted directly from the metric dimension. Diametersin Fig. 11 are approximate to nearest sixteenth inch, 45 TESTS OF REINFORCED CONCRETE BEAMS (‘OF d aay) HOvg ‘doug Aq sysoJ, HOVG ‘doug Aq 84sa J, oN’ | UOTSUAT, [BUOSIC, [V (op ‘d aay) + ‘omnpreg uorsuay, TeoIdA,— eT “D1 dA L— ZI. DI] 46 TESTS OF REINFORCED CONCRETE BEHAVIOR OF REINFORCED CONCRETE BEAM FAILING BY DIAGONAL TENSION UNDER LOAD The difference between the intensity of loading at first diagonal crack and the ultimate loading, for beams without web reinforce- ment, depends upon the strength of the concrete. Lean, or green, concrete beams fail with little or no warning, so that the load at first diagonal crack coincides with the breaking load; whereas in richer and stronger concrete beams, diagonal cracks are visible for some time before final failure occurs. Figure 12, p. 45, shows a photograph of a beam of 1:2 :3 concrete, forty-five days old, with no stirrups, after failure by diag- onal tension and slipping of the bar. The beam was loaded at one- third points. At a load of 14700 lb., the first crack developed in the middle portion of the beam. At 17 640 lb., a diagonal tension crack developed in the outside third, just beyond the load. ‘This crack increased with the load and extended diagonally upward. At 22 000 Ib., the crack extended almost to the bottom of the flange. The diagonal tension cracks were much larger than the tension crack in the middle portion. At 22000 lb., small horizontal cracks devel- oped at the level of the horizontal bar. At further loading, addi- tional horizontal cracks appeared. At failure, which took place at the loading of 28 600 lb., all previous horizontal cracks combined and formed a continuous crack extending from the support to the load, as is shown in the figure. For comparison, Fig. 13, p. 45, shows a typical tension failure of a beam of similar dimensions, as shown in Fig. 12, p. 40, but pro- vided with stirrups. All cracks are tension cracks and are confined to the center of the beam. ‘These two beams were selected from the series made by Bach. BEAMS WITHOUT SHEAR REINFORCEMENT The maximum unit shearing stress at which beams without web reinforcement fail by diagonal tension depends primarily upon the richness of the concrete and the age, and in smaller degree upon the percentage of steel and the ratio of depth to length of span. The last two items can be neglected in ordinary design. Since diagonal tension failure in beams without web reinforcement 1s sudden, a large factor of safety is necessary. (See p. 244.) In all cases quoted BEAMS WITHOUT SHEAR REINFORCEMENT 47 below, unit shearing stresses were computed by Formula (63), p. 149. Effect of Age upon Web Resistance.—The effect of age is of great importance in determining the time for removal of forms and the age at which concrete can be loaded. It is illustrated in Fig. 14, p. 47, taken from tests made by Professor Talbot at the University of Illinois.! Unit Shearing Strength Pounds per Square Inch Age in Days Fig. 14.—Effect of Age upon Web Resistance. (See p. 47.) Tests by Pror. TALBOT Effect of Richness of Mixture of Concrete upon Web Resistance.— Figure 15, p. 47, taken from Professor Talbot’s ‘tests, shows the increase of web resistance with the amount of cement in concrete. The increase is quite marked although somewhat less than the increase in compressive strength. | 1:3:6 Mix. Unit Shearing Strength Pounds per Square Inch Per Cent of Cement Fig. 15.—Effect of Proportion of Concrete upon Web Resistance. (See p. 47.) Tests by Pror. TALBoT Effect of Percentage of Horizontal Steel upon Web Resistance.— As is evident from Fig. 16, p. 48, the percentage of steel has a 12 Bulletin No. 29, January 4, 1900. AS TESTS OF REINFORCED CONCRETE marked effect on web resistance, which can be attributed to two causes. First, for smaller percentages of steel, the deflection is larger with consequently increased tendency of concrete to crack. Second, with larger percentages of steel, the tensile stresses developed near the support are smaller, consequently the appearance of the tension cracks which later develop into diagonal tension cracks is retarded. Vertical Shearing Strength Pounds per Square Inch 90.75 1.00 1.25 a ee 2.25 Per Cent Reinforcement Fic. 16.—Effect of Percentage of Horizontal Steel upon Web Resistance. (See p. 47.) Tests by Pror. TALBOT Ratio of Maximum Shearing Unit Stress Involving Diagonal Tension to the Modulus of Rupture of a Plain Beam and to the Compressive Strength.—In beams without web reinforcement, from tests by Professor Talbot,!® the ratio of maximum vertical shearing unit stress in beams failing by diagonal tension to modulus of rupture averages 0.5, and to the compressive strength, of 8 by 16-in. cyl- inders !* averages 0.09. BEAMS REINFORCED FOR TENSION AND COMPRESSION Tests prove conclusively the effectiveness of steel as compression reinforcement. Professor M. O. Withey’s Tests.!°"—The series of 1906 consisted of eight beams, 12 ft. long; breadth = 8 in.; height = 11 in.; depth to steel = 92 in., with 2.9 per cent tensile reinforcement and varying amounts of compressive reinforcement. The web reinforcement con- sisted of three bars bent up in two different places at a very flat angle. 13 University of Illinois, Bulletin No. 29, January 4, 1909. 14In determining this ratio the authors have converted the results found in cubes to a cylinder basis. 15 Bulletins of the University of Wisconsin, Nos. 175 and 197, Series of 1906 and 1907. BEAMS REINFORCED FOR TENSION AND COMPRESSION 49 The results of the tests, although interesting, do not bring out fully the value of steel as compressive reinforcement, because all beams failed by diagonal tension, with the exception of the beam without compressive reinforcement, which failed in compression. Notwithstanding this, however, the maximum load of the beam without compression steel was 22 000 lb., while the maximum load for the beam with compressive reinforcement was 29 000 lb. The series of 1907 consisted of four beams, similar in design to the beams previously described, except that they were provided at each end with 10-j-in. round stirrups. All the beams failed in ten- sion at an average load of 34000 lb., showing an increase of 55 per Wis = Round 8 7" / pees Round tp oot Fanon ten Es Sa ata Sh = H ae gu 9a” er me 1, Round i VI - yr. Vill IX x, Fic. 17.—Dimensions of Beams, Stuttgart Tests. (See p. 49.) By Pror. C. Bacw cent over the beam without compression reinforcement. Still, because of the tension failure, the full value of compression reinforcement was not demonstrated. Bach’s Stuttgart Tests.!°—Bach’s tests of beams with compressive steel consisted of six types of beams, the dimensions and arrangement of reinforcement of which are shown in Fig. 17. The results of the tests are given on p. 50. The reinforcement of Beams VII, VIII, and IX is alike except that Beam VII in the middle portion has no stirrups while Beams VIII and IX have stirrups of the shapes shown in the drawing. The amount of the compressive reinforcement in Beam X is the same as in Beams VII and VIII, but the steel is of higher elastic limit. 16 Mitteilungen tiber Forschungsarbeiten aus dem Gebiete des Ingenieurwesen, Hefte 90 and 91. 50 TESTS OF REINFORCED CONCRETE Test of Beams with Compression Reinforcement. (See p. 49.) Concrete: 1:3:4 by volume. Aggregates: Rhine sand up to 0.27 in. diameter, and Rhine gravel up to 0.79 in. diameter; 9.5 per cent of water by weight. Age of test, 45 days. Compiled from Tests by C. Bacu i Computed Unit Stresses in Steel and Concrete at Maximum Load Based on n = 15 ; és Compressive Baus Maximum Bicone Computed é Load Unit Unit Unit : Unit Fiber Specimens cre ie Cubes tensile |compressive; ’ stress Stress to lb. stress stress in Ib ._ | Strength in steel in steel concrete | — per'sd- 1"! of Cubes ts f's te lb. per lb. per lb. per sq. in. sq. In. sq. in. ie 16 860 19220 ») Mets eeu 2 220 1 590 1.40 Aig ie 20 650 14 390 28 680 2 390 1 490 1.60 Vite 27 500 17 750 29 200 2 500 1 480 1.69 VIIL.. 29 000 18 720 29 800 2 670 1610 —-1.66 iD. Ge 28 600 18 480 31 620 2 590 1.520 1.70 2 e 36 000 23 200 | 36800 3 240 1 590 2.04 Beams VI, VII, VIII, and 1X failed by compression. Beam X, in which the bond strength of the compressive steel was exceeded, failed at a considerably higher load than Beams IX, which had the same amount of reinforcement, because the compression steel in Beam X was of higher elastic limit. The table on p. 50 gives the maximum loads, the stresses in steel and concrete at the load under the assumption of n = 15, the strength of maximum cubes, and ratio of strength of cubes to figured stress in concrete in the beam. In Beams VII to IX, the compression steel reached its elastic limit first, and for further loading kept the same stress till the elastic limit of concrete was reached. In Beam X, on the other hand, the elastic limit in the concrete was reached first, and after this, stresses due to the additional loading were carried by the steel only until both materials reached the elastic limit. This points to the adjustment between compressive stresses in steel and concrete after one of the TESTS OF BOND BETWEEN CONCRETE AND STEEL ol materials passes its elastic limit. The same phenomenon was observed in the test of reinforced concrete columns. From inspection of the table, it is evident that for beams with compression steel, the theoretical unit stresses in extreme fiber in the concrete itself, computed for the maximum test load by formulas on p. 140 and on the basis of m = 15, are much larger than similar unit stresses at which the beams without compression reinforcement failed. Since the same concrete was used in all cases, it is rational to assume that this extra strength must be attributed to compressive ~ steel. This shows that the compressive steel carries larger stresses, and that its actual effect is greater, than would be expected from the formulas. It is especially noticeable in Beam X, for which the computed maximum fiber stress in concrete was 3 240 lb. per sq. in. while the crushing strength of the concrete was 1 590 Ib. per sq. in. The above tests prove conclusively that compressive steel may be relied upon to strengthen the compressive zone of a beam, and that its effect is even larger than would be expected from the formulas. TESTS OF BOND BETWEEN CONCRETE AND STEEL Bond between concrete and steel, or the resistance to withdrawal of steel imbedded in concrete, may be divided into two elements: (1) grip caused by shrinkage of concrete; (2) frictional resistance caused by the unevenness of the surface of the bar. Both elements act together until the bar begins to slip. Then the grip is destroyed and frictional resistance alone resists the pull. | In deformed bars, the grip and frictional resistance are aided by the bearing of the projections on the concrete, but this does not come into play until after the first slip. The pull-out tests are treated separately from the bond tests in beams, because the action of bond stresses is different in the two cases. PULL-OUT TESTS Pull-out test specimens consist of bars imbedded in blocks. The load is applied at the free end of the bar and is resisted by the resist- ance to withdrawal of the steel imbedded in the block. In practice, similar conditions occur in end anchors for fixed or cantilever beams where the concrete at the support corresponds to the block in the pull-out tests. ‘The maximum stress in steel at 52 TESTS OF REINFORCED CONCRETE the edge of the support, which is transferred to the support by bond, corresponds to the applied force in pull-out tests. In computation, the bond stresses are considered as uniformly distributed over the whole surface of contact between steel and concrete. (See Formula 54, p. 268.) Actually, however, the bond stresses vary from a maximum at the edge of the support to a min- imum within the support. In many cases, in fact, the bar begins to slip at the place of application of the force before the bond resistance - of the whole bar comes into play. Therefore, ordinarily the portion of the bar near the point of support offers frictional resistance only, while the farther end of the bar offers erip and frictional resistance. The variation in magnitude of bond stresses along the length of | ‘mbedded bar depends upon the length of imbedment. Hence, in basing allowable unit stresses on the tests, the effect of the ratio of the imbedded length to the diameter of bar must be taken into account. When a bar imbedded in concrete slips, the movement of the free end is somewhat greater than that of the imbedded end, the difference being equal to the deformation of the imbedded portion of the bar under stress. Effect on Bond Strength of the Ratio of Length of Imbedment to Diameter of Bar.—The average bond resistance, considered as _ distributed uniformly over the total surface area of imbedment, is smaller for long imbedments than for short imbedments. At the University of Ilinois,'” in tests by Mr. Duff A. Abrams for 1}-in. plain round bars imbedded in 1 : 2 : 4 concrete, seventy-four days old, the average bond resistance for 6-in. imbedment (4.8 diameter of the bars) was 420 Ib. per sq. in., while for 24-in. imbedment (19.2 diam- eters), it was 328 lb. per sq. in. Similar results were obtained by Prof. C. Bach.!® 7 Method of Determining Bond Resistance.—In computing the bond resistance of a bar, the ratio of the length of imbedment to diameter of bar, and not the length of imbedment, is the determining item. The required length of imbedment increases in direct ratio with the increase of the diameter of bar. Thus a 25-in. imbedment is sufficient for a 3-in. bar because the ratio of the length to diameter is 50. It would not be large enough for a 1-in. bar because the ratio then is only 25. (See p. 268.) 17 University of Illinois Bulletin No. 71, December 8, 1913, p. 39. 18(. Bach. Zeitschrift des Vereines Deutscher Ingenieure, 1911, S. 859. PULL-OUT TESTS ay) Bond Resistance for Different Slips.—Figure 18, p. 53, shows the relation between the bond stresses and slips for plain and deformed bars during the progress of loading. As is evident from this diagram, for plain bars initial slip occurred at 260 lb. per sq. in., or at about 60 per cent of the maximum bond resistance. After the maximum 700 600 500 1:2:4 Concrete Age about 2 Months = Bond Stress in Pounds per Square Inch e 14 Deformed Round Bars, Embedment Variable + 1 4 Plain Round Bars, Embedment Variable O Plain Round Bars, Diameter and Embedment Variable 0 01 202 03 04 05 Slip of Bar in Inches Fic. 18.—Relation of Bond Stress to Slip of Bar During Progress of Loading. !9 (See p. 53.) bond resistance, which corresponds to a slip of 0.01 in. was reached, the resistance to withdrawal decreased. After a slip equal to five times the slip at maximum resistance has taken place, only 70 per cent of the maximum load is required to produce further slipping. The curves for the deformed bars are discussed on p. 54. Effect of Surface Condition and Shape of Bars.—The following conclusions may be drawn from Abrams’ tests: * University of Illinois Bulletin No. 71, December 8, 1913, p. 29. 54 TESTS OF REINFORCED CONCRETE The bond resistance of square bars is only 75 per cent of the bond resistance of plain round bars. Rusted bars (with no scale) give bond resistance 15 per cent higher than similar bars with ordinary milled surface. The bond resistance of T-bars per unit of area decreases with the increase in size. For 1:2: 4 concrete, imbedment 8 in., and age 70 days,2° the maximum bond resistance of 1-in. round plain bar was 370 Ib. per sq. in.; of 1-in. T-bar, 310 lb. per sq. in.; and 2-in. T-bar, 220 lb. per sq. in. Influence of Age and Mix.—The table on p. 58 gives the effect of age and mix on bond of {-in. plain round bars and of 3-in. corrugated square bars. py et Influence of Freezing—In Abrams’ tests, specimens made out- doors in freezing weather, where they probably froze and thawed several times during the period of setting and hardening, were almost devoid of bond strength. Ratio of Compressive Strength to Bond Resistance.—The ratio of bond strength at first slip to compressive strength, of 8 by 16-in. cylinder, is about 0.13, and of the maximum bond strength, 0.19. These ratios were determined by Mr. Abrams from tests on speci- mens varying in age from two days to 25 years, and proportions from 1:1:2+t01:5:10. These values agree very well with the results obtained by other experimenters. Deformed Bars.—Results of pull-out tests with deformed bars are given 2! on pp. 53, 55, and 57. ‘The first slip for the deformed bars occurs at about the same stress as for plain bars. After the first slip, the projections help to resist further slipping. Considering all the bond stresses except those resisted by frictional resistance taken by the projections, the bearing stresses on concrete for some types of deformed bars at large slips are very large, reaching in some cases 14 000 Ib. per sq. in. of the area in contact. This high compressive stress on concrete explains the splitting of the blocks in pull-out tests. Since the allowable working stresses are only a frac- tion of the ultimate bond stress the bearing stresses on projections always are within safe working limits. The maximum bond stresses, being accompanied by large slips, cannot be utilized in construction, where only a very small slip 1s permissible; consequently, the working bond stresses must be based 20 University of Illinois Bulletin No. 71, December 8, 1913, p. 49. 21 University of Illinois Bulletin No. 71, December 8, 1913. PULL-OUT TESTS 55 Maximum Bond Stress and Bond Stress at 0.001 Inch Slip for Varying Propor- — tions and Ages.* (See p. 53.) By D. A. ABRAMS Stress in Pounds per Square Inch of Surface of Bar Proportions 1) a AS ee Pera Sees: Tee an leo Lege .o Size of Bar Age Bue) 2 le] 8 jue] 2 ld) 8 le B(S2| 2 |Se| & |E5| 8 \Es| 2 Ss Rloe| & ISB) 2 [o8l 8 |o8) & \o8 ie lew | elt le lak Viet | a ac | af a fn 2 days PLOT HOT i2a! S123 rSol Salo) S2hr e277) a7 4 days 197| 156] 231 | 195) 153} 110) 77| 43) 49) 32 7 days 246} 202} 300 | 250} 226 | 158) 165) 112) 54) 32 28 days to | 32 days plain 60 ei round i | 530| 399) 554 | 492} 452 | 363) 311) 227) 190) 135 65 ree 120 days to 666] 479| 667 | 538} 603 | 469] 536] 398] 210} 172 132 days 16 months | 779] 656} 8961] 875} 8414] 800} 372| 333] 373] 253 2 days 231| 96 205 | 92} 219] 97| 157| 44) 64| 13 4 days 368| 1761 258 | 115] 305 | 129] 239] 74] 110] 23 7 days 419| 171] 330 | 140} 459 | 187| 286| 104] 133] 35 28 ae: : | 828| 344} 560 | 281} 641 | 306] 462| 179) 273] 97 8-inch 32 ee corru- ted 60 ibe ee | 1 132] 4981 1 053%] 556} 854 | 434] 623] 280] 391] 139 square 65 hee 120 days to | 1 153] 599} 1070 | 564) 1079 | 576) 746) 326) 470; 159 132 days 16 months | 1 535) 892 ailatb, shlewer| a eee ie Me (er ene re, 9) [pee ncwre: Is ee area a 8) | t ee * University of Illinois Bulletin No. 71, December 8, 1913, pp. 82-83. + The reason for relatively low strength of Ll: 1: 2 concrete and 3-in. bars is unexplained and may be an erratic result. t Bars stressed to or beyond yield point. 56 TESTS OF REINFORCED CONCRETE on stresses at a slip not exceeding 0.01 in. rather than on ultimate bond strength. The factor of safety for deformed bars based on this slip, however, may be made smaller than for plain bars, since the high ultimate bond strength and the existence of mechanical bond reduce the danger of actual bond failure. This is of special importance during construction, when comparatively green concrete may be called upon to support a considerable construction load. Allowable Working Stresses.—Allowable working unit stresses based on the tests are given on p. 263. BOND STRESSES IN BEAMS The method of computing bond stresses in reinforced concrete beams is given on p. 262. Although the formulas do not represent the actual conditions in a beam, they form, as explained below, a proper basis for design with values for working stresses based on tests and figured for the same assumptions. The computed maximum bond stresses in a beam occur at points of maximum shear. With uniform loading, this is at the supports and decreases uniformly to zero at the center of the beam. In beams loaded at one-third points, maximum bond stresses act in the outside thirds and are zero in the central portion of the beam. Phenomena of Bond Tests.—The bond stresses in beams are caused by the change from point to point, i.e., the increase, in the stresses in the longitudinal steel. This increase in stress in steel as computed is proportional to the amount of increase in the bending moment, and therefore equal to the vertical shear. Actually, how- ever, the change in stress in steel is affected by the presence of tensile | stresses in concrete, the amount and the proportion of which to the total tensile stresses is different in different parts of the beam. The effect is smaller near the point of maximum tensile stresses (where the concrete is cracked), and larger near the support where concrete may carry stresses even at maximum load. The increment of stresses in steel is not proportional to the shear; the bond stresses which are caused by that increment are, therefore, not proportional to the shear. The table on p. 57 gives observed bond stresses and computed bond stresses for varying intensities of loading for a beam loaded at one-third points. Since the shear between the support and the point of application of the load is constant, the computed bond stresses BOND STRESSES IN BEAMS Distribution of Bond Stress in Reinforced Concrete Beams. (See p. 56.) 57 Beams 8 by 12 in. in section and 10 in. deep to center of reinforcing bar. Loaded at the one-third points of a 10-ft. span. All beams failed by excessive tensile stress in the reinforcing bars. Compiled from Tests by Durr A. ABRAmMs * Beam | Size and Kind | Age at No. of Bar Test 1055.6}One 1-in. Plain} 2 yr. Round 1055.3)One I-in. Plain} 2 yr. Round 1049 .3)One 1§ in. Cor-|13 mo. rugated Round Applied Beam Ib. 2 000 4 000 6 000 8 000 10 000 11 700 2 000 4 000 6 000 8 000 10 000 10 700 2 000 4 000 6 000 8 000 10 000 12 000 14 000 16 000 18 000 20 000 21 000 21 900 Average Load on | Computed | Over region bond stress | just outside Ib. per sq. in. ies per sq. in.|Ib. per sq. in. 38 76 114 152 190 222 38 76 114 152 190 203 34 68 102 135 170 204 236 270 306 308 359 O10 Observed Bond Stress of load points a) oye) Walt ie te” ae ie * University of Illinois Bulletin No, 71, December 8, 1913, p. 193. + These stresses are, in general, the average bond stresses developed over a length of about 12 in. in the portion of the beam about 4 to 16 in. outside the load points. t The average observed stress over a length of 9 to 151n. at the ends of the beam, 16 34 36 64 Sly 238 15 54 95 100 130 156 20 45 95 135 150 150 225 260 290 315 360 300 Near ends of beam { 58 TESTS OF REINFORCED CONCRETE are constant. ‘The observed bond stresses, however, near the sup- port are smaller than just outside of the points of application of the load until the steel reaches the elastic limit, after which a readjust- ment takes place and the bond stresses become equalized. From the table, it is evident, that in beams 1049.3, for example, the observed bond stress just outside of load points for a load of 16 000 Ib. is larger than the average computed bond stress for the ultimate load, 1e., 21900 lb. This explains why, for beams failing by bond, the aver- age computed bond stress at the ultimate load based on Formula (50), p. 262, is smaller than the maximum bond strength in pull-out tests. The bond stresses given in subsequent discussion are those obtained by Formula (50), p. 262. Effect of the Distance of the Load from the Support on the Bond Resistance.—As may be inferred from the discussion of the phe- nomena of the bond stresses, the average bond resistance is larger as the load is placed nearer the support. Prof. C. Bach,” in tests of beams of 1 : 2:3 concrete at age of forty-five days, finds values of ultimate bond strength for distances 9.8 in., 19.7 in., and 29.5 in. from the support to average 507, 325 and 308 Ib. respectively. Bond in Beams in Pounds per Square Inch Age in Days Fig. 19.—Effect of Age on Bond in Beams 1: 2:4 Concrete. (See p. 58.) Tests by Pror. WITHEY Effect of Age on Bond. — Figure 19 on p. 58, from Professor Withey’s tests at the University of Wisconsin 7° shows the increase of bond strength with age, for 1 : 2 : 4 concrete. 22 Widerstand Einbetonierten Eisens Gegen Gleiten. Einfluss der Haken, von C. Bach and O. Graf, p. 18. 23 University of Wisconsin Bulletin No. 321, October, 1909, p. 27. BOND STRESSES IN BEAMS 59 Professor Bach found for 1: 2:3 concrete the following bond strength: MR RO ee GY eee es 28 days 45 days 6 months One year Beams kept moist, lb. per sq. in... 278 308 393 435 ca aa katy, is FO. afl 319 356 363 He suggests the following formula for increase in bond strength with age: AM 6 1 (Fine 745(1 _ Vee :) Where uw = unit bond strength in lb. per sq. in; A = age in months. ) Effect of Mix of Concrete.—From tests, it is evident that the richness of mortar in concrete affects the bond strength considerably. The quality of stone is of little effect, provided pockets around the reinforcement are prevented. ‘The table below gives values for bond strength for concrete of different proportions. Bond Strength in Beams for Different Proportions of Concrete. (See p. 59.) Beams, 5 in. by 5 in. by 5 ft. 6 in. long. Reinforcement, 3-3-in. round bars. Lower bars imbedded in concrete for length of 10 inches at both supports. Beams tested on 5-foot span. Compressive tests on separate specimens. Compiled from Tests by Morton O. Wirury * : Age, | Average Bond | Compressive Strength es Days Bare jeerees te Ib. per sq. in. Ib. per sq. in. ap eae: 60 Limestone 276 1 790 | 60 Gravel 215 2 200 2326 60 Limestone » 216 830 723 DOP eee SS 267 1 600 | * University of Wisconsin Bulletin No. 321, October 1909, p. 27. Hooks as End Anchorage.—The requirements of a properly con- structed hook are: (1) it should permit the stressing of the steel to its elastic limit without appreciable movement; (2) the bearing stresses on the concrete must be within a safe limit. Since the allowable bearing stresses on concrete depend upon the properties of the con- 60 TESTS OF REINFORCED CONCRETE crete, the factor of safety against crushing must be the same as that used in determining the allowable fiber stresses in concrete. Tests show that the crushing strength of concrete, when confined, is much larger than the crushing strength of cubes or cylinders. Hence, the safe bearing stress of the hook on the concrete should be based on the crushing strength of confined concrete. In comparing, there- fore, the relative efficiency of hooks, their bearing area is of first importance. When used for end anchorage, hooks which allow stressing the steel to elastic limit, but which at the same time split or crush the concrete, have not the required factor of safety as far as concrete is concerned, because at working stresses the concrete would have only a factor of safety of 2, instead of 3 as required by rational design. Tests 24 made for the Eastern Concrete Construction Company at the Massachusetts Institute of Technology determined the capa- city of the hook, but did not determine the load at which the first movement of the hook took place. . In all the tests, 2-in. round bars were imbedded in blocks 12 in. square and 15 in. long to a depth of 12 in.; with additional bends of different lengths. Right-angle bends and semicircular bends on a 3-in. diameter were tested. Several specimens of each type were tested, and the results were extremely uniform. The following conclusions may be drawn from the tests: (1) A 4-in. right-angle bend in a {-in. round bar (5 diameters) combined with 12-in. imbedment (16 diameters) is sufficient to stress the steel to its elastic limit. This hook, however, crushed the con- crete and split the block; therefore it does not give the required factor of safety against crushing of concrete. A longer bend does not increase the security because the bearing stress is not appreciably reduced. (2) A semicircular bend with a diameter four times the diam- eter of the bar is more effective than the square bend and is prefer- able because the bearings stresses on concrete can be kept within working limits. Action of Hooks in Beams.—Beams in which longitudinal steel is provided with hooks show a much larger load-carrying capacity than similar beams with ends of bars straight. Tests at age of forty-five days, by Professor Bach, on beams of 1: 2:3 concrete, 12 in. square and 6-ft. span, reinforced with one 0.98-in. diameter : 24 Concrete, Plain and Reinforced, Second Edition, p. 466. SPLICES OF TENSILE REINFORCEMENT IN BEAMS 61 round bar provided with three different kinds of hooks, gave the carrying capacity of the beam without hooks as 14 3380 lb.; with right-angle hook, 24 250 lb.; with 45° hook, 25 800 lb.; and with circular hook, 28 060 lb. The beam with Rak Sard neele failed by Pe htening the hook. SPLICES OF TENSILE REINFORCEMENT IN BEAMS AT POINTS OF MAXIMUM STRESS Tests have been made by H. Scheit and .O. Wawrziniok 2° to determine the effectiveness of different methods of splicing steel at the point of maximum stress. The beams were 12 in. square, of spans 63 ft. and 10 ft., reinforced with l-in. bar. They were tested with two symmetrical loads spaced 3 ft. 3 in. apart for the shorter beams, and 5 ft. apart for the longer beams. Straight splices were made with a lap of 10, 20, and 30 diameters respectively for the short beams, and 40, 50, 60, 70, and 80 diameters for the long beams; in hooked splices, the hooks consisted of a semi- circle with an inside diameter of 5 diameters of the bar and an extra length of 6 diameters of the bar parallel to the bar, and the bars were lapped 10 in, 20 in., and 30 in. respectively. Results.—For straight splices, the best results were obtained with a splice of 50 diameters, with which the elastic limit of steel was reached. Hooked splices proved very effective. Even a 10-diameter lap (the smallest lap used) in combination with a hook, as described above, was sufficient to provide the same carrying capacity as the beam without the splice, DEFLECTION © The deflection in reinforced concrete depends primarily upon the ratio of the depth of the beam, or slab, to the span. It also depends upon the percentage of tension and compression reinforcement, and in T-beams, upon the width of the flange. Influence of Percentage of Steel upon Deflection.—For equal depths and widths, the deflection of beams increases with the decrease in the percentage of tensile steel. Figure 20, p. 62, shows the 25 Deutscher Ausschuss fiir Eisenbeton, Heft 14, 1912. 62 TESTS OF REINFORCED CONCRETE deflections of beams 13 ft. long, 8 by 11 in. in cross section, tested on a 12-ft. span, by twe equal loads applied at one-third points. The test was made by Messrs. Richard L. Humphrey and L. H. Losse.*® The deformations in steel and concrete for the same beams are shown (oy 0s Ono 20000 78000 16000 Ce 14000 - RRCRuGRaRecaa Oy ‘f 12000 qt 10000 : 1 8000 ee Total Applied Load in Pounds 0.42 6000 | 4000 2000 0 0.05 010 0.15 020 0.25 0.30 0.85 0.40. 0.46 0.50- 0.55 ‘ Deflection of Beam in Inches Fic. 20.—Deflection of Beams with varying Percentage of Steel (See p. 61.) Tests by Ricuarp L. Humpurey and L. H. Losse Influence of Width of Flange upon Deflection.—In Bach’s tests *7 to determine the effect of width of the flange, the results given in the table on p. 63 were obtained. It will be seen that although the percentage of steel based on the area of the stem is the same in all cases, the deflection for equal loads is smaller for beams with larger widths of flange. 6 Technologic Paper No. 2, U. 8S. Bureau of Standards, June 27, 1911. 27 Mitteilungen tiber Forschungsarbeiten aus dem Gebiete des Ingenierwesen, Hefte 90 and 91. DEFLECTION 63 Deflection of T-Beam with Varying Widths of Flanges. (See p. 62.) Span of beams, 9.84 feet; reinforcement, four 13%-in. round bars; load applied at one-third points. By C. Bacu * Deflection in Inches T-Beam, depth 9.84 in.; width of stem, 7.1 in. Total Load Rectangular | Beam Width of flange in inches 7.1 X 9.84 in. | _ 18.9 29.5 39.4 lb. in. bar in. in. 8 800 0.106 0.071 0.047 0.042 17 600 0.376 Oe. 0.110 0.097 ELE UO a a 0.368 0.188 0.161 See MR ESS es a 0.290 0. 235 2 AS ee Se ee 0.467 0.351 rare Get Eck toto, Li Pas 0.544 * Mitteilungen iiber Forschungsarbeiten aus dem Gebeite des Ingenierwesen, Hefte 90 and 91. Influence of Compressive Steel upon Deflection.—The table on p. 64 gives deflection of beams without compressive reinforcement and with different percentages of compressive reinforcement. From the figures, it is evident that for equal percentage of tensile rein- forcement the deflection decreases with the increase of compression reinforcement. 64 Deflection of Beams with Compression Steel. TESTS OF REINFORCED CONCRETE (See p. 63.) Allbeams, 7.1 X 9.8 inches; span, 9.84 feet; tensile reinforcement, four 1,%;-in. round bars; load applied at one-third points. By C. Bacu * i Deflection in Inches eee nee ee ES Ee ee Total Load Compression steel, per cent 0 0.4 1.58 1 .58-% Ib. in. in. in. in. 4 400 0.046 0.042 0.038 0: 037) 8 800 0.119 0.102 0.086 0.084 13 200 0.256 0.184 0.143 0.139 17 BOO m4 Meohenra cree 0.298 0.210 0.203 922-000. wl 52 Sa ee 0.287 0.276 | 7 | * Mitteilungen tiber Forschungsarbeiten aus dem Gebiete des Ingenicurwesen, Hefte 90 and 91. + High elastic limit steel used. TESTS OF CONTINUOUS BEAMS. Since in concrete construction beams are usually continuous over several supports, it is of the greatest importance to determine by tests whether this continuity can be relied upon. Tests of Continuous Beams by Prof. H. Scheit and Dr. Ing. E. Probst.22—These tests included the concrete beams shown in Figs. 21 to 23, pp. 66 to 68. Two beams of each type were tested to destruction. The spans and the reinforcement for the loaded spans were the same for all beams. The complete series included beams as follows: Type 1, simply supported beams. Type 2, continuous beams over two spans. | Type 3, continuous beams over three spans, end spans loaded. Type 3a, beams similar to Type 3, but supported by columns; loading same as Type 3. Type 4, beams of five spans, alternate spans loaded. 28 “Untersuchungen an durchlaufenden Eisenbetonkonstruktionen,” Berlin, 1912. TESTS OF CONTINUOUS BEAMS 65 Figs. 21 to 23 show the cracks, and Fig. 22, Type 3, the deflection at different loads. Results of tests are discussed below: Comparison of Theoretical Deflection with Actual Deflection.— To determine the efficiency of continuous beams, the ratios of deflections of continuous beams to those of simple beams obtained from tests were compared with theoretical ratios for homogeneous beams. For continuous beams of two spans, Type 2, the observed ratio of deflections was between 0.42 and 0.45, while the theoretical ratio was 0.40. For beams continuous over three spans with end spans only loaded, the observed ratio was 0.69 against the theoretical ratio of 0.74. For beams of Type 3 of the same design as the one above but in which the ends were connected with columns, the ratio varied from 0.34 to 0.37. From the theoretical figures, it appears that as far as deflection is concerned, this type is almost midway between a beam fixed at one end, for which the ratio is 0.40, and a beam fixed at both ends, for which the ratio is 0.20. Type 2.—The beams continuous over two spans failed at the sup- port at an average load of 14 240 lb. per lin. ft. The theoretical 2 negative bending moment at the support is — = The stress in steel at the support for the maximum load, figured on the basis of the above bending moment, is about 64000 lb. per sq. in. It is evident that this stress is much higher than the elastic limit of the steel used in the test, which shows that the assumed theoretical bending moment coefficient is too large. Based on the moment of resistance for the yield point of steel, we get a bending moment coefficient of 10 instead of the theoretical 8. The point of inflection was found to coincide with the theoretical point of inflection. (See Fig. 21, p. 66.) Type 3.—In the beams continuous over three spans, the first cracks appeared at the bottom of the beam in the central portion of the loaded span at a load of 2 565 lb. per lin. ft. In the unloaded span, the first crack appeared at the top of the beam at a load of 5 600 lb. per lin. ft. The sequence of other cracks and the loads at which they appeared is evident from the illustration. The failure at a load of 12 600 Ib. per lin. ft. was caused by passing of the elastic limit of steel in the loaded spans. The deflection diagram, Fig. 22, p. 67, gives a positive proof of the continuous action of the beam. As is evident from the diagram, 66 TESTS OF REINFORCED CONCRETE the deflection in the end span is positive, while in the center span it is negative. The computed stresses and bending moments at the dif- ferent loads agree quite closely with the measured stresses. The measured compressive stress for the maximum load in the middle span (which was not loaded) was found to be 1 500 |b. per sq. in., which is almost identical with the theoretical stress. The cracks in the unloaded span, which are uniformly distributed over its whole length, furnish a conclusive proof of continuous action. If no pro- | t u a ie. Hits 72 . J ‘ 4 \, Position of Loading 5l'Round °76 oun SSUES JINN. La Teo! 71" 5-10" Round km 3'- 10" phe 9 ee ivebeecseneeee Qin.) e=neeane man ane na SS SN as 51/9 3h 37X 3/ | i haat 4 G2) 9G) 4) BYIG) NC5) A Cracks in Beam During Loading Fic. 21.—Continuous Beam of Two Spans; Type 2. (See p. 65.) vision had been made for the negative bending moment in the unloaded span, failure would have been certain. Type 3a.—In the beam continuous over three spans and mono- lithic with columns, as shown by Fig. 23, p. 68, the connection between beams and columns was not rigid, as would be the case with rigid frames, but was built as in ordinary building construction. The beams were of exactly the same design as in Type 3. The compari- son, therefore, gives the effect of the connection of the beam with the column. The first cracks in the loaded span appeared at a load of 4590 Ib per lin. ft., and in the unloaded span, at a load of 10 935 lb. per lin. ft. The corresponding figures in Type 3 were 3 565 lb. per lin. ft. and 5 670 lb. per lin. ft. At a load of 13905 Ib. per lin. ft., TESTS OF CONTINUOUS BEAMS 67 the first cracks appeared at the top of the end column; and at 16 740 Ib. per lin. ft., a crack appeared at the top of the middle column. The beam failed at 17 620 lb. per lin. ft. by steel passing the elastic limit. At the time of failure, cracks were observed in the com- pressive part of the interior column. foes Ie go 7d pith Ao Idec, » 78 OS ae a ae fg tn a 3-5" Round 10-5" Round pe 7" r ane 11" y" She aie ae, ie 2 8G one (396 Round ot Round -----~ < ES) ‘, ry @ \TN a A. 2-9 et y we aN : x ati : : a0 5 i x 5 : ree nf 17g | Rou i git = pe ee - cf tz iz Section ae-7f =; ?, ees : pak 8 = 10- 6 & a : a-b ee oe re eaten sp 7 md 0 cogsee carton ieee ae Tae ee H g.! "pound '5- Up ae a, 16 ound Arrangement of Steel 2 -H Round” 0.098" Deflection of Beam 9.098" Fic. 22.—Continuous Beam of Three Spans; Type 3. (See p. 65.) After the tests, cracks were found at the bottom of the columns, located in reverse position to the cracks at the top. During test, not only the beams, but also the columns deflected, which shows that the whole construction acted as a unit. The deflection in the beams was smaller than in Type 8, as explained before. As was expected, the moment of resistance at the ultimate load does not agree with the bending moments for continuous beams based on the assumption of free ends. The construction must be con- sidered as a frame. Dr. Probst finds that the positive bending 68 TESTS OF REINFORCED CONCRETE moment coefficient in the loaded span was 12.02, which agrees very closely with the bending moment coefficient computed by him by the rigid-frame method. Type 4.—The cracks in the T-beams continuous over five spans indicate clearly that the beams acted as continuous. From the " 6 -2 Round 3 - "Round:, xe 4 3" 7 saat ”" “ J jl g Round 2-H Round... See ol Round-.,, 3-4 Round--.. “ 45.3" Round & Pate ee aft ae \ 8 y} oe “ o\ X ba 2 16 Round “Sek ‘ i ES. Y ——_—_—_——" COTA 9" ‘ } a fey ¢ I$ -F aS --S--t -2->| 11" jx -+-4 ; c 48 y 74-76 Round | ay c \ \ Xf Val, 47 Nal Vy V) V 7 YIN a? vi HT NNN ANAT PUN ~--l.5 AN = d f ‘ 1 Ley, SV SOE 4 - Ea cil be i 1 17 ' pape Ci i S ke aT : nephe p yi/ gym (5-76 Round pa AY yop Roundy] |S Ma A ROU gs 4 2 ai an ‘ ae ee ead ‘ J 8, Rounds 7 oats pouea / 4 6 Round’ "Pe Tp ia, 6 - H'Round “5 Round“ "5 ound ‘7-7 Round Section on c- d Arrangement of Steel Pe hep ned pt pete Rw ps tere tk BO. So Me ee phere oe lu Bol} hes : Teag ¢ "I <5 144 >| aa " [one oc aenedee 2] Oot=noenon-- Shee anne nes G1 Gi--so eee bf Qiasansn ies Position of Loading 49 49 49 +9 7 62 62 62 49 4949 62 23 31 26 20 26 23 23 20 Cracks During Loading Fic. 23.—Continuous Beams of Three Spans; Type 3a. (See p. 66.) comparison of the moment of resistance of the beam at the maximum load, with the theoretical bending moment obtained from ordinary continuous beam formulas, we find a very close agreement, which proves that even for five spans a continuous beam acts as contin- uOUS. DISTRIBUTION OF CONCENTRATED LOAD ON SLABS 69 DISTRIBUTION OF CONCENTRATED LOAD ON WIDE SLABS A concentrated load placed on a wide slab is carried not by the portion of the slab immediately below the load but by a considerable width of the slab. Experiments described below, conducted on slabs where the maximum width was equal to double the span length, show that within these limits the whole width of the slabs is affected by a concentrated load placed centrally, and, therefore, assists in larger or smaller degree in carrying the load. Thus, in a 32-f{t. slab tested on a 16-ft. span, the whole slab was affected by the load. This con- dition is brought about by the shearing resistance of concrete and not by any distributing reinforcement. Distribution of Stresses and Deflections.—The stresses and deflections in the center af the span are largest in the section of slab directly under the load. For sections away from the load, the center stresses and deflections decrease rapidly at first and then more slowly, until for wide slabs they become insignificant near the edges of the slab. Definition of Effective Width of Slab.—As explained above, in a wide slab under a concentrated load, along the width of the slab, there is considerable variation in stresses in the center of the span. The ordinary slab formulas assume uniform distribution of stresses along the whole width; therefore they cannot be used directly without some assumption. In order to be able to use ordinary slab formulas for the design of wide slabs loaded by concentrated loads, it is neces- sary to determine the width of slab of same depth which, stressed uni- formly along its whole width, would have the same strength as the wide slab with its varying distribution of stresses, and for which the moment of resistance, based on an accepted maximum unit stress, would be equal to the moment of resistance of the whole width of slab with a varying distribution of stress and with largest stress equal to the maximum accepted stress. This width is called effective width. It is less than the total width of the slab; and the ratio of effective width to total width depends upon the ratio of the total width to the span. Effective Width from Tests for Central Load.—From the tests made by the Office of Public Roads and reported by A. T. Goldbeck,?® the ratio of the effective width of slab to the span depends upon the * A. T. Goldbeck. The Influence of Total Width on the Effective Width of Reinforced Concrete Slabs Subjected to Central Concentrated Loading. Pro- ceedings, Am. Concrete Institute, 1917. Vol. XIII, pp. 78 to 88. 70 TESTS OF REINFORCED CONCRETE ratio of the total width of the slab to the span. The ratio may be found from the table below. Effective Width of Slab for Concentrated Load eee \| | Total Effective see Total Effective erste a Width Width “ Width Width a of Slab | of Slab Wate oe “f Slab |. OR eal eae Total Width |, Total Width Gain 0.11 1 Lait 0.671 0.61 0.21 0.21 1 Lot 0.68 1 0.57 0.37 0.281 0.93 {34 0.701 0.54 0.41 0.371 0.93 1°44 0.711 0.51 0.51 0.441 0.88 | 1.51 O78 0.48 0.61 0.501 0.83 1.61 0.721 0.45 0.71 0.551 0.79 L7H 0.721 0.42 0.81 0.581 0°72 1.81 0.721 0.40 0.91 0.621 0.69 1.91 0.721 0.38 ot 0.65 1 0.65 2.01 0.721 0.36 1 = span of the slab. From the above table it is evident that, where the width of the slab is equal to 0.4 of the span, practically the whole width is effective. When the width of span equals 1.5 1, the effective width attains its maximum value, namely 0.72 1. In such cases, about one-half of the total width is effective. The increase of the total width beyond 14 times the span has no effect on the effective width, as in all cases the effective width equals 0.72 U. From the tests made by Professor C. T. Morris, the effective width equals 0.6 times span plus 1.7 ft. for a width equal to 1.351 + 4 ft. This checks closely enough the value in the table. Effective Width from Tests with Eccentric Load.—When the load is not applied in the center of the width of the slab, the following rule may be used.®° If the distance from the load to the nearer edge of the slab is greater than one-half the effective width of the slab, considered as 30 See: Tests of Large Sized Reinforced Concrete Slab Subjected to Eccentric Concentrated Loads, by A. T. Gotppeck and H. 8. FairBANK. Journal of Agricultural Research. Vol. XI, No. 10, December 3, 1917. DISTRIBUTION OF CONCENTRATED LOAD ON SLABS ik centrally loaded, the effective width may be assumed to be same as for centrally loaded slab. If the distance from the load to the nearer edge of the slab is less than one-half the effective width of centrally loaded slab, the effective width may be taken as one-half the effective width for centrally loaded slab plus the distance of the load to the nearer edge. Example 1: Find effective width for a slab 20 ft. wide with a 14-ft. span, when the concentrated load is placed 3 ft. from the edge. Solution: The ratio of total width to span is 2¢ = 1.43. Fora slab loaded with a concentrated load placed centrally, the effective width, from table on p. 70, is 0.711 = 0.71 XK 14 = 10 ft. The distance of the concentrated load from the nearest edge is less than one-half of this width; therefore the effective width for eccentric load is 49 +3 = 8 ft. Description of Tests.—Tests were made by the Office of Public Roads and Rural Engineering on the following slabs: Reinforcement i Span | Width Ne eae aan eo meer eomerre BESS nig ae Size bars Seti sh it ft in. in. in. 835 16 oa 12 103 3 in. sq. 10 001d 930 16 oo 10 8i hte ae 8.87 O275 934 16 a2 fe 6 eee GanG O75 A19 6 12 7 6 mek | inne ORT A20 6 12 4 3 es 6.26 i ii AQ1 6 12 5 4 Ae esta 0.75 A22 6 oe 16 15 Seen. | 5 OP 75 A23 6 12 6 5 eA | 6.66 0.75 | The mix of concrete in all slabs was 1: 2:4. No cross reinforce- ment was used. All slabs were tested with their original width. Slab A25, after being tested with its original width, was gradually reduced in width by having successive sections split off one side. Method of Test.—Concentrated loads were applied in the center of the slab. Additional tests were made with two and four con- centrated loads. In some slabs the loading was carried to destruc- tion; while in slabs tested with different widths up to the working 72 TESTS OF REINFORCED CONCRETE loads, deformation of concrete and deflection readings were taken at all widths. The deformation readings on concrete were made perpendicularly to the support at regular intervals along the whole width of the slab. In several cases, deformations were also taken longitudinally with the span. Steel deformations were also meas- ured in some slabs. Conerete deformations were found to be a most satisfactory measure of distribution of stresses in the slab. In interpreting the results, however, it was necessary to take into omge mi Sat account the flow of con- Es Effective Width 49 erete under — sustained = ero loads. . 3+ Co Slab Width =10.0 L Steel deformations | = es Effective Width+ 628-1 were less satl sfactory, = ro T cH because the readings at SAE Y TCE Slab Width=-8'0 He - points where cracks oc- =e ai Effective Widtht6.24-| curred were different from ic 7 Sa) iG readings at adjoining ret | H is points without cracks. [ ce Slab Width =4'0 Deformation of Slab 1 Effective Widih}372_| and Effective Width.— iastaiek Figure 24 shows the © Slab Width- 2:0 deformations of steel in ae el Effective Width}1.s3_| Slab A19 when tested Feo : | with different widths. Span 6' Effective Depth 6. Steel 0.75 per cent The effective width is Fic, 24.—Steel Deformations and Effective marked by dash lines. Width Slab A19. (See p. 72.) Results of Tests.— Fig- : ure 25, p. 73, shows the results of all tests. In the figure, the ratios of total width to span are plotted as abscissas and the ratios of effective width to span as ordinates. The curve represents the values given in the table on p. 70 and suggested for use. To be on the safe side, the curve is made to coincide with the lowest values obtained in the tests. Breaking Loads of Slabs.—Slabs 835, 930, and 934 were loaded to destruction. The results are given on p. 73. | DISTRIBUTION OF CONCENTRATED LOAD ON SLABS 73 Breaking Loads of Slabs. (See p. 72.) ; Reinforcement Breaking Effective eu Hh emt, Lopate Number penn Load EP, p | Size | Spacing lb. 835 10} 0.0075 | fin.sq. | 10.5in. | 119.000 930 83 0.0075 | aa Ne | poy evi 80 000 934 6 fer ODOT irs ats S tery. Oe We peas OO? 40 000 All slabs were 32 ft. wide and 16 ft. span. Infliénce of Total Width on Effective Width of Reinforced Concrete Slabs Subjected to Concentrated Loading. V7 ; Beet ttt 7.0 bet nal ae 0.8 es Se nol] ae Sls 05 ++ Fy 0A-19, é'Span, 6" Depthe SI es x A-20, 3 H a4 ee ee ee Note: Symbols like those in eet AER eet, ie ae 0.3 ol left column are based on FAR-R2 16 Te via? ae | ys as ain those & S27 ae in right on steel deformations. 0.1 ma «| s iw © So 0G) 94S) Sl Re SS es Se Total Width Span Length Fig. 25.—Influence of Total Width on Effective Width of Reinforced Concrete Slabs Subjected to Concentrated Loads. (See p. 72.) a ' ' ' ' 1 ' ’ ' ’ ° & for) =) ah Gage Aeaey 74 TESTS OF REINFORCED CONCRETE It will be noticed that the breaking loads for different effective depths of the slabs are in the same relation as the squares of the depths of the slabs. This indicates that the effective width for all depths was practically the same. Fig. 26, p. 73, shows the manner of failure of a slab under a concentrated load. TESTS TO DETERMINE DISTRIBUTION OF LOAD BY SLAB TO JOISTS The tests show that if a continuous concrete slab is supported by several parallel joists of any material, and a concentrated load is placed directly above one joist, the load is distributed by the rigidity of the slab to several joists (see Fig. 27). The distribution depends upon the ratio of the thickness of the slab to the span. 100 90 80 Per Cent of Load Carried by Middle Beam Ratio of Slab Span to Thickness Fig. 27.—Distribution of Slab Load to Three Parallel Supporting Joists. (See p. 74.) TESTS BY PROF. C. T. MORRIS The laboratory test made by Professor C. T. Morris consisted of slabs, 6, 7, and 8 in. thick respectively, supported on three lines of 10-in., 25-lb. I-beams spaced 3 ft. 6 in. on centers. The span of the joists was 12 ft., and they were supported on other I-beams, which in turn rested on concrete pedestals, as in bridge construction. The load was placed directly over the middle joist at its center. In distributing the load, the slab acts as a cantilever supported DISTRIBUTION OF LOAD BY SLABS TO JOISTS 75 at the point of application of the load and loaded by the reaction on the side. The following conclusions may be drawn from the tests: (1) The percentage of reinforcement in the slab has little or no effect upon the load distribution to the joists, so long as safe loads on the slab are not exceeded. (2) If the span is ten times the thickness of the slab, or more, total load must be considered as carried by the joist under the load. The amount of load distributed by the slab to other joists than the one immediately under the load increases with the thickness of the slab. (3) The outside joists should be designed for the same total live load as the intermediate joists. (4) The axle load of a truck may be considered as distributed uni- formly over a 12-ft. width of roadway. Fig. 28, p. 75, shows the elongation in extreme fiber of the steel beam and deflection for the middle beam and for outside beams. 0.120 0.100 0.080 on in Inches ' 0.060 Deflecti 0.040 Deformation in Divisions on Strain Gage Z veal “ae HA | ee Zara 0 0 2000 4000 6000 8000 10000 12000 14000 16000 18000 20000 Load in Pounds Fre. 28.—Elongation in Extreme Fiber of the Steel Beam and Deflection of the Middle Beam and of Outside Beams. (See p. 75.) Value of One Division on the Strain Gage is 0.00019 in. Similar results would be obtained with concrete joists. The conclusions apply only to cases in which ratio of span to thickness of slab does not exceed 10. The largest ratio used was 7, but the results may be exterpolated. 76 TESTS OF REINFORCED CONCRETE TESTS OF PLAIN CONCRETE COLUMNS Professor Talbot 3! made a very comprehensive series of tests, the results of which are given in the table below. Conclusions, con- firmed also by experimenters abroad, are as follows: (1) Manner of Failure.—Plain columns fail either by shearing at a diagonal plane of fracture, or by crushing, when the material is shat- tered and cracked longitudinally. The diagonal shearing failure almost always occurs suddenly and with little or no warning, while the compressive failure 1s more gradual. Tests of Plain Concrete Columns. (See p. 76.) Materials: Portland cement, Wabash River sand, crushed limestone. Col- umns, 12.in. diameter, round, 10 ft. long. By Pror. Arraur N. TaLBot See nee rue ae Average |Maximum Y aneeeh) Minimum * ae cee we Number Broparkorns Average Slee Ultimate Pee Conk Ultimate Be Ga of of Age Unit Unit. techy Unit j Specimens Strength] Strength Strength eh Concrete Average Average Tested noo Ib. per | lb. per o% lb. per of Y* |) sqcimteslasedoan. : sq. in. a |__| mee sae 2 1 Lear 64 2 300 2480 Ges 2 120 — 8 7 125) 2844 65 1 740 2 210 at 1165 —33 2 shearers 4 9) 613 1 033 1110 ts 955 — 7 2 14 28 63 575 575: alse aa ee 575 6 122.314 192 2 025 2 680 32 1 770 —13 2 Pls 2B 801 1d moe 2 0 2770 : trees | (2) Effect of Richness of Concrete.—The strength of columns increases in nearly the same proportion, 1.e., almost as a straight line, with the increase in the proportion of cement to total dry material used. (See Volume II.) (3) Modulus of Elasticity and Poisson’s Ratio.—The modulus of elasticity of the columns is almost constant for the first third of the strength of the concrete. Beyond this point the modulus decreases till it reaches at the ultimate load about one-half of its initial value. 31 University of Illinois Bulletin No. 20, 1908, TESTS OF PLAIN CONCRETE COLUMNS fire The Poisson’s ratio, or the ratio of the lateral to the longitudinal deformation (see Vol. II) wasfound for 1 : 2 : 4 concrete to be between 0.10 and 0.17 up to a load of about one-half the ultimate. It increases with the load, reaching probably 0.25 at the ultimate load. (4) Effect of Repetition. Repetition has no effect on deformation for loads up to one-half of the breakirtg strength of the column. For higher loads, the deformation increases after repeated applications of the load. After ten repetitions of a load equal to three-fourths the normal breaking strength, for example, the deformation was increased by 25 per cent. It must be noted, moreover, that the suddenness of failure of plain concrete is increased by the length of the column. This absolutely excludes plain concrete columns from structures where they are apt to be exposed to shock or to secondary stresses due to bending, as in building construction. | Concrete vs. Brick Columns.—Tests carried out by the U. 8. Bureau of Standards on columns of common, hard, and vitrified brick laid with lime and cement mortar, indicate that the strength varies with quality of brick and mortar, while large and small columns - show about the same unit strength. A series showing the strength of piers of common, hard, and vit- rified brick, laid with different mortars, is given in the following table. The lime mortar specimens showed an almost entire lack of carbona- tion on the interior. Three piers of each kind of brick and mortar were made, with headers every other course, every fourth course, and every seventh course, but this variable appeared to have no effect. Bricks were laid flat. Two large-size columns 48 in. square and 12 ft. high, of common, hard-burned brick, one laid in 1 : 1 cement mortar and one in 1: 3 lime mortar, were tested by the Bureau and crushed at 2920 and 760 Ib. per sq. in. respectively.?? Tests made at the Watertown Arsenal and quoted by the Com- mittee of the American Society of Civil Engineers, on the Compressive Strength of Cement,?? give the ultimate strength of common brick piers about eighteen months old as ranging from 800 to 2 400 Ib. per sq. in., the results for brick laid with lime mortar averaging nearer 32 James E. Howard, Engineering Record, March 22, 1913, p. 332. 8§ "Transactions, American Society of Civil Engineers, Vol. XV, p. 717, and Vol. XVIII, p. 264. 78 TESTS OF REINFORCED CONCRETE the lower figure, and those for 1 : 2 Portland cement mortar nearer the higher figure. Compressive Strength of Brick Piers * Tests by the U. 8. Bureau of Standards. (See p. 77.) Dimensions 30 inches square by 10 ft. high a een aN I eee ee Nes Compressive Ixind of Brick Mortar Bes Strength, Months : Ib. per sq. in. ah Os (| 1:3 lime Oiben':: 170 Pe gee tad cae \ | 1:3 cement 1 575 (| 1:6 lime 4 910 15% lime Hardy uta een oe ee rae (aap ene 1 1465 1 : 3 cement 1 1650 (| 1:6 lime 4 1360 anh (15% lime Viviicd to ee ig Ee : ae 1 2.900 1:3 cement 1 2780 | | | * Engineering News, August 5, 1915, p. 242. The unit stresses allowed by the New York Borough of Man- hattan Building Code, 1916, for brickwork are, Brickwork in: Ib. per sq. in. Portland cement: Mortar... 4.<...<,.) ose ee 250 Natural cement mortar.......... 1 ies 210 Lime cement mortar .2.0. -... «. 2-5 eee 160 Lime mortar... 22.0 Jo. es oe ee 110 The first value is but little more than one-half that recommended for good 1 : 2: 4 Portland cement concrete. TESTS OF COLUMNS REINFORCED WITH VERTICAL STEEL Tests prove positively that in reinforced concrete columns, steel and concrete are effective in resisting the load carried by the columns. As explained in the chapter on the Theory of Reinforced Concrete, p. 159, the stress in steel equals the stress in concrete multiplied by the ratio of their respective moduli of elasticity. This fact also 1s borne out by the tests. TESTS OF COLUMNS VERTICALLY REINFORCED 79 Mr. Spitzer,** in Austria, and Professor Withey,*® at the Univer- sity of Wisconsin, observed that, near the ultimate load, adjustment between stresses in steel and concrete takes place, so that finally the failure occurs by both of the materials passing the elastic limit simul- taneously. ‘This adjustment may be explained as follows: Suppose that one of the two materials reaches its elastic limit, but there still exists bond between them. If the load on the column is increased, there is a tendency for the increase to be distributed over both materials. However, any increase in stress, in material which had passed the elastic limit, would produce very large defor- mations, which the other material, being still within the elastic limit, could not undergo. Since both of them must deform by the same amount, this other material takes all the stresses due to any increase of the load, till it finally reaches its elastic limit, and the column fails. The table on p. 81 gives results of tests of full-sized columns made at the Watertown Arsenal. Manner of Failure.—Contrary to expectations, in most cases failure in columns occurs near the top or bottom instead of at the center. ‘This has been explained as probably due to greater porosity of concrete at ends. In most columns, hair cracks appeared at 85 to 90 per cent of the maximum load. In some cases, however, the failures were sudden. In a number of cases, concrete split at the column reinforcement after its elastic limit was reached. The splitting effect is caused by the lateral deformation of steel, which exerts sufficient pressure on the concrete to break it. The steel, therefore, should be placed at a sufficient distance, say at least 1 in. from the face of the concrete. With proper protection, there is no danger of buckling till after the elastic limit of the steel is reached. From Spitzer’s tests, it would appear that columns with steel placed well within the cross section of the column are somewhat stronger than those with steel placed near the surface. In practice, however, columns are apt to be subjected to eccentric loading; there- fore, the placing of steel in abnormal positions must be discouraged. Factor of Safety.—A column with vertical steel only is liable to fail without notice when its ultimate strength is reached. The ultimate 4 Mitteilungen Uber-Versuche ausgefiihrt vom Eisenbeton-Ausschuss des osterreichischen Ingenieur und Architekten-Vereins, ‘“‘Versuche mit Eisenbeton- saulen,” Heft 3. > University of Wisconsin Bulletin No. 466, December, 1911. 80 TESTS OF REINFORCED CONCRETE strength of concrete columns, even if built under the same conditions, is more variable than that of steel columns. Therefore, the commonly accepted factor of safety is larger than that used in steel columns. When designed according to formulas given on p. 406 with allowable unit stresses on p. 407, columns with vertical steel only are very reliable. Modulus of Elasticity for Reinforced Columns.—The modulus of elasticity varies for different intensities of loading and for different mixes of concrete. In selecting the ratio of moduli of elasticity to be used in design, it is proper to be euided more by the required factor of safety than by the actual modulus of elasticity at any particular stage of the loading. From the tests thus far made, the moduli of elasticity given on p. 407, with the suggested working stresses, give the required factor of safety. Rich Versus Lean Mix.—As is evident from the table on p. 81, cement is very good reinforcement for the column, as the increase in strength is much larger than the cost of additional cement. Influence of Bands.—In tests made by Mr. Spitzer,2° of columns having different spacing of bands, those in which the spacing was equal to, or smaller than, the diameter of the column gave somewhat greater strength than columns in which the spacing exceeded the diameter of column. Influence of Percentage of Steel.—The tests show clearly that the effect of reinforcement in columns is the same whether the per- centage is large or small. In all cases the steel takes a stress equal to the stress in concrete times the ratio of moduli of elasticity. Watertown Arsenal Tests.—The table on page 81, from test by Mr. James E. Howard at the Watertown Arsenal, gives the relation of actual tests to theoretical computations based on a ratio of elasticity of 15. It is noticeable that the actual strength is almost always more than the theoretical, and this is especially the case with the leaner mixtures because the modulus of elasticity of the leaner concrete 1s lower, and therefore the ratio of 15 is very conservative. An excellent analytical treatment of columns reinforced with ver- tical steel is given by Professor Talbot in one of his University Bulle- tins37 The problem is discussed briefly by one of the authors in a paper before the Boston Society of Civil Engineers.?% 36 See footnote on page 79. 37 University of Illinois Bulletin No. 12, F eb. 1, 1907. 38 Sanford E. Thompson, in Journal Association Engineering Societies, June, 1907, p. 316. TESTS OF COLUMNS VERTICALLY REINFORCED Si Many of the tests at the Watertown Arsenal, for example, were made with vertical bars imbedded in columns 12 in. square and 8 ft. Jong, with absolutely no bands or horizontal steel of any kind placed around these vertical bars to hold them in place; that is, the bars 8 ft. in length were placed in the four corners of the column—in some tests only 2 in. from the surface—and simply held in place by the 2 in. of concrete." There was no sign whatever of buckling until the compression was so great that the elastic limit of the steel was passed. Strength of Plain Concrete and Mortar Columns vs. Vertically Reinforced Concrete Columns 12” & 12’. Height 8 feet. Age of Mortar and Concrete 6 months, Watertown Arsenal. (See p. 80.) Reinforced Columns Proportions| Plain Concrete or Reinforcement Computed | Reference Mortar strength | to ‘Tests Columns Actual | based on |of Metals” Actual Ratio |strength|column (4)| U.S. A. = Strength area | lb. per | and aratio a et ib:iper Description | steel to} sq. in. | of n = 15 a) e 3 Sq. in area lb. per o soiy+y O}2|R column Sq. in (1) |(2) |(3) (4) (5) (6) (7) (8) (9) 1;2]0j} 3070 | 8-2” round bars | 0.029 | 4200 4290 |1905, p. 377 3. 1:0 2380 | 8-2’’ round bars | 0.029 | 3840 3320 = |1905, p. 377 1);4/0 1520 | 8-2’ round bars | 0.029 | 3380 2120) (1905, p. 377 > Ph 1 0 1080 | 8-2’ round bars | 0.029 | 2810 1510 = |1905, p. 377 24-51) 0 1080 |13-2” round bars | 0.046 | 3 900 1780 = =|1905, p. 377 © 1 | 1 | 2*| 1720 | 4-2” twisted bars} 0.014 | 2.890 2060 |1904, p. 386 1 | 2] 3*| 1769 | 4-3” twisted bars} 0.014 | 2010] 2100 1904, p. 386 1/2/)4)] 1418 | 4-0.74"X*0.74” trussed bars} 0.014 | 1900 1689 —_|1906, p. 538 2 | 4*| 1710 | 4-2’ twisted bars| 0.014 | 1990 2050 = |1904, p. 386 2|47T| 2400 | 8-2” twisted bars} 0.029 | 3700 3360 = |1907, p. 242 1/|3{|6 | 1450 | 8-3” corr. bars 0.019 | 2290 1840 = ‘|1904, p. 379 1906, p. 535 * 1-in. to 14-in. pebbles. 7 Age 17 months 22 days. 39 Test of Metals, U. S. A., 1905, p. 344. 82 TESTS OF REINFORCED CONCRETE TESTS OF SPIRAL COLUMNS In analyzing results from tests of spiral columns, it is necessary to examine not only the strength, but also the shortening of the column. In building construction, because of the dependence of the different members upon each other, it is advisable to permit a shortening, or Stress in Pounds per Square Inch a 1 Per-Cent s Spiral Steel Ges | Ly | I (ae S - AI ; - | 6.1 Per-Cent tis Vertical Steel j 0 S S S S S S S S S S [~) S S S S S MD one ou ee ae Say Se eye es mates a ees SS im cae ee Deformation per Unit of Length Fic. 29.—Deformation Curves for Spiral Columns with Varying Amount of Vertical Steel.4° (See p. 82.) Concrete 1 : 2 : 34; 1% Spiral High Carbon Steel; 0% Vertical Mild Steel. Concrete 1 : 2 : 33; 1% Spiral High Carbon Steel; 3.8% Vertical Mild Steel. Concrete 1 : 2: 34; 1% Spiral High Carbon Steel; 6.1% Vertical Mild Steel. deformation, of not more than 0.007 of the length of the column. The factor of safety, therefore, must be based on the load causing a certain deformation rather than on ultimate strength. Although the strength of column beyond the yield point is not available in ordinary construction, the greater ultimate strength, ductility, and uniformity in strength of spiral columns reduces the danger of sudden failure, 40 University of Wisconsin Bulletin No. 466, December, 1911, pp. 92, 98, 94. By Morton O. WiTHEY.* | TESTS OF SPIRAL COLUMNS 83 and larger unit stresses may be allowed. In practice, it is more rational to increase wcrking unit stress in concrete than to compute the stress by a formula which takes into account the steel in the spiral. (See p. 421.) | Harly European tests of spiral columns were made on short columns and without deformation diagrams; and as a result, very high working stresses based on the ultimate strength were recom- mended. More recent tests, notably by Talbot and Withey in the United States, showed excessive deformations of spiral columns at high stresses. Column Tests by Prof. M. O. Withey.—The tests at the Univer- sity of Wisconsin, Series of 1910, consisted of four series, two of which will be considered below; namely, series 1, columns with varying percentage of longitudinal and lateral reinforcement, and series 2, columns with varying proportions of concrete. The table on p. 84 gives the general results of the tests. The following general conclusions may be drawn from the tests: 1. The cheapest way of increasing the strength of a column is by using a rich mix. 2. Spiral reinforcement greatly increases toughness and ultimate strength of a column, but does not raise the yield point. (See col- umns M and O, p. 84.) The strength beyond the yield point cannot be utilized in building construction; hence, the amount of steel for spirals should be made only large enough to produce required ductility and raise the factor of safety against failure. In practice, 1 per cent of spiral reinforcement seems to be sufficient. | 3. Longitudinal steel increases the stiffness of the column and raises the yield point. 4. Stress in steel at the yield point of columns is practically the same for all mixes of concrete and only a little below the yield point of the vertical steel. (See table, p. 85.) This phenomenon may be explained by the fact that for leaner concrete the ultimate strength is smaller, but the deformation at the yield point larger than for rich mixes. For rich concrete, the stress in concrete at the yield point is larger, but the ratio of the moduli of steel to concrete decreases. Since the stress depends upon the product of stress in concrete times the ratio of the moduli of elasticity, the two values evidently adjust themselves so that the product is the same in all cases. 5. Columns loaded eccentrically give results which agree closely with the formula given on p. 175, as is evident from Fig. 30, p. 86, 84 TESTS OF REINFORCED CONCRETE Tests to Determine the Effects of Varying the Percentage of Vertical By Morton Average values given. Columns C-1, 2, 3, 4, D-1, 2, 3, 4, 120 inches long; all others | | 2 Reinforcement Par Cent ss 5 Reinforce- to 4 Spiral com & ie) iw ates Area Maxi- 2 Cc ] i) Ta a mum & olumn s a ore ES Number nb z | a8 A Vertical Load - Py Pi | Ib. per | Ib. per | Ib. per jIb. per|lb. per fe Ib. per eben ae lb. P sq. in. : Sie iD Chin. | bq. in. i sd.in. ) jf. sq. in. | myiRy | my{ny | | | | ant ae oe a oe eala aout ele OOO OOOM 182 6 ree et, im 5000.88.)" 1855 Wsoe..%. 10 050 | 1855 | 1750 | 0.94 |2 350 ee 1320 127.0" 10, 127 213 000} 0.83 | 2710 | 40500} 12000; 7940 |} 1760] 0.91 |3 350000; 11.0 | 21.0 '0.110 273.000) 0.84 | 3470 | 42 000 6 450 | 1960} 2180] 0.13 |3 400000) 13.0 | 22.0 |0.110 333 000} 0.83 | 4240 | 40650; 6150 1 860 | 2055} 1.11 |4 350000} 11.5 | 22.0 |0.088 HOS000) OF 92) 1 80 Poi... 4 800; 1370} 1770) 1.29 |2 100.000] 14.5 | 19.5 |0.146 205 500} 0.67 | 2615 | 39450} 7050) 1 865|.1995} 1.07 !2 800030, 13.5 | 21.5 |0.123 265 500} 0.81 | 3375 | 38700} 6900] 1985} 1800} 0.90 |3 800 000] 11.0 | 20.0 |0.130 295 500; 0.80 | 3760 | 37 650 6 450 | 1565 | 1 685 | 1.09 |3 425000] 18.5 | 24.0 |0.101 445 500! 0.82 | 5660 | 39150; 8100); 2750] 2 365] 0.86 |5 300000} 9.6 | 14.5 |0.121 348 000; 0.68 | 4430 | 36900) 8400} 2310] 2 480/ 1.08 |4650000) 9.9 | 16.0 eae 408 000) 0.74 | 5190 | 37 500 8 250 | 2 385 | 2390] 1.005)5 250 000, On We 16). OF Oal2s 453 000} 0.82 | 5765 | 37-200 6 750 | 2 230 | 2 305] 1.03 |5 500000] 10.8 | 17.5 |0.094 Ber ene. | 2385. ). sae. lee ee Doon | cate OOM a0 s OS eels aes mene oe ete heya en coene te flea comets 280 ia 0.75 | 3 575 | AQSBOO WR. ate « 2 2238 | 2 252 | LO eas Meeasy tcl has 2 trea: DRM SiN ie care | | saet 264 500| 0.60 | 3200 | 36900) 8 250] 1115} 1770 | 1.59 |3950000|12.80 |...... 0.085 338 000)...... alrooue ds. 2). 8 400 | 4100 | 4760} 1.16 |5050000| 5.95 |...... 0.175 470 000)......}] 5700 | 30600}; 6300] 4160] 4120] 0.99 {6 700000} 5.70 |...... 0.190 318 000} 0.69 | 4050 |....... 7 350 | 4050 | 4070] 1.01 |4125 000} 7.30 |...... 0,124 453 000} 0.79 | 5760 | 37050! 7 350| 3725 | 4400] 1.18 {5 175000) 8.45 |...... 0.140 PSU OOO ONG T to G70 fon. ck. 6.750 | 3.570 | 4 880 | 1.87 |3 400000; 8.80 |...... 0.135 468 000! 0.73 | 5950 | 36450] 6750] 3970/ 4555] 1.15 |5 250000] 8.40 |...... 0.125 489 000) 5 Sao 5780 | 34750 | 7050) 5185) 6 787 | 1.80 |5625 000; 5.90 |...... 0.190 BLE OOO} ss 0.. 6 150 | 35250} 7700] 5090} 6 387] 1.25 |5 950000} 5.95 |...... 0.185 BA2'000)'-. .. .. 6 590 | 36000} 6600) 4760] 6075] 1.28 |5 150 000) 6.40 |...... 0.180 RE Pe Te eel ete y aisle js. sis, |a)4 o2e0% «| a a waite s [> 543000 (HOO ne eerie Oe oes + Column did not fail. 86 TESTS OF REINFORCED CONCRETE in which the straight lines represent figured stresses in steel and con- crete by Formulas on pp. 175 and 176, while the dots and circles show the actual stresses obtained from deformation. Action of Columns under Test.—Up to the point of breaking strength of plain concrete, the action of the columns with spirals was the same as for columns with vertical steel only. ‘The observed stress in spirals was from 6 000 to 8 000 Ib. per sq. in. For spiral columns with vertical steel, the deformation curve continues as a prac- tically straight line to the yield point of the column. The yielding is indicated by the scaling off of the protective shell and by an 2000 ‘Average Unit Stress in Pounds per Sq. In | ae | Bay Eccentricity 1 4 in. He 0 2000 4000 6000 8000 10000 0 2000 4000 6000 8000 ua 0 10000 20000 30000 40000 50000 0 10000 20000 30000 40000 : Unit Stress in Concrete or Steel in Pounds per Sq. In. 5.96 Per Cent of Vertical Steel 0.97 Per Cent of Spiral Steel nN , i if Uh ie nN Fic. 30.—Comparison of Theoretical and Actual Stresses for Eccentric Loading.* (See p. 85.) } ‘ncrease in ratio of lateral and longitudinal deformation to the applied load. ‘The yield point is more marked for columns with large percentage of reinforcement (see Fig. 29, Pp. 82). 3 Lor spiral columns without vertical steel, the deformation diagram is a curve without a marked yield point, so that the yield point is only distinguishable by scaling of the shell. After the yield point has been passed, the disintegration of the 41 University of Wisconsin Bulletin No. 466, December, 1911, p. 71. TESTS OF SPIRAL COLUMNS 87 shell progresses very rapidly. The ratio of shortening due to the applied load becomes larger and final failure takes place by buckling of the column, or, in columns with a small amount of lateral steel, by breaking of the spirals. Stresses in Steel and Concrete.—The table on p. 84 gives the stresses in steel and concrete at yield point and at maximum load. The stresses in vertical steel were obtained from the deformation by using a modulus of elasticity of 30000000. The remainder of the load assumed as carried by the concrete, and divided by the area of the core, gave the unit stress in the concrete. In figuring the stress in concrete, the area of the core was used in preference to the total area of the column, because at yield point and at maximum load, either a part or the whole of the outside shell is destroyed and is then ineffective for carrying the load. The stress in spirals obtained from lateral deformation is very small at the yield point of the column, which corroborates the state- ment that up to yield point 4? the spirals do not affect the column appreciably. The table is of interest in giving a comparison of the stresses in steel with the stresses in concrete for different mixes and also in giving the values of the ratio of moduli, n. Although the value of this ratio was variable, the stress in steel at the yield point of the column was about the same for all columns, which seems to show that in a column under load an adjustment of stresses takes place. From the table, also, it is evident that the stress in concrete at the yield point was the same irrespective of the amount of vertical reinforcement. An interesting experiment was tried in connection with tests by Wayss and Freytag? The columns, after reaching the maximum stress, were again loaded after nine months and after one year, and showed a large increase of strength over the original maximum strength, in some cases reaching 50 per cent increase. After these two loadings, the column spirals were removed and the core tested again. In no case was the core, upon removal of the spirals, found to be dis- integrated; in fact, each core showed a considerable strength, which tends to disprove a contention previously held by several authorities that the concrete in hooped columns becomes disintegrated after a certain point in loading is reached and is simply prevented from flowing by the hoops. 42 See also Professor Talbot’s Bulletin No. 20, University of Illinois. 43 Morsch, 4th edition, p. 117. 88 TESTS OF REINFORCED CONCRETE TESTS OF SQUARE COLUMN WITH RECTANGULAR BANDS Tests were made by Wayss and Freytag 43 on eleven types of square and rectangular in cross section with closely spaced hoops and bands. These tests prove that the bands in square columns are not very effective and can not be compared as to effi- ciency with spirals in round columns. The outside shell started break- ing off as soon as the concrete reached its maximum crushing strength. TESTS OF COLUMNS WITH STRUCTURAL STEEL REINFORCEMENT The size of column can be reduced by the use of structural shapes, rigid enough to serve as a structural steel column, 12-in. columns, a = om imbedded in concrete. Below are given results from tests of columns with two types of structural steel. The results must not be considered as apply- ing to all steel sections and must be used with caution where the structural mem- bers differ materially from those in the tests. _--- Outline of spiraled type _16 inches” Ge gar ler Outline of H core type ON. yor Outline of dh bes 5 fireproofed type 11, or 14 spaces. ’ NOTE a” ” Angles: 2% 24x+ fie sLRIKO Et ea coe} z 5 au . Rivets: 3 diameter Fig. 31.—Dimensions of Test Columns. ** (See p. 88.) 43 Morsh, 4th Edition, p. 117. Talbot-Lord Tests of Columns.’—The Talbot- Lord tests consisted of thirty-two columns dl- vided into four groups: 1. Plain steel columns. 2. Core-type columns, i.e., columns in which the portion within the struc- tural steel members was filled with concrete. 3. Fireproofed columns, i.e., core-type columns having a 2-in. protective covering. 44 University of Illinois Bulletin No. 56, Fig. 1,-ps 4.4 45 University of Illinois Bulletin No. 56. TESTS OF COLUMNS WITH STRUCTURAL STEEL 89 4, Spiraled columns, i.e.; core-type columns enclosed in close- fitting spiral and filled with concrete to outer surface of spiral. The cross section of the structural steel, which is shown in Fig. 31, p. 88, was the same for all columns. Ratio of length to minimum radius of gyration varied from 6.1 to 59.5. The results of the tests are given on p. 90. Plain Steel Columns.—No bending was visible to the eye at the maximum load. After the maximum load was passed, bending developed very gradually. The average deformations per unit of length are shown in Fig. 32, below. ; 400000 200000 Applied Load in Pounds 100000 0 ,0002 ~=.0004 .0006 0008 .0010 0012-0014 0076 .0018 Deformation per Unit of Length Fig. 32.—Average Deformations of Plain Steel Columns.‘* (See p. 89.) The effect of the length of the column was more marked at high than at low loads. According to Professor Talbot, the ultimate stress in column for different ratios of ! may be represented by a Beet tine.tormuls, = = 36 500 — 155 : Core-type Columns.—The columns of core type were very tough and failure was slow. For short columns, the failure was caused in most cases by crushing of the concrete; for longer columns, by bend- ing and crushing of concrete. No bending visible to the eye was observed until maximum load was reached. The effect of mixture 46 University of Illinois Bulletin No. 56, 1912, p. 19, 90 TESTS OF REINFORCED CONCRETE Tests of Steel Columns, Reinforced with Concrete By Tausor and Lorp.* Age of concrete, 59 to 61 days. (See Fig. 31, p. 88.) P.S. = Plain steel column of 8 — 3’ X 23’ X 3” angles with no concrete core. GC. T. = Core-type column of 8 — 3” angles with concrete core. F. = Fireproofed column same as core type with 2” of extra concrete outside of structural steel. No spiral. S. = Spiraled column same as core type with spiral steel and concrete core filled out to conform with diameter of spiral. = 5 Stresses in 3 « |‘82°8| Total Load in lb. sn al ee lb. per sq. in. =e 5 |eeor E rah ope ia Se NS B 2s. je) Column | 0 15 aaa ae ir an Load | Load ae Con- ogs Number os ci ae con- con- ‘, crete | & |-="6 3 : In 5/8 2 ls0e Column! sidered | sidered | In a aha) Hm load | carried | carried | steel 1 b i; ,crete sq: : x y im. r steel {concrete Lite a ee eee ee —_—_|—_—_—— EN ee eee omen ft. in : 8 902 P.S.| 13} none | 2-0) 6.1 | 487 300). Oke eee BL DOON ey wc. 8 905-6 P.S.| 13} none | 4-8} 14.4 |444600).......]....... 34200), 221. 8 907-8 G.T 12011 22 24) 4-8in 589 500) 444 600} 144 900) 34 200] 1 355 8 910-11-14|P.S. | 13} none |10-0| 30.8 | 420 000).......|....... 32 400]. ..2. 8 912-13 “1 CrT 201 28s 410-0) eee 547 350| 418 000] 129 350) 32 150} 1 210 8915-16 |P.S.| 13} none |15-4| 47.2 |372 000).....-7).. 22. 28 GUO = ser 8 917-18 © |C.T.120)1 224/154) 500 350| 372 000} 128 350] 28 600} 1 200 8 920-21 |P.S.| 13} none |19-4| 59.5 | 359 600|....... Sige hee 27-B501 As § 922-238 (CATA120/1 224110 4) ee 493 450] 359 600] 133 850} 27 650) 1 250 8'925-26. 1G. T. 120i tlees2 110-0 645 500] 418 000} 227 500} 32 150] 2 125 8 027-28 | (Oc 12011 S55 O11 O30 ae vies 523 250! 418 000) 105 250) 32 150) 985 8 930-31.) JF Se 4213102 2410) eee 633 200] 418 000} 215 200} 32 150) 1 075 8 933 S.7 \153i1 22s A0-O18 aa 600 000} Applied 5 times; not broken. 8 934 Sit 115311 32 24/100) 856 000) Not broken; near ultimate. 8 935 Sit [P53ite 22 LOO a 600 000] Applied 3 times; not broken. Sire good wechcang | eA eae ae ree leet ce 830 000] Second test; near ultimate. 8 936 S. 4 (15302 4 10-0 ene. 625 000; Not broken. 8 937 S/t ise Ie 200 uae): 830 000 | oh: Pam el eae 714 000} Second test with spiral and * University of Illinois Bulletin + 0.75% of spiral reinforcement. t 1.00% of spiral reinforcement. outside concrete removed. No. 56, 1912, pp. i4 and 105. TESTS OF COLUMNS WITH STRUCTURAL STEEL | il of concrete on the strength was small, because the strength of the column was governed by the steel rather than by the concrete. In Fig. 33, p. 92, are shown deformations per unit of length of core-type column. Deformations of plain steel columns are also shown for comparison. (See also Fig. 32.) | The relative loads carried by the concrete and by the steel, respectively, are given in the table on p. 90. The amounts were determined by assuming that for equal deformations the structural shapes in the core-type column carried the same load as a similar plain steel column and the balance was carried by the concrete. Fireproofed Columns.—The behavior of fireproofed columns of the core type was about the same as that of the column without fireproof- ing. The concrete shell outside of the structural shapes, however, remained intact until the ultimate deformation of the column was nearly reached, but its effective unit strength was lower than the unit strength of the concrete core, probably because the shell failed before the maximum stress in steel and concrete core -was reached. Even if the protective covering is not relied upon as adding to the strength of the column, it is advisable to tie it by means of hoops or spirals of large pitch so as to prevent spalling in case of fire. Spiraled Columns with Structural Steel—The core-type column with surface spirally reinforced exhibited larger strength and tough- ness than similar columns without spiral. The thin protective cover remained intact. The strength of the columns exceeded the capacity of the Illinois and Lehigh University testing machines, which is 830 000 Ib. Because of the large deformation, the increase in strength afforded by the spiral is not available in ordinary building construction; hence a large percentage of spiral is not justifiable. One per cent of spiral reinforcement, however, makes the column tougher and safer and also prevents the outer shell from spalling. Since the danger of sudden failure is removed, such columns may be designed with some- what higher working loads than are allowed for the fireproofed type. Tests by Professor Withey.‘7—The structural steel reinforcement consisted of four angles 2 < 2 X 3; in. placed in the four corners of a square column. The out-to-out dimensions of the steel core were 8 in. square. The result of this test agrees with the tests previously described. 47 University of Wisconsin Bulletin No. 300, ‘‘Tests of Plain and Reinforced Concrete Columns,” 92 600000 600000 400000 300000 200000 coe 100000 4 Numbers 8907 - 8 500000 Applied Load in Pounds 400000 300000 200000 100000 / Numbers 8917 - 18 0 0005 0010 .0015 0 TESTS OF REINFORCED CONCRETE Length 10'-0” Numbers 8912 - 13 Length 19'- 4" Numbers 8922 - 23 0005. ~.0010.~—=«w00785 Deformation per Unit of Length Fic. 33.—Load Deformation Diagrams for Plai ing Core-type Columns.** Norr.—Curves labeled Column refer to core ty steel type. See also Fig. 89. 48 University of Illinois Bulletin No. n Steel Columns and Correspond- (See p. 91.) pe; those labeled Steel refer to plain 56, March, 1912, p. 24. TESTS OF LONG COLUMNS 93 Recommendations for Design of Core-Type Columns.— 203" my SAS eas eZee 976 SoS > weary O 1 2 Wom 213 3-" 2529" { 1 iy 71 Ns. 4'%721.8% 3 HEE es 4 ! ‘dae 3%" 98's i ! 2-7 25+9 Fe: ee tn Vell ' 1 os 1 a 2 ibe 1 { ' { ee ee alse oe 2-34'°928"5 ; ' a 4,"0 Sa PS ee Nest |5- "92.8% yy @ at ll9; o AS 5-" 2549" 3 + r H soe tye ee oy 2.8 ae roc ry 3-3"° 9-8" = a bel date tee Fier GA] 5-34" 72" re ae 5-9 2753" We ee 1 Sy a ow ia i ' ! ( ' tis’ | at ——3|3-76°"/2"8"s 2-99"°9-8"s ' 2-” 21-3"b 3-9) 25UQM fe by | t TH (% 7794 a" Ure A wy Wore ; 5-2 9°8"s 4 a eee 263") 5" 25°9"% Iixie 10-129 4-6"6 TB. ays ‘ oN 2-3" °272-8"s 6-72 ©916"6I, 7 wo Ke) jo tangas [atease twas ae TNT "7 122 8"3 SSN SH T- BS SAS Viv |) Sate 223 bSadead Odo eel % ine 3 hewn: 3 (lieeos es Sonn nnanan- a a ee 186 Seger tage ag 2s enn Ro ROSH ae agen ene eli hy HG, A South Fig. 35.—Steel Layout, Slab J. (See p. 99.) At load of 600 Ib. per sq. ft. the stresses were: Average stress in band that was stressed to the greatest degree, 56 000 lb. per sq. in. Maximum compressive stress, 1 800 lb. per sq. in. Ratio of yield point of steel to stress, 1.40. With poorer concrete, the compressive strength of the concrete and the yield point of the steel would have been reached earlier. 102 TESTS OF REINFORCED CONCRETE Steel Stress and Moment Coefficients in Slab J—Interior Panel—Live Load Band Average.... Average... « Average.... A |Positive N-S.. Positive E-W. Positive N-S.. Positive E-W. Positive N-S.. Positive E-W. m = moment coefficient 150 Stress 1 470 3 110 2 290 1 340 2 510 1 925 1970 1 890 1 930 lb. 300 lb. 450 lb. | 600 Ib. m Stress m Stress | ~™ Stress | m 5 820 15 580 28 780 9 620 16 040 30 650 0.0100| 7 720/0.0166; 15 810/0.0225 29 71510.0315 6 000 16 450 32 940 10 590 20 490 36 740 (0.0048! 8 29510.0106| 18 470 0.0154| 34 840)0.0220 5 600 15 650 31 820 10 190 18 870 33 420 0.0024! 7 895)0.0048)| 17 260/0.0070 om 620,0.0100 0.0130 32 300|0.0184 | | Positive, Whole Av. | 2 048|0.0048 7 970|0.0095| 17 180 35 040 33 020 45 740 55 810 50 028/0.0480 | 56 290 55 500 A |\Negative N-S.| 2 660 8 617 20 880 Negative E-W| 3 590 10 495 21 778 | Average....| 3 125)0.0228) 9 556 0.0349] 21 328/0.0515} 34 030\0.0655 B \Negative N-S.| 2375 8 275 24 320 Negative E-W| 3 680 11 450 29 675 Average....| 3028 9 863 26 748 C |Negative N-S.| 2170 TAGS 33 890 Negative E-W| 3 110 8 010 29 370 Average....| 2640/0.0027| 7 590 0.0041) 31 630/0.0114 | | | Negative, Whole Av.| 2 931 0.0100; 9 000/0.0151) 26 570 55 895)0. 0150 0.0270) 46 650/0.0355 Pntires. 2) 22 2 478 0.0148) 8 480/0.0246) 21 870 (0.0400) 39 500 0.0539 TESTS OF FLAT SLAB—TWO-WAY SYSTEM 103 Distribution of Stress —The ratio of negative to positive stress is as follows: Band Load 450 Load 600 Ee ert ee a. ee owak sla. Lo 1.14 ee og che ky bs rr 1.45 1.43 eM ee es ok haw na 1.86 1.72 Evidently, the negative steel may well be increased, especially in the margins or mid-section. The uniformity of stress is shown in the following, where the average = 100. Range of minimum to maximum band stress: Load 450 Load 600 ORES Se fo OE a Se 92-107 92-107 VASO. tees Duce, Tensei an a i aa 82-120 73-120 Deflections as a Fraction of the Direct Span (16 ft.)— Load | Direct Span Diagonal Span pote oms ere a). te Mie cas, 1/3800 1/1920 Fem Bo Fdieaps cys goo «GS ave os 1/1370 1/735 OE RS ES OS be ae 1/580 1/315 Fy me ee Ne gee ek s 1/375 1/194 Distribution of Moment Coefficients—The total coefficient is divided in per cent. Slab J Positive 450 lb. 600 Ib. Gh . 43.0 43.2 | See oot 29.6 beatin. 3 BILE we ELD) s2baselO0. 0-99 34,0 Negative nor: ack |. 3 645.2 Tee 3109 34.1 CR ie . 20.6 20.7 1002085 67-7 100.0%" (66.0 Load Carried by Lintel Beam.—'The steel stresses at the center and at the support of the lintel beam allow computation of the total moment of resistance of this beam. This may be equated to $W1, and the value of W, the load causing the flexure of the beam, may be computed. In Slab J, at 600 lb. per sq. ft. loaded on the panel, the indicated load on the lintel beam was 0.121 of the total panel 104 TESTS OF REINFORCED CONCRETE load. Another slab, with more highly developed steel stresses, yielded a fraction of 0.16. The moment in the wail column was 0.02 WI where W=load on panel and /=panel length. Comparison of Wall and Interior Panel.—The ratio of positive stress in wall and positive stress in interior panel was: Loading 450 lb. 600 lb. Acoycia bias See NS Fak OE ae ee ee OFS 0.73 Bw begins ath Sead Cen ek ean ae 0.91 0.82 Ca ene MMR PR UN in Ae rae ey 1.01 0.93 Average.) Jueed. «os en Fee ae ee 0.89 0.83 The 20 per cent additional steel in the wall panel positive sections is evidently sufficient. The stresses in the negative steel running into the wall beam, 2 680 Ib. per sq. in., and into the head of the wall column 10 230, are, at 600 lb. load, low compared to the negative stresses in the interior panel. Of course, this is only an expression of the yielding of the wall beam and wall column. Measurements on the web of the lintel beam gave consistent indications of torsion. The distribution of compressive stresses over the drop panel shows approximately that the concrete compressive stress at the edge of the drop is only about 25 per cent of the stress at the edge of the head of the column. In considering the allowable unit stress computed as an average over the drop panel, the maximum that fixes the strength of the slab must be considered. Various diagrams are reproduced, showing distribution of steel and concrete stresses and photographs of the slabs. TEST OF FLAT SLAB—SMULSKI SYSTEM The test was made by Professor Hatt for Mr. Edward Smulski and was reported in the Proceedings of the American Concrete Institute, 1918, p. 206. The information given below is taken from this report. The test slab consisted of four panels 16 it. square with overhang on three sides and a wall beam on the fourth side. The dimensions of the panel are shown in Fig. 36, p. 105. The overhang was intended to balance the bending moment at the column and thereby reproduce in the panels the conditions that TEST OF FLAT SLAB—SMULSKI SYSTEM | 105 exist in interior panels. Test showed that the 4 ft. 6 in. overhang was not sufficient. In a subsequent test made by the Corrugated Bar Co., and described on p. 99, an overhang of 6 ft. was used. _ The reinforcement is shown in Fig. 37. For fuller description of the system see p. 367. yeaa sci ipte ate oe SB0 Oey s eh aec sd ele ea ores ee tees: ie) ee a rie a! © ae ee a Section A-A Fic. 36.—General Dimensions of Smulski Test Slab. (See p. 104.) The design load was 150 Ib. per sq. in. and the amount of steel was that necessary to meet the requirements of the moment coeffi- cients recommended in the Joint Committee report, 1916. The two panels adjoining the overhang were treated as interior panels. Steel of structural grade was ordered. From test coupons it was found that a large part of the steel was of intermediate grade with 106 TESTS OF REINFORCED CONCRETE elastic limit of 42,200 lb. Slab was built Nov. 30, 1916. The test began Jan. 29, 1917 and extended over a period of forty-six days. Loading —The slab was loaded with-brick. The load was applied in the following increments: Load, lb. per sq. ft. East panel....... 150 300 450 450 3875 475 475 West panel....... 150°” 300 450. 625 700— 700 to, 950 70 Bey a = y K H = yy oO s 1 / ie (saan MMC inte ernie WATS OOd wae? Lp Ss Ks SED YY I Zs eS Ae: ean Mainwa Nes HME te oa 1 Oe, x Se PSE em, at | Cae Ne ee CL ee jh e=—- \K se am 8, Ss ‘< P y, 7 8 he L | ~ e, s as =" = ery Jes: KS iS SES ee 5X @ .] S ox ries SA t+) SS SE a aps aay Kt a 3 s. Ss =] at r | Tit <5) OY wal wane a oe oe LY Ny — wa eet os Ome Se 04, Sas Papas Sn Bee ign | > N= est ae eau OO SAT ett: {J eS Ore: Oo r= 1) <1 “s o™, e o OF : vd os (_] a} 4 aye a L ji CY SS IO % Nowe ea KS eS ee GE ~. SS uf As SLES To ry ines real Ws Aa ROSS RR ee a % BH FSS ies nn ez es 7) oe Ke S Mi AN SRE es Mn ents KX XS KR RN INVERN S VRS HCO | NS ec rrT ae Wine: nit Tw--? Init Tv : North Haff of Unit Ce same as Unit C V Le \N AA / a SHE ey LS <> oo SY, LL a, o, ay see SP, [? aA Sr Pk Sree ey, 3 waa Seraeeea a ca Spandrel Beam 9"x16" —Radials _---Center Rings Reh Mame earn ‘Truss Bars Section Thru Colurnn Head Fic. 37.—Arrangement of Reinforcement in Test Slab. (See p. 105.) The loads in the fourth to seventh increments were unbalanced and were much more severe on the two west panels than would have been the case with uniform loading. TEST OF FLAT SLAB—SMULSKI SYSTEM 107 Steel stresses were measured with 10-in. Berry Extensometer. The results are given in the tables on p. 107 and 108. Summary of Steel Stresses, Top of Slab, in Lb. per Sq. In. (Zero at 150 Lb. per Sq. Ft.) Columns Position of Gage Lines Loading of Panels, lb. per sq. ft. Quadrant Between|West 300; 450 625 700 | 700 rings |East 300} 450 450 | 950 | 475 | 700 Radials, Unit C Center column. . |Northwest... 1-2 3 360 5 760} 8 980] 33 000) 42 800 2-3 2 460 3 060} 7 180} 26 400) 42 700 3-4 PAG Tee eg eee ee 13 200} 35 700 4-5 BELO. Pate et cls s 1h Oe se 16 800 Southwest... 1-2 SABO GRY o.: 8 080) 34 200] 41 100 2-3 2640 | 2280] 7 240] 26 400] 31 500 3-4 1 440 900} 3 220] 14 400} 21 900 4-5 L260 eee 2980) fos 12 600 North column.../Southwest. . . 1-2 4860 | 5040} 10 880] 39 600/ 42 600 2-3 5 580 | 6900} 9 700) 22 200] 48 300 3-4 3 960 | 2400) 7 360} 24 000] 42 600 4-5 4240 | 3420] 7 780/19 800} 21 600 Southeast... . 122 7 260 | 8580} 6 520/17 400} 44 700 2-3 7800 | 7680} 6 460) 19 200] 40 800 3-4 4620 |11 340} 6 700) 19 800/31 800 Be sa cent 2 5 400} 9 280} 12 000! 36 000 Rings, Unit C Center column. . |Northwest... 2-5 D 460724. 54 es 8 110} 24 600; 40 600 Southwest... TOGO Tes es 8 040} 27 320) 38 400 Southeast... . 4900 | 5280} 5 920) 16 140} 33 300 Northeast... . 5700 | 4800] 6 640; 21 500) 41 940 North column...|Southwest. . 5 3800 | 6120) 8 980} 18 780) 34 000 Southeast . 4380 | 5600! 5 380/13 200] 34 000 South column... |Northwest... OOO wate 7 560} 14 340) 28 700 108 TESTS OF REINFORCED CONCRETE Summary of Stresses, Bottom Reinforcement, in Lb. per Sq. In. (Zero at 150 Lb. per Sq. Ft.) Position of Gage Lines Loading of Panels, Lb. per Sa. ft. Unit West 300 | 450 | 625 | 700 | 700 Ring Number Quadrant or Panel 950 East 300 | 450 | 450 475 700 Deena ee Pee aa Re a ne i Unit A ANGOvA ei aa oe ae Northwest....-<5- 3 240] 6 O60!..... Southwest......--- 3 000) 5 340|..... Southeast.....>. 14 040) 49 800 43 680) 49 800 12 580) 23 640 Unit, Aw.h opie cas tee Sia: “ance West panel........ 26 240| 34 620 Hast panel. vs ws. - PUTA beni Pete cts ics cil ids oo 18 840 ( (East panels) | Wall panel........ 3 480/001 thie 2 eee 22 500 Units B | 5456 Intermediate panel. 5 B80) Jas Teak eae 34 740 (West panels) | Wall PAUEl dy e-Mat 3 960] 6 360| 7 420] 30 540) 36 720 Intermediate panel. 3 900| 5 400] 6 070} 42 120; 49 800 (i(ast). eee2h Wall panel........ 2 S20! Col.C2 | oh Ae |. if ys Col A2 eeeUPAe Re) MA Col BI. a yh iN) osu live lates Sj >: Y Unit Ce-Ring No.6 Fig. 38.—Distribution of Stress in Rings of Column Head Unit C. (See p. 110.) in earlier test, did not take place. Therefore, the results of the two tests are comparable only for the loads prior to those producing cracking of wall columns in the Smulski System. (3) As evident from compressive tests of cylinder, the concrete 112 TESTS OF REINFORCED CONCRETE in the Two-way System was of better quality, presumably because +t was cured under more advantageous conditions. (Smulski slab was built in winter while the Two-way slab was built in the summer.) (4) The distribution of test load as adapted by Prof. Hatt for the two tests differs to some extent. Fic. 39.—Distribution of Stress in Rings of Unit Be. (See p. 110.) In the earlier (Smulski) test, open spaces were left at two columns for observation purposes, and extra brick was placed in the central portion of the slab to compensate for the omitted load and to pro- duce same bending moment as would have been produced by a full uniform load. In the second (Two-way) test practically the whole area was loaded and observations were made in tunnels. Due to this difference in loading, for equal unit load in the two TESTS OF REINFORCED CONCRETE BUILDINGS 113 tests, the Two-way slab was subjected to larger external shears, although the difference in bending moments was not appreciable. Keeping these differences in mind, the stresses as given in table on p. 102 may be compared with the stresses for equal loading as given in table on pp. 107 and 108. The stresses for the Smulski test are given more in detail while for the Two-way System as averages for whole bands. TESTS OF REINFORCED CONCRETE BUILDINGS UNDER LOAD One of the most important developments in testing within the last few years is the testing of complete structures under load. It has been customary to make deflection tests of engineering structures to determine whether the structure can safely carry the load for which it has been built. Such tests, however, are of little scientific use as they do not give the stresses in the structure. Sometimes they are actually misleading, because with small deflection there may exist in certain parts of the structure stresses that are much higher than allowable. Moreover, such tests do not determine _ weakness of details or the effect of continuity. The recent tests on completed structures are much superior to the old deflection tests, as they measure not only deflection of the structure, but also the stresses in various parts of the members. These tests were inaugurated by Prof. Arthur N. Talbot of the University of Illinois, with the assistance of Messrs. A. R. Lord and | W. A. Slater. The first building tested in this way was the Deere and Webber building in Minneapolis, in October and November, 1910. Following this test, several other tests were made under the auspices of the Reinforced Concrete Committee of the American Concrete Institute. The instruments used in such tests are (1) extensometer for measuring the stretch or compression of the materials, (2) deflecto- meter, for measuring deflection. The extensometer consists of a framework (which in the best instruments is made of invar steel to prevent appreciable changes in length due to the changes of temperature), two movable legs attached to the framework and provided with sharp points, and a device for measuring accurately any changes in distance between the points. In order to find the stretch in steel, the bar to be tested is uncovered in two places a few inches apart, and small holes, called gage holes 114 TESTS OF REINFORCED CONCRETE (0.055 in. in diameter) are drilled. An observation on the gage line is taken before the structure is loaded and again at each increment of the load. The difference between the original reading and the reading at any load gives the stretch of the steel due to that load. The stress is then found from the known relation between the defor- mation and the stress. The compression in concrete is measured by making small holes in the concrete, inserting metal plugs, and then marking the gage holes in these plugs as was done for the steel. Readings and stresses are obtained in the same way as for steel. Since concrete flows grad- ually under heavy loads, the readings must be made immediately after each loading. For measuring deflections, a rigid scaffold is built directly under the members to be tested. In the place in which a deflection reading is desired, a steel plate is fastened by plaster of Paris to the under side of the beam or slab. On a vertical line below this steel plate, a steel rod is fastened to the scaffold. Before the test is begun and at different stages of loading, the deflectometer is placed between the plate and the rod below, readings are taken, and the difference between the original reading and the readings under the load give the deflection. For loading the panels, there may be used: (a) brick, (0) cement in sacks, (c) loose sand in boxes or in sacks, and (d) pig iron. In making tests, care always must be taken that the material does not arch. The whole floor cannot be covered with the load, because places must be left uncovered for the taking of measurements, and aisles must be left to make the points accessible. It is important that the test load should cover a sufficient floor space to insure that certain parts of the floor resist nearly the full load which, in the calculations, they are considered to take. Wenalden Building Test.®*—The floor panels in this building are 15 ft. by 20 ft. The slab is 33 in. thick; the girders, placed between columns in the short direction, 74 in. by 20% in., reinforced with four 3-in. square bars in the middle. The longitudinal beams, 5 ft., apart on centers, are 6} in. by 18% in., reinforced with four {-in. square bars in the middle. Half as much steel was used over the sup- ports as in the center. The floor was designed for a live load of 200 lb. per sq. ft., and the total test load was made 400 lb: per sq. ft. and placed in layers of 63 University of Illinois Bulletin No. 64, January 13, 1913. TESTS OF REINFORCED CONCRETE BUILDINGS 115 80 Ib. per sq. ft. A set of observations was taken after every addi- tional loading. The measurements were taken in steel at the support and at the center of the span; also measurements of stresses in concrete at the support and at the center of the span. This test proved conclusively that the beams and the girders act as continuous ones. While the stresses in steel in the center and at the support were not excessive, the highest stress being 17 000 lb. for the total test load, the stresses in concrete at the supports of the _ beams were high and in some places even reached a stress of 2 200 lb. per sq. in. Even under the working load, stress in concrete was 1 150 lb. per sq. in. The compressive stresses in the center of the beam were low, and it appeared from the test that the total slab acted as a compressive flange of the T-beams. It must be noted that in this case the overhang of the flange was seven times the thick- ness of the slab, while in practical design, we consider only an over- hang of six times the thickness of the slab as effective in taking compression. The total compression in a beam, figured with the assumption of a straight line distribution of stress and no tension in concrete, was much larger than the total tension, showing that either arch action existed in the beam or considerable tension was carried by concrete. ‘The difference was especially large at the supports, where the tension must have distributed itself over the entire slab. Test Cracks.—Tensile cracks were observed in the middle portion of the bottom of the beams. They formed at the same stress in steel as is usually found in the laboratory. Diagonal cracks developed in the girder which carried a very large shear (V = 40 000 lb. and v = 360 lb.), just outside the junction with intermediate beams. The cracks were inclined at about 45°. They did not close entirely after removal of the load. It is supposed that the restraint at the ends prevented fuller development of the cracks. (See also p. 66 on formation of cracks in continuous T-beams.) Deflections.—The deflections offered further proof of the contin- uity of the beams, being much larger, in the middle panel, for one panel loaded than for three panels loaded, as would be expected from a continuous beam. With three spans loaded, deflection of intermediate beam was 0.09 in., and for one span loaded 0.15 in. Turner-Carter Building Test.—The panels in this building are 17 ft. 4 in. by 19 ft.6in. The girders are placed in the short direction and their dimensions are 10 by 24 in., with two 1-in. square and three 116 TESTS OF REINFORCED CONCRETE Zin, square bars at the middle, and two 1-in. square bars over the support. Beams, 7 by 18 in., reinforced with one 1-in. square bar and two 2-in. square bars at the middle, and one 1-in. square bar (plus ten 3-in. round bars in the slab) over the support, are placed between the columns and at one-third points of the girder. The thickness of slab is 4 in. The structure was designed for a live load of 150 lb. per sq. fis and the beams and girders were figured as simply supported, but reinforcement was supplied for continuity. The test load was 300 lb. per sq. ft. or double the designed load. Results of Test—The beams and girders acted as continuous. The stresses in steel in the beams were comparatively low, the max- ‘mum observed for the test load being 11 000 lb. The stresses in concrete, however, at the end of the beam reached 1 100 lb. per sq. in. At the middle the compression in concrete reached only 350 lb. per sq. in., which shows that the. compression there must have distrib- uted itself over a large portion of the slab. In the girders the tensile stresses at the middle reached only 8 000 lb. per sq. in. At the supports no measurements were taken because the steel was not accessible. ‘The compressive stress at the end of the beam in the bottom was 900 Ib. per sq. in. and was very low at the center in the top surface. In both beams and girders, the total compression was much larger than the total tension, a condition that was found in the previous test. As far as observation shows, the entire slab acted as compression flange of the T-beam. General Conclusions.—In drawing conclusions from tests on com- pleted structures, it must be remembered that low stresses in the steel do not indicate a large factor of safety. The conditions are the same as were explained in connection with laboratory beam tests (see p. 27) in which the stresses at half the maximum load were small, while the maximum load stressed the steel to the elastic limit. The results of such tests must be used with caution. TESTS OF OCTAGONAL CANTILEVER FLAT SLABS An interesting test of cantilever flat slabs supported on a central column, as shown in Fig. 40, p. 118, was made in the year 1914 by Mr. Edward Smulski under the supervision of Sanford E. Thompson. Eight specimen slabs were made octagonal in shape, 6 ft. 6 in. in TESTS OF OCTAGONAL CANTILEVER FLAT SLABS Die small diameter, and with an octagonal column head in the center built monolithic with the slab and having an inside diameter of 2 ft. The slab was 4 in. thick. The reinforcement of Specimens 1 to 4, arranged as shown in Fig. 40, differed in the diameter of bars used for the five outside rings, as shown in the table on p. 120. Speci- mens 5 and 6 were similar to 3, except that 10 and 5 radials respec- tively were used instead of 20. Specimen 7 was reinforced by four layers of bars running in four directions, each layer consisting of nine ;%-in. round bars. Specimen 8 was reinforced with steel in top and bottom; the tensile reinforcement consisted to two layers placed at right angles, with twelve 3-in. round bars per layer, and the compressive reinforcement consisted of two layers with eight $-In. round bars per layer.*4 Purpose of Test.—The behavior of a circumferential cantilever is similar to the behavior of a flat slab floor system at the support. The purpose of the test was to compare the effectiveness of circum- ferential reinforcement with that of band reinforcement and to determine the most effective distribution of steel between rings and radials. The results of the test are directly applicable to flat slab construction. Materials of Construction—Concrete in the proportion of 1:2:4 was used. The compressive strength of 6-in. cubes, tested at fifty-two days, was 1 100 lb. per sq. in. Reduced to eight 16-in. cylinders and to twenty-eight days, the strength of the con- crete was about 1 400 Ib. per sq. in., or lower than that of first-class 1:2: 4 conerete. ‘ Plain round bars with an average elastic limit of 35 000 lb. per sq. In. were used. , Method of Testing.—In testing, the slabs were placed on a wooden column resting upon a base which distributed the load to the soil. The load, consistiag of pig iron averaging 56 lb. per pig, was placed on swings arranged along the circumference of the can- tilevers as shown in Fig. 40, p. 118. By this method the point of application of the load, and therefore the moment arm, was posi- tively fixed. | Deformation Readings.— Deformations in steel, due to the loading, 4 Design of specimens 1 to 6 were Smulski system, specimen 7, four-way System, and specimen 8, two-way system. Specimens 5 and 6 were of pre- liminary type. Their results are not given because they are not comparable with other specimens, 118 TESTS OF REINFORCED CONCRETE Fra. 40.—Cantilever Slab and Loading Platform. (See p. 116.) TESTS OF OCTAGONAL CANTILEVER FLAT SLABS 119 were measured by a Berry Extensometer on 8-in. gage lines. For this purpose, gage holes 4 about z5-in. diameter 2a Bee nemaeeee lag eal ie i were drilled in the steel. Each ring and each bar / was provided with at ee |p [Re least four gage lines to a eliminate the possibility of erratic results. Aver- - age stresses were plotted on a deformation dia- | gram. Fig. 41, p. 119, shows the deformations Si NS SS = & @© Load in Hundred Pigs (Average Weight of Pig, 56 Lb.) at different loadings for 4 Specimens 1 and 2, and nae mig 42, p. 120,: for ; ae ee Specimen 7. The curves al Ae a vl Ase, for Specimen 8 are sub- 0 10 20 30 40 10 20 0 10 20 stantially like those for Unit Stress in ridlanal Pounds SPECIMEN 2 Specimen 7. Results of the Tests. —The results of the tests are shown in the table on p. 120, which gives the dimension of speci- mens, total loads, and load per pound of ten- sile steel. The measured stress and the load at. first visible crack also are given in the table, from which it is evident that the first visible cracks occurred at about two-thirds of the load at the elastic limit. Judging from the stress 0 10 20 30 40 0 10 20 30 40 0 10 20 Unit Stress in Thousand Pounds diagrams, hair cracks SPECIMEN? invisible to the eye must Fia. 41.—Deformation Diagrams for Slab Speci- have appeared at a mens 1 and 2. (See p. 119.) Load in Hundred Pigs (Average Weight of Pig, 56 Lb: ) 120 TESTS OF REINFORCED CONCRETE smaller load corresponding to the break in the deformation curve. Summary of Results of Tests of Octagonal Cantilever Flat Slabs Octagonal slab 6 ft. 6 in. inside diameter; column head 2 ft. diameter; 1:2: 4 concrete; mild steel. Specimens 1 to 4 Radial; Specimen 7, 4-way; Specimen 8, 2-way. All slabs 4 inches thick. ee on moh | Specimen Number =I Number and Diameter ais fo 4 4 a 33 cr of Round Bars ets aa » : Roo ne © : Hg =! We) on sa £ ra] o oF in Inches hh o fe) a 2 2 - © mee a © 7H te) a7) ae rs 65 ee fs a S|: ox Do a aey| 35 42 glee melee 235 5 -6= A (oP) ~ am Aa. +o if NM — ne) = o sao s lover tsiitel G > eZ A : 5 s ra 2 * a ee Rae Te es @ © |Radial|Outside|Straight| {3 . te © = oO Py a) Q = j Sno O.8 ep I Qu ars Bo oO oD bars | rings bars’ ig.= & . Oh si oe aoe 3 $ 38 Be g | 3 s84| s¢ |88| $4 | 22] 85 | gee a < < < H 4 4 = AT \ 4226 120 —$) 0 "ye lee 29.2 | 0.368t |2.56) 18 800 | 442 12 400 | 14 000 44. \ 5022) 120 — "215 9 eee 2.2 | 0.484t |2.62| 22500 | 448 | 15 000 | 15 000 AD FT 2 Ot Sa ae Ulcer 2.2 | 0.901¢ 12.60} 32 000*| 447 | 21 200 | 16 000 25 (101.2 120 — 4 Sr eee 22 NPI 3s 2.50| 42 500t| 421 | 26 600 | 15 000 Bt W600 accel aaeeaetete 36 — | 5.5 |........ 2.25| 12600 | 210 8 950 | 15 000 50 | 56.0 Bee a: 24— 3 Mase pine we QESS 12 BOUT eee 8 900 | 15 000 Compression |16 — 3 So | ee ee * Estimated from stress diagram. Broken by accident at 29 500 lb., before elastic limit was reached. + Estimated from stress diagram. Elastic limit not reached at maximum applied load. + Only part of the area of Ring 5 was considered as effective because it was placed too near the column head and therefore carried smaller stress than the other rings. Load in Hundred Pigs S to El peels ls] [eal 7 ea ce ‘creda Ae eA 1020.30 40.0 10 20 30 40 50 0 10-20 30 40 0 10 20 30 40 | | DiDe| Dg Ds a] sae | | Beebe [Or] oe Las gen FA —— S4aenan aa if | y| roe | Vi) ela ee 0 10° 390° 30. 40 1507-07 10 20 Fal, e 10 20 30 40 50 0 10 20 30 40 Unit Stress in Thousand Pounds %> ~— SS & Fic. 42.—Deformation Diagrams for Slab Specimen 7. (See p. 119.) The first crack, at first hardly noticeable, extended all the way around the circumference of the column head, several inches from its edge. Upon additional loading, this crack opened slowly and other circumferential and radial cracks appeared. The test was TESTS OF OCTAGONAL CANTILEVER FLAT SLABS 121 discontinued after the steel had reached the elastic limit with the exception of Specimen 4, in which the elastic limit of the slab was not reached on account of the difficulty of applying further loading. No cracks developed within the column head although the radials were stressed to elastic limit and the hooked portions did not bear against the center ring. Of interest is the fact that the cracks in Specimens 7 and 8, reinforced with bands of bars, were also radial and circum- ferential. From the stress diagrams, it is noticeable that the outside Rings, 1 to 4 in Specimens 1 to 3, and Rings 1 to 3 in Specimen 4 were equally effective in resisting the bending moment, the stresses at different loads being almost equal. The stress in Ring 5 was smaller than in the other rings, which can be accounted for by the fact that the ring was placed too near the column head. ‘The stresses in Rings 6 and 7 within the column head were very small, showing that very little stress is transferred by the radials to the center rings. Evidently, most of it is transferred to concrete by bearing. Splicing of Rings.—All rings were spliced with a 50-diameter lap During testing, special attention was paid to the behavior of the steel at the splices and it was found that the elastic limit was reached without any movement being observed at the splices. Conclusions.°°—(1) First crack occurred at substantially the same measured stresses in the steel, irrespective of the arrangement and amount of reinforcement. The load at first crack increased with the increase of reinforcement. (2) The actual load sustained in all specimens was larger than would be expected from ordinary methods of computation, proving the effect of Poisson’s ratio. The reduction of bending moment coefficients suggested for flat slabs on p. 331 is justified. (3) The relative effectiveness of the various arrangements of steel can be obtained by comparing the load per pound of tensile steel, which for Specimens 1 to 4 varied between 420 Ib. and 450 Ib. and for Specimens 7 to 8 between 210 and 225 lb. (4) In specimens reinforced by rings, the stresses were uniformly distributed over all rings. (See stress diagrams, p. 120.) (5) The lap of 50 diameters of a plain bar was sufficient to develop the elastic limit of the rings. . 55 Taken from report by Sanford E. Thompson. CHAPTER IV THEORY OF REINFORCED CONCRETE Reinforced concrete is concrete in which steel or other reinforcing metal is imbedded to increase its strength. The reinforcement in general consists of bars of small diameter connected with the con- crete by bond. Since the reinforcement, has little stiffness in itself, in compression members 1t relies upon the concrete and the ties for lateral support. When properly designed the reinforcing bars are capable of developing full tensile or compressive resistance. Some- times rigid structural shapes are used as reinforcement. - Not all construction consisting of steel and concrete may be classed as reinforced concrete. The function of the two materials in a completed structure determines whether it is a reinforced con- crete structure or a steel structure fireproofed with concrete. An arch of the Melan type, for example, consisting of rigid metal ribs capable of carrying some of the load unassisted, but depending largely for its stiffness and ultimate strength upon the concrete, (see Vol. III) may be considered a reinforced concrete structure. A similar arch, consisting of metal ribs connected laterally by metal bracing and strong enough to carry the entire load, although encased in concrete, cannot be considered as a reinforced concrete structure, because the concrete does not serve the purpose of resisting stresses, but only the auxiliary purpose of protecting the steel against cor- rosion and giving the arch the appearance of a masonry structure. Any increase in strength of the structure due to the concrete is only incidental. Similarly, conerete columns and other mem- bers with structural steel shapes may be considered as reinforced concrete when the concrete serves to increase the strength of the member. When, however, as in steel frame structures fireproofed with concrete, the steel is self-supporting and designed to resist the whole of the stresses while the concrete serves chiefly for pro- tection, the member is not reinforced concrete. The theory of the design of reinforced concrete is definitely estab- 122 GENERAL PRINCIPLES OF REINFORCED CONCRETE BEAMS 123 lished. The action of the combination of steel and concrete in ten- sion, compression, and shear is well understood, so that a thor- oughly rational treatment is possible. In practice, in beam design, the straight-line theory, as it is termed (see p. 127), which was selected and adopted by the authors for the First Edition of this treatise, In 1905, has since that time been accepted universally as affording the simplest method of computation and as giving results which may be used in design with safety and economy. In this chapter is presented the analysis of this straight-line theory of stresses for rectangular beams (p. 129), followed by the application of the same theory to T-beams (p. 133). The analysis of a beam with steel in top and bottom is given on p. 137. Wedge- shaped beams are treated on p. 140 and unsymmetrical T-beams on p. 149. The analysis of shear and diagonal tensionwill be found on p. 143. The theory of columns of concrete reinforced with vertical steel bars is treated on p. 159, and that of columns reinforced with ver- tical steel bars and spirals on p. 160. Analyses and formulas are presented for the distribution of stresses in reinforced concrete under combined thrust and bending moment (p. 164) for use in arch design and in the design of columns and beams with eccentric load or thrust. The theory of members subjected to direct tension or pull, p. 162, and also to pull and bending moment is given on p. 189. The theory of reinforced concrete chimney design is treated in Chapter XX. Formulas for use in practical design, with illustrations of methods of treatment, will be found in Chapter V. Tests of reinforced concrete, covering all usual features of design, are taken up in Chap- ter III. - GENERAL PRINCIPLES OF REINFORCED CONCRETE BEAMS Concrete is very strong in compression, but is brittle and unreli- able in pull or tension. Therefore, it cannot be used safely nor economically where tensile stresses have to be resisted. Steel, on the other hand, being a comparatively ductile material, is well adapted for resisting pull, but is more costly than concrete for resist- ing compression. The economy in the use of reinforced concrete is obtained by placing concrete where compressive stresses are to be resisted, and steel where tensile stresses are to be resisted. The bond and shearing resistance of concrete holds the steel and concrete together so that they act as a unit. 124 THEORY OF REINFORCED CONCRETE Requirement for Formulas for Reinforced Concrete Beams.— The behavior of reinforced concrete beams under load, as discussed on p. 20, is different from that of homogeneous beams. The location of the neutral axis for varying intensities of loading is not constant. ‘The sum of compression stresses in the concrete is nearly proportional to the load, but the tensile stresses in the steel are not proportional to the load because of the variable proportion of ten- sion resisted by concrete (see p. 97). It is impossible to make formulas which would represent actual stress conditions for all loads within elastic limit. This, however, is not the purpose of formulas used for design. Their purpose is to supply a construction to carry a load equal to the design load multiplied by the factor of safety, without either steel or concrete exceeding the elastic limit. T his being the case, the design formulas must be based on conditions existing at the elastic limit. In practice, instead of multiplying the design load by the factor of safety and working with stresses at elastic limit, the work is simplified by establishing working stresses equal to the stresses at elastic limit divided by the factor of safety, and using them with bending moments for design load only. It is important to keep the purpose and the origin of the formulas con- stantly in mind, to avoid using them for purposes for which they are not intended. Formulas for reinforced concrete beams must satisfy the following requirements: | (1) The compressive stresses in concrete for working loads must not exceed the allowable unit stress; (2) The beam must have the required factor of safety based on the elastic limit of steel. , The first requirement fixes the unit stresses for concrete. To satisfy the second requirement, it is necessary in the analysis to eliminate the variable amount of tensile stress carried by the con- crete (see p. 30), by assuming that all the tensile stresses are resisted by the steel. Analysis based on this assumption will not repre- sent the actual conditions in a beam under working loads because the actual stresses in steel will be less than the computation would show, but it will give the required factor of safety and, therefore, be correct for design. On p: 27 is given a comparison between actual stresses and stresses computed by the accepted formulas for beams with different ratios of steel to concrete. It is seen that, GENERAL PRINCIPLES OF REINFORCED CONCRETE BEAMS 125 for earlier stages of loading, the actual stress is much less than the computed. ‘This difference, which is due to the tensile resistance of concrete, decreases with the increase of the load. At the elastic limit of steel, the stage upon which the factor of safety would be based, the computed stresses agree fairly well with the actual stresses. - The action is shown by the tests, illustrated in Fig. 9, p. 29, which give the deformations of concrete and steel at various loads. Stresses, of course, are proportional to the deformations. Factor of Safety.—By the factor of safety is understood the ratio between the load at which the construction would fail and the load which it is intended to carry. The words “ fail’ and “ failure ” do not mean actual collapse, but a state where the structure ceases to be useful. If the material of which the construction is composed is equally strong in tension and in compression, the factor of safety is the same against both tensile and compression failure. Ina structure of a composite material, such as reinforced concrete, the factor of safety against compression failure may be different from the factor of safety against tensile failure. The factor of safety of the structure as a whole is governed by the smaller factor. In reinforced concrete construction, steel is considered as the more reliable of the two materials, because it is manufactured under standardized and uniform conditions and the variation in its elastic limit and ultimate strength is, therefore, small. The manufacture of concrete, on the other hand, as well as the strength and character- istics of the materials composing it, varies from job to job so that the ultimate strength of the concrete in different constructions may vary to a great extent. Also, during construction, concrete may be called upon to resist considerable stresses before its full strength has been attained. The allowable unit stresses in compression are so selected that the construction has a larger factor of safety against compression failure than against tensile failure. The structure is, therefore, governed by the factor of safety of the tensile steel. The factor of safety, governed by the reinforcement, is based upon the elastic limit of the steel and not upon its tensile strength, because failure of the beam follows closely after stresses in steel reach the elastic limit., The ultimate strength of steel in tension is of no significance as far as the strength of reinforced concrete is con- cerned. Formulas Considering Tensile Strength of Concrete.—In analyz- ing the results of the tests in-the early stages of the loading, it is 126 THEORY OF REINFORCED CONCRETE sometimes necessary to consider the tensile stresses in concrete. Formulas for such a case are given on p. 140. ASSUMPTIONS In the analysis of beams, the following assumptions will be made: (1) A section that is plane before bending remains plane after bending. (See p. 127.) (2) Modulus of elasticity of concrete is constant. (See p. 201.) (3) Tension is borne entirely by the steeel. (See p. 124.) (4) Adhesion of concrete to steel is perfect within the elastic limit of the steel. (5) There are no initial stresses in the steel. Reasons for selecting these assumptions are as follows: (a) Beams designed by formulas based on them have the required factor of safety. (b) The method of design based on them is the simplest. (c) They have been adopted by the highest authorities in America and Europe. Assumption (2) is not actually correct. The modulus of elas- ticity varies with the variation in intensity of the stresses; but the allowable compression stresses and the modulus of elasticity have been selected in such a fashion that the results obtained by the simple formulas give the same dimensions as would be obtained with the more complicated formulas using the variable modulus of elasticity. The maximum fiber stress obtained by formulas is larger than the actual stresses. In the selection of the working stresses, this has been taken into account. Working stresses recommended in this book should not be used with other than “ straight-line ”’ formulas. ANALYSIS OF RECTANGULAR BEAMS Bending Moment and Moment of Resistance,—In a beam sub- jected to bending, the bending moment due to the external forces, or loads, is resisted by the moment of the internal resisting forces, which will be called stresses. Since, by the simple mechanical laws of equilibrium, the bending moment must be equal to the resisting ANALYSIS OF RECTANGULAR BEAMS 127 moment of the internal forces or stresses, the unknown stresses in the materials may be found by equating the known external bending moment to the internal resisting moment. Straight-Line Formula.—The stresses cause deformation in the material and the consequent deflection of the beam. At any ver- tical section through the beam, the compressive stresses above the neutral axis cause shortening of the fibers, and the tensile stresses below the neutral axis cause lengthening of the fibers. According to the first assumption above, a section that is plane before bending remains plane after bending. This means that when the beam is bent, a plane section through the beam simply swings into a new position, without warping. ee ee eee Before Bending-------- me tiges. pI 27 for Oaihe pads example, section AB after f | bending assumed the posi- A_ Ay tion A’B’ by turningaround Compression point O at the neutral | nics axis. Therefore, the de- Ae OF WO formation, or change in \ ee length, in any fiber is pro- ‘ portional to its distance /__| ney from the neutral axis and Ebina pea 98 Puna J the variation of deforma- E ie ices ok After Bending 2—-->2--22—<2 tion of different fibers pyc. 43—Plane Section before and after may be represented by a Bending. (See p. 127.) straight line. According to the second assumption, the modulus of elasticity is constant for all intensities of stresses. With a constant modulus of elasticity (see p. 126), the ratio of the deformation to the stress is constant. ‘Therefore, the variation of the resisting stresses from the neutral axis upward can be represented by a straight line, in the same manner as the variation in deformations. The compressive stresses then form a triangle having for its base the stress in the extreme fiber, f., and for its height the distance from the extreme fiber to the neutral axis, kd. The total compression may be con- sidered as concentrated at the center of gravity of the triangle, which Is distant $kd from the extreme fiber. (See Fig. 44, p. 129.) According to the third assumption, all tensile stresses are resisted by the steel. They may be considered as acting in the center of gravity of the steel, which is at the center of the bar if there is only 128 THEORY OF REINFORCED CON CRETE one layer of bars, and at the center of gravity of the set of bars, if there is more than one layer. For equilibrium, the sum of all forces must equal zero; that is, the total compression must be equal to the total pull. The total tension and the total compression, which are equal and act in oppe- site directions, form a couple with a moment arm equal to the dis- tance between the center of steel and the center of gravity of the triangle of compressive stresses. ‘The moment caused by this couple is the resisting moment. For equilibrium this must equal the bending moment due to exterior forces. The discrepancy between the actual conditions in a beam and the assumption of straight-line distribution of stresses is discussed on p. 142. | : NOTATION For this and succeeding analyses, let h = total depth of beam; t = thickness of T-beam flange, i.e., thickness of slab; b = breadth of rectangular beam or breadth of flange of T-beam; b’ = breadth of web of T-beam; A, = area of cross section of steel = pbd; p = ratio of steel in tension to area of beam, bd e ° e A, e Pm = Maximum ratio of steel in T-beams, ba for which compres- sion stress in concrete equals maximum allowable stress. In beams with steel in top and bottom: pi = ratio of tensile steel to area of beam, bd; ratio of compressive steel to area of beam, bd; f, = compressive unit stress in outside fiber of concrete; f, = tensile unit stress in outside fiber of concrete; f; = tensile unit stress, or pull, in steel; ; = compressive unit stress in steel; E, = modulus of elasticity of concrete; E, = modulus of elasticity of steel; n= _ = ratio of moduli of elasticity; d = depth from outside compressive fiber to center of gravity of | steel; . Ss I ANALYSIS OF RECTANGULAR BEAMS 129 a = ratio of depth of compressive steel to depth, d, of beam; k = ratio of depth of neutral axis to effective depth of raty d; Z = total amount of shearing stresses in distance, s, and cantik A in any horizontal plane below neutral axis. | Total Compression Moment Arm Total Tension , bd, va i pdf; n Fig. 44.—Resisting Forces in a Reinforced Concrete Beam. (See p. 129.) | ! | 3 | | ! =o es kd =depth of neutral axis below the compressive surface in a beam; j = ratio of lever arm of resisting couple to depth d; jd = moment arm, i.e., distance between centers 6 tension and compression ; M = moment of resistance, or bending moment in general; C= constant i in ‘Tables, pp. 880 and 881. Since it is assumed that a section which is plane before bending remains plane after bending, we have the proportion stretch in steel ms Clk) deformation in outside compressive concrete fibers kd stress per square inch ff. = and unit and since unit stretch in steel = chad ain of eld Stik ipmae ns stress per square inch __f, deformation in concrete = modulus of elasticity = j we have fe deny be on Z = i or ahve Hise Je E, fe ar 1 #2 a) era Za ae Ae 130 THEORY OF REINFORCED CONCRETE Solving Formula (1) for fe k Lee locTaeRy: ale te AS Now, as stated above, for equilibrium the total tension in the steel must be equal and opposite to the total compression in the concrete. The total tension in the steel is its unit stress, fs, multiplied by the area of the steel, pbd, and the total compression in the concrete is rep- resented by the area of the pressure triangle, 3f.kd, times the breadth of the beam, 6. Equating these two forces and canceling out the bd which occurs in both, of 1, If the value of k in Formula (2) be substituted for the k in Formula (4) we have “ 1 ods i) aE (t+ ag For any given percentage of steel the values of f, and f. cannot be assumed independently, as they bear a constant ratio to each other. Substituting the value of f. in Formula (3) for fc in Formula (4) we have 7 (5) k k ie ase GER a; _ previo, Gein Solving this quadratic equation and adopting the positive sign before the square root, kz = np + Vinp + (ip) (7) From Formula (7) the location of the neutral axis may be deter- mined for any percentage of steel, p, and any assumed ratio of mod- uli of elasticity, n. Values for k are given in Tables, pp. 880 and 881. The center of gravity of the compressive stresses is distant 3kd from the top of the beam, so that jd = d — 3kd = (1—4k)d. Since the total compression equals the total tension, the moment of resistance of the beam may be obtained by multiplying either the ANALYSIS OF RECTANGULAR BEAMS 151 total tension, pbdf; = Asf;, or the total compression, 5febkd, by the moment arm jd. M M eae: A.f.jd ° e (8) and he — A,jd’ ° e (9) = fkjbe _ 2h Mie pain « (10) and fo= op + + (1) For a given quality of concrete and steel, the values of Va! ote and j, are constant, so that we may consider the terms Pisj = sfkq equal to a constant = - This changes Formulas (8) and (10) to By M =. R ©or(19) al pCa (13) The constant values in Formulas (8) and (10) may also be called $049 = pi.) = R. The formula for depth of beam may be also expressed by d = (14) bR’ The area of steel may be obtained by solving Formula (8) for As. It is M Sacer SHO Rat Le couse | ELE) For balanced design the area of steel equals oka DOU, Lik hi Ae el oe tT hy where p is the ratio of steel corresponding to the allowable stresses in steel and concrete and d is the depth computed on the basis of the Same stresses. To obtain the total depth of beam, h, a value e must be added to the theoretical depth, d. Then, h =d-+e. For the use of these formulas in design, see pp. 203 to 214. 132 THEORY OF REINFORCED CONCRETE Values of constants j, k, and p are given on pp. 880 and 881. Values of constants C and R are given on pp. 880 and 881. Balanced Design of Rectangular Beam.—As is evident from Formulas (9) and (11), the stresses of fc and fs are practically inde- pendent of each other. The stress in concrete depends upon the depth and breadth of beam, while the stress in steel depends upon depth of beam and the amount of reinforcement. A balanced design of a rectangular beam is one in which the stress in steel and the stress in concrete reach their maximum allow- able values at the same time. The moment of resistance of concrete, figured by Formula (10), is equal to the moment of resistance of steel, figured by Formula (8). In a balanced design for fixed stresses fs, fe, and n, there is a fixed relation between the area of steel and area of concrete. Thus, for a stress in steel, f= 16 000, and stress in concrete, fe = 650, the ratio of steel for a balanced design is p = 0.0077. Unbalanced Design of Rectangular Beam.—An unbalanced design of a rectangular beam is one in which either the amount of steel or the amount of concrete. is larger than required by the stresses. In such a case, the moments of resistance figured for the two mate- rials, based on their respective working stresses, will not be equal. The available strength of the beam will then be governed by the material giving the smaller moment of resistance. Thus, if the amount of steel is larger than that obtained from the formula As = pbd, where p is the ratio of steel for a balanced design corresponding to the selected unit stresses, the design has more steel than is required by tensile stresses. Therefore, the bend- ing moment must be based upon the compressive strength of the con- crete, and the steel cannot be stressed to its full working value. For a bending moment based on the full working value of steel, the concrete stresses would be excessive. If such a condition should occur in a design, the amount of steel should be reduced, unless the increase in the amount of steel was necessitated by bond stresses. (See p. 264.) On the other hand, if for any reason an amount of steel smaller » than A, = pbd is used (p being the ratio for a balanced design), the moment of resistance must be based on the reinforcement. The stresses in concrete will be smaller than the allowable fiber stresses. To stress the concrete to its working value would over- stress the steel. | FORMULAS FOR T-BEAMS 133 FORMULAS FOR T-BEAMS If a reinforced concrete beam is built monolithic with the slab, the beam may be considered as a T-beam in which a portion of the slab acts as a flange. (See Fig. 45, p. 133.) The formulas for T-beams given below are based on the same assumptions as those for rectangular beams. Tension is considered as taken entirely by the steel, and the variation of stresses in con- crete is according to a straight-line. Unless the slab is very thick, the neutral axis is located below the flange. For notation, see p. 128. Case I.—Neutral Axis below Flange, kd > t. Formulas Neglecting Compression below Flange.—In ordinary T-beams, the amount of compression stresses resisted by stem below the flange is insignificant and may be neglected. This simplifies the formulas without causing any appreciable error. Neutral Axis ae Fig. 45.—Resisting Forces in T-shaped Section of Beam. (See p. 133.) The location of the neutral axis is determined in the same man- ner as for rectangular beams. The ratio k may be expressed by ae 1 1+ 4 (17) The total tension is equal to the unit stress in steel, f,, multiplied by the area of steel, A,. The total compression in the concrete is ; : kd -— represented by a trapezoid, the sides of which are f, and fou s a 134 ‘THEORY OF REINFORCED CONCRETE and the depth is equal to t. The total compression, therefore, equals 2k 5 2kd — t Vic kd bt, also ‘eo imi bt. By equating total tension to total compression acting on the section Dh st Asfs came” v Id bt. ° ° e ° ° ° ° (18) Solving the two above equations for kd and eliminating f. and fs, we get Position of neutral axis 4 2nd A, + bt? Keo As oe (19) The distance of the center of compression from upper surface of beam 3kd — 2t/t ,- MAG). ictroa a’ se aU otea Re (20) Arm of resisting couple 90. chi Moment of resistance Qkd — 1, ,. Meat, 70fe! i) Bree (21) and M= Phd btjdf.. (21a) Fiber stresses M L-a oe Mkd » hci fe = bikd — Bhd (2); sg beter ceg, PE Area of steel M A= 2 fajd ce The form of this equation is the same as for rectangular beams (Formula (15) p. 131). The difference is in the value of 7. As in the case of rectangular beams, the moment of resistance of a T-beam may be governed either by the allowable tensile stresses in the rein- forcement or by the allowable compressive stresses in concrete, whichever gives the smaller value. For a T-beam of fixed dimen- FORMULAS FOR T-BEAMS 135 sions there is one value of p for which the stresses in steel and con- crete reach their allowable stresses simultaneously. This is the maximum ratio of steel that can be fully utilized without over- stressing the concrete. Its value is given in the diagram, p. 894, for different unit stresses and different dimensions of the beam. Minimum Depth of T-Beam.—In a T-beam design, the depth of slab to be used for the flange is designed first. This fixes the value ¢. ‘The width of the flange depends largely upon the thick- ness (see p. 217), so that for fixed ¢, the value 6 is practically fixed. Since the area of compression flange is fixed, for specified working stress in concrete, fc, the available compression is practically fixed. The moment of resistance being equal to total compression multiplied by moment arm jd, it is clear that with total compression practically fixed there corresponds to each moment M, a definite value d, for which the stresses in the flange are equal to the maximum working stresses. This depth is called minimum depth. The formula for minimum depth is Minimum d = a ae 2k — 5p q)ae Calling Ca = ( oy , the formula becomes Minimum depth M Minimum d = Cot Ho (20) Values of Cz depend upon the ratio by and upon, k, 7 and f., which are d functions of the allowable working stresses. The diagram on p. 894 gives values of Ca for various assumptions. Maximum Area of Steel in T-Beam Permitted by Compression.— In a T-beam of definite design, the compression area is fixed as explained above. For a definite depth d and working stresses f., f, siotbak d ay The total tension available equals A,f., in which only f, is fixed. The available moment of resistance of the beam equals either total tension or total compression multiplied by the moment arm, which- and n, this fixes the available total compression, which is f, 136 THEORY OF REINFORCED CONCRETE ever gives the smaller value. If the total available tension, A.fs, is smaller than the total available compression in the flange, then the tension governs the bending moment and it is M=A,f,jd. When the total available tension, A.f., is larger than the available total com- pression in the flange, the compression governs the bending and the area of steel is not fully utilized. It follows that, in a rational design, such an area of steel should be used that the total tension is smaller than, or equal to, the total compression. The maximum area of steel which can be used economically is that for which the total tension equals the total compression, of Maximum A,f, = fo oot Calling Maximum A, = Pmbd in substituting in the above equation, we have 2k — 5 pnbdf, = f, es es or finally Maximum ratio of steel | t 2k— 5 Pn = ope oot . > See and Maximum area of steel Maximum A, = Dubd: 2. ae (27) The ratio pm depends upon the values k, f, and f., all of which : ne : depend upon the working stresses and the ratio a The diagram on p. 894 gives values of pm for various assumptions. Maximum Moment of Resistance of T-beam.—In a T-beam of definite concrete dimensions, the maximum moment of resistance is the one for which the compression stresses in the extreme fiber in concrete reach the maximum working value. 2k — 5 Maximum M = ah af bid. (2% — 5)it Calling, as on p. 135, Ca = Remy. the formula becomes Maximum Mo = dCqbt- 2) (28) BEAMS WITH STEEL IN TOP AND BOTTOM 137 The diagram on p. 894 gives values of Cg for various assum ptions. Formulas Considering Compression below Flange.—For large beams where the stem forms a large part of the compression area, the above formulas do not give results accurate enough for practical purposes. or such cases, formulas given below may be used, which take into account the compressive stresses in the stem as well as in the flange. The following formulas are derived by the same prin- ciples used in derivation of formulas in the previous analysis. Depth to neutral axis 2ndA, + (b — b’ en) nA,+(b—0b’)t\? nA,+(b—d't b’ +( b’ ore b/ (29) _ (kdi? — 3t°)b + [(kd — £)2(t + 4(kd — £))]b" ‘eo t(2kd — t)b + (kd — 2)?’b kd = (30) Arm of resisting couple De ee Rr es Boge Nc ee et (3.1) Moment of resistance Pe Aad. (32) Me sgl ht — bt -+ (kd — £)2B’]jd. (33) Fiber stresses M cAal 2Mkd eee ee Odie = op ‘bt + (kd — 8)2b"Vjd" (35) A simplified. method is given on p. 224. This can be used in design for determining dimensions of the beam for known stresses in steel and concrete. It cannot be used for computing stresses, how- ever. Case II.—Neutral Axis in Flange or at Underside of Flange, kd < t. In this case, which occurs only with slabs very thick in pro- portion to the depth of the beam, use the rectangular beam formula, considering the T-beam as a rectangular beam of the same depth, the breadth of which is the breadth of the flange. The percentage is then based on the total area, bd. REINFORCED CONCRETE BEAMS WITH STEEL IN TOP AND BOTTOM In beams reinforced with steel placed both in the compressive and tensile portions of the beam, the steel in the compressive portion, as in reinforced concrete columns (p. 159), may be considered as resisting compression stresses equal to the stresses in the concrete 138 THEORY OF REINFORCED CONCRETE which it replaces, multiplied by the ratio of moduli of elasticity, 7. Same assumptions are made as for simple rectangular beams (p. 126). The tension in concrete may be neglected, as in beams without compressive steel, and all the tension may be considered as resisted _ by the bottom steel. Referring to Fig. 46, p. 188, the total com- pression consists of the compressive stresses in concrete represented by a triangle and the compressive stress in steel. The compressive unit stress in steel equals the unit stress in concrete at the same level, multiplied by the ratio of their moduli of elasticity. ompression in Compression Reinforcement, “f-> -C ment 7 ‘Steel p'bdfg ; M. Arm for Comp Compression in | Concreté rah kbd | --d (1-4}-- mp.\Steel aS (-a)- JES M.Arm for Co ; p, bd fs Py bd -J->| Total Tension Tensile y Reinforcement” Fic. 46.—Resisting Forces with Steel in Top and Bottom of Beam. (See p. 1388.) The sum of all the horizontal stresses acting on a cross section must equal zero;! therefore, the total tension in steel must be equal to the compression in concrete plus the compression in the top steel. The resisting moment, that is, the moment of the internal stresses, may be obtained either by multiplying the total tension or com- pression by the distance between center of tension and center of compression, or by taking moments about the center of tension steel, ° the center of compression in concrete, or the center of compression in steel. The moments of rec'stance obtained by either of the four methods must be equal. To find the stresses for a certain loading, the moment of resistance taken in any one of these ways is equated to the known bending moment. Formulas.—Deformations, as usual, are assumed to vary directly as distance from neutral axis, hence from Fig. 46, using notation on p. 138, dda Le ee Ey OUR eh ae Whence k = 1 7 Ste dk k 1+ pis E, nf c 1 This is a law of simple statics. Otherwise there would be a movement in the beam, (36) BEAMS WITH STEEL IN TOP AND BOTTOM 139 By comparing the above equation for k with that given for simple fs nfo’ neutral axis is the same irrespective of whether the beam is provided _ with compressive steel or not. By similarity of triangles in Fig. 46, p. 138, the following relations between the unit stresses may be obtained: beams, p. 129, it is evident that for any ratio of —, the position of the f's (37) and fi =inf, {, = op (39) and ie Hide Sere hy The total tension in steel equals bdp,f, and the total compression in steel and concrete is bd + dap'f’, = df. + vf). Since the sum of all the stresses must equal zero, the total compression acting on the cross section of the beam equals the total tension, or ba( nip J .)= bdpifs. ma / fo Poel ffeil tk? ka Pi =7(5 rays} = 7(# eeen: oo De makige is ‘) kK ka aOR —R\at Oe h: Solving equation (41) for k, | k = V2n(pi + pa) + 21 + p')? — n(pit yp’). . (42) Taking moments about the center of compressive stress in the steel, we have A ba?| fart —a)— BG - a) | or by eliminating f. | pill — a)2n(1 — k) — ae — : M-—-f,bd? ius fr CAs eee .. (48) Whence Hence (41) From which es M 6n(1 — k) 4 (bP 6npi(1 — k)(1 — a) — k2(k — 83a) (44) 140 THEORY OF REINFORCED CONCRETE By substituting this value of f, in equations for fc and f’, respec- tively, we get _M oh Se = 5@2 @npid = EC — a) — Ph — 3a)’ eo and ell. 6n(k — a) eat 6npidl — k)(1 — a) — KAR — 3a)” ee It may be noted that the denominator in the three above equa- tions is the same. A simplified method for designing beams with steel in top and bottom 1s given in the chapter on Design. Tables of constants are given on pp. 904 to 910. WEDGE SHAPED RECTANGULAR BEAMS Under this heading come beams in which one or both faces are not at right angles to the section of bending moments. They are found in retaining walls, footings, dams and also at the support of beams provided with haunches. When the angle of inclination of the faces with the horizontal does not exceed 10 degrees, ordinary formulas may be used. For larger angles the following formulas suggested by Professor William Cain ** will give closer results. These formulas are approximate. The number of experiments made on wedge shaped beams is very small so that it is not possible to check the accuracy of the formulas. Figure 47 represents part of a wedge shaped beam, both faces of which are sloping. The stresses produced by a bending moment acting on section AB will not be perpendicular to the section. In the steel they act in the direction of the reinforcement, while in concrete they are parallel to the compression surface. The direc- tion of stresses is marked in the figure. To get a moment of resist- ance acting on section AB ‘t is necessary to take components of compression and tension at right angles to the section. Following formulas are based on condition represented in Fig. 47, p. 141. Netaticn.—See pp. 128, 246 and 262. Both Faces Inclined. cos B [oes COS? a lee np + 2m mes + inp)? | vo eae 14 See Professor William Cain “Barth Pressure, Retaining Walls and Bins.” STEEL IN BOTTOM OF BEAMS 141 hs NE i ecg (48) ee M ° A, cos Bjd * Fae ° kjbd? cos? a _ ple costa (tat if, bcos Bi” (49) (50) (51) _ V _ M(tana + tan 8) (52) bjd bjd? mane nA * M (tana + tan 8) u= Sojd Sop Aree User ee {(O8) Component 1 “a bkd cos*x bkd cos « 3 A, fy cos 3 Component Part Elevation Section Fig. 47.—Resisting Forces in Wedge Shaped Beam. (See p. 140.) Compression Face Inclined. Steel at Right Angles to Section.— p= 0°. = <3 (— mp + V2np cos? a + (np)?) te tos) M ee ee ee ey ean) 55) 2M CMa ye ce ey ae tO) A cor? oe. EEE are tee are seen Mier dy Lelia fe V Mtanea Cpa ia oe V M tana (59) aaa Lojd Sojd? ° 142 THEORY OF REINFORCED CONCRETE T-BEAMS WITH FLANGES ON ONE SIDE ONLY T-beams with flange on one side only are found where the slab extends only on one side of the beam as at the wall and at openings in the floor. It is common practice to treat these beams in the same manner as symmetrical T-beams, assuming that the neutral axis will be parallel to the compression flange and that the distribution of stresses will be the same as accepted for symmetrical T-beams. Tests by Professor Bach on Unsymmetrical Beams prove that the neutral axis, instead of being parallel to the flange, slopes upward toward the side provided with the flange and may even intersect the flange. The maximum stresses in concrete occur at the rec- tangular corners. The ‘stresses in steel are not affected much by the unsymmetrical arrangement of the flange. While formulas have been proposed for this condition,” there are not sufficient data to warrant their adaption. Until additional information is obtained the stresses should be figured in the same manner as for symmetrical beams, except that the allowable working stresses in compression should be reduced by about 30 per cent. COMPARISON OF STRAIGHT-LINE FORMULAS WITH ACTUAL CONDITIONS Tests prove that the assumption (2) on p. 126, that the modulus of elasticity of concrete is constant for all stresses, is not correct. Actually the modulus of elasticity is a maximum for small stresses and decreases with the increase of the intensity of the stresses. (See Volume II.) If a plane section before bending remains plane after bending, the deformations vary according to a straight line, as shown in Fig. 48a. The unit stresses are equal the deformation multiplied by the modulus of elasticity. For constant modulus the variation of stresses will be the same as the variation of deformations. This gives the “ straight-line ” formula, Fig. 480. Actually the variation in stresses is different from the straight- line variation, because, to get the stresses, the deformations must be multiplied not by a constant modulus of elasticity but a varying modulus of elasticity, which is a maximum for small stresses near 10 Gog Karl Hager ‘‘ Vorlesungen wber Theorie des Eisenbetons.”’ SHEARING STRESSES IN A BEAM OR SLAB 143 the neutral axis and decreases for larger stresses. Figure 48¢ shows parabolic distribution of stress, as proposed by Prof. Arthur N. Talbot. Special attention is called to the fact that all formulas are based upon the law that plane section before bending remains plane after bending. In the parabolic formula it is not the plane that assumes a curved shape, but the variation of stresses acting upon it. Since in the straight-line formula an average value for the modulus of elasticity is used, the actual stresses are larger for the small stresses (near the neutral axis) than obtained from straight-line formulas. For stresses near the extreme fiber the actual stresses (based on Deformations, Stress s Deformation x Constant Ee Stress-= Deformation x Variable E', pe i Variation in Modulus Eg Tension Tension a ay a - bis aplane before bending Straight - Line Parabolic a, b,is same plane after bending Variation of Stress Variation of Stress (a) (0) (c) Fic. 48.—Different Assumptions as to Variation of Stresses. (See p. 142.) varying modulus of elasticity) are smaller than in the straight-ling formula because the actual modulus of elasticity is smaller than the assumed. ‘To compensate for this, the allowable working ‘stresses to be used with straight-line formulas were made larger than would be permissible for exact formulas. In formulas with parabolic dis- tribution of stress the working stresses for straight-line formulas would give unsafe results. SHEARING STRESSES IN A BEAM OR SLAB The bending of a beam produces a tendency of the particles to slide upon each other. This tendency is called shear. The resistance to sliding offered by the material, called shearing strength, is a property of the mate- nal. The effect produced by bending is called shearing stress and its magnitude depends upon the external loads and upon the dimen- sions of the beam. The shearing stresses are produced by external 144 THEORY OF REINFORCED CONCRETE shear. This depends only upon the magnitude and character of the loading and is independent of the dimensions of the beam. These terms are used in the discussion below. In a beam, shearing stresses are developed in two principal direc- tions. It is, therefore, necessary to study (1) vertical shearing stresses, (2) horizontal shearing stresses. Vertical and Horizontal Shearing Stresses.—Vertical and hori- yontal shearing stresses are equal. They may be computed from formulas on page 149. The resistance of concrete to direct shear is large. (See Vol. II.) Itis safe to use a working shearing stress of at least 200 Ib. per sq. in. Therefore, a concrete girder, beam, or slab will always have sufficient area of section to withstand this direct shearing stress. However, since the direct shearing stress is a measure of the diagonal tension (see p. 148), which is excessive when the direct shearing stress is comparatively low, it must always be computed in a beam or girder for use in the designing of web rein- forcement. Magnitude of External Shear.—The external shear is a maximum at the support, where it is equal to the reaction. For simply sup- ported beams, the external shear may be determined by statics. In continuous beams, the actual external shears differ from the ‘statie shears, as is evident from the formulas given in the section on continuous beams. While, with uniform or symmetrical loading, the reaction, and therefore the maximum shear, in simply supported beams is one-half the total load upon the beam, it will be noticed from the diagrams that in the end spans of continuous beams with free ends the external shear at the first support away from the end may be 25 per cent greater than the static shear, and should be spe- cially provided for in cases where the full live load is likely to be constantly maintained. For complete treatment of external shears ‘n continuous beams see Volume Il of this treatise. Longitudinal Shear in T-beam.—The projections of the flange of : a T-beam on each side of the stem are subject to compressive stresses. These are zero at the support or at the point of inflection and increase with the increase in bending moment. T he increase in compression stresses in the projections of the flange must be transferred from sec- tion to section by the resistance to shear at the junction of the flange projections and the stem. These shear sections are marked AB and A,B; in Fig. 49, p. 145. When the shearing stresses are larger than SHEARING STRESSES IN A BEAM OR SLAB 145 the resistance to shear, the flanges become separated from the stem, the compression area becomes reduced, and the beam fails by crushing of the stem. Such failures are described in the chapter on Tests, p. 36. An analogy may be found between the shearing stresses along the flanges of a T-beam and the shear in the rivets of a built-up steel beam, which connect the flange angles to the steel web plate. The magnitude of the shearing stresses may be found in the fol- lowing manner: Consider two vertical cross sections of a beam, one inch apart. Let M, be the moment at the section nearer the sup- port, M2 the moment at the other section, and V the external shear, and assume for the sake of simplicity that there is no loading between eens a — Fig. 49.—Longitudinal Shear in T-beam. (See p. 144.) the two sections. Then, from simple laws of statics, the moment M2=M,+VX1. The total amount of compression stresses produced at each of the two sections is Ms and ike = Ms + Me qd ad Jdaks 44 If the part of the beam between the two vertical sections is detached from the beam, the compression stresses act in opposite direction, as shown in the figure. Since the amount of compression on the two sections is not equal, only a part of the compression stresses will balance. The unbalanced difference between the compression stresses at the two vertical sections produces shear in the beam. This difference is Mz — M, = a This increment of compression stresses acts on the whole flange. To get the amount of compression on the projections only, it is necessary to multiply the total incre- ment by the ratio of the width of the projections b — b’ to the width of flange, b. Thus the amount of compression stresses producing 146 THEORY OF REINFORCED CONCRETE longitudinal shear at the juncture of the flange projections and the stem equals ree ey) Ate Ue The intensity of the compression stresses on the flange is not uniform, but is a maximum at the extreme fiber and a minimum at the bottom of the flange. The unit increment producing shear is proportional to the compression unit stresses; therefore, the longi- tudinal shear will be a maximum at the top and a minimum at the bottom of the flange. If the shearing stress at the top is called v,, the minimum shearing stress at the bottom is u a : The total shear on two sections AB and Ai Bi, for a depth of flange equal t 2k — = to t 1s a(o, + bile = *) 2 = ppt kd k stresses producing shear, found above, By equating this to the t ab dt Vib ee poe Gd tbe Val from which maximum longitudinal shearing stress in flange Dagieiiies” (aaa ijd ‘ b t « ° ° ° ° e (60) 2h — 5 Vh From this formula it is evident that the shearing stresses increase with the ratio of the projections to the total width of flange. Ordinarily, the shearing stresses are so small that the concrete alone can resist them. Excessive longitudinal shear may be devel- oped only in short beams with heavy loads. As evident from the chapter on Tests, the strength of a T-beam is increased by fillets at the junction of the slab and the stem. Also, cross bars placed across the beam near the top of the slab increase the strength of the T-beam. Where the tensile reinforce- ment in a slab is placed across the beam, it strengthens the connec- tion between the stem and the flange. Where the slab reinforce- ment is parallel to the beam, special short bars should be placed across the beam. ‘These serve two purposes: first, they strengthen the connection between the stem and the flange; second, they resist negative bending moment stresses which are developed by the load & DIAGONAL TENSION 147 in the slab. Without the tensile bars, tensile cracks often form at the junction of the flange and the stem and materially reduce the effect- iveness of the flanges. DIAGONAL TENSION In any beam, besides direct horizontal tension and compression and direct horizontal and vertical shearing stresses, there exist also stresses acting in diagonal directions. The maximum diagonal stress composed of the tension and the shearing stresses is called diagonal tension. In steel and other homogeneous beams, diagonal stresses need no attention. In reinforced concrete, however, it has been shown in beams tested to destruction that, beside tensile cracks at the points of maximum moment, diagonal cracks, caused by diagonal tension, develop near the supports. (See tests on pp. 38 to 46.) These cracks have often been the cause of failure, frequently without warn- ing, especially in beams reinforced with straight bars only or pro- vided with insufficient web reinforcement. The need of effective web reinforcement is discussed in the paragraphs which follow. Diagonal Tension in Homogeneous Beams.—The magnitude and inclination of the diagonal tension in homogeneous beams may be found from the following formula: Let fa = diagonal tensile unit stress; f: = horizontal tensile unit stress; v = horizontal or vertical shearing unit stress. Then 2 Meat iy tere ee aT The direction of this diagonal tension makes an angle with the ee horizontal equal to one-half the angle whose cotangent is sh As is evident from the formula, the value of diagonal tension depends upon the values of f, and v. Since the values in f, and v vary in different parts of the cross section of the beam, the value of the diagonal tension and its angle: of inclination also vary. At the bottom of the section, where v equals 0, fa equals f: and acts hori- zontally. At the neutral axis, the direct tension, fi, equals 0, * For derivation, see Merriman’s “Mechanics of Materials,” 1905 edition, p. 265. 148 THEORY OF REINFORCED CONCRETE which reduces the formula to fa = v and the angle of inclination to 45°. Measure of Diagonal Tension for Reinforced Concrete Beams.— In homogeneous beams, the diagonal stresses can be determined easily by means of Formula 63, p. 149. In reinforced concrete, how- ever, the diagonal stresses cannot be computed with exactness because, as may be seen from the formula, they depend upon the horizontal tensile stresses in concrete, fi. The action of concrete in tension cannot be computed. It varies for different stages of loading. Tor heavier loadings, concrete cracks, thus decreasing the tensile stresses carried by it. The tensile strength of concrete, which may be dis- regarded in figuring the moment of resistance of the beam, affects the magnitude of the diagonal tension to a great extent, especially near the ends of simply supported beams, where the stresses due to the bending moment are low and the tensile stresses in concrete may not exceed its breaking strength in tension. While the exact determination of diagonal tension is impossible, tests show that the shearing ‘unit stress, figured as given on p. 149, may be accepted as a convenient measure of diagonal tension; that is, the diagonal tension may be assumed as proportional to the direct shearing stress. Therefore, by adopting proper working stresses based on tests pro- ducing diagonal tension failures, formulas for shearing stresses may be used for diagonal tension. This measure has been universally accepted and, in subsequent discussion, diagonal tension is expressed in terms of shearing stresses. | Formulas for Shearing Stresses and Diagonal Tension.—A convenient and safe method of determining the diagonal tension is by accepting for its measure the unit shearing stress as discussed above. | Let V = total external shear at section considered. (Reaction minus the loads between the support and the section.) » = horizontal (or vertical) shearing unit stress at section considered ; b = breadth of beam; b’ = breadth of web of T-beam; jd = moment arm or distance between center of compression and center of tension (approximately, in a T-beam, distance between center of slab and steel). 7, = total shearing stress or diagonal tension in a given length of beam, s. DIAGONAL TENSION 149 The following general principles and formulas are discussed in the paragraphs which follow: (1) Vertical shearing unit stress is equal to the horizontal shear- ing unit stress, and acts at right angles to the plane of horizontal shearing stress. The distribution of vertical shearing stress over a vertical section is shown in Fig. 50, p. 150. (2) Horizontal (or vertical) shearing stress is zero at the top of the section and changes according to a parabola till it reaches its maximum at the neutral axis. (See Fig. 50.) (3) If tension in concrete is neglected, the horizontal (or vertical) shearing stress is constant below the neutral axis. (4) The total amount, then, of horizontal (or vertical) shearing stresses developed below the neutral axis at any horizontal plane in a distance, s, and width, b is Z eae aii « Gite Be peak med) vane ee” (62) (5) Shearing unit stress, the measure of diagonal tension, is total horizontal shearing stress, Z, divided by the horizontal area, b X s, 7 that is v = Vs + bs, ad Hence (63) For T-beams, v = at ire tt BA) If the external shear, V, changes in the distance, s, the same for- mulas may be used except that V in the formula is the average shear in that distance. | (6) Diagonal tension may be expressed in terms of the shearing stress and the above formulas may be accepted as its measure. (7) If the width of the section below the neutral axis is not con- stant, the shearing unit stress will vary with the width, 6. The smallest b must be taken in figuring maximum shearing unit stress and the maximum diagonal tension. (8) In continuous T-beams near the support, the maximum shearing unit stress will be in the stem directly under the flange. The shearing stress and diagonal tension in the plane of tension steel (at top of beam) is small, because the width, b, being the total width of the flange, is large. 150 THEORY OF REINFORCED CONCRETE The action of horizontal shearing stresses is illustrated in Fig. 50, in which is shown a section of a beam between two vertical sections spaced a distance s apart. If the bending moment at the left is M,, the external shear V, and the distance between the sections under consideration s, then, from the principles of mechanics, the bending moment at the right is M, = M, + Vs. Since the bending moment at the right is larger than the bending moment at the left, the unit compressive and tensile stresses at the right section are larger than at the left. Consider an arbitrary longitudinal plane, ef, above the neutral axis. The compressive stresses above this plane are represented by the shaded portions of the triangles. They act in opposite directions. At the left they are ylt A Cyl hi— hi eee / Neutral Axis___ “a \ Mt, = seed oe Section Distribution of shearing stresses Elevation Fic. 50.—Horizontal and Vertical Shearing Stresses in Beam. (See p. 150.) equal to Cy, and at the right to Cy = Cn+AC,. The difference between C,, and Cyr, which act in opposite directions, is AC je. jethis tends to move the upper portion of the beam along the plane eff 1€1, but is kept in equilibrium by the horizontal shearing resistance in the beam on that plane. The shearing unit stress produced by the force AC; is equal to the force AC, divided by the area, bs, of the plane, effier. At the top of the beam, the value of AC,, and also the total shear- ing stress is zero; it increases steadily according to a parabola till it reaches its maximum at the neutral axis. There its value equals the difference between the total compression on the right and the total compression on the left. From the ordinary beam formulas, p. 130, we know that the total compression may be found by dividing the DIAGONAL TENSION 151 bending moment by the moment arm: thus, at the left, the total. : : bald M. ast M, eee Vs compression is C, = id and at the right, C, = hiv Maia di The difference between C, and C, is 1 Therefore, the total amount jd of horizontal shearing stress at the neutral axis for the length, s +r and width, b, is Z = a This has to be resisted by the horizontal plane of the beam, bs, so that the shearing unit stress, v, equals Z divided by bs. It is Shearing Unit Stress AG Diz ai (65) So If there is no tension in concrete, the difference between the stresses acting above any plane located below the neutral axis is the same as the difference at the neutral axis. Consequently, the sum of horizontal shearing stresses is uniform at all planes below the neutral axis. As the shearing unit stress depends upon the width, b, it is constant for rectangular sections, but various for sections with variable widths, b. At the plane of reinforcement, the stresses in steel at the left are Sie M M. _M, V Ss t= ae and at the viel ates Shee ns ad auld and the difference, T, — T, = = This shows that the total horizontal shear, or the tendency to move the upper portion of the beam, is the same at the plane of the bars as at the neutral axis. If there is tension in concrete, the total horizontal shearing stress, Z, on any plane below the neutral axis will be decreased by the difference in tension at the two vertical sections above that plane. r Bond Stresses.—The increase in the stress in steel, equal to ea between the two sections considered, must be transferred from the beam to the steel. Therefore, bond must exist between steel and concrete, or else the beam will slide on the steel instead of increasing its stress. Tests of bond, or resistance to slipping of bars, are treated on p. 56. Discussion of bond stresses and their importance is given on p. 260. 152 THEORY OF REINFORCED CONCRETE Diagonal Tension Acting on an Element of a Beam.—Figure 51 represents the stresses to which any element-of a beam is subjected. In Fig. 51 is shown a rectangular element of the beam, the sides of which are dz and dy. This element is kept in equilibrium by six forces: two forces, fidy, acting in opposite directions and being either direct tension or compression; and four shearing stresses, vdx and vdy, respectively, caused by the increment of the bending moment, as explained in the preceding paragraphs. The two horizontal shearing stresses, vdv, form a couple, vdx X dy, which is resisted by a vertical couple, ody X dx. The moments of the two couples are Fic. 51.—Stresses Acting on an Element of Beam. (See p. 152.) equal; therefore, the horizontal shearing unit stress must be equal to the vertical shearing unit stress. Illustration of Action of Web Reinforcement.— Figure 52, p. 153, illustrates the action of vertical stirrups in a simply sup- ported beam. Stirrups do not act until minute cracks open. After a, crack forms, as in the figure, the reaction and the shear, V, tend to widen the crack and cause failure of the beam. This tendency is resisted by the stirrups, acting in tension. Figure 52 also rep- resents what would happen if there were no web reinforcement. Figure 53 represents the action of stirrups in-a continuous beam. It may be noticed that the stirrup gets its stress at the top instead of at the bottom. , Distribution of Diagonal Tension to Concrete and. Stirrups.— Tests prove that in beams with web reinforcement, the diagonal DIAGONAL TENSION 153 tension found by Formula (65), p. 151, is resisted by both concrete and steel. The relative proportion of stress taken by the concrete _ Ends of Stirrups must be late / /“ Anchored in Compression Zone — External Shear ~~ Direction of Force Acting on Stirrup ‘Ends of Stirrups Looped around Tension Reinforcement if External Shear Reaction Reaction Beam with Vertical Stirrups Beam without Web Reinforcement Pia. 52.—Action of Vertical Stirrups in Simply Supported Beams. (See p. 152.) Ends of Stirrups Looped around K Tension Reinforcement Column Tension Reinforcement. (epee eae EE CW, ) le | paid EE LA ~Extédynal Shear i ee of Stirrups Anchored in Compression Zone Diaganal Crack Compression Reinforcement aa Fig. 53.—Action of Stirrups in Continuous Beams. (See p. 152.) and steel cannot be determined by computation. Assumptions variously made are: (1) Web reinforcement takes all the diagonal tension with no reliance on concrete. The web reinforcement therefore resists, in the length s, a total amount of shear equal to Vs jd’ (2) Web reinforcement takes two-thirds of the diagonal tension and the concrete the remainder. Web rein- forcement resists in the length, s, the force sor Where the shearing unit stress does not exceed the allowable unit for plain concrete, v’, all stress is taken by the concrete. 154 THEORY OF REINFORCED CONCRETE (3) Concrete resists a certain definite unit stress per square | inch, v’, the whole length of the beam, and the stirrups resist the remainder. Then, in a length, s, the concrete resists v’bs, and the web reinforcement resists, ae zy = 1 — vos = VS — sfos = Pe The first assumption corresponds to that made in ordinary beam design, where the tensile strength of the concrete is disregarded. Tests have shown, however, that the actual stresses in the stirrups are less than would be obtained with this assumption. (See p. 40.) The second and third assumptions are most commonly used in practice. Hither of them will give satisfactory results, especially ¢ the rule about maximum allowable spacing is followed. The Joint Committee on Concrete and Reinforced Concrete, 1916, recom- mended the second assumption, while in the later, 1924, reeommenda- tions the third assumption is accepted. The authors recommend the use of the third assumption. Area and Spacing of Vertical Stirrups.—The area of steel and the spacing of stirrups may be found by making the force to be resisted, as given above, equal to the working strength of the stirrups in tension. . 3 Let V = total vertical shear in pounds at section « feet from left support; v = total shearing unit stress at section in pounds per square inch; vy’ = allowable shearing unit stress (or diagonal tension) on concrete alone; A, = cross-sectional area of all legs of a vertical stirrup in square inches. (In a U-stirrup this is the sum of the area of the two legs) ; fs = allowable unit stress in stirrups in pounds per square inch. jd = moment arm or distance in inches from center of com- pression to center of horizontal reinforcement. (In a T-beam, this may be taken as distance between center of slab and steel; in a rectangular beam, as 0.87 of the total depth to steel) ; b = breadth of beam in inches; b’ = breadth of web in T-beam in inches; I DIAGONAL TENSION 155 $ = spacing of stirrups in inches at a place x Hey from left support; x = distance in feet from left support to point at which required spacing is desired; z; = distance in feet from left support to point beyond which stirrups are necessary ; ! = span of beam in feet; w = uniform load in pounds per foot. Since A, is the area of a stirrup resisting diagonal tension in a dis- tance, s, and f, is the tensile strength of steel, the strength of the stir- rup in pull is A,f,. The area of stirrups and the spacing for different assumptions of distribution of diagonal tension between stirrups and concrete may be found as follows: (1) Area and spacing if stirrups take all the diagonal tension. The diagonal tension to be resisted is a Hence A,f, = “aa nd ere _ fajd WAS ane ee Ob) and $= “yy Ae oa) (2) Area and spacing if stirrups take two-thirds of the diagonal tension. 2Vs one Diagonal tension to be resisted is 3 id’ Hence A;f; = 3 Gd? and we get by solving for As and s, 2 _ 3f.jd As S 3 iad pags (68) HOTS he) 5°) di Rpt ested REED (3) Area and spacing if concrete takes a definite amount of shear, v’, and the stirrups, the remainder. The stress resisted by concrete in the distance, s, equals v’bs. As the total stress is Z = “a! the stirrups must carry the difference, ase ec, V — v’bjd ARR pbSy OF. tegriee Equating this to the resistance of the stirrup, A,f,, and solving for A, and s, we get (V—v jd) . . fsjd A, = 3a (70) SCs (V— vijay" : Cub) In T-beams, use width of web, b’, in place of b. 156 THEORY OF REINFORCED CONCRETE Area and Spacing of Bent Bars.—The spacing of bent bars, s, (or the distance in which they are effective) is usually measured along the neutral axis of the beam. The diagonal tension stresses to be resisted by the bent bars may be taken, as equal to the stresses to be resisted by the stirrups, multiplied by the sine of the angle of inclination of the bent bars with the horizontal. Since, however, tests indicate that there is practically no difference in effectiveness between bars bent at angles between 25 and 50°, it is near enough to accept a constant value of 0.7 for the sine for these inclinations. Using same derivation as for the stirrups, the formulas for required area of bent bars and spacing become as follows: (1) Area and spacing if bent bars take all the diagonal tension. ae Wee A; rae: wg SEED) GS et ye Ma eos (2) Area and spacing if bent bars take two-thirds of the diagonal tension. EN an _ 9 13434 tinal ee ot (hl O74) nds hae ieee y oe - (75) (3) Area and spacing if concrete takes a definite amount of shear, »’, and the bent bars, the remainder. (V-—v' bjd) . fsjd ; Uniformly Distributed Loading.—The distance from the support to the point where no stirrups are required, for uniform loading, 1s é m= 5(1- 5). oy Ral eeu Dae Maximum Spacing of Stirrups.—Tests tend to prove not only that a sufficient amount of web reinforcement should be provided but also that the maximum spacing of stirrups or bent bars should be limited. (76) and s = 1.43 | ee ee (77) ra (V — v'bjd) 3The diagram of shearing unit stresses is a triangle from which the distance l a, may be obtained by the known rule 5 = (5 ae a) my +’. This equation solved for a gives formula (76), above. DIAGONAL TENSION AIBY. If the spacing is too large, cracks may develop between the stirrups. A recommendation for maximum spacing is given on p. 250. Usefulness of Web Reinforcement.—Numerous tests have demonstrated that a beam properly reinforced with stirrups or bent bars sustains three or four times as much load as the same beam without web reinforcement. The same tests, however, show that the web reinforcement retards the appearance of first diagonal cracks very little, and that the web reinforcement does not get any stress until the first crack appears. It has been noticed * also that under working loads (that is, before the diagonal tension exceeds the tensile strength of the concrete) the beam acts like a homogeneous beam, and as would be expected, the stress in the stirrups is some- times compressive instead of tensile. This is, nevertheless, no argument against the use of web rein- forcement, because in beams without stirrups, final failure follows closely the appearance of the first crack, while with beams having web reinforcement, stirrups and bent bars represent a factor of safety which allows stressing of concrete in diagonal tension nearly to its ultimate strength without any danger to the stability of the structure. Under working loads the stirrups may not act; but in case of overstressing, due to faulty construction or to occasional excessive loading, although the concrete cracks, the stirrups not only prevent the failure of the beam, but enable it to develop its full strength. The minute cracks that may develop are not dan- gerous and in many cases are hardly visible. After the excess load is removed, the cracks close up. Tests showing conclusively the value of stirrups are shown on p. 88. Web Reinforcement for Continuous Beams.—The formulas given above are based upon results of tests of simply supported beams. ‘Their use for continuous beams allows a margin of safety. In continuous beams, several conditions tend to prevent, or at least to retard, the formation of diagonal cracks. The compres- sive force, due to the reaction, tends to close the developed cracks. There exists also, almost invariably, some arch action, which de- creases the direct. and diagonal tension. In continuous T-beams, the horizontal shearing unit stress is zero at the bottom and increases till it reaches a maximum at the neutral axis. From there it is constant till it reaches the bottom of the flange. As the width of the flange is much greater than the width of the stem, the shear- 4 University of Illinois, Bulletin No. 64, January 13, 1913. 158 THEORY OF REINFORCED CONCRETE ing unit stress in the flange is much smaller than in the stem. Diag- onal cracks, therefore, tend to develop in the portion between the neutral axis and the bottom of the flange, and larger unit stress is required to open them than in simply supported beams. This is offset by the fact that at the support of continuous beams, both the tensile stresses and the diagonal tension stresses are a maximum. As there have been comparatively few tests on continuous beams, the for- mulas for web reinforcement given above should be used. Web Reinforcement for Cantilevers.—The conditions affecting web reinforcement are the same for cantilevers as at the supports of continuous beams. In cantilevers supporting vertical loads, vertical stirrups must, therefore, be attached to the tension steel (at the top) and the free ends hooked in the compressive portion of the beam (at the bottom). In cantilevers carrying loads acting in other directions, the stirrups must be placed. parallel to the direc- tion of the force and attached in the manner suggested above. COLUMN FORMULAS For reinforced concrete columns centrally loaded, the following formulas may be developed: Let jf = average compressive unit stress upon the reinforced column, equal to the total load divided by the effec- tive area; f. = compressive unit stress upon the concrete of the column. f’.= ultimate strength of concrete cylinders. f; = tensile unit stress in spiral, New York Code. f’,= compressive unit stress upon the vertical steel in the I column; n= ioe ratio of modulus of elasticity of steel to modulus of elasticity of concrete; P = load to be sustained by the column; A = area of total effective cross section of column (see pp. 272 and 406) ; , = area of concrete in effective cross section, equal to Age Ay, A A, = area of steel in cross section; p = — = ratio of area of steel to total effective area of eolumn. | ’ = ratio of vol f spiral t ] f losed t p ratio of volume of spiral to volume of enclosed concrete. COLUMN FORMULAS 159 Since, as is evident from tests, a reinforced concrete column under load acts as a unit, the deformation or shortening of steel in the column is the same as the deformation or shortening of the concrete. stress per square inch modulus of elasticity From mechanics, = unit deformation, hence M4 a = unit deformation of steel and A = unit deformation of concrete. The deformation of steel in a reinforced column is the same as the ° e E deformation of concrete, and since i = n, we have: c fe_ fe and’ if/s = nf; E, EK. s ce The stress in steel is therefore equal to the stress in concrete multiplied by the ratio of the moduli of elasticity, n. If a column sustains a load P, stresses in steel and in concrete must be equal to the load. Hence: P = f.A, + PuAPorP: fA) +onfAy Since A, = A — A,, we have P = fA(A — As) + nA,l. Finally, 1: ied al VA to P=f{A+(n—1A] (79) and f= The area of steel, A,, may be expressed in terms of A, by substituting A, = pA, which changes the above formulas to fr Asso) mock, P=f.A{1+(n—1)p] (81) and iis Knowing the stress, f,, and the steel ratio, p, we may find the required area from P — fdl+(n—Wpl * Knowing the stress, f., and the total area, we may find the required area of steel, A,, and the percentage from A (83) pe DE RS vah (85) The average unit stress, which is the total force, P, divided by the effective area, A, ‘2 Pe ai, and Nie? 160 THEORY OF REINFORCED CONCRETE The relation between f and f, may be found by substituting for P in the above equation its value from Formula (81), giving f= fii + (n— Vol. OS eee Values of f for different percentages of steel are given on p. 916. Columns with Spiral Reinforcement.—The ultimate strength of a column with spiral reinforcement depends upon (1) the amount of vertical steel, and (2) the amount of spirals. Therefore, in formulas for the breaking strength of a spiral column, the amount of spirals must be considered. In design, however, the elastic limit and not the breaking strength of the column is the determining value, as explained on p. 82. } There are differences of opinion as to the proper formulas for spiral columns. Practical design is treated on p. A419. The following formulas are in use: (1) Columns with spiral limited to 1 per cent of the volume of concrete within the spiral, and variable amount of vertical steel. Same formulas are used as for columns with vertical steel only. To allow for the effect of the spiral, the unit stresses in concrete and vertical steel are increased. | This method is specified by the Codes of Boston, Cleveland, and, with some modification, Philadelphia. It was recommended by the Joint Committee in 1916. Although it does not fully represent the actual conditions, the authors consider this method as the most rea- sonable thus far developed. ; (2) Columns with spiral varying from one-half to 2 per cent of the volume of concrete (p’ = 0.005 to 0.02) and variable amount of vertical steel. Formula for strength of column (Building Code of New York). ig aa Arie me Te ele se 27 Ae In this formula, the stress in concrete and vertical steel is assumed to be the same as for columns with vertical steel only. The spiral is assumed to increase the strength of the column by its strength in tension. (Any section cuts the spiral at two places, which explains the value 2 in the term 2p'A fs.) (3) Columns with spiral from 3 to 14 per cent of the core (p’ = 0.005 to 0.015) and variable percentage of vertical steel, but not to exceed the percentage of spirals. COLUMN FORMULAS 161 Formula for strength of column (Building Code of Chicago). Seda Wt 1p + 28(n — yp]. (87) In this formula, the effect of spiral is considered as two and one- half times as large as the effect of same amount of vertical reinforce- ment. In addition, the allowable stress, f., is larger than for columns with vertical bars only. (4) Joint Committee 1924 formula. Percentage of spiral shall not be less than one-fourth the percentage of vertical steel. Vertical] steel shall be from 1 to 6 per cent of the core. The formula for strength of column is of the same form as that for columns with vertical steel only; the difference is in the allowable unit stress in concrete, which is a function of the ultimate strength and of the steel ratio, :. Pew a hi let (i alpen (88) ~ Where the allowable stress is fee 200 + (O10 anyfie Oy (89) The value of f’. is the ultimate compressive strength of concrete at age of twenty-eight days, based on 6 by 12 in. or 8 by 16 in. eyl- inders. In this formula, an attempt is made to take into account the stresses in steel due to shrinkage and yield of concrete, in accordance with the ideas advanced by Mr. Franklin R. McMillan in his paper, “Study of Column Test Data.’ 5 The authors consider that the refinements and the change from present practice required by these new formulas are not yet war- ranted by our present knowledge of the subject. The formulas are not well adapted for use without tables, because a change in per- centage of steel not only affects the amount of total load carried by steel, but also the unit stress in concrete. For instance, to find the amount of steel for an accepted size of column (a very common problem) requires solution of two simultaneous equations, one of which is a second-degree equation. Moreover, the cost of columns required by these formulas is larger than by the formula of the previous Joint Committee recommendation, while there is not yet sufficient proof that the previous formula did not give safe results to warrant the increase in cost of construction. ° Proceedings, American Concrete Institute. Vol. XVII, 1921, p. 150. 162 THEORY OF REINFORCED CON CRETE MEMBERS SUBJECTED TO DIRECT TENSION In concrete construction, there very often occur members sub- jected to direct tension, as for instance, stair hangers, floor hangers in arch bridges, tie-rods in arches, walls in circular tanks, conical hoppers in bins. Such members are subjected to direct tension only. Any bending in the direction of the tensile stress is only incidental. Members like the bottom chord in concrete trusses, particularly when the truss is without diagonals as in Vierendeel trusses, are subjected to tension and a bending moment. The same is true of the sides of square or rectangular reservoirs. The stress conditions are discussed below. Members under Direct Tension Only.—When a member is sub- jected to direct tension only, all stresses must be resisted by steel and no allowance should be made for the strength of concrete. The required area of steel is (90) where T is the total tensile stress and f; the allowable unit stress in steel. The unit stress, fs, for which the reinforcement is to be designed, depends upon conditions. Where the stretch due to the stress is not harmful, a stress in steel of 16.000 Ib. per sq. in. may be used. In circular tanks designed to contain fluids, the stresses in the rings must be limited to from 10000 to 12 000 lb. per sq. in. to limit the stretch as much as possible and thus avoid open cracks in concrete and the consequent leakage. Also, deformed bars are recommended to properly distribute any cracks. The hair cracks which may occur for these recommended stresses are not likely to extend the full thickness of the wall and therefore would not affect its water-tight- ness. | In designing arches with tie-bars, their stretch must be considered (See Arch Design, Volume III.) Composition of Tension Members.—When a tension member is not likely to be subjected to secondary bending, it may consist of a single bar. When imbedded in concrete as a protection against rust and fire, the bar is placed in the center of the concrete section. In a tension member likely to be subjected to secondary bending, as, for instance, the hanger for floor beams in arches which may be MEMBERS SUBJECTED TO DIRECT TENSION 163 struck by a vehicle, the reinforcement should consist of four bars placed in the corners of the concrete section. 'The member is then able to resist any such bending moment. Since the stresses due to bending moment increase the tensile stresses due to direct tension, it is necessary to use a larger area of steel than would be required _ for direct tension alone. Sometimes the bars for direct tension are placed in the center of the member and small bars are added in the corners of the concrete section for bending. Details of Tension Members.—It is important to keep the tension members straight. It is obvious that a member with an accidental curvature would straighten under tension, and the move- ment due to this straightening would be injurious to the rest of the construction. A sag in horizontal ties also should be avoided. For these reasons, in arches, the tie is provided with a turnbuckle for the purpose of tightening it after installation. The tie-bar is also supported by suspenders at proper intervals so as to prevent sagging under its weight. If the tension member consists of a number of bars, they should be held properly in place by ties or hoops in the same manner as column reinforcement. The bars composing the tensile reinforeement must be anchored at the ends, so as to develop their full strength. The anchorage may consist of straight imbedment, of straight imbedment with a hook, or of a screw thread at the end with a nut and a plate. In circular tanks, the rings are usually spliced by lapping. It is important that the splices of the rings be staggered. With straight imbedment only, the bar must be extended into the anchoring member a sufficient distance to develop by bond the strength of the bar. (See p. 268.) _ With straight imbedment and hook, the straight imbedment may be made shorter because of the effect of the hook. In important members, the straight imbedment is made long enough to satisfy bond requirement and the hook is added as an additional factor of safety. The bearing stresses of the hook in the concrete are high, although with properly designed shape of hook (see p. 269) they are not excessive. ‘To relieve the stresses, the hook may engage a cross bar; this distributes the compression stresses and also tends to pre- vent any tendency to splitting. The plate method is illustrated in Fig. 218, p. 679. In this method the bars are threaded at the end, a plate of proper dimensions is placed at the end, and this is kept 164 THEORY OF REINFORCED CONCRETE in place by nuts, one for each bar in front, and the other back of the plate. When ‘mbedded in concrete, the plate distributes the tensile stresses on the concrete. ‘The area of the plate is determined by dividing the total tension in the bar by the allowable bearing stress on concrete. If more than one bar is used in the member, each bar may be provided with a separate plate or one plate may be used for. all bars. The thickness of the plate should be determined by con- sidering it supported by the bars and loaded uniformly over its: whole surface by the reaction of the concrete. When a plate is used for a single bar, it should be treated as a cantilever. When one plate is used for two bars, it may be considered as a simply supported plate uniformly loaded. The plate must be imbedded far enough beyond the end of the tensile member to prevent shearing of the concrete. Thickness of Concrete in Tension Members.—In fireproof con- struction, the tensile members should be imbedded in concrete to protect them from fire and rust. If subjected to tension only, the required concrete section is governed mainly by the required thick- ness of the protective cover. Where lateral stiffness of the member is desirable on account of the possibility of accidental stresses, the dimensions should be adopted accordingly. In circular tanks, the thickness of concrete should be sufficient to make them watertight. MEMBERS UNDER FLEXURE AND DIRECT STRESS The formulas given below apply when any member of a structure is subjected to compression (or tension) due to a central force and to bending or flexure. This takes place when the member is sub- jected: | (1) Simultaneously to a bending moment and a central force or thrust; (2) To an eccentrically applied force or thrust. The first condition occurs in columns carrying the column load and in addition—particularly in wall columns—subjected to a bend- ing moment caused by rigid connection between beam and column. Another instance is that of columns with brackets for crane runways. The crane load produces a bending moment in the col- umn in addition to the central force. All members in rigid frames, particularly the vertical members, are subjected to central force and bending. In trusses, the compression members must resist, in addition to the central force, a bending moment, which in trusses DIRECT STRESSES AND FLEXURE 165 with diagonals is produced by the rigidity of connection at the joints, and in trusses without diagonals by the shear, as explained in Volume III of this treatise. The second condition occurs in arches and dams when the line of pressure intersects the normal section at a distance from the arch axis, in other words, when the thrust is applied eccentrically. The thrust may be normal to the section or, as is usually the case, it may be inclined to the section. Before determining the stresses, the inclined thrust, 2, is resolved into a component normal to the sec- tion, V, and a component parallel to the section, V. The normal component produces both direct stresses on the section and bending, while the parallel component produces only shearing stresses which are usually small and do not need to be considered in design. An eccentrically applied force is found also at the base of retaining walls where the resultant pressure is composed of the horizontal earth pressure and the vertical weight of the structure. In this case, the formulas below are used to find the distribution of pressure over the foundation. | Sign of the Thrust.—The thrust is either positive, when it pro- _ duces compressive stresses on the section, or negative, when it pro- duces tensile stresses. Usually, the positive thrust is called “ thrust,” and the negative thrust, ‘ pull,’”’ although the desiginatons “‘ positive and negative thrust ” are often used. Relation between Bending Moment and Eccentrically Applied Thrust.—The stresses produced by a combination of a central thrust and a bending moment are the same as those produced by an eccentrically applied thrust. Thus, a central thrust, V, and a bend- ing moment, M, may be replaced by an eccentric thrust, N, acting at a distance from the axis of the section equal to e = NV" In turn, the eccentric thrust may be replaced by a central load of the same intensity and a bending moment equal to the thrust multiplied by the eccentricity.° Therefore, both cases can be solved by the same 6 Proof of the above statement. Assume that a section is submitted to an eccentric thrust N acting at a distance e from the axis. The stress conditions will not be altered if in the center of the section are added two equal forces, but acting in opposite directions, such as + N and ~ N. The section is then exposed to three forces, namely + N at a distance e from center of section and — N and + N in the center of the section. The eccentric force with the negative central force forms a couple, Ne, and may be replaced by a bending moment M = Ne. ‘There remains in the center a positive force N. Thus the eccentric thrust is replaced by a central force, N, and a bending moment M = Ne. 166 THEORY OF REINFORCED CONCRETE formulas. The case of an eccentrically applied thrust gives simpler formulas. Relation between Position of Eccentric Thrust and the Sign of Bending Moment.—A _ positive bending moment, producing in a horizontal member,’ compressive stresses at the top of the section and tensile stresses at the bottom, may be replaced by a positive thrust acting above the axis. A negative bending moment, producing compressive stresses at the bottom of the section and tensile stresses on the top, may be replaced by a positive thrust acting below the axis. Gravity Axis __ _ Gravity Axis P Eccentric Thrust applied cnut : Central Thrust and above Gravity Axis- puivniest te Positive Bending Moment Fic. 54.—Positive Bending Moment and Central Thrust. (See p. 166.) _ Gravity Axis 4 Eccentric Thrust applied Ea enee Central Thrust and below Gravity Axis Shahan ae ne Negative Bending Moment Fic. 55.—Negative Bending Moment and Central Thrust. (See p. 166.) Conversely, a positive thrust acting above the axis produces maximum compression at the top and minimum stresses at the bot- tom. A positive thrust acting below the axis produces reverse results. Negative Thrust, or Pull—tIn the discussion, that follows N is assumed as a positive thrust, ie., a thrust producing compression, and therefore the maximum stresses are compressive. If N is a negative thrust, or pull, as in arch design with the thrust due to 7 Similar action, of course, is true in a vertical or inclined member. DIRECT STRESSES AND FLEXURE 167 rib shortening or to fall of temperature, it will produce tension over the section. A negative thrust, applied below the gravity axis, produces positive bending moment, while, if applied above the AXIS, it produces a negative bending moment. Conversely, positive bending moment and pull can be replaced by an eccentric pull acting below the gravity axis. The reverse is true of negative bending moment and pull. Negative thrust (or pull) combined with positive bending moment (or negative thrust acting below axis) produces maximum tension at extreme fiber below the axis and minimum tension (or small com- pression) at the extreme fiber above the axis. Pull combined with negative bending moment (or negative thrust acting above the axis) produces reverse results. The formulas given in subsequent pages may be used for negative thrust (or pull) by substituting —N for N. This changes the compression unit stresses due to the thrust to tensile unit stresses and reverses the position of bending stress in the section. A question arises as to whether, with negative thrust, it is per- missible to use the formulas given below for materials such as con- crete, which is not capable of resisting tensile stresses. If the nega- tive thrust and bending moment act alone, then, of course, the formulas are not applicable and the formulas on p. 189 to 194 should be used. When, however, the tensile stresses due to the negative thrust are counteracted to a great extent by a positive thrust, the formulas in this section, after the sign is changed before N, are appli- cable. For instance, in arch design, the negative thrust due to fall of temperature is counteracted by the positive thrust due to dead load. In such a case, the stresses due to both thrusts are computed by the same formulas separately and the results added. NOTATION Let #& = inclined thrust acting on a section; N = normal thrust, a component of thrust R normal to the section; also a central load; V = external shear, the component of 2 parallel to the sec- tion; T = normal pull, in members subjected to direct tension or tension and bending; b = breadth of rectangular cross section; 168 €0 itn THEORY OF REINFORCED CONCRETE height of rectangular cross section; eccentricity, that is, the distance from axis of gravity of the section to the point of application of the thrust; value of eccentricity which produces zero stress in con- crete at outer edge of rectangular section; bending moment on the section; perpendicular distance to any point in the section from eravity axis of section; moment of inertia of cross section of concrete about the gravity axis; moment of inertia of cross section of steel about the horizontal gravity axis; moment of inertia of transformed section; total effective area of cross section of concrete; total area of section of steel; — area of transformed reinforced concrete section; perpendicular distance from outside fiber to gravity axis of unsymmetrical section having maximum compression ; perpendicular distance from center of gravity of unsym- metrical section to outside fiber having maximum tension or minimum compression ; maximum unit compression in concrete; maximum unit tension in concrete, or minimum com- pression ; ate maximum unit compression in steel; maximum unit tension or minimum unit compression in steel; ratio of steel to total area of section; es ~ yatio of moduli of elasticity of steel and concrete; ratio of depth of neutral axis to depth d; distance from outside compressive surface to neutral AXIS; depth of steel in compression; depth of steel in tension; distance from center of gravity of symmetrical section to steel; constants. PLAIN CONCRETE UNDER COMBINED STRESSES 169 PLAIN CONCRETE SECTION UNDER DIRECT STRESS AND BENDING MOMENT General Formula.—For members of any cross section subjected to a central load, N, and a bending moment (or to an eccentric thrust), the stresses may be obtained by computing separately the stresses caused by the central load and by the bending moment. The sum of the results then gives the actual stresses. The unit stresses produced by a central load, N, obtained by dividing the load by the effective area of the section, are equal to _ The unit stresses produced by the bending moment, M, at any point at a distance, y, from gravity axis, from simple mechanics, are equal to + ee The sign + is used for points above the axis, and — for points below the axis. The combined stresses, therefore, at any distance, y, from the axis of gravity, are equal to the sum of the two above expressions. Stresses at any point. NM io A= ie (91) The second term of this expression, depending upon the value y, varies with the position of the point in relation to the axis of gravity, therefore, the intensity of stresses varies from a maximum at one edge of the section to a minimum at the opposite edge. Gravity Axis aie Section Stresses due to Stresses due to Combined , Bending Moment. M. Central Thrust, N. Stresses Fig. 56.—Section Subjected to Direct Stress and Bending. (See p. 169.) Usually, only the maximum and minimum stresses in the extreme fibers are of interest. If, as given in Fig. 56, y: is the distance of 170 THEORY OF REINFORCED CONCRETE the extreme fiber above the axis and y2 below the axis, the combined stresses are: : Maximum Stresses, = N My he aE. A + yi ° . e ° ° e » e (92) Minimum Stresses, oe N oy Mype This condition exists for a positive bending moment, or when the eccentric thrust is applied above the axis. For negative bending moment or a thrust acting below the axis, the sign before the second term in the formulas changes. The maximum stress acts below the axis and the minimum stress above the axis. Formulas for Rectangular Plain Sections.—In a rectangular sec- tion with breadth, b, and height, h, the area A = bh, the moment of inertia [ = te and the distance from axis to extreme fibers, UIE AS These values substituted in the general equation give the following expressions for rectangular section: Maximum Compressive Stress, N , 6M Se a bh + Bh2’ = F 3 ° . : 5 (94) Minimum Stress, N 6M fi = pp — Bye (95) If the bending moment M is replaced by Ne the above formulas change to: Maximum Compressive Stress, Minimum Stress, f= (+7). We h hisae (1 . i) (97) PLAIN CONCRETE UNDER COMBINED STRESSES 171 For positive bending moment (or thrust applied above the axis), maximum stresses act above the axis. The position of maximum stresses is reversed with negative bending moment or with thrust below the axis. When 0, fh and e are in inches and M isin inch-pounds, and N in pounds, the unit stresses are in pounds per square inch. For M in foot-pounds, multiply the second term by 12. To get the stresses in pounds per square foot, as is customary in dam design, the dimensions b and h must be in feet and M in foot- pounds. | Force at Edge of the Middle Third. Middle Third. Fic. 57.—Stresses Caused by Eccentrically Applied Thrust. (See p. 171.) £4 et S| eth 3 es Via. 58.—Stresses Caused by a Force Acting Outside the Middle Third of Plain . | Concrete Section. (See p. 172.) For positive thrust, the maximum stress is always compression. The minimum stress from Formula (97) is compression when - is _ smaller than unity, that is, when the eccentricity, e, is smaller than : When the force acts at the edge of the middle third of the section, the minimum stress equals zero, and the maximum stress equals double the stress caused by a central load of equal intensity. (See mie. of, p. 171.) 172 THEORY OF REINFORCED CONCRETE When e¢ is larger than i and the load acts outside of the middle third, the minimum stress is negative, and the section is subjected to tension. In such cases, this formula can be applied only when the material is capable of resisting tensile stresses. If the material cannot resist tension, as in masonry, Or in con- crete where tension exceeds the allowable stress, it is necessary to assume that the pressure is distributed only over a section equal to three times the distance of the point of application of the load from the nearest edge. (Sce Fig. 58). If that distance is g, the total effective depth of the section is 39. Substituting in Formula (96), e = | andh = 39, we get for plain concrete and masonry. Maximum Compression, ate a 3b an ble SRR che ae eens (98) Plain Circular Sections.—lor circular sections where d is diameter 4 of circle, the area of moment of inertia is J = ie and the section is A = ae — (,785d?, and the distance to extreme fiber yi = Yy2 = = These values substituted in equations 92 and 93 give Maximum Compression in Concrete, N 10.2M (99) Se = D751 7 de Minimum Stress, N 10.2M i= Tye (109) Substituting M = Ne. Maximum Compression Stress, ne zp( 1.275 ab 0). i Minimum Stress, . fz (1-275 a 02). ON eee The stresses at one extreme edge become zero for e = 0.125d. For larger eccentricity, tensile stresses will be developed. DIRECT STRESSES AND FLEXURE 173 REINFORCED CONCRETE SECTION SUBJECTED TO DIRECT STRESS AND FLEXURE General Formulas.—The stresses in a reinforced concrete section subjected to direct stress and flexure may be obtained by replacing the reinforcement by an equivalent area of concrete and then apply- ing to the transformed section the general Formulas (92) and (93) developed for plain concrete sections. To get the equivalent area of concrete, the area of reinforcement is multiplied by the ratio of moduli of elasticity of steel and concrete, n. The area of the transformed section is A: = A+ (n—1)As. The moment of inertia of the transformed section about the axis of gravity equals the moment of inertia of the concrete section plus the moment of inertia of the concrete replacing the steel. The substitute area must be placed at the same distance from the axis as the steel area, so that the moment of inertia of the substitute area equals (n — 1) times the moment of inertia of the steel area about the gravity axis of the sec- tion. The total moment of inertia, I, =I +(n—1)I;. The value of (n — 1) is used in the formulas instead of n because, in taking the concrete area, one A is already included. Substituting these values in Formulas (92) and (93), the formulas for maximum and minimum stress become Maximum Compressive Stress, N My Peas, PTE Gh) 40) Minimum Stress, | my, N Mye Meena =A, Ton DL (104) The stress in steel equals the stress in the adjoining concrete multiplied by n. As evident from the formulas, for positive bend- ing moment, maximum stresses act above the neutral axis while minimum stresses act below the neutral axis. For negative bend- ing moment, the sign before the second term of the equation changes, and the maximum stresses act below the neutral axis while minimum stresses act above the neutral axis. | 174 THEORY OF REINFORCED CONCRETE RECTANGULAR SECTION WITH SYMMETRICAL REINFORCEMENT The formulas for a rectangular section subjected to direct load and bending moment depend upon the arrangement of reinforcement. The bars may be placed in the section as shown in Fig. 59(a) or as shown in Fig. 59(0). _-Bars f, Per, * + ‘eae Tage an ae Eccentric Pull applied Equivalent Central Pull and below Gravity Axis to Positive Bending Moment Fic. 64.—Section Subjected to Direct Tension and Bending. (See p. 190.) tension. Since conerete cannot resist any tension, all stresses must be resisted by the steel. For equilibrium, the sum of all the stresses must equal 7. Therefore, —T=—(A\ fi +a From the figure, it is found that the stress in the reinforcement nearest the force, N, equals | ate Pj Aad. Af,=T (136) MEMBERS SUBJECTED TO DIRECT TENSION AND BENDING 191 The stress in the reinforcement away from N is nsf. af UAT ite 4 en SG re The two equations are sufficient to determine the stresses carried by the respective tensile members. Since in each equation two values, namely, A, and f,, are unknown, it is necessary to make some assump- tion. Either the area of steel is assumed and the stress determined, or the stresses are accepted and the areas computed. If A, = A’,, the required area of steel is obtained from the follow- ing equation, Tate oh la (138) in which f, is the maximum allowable tensile stress in steel. The stress in the other section is f’, = are It is smaller than the allowable stress. If it is desired to use the same unit | 8 Bending Sketches 3 al Q wm . ' Q, 2 @ } 2 ASwwel \ BS a ES j ara he SS re We--- J ---- Sisco nae ne Ebr a | ots NS | ©), ee oS po NIE Bia ee Nis Nig \ Bs l 3! ioe &| l | | Aig i ! \ Bent | | 1 accent ears, i] i} | <1--Beam “Straight | K-Sf 0.214 6" Scheme 4: Long Straight Bars on Bottom Short Straight Bars’on Top. Fig. 69.—Arrangement of Slab Reinforcement in Continuous Slabs. (See p. 209.) DESIGN OF SLABS PAM Cross Reinforcement of Slabs.—Cross reinforcement, that is, bars at right angles to the principal reinforcement, is customarily used to prevent shrinkage and temperature cracks. The amount of steel to use for this is usually selected somewhat arbitrarily, a cross- sectional area of bars equivalent to 0.2 per cent to 0.3 per cent (p = 0.002 to 0.003) of the cross section of the slab being usual in practice. Reinforcement over Girders.—If a girder is parallel to the prin- cipal reinforcement of the slab, short bars should be placed at the girder transversely, near the top of the slab, not only to stiffen the T-beam (see p. 38), but also to provide for the secondary negative bending moment produced by the bending of the slab next to the girder. ‘This reinforcement is necessary even when the beam is simply a small stiffener. (See p. 223.) The authors recommend that the amount of such reinforcement should not be less than 0.3 per cent of the effective area of the slab. The bars should be placed near the top of the slab, not farther apart than 18 in. Their length should not be less than two-thirds of the total effective width of the flange as given on p. 218. This recommendation agrees substantially with the 1924 Joint Committee recommendation. : Bar Spacers.—To keep the bars in place and a proper distance above the forms, the use of proper spacers is recommended. ‘There are numerous types of spacers on the market, which serve not only to keep the bars in place, but also to keep them a proper distance above the forms. If no spacers are used, the bars should be securely tied to the transverse reinforcement. Computing Ratio of Steel in Slab.—The ratio of steel in a slab may be found by dividing the cross section of one bar by the area between two bars, this area being the spacing of the bars times the depth of steel below top of slab. For example, a slab with steel 4 in. below the top and 4-in. round bars spaced 6 in. apart has a ratio, p = ie = 0.0082, or 0.82 per cent steel. Square and Oblong Slabs Supported by Four Beams.—When a slab is supported by four beams and its length does not exceed 13 times its width, the loads will be carried by the slab to all four beams, and therefore the slab must be reinforced in two directions, as shown below. 212 REINFORCED CONCRETE DESIGN The load carried in either direction can be determined from the following formulas: Let w = total unit load; we = unit load carried by the short span; wz, = unit load carried by the long span; L = length of slab; B = width of slab; Then apie (5 si 5) : (14) (ey ? , and Wr = (1.5 — 5) . (15) B The following table gives the ratio of the unit load, we, carried by the short span for different ratios of ak Load Carried by the Shorter Span. (See p. 212.) Ratio of Length | qoad Carried by Ratio of Length | 149d Carried by to Breadth of : to Breadth of L the Shorter Vk the Shorter Slab B Span, we Slab B Span, we Jet Se ert | 4 ee Ee ee 1.00 0.50w 1.30 0.80w 1.05 0.55w a oe te 0.85w 1AM, 0.60w 1.40 0.90w VO: 0.65w 1.45 0.95w 1.20 0.70w 1.50 1.00w de 0. 75w It must be noticed that the shorter span carries the larger pro- portion of the load. After the proportion of the load is determined, the bending moments are found as for slabs reinforced in one direction, and the dimensions or stresses are found by the ordinary formulas. The thickness of the slab, of course, is governed by the larger bending moment of the two. The above proportion is based on the assumption that the con- ditions of restraint on all four sides are the same. It does not apply if the slab is continuous in one direction and simply supported in the other direction; in this case the slab in the continuous direc- DESIGN OF SLABS 213 tion, will resist a much larger proportion of the load, because it takes a larger load to deflect a continuous slab an spueTemi ahs equal to the deflection of a simply supported slab. Most Economical Design of Rectangular Beam.—To get the most economical design, the ratio of width of beam to depth should be as large as practicable. For any given width of beam best economy is obtained by the use of the balanced design of beam (see p. 203). No economy results from using a larger depth than required by’ Formula (1), p. 204, because the cost of additional concrete is larger than the saving in the cost of steel. Economy of Rich Mix of Concrete in Slabs.—Under ordinary conditions no economy results from the use in slabs of concrete of richer mix than the leanest permissible in reinforced concrete, for the reasons given below. (See p. 4 for permissible mix.) The use of richer mix of concrete results in smaller depth of slab with consequent reduction in the amount of concrete, but it requires a larger amount of reinforcement than necessary for the leaner mix. While the amount of concrete in cost for rich mix is smaller than for lean concrete, the difference in cost due to higher unit cost of rich _ mix, is small. The saving in cost of concrete is more than offset by _ the extra cost of steel. As an illustration, compare the cost of slab required by M = 40000 in.-lb. using 1: 2:4 and 1: 14:3 concretes with the unit stresses given in table on page 205. Following unit prices ? should be used: 1: 14 : 3 concrete—49 cents per cubic foot; 1: 2:4 concrete—45 cents per cubic foot; steel—5 cents a pound. The concrete dimensions and the required area of steel are found from Formulas (11) and (3). Mix Effective Depth Total Depth Area of Steel Gh es d=0.021V 40 000=4.2 in. h=4.2+1=5.2in. A,=0.0133 X4.2 X12 = 0.672 sq. in. fee 2 4 d =0.024-V 40 000=4.8in. h=4.8+1=5.8in. A,=0.0107 X4.8 X12 ! =(0.615 sq. in. Volume and cost of concrete: Mix Volume X Unit Cost =Cost per sq. ft. of floor B.2 X 12 | Pig 3 ——_— = 0.48 cu. ft. X 49= 21 cents 144 2 Ree 34 cee = 0.48 cu. ft. X 45=21.6 cents Saving due to rich mix = 0.6 cents per sq. ft. 2 Prices prevailing in year 1925. 214 REINFORCED CONCRETE DESIGN Weight and cost of steel: Mix Area of Steel Weight of Steel X Unit Cost = Cost of steel per sq. ft. of floor 1: 12:3 0.672 sq. in. 2. 3ilbr Xx So = Ts eo 1s 2.3 46280,.615. seein: 2.1 lb. XS 10centa Extra cost of steel = 1 cent per sq. ft. Since there is a saving in concrete due to rich mix of 0.6 cent and an extra cost of steel of 1.0 cent, the difference is 0.4 cent per sq. ft. in favor of leaner mix. Hence rich mix not economical. Economy of Rich Mix of Concrete in Rectangular Beams.— Under some conditions the use of rich mix in beams is economical. It is particularly advantageous for very long spans or where heavy loads are to be carried with limited headroom. Often, the reduction in dead load resulting from the use of rich concrete is of advantage. Rich mix of concrete permits reduction of the concrete dimensions of a beam. Either the depth may be reduced, keeping the width same as for leaner mix, or the depth may be used same as for leaner mix and the width reduced. In the first case the conditions are identical with those explained in connection with slabs and no economy would result. In the second case distinct economy may result from the use of rich mixture as evident from example below. As an illustration compare the cost of beams designed for 1: 15: 3 and 1: 2:4 mixes for a bending moment M = 720 000 in.-Ib. when the depth of beam is limited to h = 22 in. using stresses and constants in the table on p. 205. | Since the total depth is h = 22 in., the effective depth d = 22 — 3.5 = 18.5 in. This allows for 2 in. fire proofing and two layers of steel. Since the depth is fixed, the width of beam will be obtained from Formula (38). Mix Width of Beam Volume of Concrete per lin. ft. 720 000 11.4 X 22 1:14: 36 = ——— = 11.4in. ——— = 1.74 cu. ft. : 184 X 18.52 144 et 720 000 14.3 X 22 1:2:4 b= —— = 14.3in. ———— = 2.18 cu. ft. 147 X 18.5" rE 144 ee Comparison of cost: Mix Volume of Concrete X Unit Cost = Cost per lin. ft. Lely 1.74 cu. ft. X 49 cents = 85.3 cents 122%4 2.18 cu. ft. X 45 cents = 98 0 cents DESIGN OF T-BEAM 215 The saving in favor of rich mix is 98.0 — 85.5 = 12.5 cents per lineal foot. The amount of steel is the same in both cases. An additional saving will be due to the reduction in dead load which amounts to (2.18 — 1.74) x 150 = 66 lb. per lin. ft. DESIGN OF T-BEAM The formulas given below are sufficient to design or review a T-beam for any given span and loading. The derivation of the - formulas is given on p. 134. Formulas for bond stresses and for diagonal tension are given in subsequent pages, and the design at supports for negative bending moment is also treated. | How T-Beams are Obtained.—If a beam is built at one pouring with the slab, and precautions are taken to prevent separation of the slab and beam, the portion of the slab near the beam may be considered as part of the beam. The beam then assumes the shape of the letter T and is called a T-beam. Figure 70 shows a cross section through a floor. The section of the slab considered as part of the T-beam is indicated. : atgle Fig. 70.—Section Showing T-Beam in Floor. (See p. 215.) - NOTATION Let 6 = breadth of flange of T-beam in inches; b’ = breadth of web of T-beam in inches; d = effective depth of T-beam in inches; jd = moment arm in inches; t = thickness of flange in inches; r = ratio of cost of steel to that of concrete; f, = compressive unit stress in concrete, pounds per square inch; f,; = tensile unit stress in steel, pounds per square inch; n = ratio of modulus of elasticity of steel to that of concrete; A, = area of tensile steel in square inches; 216 REINFORCED CONCRETE DESIGN Dm = ratio of steel, a in T-beam for which stresses in steel and concrete simultaneously reach their maximum working value; constant for figuring minimum depth. Ca Formulas used in Design of T-beams.— Area of Web required by Diagonal Tension, I v(d — 5) equal or larger than 7 LEP) 2 Minimum Depth Required by Compression, M (Luz Ct (17) where C, is constant on p. 894 for fi, fs and -: For explanation see p. 219. Economical Depth, rM 6k where r is cost ratio of steel and concrete. Area of Steel, M As = a Ae As 19 jaf a) where j is as given on pp. 221 and 897. This area of steel must not exceed the value found from the formula below, else compression stresses will be exceeded. Maximum Area of Steel Permitted by Working Compression Stress, fc, . ? max. A, = Pmbd,.<-) G. yeiteleeee enn eneE on where pm is constant on p. 895 for f., fs, m and 7 For explanation see p. 136. Area of steel used in design must not be larger than the above value, else compression stresses will be exceeded. Maximum Moment of Resistance of Beam Based on Working Stress in Concrete, fe, | . M = dCabt,~ 220 Be ee t where C, is constant on p. 894 for f., fs and 7 For explanation see p. 186. DESIGN OF T-BEAM 217 The maximum area of steel given in the formula just above corresponds to this maximum moment of resistance. The uses of the above formulas are explained below under proper headings. To Design a T-beam.— Design the slab. Determine bending moments and end shears in beam. Determine width of flange (see p. 217). If headroom is not limited, determine most economical depth (see p. 220). (In design- ing a number of similar beams, the eccnomical depth needs to be figured only for one beam, and estimated for the remainder.) Before selecting final depth of beam and breadth of stem, b’, see that the compression in concrete in the center and at the support and the shearing unit stresses do not exceed the allowable working stress (pp. 244 and 879). For continuous T-beam, the depth required at the support, where the T-beam becomes a beam with steel in top and bottom, may be the governing depth. r | dl : uy 2 Bhd... Total Compression Moment Arm : Total Tension nS Fig. 71.—Section of T-Beam. (See p. 216.) Figure amount of tension steel by Formula (19), p. 216. Design web reinforcement. For large T-beams, a saving in concrete may be effected by using the more exact formulas, which take into account the compression in stem below the flange. In such cases, preliminary A, for k and z may be found from Formula (19), and final A; from Formulas (19) to (24), p. 134. The problem is more easily solved by the authors’ simplified method given on p. 224, which gives exact results and requires much less work. Width of Flange.—The width of the slab, b, to use for the flange of the T-beam is selected by judgment on the basis of results from tests. In no case, of course, can it be greater than the distance between beams, The Joint Committee has recommended the 218 REINFORCED CONCRETE DESIGN following rules, which are approved by the authors, for the effective width of flange: (a) The effective width shall not exceed one-fourth of the span length of the beam; (b) Its overhanging width, on either side of the web, shall not exceed eight times the thickness of the slab, nor one-half the clear distance to the next beam. ‘For beams having’a flange on one side only, the effective flange width to be used in design shall not exceed one-tenth (7'y) of the span length of the beam, and its overhanging width from the face of the web shall not exceed six (6) times the thickness of the slab nor one-half (4) the clear distance of the next beam.” This practice is conservative. (See tests, pp. 36 to 38.) Isolated beams in which the T-form is used only for the purpose of providing additional compression area of concrete should prefer- ably have a width of flange not more than four times the width of the stem and a thickness of flange not less than one-half the width of the web of the beam. Width of Flange Required by Different Building Codes.—In large cities the Building Codes must be followed. The dimensions of the flange required by the Codes of the Cities of Boston, Chicago, Cleveland, New York, and Philadelphia are given in Fig. 72, p. 219. Cross Section of Web as Determined by the Diagonal Tension.— The width of the web of a T-beam is governed by the shearing stress (see p. 244) and by the layout of the tension bars (see p. Zia) The area of the web required for shear involving diagonal ten- sion, using notation on p. 215 and letting V = total external shear, and v = shearing unit stress, may be found from the formula (see also Formula (37), p. 247). Area of Web Required by Diagonal Tension.— v(d ~ 5) oe 4 (approx: fer eee) or e|< Wid S 71 ee a This means that the area of web at any point in the beam (con- sidering this up to the middle of the slab) must not be less than the DESIGN OF T-BEAM 219 total shear divided by the maximum allowable unit shear for the reinforced beam. | The vertical unit shearing stress, v (used as measure of diagonal tension), in a beam effectively reinforced with bent bars or stirrups, or both, should be limited to 120 lb. per sq. in. for ordinary con- crete having a compressive strength (in cylinders) of 2000 lb. per sq. in. at twenty-eight days. Ce Boston & Cleveland \~«----- Gi-----: --b!--3 ceaOberan == b not to exceed d l New York & Tulare a a Gt ----->p<-- b'-->he---- 6t-------> @ UMax. + L Max. Chicago ~ <---------- 4 Spacing of beams------------ b not to exceed i l : l =Span of beam Fig. 72.—Width of Flange Required by Building Codes. (See p. 218.) Minimum Depth of T-Beam at Center of Span.—A minimum depth is the depth at which concrete and steel are stressed simul- taneously to their working limits. It is governed by the law that the compression in the flange must not exceed the working com- pressive stress in the concrete. Greater depth than the minimum, with proportionally decreased amount of steel, is generally used for economy. A smaller depth would give excessive compressive stresses. To find the minimum depth, the rectangular beam formula (1) p. 204, may be used where the depth of the beam is not greater than four times the thickness of the slab, the breadth of the flange of the T-beam being used in this formula for the breadth of the rectangular beam. For ratios of depth of T-beam to thickness of slab larger than four, the rectangular beam formula gives unsafe results and the following formula should be used: 220 REINFORCED CONCRETE DESIGN Minimum Depth of T-beam at Center of Span, M In the above formula, Cy is a constant from pp. 894 and 895, depending upon the stresses f, and f, and the ratio _ Since the depth of beam, d, is not yet known, the depth required by shear may be taken as a close approximation, in computing the ratio of > The use of the formula is illustrated in the example on p. 896. Economical Depth for T-beam.—For T-beams, a depth greater than the minimum depth required by compression stresses will usually be found economical. By increasing the depth of the stem, the area of the tensile reinforcement is decreased, with a consequent reduction in cost of steel. The volume of concrete in the web, and as a consequence the cost of concrete, is increased; but, up to a cer- tain point, the reduction in the cost of steel is larger than the increase in the cost of concrete, and the cost of the beam as a whole is there- fore reduced. The most economical depth is the depth for which the ‘nerease in cost of steel is equal to the decrease in cost of concrete, but beyond which any further increase in the depth would cause an increase in cost of concrete larger than reduction in cost of steel. The following convenient formula for the economical depth of the beam was developed by Professors Turneaure and Maurer? Using the notation given p. 215 and y = ratio of cost of cubic foot of steel in place to cost of cubic foot of concrete in place. Economical Depth of Beam, rM t oe ay + (24) From this formula the economical depth for different widths of stem may be computed, and the most suitable combination of depth of beam and width of stem may then be selected. The value of the ratio r ranges from 35 to 75. It is computed Aig 3 Turneaure and Maurer, “Principles of the Reinforced Concrete Construc- tion, Second Edition, p. 238. DESIGN OF T-BEAM 221 as follows: If concrete in place costs 40 cents per cubic foot and steel in place 4 cents per pound or 4 X 490 = 1960 cents per cubic foot, the ratio is r = +38° = 49. In continuous beams add from 25 to 40 per cent to the unit cost of steel to allow for the overlapping of bars at the supports. Value r as Affected by Cost of Formwork.—In computing the value r above, only cost of materials was considered because the actual cost of formwork is little affected by small changes in depth of beam. The supports for the forms and the number of joints remain the same and the labor of making, erecting and stripping the form also is little affected by changes of depth of beam. The only effect caused by changes in depth is a small difference in cost of lumber for the sides, which is negligible, specially when forms are reused several times. Often, however, in practice the cost of formwork is computed by multiplying the area of the bottom and the sides of the beam by a certain unit cost. In such case the computed cost of formwork is affected appreciably by changes in depth of beam. The economical depth of beam, when estimated on that basis, may be obtained by Formula (24), p. 220, in which value of r is found as follows: Com- pute volume and cost of concrete for a section of beam, 1 ft. long and 1 ft. deep with width of stem equal to 6’. Add to this the cost of 2 sq. ft. of formwork based on the units used in estimating. Divide this total cost by the volume of concrete previously computed. Substitute this new unit cost for the cost per cubic foot of concrete in figuring the value 7. . Area of Steel in a T-beam.—The area of cross section of steel in tension may be obtained very closely by the following formula: ' Required Area of Steel, M A, an rar jd The value of 7 may be taken from the following table: (25) Value of j for Different Ratios of Depth of Beam to Thickness of Slab : a i = ee ae 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 » Poel 0.95 0.94 0.98 0.98 0.92 0.91 0.90 0.90 0.89 0.88 0.88 Sometimes jd is assumed to equal 2d, or d — but the error is 9? larger than in the above table. 222 REINFORCED CONCRETE DESIGN More exact values of 7 may be found from the table on p. 897, or by the use of formulas on p. 184. Minimum Depth at Support in Continuous Beams.—At the sup- port of a continuous beam, owing to the change of the bending moment from positive to negative, the compression flange of the T-beam is in the tensile zone and is therefore ineffective. ‘The com- pression zone consists only of the width of the stem, and the T-beam changes into a rectangular beam with a depth equal to the depth of the beam and a width equal to the width of the stem. The com- pression stresses, therefore, are much larger than in the center and may require special provision. In all cases, some of the reinforcement used in the center for positive bending moment is carried straight at the bottom, through- out the full length of the beam. ‘This reinforcement, if carried a sufficient distance into the column to develop the required com- pression stresses, may be utilized as compression reinforcement. The beam then becomes a rectangular beam with steel in top and bottom. On account of the great reduction in the compressive area, the compression stresses at the support may be excessive and the con- dition at the support may determine the depth of the beam. The required depth at the support will depend upon the amount of com- pression reinforcement used. Usually, about one-half of the bottom steel is available as compression reinforcement. When the required area of tensile steel is the same at the support as in the center and one-half is bent up and carried over the support, the ratio of the available compression steel to the tensile steel at the support is oe This is the most desirable ratio to use. If this ratio is not sufficient, additional compression steel may be obtained by extending the bottom steel from the adjoining span a sufficient length to make it available as compression reinforcement in the span under consider- ation. With such design the ratio of compression steel to tensile steel may be increased to one. Larger ratios than one can be obtained only by introducing additional short bars at the bottom. After the ratio of the area of compressive steel to the area of tensile steel is selected, the required minimum depth of beam may be found from the following formula: Minimum Depth at Support, M . d — 1.054). o (leu, evoke See oFeRes (26) DESIGN OF T-BEAM 223 In this formula, M is bending moment at support, b and d are dimensions of rectangular beam, f; is tensile stress in steel, and p; ; Oe : is the tensile steel ratio, bd? the magnitude of which depends upon the accepted compression stresses in concrete and also upon the ratio of the area of the compression steel to that of the tensile steel. The value of p; may be taken from pp. 904 to 907. The use of the formula is illustrated by the example given below. The allowable compression stress in concrete at the support is larger than in the center of the beam, as explained on p. 282. Example 3.—Given the bending moment at support, M = 1 200 000 in.-lb.; allowable working stresses, fe = 750, fs = 16000, n = 15; width of stem, 6 = 12 in. Find minimum depth if it is desired to limit the area of compression steel to one-half the area of tensile steel, so that - = 0.5. i Solution.—Assume the ratio of depth of compression steel to depth of beam, a=0.1. From Diagram, p. 904, in the section for f, = 750 and f,; = 16 000, and in the column for a = 0.1, we find by interpolation that a value of p’ = 0.006 corresponds to a value of pi = 0.013. This gives substantially the desired ratio between the compression steel and the tensile steel. The value of p1 = 0.013 will be used in Formula (26), p. 222. Thus, minimum depth —— ey d= 1.054|-™ ~ 1.054] mg O00 = 1.05V480 = 23 i eee NS pee N12 «0.018 X 16.000,” ari ea Details of Design.—In designing T-beams, diagonal tension must be investigated and web reinforcement provided where necessary (see p. 247). Bond stresses must be computed (see p. 262). In continuous beams, negative reinforcement must be provided (see p. 281). To insure proper T-beam action and prevent cracks through the slab adjacent to the stem, reinforcement at right angles to the beam or girder is necessary. When the principal slab reinforcement is parallel to the girder, short bars should be placed at the top of the slab, transversely over the girder and extending on both sides well into the slab. The ratio of the area of this reinforcement to the cross section of the slab should be at least 0.003. The example on p. 578 illustrates the use of the formulas and principles of design. The selection of bending moments is treated on p. 275. 224 REINFORCED CONCRETE DESIGN To Find Stresses f, and f, in T-beam.—TIf the dimensions of beam, the amount of steel and the bending moment are known, the stresses may be found as follows: Find k and j from Formulas (19) and (20), p. 134. Find f, and f, from Formulas (22) and (23), p. 134. The work may be simplified by use of the table on p. 896, giving values of k, j andC,. Compute ratios ae and M For these find k from the Table of k. Knowing k and a find value of j from the Table of j. The stress in steel is then founed from M Je = Asjd The stress in concrete is found from fot A, = Crh PUT Sawet (27) where C7 may be taken from table on page 896. T-beam Design Considering Compression below Flange.—In a large T-beam, it is often desirable to take advantage of the com- pression stresses in the web below the flange, especially when the compression stresses, as computed by the approximate formulas, are excessive and it would be necessary to increase the compression area. Computed by the formulas that take account of the com- pression below the flange, the compression stresses are smaller. In instances where the thickness of the flange depends upon com- pression stresses in the T-beam, the use of these formulas reduces © the amount of concrete in the flange and thereby the dead load. The formulas on-p. 137 are too complicated and may be replaced by the following simple method, which gives equally exact results. Simplified Method.t—This method, which gives exact results, need be used only when the compression stresses in the beam are practically equal to the maximum allowable stress. Where the com- pression area is in excess, the approximate formulas on p. 134 give satisfactory results. The simplified method is most useful in large T-beams, in isolated T-beams, and for girders in joist-constructed floors. Its use frequently saves appreciable amount of concrete. Consider a T-beam section, as shown in Fig. 73,.p. 225. If the stress in concrete in such a beam equals the maximum allowable, fe, 4'This method was devised by Edward Smulski. DESIGN OF T-BEAM 220 and the stress in steel is fs, the position of the neutral axis of the T-beam is located in the same place as in a rectangular beam of the same depth and with the same unit stresses. The T-beam may be considered as consisting of two parts: (1) a rectangular beam with a width equal to the width of stem, 0’, and a depth equal to the depth, d, of the original beam; and (2) an auxiliary T-beam, the width of flange of which is equal to the sum of the projections of the T-beam and the depth equal to the depth of the original T-beam. The dimensions of each part and the area of steel required for each may be determined separately and the results added. The rectangular beam may be designed by Rectangular Beam Auxiliary T-Beam Fig. 73.—Illustration of Simplified Method of T-Beam Design. (See p. 224.) rectangular beam formulas. In the auxiliary T-beam there is on concrete below the flange; therefore, the formulas on p. 216 give exact results. This method is particularly advantageous if it is necessary to provide compression reinforcement in the T-beam, as may be the case in theater construction, where not only the depth of the beam but also the width of the flange is limited. In such cases, the T-beam is considered as composed of a rectangular beam with double rein- forcement and an auxiliary T-beam without concrete below the flange, and the results are combined. Case 1.—The procedure with ordinary beam and slab may be seen from the following examples: 226 REINFORCED CONCRETE DESIGN Example 4.—Design a T-beam for an external shear, V = 134 000 Ib. anda bending moment M = 16 000 000 in.-lb., when the depth is limited to 60 in. and the thickness of slab is 5 in. The allowable stresses are fe = 650, fs = 16 000, i =oela. Solution.—The required width of stem for shear will be determined first. Since h = 60 in., d = 60 — 5 = 55 in, andv = 120 and, assuming j = 0.9, the required width of stem is 134 000 ~ 0.9X 55 xX 120 This value of b’ is accepted and the cross section of the T-beam is then as shown in Fig. 74, below. b’ = 92.6 in. Usebenz3 1a ! ! ‘ I | ! j | | ! tee ecceee phos Saas I “ye eoccce! : Loge Aar= 78d=9.9 89: In: ‘Agg = 12.0 89. In T-Beam Rectangular Beam Auxiliary T - Beam. Balanced Design ; Fic. 74.—Cross Section of T-Beam and Auxiliary Beams. (See p. 226.) Divided into two parts, the beam consists of a rectangular beam with b = 23 in., h = 60 in., d = 55 in, and an auxiliary T-beam with b = 80 in., = 5 in., and d = 55 in. For the specified stresses, for rectangular beam, the ratio of steel isp = 0.0077, | andj = 0.874. The corresponding area of steel is Ag, = 0.0077 X 23 X 55 = 9.7 sq. in. The moment of resistance of the rectangular beam, from formula M = idAsfs, iS M = 0.874 X 55X 9.7 X 16 000 = 7 462 000 inh. The auxiliary T-beam must resist the difference between the total bending moment and the moment resisted by the rectangular beam, or 17 500 000 — 7 462 000 = 10 038 000 in.-Ib. : For d = 55 in., t = 5 in, the ratio of thickness of slab to depth is b ty Pang Bo Tyee a for which, from table on p. 221, 4.=.0;,95. » The required area of steel, from formula As = af is jdfs 10 038 000 4. De ae ee >= 9 95 X 55 X 16 000 al DESIGN OF T-BEAM 227 To determine whether compression stresses are satisfactory, compare the area of steel, just computed, with the limiting area of stecl. The limiting area of steel corresponding to the allowable working stresses, from Formula (19a), p. 216, is As = bdpm = 80 X 55 X 0.0032 = 14.1 sq. in., where p» is taken from diagram, p. 894, corresponding to the stresses fe = 650, t fs = 16 000, n = 15, and NT 0.09. This is larger than the value found above; therefore, the design is satisfactory, as far as compression stresses are concerned. The total area of steel in the T-beam is A, = Asi + Ay ee ++ Teer? le7 Sq. in. Comparison of Result of Exact Formula and Approximate For- mula.—To get a comparison between the areas of steel and com- pression stresses computed in the previous example by the exact (simplified) method and by the approximate method, it is now necessary to compute the steel areas and compression stresses by the approximate method. Steel Areas.—If the compression stresses below the flange are not considered, the area of steel, from Formula (19), p. 216, would be 17 500 00 = ron x ba x16. 000 = A, It will be noticed that this area of tension reinforcement is smaller than the area A; = 21.7 sq. in., obtained previously by the exact (simplified) formula. This is easily explained by the fact that in. the approximate formula the moment arm, jd = 0.95d was used for the total bending moment, while in the exact formula the moment arm jd = 0.95d was used for the T-beam part and jd = 0.874d for the rectangular beam part of the T-beam. Compression Stresses.—The compression stresses, for the case where compression in the stem below flange is not considered, will be found from formula f, = C7f,;. The value of Cy will be taken from Table 13, p. 896, for corresponding In the present case, eA nla X 20.9. ae hee , is ee maven 0.6 and aoa 0.09, for which k = 0.4 and Cp = 0.044. The figured stress in concrete, therefore, would be f,; = 0.044 X< 16 000 = 704 Ib. per sq. in. This stress is about 8 per cent larger than the allowable stress, f. = 650, and to reduce the stresses to the 228 REINFORCED CONCRETE DESIGN allowable working stress it would be necessary either to increase the depth of the slab (since the depth of beam is limited) or to add compression reinforcement. By the use of the exact formula, how- ever, which correctly utilizes the compressive resistance below the flange, this necessity is avoided and an appreciable saving made in materials. Case 2.—Another use for this (simplified) method is for deter- mining the required flange area for a beam where the flange is pro- vided solely to increase the compression area of the beam. Such is the case in beams carrying joist construction, where the floor is thickened on both sides of the beam, and in isolated beams with projecting flanges. This method is not restricted to cases where the flanges are of uniform thickness, but may also be used with unusual flanges, as shown in Fig. 75, below. Lyte] As" Asr*Asz tAss|. ys{As1= ped ‘As2 As3-” T-Beam. Rectangular Beam : Auxiliary T - Beams Balanced Design Fic. 75.—T-Beam with Unusual Flange. (See p. 228.) In such cases, the T-beam is replaced by a rectangular beam and two T-beams, one of which has the thicker flange and the other the thinner flange. The beams are shown in Fig. 75. Example 5.—Design a beam to support joist construction consisting of joists 8 in. deep, for which external shear, V = 100000 lb., and bending moment, M = 15000000 in.-lb. The total depth of beam is limited to 54 in. The allowable stresses are fe = 650, fe = 16 000, n = 15,9 = 120. Solution.—Total depth of beam, h = 54 in., allowing 2 in. for fire proofing and assuming three layers of one-inch bars with one-inch separators between layers. The distance from bottom to center of steel is 2 + 2.5 =4.5in. The effective depth of beam is d = 54 — 4.5 = 49.5 in. For V = 100 000 lb., the width of beam required for shear is Va 100 000 “0 874 < 49-5 120 b = = 19.2in.:, Use 20 m: DESIGN OF T-BEAM 229 For M = 15 000 000 in.-lb. and the depth d = 49.5 in., the required width of a rectangular beam, from Formula 2, p. 204, is M 15 000 000 b = oh ee tet mantic Rd? 106 X 49.52 - in. This width required by bending moment is much larger than the width required by shear. A rectangular beam would be too expensive and too heavy. It will be cheaper to use a T-shaped beam in which the width of the stem is equal to the width required by shear and compression is supplied by a flange. The required dimensions of this T-beam will be found by considering the compression below the flange. The simplified method will be used, assuming the T-beam replaced by a rectangular beam and an auxiliary T-beam consisting of the two projecting flanges on each side of the stem only. The dimensions of the rectangular beam are already known. They are: b = 20 in., d = 49.5 in. (6 of rectangular beam becomes 6’). For f, = 650, fs = 16 000, the ratio of steel is p = 0.0077 and 7 = 0.874 (see table, p. 205). The area of steel for the balanced rectangular beam, from formula A, = pbd, is map = 00017 X 4075 xX 20 = 7°6 sq. in. The moment of resistance of the beam, from M = A,jdf,, is M, =7.6 X 0.874 X 49.5 X 16 000 = 5 030 000 in.-lb. The bending moment to be resisted by the auxiliary T-beam is obtained by subtracting from the total bending moment the moment of resistance of the rectangular beam, : M, = 15000 000 — 5 030 000 = 9 970 000 in.-lb. It is now necessary to decide upon the thickness and the width of the flange. To simplify the formwork, a thickness of flange equal to the depth of the joist will be accepted. Therefore, ¢ = 8 in. The width (6 — 0b’) will be computed. t For d = 49.5 in. and t = 8 in., the ratio Ee 0.16, for which, from table on ) M p. 221,7 = 0.93. ‘The area of steel, from A, = —, is jdfs 9 970 000 hi Sei a * 0.93 X 49.5 X 16 000 = 13.5 sq. in. The width of flanges of the auxiliary T-beam is obtained from the fact that the above area of steel is equal to the maximum allowable steel area permitted by compression stresses. The equation for maximum allowable steel area in a T-beam is As. = bdpm, : t in which pm is a constant obtained from diagram p. 894, for known f;, fs, and 7 In this case, pm = 0.0051 and the steel area A,z = (6b — b’) X 49.5 X 0.0051. 230 REINFORCED CONCRETE DESIGN Equating this to the steel area computed above, 13.2 = (b — b’) X 49.5 X 0.005, from which 13.5 b— db! = —— = 53 in. 0.253 To get the total width of flange, it is necessary to add to the above value the width of stem b’ = 20 in. Hence b = 52 + 20 = 72 in. Final Design, Concrete dimensions, b = 72 in.; b=" 20 im.; a’ =49.5 ins h = 54. Area of steel A =Agi t+ Ag = 7.6 + 12.9 = 20.5 8q. in. T-BEAM WITH COMPRESSION REINFORCEMENT Use of Compression Reinforcement in T-beams.—Compression stresses are seldom a controlling factor in T-beam design. Occa- sionally, however, the stresses may be larger than can be taken care of by the concrete, and compression reinforcement may then be advantageous. | Compression reinforcement is required when not only the depth of the beam but also the available width for the flange is limited. One such case is that of beams supporting balconies in theater con- struction. | Compression steel may be useful also in a series of T-beams of similar design, of which one beam carries a larger load than the others. If the compression stresses in all but the heavy beam are satisfactory, it is cheaper to add. compression steel rather than to change the concrete dimensions. Formulas to Use.—In designing T-beams requiring compression reinforcement, formulas taking into consideration the compression in the stem below the flange should be used. The simplified method, in which the T-beam is considered as a combination of a rectan- cular beam and an auxiliary T-beam, may be used to advantage. The rectangular beam will be one with compression reinforcement. Problem to be Solved.—The problem in this connection usually is: Given the depth of beam, d, and the width and thickness of flange, b and t, and the width of stem, 6’. Determine the required tensile and compression reinforcement for a given bending moment, M, and specified unit stresses, fs and fo. (See Fig. 76, p. 231.) The problem is solved by computing the moments of resistance, M,, of a balanced rectangular beam without compression rein- T-BEAM WITH COMPRESSION REINFORCEMENT | 231 forcement. For the specified stresses, the steel ratio, p, is taken from the table on p. 880. Then the area of tensile reinforcement equals As; = pb/d, and the moment of resistance is My = Asijdfs. Next, the moment of resistance, M2, is computed for the aux- iliary T-beam. ‘The tensile area is A52 = Pmbd, Compression Steel Ag Tae a “Vv -~Tension Steel o , 1 A,=A,, + Ago t Ags hese T - Beam with Compression Reinforcement | = AR eer arr RK baa shir | ' ‘ ' a Vibe tices } 8 1 bo ! ! i! a=: an he Luo aL Leet ee Se Ae (see f Aggy Rectangular Beam Auxiliary T-Beam Steel Resisting Couple Balanced Design Elements Replacing T - Beam with Compression Reinforcement Fic. 76.—T-Beam with Compression Reinforcement. (See p. 230.) where pm is taken from diagrams 3 or 5 for the specified stresses and the proper value of . The moment of resistance 1s Mz = A.ojidf., where 7; 1s value from Table 13, p. 897. Finally, the bending moment to be resisted by the compression steel is equal to the difference between the total bending moment to be resisted, M, and the sum of the bending moments, M; and Mo, resisted by the rectangular beam, and the auxiliary T-beam, respect- ively. It is, therefore, M3 = M — (M; + Mo). 232 REINFORCED CONCRETE DESIGN This bending moment may be considered as resisted by a couple, composed of compression resisted by compression steel and tension resisted by tensile steel. The moment of this couple is Mz = A,3d(1 — a), where d(1 — a) is the moment arm. From this : ee _ M — (Mi + M2) Asay 9) oT dl (28) The total amount of tensile steel is A, = Aa + Ait Ale pt (29) The compression reinforcement may be found from formula Teil 1—k Alm Mesias) tile i eee where k is the ratio of neutral axis and a the ratio of depth of com- pression steel. RECTANGULAR BEAM WITH COMPRESSION REINFORCEMENT Use of Compression Reinforcement.—When, in a beam of fixed concrete dimensions, the compression stresses in the extreme fiber are larger than the allowable unit stresses, the beam may be strength- ened by the introduction of reinforcement in the compression zone. This compression reinforcement resists fifteen times the stresses (for n = 15) which were resisted by the replaced concrete and thus reduces the stresses in concrete. To be effective, the bars composing the compression reinforcement must be straight and they must extend a sufficient distance beyond the support to transmit the stresses from the bar to the support. PROBLEMS IN CONNECTION WITH CoMPRESSION REINFORCEMENT The following problems may occur: Problem 1.—Concrete cross section fixed. Determine area of tension and compression reinforcement for given bending moment and specified working stresses. Problem 2.—Ratio of area of compression steel to area of tension steel fixed. Find required dimensions of beam for given bending moment and specified working stresses. RECTANGULAR BEAM WITH COMPRESSION STEEL 233 Problem 3.—Concrete dimensions and areas of steel given. Find stresses in steel and concrete for any given bending moment. Problem 4.—Concrete dimensions and areas of steel given. Find moment of resistance of beam, and from this the load the beam is capable of carrying for specified working stresses. Solution of the Problems.—All these problems may be solved by using formulas given in the section on theory (see p. 137). These formulas, however, are too complicated. Easier Solutions are Provided by Two Simplified Methods, Devised by Mr. Edward Smulski and Given Below.—These methods give exact results, require less work, and are easier to understand and to remember than the elaborate formulas. By the first method, the required areas of compression steel may be obtained directly. Problem 1 above may be easily solved by this method. The method is especially valuable when handbooks (such as this volume) with tables of constants are not available. The second method, a modification of the first method, gives the required ratio of compression steel, p’, when the computed ratio of tensile steel, p1, exceeds the value, p, permissible for a simple beam; i.e., when the area of concrete is too small to resist safely all of the compression. This method may be used for solving Prob- lems 1 and 2, either directly or in conjunction with diagrams, p. 904. It is particularly useful for determining compression steel needed for negative moment in a continuous beam. Problems 3 and 4 are best solved by using the diagrams, pp. 908 and 909, as explained on p. 228 and illustrated by examples opposite each diagram. Let A, = area of cross section of tensile steel in beam under consideration ; Asi = pbd = area ® of tensile steel in balanced: simple beam in which steel and concrete are stressed to specified allowable unit stresses, f, and f.; p = steel ratio for balanced design of beam with tensile steel only; = ratio, in beams with stecl in top and bottom, of total cross section of steel in tension to cross section of beam, bd; ) as 5 The limiting value for which no compression steel is required. 234 REINFORCED CONCRETE DESIGN p’ = ratio of cross section of steel in compression to cross section of beam, bd; k = ratio of depth of neutral axis to depth of beam, Jie a = ratio of depth of compressive steel to depth of beam; Awe = A, — As = extra tensile steel, in beam with steel in top. and bottom, to be balanced by compression steel ; A’, = area of compression steel required to balance extra tensile steel = p’bd; M = moment of resistance or bending moment for beam with steel in top and bottom; M, = moment of resistance of beam without compressive reinforcement. First Method. Simplified Method of Design of Beam with Steel in Top and Bottom.—As evident from Fig. 77, the position of the neutral axis for a beam with steel in top and bottom is the same as for a rectangular beam of the same dimensions and subjected to the same stresses, f, and fs. The beam with steel in top and bottom may, therefore, be considered as a combination of (1) a rectangular beam without compression steel and (2) a couple of equal forces, one of which forces is the compression resisted by the compression steel and is equal to A’s f’s, while the other force is the tension resisted by tensile steel and is equal to Asofs. The moment of resistance, M, of a beam with steel in top and bottom may therefore be separated into two moments: (1) the moment of resistance, Mj, of a rectangular section of balanced design with bottom reinforcement only, and (2) Me, the moment of resistance of the couple of forces. By a beam of balanced design is understood a beam with an area of tensile reinforcement, As = pbd, such that both the maximum tensile and the maximum compression stresses are reached at the same time. The moment of resistance of a simple beam of balanced design is M, = f:Asjd in which Asi = pbd. The value, p, is a definite ratio depending upon the specified values Ofte a... a0 J The moment of resistance of the couple of forces is obtained by multiplying the tensile force, As2fs, by the moment arm d(1 — a), Ms: — Af dl = a). RECTANGULAR BEAM WITH COMPRESSION STEEL 235 The value of M; is fixed for any size of beam and specified stresses by the strength of the concrete. The value of As; is also fixed. The value of Mz depends only upon the tensile area As2 and the corresponding compression area A’;. Within practical limits it may be made of any magnitude. The moment of resistance of a beam with steel in top and bottom is M = M, + Mz. t ] Res a: Compression Steel A, wins - f- ; | \ ro loo Cc _ Compression Taare ed eee = in Steel hs | Potts Neutral ~-Total eae 3 lage \ Compression LS WNW X1S \ : Sere ‘Comp. in Concrete | | 2 Arr eA AE es --Tension Steel -- b-->| A,= pbdtA,, Beam with Steel in Top and Bottom d(1-k) Neutral = Axis ea ~~ d (1 - k) rs | Benin steel) ee Na- He in podf, ee : Ago f, Rectangular Beam Balanced Design Steel Resisting Couple Elements Replacing Beam with Steel in Top and Bottom Fic. 77.—Beam with Steel in Top and Bottom. (See p. 234.) The above relation may be used for solving Problem 1, in which the dimensions, b and d, are given and it is desired to find the areas of tensile and compression reinforcement to resist a bending moment, M, with specified stresses fs, fc, and n. The ratio of steel, p, corresponding to the specified stresses, is found first, either from table on p. 880 or from Formula (5), p. 130. . (For instance, for f; = 16000, f. = 800, n = 15, the steel ratio is p = 0.0107.) The area of steel is Asi = pbd and the moment of resistance of a balanced rectangular beam without compression reinforcement is M, = jdAaf, = jdpbdf,. 236 REINFORCED CONCRETE DESIGN | Subtract this value from M: The remainder, M — Mi, must be resisted by the steel couple whose moment of resistance 1s Mos = Asefd aa a). Thus, A,fd(1 —a) =M—M, and M— M, The total area of tensile steel is 2 oe M—M, A, = AA -- Ag — pbd ct. aera este (32) The area of compression steel is Be (Neco x < Aa Aw ee (33) This method is exact. It does not require any tables. The values of p and k corresponding to the specified stresses are the only constants required. Second Method. Determining Ratio of Compression Steel, p.— The method given below may be used for determining the required ratio of compression steel, p’, if the ratio of tensile steel, 1, is larger than the allowable ratio, p, for a balanced simple beam. ‘This is a modification of the previous method. It is particularly useful in continuous beams at the support. For any accepted stresses, fc and fs, and the ratio n, the location of the neutral axis is the same for beams with compression rein- forcement as for simple beams. The formula for tensile steel ratio for beam with compression reinforcement is (see Formula 41, p. 189). ke? k—- a aso 2n(1 — k) TP ocae ° ke Pweg : : The first term in this formula, nd — is identical with 2 Formula 6, p. 130, p = mF in which p is the ratio of tensile * The formula for the area of compression steel is obtained from the require- — ment that the total tensile stresses in the couple of forces (see Fig. 77) must be equal to total compression stresses. The ratio of unit stresses in tension and compression is equal to the ratio of their distance from neutral axis. RECTANGULAR BEAM WITH COMPRESSION STEEL 230 steel for a balanced simple beam. This may be substituted in the above equation and jk—a Php Palliy?.qeaty From this: | Required ratio of compression steel, ll (34) = (p This formula gives the ratio of required compression steel if the ratio of tensile steel, p:, necessary to resist a bending moment, is larger than the value of p corresponding to the specified stresses in a balanced beam without compression steel. For instance, assume that the computed ratio of tensile steel to resist bending moment is p; = 0.0125. If the allowable stresses are f. = 750, fs = 16 000, and n = 15, the allowable ratio without com- pression reinforcement is p = 0.0097, and value of k = 0.414. The required ratio of compression steel is then 0.414 = (0.0125 — 0. UND eres The value of a is the ratio of the distance of the center of com- pression steel from the surface to depth of beam. The amount of ee steel (and from this the value pj) is found from formula A, = = ip in which j is assumed as 0.9. This value of j is near enough for practical purposes. If desired, after computing the ratio of compression reinforcement, the correct ratio of j may be taken from the diagram on p. 910 ciate the values of pi and p’ revised accordingly. Solving Problems 1 to 4 (p. 232) by Means of Diagrams.— Formula (34) for the required compression steel was used in _pre- paring Diagrams, p. 904. From these the required ratio of com- pression steel may be taken directly for any ratio of tensile steel and different allowable stresses, f. and f,. Problem 1 may be solved by using Diagrams as follows: iki this problem it is required to find the amount of tensile and compression steel for given moment and stresses. For the given bending moment, the area of tensile steel is computed by formula A, (assuming M ~ 0. 9df, 238 REINFORCED CONCRETE DESIGN 7 = 0.09). From this p: is found. The value a is estimated. Then the required ratio of compression steel is taken directly from the diagram corresponding to the specified stresses, fs and f-, and to the value a. Diagrams, p. 908, may also be used to solve the problem. Problem 2 also may be solved by means of diagrams. In this problem, dimensions of beam are required, with the ratio of 7 fixed 1 and stresses f. and f, specified. In solving, the value of a is estimated first. Then reference is made to the section for the specified stresses and the column for the nearest value of a. From this, a value of p’ is found which has the required relation to p:. For instance, if for f. = 16 000, f. = 750, n = 15, it is required to get a design in which / ratio of aa 1, when a = 0.08. In the section for the specified f, and f, and on the curve for a = 0.08, it is found by inspection that a value of p’ = 0.016 corresponds to pi = 0.016. These values satisfy our requirement and may be used in design. (See example, p. 903.) For solving Problems 3 and 4, with given concrete dimensions and steel areas, where the relation between the stresses f. and f; 1s not known, the diagrams on p. 904 cannot be used advantageously. For this purpose, the diagrams on pp. 908 and 909 were prepared; from these the relation between the steel ratios, pi and p’, may be found for any ratio of unit stresses of ie and any value of a, and a 15fe of steel ratios, pi and p’, and any value of a. In solving Problem 3, where it is required to determine the stresses for known dimensions of beam and known areas of steel, the values of p1, p’, and a are computed first. The tensile stresses may conversely the ratio of stresses may be found for any combination then be computed from the formula f; = pee (assuming j = 0.9). s The ratio of unit stresses, oe is found from the diagram for nearest value of a and for the known values of pi and p’. The tensile stress, fs, and the ratio, 7 being known, the compression stress is easily found by dividing the stress f, by 15 times the determined value for & RECTANGULAR BEAM WITH COMPRESSION STEEL 239 Problem 4, in which moment of resistance is required for known dimensions and specified unit stresses, may be solved as follows: The moment of resistance may be governed either by the steel stresses or by the concrete stresses, whichever are smaller. The simplest method is to determine the moment of resistance governed by steel stresses and then find whether the compression stresses for this moment are within the specified working limits. The moment of resistance governed by steel may be found directly from formula M = 0.9dA,f., in which all values are known. (The value of moment arm is assumed j = 0.9.) To find compression stresses correspond- ing to this moment, compute, from known areas of steel and concrete, the steel ratios p; and p’ and ratioa. From the diagram for value of i corresponding to the steel ratios, p1 and p’. The compression stress in concrete may now be found by dividing the stress in steel, f,, by 15 times the a nearest to the computed value, find the ratio ratio ee, which was found from the diagram. If this compres- sion stress is smaller than the specified stress, the moment of resist- ance of the beam, as determined by tensile steel, is satisfactory. If the compression stress in concrete is larger than the allowable stress, it signifies that the moment of resistance of the beam is governed by compression stresses; hence the moment of resistance determined from tensile steel is too large, and the load the beam is able to carry is lighter than would follow from this moment of resistance. If the beam is already built, the moment of resistance governed by the concrete is computed, and from this is found the load the beam is able to carry. For this moment of resistance, the stress in if aie 15, is already known. The reduced stress in steel may be computed by multiplying the allowable stress in concrete by n and by the ratio fs 15f. is the moment of resistance as governed by compression stresses. If the beam is not built, the moment of resistance as governed by concrete should be increased by adding compression steel. Or, if the moment of resistance is sufficient, the amount of steel should be steel is smaller than the maximum allowable. The value The moment of resistance based on this reduced stress in steel reduced. The tensile stress may be found from the diagram oe 7 , by 240 REINFORCED CONCRETE DESIGN substituting the value of f.. The value of the reduced f, then follows, and this will give the required moment of resistance. Moment Arm.—The magnitude of the moment arm for a beam with compression steel is different from the magnitude of the moment arm for a simple beam. Its exact value is Moment arm, Beam with Steel Top and Bottom, ae k op’ (k x —a This value may be obtained from the diagram on p. 910 for differ- ent ratios of stresses, z, and ratios of * In practice, 7 = 0.09 aE 1 may be accepted. After p’ and pi are computed, the value may be corrected by referring to the diagram. EXAMPLES OF BEAMS WITH STEEL IN TOP AND BOTTOM Example 6.—Given size of beam, b = 10 in. andd = 18 in. as determined by architectural requirements; bending moment, M = 500 000 in.-lb.; allowable unit stresses, fs = 16 000 Ib. per sq. in., fe = 700 1D. per eq. in., 7 = 16. Find required amount of tensile and compressive steel. Solution.—For f; = 16 000 lb.; f¢ = 750 lb. y trom table on p. 880, k = 0.414; j = 0.862; and p = 0.0097. The area of steel in a beam without compression steel is) As: = LOX 18 >< 0.0097 = 1.75 sq. in., and the moment of resistance is M, = jdAeifs = 0.862 X 18 X 1.75 X 16 000 = 435 000 in.-lb. By com- paring this moment with the bending moment of 500 000 in.-lb., we find a differ- ence of M — M1 = 500000 — 435 000 = 65 000 in.-lb. This must be provided for by extra tensile steel and by the corresponding amount of compressive steel. Since center of compression steel is distant ad = 2.25 in. from the top of beam, a = 0.126 and distance between tensile steel and compressive steel is d(l ua): S186 Ors ite 15.73 in. This is moment arm of couple composed of compression stress in steel and tensile stress of extra amount of tensile steel. Required extra amount of tensile steel, Ago, will be found by dividing, extra bending moment, M — M,, by dl — a)fs; 65 000 hence As: = = 0.26 sq..in. The total tensile steel therefore 15.73 X 16 000 equals As = Asi + Ag — 178 + 0.26 = 20 sq. in. From Formula (33), required amount of compression steel is —& 1 — 0.414 se | tad If eer 126i 3 = 0.53 sq. in. gray (eam A as) has DIAGONAL TENSION 241 Example 7.—Find required amount of compression steel for a T-beam at the support, where b’ = 12 in., d = 22 in., As = 4.0 sq. in. and the specified stresses are fs = 18000 and f, = 900, n = 15. Distance from top of beam to center of ¢ . e compression steel, dj = 2} in. and a= = 0.102. Solution.—At the support, the T-beam changes into a rectangular beam (see p. 282). By dividing area of steel by b/d, ratio of tensile steel, 4.0 = ——— = ().0152. 12 X 22 Pi From table on p. 880, the ratio of steel for beams without compression steel, for stresses, fs = 18 000, and f, = 900, is p = 0.0107 and k = 0.429. Since ratio pi = 0.0152 is larger than p, compression steel is required. Ratio of required compression steel, from Formula (34), p. 237. — 0.429 0.571 | ean ’ = (0.0152 — 0.0107 = 0.0045 x —— = 0.0079. Beil * 0.825 ui 0.429 — 0.104 Therefore, an area of compression steel A’, = 0.0079 x 12 x 22 = 2.1 sq. in, is required to keep compressive stress in concrete within working limits of 900 lb. per sq. in., when the stress in steel is 16 000 lb. per sq. in. DIAGONAL TENSION Diagonal Tension Stresses.—In reinforced concrete members subjected to bending, diagonal tension stresses must always be com- puted, and if they exceed the allowable unit stresses for plain: con- crete, web reinforcement must be provided. All members subjected to bending develop, in addition to the tension and compression stresses due to the bending moment, ten- sion stresses in the web acting at an inclined plane and mainly caused by shear. In homogeneous beams, these stresses do not require any special attention. In reinforced concrete, on the other hand, they assume considerable importance. Full discussion of diagonal tension is given in the chapter on Theory, with which the designer should familiarize himself thoroughly. Formulas to be used in design and recommendations for their practical application are given below. Vertical Shearing Stresses as Measure of Diagonal Tension.— Concrete is strong in direct shear and capable of withstanding (in 2000-lb. concrete) a working shearing stress of at least 200 Ib. per sq. in. Concrete beams or slabs, therefore, always have suffi- cient area to withstand direct shear. 242 REINFORCED CONCRETE DESIGN However, since shearing stresses have been accepted as a measure of diagonal tension stresses, and the resistance of concrete to diagonal tension is small, the unit shearing stresses must always be com- puted. If the stresses exceed the values for plain concrete, web reinforcement must be provided to take care of the excessive diagonal tension stresses. Usefulness of Web Reinforcement.—Numerous tests have demonstrated that a beam properly reinforced with stirrups or bent bars sustains three or four times as much load as the same beam without web reinforcement. This effectiveness in increasing the strength of the beam is unquestionable. There is a peculiarity in the action of the web reinforcement, however, particularly of the stirrups, which has sometimes created confusion and doubt—not as to their effectiveness for high load, but as to their usefulness in beams designed for ordi- nary working conditions. ‘Tests show that, until diagonal cracks appear, a reinforced concrete beam acts like a homogeneous beam. Moreover, the measured stresses in web reinforcement are small, and in some cases compression stresses, instead of tensile stresses, have been found in stirrups. The stirrups do not come into action, in fact, until after diagonal cracks develop. Since the allowable unit stress for diagonal tension is usually somewhat lower than the tensile strength of concrete, it can easily happen that the stresses in a beam will never be sufficient to crack the concrete and bring the stirrup into action. This fact, then raises a question as to whether there is any obiect in providing web reinforcement which may never be called upon to carry stresses. The answer is obvious. Reinforced con- erete beams are not designed for the condition at working loads, but for the condition at working loads multiplied by the factor of safety. (See also p. 125.) Web reinforcement should always be used, therefore, when unit stresses exceed the allowable stresses for plain concrete, irrespective of whether or not diagonal cracks are likely to occur under working loads. In beams without stirrups, final failure closely follows the appearance of the first crack; while with beams having web reinforcement, stirrups and bent bars represent a factor of safety which allows stressing of concrete in diagonal tension up to its ultimate strength. Under working loads, the stirrups may not act; but in case of overstressing, due to faulty construction or to occasional excessive loading, the stirrups prevent the failure of the DIAGONAL TENSION 243 beam. The minute cracks that may develop are not dangerous and in many cases are hardly visible. They close after the excessive load‘is removed. Description of Web Reinforcement.—Web reinforcement may consist of (a) vertical or inclined stirrups; (6) bent-up tension bars; (c) a combination of stirrups and bent-up bars. Stirrups are steel members, usually U-shaped, made of bars of comparatively small diameter and independent of the main rein- forcement, extending from the tensile zone into the compression zone and anchored at both ends. Fig. 78, p. 243, shows different types of stirrups. ald to keep Stirrups in a _--Bars to keep Stirrups i. in Position.. Toor t Fig. 78.—Types of See (See p. 243.) Vertical Stirrups.-—Vertical stirrups are most commonly used. Although not placed in the theoretical direction of diagonal tension, they resist the stresses effectively. They are secure against slipping in a horizontal direction and are easy to keep in position during construction. As contrasted with bent-up bars, the number and spacing of stirrups can be varied to suit conditions. Inclined Stirrups.—Theoretically, the most appropriate rein- forcement to resist diagonal tension stresses would consist of inclined stirrups placed in the direction of the stress. The inclination of diagonal tension varies in different parts of the beam, being steepest near the end of the beam and flattening out toward the center of the beam, so that to follow the direction of the stress the inclination of the stirrups would have to be varied throughout the beam. This is impracticable. If used at all, the inclined stirrup should be inclined about 40° with the horizontal. The objection to inclined stirrups is that they are difficult to attach to the tensile reinforce- ment and to keep in place. Therefore, in spite of the superiority resulting from a closer coincidence between the direction of the stress and the direction of the bar, for practical reasons they are seldom used. Bent-up Bars.—Bent-up bars, if properly distributed and inclined, _ form excellent web reinforcement. Like inclined stirrups they have the advantage resulting from closer coincidence with the direction 244 REINFORCED CONCRETE DESIGN of the diagonal tension stresses, and in addition they have the advantage of being rigidly connected with the main reinforcement. When bent horizontally at the top of the incline, and extended into the adjoining span or into the support, the bent bars have suf- ficient anchorage, so far as their function as web reinforcement is concerned. If not so extended, they must be provided in the compressive zone with hooks of proper dimensions to develop their tensile strength.- The only disadvantage of bent-up bars lies in the fact that their number in a beam is small and that the points of bending up may be governed by the bending moment requirements which, in turn, do not coincide with the requirements of diagonal tension reinforcement. Stirrups and Bent-wp Bars.—The most effective diagonal tension reinforcement consists of a combination of bent-up bars and prop- erly placed vertical stirrups. Anchorage of Web Reinforcement.—In the tensile zone of the beam, the web reinforcement must be continuous with the tensile reinforcement, as in bent-up bars; or, if stirrups are used, it is neces- sary to anchor them to the tensile bars by bending the stirrup around the lowest layer of tensile reinforcement in such a way as to develop the tensile value of the stirrup at the level of the tensile reinforce- ment. In the compression zone, the web reinforcement should be brought as near the compressed surface as possible. It should be provided with hooks of sufficient dimensions to develop, in con- nection with straight imbedment, the full tensile value of the bar in a distance 0.3d from the compression face of the beam. A semi- circular hook with a radius not less than four times the diameter of the bar will be found satisfactory. In continuous beams at the support, the tensile zone is at the top and the compression zone at the bottom of the beam. | Allowable Unit Stresses for Diagonal Tension.—The following unit stresses are recommended for concrete testing 2 000 Ib. per sq. in. at the age of twenty-eight days. This strength corresponds to the | : 2:4 mix used in general practice. (a) In beams without web reinforcement, the maximum unit shearing stress (being measure of diagonal tension) computed from - Formulas (36) and (37) must not exceed 40 Ib. per sq. in. (b) In beams provided with web reinforcement, the maximum unit shearing stress (being measure of diagonal tension) computed © , as before must not exceed 120 Ib. per sq. in. Not more than 40 lb. DIAGONAL TENSION 245 per sq. in. of this stress may be considered as resisted by concrete, the balance being considered as resisted by web reinforcement. (c) In beams without web reinforcement, but where tensile reinforcement is anchored at the ends,® the maximum unit shearing stress, computed as before, may be 60 lb. per sq. in. (d) In beams with web reinforcement, where at least 50 per cent of tensile reinforcement extends the full length of the tensile zone and is anchored at the ends,° the maximum unit shearing stress may be increased to 180 lb. per sq. in. Not more than 60 lb. of this stress may be considered as resisted by concrete, and for the rest web reinforcement must be provided. The increased unit stress must not be used where too thin members would result. In continuous beams in the region of negative bending moment the top reinforcement must be anchored by extending beyond the point of inflection, to permit the increased unit stresses under (c) and (d). The 1924 Joint Committee allows a unit shearing stress of 240 Ib. per sq. in. for beams with end anchorage. This stress is considered by the authors as excessive, because it is likely to give thin members so filled with reinforcement as to interfere with proper placing of concrete. Moreover, such members would be particularly vulner- able in case of fire. The allowable unit stresses recommended above may _ be expressed in terms of the ultimate compressive strength of con- crete,’ f',, as follows: 40 lb. = 0.02 f’.; 60 Ib. = 0.03 f’.; 120 lot 0.06.7 <5 180: 1b. = 0.09’; and. 240 Ib. = 0.12 f'.. Thus, for concrete of greater compressive strength, f’., than 2 000 Ib. per sq. in., the allowable stresses may be increased proportionally by using the corresponding ratios. The allowable unit shearing stresses should be used in deter- mining the smallest allowable area of a beam. In beams of large proportions carrying considerable load, also in beams subjected to dynamic loads, such as those in bridges, the whole length of the beam should be provided with web reinforcement even if not required by unit stresses. Distribution of Diagonal Tension between Concrete and Web Reinforcement.—Tests indicate that diagonal tension is resisted 6 Required anchorage is same as specified under “ Working Stresses, Bars with Anchored Ends,” p. 263. 7 Cylinder strength of concrete at 28 days. 246 REINFORCED CONCRETE DESIGN by the concrete in the web and by the web reinforcement. The proportion resisted by each material is not definitely known; there- fore, in determining the amount of web reinforcement, part of the stresses to be carried by the steel must be assumed as discussed in the chapter on Theory of Reinforced Concrete. Two methods of distributing the stresses are in common use, namely : Method 1.—Conerete takes all stresses when the unit stresses are equal to or less than the working unit stress specified for plain concrete (same as in method 2). Where the working unit stress for plain concrete is exceeded, the concrete is assumed to resist an amount of shearing stresses equal to the allowable stresses for plain concrete, and the steel is assumed to resist the balance. Concentrated Loads ~. b “I Weight of Beam | Vo Stirrups | 5 = Ere ra & { . No Stirrups 1S : S Required ae NGO ane % | by Web sae chal (S = Reinforcement © iS ment 5 | emer ey |S g jue Resisted by Concrete ia ae ihe Aesiste Concrete 1S Woe ne Ss ee | a Uniformly Distributed Loading Concentrated Loading Fic. 79.—First Method. Distribution of Diagonal Tension between Concrete and Web Reinforcement. (See p. 246.) Method 2.—Concrete takes all stresses when unit stresses are equal to or less than the working unit stress specified for plain concrete. For sections where this working unit stress for concrete is exceeded, the concrete resists one-third and the steel two-thirds of the diagonal tension. The first method requires somewhat less web reinforcement than the second method. The required spacing of stirrups at the points of maximum shear is fairly alike in both methods, since here the stresses attributed to concrete are about the same in each case. The difference is more noticeable at points of small shear, where the first method would permit larger spacing of stirrup; but here it is apt to be offset by the rule which limits the maximum spacing of stirrups. FORMULAS FOR DIAGONAL TENSION 247 Either method may be used with satisfaction. The first method is recommended by the authors. The 1916 Joint Committee recom- mended the second method, while the 1924 Joint Committee recom- mends the first method. FORMULAS FOR DIAGONAL TENSION NOTATION V = total external shear, in pounds, at section considered; v = total unit shearing stress at section, lb. per sq. in. , alone, lb. per sq. in.; 1 zx = { S 7 No Stirrups EB ae Ss Required ~.. > ! "8/9, 4 5| © 1 ee ei ee 5 1 Con> v' 3 H Resisted by . S Web Reinforcement Uniformly Distributed Loading niles Resisted v’ = allowable unit shearing stress (or diagonal tension) on concrete Concentrated vw boads-. Weight of Beam, < by Concrete Wo Stirrups A Required} 5 mae £ Center Line of Span raed l 3,"2 by ‘orcement Concentrated Loading Resisted Wed Rei. Fic. 80.—Second Method. _ Distribution of Diagonal Tension between Concrete and Web Reinforcement. (See p. 246.) jf’. = ultimate unit compression stress of cylinder of same concrete at 28 days; o | = allowable unit stress in stirrups, lb. per sq. in.; A, = cross-sectional area of all legs of a vertical stirrup in square inches. (In a U-stirrup this is the sum of the area of the two legs) ; gd = moment arm, or distance in inches from center of compression to center of horizontal reinforcement. (In a T-beam, this may be taken, for diagonal tension computation, as distance between center of slab and steel; in a rectangular beam, as 0.87 of the total depth of steel) ; ~ I breadth of beam in inches; 248 REINFORCED CONCRETE DESIGN b’ = breadth of web in T-beam in inches. s = spacing of stirrups, in inches, at section considered (Figs. 81 and 82). | Unit Stresses.—The unit shearing stresses at any point, where the external shear is V, may be found by formulas given below. These values should not exceed the allowable stresses given in pre- ceding paragraph. Shearing Unit Stress in Rectangular Beam, Vv pid (36) Value of jd may be ordinarily taken as id. Shéaring Unit Stress in T-beams, Vy i b’jd a peel ee Pio ck toate ann (37) Value of jd may be taken as d — ; but not less than 7d. In irregular sections, for b and b’ in above formulas, the smallest dimensions should be used. Area and Spacing of Web Reinforcement.—Formulas are given below for area of cross section and maximum spacing of web members at any point in the beam where the external shear is V. Two alternate assumptions are given as to the distribution of the diagonal tension between the concrete and the web reinforcement. In both cases the external shear, V, is the average value in the length of beam equal to the spacing of stirrups. 1. Concrete assumed to resist a definite amount of diagonal ten- sion, v’, and the web reinforcement the rest. (See Fig. 79, p. 246.) Required Area of Web Reinforcement in a Distance s, Sennen ine! A; fs s for vertical stirrups. . . . . (88) Ales j s for inclined members. . . (89) ° FORMULAS FOR DIAGONAL TENSION 249 Spacing of Web Reinforcement having a Cross Section, A., 42518 | ; Seis. FOR VENLICAl SUIsTUDS. | sy + yy (40) ee ae v’b qd A sf s= 1.435, for inclined members: . . . (41) ——vb jd 2. Concrete assumed to resist one-third of the diagonal tension and the web reinforcement the rest. (See Fig. 80, p. 247.) Oe 1a. Stirrup Supports Z \ s ~Tensile Bars Spacings of Stirrups Fic. 81.—Spacing of Stirrup. (See p. 248.) Neutral Axis Spacing of Bent Bars _ Fig. 82.—Spacing of Bent Bars. (See p. 248.) Required Area of Web Reinforcement in a Distance, s, ee 1 aad SOT VOEUICa) SULT DS auld ear ee ee A, = 047 i, for inclined members... . . . (48) Spacing of Web Reinforcement having a Cross Seciion, Az, ie a ‘for vertical stirrups. . . . . (44) jd yee a eas ols for inclined members. . . . (45) jd 250 REINFORCED CONCRETE DESIGN The area of reinforcement, As, in these formulas is the sum of the areas of all prongs of the stirrup or all bars bent up at one place. The above formulas may be used for angles of inclination of the bar with the beam axis of from 25 to 50°. For larger angles use formulas for vertical stirrups. Tests discussed on p. 42 prove that there is no difference in effectiveness between bars bent at different angles within this limit. The 1924 Joint Committee recommends more elaborate formulas based on the use of actual angles. Different: formulas are used for small and large angles. This refinement is hardly warranted. Limiting Spacing of Stirrups.—In the section of the beam where the unit shearing stresses exceed 60 per cent of the maximum allow- able unit stress, the spacing of stirrups should not exceed 3d. Else- where the spacing should not exceed {d. Diagonal tension in this section not provided for Fic. 83.—Limiting Spacing of Bent Bars. (See p. 2505) Limiting Spacing of Bent-up Bars.—To make the bent bars fully effective as diagonal tension reinforcement, the spacing of the bars, measured along the axis of the beam, should not be larger than the effective depth of the beam. For larger spacing, or where only one bar in a beam is bent up, the region of effectiveness of the bent bar as diagonal tension reinforcement extends, on both sides, from the point of intersection of the bar with the neutral axis for a dis- tance along the beam equal to one-half of the depth of the beam. The effect of the bent bar in resisting diagonal tension cannot exceed the amount of shearing stresses to be resisted within the spacing specified above, irrespective of the strength of the bar. A bar bent up at a flat angle (around 15° with the horizontal) may be considered as effective for a distance equal to twice the dis- tance given above. FORMULAS FOR DIAGONAL TENSION 251 Procedure in Designing Web Reinforcement.—Determine the maximum total shear, V, and from this the unit shearing stress, v. See that v does not exceed the maximum allowable stress. Vertical or Inclined Stirrups.—If vertical or inclined stirrups only are used, select the diameter and shape of the stirrup so that the minimum spacing is not too small (preferably not less than 4 in.), and the total number of stirrups in a beam not too large. Remem- ber that the maximum spacing of stirrups in the part of beam where stirrups are required must not exceed three-quarters of the depth of beam. Use the same size of stirrup and the same design for the whole length of the beam and, if possible, for all similar beams in the entire structure, as a variety of designs may lead to errors and confusion. Common types of stirrup consist of {-in., 3-in., ~4-in, or 4-in. diameter round bars in the shape of a U with the free ends hooked. For uniformly loaded beams, the number and the spacing of stirrups for different conditions may be taken from the table on p. 900. For concentrated loads, a semi-graphical method of determining the spacing of stirrups is the most satisfactory. (See p. 252.) Vertical Stirrups and Bent-wp Bars—Determine the maximum total shear, V, and shearing unit stress, v, as in previous case. Determine the number of bars to be bent and the places where the bends can be made. (See p. 287.) Mark the distances within which the bent bars may He considered as effective as web reinforcement. Compute the shear to be resisted in these regions and also the available strength of bent bars. Ifthe strength of bars is equal to or larger than the shear to be resisted, no stirrups are required in these regions. Select proper diameter of stirrup, as suggested in previous case. Provide stirrups where stresses are not resisted by bent bars. If bent-up bars are bent in places where they cannot resist diag- onal tension or are bent in one or two places only, their full value as web reinforcement may not be available. Bent-up bars may be considered as effective web reinforcement for a distance equal the depth of the beam. 252 REINFORCED CONCRETE DESIGN SEMI-GRAPHICAL METHOD OF SPACING STIRRUPS The semi-graphical method suggested below is well adapted for practice. The diagrams may be drawn free-hand on computation sheets, which, to simplify the work, should consist of cross-section paper with squares of about 1_in. diameter. In drawing the diagram, each horizontal division may be taken to represent 1 ft. of the span and each vertical division 10-lb. of the unit shearing stress. The fractions of a foot may be gaged by the eye, so that no additional scale is necessary. Method of Procedure.—Lay out length of beam to a convenient scale. When beam is loaded by concentrated loads, indicate points of application of load; for uniform or symmetrical loading, only one-half of beam needs to be considered. Compute unit shearing stresses (to be considered as measure of diagonal tension) at points of maximum shear and, in case of con- ‘ centrated loads, also at points of concentration. Plot the unit shearing stresses as ordinates, to scale, at points of maximum shear and also at points of concentration, if any. The resulting diagram will be similar to the external shear diagram. Mark off the portion of diagonal tension resisted by concrete. If concrete is assumed to resist at all sections a definite amount of diagonal tension (say 40 lb. per sq. in.), draw a horizontal line at a vertical distance from the bottom equal to 40 lb. Then the ° upper part of the diagram represents the shear to be provided for by stirrups. If concrete is assumed to resist a percentage of total shear (say one-third of the shear) mark off, starting from the top, the portion resisted by the concrete. The lower part of the diagram will represent shear to be provided for by stirrups. . Decide on diameter of bar for stirrup and on the number of legs. Find spacing of stirrups as follows: From the sketch, scale at the support the shear to be resisted by stirrup. From the table on p. 899 for proper width of beam, read the spacing corresponding to the selected diameter of stirrup and the scaled shear. Mark this length on the sketch, starting from edge of support. This is the © first division. Next, scale the shear at the end of this first division § and proceed in the same manner until all diagonal tension to be resisted by web reinforcement 1s taken care of. The stirrups should then be placed in the center of each division. 8 Actually, the shear should be scaled at the center of each division. The error of the recommended procedure is slight and on the safe side. SEMI-GRAPHICAL METHOD OF SPACING STIRRUPS 253 In practice, the first spacing is repeated several times, then the next spacing is duplicated, so that actually only three or four spac- ings need to be determined by reference to the table. When bent-up bars are used, mark on the sketch their position and the distance in which they are effective. Compare the available tensile strength of bent bars with the amount of diagonal tension to be resisted in the distance in which the bars are effective. Ten- sile strength of bar is area multiplied by unit stress in steel, fs, and by 1.48. The amount of diagonal tension equals shear scaled at point of application of the bar multiplied by width of beam and by the distance in which bent bar is effective.. If the tensile strength is equal to or greater than the amount of diagonal tension to be resisted, no stirrups are then required; otherwise, the diagonal tension not taken care of by the bent bars should be provided for by stirrups. Elsewhere the same method should be used as when no bent bars are used. (For example, see p. 582.) Spacing of Stirrups for Uniformly Distributed Loading.—For uniformly distributed load, the shear is a maximum at the support and zero in the center. The shear diagram to be used in determining the spacing of stirrups for one-half of the beam is a triangle, as shown in Fig. 253. The required spacing of stirrups obtained in the man- ner described above is also shown. a Resisted by Stirrups _-- Shear Diagram -----------)7-------- — Seen ea Se eee wat : Fig. 84.—Spacing of Stirrups for Uniformly Distributed Load. (See p. 253.) Spacing of Stirrups.—For concentrated loads, such, for instance, as are produced by beams running into a girder, the shear diagram used for determining the spacing of stirrups will not always be a regular figure, as it depends upon the number of concentrated loads and their position. For symmetrical position of concentrated loads, the spacing of stirrups is symmetrical at both sides, and one- 254 REINFORCED CONCRETE DESIGN half of the diagram is sufficient. For unsymmetrical position of concentrated loads, the spacing will be different at both ends. Particular attention is called to the fact that for two beams of — equal length, equal concrete dimensions, and same maximum shear at the supports, but one of which is uniformly loaded and the other loaded with concentrated loads, the spacing of stirrups throughout the beams will be different. This is evident from a comparison of Figs. 84, 85, and 86, giving spacing for different position of con- centrated loads. Shear Diagram Eq ee Se Resisted by Stirrups ‘ H i] 1 ' ‘ ss ( ' > ahaa haat ress ieee ees ere, EMG Prize, Fic. 85.—Spacing of Stirrups for Load Concentrated in the Center of Beam. (See p. 254.) | _ Shear Diagram-~., Constant PStirrups > quagt Bape ee ee ~~ Constant Spacing,. Resisted by Stirrups \ — aa ee owe )) Hen nn es Vv ae ae ok ee aw es ae Fic. 86.—Spacing of Stirrups for Loads Concentrated at Third Points. (See p. 254.) One Load Concentrated in the Center.—In this case, the shear will be uniform for the whole length of the beam (except for the small difference caused by varying shear due to dead load of the beam). After computing the shearing unit stress and selecting the kind of SEMI-GRAPHICAL METHOD OF SPACING STIRRUPS 259 — web reinforcement, determine the spacing by dividing the tensile value of one stirrup by the shearing unit stress to be resisted by the stirrup times the width of the beam. ‘The spacing of stirrups will then be uniform throughout the beam. Loads Concentrated at Third Points.—The shear due to concen- trated loads is constant between the supports and the loads. The spacing of stirrups there is uniform. No stirrups are theoretically necessary in the middle third of the beam. For important beams, stirrups are placed at intervals throughout the whole length of the beam. Concentrated Loads, Irregularly Spaced with Uniform Load.— Lay out the span and, at points of concentration, mark off to scale 2000 16. 15000 1b. 20000 Ib. 5000 1b. 25000 16. 23000 1b. } bay OS) =~ S bs Resisted by 2 Concrete 174 lb.-- © miles eee a b ’ ' j K oe ley spaces @ 10- | vee spaces @ 10~--- @ os | & ace amas Resisted by Stirrups Fee 2 dq e1 Ss . Resisted by *5 on = de 3 Concrete 14 ? ey A Fic. 87—Spacing of Stirrups for Concentrated Loads. (See p. 255.) the unit shearing stresses (as the measure of diagonal tension). Mark off the amount of shearing stresses to be resisted by the con- crete. Then, starting at the support, determine and mark off the area that can be resisted by one stirrup, and place the first stirrup in the center of gravity of that area. Next, mark off the area for the second stirrup and place the stirrup and proceed thus until the total area of the diagram is provided for. Where the shear does not change radically, the same spacing may be repeated several times. Table 14 on p. 899 will be found useful in determining the spacing of stirrups. 256 REINFORCED CONCRETE DESIGN To illustrate the method more clearly, detailed computations are given below for determining the number and the spacing of stirrups for a beam 20 ft. long, the dimensions of which are 6 = 10 in., d = 25 in., jd = 22 in., and the loading as shown in Fig. 87.. The reaction is found first. Then the external shears are com- puted at points a, 6, ¢, and d. The external shears divided by bjd, which in this case is 10 X 22 = 220, give the unit shearing stresses at the respective points, as follows: External Shear Shearing Unit Stress Points Left Right Left Right CSA ORI ee 20005) Meee aes 114 b 24 200 22 000 110 100 Cc 20 200 5 200 92 23 d 4 000 —16 000 18 —73 e —18 800 —21 800 —86 —99 sf +93. 000. aie 5 ene —104 The shearing unit stresses are plotted as shown in Fig. 87. The length of the beam is laid out first, and the location of the points of application of loads marked. The shearing unit stresses are then plotted as ordinates at the respective points. After the points are connected, an area is obtained which represents the total amount of shearing stresses acting on a beam 1 in. wide. The shearing stresses at the left are plotted above the line af, and those at the right below the line. The section in which concrete alone can resist the total diagonal tension is found by drawing horizontal lines above and below at a distance equal, in the accepted scale, to vo’ = 40 lb. At the left end of the beam, this line strikes the outline at point g, and at the right end at point di. To the right of point g and to the left of di, all diagonal tension is resisted by the concrete and no stirrups are neces- sary. The shearing stresses outside of points g and d; must be provided for by web reinforcement. Assume that steel resists two-thirds of the shearing stresses and concrete one-third. Then, to the left of point g and to the right of di, one-third of the area is resisted by concrete and two-thirds by SEMI-GRAPHICAL METHOD OF SPACING STIRRUPS 257 web reinforcement. The area resisted by concrete is marked off on the diagram. The balance of the area represents the stresses to be resisted by steel. The total amount of diagonal tension to be resisted by steel at the left end of the beam may be found by computing the areas aa;beb and 6bicc; and multiplying them by the width of the beam yb =" 10in. Areas aajbeb + bbicic = 2A TOV ie. 2/100 +72\ | A : ) x 24 + 3( ) x 60 = 1792 + 3 440 = 5 232 Ib. Multiplying by 6 = 10, we obtain 52 320 Ib. as the total amount of diagonal tension to be resisted by the web reinforcement. For 3-in. stirrups with two legs, the area of one stirrup is A; = 2 X 0.196 = 0.392 sq. in. and the stress resisted by one stirrup is Asfs = 0.392 X 16000 = 6 272 lb. Divide the total amount of diagonal tension by the stress resisted by one stirrup, to get the total number of stirrups; or, _ 52320 N; — 6 272. a 8.3. Use 9 stirrups. The spacing of stirrups may be found as follows: By trial, start- ing at the support, find shear areas in Fig. 87 equal to the resisting O74 = 627.2 1b. This may be done by scaling the ordinate above the line af, approximately in the middle of the first division, and dividing 627.2 Ib. byt. Scaling the ordinate at a distance of 4 in. from a, we get 73 Ib. as the average unit shearing stress for the first division. The length we = 8.6 inches. Lay off this dis- tance and place the stirrup in the middle. Next; scale the ordinate 4 inches from the end of the first division and find the next spacing. In this example, as the effect of the uniform load is small, the matter is simplified by considering the ordinates between a and b and between b and ¢ as constant and finding the spacings for these two constant values. Thus, we find the spacing in portion ab to be 8.5 in., and in portion bc, 10.5 inches. So we may use three spaces value of one stirrup divided by b = 10, or of the first division, then, is 258 REINFORCED CONCRETE DESIGN at 8 in. and six spaces at 10 in., giving the required number of stir- rups. The same method may be used in determining the spacing in the right end of the beam. The shear here is somewhat smaller than at the other end, but it facilitates the erection to adopt the same spacing at both ends. Of course, the stirrups must extend to the point d. Table 14 on p. 899 will be found useful in spacing stirrups. DIAGONAL TENSION FOR MOVING LOADS For beams carrying moving loads, as bridges and crane runways, it is necessary to determine for every section the maximum total shear, draw the shear diagram and space the stirrups as suggested above. | For heavy moving loads, it is advisable to make allowance for impact. The magnitude of the impact depends upon the character of the structure, the loading, and the re- ; BBs Bs lation of the weight of the structure to the moving load. Fig. 88.—Shear Diagram for Uniformly Distributed For railroad bridges Moving Load. (See p. 258.) and bridges carry- ing electric cars, the ordinary impact formulas may be used; for crane runways and highway bridges, from 29 to 50 per cent of the live load should be added to allow for impact. Shear Diagram for Uniformly Distributed Moving Load.— Assume moving load, w, per lin. ft.; then the maximum positive shear at any section occurs when the load extends from the right - support to the section under consideration and the portion between the left support and the section is unloaded. (See Fig. 88.) The --Dead and Live Load-- DIAGONAL TENSION FOR MOVING LOADS 259 general equation for the maximum shear due to a moving load approaching from right to left then isan = saa This is an equation of a parabola. | wl al! mee ‘a a ior. = 0, oe for x = 5, Verte; Foro le VV iO): To the shear due to moving loads, the shear due to fixed (dead) loads must be added. In Fig. 89, the diagrams for dead and live loads are drawn. The first line gives the diagram for dead load only; the second line for moving load plus impact only; and third line for the sum of the dead and live load. It will be noted that there is con- siderable shear at the center of the beam. Stirrups must be provided for the sum of shears. (6) The diagram Fia. 89.—Shear Diagram for Two Moving Loads a Con- should be used stant Distance Apart. (See p. 259.) only for half of the ‘span. For the other half, the moving load should be considered as the left. The spacing of stirrups must be made the same for both ends of the beam. Shear Diagram for Two Equal Moving Loads a Constant Distance Apart.—This case occurs in a beam carrying cranes and in highway bridges (see Vol. III). The maximum shear at any section is obtained by placing one load at the section considered. A general equa- tion is: ; P = concentrated moving load; e = constant distance between the loads, P. | --- Dead and Live Load------ —------Live Load------- 260 REINFORCED CONCRETE DESIGN i— (e435) A ees: Ve ea or em (46) V =P * tors >t—« en The variation in shear is a straight line for both equations. It is therefore sufficient to determine shear at two points only, namely for ¢ a Qiatid Col =e: (eee Migs soap Maximum shear for x = 0, Vinax = an : | (48) and Shear for fo madi ne, V= PS. pvt 49) With these two values, the shear diagram can be drawn. To this must be added the shear due to the dead load. The shear diagram having been drawn, the spacing of web reinforcement is determined as in previous cases. The diagram is used only for one- half of the beam. The stirrups in the other half are made sym- metrical. . ; For bridge design, the shear may be found as directed in the chapter on Bridge Design, Vol. III. The shear diagram may then be plotted and the spacing of web reinforcement determined as explained before. BOND OF STEEL TO CONCRETE Importance of Bond Resistance.—It is of the utmost importance to design reinforced concrete construction so that the bond stresses between the steel and the concrete will not exceed the safe bond resistance. The safe bond resistance is equal to the unit stress at which the bar begins to slip, divided by the factor of safety. If bond stresses exceed safe working values, the factor of safety is low- ered and there is danger of separation of concrete and steel and a con- sequent failure of the construction from pulling out of the steel. Bond stresses do not always receive proper consideration. Many designers consider them of secondary importance and often do not even compute them. As a result, one often finds designs with adequate areas of cross section of longitudinal steel, but with no means of bringing the reinforcement into action. BOND OF STEEL TO CONCRETE 261 Bond Stresses Should Always Be Computed.—It is particularly important to compute the bond stresses in short beams with heavy loads, footings, and retaining walls, where they are likely to be excessive unless properly provided for. Bond stresses are often excessive also in joist construction, where, because of the narrowness of the joist, bars of large diameter are used. The importance of proper provision for bond stresses will be péalized if it is considered that, ordinarily, the bond is the agency which causes cooperation between concrete and steel, and that the resistance of steel to tension can be utilized only through the bond between concrete and steel. When the bond between steel and con- crete is not sufficient, the bar will slip instead of resisting tensile stresses. The function of bond in reinforced concrete is the same as the function of rivets in built-up structural steel beams. In a structural steel beam consisting of web plates and flange angles, the angles will not act unless they are properly riveted to the plates. When the number of rivets is not large enough to hold the plates and angles together, the rivets shear off and the angles become separated from the plates. In the same way, in a reinforced con- crete construction, when bond stresses are too large, the bars pull out instead of acting as tensile reinforcement, and the beam will fail even if it is provided with a sufficient amount of tensile rein- forcement, because there is no medium which will bind the two materials. It would be just as illogical, in designing a reinforced concrete beam, to supply tensile reinforcement without proper bond as it would be, in steel construction, to supply plates and angles for a built-up section without a sufficient number of rivets to make the component parts act as a unit. Bond is a Function of External Shear.—Consider, in a horizontal beam subjected to bending, a portion of beam between two vertical sections. Assume that the depth and the amount of steel. between _the two sections are constant, but that the bending moment at one section is larger than at the other section. The unit stresses in steel, being proportional to the bending moment, are larger at the section with the larger bending moment. The increase in stresses in steel between the two sections must be developed in the bar by the bond between the steel and concrete acting in the distance between the two sections. From mechanics, the increase in the bending moment between two sections is a function of ‘the external shear. The increase in stresses in steel will also be dependent upon the magni- 262 REINFORCED CONCRETE DESIGN tude of the external shear. Therefore, finally, the bond stresses are a function of the external shear. Formulas for Desiga. Let V = total external shear at section considered, in lb.; y = shearing unit stress in lb. per sq. 1n.; u = bond unit stress in lb. per sq. in. of surface area of tension steel; o = perimeter of one bar in inches; So = sum of perimeters of all horizontal tension bars at section considered; jd = distance between centers of tension and compression (approximately $d) ; d = effective depth of beam, in inches. Then Bond Unit Stress, mA Required Perimeter of Tensile Reinforcement, y =< ic 20m idu’ 2 tel STS oie Pane (51) If a portion of tensile reinforcement is bent up, only horizontal bars, and those bent-up bars which are not farther up than one- third of the effective depth of the beam, should be considered as resisting bond at the section under consideration. Special attention is called to the fact that in computing bond stresses in continuous beams, in the region between the support and the point of inflection subjected to the negative bending moment, the top bars must be used in the formula. In the balance of the beam, the bottom steel must be used. Bond Stresses for Small Bars are Smaller than for Large Bars.— The ratio of the perimeter of a bar to its sectional area is not constant for all sizes of bars. It is larger for small bars than for large bars. This means that, for each square inch of sectional area, the sum of perimeters is larger for small bars than for large bars. When the bars are subjected to equal tension, those having larger perimeter will have smaller bond resistance than those having smaller perimeter. For this reason, where subjected to equal tensile unit stresses, the bond stresses for small bars are smaller than for large bars of the BOND OF STEEL TO CONCRETE 263 same sectional area. This property is often utilized for reducing excessive bond stresses. The above principle can be clearly seen from the following comparison. For instance, 4-in. round bars have an area of cross section of 0.196 sq. in., and a perimeter of 1.57 in. The ratio 1.57 0.196 that for 3-in. bars of an aggregate area equal to one unit (say one square inch) the total perimeter is equal to 8 units (8 in.). For one-inch bar, on the other hand, the area of bar is 0.785 sq. in., the perimeter is 3.14 in., and the ratio of perimeter to area is 3.14 0.785 unit has only 4 units of perimeter. From the above figures, it is obvious that for equal area of cross section the $-in. bars have twice as large a perimeter as the one-inch bars. The tensile stresses are governed by the area, and the bond stresses by the perimeter. The larger the ratio of the perimeter to the area, the smaller are the bond stresses for equal tensile strength. Therefore, for equal tensile stresses, the bond stresses for 4-in. bars would be only half as large as for one-inch bars. Conversely, the bond resistance of 4-in. bars is twice as large as the bond resistance of one-inch bars. Working Unit Bond Stresses.—The bond stresses computed by Formula (50), p. 262, should not exceed the following values. For plain bars, u = 0.04f",, or 80 lb. for 2 000 lb. concrete. For deformed bars, u = 0.05f’c, or 100 lb. for 2 000 lb. concrete. u — 0.06f’, may be allowed on specially efficient deformed bars. Working Stresses, Bars with Anchored Ends.—When the ends of the bars, for which bond stresses are computed, are. provided with hooks of proper proportions or are extended beyond the point of zero bending moment a sufficient length to develop one-half of the strength of the bar, the allowable bond stresses may be increased by 50 per cent. To comply with this requirement, in simply supported beams the bars must be extended the proper length beyond the support. In continuous beams, if bond stresses for the bottom reinforcement are considered, the bars must be extended beyond the points of inflection governed by the positive bending moment, as shown in Figs. 98 to 103. If bond stresses for the negative reinforcement are considered, the top bars must be extended the proper length beyond the points of inflection governed by the negative bending between the perimeter and the area is = 8. This means = 4. The aggregate area of one-inch bars amounting to one 264. REINFORCED CONCRETE DESIGN moment. The bent bars which are continuous with the bottom reinforcement may be considered as fulfilling the above require- ment. In cantilever footings, to allow the increased bond stresses, all bars must be provided with hooks at their ends. Problems in Bond.—In design, the following problems may have to be solved: Problem 1—To determine whether the bond stresses for the tentatively selected bars are within working limits. The external shear, V, is known. The required area of steel is found from bending moment. Bars giving the required area of steel are tentatively selected. To solve this problem, the perimeter, 0, of one bar is found and multiplied by the number of bars. This gives the value 2o. If eroups of different sizes are used, the perimeter of each size is found and multiplied by the number of bars in the respective group. The sum gives the value of Zo. T he bond unit stress is now found from Formula (50). If the bond stresses are within working limits, the selected bars are satisfactory. IH the bond stresses are larger than the allowable working stresses, either of the following methods may be used. (a) A larger number of bars of smaller diameter, giving the same area, may be substituted. This is the most economical method of reducing bond stresses. Small bars have a larger perimeter in pro- portion to the area than larger bars. This difference may be sut- ficient to bring the bond stresses within working limits. For instance, six l-in. square bars have an area, As= 6 sq. in. and a sum of perimeters, Zo = 24 in. Ten “in, bars have the same area, but their perimeters are Lo = 27.5 in. The bond stresses for the Zin bars would be about 15 per cent smaller than for 1-in. bars. (b) The depth of the beam may be increased sufficiently to reduce the bond stresses. The area of steel found for the smaller depth, of course, must be retained. This method, since it requires more concrete, should be used only when, for any reason, the first method is not possible. For example, if the bond unit stress in a beam, as found by formula, 1s vu = 120 lb. per sq. in. and the allow- able bond unit stress is 100 lb. per sq. in., the necessary increase by this method in the depth of the beam would be in the ratio +2% = 1.2, keeping same reinforcement to reduce the bond stresses to 100 lb. per sq. in. ANCHORAGE AND SPLICING OF REINFORCEMENT 265 (c) The depth of beam and size of bars may be kept the same, but the number of bars may be increased. ‘Thus, in the example mentioned under (b), the bond stress may be reduced by multiplying the number of bars by 1.2. This also is evidently uneconomical. (d) A combination of (a) with either (b) or (c) may be used. In such a case the bond stresses should be decreased as far as possible by the use of bars of small diameter. After that, further reduction may be obtained by increasing either the depth or the number of bars. Problem 2.—To find the bars giving bond stresses within working limits. The external shear, V, and the required area of steel are known. 3 | In solving this problem, the smallest allowable sum of perimeters, o, is computed from Formula (51), p. 262. Bars having an aggre- gate area of cross section equal to the required area, and having also a sum of perimeters equal to or larger than the value, o, obtained from the formula, are then selected. The required area and the required minimum perimeters being known, the bars can be easily selected. | ‘Problem 3.—To review a design already made, for known load- ing, to determine whether the working stresses are exceeded. This problem is solved in the same way as Problem 1. ANCHORAGE AND SPLICING OF REINFORCEMENT Where Anchorage and Splicing is Required.—Anchorage of reinforcement is required when the reinforcement at the face of the support is subjected to stresses which must be transmitted from the bars to the support. The stresses from the bars may be transferred either to the concrete of the support or to another set of bars within the support. As an example, we may consider the bars at the sup- port in continuous and restrained beams and in cantilevers, because there the maximum tensile stresses in steel, produced by the bending moments, exist at the face of the support and must be transmitted to the support. Another instance of anchorage is seen in the splicing of two bars at a point where the bar to be spliced is subjected to considerable tensile stress, and it is desirable to transmit the stress to the other bar. Means Used for Anchoring Bars.—The simplest means of anchor- ing a bar is by extending it from the beam into the concrete of the 266 REINFORCED CONCRETE DESIGN support, for such a distance that the stresses from the bar are trans- ferred gradually, by bond, to the concrete of the support. (See Fig. 90.) | Another method of anchoring a bar is by providing it at the end with a hook of sufficient dimensions. The hook is placed within the support. The best method of anchorage is to extend the bar some distance into the support and also to provide it with a hook at the end. (See Fig. 90.) -Full Strength of bar at this Section at Crack may open Correct Anchorage as Wrong Anchorage é Bars should overlap Nor be continuous PE i PEE A | yi Grack may open ED BLO natn a vee semiais ERE S At CU tsi an sake oe Ti ecu dene fetus ost POLS Me We eee es Ver oe ~ eaeeie SSeS ie Correct Anchorage Wrong Anchorage Fra. 90.—Method of Anchoring Bars. (See p. 266.) Method of Splicing Bars.—The stress from one bar may be transferred to another bar by splicing. The splice may be formed by lapping the bars for a length equal to the length of imbedment, 11, required for anchorage. A splice may also consist of a straight lap with hooks at both ends of bars. The length of the lap may then be shorter than for straight ends. Sometimes it may be necessary to splice two bars by means of a third bar as shown in Fig. 91. The third bar must lap each bar a length sufficient to develop the strength of the bar. The length of the short bar is then equal to twice the length of imbedment, [1, required for anchorage. When several bars in a beam or slab are spliced, the splices ANCHORAGE AND SPLICING OF REINFORCEMENT 267 should be arranged so that not more than one bar is spliced at any one place. Splices at points of maximum stress, for example, in the middle of beams, should preferably be avoided. This does not mean, however, that they may not be permissible with proper staggering. When tensile bars are spliced by lapping, the length of the lap- should be based on the maximum tensile stress in steel, even if the stresses at the points of the splice are smaller. é fs -- Length of Splice tic | @ ) o 2SGe S50 TT | Straight Splice Splice with Hooked Ends - . Length of Length of | fadeSnlice's i ay a Splice" Splicing two Bars by a Third Bar Fig. 91.—Method of Splicing Bars. (See p. 266.) Problems to be Solved in Connection with Anchorage.—The following problems may have to be solved in connection with anchor- age of the bar: , 1. Diameter of bar and stress are given, and it is necessary to deternmrine the means of anchoring the bar. 2. Stress, diameter of bar, and length of imbedment are given, and it is necessary to determine the bond unit stress. 3. Length of imbedment, diameter of bar, and bond unit stress are given, and it is necessary to determine the safe tensile stress to which a bar may be subjected without exceeding the allowable bond stress. 4. The length of the lap or splice for a bar may have to be deter- mined. These problems are solved by use of formulas and _ principles given below. Formulas for Determining Anchorage by Straight Imbedment.— Let f; = tensile or compressive stress per square inch in bar; 7 = diameter of bar in inches; u = bond in pounds per square inch of surface; 1; = necessary length of imbedment of bar in inches. 268 REINFORCED CONCRETE DESIGN Then Length of Imbedment of Square and Round Bars, borspl Hevea im iis ay eye . a Bond Unit Stress, Shoes w= at 0 | ph panera ee Allowable Tensile Stress in Bar, fa = Ale. ee Conditions in Beams at Support.—At an interior column ‘in con- tinuous beams, the tensile bars at the support are extended from one span into the adjoining span. The extended portion acts not only as anchorage for the stresses developed in the first span, but also as tensile reinforcement in the adjoining span. The length of the extended portion of the bar is governed not by anchorage require- ments, but by the moment requirements in the adjoining span. Good practice in such cases requires that the bars be extended 12 in. beyond the point of inflection of the adjoining span. If a beam extends beyond the column, forming a cantilever, the same conditions exist. In the end spans of continuous beams, and in restrained beams where it is not possible to extend the bars into the adjoining span, the length of imbedment is determined solely by the requirement for anchorage. It is often impossible to get sufficient length of imbed- ment, and the bar is provided with a hook. The support in which the bar is anchored must be strong enough to withstand the tensile stress transferred by the anchored bar. If the support is a column, provision may be required for tensile stress in the column, due to bending. The tensile reinforcement for the beam and column must be arranged so as to prevent cleavage between the two members. Allowable Bond Unit Stresses for Anchorage and Splices.— The allowable bond unit stresses to be used in computing the anchor- age and splices are the same as recommended on p. 263 for beams. Anchorage of Compression Reinforcement.—In continuous beams at the support, bottom reinforcement. is subjected to compressive stresses. If the bars are intended to serve as compressive reinforce- ANCHORAGE AND SPLICING OF REINFORCEMENT 269 ment, provision should be made for transferring the compressive stresses in the bars at the face of the support into the column. For this purpose, the bars are extended a sufficient length into the sup- port to gradually transfer the compressive stresses, by bond, from the bar to the concrete of the column. These compressive stresses act at right angles to the main column stresses and therefore they do not affect its capacity as a column. The necessary length of imbedment, 11, may be found by Formula (52). The stress, f,, to be used in computing the length, is the compressive stress in the bar, which is usually much smaller than the allowable tensile stress in the bar. Sometimes, to increase the amount of compression reinforce- ment at the support of a beam, the bottom bars from the adjoining beam are extended into it. The extension then serves both as anchorage in one span and as compressive reinforcement in the other. The length in the adjoining span will be governed by the moment requirements. Anchorage of Column Reinforcement.—Compression reinforce- ment in columns carries considerable stress. Since the bars ordi- narily are only one story high, it is necessary to transmit the stresses from the bars in the column above to the bars in the column below. The requirements are treated in the column section. Use of Hook as Anchorage for Reinforcement.— When it is not possible to extend the bar sufficiently into the support to develop its strength by straight imbedment, the anchorage is accomplished by a hook. The effectiveness of the hook depends upon the strength of the concrete, as the hook transfers the stress from the bars to the concrete by bearing on the concrete. The dimensions of the hook must be such that the bearing stresses do not exceed allowable working values. Where considerable stress is transmitted by a hook, it is advisable to place a cross bar of proper length inside of the hook, to distribute the bearing stresses to a large area of con- crete and to prevent splitting of the concrete. It is not essential that the hook should bear perfectly on the cross bar. Even if there is no contact between the hook and the bar, the cross bar will pre- vent splitting of the concrete. The conditions are still more improved if the cross bar, instead of being straight, is inclined at an angle to the direction of the bar to be anchored, as shown in Fig. 92, p.270. Then part of the pull may be taken by the cross bar in tension and transferred gradually to the 270 REINFORCED CONCRETE DESIGN concrete. Another instance of effective anchorage is found in the Smulski Flat Slab System at the column head, where the pull from radials is transferred partially by bearing on the concrete and par- tially by tension in the center ring. When the reinforcement in another beam is at right angles to the reinforcement requiring anchorage, it may be considered as assist- ing in the distribution of the bearing stresses on the concrete. The best anchorage is obtained by a combination of straight imbedment equal to 15 diameters of the bar and a semicircular hook, the inside diameter of which is at least four times the diameter of the bar. (See Fig. 92, p. 270.) With such a design, the elastic limit of the steel can be reached without causing excessive secondary pecan cal eae 15 Bar Diameters a ae _--~;--10 Bar. Diameters y- Section of Maximum ta . —“~4 Bar Diameters a ension Elevation Elevation ; Cross Bar } Plan Plan Fic. 92.—Hook Recommended for Anchorage. (See p. 269.) stresses in the concrete. When cross bars are used, the straight imbedment may be reduced to 10 diameters of the bar. BEARING STRESSES Let P = superimposed load in pounds; A = total area of pedestal or footing (subjected to limitation according to depth) ; A, = bearing area, in same units as area A; fy = allowable bearing unit stress, lb. per sq. im? fri = allowable bearmg unit stress on wall, lb. per sq. in.; f’. = ultimate compressive strength of cylinder, Ib. per sq. in. Explanation of Bearing Siresses.—Bearing stresses are produced by a supported member, resting upon a supporting member, at the surface of contact of the two members. Bearing stresses are pro- duced under the bearing plate or base of a steel beam or column BEARING STRESSES 271 resting on concrete. A concrete column resting upon a_ pedestal also produces bearing stresses in the concrete of the pedestal, the magnitude of which may be computed by dividing the column load by the gross area of the column. When the stress from vertical column reinforcement is transferred to the pedestal or footing by a plate, the stresses produced under the plate may be called bearing stresses. The magnitude of bearing stresses equals the total superimposed load P divided by the area of contact, 41. Therefore P fo = Ge (55) Allowable Unit Stresses.—The allowable bearing unit stress depends not only upon the strength of the material composing the supporting member, but also upon the relation between the size of the bearing area and the total area of the top of the supporting member. If the bearing area equals the area of the top of the supporting member, the allowable working unit stress over the bearing area must not exceed 0.25f’.. For 2000 lb. concrete the allowable bearing unit stress is 500 Ib. per sq. in. If the area of the pedestal is larger than the bearing area and projects on all sides beyond the bearing area, the allowable unit stresses may be found from formula below. Allowable bearing unit stress, when area A extends on four sides outside area A1. / = A fo= 0.25f". Aj ° e ° ° ° ° ry (56) . but not more than 0.7f’c. The thickness of the footing, pedestal or pedestal and the footing must not be less than six-tenths of the maximum projection of the area A, beyond the bearing area. If a column rests directly on a shallow footing, the total area of the footing can not be used in the formula. Instead an imaginary area should be used, the sides of which equal the sides of the bearing area plus 12 times the depth of the footing. This imaginary area must not in any place come outside of the actual area of the footing. Formula 56 applies if the bearing area is surrounded on all sides by concrete of the pedestal or footing. If the bearing area on one or two sides coincides with the sides of the supporting area, as in 272 REINFORCED CONCRETE DESIGN case of a bearing plate resting on a wall, only one-third of the mcrease ‘n allowable bearing stresses over 0.25 f’: as given by Formula (57) may be used. The formula changes to Allowable bearing unit stress when side of bearing area A1 coincides with side of supporting area A, 5 yet 9 0.25/'.(0.67+] 4). gia ie but not more than 0.5f’c. PROTECTIVE COVERING FOR REINFORCEMENT Purpose.—The concrete outside of the reinforcement of concrete members serves several purposes. First, it bonds the concrete and the steel. In computing bond resistance, the whole perimeter of the bar is considered as resisting bond stresses. This is only possible where there is sufficient concrete around and below the bar. Second, it protects the reinforcement from fire. Third, it protects ‘the reinforcement from weather and moisture and therefore from rusting. Since the amount of concrete that must be placed below the steel for protection is always sufficient for bond, only the pro- tective purposes will be treated. Fire Protection—The thickness of concrete necessary for pro- tection of reinforcement against fire depends upon the quality of the aggregates, and upon the severity of the possible fire to which the structure may be subjected. Aggregates having low coefficient of expansion, such as limestone and traprock, are desirable for fire protection. Gravel and granite should not be used as an aggregate where fire hazard is large, because the quartz, of which they largely consist, has a high coefficient of expansion and, under high heat, will expand and crack the concrete. Limestone and Traprock.—With aggregates such as limestone or traprock, the protective concrete, measured from the surface of the concrete.to the surface of the lowest bar, should be, for large fire hazard: Slabs and walls, 1 in. Beains, girders and columns, 2 in. Where the fire hazard is small, the protective covering may be as follows: Slabs and walls, 2 in. Beams, girders, and columns, 13 in. CLEAR DISTANCE BETWEEN PARALLEL BARS IN A BEAM 273 Gravel and Granite.—When gravel or granite aggregates are used in buildings with small fire hazard, the larger protective covering recommended for the traprock should be used. ‘When it is absolutely necessary to use these aggregates in a build- ing exposed to severe fire hazard, the protective cover should be 1 in. larger than recommended for traprock and should be reinforced with metal lath placed 1 in. from the surface. The openings in metal lath should not exceed 3 in. Moisture Protection.—Reinforcement in footings should be pro- tected with a minimum covering of 3 in. Where concrete is exposed to weather, the concrete covering for the reinforcement should be 2 in. thick. In all cases, the thickness of the protective covering is measured from the face exposed to moisture to the nearest surface of the bar. CLEAR DISTANCE BETWEEN PARALLEL BARS IN A BEAM Parallel bars in a beam must be placed so far apart that the amount of concrete around each bar is sufficient to provide proper bond between steel and concrete. Proper clear spacing between bars, depending upon the maximum diameter of coarse aggregates, is also required to allow the concrete to flow below the bars. The width of the stem ‘of a T-beam is often governed by the distance required for spacing of bars. | Spacing Recommended by the Authors.—The minimum clear distance between parallel bars should not be less than 1 in., or less than 14 times the diameter of the larger round bar and 2 times the side of a square bar. The minimum clear distance of the bar from the side of the beam should not be less than required for protective covering, nor less than 14 times the diameter of the bar. As a further requirement, the minimum spacing of the bars should be larger than the maximum diameter of the coarse aggregate. If aggregates cannot pass freely between bars, there is danger that arching of the aggregates, will produce voids in the concrete at the level of reinforcement. Such voids are dangerous because they reduce bond and also reduce the effectiveness of the protective covering. The table on p. 274 gives the required width of the beam for dif- ferent number and size of bars. 274 REINFORCED CONCRETE DESIGN Width of Beam Required by Different Number of Round and Square Bars Assumptions: Fireproofing 13 In. to Face of Bar. Clear Space between Bars Equal to 1.5 Diameters of Round Bars and 2 times Side of Square Bars. ne aS Width of Beam in Inches Diam. Bars, Number of Bars in One Layer In. 1 2 3 4 5 6 7 8 9 10 a eee Round Bars I 7 | BPH ia pee eet 6.01 7.2 | 8.5 | 9.7 | Si:O8 ee roe 3 3.02 | toe 6.7 | 8.3 | 9.9 | 11.4°1-18,0 ) Oe SiGe etree 3 Seen po 7.5.12 9,.4 | 11,2218: 14) 15209 2620) eae 4 3:9) ie orl 8.2 110.4 | 12.6 | 14.8 | 17.0 | 19.2) 284 2576 1 470 | 4625 9.0 | 11.5 | 14.0 | 16.5 | 19.0 | 21.5 | 24.0 | 26.5 1} Aoi Goo 9.7 | 12.6 |.15.4 | 18.2 | 2130 1 2358 e20rea ees 1} 4.2 | 7.4°| 10.5 |13:6°) 16-7 | 19199792300 (226s 2o ee 13 4.5 | 8.2 | 12.0 | 15.7 | 19.5 | 23.2 | 27.0 | 30.7 | 84.5) 38:2 ao ee a ee Square Bars eee 3 Bb aioed 6.5 | 8.0:1° 9.5 | 11. ON R22 S50 SOR) Tose 5 3 Os) Pose 7.41 9,2 111.1 | 18.0 171479 yl Gay Ste eee 2 ae eieieO10 8.2 | 10:5 | 12.7 | 15.0 | 17.29) T9050 aie aieeee rf 3.9 | 6.5 9.1 112.7 | 14.4 1 17.0 | 190622025 eae ee 1 4.0:| 7.0 | 10.0.| 13.0 | 16.0 | 19.0 22.05 257 0S eet ete 1} A.1 | 7.5 | 10.9 | 14:0 | -17.6°| 2150) 245s S27 aera 4 4.2 | 8.0 | 11.7 | 15.5 }-19:2 | 23.0 | 2657) ))300 "eee eee 13 4.5 | 9.0 118.5 | 18.0 | 22.5 | 27.0 NSLeSpiaeae0 es eeltoee Vertical Spacing of Layers of Reinforcement.—Whenever pos- sible, bars should be placed in one layer. If more than one layer is necessary, the vertical spacing of the layers in the clear should be not less than 1 in., or less than the diameter of the largest bar. If the design requires the extreme number of three layers of bars, the bars in the lowest layer should be properly anchored at the ends , BENDING MOMENTS FOR USE IN DESIGN 275 by hooking or extending into the supporting column. At least one- half of the bars, by area, should be bent up and anchored in the com- pressive zone, not only to serve as diagonal tension reinforcement, but also to develop the tensile stresses by anchorage in the com- pressive zone and thereby to insure proper action of the reinforce- ment. Keeping Bars in Place.— When the spacing of bars is the minimum allowable, it is important that the bars should not be even slightly misplaced during construction. For this purpose, some positive means of separating the bars, such as spacers, should be used. This is particularly important in narrow beams. Layers of bars must be separated vertically by mechanical means. One of the accepted means consists of placing across the reinforce- ment short pieces of bars of proper diameter, spaced not more than 3 ft. on centers. : Bars should be kept the proper distance above the forms. Cer- tain mechanical spacers keep the bars a proper distance above the form and a proper distance apart. Their use is recommended. BENDING MOMENTS FOR USE IN DESIGN OF REINFORCED CONCRETE BEAMS The bending moments for use in design of reinforced concrete beams depend not only upon the length and the loading of the span under consideration, but also upon whether the beam is a single span or is built continuous with a number of other spans. In a single span, the question to be considered is whether the beam is simply supported, or restrained at the supports. The qual- ity of the restraint is also to be considered. The beam may be fully restrained, 1.e., fixed, at the support; it may be only partly restrained by a concrete column; or it may be slightly restrained by masonry. In a continuous beam consisting of a number of spans, it is impor- tant to determine whether the spans are equal or not; whether the beams are connected with intermediate columns; whether the loading of all spans is the same; and finally, whether the ends of the beam are freely supported, restrained, or partly restrained. Quality of Restraint at Ends.—When an end of a beam rests on the support and is free to turn, it should be considered as freely supported, and no provision is necessary for negative bending moment. 276 REINFORCED CONCRETE DESIGN When an end of a beam runs into a masonry wall and is encased in it, especially when additional masonry is placed upon it, the end is not free to move as in the previous case. The masonry offers restraint to the beam, which should be considered as slightly restrained. The amount of such restraint is uncertain and unre- liable, however, and it cannot be counted upon as reducing the posi- tive bending moment. To prevent cracks at the support, some pro- vision should be made for a possible negative bending moment by using negative bending moment reinforcement. The amount of reinforcement should be governed by the actual conditions, but should not be smaller than required by formula on p. 278. The bars must be anchored at the end, and at least one-third of the reinforcement should extend to a point distant 0.2 / from the edge of the support. When the end of a beam runs into a flexible support, such as a column, and is rigidly connected with it, the end should be con- sidered as restrained. The magnitude of this restraint depends upon the relative ratio of stiffness of the beam and the column, as fully explained in Volume III. Such restraint is reliable and may be counted upon in determining bending moments. When the end of a beam runs into a support of much larger stiff- ness and is rigidly connected with it, it may be considered as fixed at that end. A beam may be considered as connected with a column, wall, or other support, when the connection is capable of resisting the bending moment produced at the support. Difference between Continuous Beams and Building Frames.— When a continuous beam simply rests upon the intermediate sup- ports—as, for instance, a continuous bridge simply resting upon the piers and not monolithically connected with them,—or when the beam is supported by small intersecting girders, full bending moment from one span is transferred directly to the adjoining span. Such a continuous beam is sensitive to unequal loading. ~The positive bending in such a case should be based upon the possibility of unequal loading of the adjoining spans, and the Formulas (63) to (68), under the heading “Continuous Beams of Equal Spans,”’ should be used. Continuous beams are shown in Fig. 93, p. 277. When a continuous beam is connected with intermediate columns, as shown in Fig. 94, the bending moment from one span is not transferred directly in full to the adjoining span, but a part of this bending moment is transferred to the columns, so that the effect of BENDING MOMENTS FOR USE IN DESIGN 277 the loading in one span upon the adjoining span is much smaller than for the continuous beams just described. Smaller provision is necessary for unequal loading and, for the positive bending moment, the Formulas (69) to (74) under ‘‘ Continuous Beams of Equal Spans Running into Columns,” may be used. Continuous Beam - -, Fic. ¢3.—Continuous Beams. (See p. 276.) The subject is thoroughly treated in Volume III, to which refer- ence should be made for full explanation of the subject. Span Length.—The span length for a freely supported beam or slab should be taken equal to the clear distance between the sup- ports, plus the depth of the beam or slab. The span length for a continuous or restrained beam should be taken as the clear distance between the faces of the support. Fic. 94.—Building Frame. (See p. 276.) Use of Formulas.—The formulas given below may be used for equal or nearly equal spans and for uniform loading. Where spans are not equal or the loads on different spans are not uniform, formulas given in Volume III should be used. NOTATION Let M = bending moment; w = uniformly distributed load, lb. per lin. ft.; 1 = span length of beam in ft., as defined above. 278 REINFORCED CONCRETE DESIGN FORMULAS FOR SINGLE SpAN BEAMS Ends Freely Supported.—Beam free to turn at support. Maximum Positive Moment at the Center, wi? ; mire ft.-lb. = 1.50l? inelbs ee No Negative Reinforcement is Required, Beam Slightly Restrained at Ends.—Ends not free to turn. Mazimum Positive Moment, wilt oe M = ce {tIb. ='1/5ul? ih -be Negative Bending Moment at End Supports, 'F F not less than M = oe ft.-lb.-= 0.60wl? in.-lb.3 0 2) OU) Beam Fixed at Ends. Maximum Positive Moment at Center, wl? ; : M = 17 ft.-lb. = wl? in.-lbs. of ak ee Negative Moment at Supports, wl? = a ao, ft.lb. = — wi? in-lb. |. eee CONTINUOUS BEAMS OF EQUAL SPANS These formulas are adapted to such constructions as ordinary slabs in beams and slab floors, beams supported by girders, and roof construction, as shown in Fig. 93, p. 277. The ends may be considered as slightly restrained. Beams and slabs continuous for two spans only. Mazimum Positive Moment near Center, wl? , Me 0 ft.-lb. = 1.2wi2anelbp a eee Negative Moment over Interior Support, wl? : 1 = mp ft.-lb. = 1/5wl? in loan ee CONTINUOUS BEAMS OF EQUAL SPANS Negative Moment at End Supports, 2 M = not less than Oe {t.-lb. = 0.60wi? in.-lb. Beams and slabs continuous for more than two spans. (a) Interior Spans Maximum Positive Moment near Center, Negative Moment at Support, pean" rie eee (ie 19 ft.-lb. = wi? in.-lb. (b) End Spans. 279 (65) (66) Maximum Positive Moment near Centers of End Spans, Negative Moment at First Interior Support, 2 dee ee pele ale ink inelb: Negative Moment at End Supports, (67) 2 M = not less than ft-lb. = 0.60wil?in.-lb. (68) (Building Frames) CONTINUOUS BEAMS OF EQUAL SPANS, RUNNING INTO COLUMNS Beams and slabs of equal spans built to act integrally with (a) Interior Spans. Negative Moment at Interior Supports except the Furst, ae pacts Mex 12 ft.-lb. = wi? in.-lb. intermediate and end columns, walls, or other restraining supports, and assumed to carry uniformly distributed loads, as shown in Fig. 94, p. 277, should be designed for the following moments at critical sections: (69) Maximum Positive Moment near Centers of Interior Span, wil? ; ve 16 ft.-lb. = 0.75wl? in.-lb. (70) 280 REINFORCED CONCRETE DESIGN (b) End spans of continuous beams for which the ratio of stiffness, 9 is less than the sum of the ratios of stiffness for the exterior columns, above and below a rigidly connected beam. Maximum Positive Moment near Center of Span and Negative Moment at First Interior Supports. wl? ; M 12 ft.-lb. = wi? in.-lb.3, Wiees ee Negative Moment at Exterior Supports, wil? Mi= 17 ft.-Ibs= wi? inelb. eee ee (c) End spans of continuous beams for which the ratio of stiffness, - is equal to or greater than the sum of the ratios of stiffness, a for the exterior column above and below a rigidly con- h nected beam. : Maximum Positive Moment near Center of Span and Negative Moment at First Interior Support, wl? me | Mo= Tz ft-lb. =] lw? m-th eee Negative Moment at Exterior Support, 2 M =“ ft-lb. = 0.7502 inlb, - . 12 (7) The formulas under (c) are adapted to ordinary floor construction where the beams run into column and concrete walls. Continuous Beams or UNEQUAL SPANS OR WITH NON-UNIFORM LOADING Continuous beams with unequal spans or with non-uniformly distributed loading should be designed for actual bending moments as directed under proper headings in Volume III. With unequal spans, provision should be made for bending moment in the columns. 9'Table on p. 945 gives formulas for moments of inertia, J for various cross sections. DETAILS OF CONTINUOUS BEAMS . 2381 Bending moments for concentrated loads are also fully treated in Volume III. EFFECT OF VARYING MOMENT OF INERTIA UPON THE BENDING MOMENT As the bending moment in a continuous beam depends upon its moment of inertia, it will vary with the variation in the moment of inertia. The moment of inertia is seldom constant for the whole length of a continuous reinforced concrete beam. - This is especially true of T-beams and of beams provided with haunches. However, a thorough study of different conditions, by the authors, shows that the greatest variation in moment of inertia in a beam that is possible in ordinary design causes a variation in bending moment of less than 10 per cent. Under most conditions, the varia- tion is even smaller. Consequently, a continuous beam may be designed safely with the bending moments recommended on p. 279. The effect of varying moment of inertia is fully treated in Volume III. DETAILS OF CONTINUOUS BEAMS Continuous Beam at the Supports.—In the past, too little atten- tion has been paid to the details of reinforced concrete beams at the supports, with the result that a number of reinforced concrete structures have been built with beams and girders containing insuf- ficient steel at the top of the beam over the supports to take the pull caused by the negative bending moment, and insufficient area of conerete to take the compression.!° Not only do these beams fail to have the required factor of safety, but frequently even the working loads cause cracks that are always unsightly and sometimes dan- gerous. Moreover, moisture is likely to penetrate the open cracks and rust the steel. Just as much care, therefore, is necessary in designing the rein- forced concrete beam at the supports as in the middle of the span. Not only the tensile stresses in the steel, but also the compressive stresses in the concrete must be computed and properly provided for. In T-beams particularly, there is a likelihood of the concrete being overstressed at the supports. At the center of the span, the portion of the slab forming the flange is available for resisting compression. 10 See p. 282. 282 REINFORCED CONCRETE DESIGN At the supports, however, where the bending moment is negative, the flange is in the tensile part of the beam and the compressive area consists merely of the part of the stem below the neutral axis. Usu- ally, the compression area is not sufficient and it is necessary to increase it by using compression reinforcement. Allowable Working Stresses at the Support.—As is evident from the curves of bending moments, the negative bending moment decreases very rapidly, so that the extreme fibers are under maximum stress only in a very short length of the beam. It is therefore per- missible, at the supports, to use a higher compressive working stress in the extreme fiber than is used in the center of a beam where the rate of decrease in bending moment is much smaller and a larger part of the beam is under high stresses. The 1924 J oint Committee recommends an allowable working stress in concrete at the support of 900 lb. per sq. in. for 2 000 lb. concrete, when the working stress in the center of the beam is 800 lb. per sq. in. The authors indorse this recommendation. If other unit stress is used in the center, it may be increased at support by 15 per cent. For a stress in the center of 650 Ib. per sq. in. as specified by most Building Codes, a stress of 750 lb. per sq. in. is allowable at the support. Even with this allowance, it is often impossible, without special provisions, to keep the stresses within working limits. Compression Steel at the Support.—Frequently, additional com- pressive area is provided by reinforcement in the compressive part of the beam so that the beam must be considered and computed as reinforced at the top and bottom. The effectiveness of steel in com- pression has sometimes been questioned, but the results of tests with reinforced concrete columns, as well as with beams with steel in top and bottom (see p. 49), have proved conclusively that compression steel not only takes its share of the stress, but also stiffens the beam. First Method. (See Fig. 95).—In a reinforced concrete beam, part of the bottom steel is bent up, and the rest, usually composing one-half of the steel area, is carried horizontally to the supports. This steel may be utilized as compression reinforcement. As the bars get their maximum compression at the edge of the support, to be effective as compression reinforcement they must be extended into the support, a sufficient length beyond the edge to develop the stress by bond. (See p. 268.) A shorter length of imbedment for bond is required for compression than for tension, because the compression DETAILS OF CONTINUOUS BEAMS 283 stresses are smaller. This is the simplest method of providing com- pression reinforcement and should be used where possible. It is illustrated in Fig. 95, p. 283. The ratio of compression reinforce- ment, p’, equals about one-half of the ratio of the tensile reinforce- fe ment, or Beh, Pi -+:2| _----Length of imbedment for compression----~.._ «<— Fig. 95.—Compression Reinforcement at Support. First Method. (See p. 283.) Second Method. (See Fig. 96.)—When the amount of compres- sion reinforcement provided, as explained above, by the straight bars is not sufficient, additional reinforeement may be obtained by extending into the span under consideration the straight bars from the adjoining span. The length of the extended bar, measured from the edge of the support, will depend upon the bending moment, but NN, SS So —_—= = oe oes ee oe “\Z_____---Compression Steel from-- i Adjoining Span pie: Fig. 96.—Compression Reinforcement. Second Method. (See p. 283.) it must not be smaller than required to develop by bond, the com- pression stresses at the edge of the support. In turn, the straight bars from the span under consideration are extended into the adjoin- ing span so that the compression reinforcement at both sides of the support consists of the sum of the horizontal bars in both spans. Such an arrangement is shown in Fig. 96, p. 283. The ratio of the compression reinforcement at the support, p’, is thus about equal to , the ratio of tension reinforcement, pi. Thus a = |. 1 Since each horizontal bar is subjected to compression stresses ‘in the original span and also to equal compression stresses in the adjoining span, a question may be raised about the amount of com- 284 REINFORCED CONCRETE DESIGN pression stresses carried by the bar. The answer can be had by referring to Fig. 97. From this, it is obvious that the compression stresses in one span act in the opposite direction to the compression stresses in the adjoining span. Thus, the compression stresses in the bar from the two spans are not added but balanced. The compres- sion stress in the bar in one span acts as a reaction to the compression stress in the other span, and the stress 1s the same as though only one span were acting. Third Method—Sometimes the amount of compression reinforce- ment obtained by the second method is not sufficient. Ad- ditional reinforcement then can be supplied by short bars, placed at the support at the bottom of the beam. The length of the short bars will <--- Column Direction of Direction of ‘be governed by the bending Compression Compression ae in Pe ban moment, the minimum length Span being such that the projections Fic. 97.—Compression Stresses in Bar at beyond the column faces on Support. (See p. 284.) each side are equal to the length of imbedment required by bond for the expected compression stresses. The ratio of com- pression steel, p’, is then larger than the ratio of tension steel, p1. Depth of Beam Increased by a Haunch.—Another method of providing additional compression area is by increasing the depth of the beam at the support by a shallow haunch. Under ordinary conditions, this method is less desirable than increasing compression area by compression reinforcement. While material is saved, the excess cost of forms is apt to overbalance this saving. In many cases also, the haunch is objectionable from the architectural stand- point. A haunch is only useful, therefore, where it is desirable to avoid excessive compression reinforcement, or where more lateral stiffness is required for the structure. The required depth of haunch may be obtained by trial. A depth is assumed and then, with the aid of the diagrams on pp. 292 to 297, the stresses are computed and a new depth tried out if necessary. The depth is satisfactory if the compression stresses do not exceed the specified working stresses. The depth of haunch may also be found from the rectangular PROCESS OF DESIGNING CONTINUOUS BEAM 285 beam formula (Formula (1), p. 204). In this case, the effect of compression reinforeement may be neglected, because for deep haunches the bars are placed some distance from the extreme fiber. In any case, the error is on the safe side. Under ordinary conditions, the computation need be made only for the maximum bending moment. The point at which the slope should be ended can be readily found from the following formula: For a uniformly loaded beam, let M, = negative bending moment next to the support; M, = moment of resistance of the inverted T-beam without the haunch, governed by the concrete; a = length of haunch measured from face of column; 1 = clear span of beam. Then I — M, l ie r 11 2 ea (approximately). Evie tee Reaetey or) This formula errs slightly on the safe side. PROCESS OF DESIGNING CONTINUOUS BEAM _ After the floor is laid out and the spacing of beams fixed, the slab is designed first. Next the dead and live load per lineal foot of beam is computed by multiplying the sum of dead and live load by the spacing of beams and adding the estimated dead load of the beam. Knowing the condition of restraint at the ends, the shears and bending moments of the various spans are computed. | The depth of the beam is then decided upon. In a continuous beam of equal spans, it is desirable to use the same depth for all spans although the bending moments in the end spans are larger than in the center spans. This formula is based upon the fact that the point of zero moment is at approximately 4 of the span. The variation in the moment between the support and the } point is assumed to be a straight line. (This assumption is correct enough for the purpose.) Hence, the difference between the bending moment and the moment of resistance is in approximately the same ratio to the bending moment as is the ratio of the distance from the point where the haunch is needed to the point of zero bending moment. When the point of zero moment is not approximately at 1 span, the fraction may be altered accordingly. 286 REINFORCED CONCRETE DESIGN When the beam is rectangular, the problem is simple. The depth is determined for the largest bending moment by Formula (1), p. 204, and for the largest shear by Formula (86), p. 247. For a T-beam, the problem is solved as aN for T-beam design on p. 217. The selection of the proper depth for a continuous beam is some- what more complicated than for a single beam because the depth must be satisfactory for all spans. Experience teaches that in most continuous beams, the depth is governed by the compression stresses at the support. The proper solution then would be to select the ratio of compression to tension J steel, a and then find the required depth for the largest negative 1 bending moment as described on p. 237. It is desirable, if possible, to arrange the reinforcement as described in “ First Method” on , p. 283, in which case the ratio i = Qa: 1 If, in a series of beams, a few beams are. subjected to a dispro- portionately large bending moment at the support and if, in spite of this, it is desirable to retain the same depth for all beams, it will be found economical to use a large amount of compression reinforce- ment for these few beams. Considered by itself, such a design would not be desirable or economical. Taking the series of beams as a whole, however, the design will be more economical than if a deeper beam were used throughout. Having tentatively decided upon the depth of beam and width of stem, diagonal tension stresses are computed to see that they do not exceed the allowable working stresses. After the depth of beam is decided upon, the areas of reinforce- ment required in the different spans of the beam for negative and positive bending moment are computed, as discussed below. For main reinforcement, such bars are selected as will satisfy the positive bending moment in each span. Before these bars are finally accepted bond stresses should be investigated. Bending of bars is then decided upon, and reinforcement for negative bending moment investigated. After the points of bending of bars is decided upon, the required web reinforcement must be designed for beam in which the shearing unit stresses exceed those permitted for plain concrete. PROCESS OF DESIGNING CONTINUOUS BEAM 287 Tension Steel in Center of Span.—The size and number of bars necessary to make up the required cross-sectional area are selected for each span independently. If possible, bars of one diameter, or of diameters that do not differ by more than 4 in. should be used. It is desirable to use not less than two bars in small beams and not less than four bars in larger beams. If a larger number is required, bars of the largest diameter permissible by bond should be selected, as this reduces the number of bars and therefore the cost of handling. The number of bars should be chosen, also, to permit the required proportion of bars to be bent. up. When bending of one-half of the bars is contemplated, an even number of bars should be used. Whenever possible, the bars should be placed in one layer. Often it is cheaper to widen a beam sufficiently to have all bars in one layer than to introduce two layers of bars and thereby increase the the depth of beam. With two layers of bars, a strip of concrete, the width of the beam, and equal in depth to the diameter of bar plus 3 in., is wasted. This may be more than is required for widening the beam. In addition, with wider beam the extra concrete is used to resist shear, thus reducing the number of stirrups. Bending of Bars is Now Decided upon.—The bars should be bent with due regard to the bending moments. ‘The points of bend- ing should be far enough from the center of the span, where the bending moment has decreased sufficiently to permit the reduction of reinforcement. The location of points of bending is not the same for uniformly distributed loads as for concentrated loads. Diagrams, pp. 292 to 297, give the location of points of bending. Before using the diagrams, read the notes carefully. | In small beams, bars should be bent at one point only at each end. Many constructors favor bending the bars at one point only for beams of all sizes, because it reduces the number of types of bar on the job and thereby facilitates construction. The authors disagree with this practice and recommend bending bars in larger beams at several points. The disadvantage in the field is more than balanced by better distribution of reinforcement in the beam and by the increased effectiveness as diagonal tension reinforcement. If a large number of closely spaced bars are bent up at one point, they form ina beam an inclined plane which impedes the flow of concrete. This tends to form a plane of weakness, especially when the bars are not kept exactly in position. Proper distribution of the points of bend- ing prevent such conditions. 288 REINFORCED CONCRETE DESIGN When bending bars, there is a natural tendency to bend them symmetrically in the cross section. While symmetrical arrange- ment in a section is preferable, it is not absolutely necessary. Bend- ing bars in two points with unsymmetrical arrangement in cross sec- tion is usually preferable to bending all bars at one point. Tension Steel at the Support.—After the number and the size of the bars required in the center of the span are decided upon, the requirement at the supports is studied. - If Maximum Positive and Negative Bending Moments are Equal.— If the negative bending moment at the support and the maximum positive bending moment in the center of span are equal, and the beam consists of a number of equal spans and is of uniform depth, the arrangement of steel is simple. The positive reinforcement is designed so that it can be divided into two equal parts. Then one- half of the bars are laid straight and the other half bent up and carried across the support into the adjoining span to resist one-half of the tension there. An equal amount of steel is brought across from the adjoining span into the span under consideration. The sum of the effective bars at the support is then equal to the sum of the bars in the center. The requirement as to the amount of steel is satisfied. The only problem is to decide upon the points of bending of bars. The condition is somewhat different at the first interior support. Assuming again that maximum positive and negative bending moments in a span are equal, the bending moment at the first interior support is equal to the positive bending moment in the center of the end span, but is 20 per cent larger than the bending moment in the center of the interior span. If the depths are equal, the required area of steel is proportional to the bending moments. When one- half of the steel in the end span and one-half of the steel in the interior span are bent up and carried across the first interior support, the available area of steel for the negative bending moment will be about 10 per cent less than required. ‘The difference must be supplied by extra bars of proper length placed in the top of the beam at the support. At the wall support, where negative bending moment equal to 2 a is required, those bottom bars which are bent up, and which are half the total number of bottom bars, will not give sufficient reinforce- ment for the negative bending moment; neither will the reinforce- PROCESS OF DESIGNING CONTINUOUS BEAM 289 ment extend far enough from the column to take care of the tensile stresses at all points. It is necessary to add there, in the top of the beam, short bars provided with a hook at the support and extending to a point distant 0.27 from the edge of the support. If Maximum Negative Bending Moment is Larger than Maximum Positive Moment.—For continuous beams running into columns, wl? where the positive bending moment is 6 while the negative bending pol , moment is 75 to supply the required amount of tension reinforce- ment at the support, two-thirds of the bottom steel must be bent up and carried across the support. In other respects, the condition is the same as in the previous case. It may be found that with such an arrangement the straight bars are not sufficient to take care of the compression stresses. In such ease, it is a good practice to bend up a smaller proportion of bars. and supply the deficiency in tensile reinforcement at the support by extra short bars. If no Bars are Bent Up.—Sometimes, it is not desirable to bend _ up any reinforcement, but to use only straight bars for positive and negative bending moment. This method is used in short, deep beams where bent-up bars would not reach the top at the proper place to be effective as negative bending moment reinforcement. The positive bending moment reinforcement is then made up of straight bars, some of which extend the full length of the beam while the rest extend on both ends beyond the points of inflection. It is not per- missible to stop any bars short of the point of inflection, unless the bars are provided with a hook of proper dimension to develop the existing stress in bars at the point where the bar was stopped short. Otherwise there would be no means of inducing any stresses into the short bar without slipping of reinforcement. The top reinforcement at the support consists of short bars of proper area. These should extend on both sides of an intermediate support beyond the points of inflection. At the wall column, one end should be hooked at the wall and the other should extend beyond the point of inflection. A design in which the spans are not equal must be studied with more care. If the difference in span is large, the bending moments must be determined as explained in Volume III. In this case, it will not be possible to make up the area of tensile reinforcement at 290 REINFORCED CONCRETE DESIGN the support by the bent-up bars, and additional top bars will be required. Top Bars to Hold Stirrups.—To keep stirrups in position, two straight bars of small diameter are placed at the top, as shown in Fig. 81, p. 249. The stirrups are hooked on these bars and wired to them. With such an arrangement there is some assurance that the stirrups will not become disarranged in process of construction. POINTS AT WHICH HORIZONTAL REINFORCEMENT SHOULD BE BENT In a beam, the bending moments decrease from a maximum to zero at the points of inflection in continuous beams (and at points of support in simply supported beams); therefore, the required amount of reinforcement at intermediate points, for a beam with constant cross section, decreases proportionally. Consequently, it is per- missible to bend up a part of the positive moment reinforcement at proper points and utilize it either as diagonal tension reinforce- ment or as negative bending moment reinforcement. Ag the reduction of the required amount of steel depends upon the reduction of bending moments, the bending moment curves should be used as a guide in preparing bending sketches for the bars. To facilitate the work, bending moment diagrams for uniformly dis- tributed loading are given on pp. 292 to 297 for different conditions of restraint at the support. Under each moment diagram, corre- sponding bending details are given. It should be noted that in the bending moment diagram the maximum bending moment is assumed as a unit and the intermediate bending moments are ex- pressed as a fraction of the maximum moment. Where the positive and negative bending moments are equal, the assumed units for both eases are equal. Where the maximum positive bending moment is different from the maximum negative bending moment, the mag- nitudes of the units are also different. To facilitate the use of the diagram, horizontal lines divide the maximum bending moment into ten parts. The span is also divided. In this way, the bending moment at any point may be found from the diagram in terms of the maximum moment. In all diagrams for continuous beams, the distance between the supports was assumed as the span, in accordance with the rec- ommendation given on p. 277. The diagrams may be used for other assumptions as to the length of the span. However, if for any POINTS AT WHICH REINFORCEMENT SHOULD BE BENT 291 reason it is required to assume the point of support to be within the column, the distance of points of bending of negative reinforcement should be based on the net span and measured from the edge of sup- port (and not from the theoretical point of support). Explanation of Max. and Min. Dimensions in Bending Diagrams. —In the bending diagrams for each bend, two hmiting dimensions are given, one at the bottom and the other at the top. ‘This does not mean that each bent bar will be bent at both these points. The two points are limiting points. The bottom dimensions are given as, say, Max. 0.251. This means that the distance of the bent of the bar from the edge of the support must not exceed 0.251. As far as positive bending moment is con- cerned, however, the bent may be moved nearer towards the support and the distance of the bent from the support may be smaller than the allowable maximum value. The top dimensions are given as, say, Min. 0.15/. This means that the bent must not be placed nearer the support than 0.151. As far as negative bending moment is concerned the position of the top bent may be moved away from the column. Any bend between these points is satisfactory. In selecting the actual points of bending, the diagonal tension should also be con- sidered. In fixing the points of bending, it is always advisable to take into account the possibility of small inaccuracies in bending of bars and provide for it by placing the points of bending a few inches from the theoretical point. Thus, the point at the bottom should be a few inches nearer the support, and the point at the top a few inches farther from the support, than the theoretical values. In practice, the point of bending at the bottom will be assumed, and the bar bent at an angie of 35 to 45°. The point of bending at the top is obtained automatically. If the distance from the support to this top point is equal to or larger than the minimum given in the bending diagrams, the design is satisfactory. If the point at the top comes too near the support, a steeper angle should be used. Although the angles recommended above are desirable, as then the bar is most effective as web reinforcement, angles of any steepness may be used if necessary. The use of the diagrams is shown in Example 1, on page 578. 292 REINFORCED CONCRETE DESIGN (4M, 7M, : jie Ot _ 0.3 M,.. 1 = 0.2 M; a|S Ae ae Se 01M & eee 6.20 Bending. Moment Curve ns OLS 12 4 SS —_—--- ~~ A,=Total Area Pos. Reinf. omar. 0.221 --------- Bending Details Fic. 98.—Beam Freely Supported at Both Ends. (See p. 290.) Note: For formula of bending moments see p. 278. For explanation of terms Max. and Min. used in bending diagrams see p. 291. The limiting dimensions for bending do not always coincide with selected points for bending of bars. POINTS AT WHICH REINFORCEMENT SHOULD BE BENT 293 posal bone de EG PIS OE) EME us mer 0.151412; Bending Moment Curve seencb nce Weg. Reinf=Ag at Right ee. -28 A,= Total Area Pos. Reinf. Trax pom eS dary Bars Bent at One Point ~Brick Neg. Reinf=A, at Right 0.21412 1 ‘ SA 0.09A,. A,=Total Area geseteeteen sy ha Pos. Reinf. e-Mae 028 ~Max. 0,131 1 ee er eee ee ere ee ree Bars Bent at Two Points - ‘Brick Bending Details Fic. 99.—End Span of Continuous Beam, Freely Supported at End. (See p. 290.) Note: For formula of bending moments see p. 279. For explanation of terms Max. and Min. used in bending diagrams see p. 291. The limiting dimensions for bending do not always coincide with selected points for bending of bars. 294 REINFORCED CONCRETE DESIGN pan 2 “a Miki Bending Moment Curve Min. 0.101------ <0. St Neg. Reinf.=A, each End |<0.21+12- Min. 0.051 : At As | 4A,-~- Bars Bent at Two Points -- Max. 0.3l-- A xTotal Area of Pos. Reinf. Nee Point RN ae Re See a a yor Bars Bent at Ee 0.21 Bars Bent at Three Points Bending Details Fic. 100.—Interior Span of Beam Continuous for More than Two Spans. (See p. 290.) Note: For formula of bending moments see p. 279. For explanation of terms Max. and Min. used in bending diagrams see p. 291. The limiting dimensions for bending do not always coincide with selected points for bending of bars. POINTS AT WHICH REINFORCEMENT SHOULD BE BENT 295 on VL eee 0.551 : Do Site = AS ie. | SP Re ae ee COE ee I 1 | OUT BS ie i ae ee ee Pun SE 2a “seo Ah sa SIS pea se Tah Sled st J Se Be Bs UR LI ee Be : S Nisioer 1 1) | Se Oma ae ude! Sib. ea eters es a ES | oN ae ck) maAN Pia = 3 , S E - m2 = ----- - --------- -----] ---- -------- Bending Moment Curve Neg. Reinf. =A, at Right 0.7A, at Left. _ars Bent at One Point ke.-9 Q]->4 Neg. Reinf.=A , at Right yr 0.212: m OD Alect letter ear Max ae A, = Total Area Pos. Reinf. 0.25; <- Mas. 0,321 Bars Bent at Two Points Bending Detatls : Fic. 101.—End Span of Building Frame End Partly Restrained. (See p. 290.) Note: For formula of bending moments see p. 280. For explanation of terms Max. and Min. used in bending diagrams see p. 291 The limiting dimensions for bending do not always coincide with selected points for bending of bars. 296 REINFORCED CONCRETE DESIGN NEGIMA EST SRI a0, CHE eae REBUMIADE GSES Ne Ra ged So eee Srey MANES Be a eel | ee oy SEA a Sols be a mmbety 1 ENE ss DNS) eee a Bl ee SAAS PEN afer) an Se Pe ep Lo Vee a Ae Sees Ni ee Pe I AW a Gea Py SG GNM mile aay SS Rey me BLE ee GEN Es har FN waele 5, Ses muy ~en maw eer eee am — eens == Bending Moment Curve 0.21412 As=Total Area Pos. Reinf. \_Ma : witene-- +--+ --------------] ecw ewan cana ness s- ses ee Bars Bent at One Point 0.21412" 0.21412" ~+--6 “ Neg. Reinf=A, Each End “>f~--~4 Min.0.101 Min.0.101-7 : Min. 0051-4" ---> -7 = a -- 5 ween Tage NOX 0. ‘As A =Total Area Le-Max. 0.31---- _ Pos. Reinf.___._. ee ee D Bending Details Fic. 102.—End Span of Building Frame. End Fully Restrained. (See p. 290.) Note: For formula of bending moments see p. 280. For explanation of terms Max. and Min. used in bending diagrams see Pp. 291. The hmiting dimensions for bending do not always coinciae with selected points for bending of bars. POINTS AT WHICH REINFORCEMENT SHOULD BE BENT 297 . == ane : Bending Moment Curve 0.21 + 12" Neg. Reinf.=1 3A, each End Str. Bar <~-Min. 0.071 ; gd Min. 0.081 t int Pare Pea Ot Tie foints Bars Bent at One Point One - Half Positive Reinforcement Bent » Neg. Reinf. =1 +A, each End 0.21 + 12> PIR cha A, = Total Area Pos. Reinf, | Max. 0.21 ----2----->| > Bars Bent at Two Points Bars Bent at One Point Two-Thirds Positive Reinf. Bent Bending Details Fic. 103.—Interior Span of Building Frame. (See p. 290.) Note: For formula of bending moments sce p. 279. For explanation of terms Max. and Min. used in bending diagrams see p. 291 The lmiting dimensions for bending do not always coincide with selected points for bending of bars. | 298 REINFORCED CONCRETE DESIGN REINFORCEMENT FOR TEMPERATURE AND SHRINKAGE STRESSES All masonry is subject to temperature cracks, but when they are distributed in the many joints between bricks or stones they do not show so plainly as on the smooth surface of concrete. Expansion due to a rise in temperature rarely causes trouble except at angles where the lengthening of the surface may produce buckling or sliding of one portion of the wall past the end of the other. Ina building, the walls and floors are generally so well bonded together and free to move as a unit, that no provision need be made for expansion. In a structure like a square reservoir, the effect of expansion must be taken into account in the design, to prevent failure at the corners. Contraction is often more serious, although cracks are by no means necessarily dangerous. ‘Temperature reinforcement is ordi- narily used to prevent excessive cracking due to the shrinkage of the concrete in hardening or to the lowering of the temperature (see Volume II. Also, expansion joints are used in long structures. Temperature reinforcement, placed in the direction of the expected contraction, does not prevent cracking entirely; but it distributes the cracks in such a way that numerous very small cracks, which are often invisible to the eye, are formed instead of a few large ones. To accomplish this purpose, the temperature reinforcement must be sufficient in quantity, and it should consist of bars of small diameter placed as close as practicable to the surfaces. Deformed bars, that is, bars with irregular surfaces which provide a mechanical bond with the concrete, are more effective in distributing cracks than smooth bars. Steel of high elastic limit also is advantageous. The area of temperature or shrinkage reinforcement used in practice varies from 0.2 to 0.4 of 1 per cent of the cross section of the concrete (p = 0.002 to 0.004). ; The amount of temperature steel to be used depends upon con- ditions and the degree to which the cracks should be reduced. Ifa water-tight job is required, the higher value should be used. The tensile strength of concrete is so low that a small change in temperature will crack it. For example, the coefficient of expansion of concrete is 0.0000055 (see Volume IT) and the modulus of elastic- ity is generally assumed as 2,000,000; therefore, the stress per degree Fahrenheit is 0.0000055 X 2,000,000 = 11. Ib. peresanaie, REINFORCEMENT FOR TEMPERATURE AND SHRINKAGE 299 and a fall in temperature of 32° = 27° is sufficient to crack a con- crete the tensile strength of which is 300 lb. per sq. in. It is evident, and it has been proved by experience, that there is less cracking in concrete laid in cold than in warm weather. Longitudinal reinforcement is especially necessary in conduits, which must be water-tight. Cracks due to shrinkage in hardening of the concrete may be pre- vented by keeping the concrete wet. (See Volume II.) It has been suggested by Mr. Charles M. Mills that the relation between the tensile strength of the concrete and the bond with the bars is an important factor in governing the size of the cracks; and the following analysis, based on his suggestions, gives a means of esti- mating the size and distance apart of the cracks, thus forming a basis for judgment as to the sizes and percentage of steel to use. The tensile stress in the steel at a crack tends to pull out the bars from the concrete, and, referring to Fig. 104, the bond stress of the bar in the length ab must equal the tensile stress in the whole cross section of the concrete at b caused by the contraction of the concrete. Let x = distance apart of cracks inches; D = diameter of round bar or side of square bar; p = ratio of cross section of steel to cross section of concrete. Then,!2 if, as is sufficiently accurate for practical purposes, the strength of concrete in tension is assumed to be equal to the bond between plain steel bars and concrete, the distance apart of cracks is CL = ; “ for square orround bars. . . . (76) The distance apart is inversely proportional to the unit bond,” so that a deformed bar having twice the bond strength would space 12 In addition to above notation, let A = area of section of concrete; u = unit bond between plain steel and A; = area of section of steel; concrete; o = perimeter of steel bar; f,; = unit tensile stress in steel; fi = tensile stress in concrete; D = diameter of bar. 2Af 2A Then Af; = juor, or x = =! If fp =u, 2 = —, and since p = —, uo 0 a. 2A A D 1D ~=-——. Also, — = — for both round and square bars, hence x = 5 —. op O 2p | 300 REINFORCED CONCRETE DESIGN the cracks one-half as far apart and allow them to be only one-half as wide. _ It is evident that the distance apart of the cracks is proportional to the diameter of the reinforcing bars, and inversely proportional to the percentage of steel. a 6b Fig. 104.—Reinforcement for Temperature Stresses. (See p. 299.) From this formula is tabulated the estimated percentage of rein- forcement for different spacing of cracks and different sizes of bars, assuming the bonding strength of the steel to the concrete to equal the tensile strength of the concrete. Estimated Percentage of Reinforcement for Different Spacing of Cracks Distance Apart of Cracks Pisin Barsey so. oso Ve 18” 24% 36” 48". 60”’ Deformed Bars *..... 8” jee 16” 24” oo 40” % % % % % % 1) 1.04 0.70 0.22 0.35 0.26 0.21 Diameter of round | 2”) 1.56 1.04 0.78 0.52 0.39 0.31 or side of square | 3”) 2.08 1.39 1.04 0.69 0.52 0.41 bar: Ae Bie 5" 2.60 1.74 1.30 0.87 0.65 0.52 Boeke 2.08 1.56 1.04 | 0.78 0.62 i”) 3.65 2.44 1.82 122 0.91 0.73 Pe er. 1.04 0.83 2.78 2.08 Ber: Norte: To express the steel as the ratio of area of cross section of steel to cross section of concrete, divide the percentages by 100; thus 1.04 becomes p = 0.0104. * Assuming the bond of deformed bars to be 50 per cent greater than plain. The size of the crack is governed by the amount of shrinkage and for cracks due to temperature changes may be estimated as the product of the coefficient of contraction (0.0000055) by the number of degrees fall in temperature by the distance between cracks, REINFORCEMENT FOR TEMPERATURE AND SHRINKAGE 301 Estimated Width of Cracks for Different Distances Apart Distance Apart Change of Temperature ja 18” 24’ 36” 48’ 60’ CT OSL SE ited ir 0.0020 | 0.0080 | 0.0040 | 0.0059 | 0.0079 | 0.0099 ae ee 0.0033 | 0.0050 | 0.0066 | 0.0099 | 0.01382 | 0.0165 CO SD i ora 0.0046 | 0.0069 | 0.0092 | 0.0139 | 0.0185 | 0.0232 * 30° corresponds to a shrinkage of 0.017 per cent; 50° to 0.028 per cent; 70° to 0.038 per cent. From this, if it can be determined how large a crack will be allow- able, the corresponding spacing can be obtained. To avoid large cracks it may be necessary to use enough steel to prevent its passing its elastic limit. If the bars are continuous for such a length that the ends are practically immovable, as in a long retaining wall, a drop in temperature, tending to shorten them, pro- duces a tensile stress which is independent of the distance between the restrained ends. Assuming the coefficient of expansion of steel to be the same as that of concrete and the modulus of elasticity of steel as 30 000 000, this stress is 30 000 000 X 0.0000055 = 165 Ib. per sq. in. per degree of temperature, or for 50° F. is 8 250 lb. per sq. in. This is well within the elastic limit of the steel and would not, of itself, cause the steel to take a permanent set. However, since the concrete surrounding the steel will be continuous except at certain cracks, the stretch in the steel may be unevenly distributed and largely confined to the immediate vicinity of the cracks. If cracks occur while steel is unstressed, through the concrete shrinking, the steel tends to resist the shrinkage by tension at the crack and compression at the center of the block of concrete, and the tensile stress will be equal to the compressive and each equal to one-half the tensile strength of the concrete. This may be expressed by the following formula, in which the foregoing notation is used.!3 1 fe = gfe 13 Af, : fad, A ; toy 15 iil on NR Pore Maaco 302 REINFORCED CONCRETE DESIGN See the tensile stress in the concrete is liable to be low at the time shrinkage cracks are formed, it may be assumed, for illustration, as 200 Ib. per sq. in., making LO) I's a This represents the stress due to local cracks, which is additional to the temperature stresses above described. The total stress is, therefore, for 50° change of temperature 8 250 +- f’, or 8 250 + If the clastic limit of the steel is 40 000 Ib. per sq. in., and we must _ keep below this, 40 000 = 8 250 + - and p = 0.0031. For steel, the elastic limit of which is 50 000 Ib. per sq. in., 50 000 = 8 250 + ~" and p = 0.0024. These values of p represent the lowest theoretical ratio of area of cross section of steel to area of cross section of concrete which can be used without the steel passing its elastic limit at certain of the cracks when the ends are restrained or the length is so great that inter- mediate parts are practically restrained. In view of the very slight stretch required to relieve the stress in the bars when the elastic limit is exceeded, and the probability of its distribution by the restraint to movement by the mass, it is not always essential to consider the elastic limit. CHAPTER VI DESIGN OF FLAT SLAB STRUCTURES The aim in this chapter is to present a complete and logical treat- ment of the design of the flat slab. Precedent and common practice are followed when they agree with sound engineering. A number of rules of thumb, however, which have accumulated during the early stages of flat slab development have been eliminated, and rules based on logic and on accepted principles governing other types of concrete design have been substituted. In this chapter, besides covering the general principles of flat slab design, the authors aim to treat the details in such a way as to give the engineer, and also the student, the basic information and instruction required to design a safe and economical slab adapted to any required conditions. Design of interior columns is treated on p. 305; that of exterior columns on p. 312; column heads and drop panels, p. 319; rules for thickness of slab, pp. 325 and 337 ; bending moments to be used in design, p. 328; formulas for compression stresses in slab, p. 341; use of compression reinforcement in flat slab design and formulas, p. 345; shearing stresses in flat slab, p. 347; method of computing tensile reinforcement and arrangement of bars in flat slab, p. 351. For computing bending moments in flat slabs, the authors have adopted the widely accepted method based on the division of the slab into parallel strips. This permits a fairly well balanced design which, with proper tables, can be worked out without excessive labor. For a method more logical in theory and still simpler in practice, the reader who desires to investigate theory is referred to the treatment in the Second Edition of this book, which was worked out by the authors. The strip method, however, has become established by precedent and gives workable results; for the present, therefore, it is accepted by the authors in preference to introducing a different type of formulas. Further experiment and analysis will probably throw new light on the subject. 303 304 DESIGN OF FLAT SLAB STRUCTURES The formulas for bending moments, recommended for use, agree in general with those given by the 1924 Joint Committee, which, in turn, correspond with those now in common use. Modifications have been made by the authors, however, where necessary for uni- formity of design. The formulas may be safely used for rows of equal or substantially equal panels. Unusual designs should be analyzed according to formulas given in Vol. III. In all respects, the material in the chapter has been based upon the considerable experience of the authors in design of flat slab con- struction, and upon various tests, many of which were made by or for the authors. Advantages of Flat Slab Construction.—In flat slab construction, concrete is used more logically than in beam and girder construction. Whereas, in beam and girder construction, the load is transmitted from the slab to the beam, then from the beam to the girder, and finally from the girder to the column, in flat slab construction the load is transmitted directly by the slab to the column. The load, irrespective of its position on the slab, is carried by the whole slab; therefore, overstressing, due to concentrations of load, is impossible. The under side of a flat slab is practically flat. Because of the absence of exposed corners, flat slab construction is less vulnerable in case of fire than beam and girder construction. The formwork for flat slab 1s simple. Forms for column heads and drop panels, usually made of metal, are adjustable and can be rented at a small cost. - Ordinarily, scarcely any adjustment of forms is necessary for different floors designed for different loadings, since the change in thickness of slab does not require change in formwork. Change in thickness of drop can be readily accomplished. Universal Adoption of Flat Slab Construction Owing to its many advantages, flat slab construction has practically superseded beam and girder construction for spans Up to 30 ft. and live loads over 100 lb. per sq. ft. For smaller live loads, light-weight floors are’ often more economical. For more detailed discussion, see Chapter and also discussion of various types of buildings. Description of Flat Slab Construction.— Reinforced concrete flat slab construction is distinguished from all other types of slabs by the absence of beams and girders. The slab carries the load and is supported directly by columns, which are provided ordinarily with enlarged heads, called column heads or capitals. For heavy loads, the slab is thickened at the column by drop panels or plinths. The COLUMNS IN FLAT SLAB CONSTRUCTION 305 _ elements of flat slab construction are, therefore: columns, column heads, slabs, and drop panels. For light loads, the drop panels, and sometimes even the column heads, are omitted. The functions of the elements are discussed under separate headings. Flat slab construction is illustrated by photographs in Figs. 105 and 106. In Fig. 105 is shown light flat slab construction without drop panels. The pleasing appearance of the room, which is used as an office space, is noticeable. Attention is called also to the dis- tribution of the light. Fig. 106 shows a heavy flat slab construction with drop panels, as used in a warehouse. Reference is also made to the photograph of flat slab building, Fig. 185, p, 572, and also to the perspective view, Fig. 187, p. 574. COLUMNS IN FLAT SLAB CONSTRUCTION In flat slab construction, distinction is made between interior and exterior columns, not only on account of their shape but also because of the difference in function. INTERIOR COLUMNS. Interior columns may be round, square, octagonal, rectangular, or oblong in cross section, as required. In most structures, the only consideration affecting the shape of the column is economy of the formwork for the column and the column head. Where metal forms are easily obtainable, round columns and round column capitals are most economical, especially with spirally reinforced columns. Octagonal or square columns are sometimes specified. Octagonal columns are considered by some architects to be more pleasing in appearance than round columns. Square and _ octag- onal columns are also preferred, in some cases, because it is easier to fit partitions to the flat surfaces of the columns. Square and octagonal columns are also used where metal forms are not readily obtainable, since wood formwork for either square or octagonal columns and column heads is cheaper than for round columns. Rectangular or oblong interior columns are used where it is desir- able to keep the clear span in one direction as large as possible. Such may be the case in manufacturing establishments, such as shoe factories and textile mills, in which the manufacturing processes or machines run in one direction, and a strip of the building of the width 306 DESIGN OF FLAT SLAB STRUCTURES Fic. 105.—Interior of Office Building. (See p. 305.) Densmore LeClear and Robbins, Architects. 8. M. I. Engineering Co., Consulting Engineers. Fic. 106.—Interior of Warehouse. (See p. 305.) Chas. H. Way, Architect. §S. M. I. Engineering Co., Engineers. COLUMNS IN FLAT SLAB CONSTRUCTION 307 of the column is unoccupied space. Rectangular columns are some- times used in garages and in printing establishments, to make more room for the machines between the columns. Design Formulas.—lormulas for design are given in the chapter on Column Design. The economy of design of the columns is also discussed there. INTERIOR COLUMNS FOR SYMMETRICAL ARRANGEMENTS OF PANELS When the arrangement of panels is symmetrical about the center line of the column and the load is uniformly distributed over the whole area, the interior column is subjected to vertical load only, and no bending moment acts in it. However, in the actual use of the building, occasional unbalanced loading is unavoidable. This is caused by live load placed in panels on one side of the column line, while the opposite side is not loaded. Such loading produces bending moments in the interior columns. Ordinarily, the possibility of such bending moment is taken care of by limiting the minimum size of the interior column to a definite ratio of the span. The columns are then designed for the vertical load only. Limitation of Minimum Size of Interior Column.—Most specifi- cations and building codes require that the diameter of the interior round column should be not less than a certain fraction of the span measured center to center of columns. The requirements of various building codes are given in the table on p. 308. Authors’ Recommendations.—The authors recommend the fol- lowing limitations of interior columns: ‘ Limitation of Outside Diameter Design Live Loads, Ib. per sq. ft. of Interior Round Column 60 Ib. and under No limitation 70 to 100 Ib. zis of span 110 to 200 Ib. z's of span 210 to 300 lb. 5 of span For higher loads, computations of bending moments should be made. The above limitations apply when the bending moment due to unbalanced load is resisted by a column above and below the floor. The columns should be provided with vertical bars of an area equal to at least 1 per cent of the effective area of the column, but not 308 DESIGN OF FLAT SLAB STRUCTURES less than four 2-in. round bars should be used in square columns nor less than six -in. round bars evenly spaced around the circum- ference in round columns. If no column extends above the floor, the minimum diameter of column should be increased by 15 per cent, Limitation of Diameter of Interior Columns for Flat Slabs With Symmetrical Arrangement of Panels (In Effect in 1925) Cit matte Diameter| Other Diameter Provision for y in Terms of Span Limitations Unbalanced Loading Bostonians. eaters z's span in longer direction Chicago ais 0" «3 zis average span zis clear height (| =}; average. span A Ate Cine | é for floors Min diameter, qs average span 14 in. for roof New YOrk, 2. twa. z's average span Min., 16 in. for | Moment of zoW il* to round, and 14 be resisted by column in. for square above and below in a proportion to their ‘ Philadelphia. 2.2)? sak 6 04 = 29k 4: Bee ee Interior columns should be capable of resist- ing unbalanced bend- ing moment pro- duced by panel with live load adjacent to a panel without | live load Joint Committee, 1924) 3 eee No special |requirement given * W , = total live load on panel in lb.; 1 = average span center to center of columns in feet. Moment is in ft.-lb. : or the amount of steel increased beyond one per cent, in accordance with the discussion given below. The limitation of the column size serves the purpose of supplying sufficient rigidity for the column to take care of the bending moment COLUMNS IN FLAT SLAB CONSTRUCTION 309 produced by occasional unbalanced loading. The limitations given are based on experience and are satisfactory with average spans and average heights of columns. In special cases, the bending moment due to unbalanced live load in the interior column should be com- puted in the manner given below. This should be done for unusual heights of columns, special sections, and especially heavy live loads. Stresses due to bending should be computed when the loading in the panels is not uniformly distributed, for instance, if one panel carries much heavier machinery than the adjoining panels. The authors’ recommendations are based on the following rea- soning. Since the limitation of the diameter of interior column is for the purpose of supplying in it sufficient resistance to bending caused by unbalanced loading, the required resistance should depend not only upon the span but also upon the magnitude of the live load and upon _ the possibility of unbalanced loading. If a column with a diameter equal to, say, +51, is rigid enough for a warehouse designed for a live load of 150 Ib. per sq. ft., where it may have to resist large bending moments caused by one-sided loading of panels, a smaller diameter may be used with equal satisfaction for an office or a hotel with live loads from 40 lb. to 70 lb. per sq. ft. and where heavy unbalanced loading is practically impossible. The limitation of the diameter of column to a fraction of the span, say ;';l, applies directly to round or octagonal columns provided with a minimum amount of vertical steel. An increase in the amount of steel increases appreciably the magnitude of the moment of inertia. Therefore, if it is necessary for any reason to use a column having a smaller diameter than the required rigidity may be l 15’ obtained by using an increased amount of vertical steel, so that the moment of inertia of the column is equal to that of a column with a diameter equal to “ but with the minimum amount of steel. If square or rectangular columns are used, the limitations in the table do not apply directly. In such cases, the column selected should have a moment of inertia equal to that of a round column of a diameter equal to the required fraction of the span. The moment of 4 ° . inertia of a square with a side, d, is is or 0.082d*, while for a circle with a diameter, di, it is 0.049d;4.. For equal moment of inertia, 310 DESIGN OF FLAT SLAB STRUCTURES 4 = = 0.049di4, from which d = 0.87di. This means that, for equal moment of inertia, the side of a square may be only 0.87 of the diameter of the round column. When the minimum diameter of round column is required to be 7s!, the minimum side of a square column 1 should be 17.2" For a wide rectangular column equivalent to a round column with a diameter equal to ;';/, the thickness of the column, for rigidity, may be much smaller than jl, as illustrated below. Example of Square and Rectangular Columns.—Tlor a specific example, assume a 25-ft. square panel and a limiting fraction j'sl.. The minimum allowable 25 X12 15 -the side of the square may be made equal to 20 X 0.87 = 17.4 in. Now assume that for a special reason it 1s desirable to reduce the thickness of the column in one direction by using a rectangular column 30 in. wide. The moment of inertia of arectangular column, where b is width and d thickness, is 7,bd*. For b = 30, this is 39d? = 2.5d%. The moment of inertia of a 20-in. round column is 0.049 X 204. Comparing the two values, 2 5d3 = 0.049 X 20!, from which d= 14.6in. Thus, in this case, a 20-in. round column, a 17}-in. square column, or a 14.6 by 30-in. rectangular column may be used. diameter of a round column is — 20 in. If a square column is used, In a rectangular panel with a round or square column, the min- imum diameter is based on the average span. If a rectangular or oblong column is used, the rigidity in the long direction may be made equal to that of a round column of a diameter equal to the specific fraction of the long span, and the rigidity in the short direc- tion may be made equal to that of a round column with a diameter equal to the same fraction of the short span. , The rigidity of the column depends not only upon the moment of inertia but also upon the height of the column. Rigidity is expressed by the ratio 7 The limitation given above applies to columns of ordinary height, say up to 14 feet. The rigidity decreases with the increase in height of the column. Therefore, for higher columns, the diameter of the column should be increased above 4'5/ if it is desired to make its ratio of stiffness equal to the ratio of stiffness of a 14-foot round column with a diameter ;'5!. Bending Moment in Interior Columns.—Some building codes, notably those of the cities of Philadelphia and New York, and the 1924 Joint Committee recommendations require that interior columns COLUMNS IN FLAT SLAB CONSTRUCTION oll be computed for the bending moment produced by the live load placed on one side of the column with adjacent panels without live load. This bending moment may be computed by the following formulas, which apply when the column extends above and below the floor under consideration. Let W;= total live load on the loaded panel, Ib. ; ! = panel length, ft.; M, = bending moment produced by unbalanced live load. Then Bending Moment due to Unbalanced Live Load, 1 M, = 40 W, HR By ee ss ‘ 5 é (1) PEON Vapi ith lain wh 2069} This bending moment should be considered as resisted by the columns above and below the slab in proportion to their ratios of I, stiffness i: If the stiffness of the two columns is equal the moment M M to be resisted by each column equals sat When the difference between the ratios is appreciable, the proportion of the moment resisted by each column should be computed. Usually it is near enough to assume that the lower column resists 0.61 and the upper column 0.4M;. The points of application of the bending moment are evident — from Fig. 107, p. 311. This bending moment is combined with the column load, and stresses are computed by formu- pe las for direct load and moment Panel». (see p. 163). The combined compression stresses due to the direct load and _ Tension--f the bending moment should not exceed the stresses recommended on p. 463. Of course, the column must be strong enough to resist the maximum direct stresses without exceeding the stresses specified for Fig: 107:-ePendineemromene: aise columns. terior Columns. (See p. 311.) If the column steel is not suffi- cient to resist the tensile stresses, additional bars should be added. Loaded Panel 312 DESIGN OF FLAT SLAB STRUCTURES Since, ina square panel, some bending moment may occur at any side of the column, the required additional reinforcement must be’ so placed as to be effective on all four sides of the column. It is best to place such bars in the corners, since then they may be con- sidered as effective for bending moments acting on either of the adja- cent sides. In rectangular panels, bending moments in both direc- tions may have to be computed. It is not necessary, in columns, to consider unbalanced live loading due to the roof load. The most unfavorable effect of the bending moment on the column will be found in the top story column, where the bending moment due to floor load is essentially the same as in other floors, but the size of the column and the amount of reinforce- ment is smaller than in any other story. Unsymmetrical Arrangement of Interior Columns.—For unsym- ‘metrical arrangement of interior columns, the bending moments in columns exist even for uniform loading. The required bending moments in columns and slabs should be determined as explained in Volume III. EXTERIOR COLUMNS. Exterior columns are usually rectangular in shape. A rectangular - column imbedded in the wall projects less into the building than a square column carrying the same load. Also, exterior columns often serve as pilasters between windows, in which case their width depends upon the width of the window sash. For this and other reasons, it is advisable to make the width of the column constant for the whole height of the building. Some specifications (notably the Flat Slab Regulation of the City : : : beeeens: aa) OL Oa ete quire large mini- mum dimensions Typical Exterior ee Special Exterior Columns for wall columns. Fic. 108.—Cross Sections of Exterior Columns. In such cases, rect- (See p. 312.) angular columns Note: Reinforcement not shown. may not be eco- nomical. The sections shown in Fig. 108, p. 312, are often used under these conditions. Design of Exterior Columns.—Exterior flat slab panels, when supported by concrete columns, are designed as partly restrained at the ends. To get this favorable condition for the slab, the exterior COLUMNS IN FLAT SLAB CONSTRUCTION 313 columns must have sufficient rigidity to produce in the slab the required restraint. Also, they must have sufficient reinforcement to resist the bending moment transferred from the slab to the column. The fact that exterior columns are subjected to bending and that the strength of the slab in the wall panels depends upon the strength and rigidity of the exterior columns cannot be overemphasized. Outside columns of light or improper design are the most frequent cause of weakness in flat slab design. Required Rigidity of Exterior Column.—The rigidity of a member depends directly upon its moment of inertia, J, and inversely upon its height, h, or length, 1. It is represented by the ratio - for columns, and by : for beams or slabs. Exterior columns should be made rigid enough to develop in the slab, at the wall columns, at least eight-tenths of the negative bend- ing moment in an interior panel. By the theory of least work, the authors find that for this purpose the relation between the ratio of rigidity of the column and that of the slab is expressed in the formula below. | Let J = moment of inertia of slab; 1 = length of span perpendicular to the wall; I, = moment of inertia of column; h = height of column. Then . mes Sy reenter aetna ee eentCy From this relation, the required dimensions of the column may be computed as given below. The moment of inertia of the slab should be based on a thickness equal to the thickness of the slab plus one-third of depth of drop panel, and a width equal to the spacing of the exterior columns. The minimum thickness of a rectangular wall column for a given width, 6, determined by the required rigidity, may be easily found from the following equations: Let J = length of span of slab at right angles to the wall, from center to center of columns, ft.; 1, = length of span of slab parallel to the wall, from center to center of columns, ft.; 314 DESIGN OF FLAT SLAB STRUCTURES h = height of column from center to center of slab, ft.; b = width of column, ft.; d = depth of roof slab or top floor slab, in.; d, = minimum thickness of column, in.; Then I dy = 1.154) a rs For slab with drop panels, the value d in the above formula equals the thickness of the slab plus one-third of the depth of the drop panel. Bending Moment in Slab at Wall Column:—The magnitude of the negative bending moment in the slab at the wall column, which is transferred to the columns, depends upon the relative rigidity of the column and the slab. This bending moment should not be taken as less than eight-tenths of the negative bending moment at the column in an interior panel of the same dimensions. For this condition, the relation between the ratios of stiffness of the column and the slab should be as expressed in equations (3), p. 313. Where the wall columns are much stiffer, negative bending moment in the slab at the wall columns should be increased. Bending Moments in Exterior Columns.—The negative bending moment developed in the slab at the wall, the magnitude of which is discussed above, is transferred to the exterior columns and must be resisted by them. In the roof, the total bending moment is resisted by the column below. In lower stories, the bending moment is resisted partly by the column above and partly by the column below. The proportion of the moment resisted by the two columns depends upon their ratio of stiffness, £ The stiffer column will resist a proportionally larger share of the bending moment. The bending moment in each column may be found by dividing the bending moment in the slab by the sum of the ratios of stiffness (the moment of inertia [., divided by the length h, of the two columns), and multiplying the result by the ratio of stiffness of the column under consideration. For practical purposes, it is accurate enough to assume that 60 per cent of the bending moment will be resisted by the lower column and 40 per cent by the upper column. It should be noted that the bending moment in the upper end of the column below the slab is negative, i.e., it produces tensile stresses COLUMNS IN FLAT SLAB CONSTRUCTION 315 on the outside face of the column. The bending moment in the lower end of the column above the slab is positive and produces tension on the inside face of the column. Points of Application of Maximum Moments in Columns.—The maximum negative bending moment in the column, as determined in the above paragraph, may be considered as acting at the bottom of the bracket, or, when no brackets are used, at the bottom of the slab or drop panel. The maximum positive bending moment acts above the slab. Plotting the maximum negative moment at the top and the maximum positive bending moment at the bottom, and connecting the points by a straight line, the moments at intermediate points may be determined. Steps in Design of Exterior Columns.— After the column load and bending moments are determined as described above, it is necessary to compute the maximum compression stresses for the combination’ of the concentric vertical column load and the bending moment, and to determine the amount of tensile reinforcement required to resist the tensile stresses for the most unfavorable combination of the con- centric column load and the bending moment. The first step in this process is to determine the dimensions for a column required by the concentric column load alone. This con- sists simply in applying the ordinary rules for column design, pp. 403 to 469. Next, the maximum compression unit stresses in this column are computed for the combination of maximum concentric column load and the maximum bending moment, either by formulas on p. 175, or formulas on p. 182, depending upon the amount of tensile stresses developed for this condition of loading. The maxi- mum unit stresses thus obtained should not exceed the maximum allowable stresses as recommended on p. 463. In computing the combined stresses, the full section, including the fireproofing, may be used in the same manner as in slab at the support. If the computed unit stresses in concrete exceed the allowable unit stresses, the sec- tion should be increased. The final step is to determine the required amount of tension steel to resist the bending moment. For this purpose, that combina- tion of the vertical column load and the bending moment which produces maximum tension should be used. This combination will be different from that used for determining maximum compression. There is a difference of opinion as to the method of determining 316 DESIGN OF FLAT SLAB STRUCTURES the required amount of reinforcement to resist the bending moment in the column. The following method, developed by the authors, is recommended for use. For the top columns in a structure, where the vertical column load is small in comparison with the bending moment, the amount of tensile steel should be determined for the bending alone, the effect of the vertical load being disregarded. The amount of steel is determined from the beam formula A, = ai where d is the distance from center of tension steel to the face of the column. If the amount of steel required by the column design is not sufficient, additional bars should be used. These do not need to extend the full length of the column, as the bending moment decreases rapidly. For columns in lower floors, where the total dead load carried by the column is appreciable, the following method for computing tensile stresses in column is recommended by the authors. Assume that the structure above the floor under consideration is loaded with dead load only, but that the floor under consideration is loaded by dead and live load. Under such conditions, provide sufficient tensile strength in the column to give a factor of safety of two in case of overloading of the floor (with the structure above not loaded). For this purpose, compute the dead load on the column above the foor under consideration. Add to it double the reaction due to live and dead load from the floor. Compute the bending moment in the column produced by the dead and live load on the floor under consideration. Multiply this moment by two and combine it with the column load computed above. Provide such an amount of steel in the column that the tensile stresses will not exceed double the allowable tensile stresses in steel. If the amount of reinforcement required by column design is not sufficient additional bars must be added. This method gives the same factor of safety, against cracks in the column, as is used in the floor construction. The method will be understood from the following explanation. It may easily happen during the life of a structure, and even during construction of the structure, that the wall panels of one floor are overloaded, while no appreciable live load is placed on the floors above. With such loading, only the dead load above this floor can COLUMNS IN FLAT SLAB CONSTRUCTION 317 be counted upon as reducing the tensile stresses produced by the bending moment. This dead load is constant. It will not increase with the overloading of the slab. When the slab is overloaded by double the design load, the column load below the slab will be increased by the reaction of the slab, but the column load above the slab will not be changed. | | Some engineers and some building codes (notably that of the City of Chicago, Flat Slab Ruling) require that the amount of steel for the bending moment in the column be computed for the bending moment alone, and that the regular beam formulas and the same allowable tensile stress in steel be used here as for parts of the build- ing subjected to bending only. The amount of steel thus obtained is required in addition to the regular column steel. The steel may be arranged as shown in Fig. 109, p. 318. This method is too wasteful. 7 Other specifications permit the maximum direct column loads to be combined with the bending moment, and if the resulting tensile stresses in steel do not exceed the allowable unit stresses, the column is considered satisfactory. This method is not safe. It depends upon the total dead and live load on the column to balance the tensile stresses, although it is clear that the total column load is not always there to balance these stresses. Also, in case of overloading of one floor, the bending moment in the column is increased but the col- umn load above the floor remains the same. The method already described, on the other hand, safely provides for such contingencies. Anchoring of Steel Resisting Bending in Wall Column.—The steel designed to resist bending stresses in the column should be properly anchored, else the bars may pull out instead of resisting tension. In the top of the top story columns, the outside bars should extend, above the points of maximum bending moment, a sufficient length to develop the strength of bar in tension. If this is not possible by straight imbedment, the bars should be properly hooked at the top in the slab. The inside bars are compression reinforce- ment. In intermediate stories, the outside bars at the top of any column are sufficiently developed when they are extended into the column above, the proper distance from points of maximum moment to develop the strength of the bar in tension. The inside bars of any column serve as compression reinforcement below the slab and as tension reinforcement above the slab. The length of imbedment of 318 DESIGN OF FLAT SLAB STRUCTURES WL __--- Bend. Mom. Steel ~35- Roof Slab Regular Vertical and Spiral Reinforcement i h 3 SECTION A-A Not less than v Floor Slab eet oe Bend. Moment Steel ae ar wor 1-2] 05 | Ae ua A} 2 hes Soi ™ Slow | i=) a= , ' H ' Frc. 109.—Wall Column Reinforcement, Chicago Ruling. (See p. 317.) COLUMN HEADS 319 these bars in the column above, measured from the top of slab, should be such as to develop the full tensile strength in the bar. In all the above cases, an imbedment equal to 40 diameters of the bar for deformed bars, and 50 diameters for plain bars, may be con- sidered ample. | COLUMN HEADS The column head usually has the shape of a truncated cone or a pyramid. Its cross section depends upon the cross section of the column. For round columns, the cross section of the head is also Round Column Head Octagonal Column Head for Square Column Fig. 110.—Typical Design of Column Heads. (See p. 319.) round; for square columns, it is sometimes square. A more pleasing effect, however, is obtained by making the section octagonal, as shown in Fig. 110. Sometimes, when metal forms are available, the column heads for square columns are made circular in section. The effect is not pleasing, as the joint between the column and the head does not look neat. For octagonal columns, octagonal heads are used; for oblong columns, the heads may be as shown in Fig. 111. The forms for such heads may be made of sheet metal, the two halves of a circular column head being joined with two flat pieces of sheet metal. 320 The column heads may be made ornamental. DESIGN OF FLAT SLAB STRUCTURES Metal forms for the column head shown in Fig. 112 are readily obtainable. Fig. 111.—Column Head for Oblong Column. (See p. 319.) At wall columns, when spandrel beams are used, brackets are substituted for column heads, as shown in Fig. 113. Some codes, how- ever, require that a section of a regular column head be provided at the wall column. Designof Column Heads.—The function of the column head is two- fold: first, to decrease the shear; second, to reduce the net span and thereby the critical bending moments. It is a well-established rule that, in order to fulfill the second function, le., to decrease the span, the sides of the column head should form with the horizontal an angle of not less than 45 degrees. Also, the diameter of the column capital to be used in calculations should be taken where the vertical thickness of the column head is at least 14 in. The theoretical shape of the column head is shown in Fig. 110. All rules are based on round column heads. When square column heads are used, they may be considered as equivalent to a round column head with a diameter 20 per cent larger than the side of the square. If it is desirable, for architectural reasons, to use for the column head a flatter angle than 45 degrees, as shown in Fig. 114, p. 321, the effective at mike Diameter---- Fig. 112.—Ornamental Column Head. (See p. 320.) column head is then governed by lines drawn at 45 degrees, as shown in the figure. If a block is placed above a column, as in Fig. 115, it may be con- COLUMN HEADS 321 sidered as a column head and its diameter determined as shown in the figure. --Wall Column \~--. Bracket Section A-A Section B-B Fig. 113.—Wall Bracket. (See p. 320.) Early rules (some of which are still in use) specified a definite relation between the span of the panel and the diameter of the Fig. 114.—Flat Column Head. (See ps 320,) column head, without regard to the design load. In such specifi- cations, the bending moments were computed on the basis of the specified’ size of the column and were expressed as functions of the gross span and the load. Modern rules do not im- pose any arbitrary restrictions as to the minimum diameter of the column head. Its size is only limited by the shearing stresses in the slab. The selection of the proper size depends upon economy of design and also upon the architectural pig 115.—Block Serving as requirements. As the bending moments Column Head. (See p. 320.) are functions of the net span, the increase in the diameter of the-column head decreases the bending moments in the slab, and vice versa. 322 DESIGN OF FLAT SLAB STRUCTURES The minimum diameter of the column heads specified by the Chicago and New York City regulations equals 0.2251 as given on page 397. The authors recommend that the size of the column head be varied with the design load. For factories and ware- houses, the diameter of the column head may be equal to 0.225 to ().251, where J is the span in square panels and the average span in rectangular panels. For light manufacturing buildings, 0.22 may be used. In buildings carrying light loads, such as office buildings and apartment houses, the column head may be greatly reduced or even omitted if desirable for architectural reasons. For the method of computing shearing stresses at column head, see p. 346, under Shearing Stresses. DROP PANELS By the term drop panel, is understood a thickening of the slab around the column. ‘This portion is sometimes called the plinth, or simply the drop. The drep panels in square or nearly square panels are usually square. In rectangular panels, rectangular drops may be used, with a ratio of dimensions of drop equal to the ratio of the corresponding sides of the panel. Sometimes, for the sake of appear- ance, the drop panels are made round or octagonal. This compli- cates the formwork, and the effect does not warrant the departure from general practice. Advisability of the Use of the Drop Panels.—In early flat-slab designs, no drop panels were used. This gave unsatisfactory results inmany cases. The shearing and compression stresses at the column, in many designs, were too high and the deflection of the slab exces- sive. In recent years, the practice went to the other extreme of using drop panels indiscriminately even in cases where they could be omitted with economy. The designer should keep in mind that, from the standpoint of the user of the building, drop panels are not desirable. This is particularly the case where shafting is suspended from the under side of the slab. As a general proposition, then, drop panels should not be used except for economy, or when it is not possible to satisfy the stresses at the column without them. In deciding whether or not to use drop panels, comparative estimates of cost should be made of designs with and without drop panels, taking into account not only the cost of the materials but also the extra cost of formwork for the drop panels. DROP PANELS 020 It will be found that for live loads over 150 Ib. per sq. ft., drop panels are economical. Tor smaller loads, down to 100 Ib. per sq. ft., drop panels are economical only for long spans and for specifications requiring bending moment coefficients in excess of those recommended by the authors (such as 1916 Joint Committee Recommendations or the Boston Code). In many cases, the use of larger column heads, with some compression reinforcement in the slab, may be found more economical than the drop panels. The cost of extra steel may be more than balanced by the saving in concrete and in the cost of formwork for the drop. For loads of 100 Ib. per sq. ft. and under, drop panels are hardly ever economical. Dimensions of Drop Panels.—The dimensions of the drop to be determined are the depth and the width. These are to some extent interdependent, as will be explained below. The purpose of drop panels is to reduce the shearing unit stresses, both at the column head and at the edge of the drop panel ; also to reduce the compression stresses and the amount of tensile steel at’ the column head. The magnitude of the shearing and compression — unit stresses at the column head depend upon the depth of the drop panel. The most economical depth is one for which the compression unit stresses in the concrete are about equal to the maximum allow- able unit stress. Additional depth saves steel but increases the amount of concrete in a larger ratio. Instead of adopting an arbi- trary depth for the drop panel, the designer is advised to compute it by formulas on p. 389. Since it is advisable to use the same depth for all drop panels throughout the floor, the depth should be com- puted for the largest bending moments. Ordinarily, these will occur in end panels, i.e., at the first interior columns next to the wall. ‘The compression stresses for interior panels will then be some- what smaller than the maximum allowable. The width of the drop panel depends upon three factors, the first being the compression stresses at points of maximum bending moment. As evident from the discussion on p. 341, under “ Com- pression Stresses in Concrete,” and from Formula (29), p. 344, a large part of the compression stresses is resisted by the drop panel. If the compression stresses in concrete are nearly equal to the maxi- mum unit stresses, the width and the depth of the drop are inter- dependent. ‘The width in this case cannot be reduced without over- stressing the concrete in compression. The second factor is the diagonal tension stresses at the edge of 324 DESIGN OF FLAT SLAB STRUCTURES the drop, as figured by Formula (41), p. 349. If, for a certain width of the drop panel, the shearing stresses are excessive, the width of the drop panel must be increased. The third factor is the compression and tensile stresses at the edge of the drop panel, where the depth of the flat slab is suddenly reduced by the depth of the drop. This reduction is permissible only when the bending moment becomes so much smaller that, with the decreased depth, the compression and tensile stresses do not exceed the allowable unit stresses. The location of the section where the drop panels may be stopped will depend mainly upon the ratio of the depth of the drop panel to the total depth at the point of maximum bending moment. For larger ratios of the depth of the drop to the depth of slab, the required reduction in bending moments is larger than for smaller ratios. The points where the drop can be stopped are, therefore, farther from the edge of the column head for large ratios than for small ratios. Large ratios of the depth of the drop to depth of slab, therefore, require larger width of the drop than small ratios. : n various specifications, including that of the 1924 Joint Com- mittee, the width of the drop is arbitrarily fixed, irrespective of its depth, and varies from one-third to three-eighths of the span. This is not consistent. The authors recommend the following relation between the width of the drop panel and the ratio of the depth ‘of the drop pane! to the depth of the slab: Let d = effective depth of slab outside the drop panel; ty = depth of drop below slab; 1 = span, center to center of columns, in the direction of the width under consideration. Then the width of the drop will be as civen in the table below. Ratio 2 Width of d Drop Panel 0.1 0.281 0.2 0.301 0.4 0.331 0.6 0.361 0.75 0.381 Drop Panels at the Wall Columns.—lIf drop panels are provided at the wall columns, they should be of the same depth as the interior drop panels. The width of the drop panel, parallel to the wall, should be the same as for interior drops in the same direction. The THICKNESS OF SLABS 325 width at right angles to the wall, measured from the center of the column, should be equal to one-half the corresponding width of the interior drop panel. . In many instances, the shearing unit stresses at the wall columns are much smaller than at the interior column, on account of the larger circumference of the wall column, and also because part of the shear is resisted by the spandrel beam. Also, the bending moment is smaller. For these reasons, the drop panel may often be omitted at the wall column. The required amount of negative reinforcement at the wall must then be computed on the basis of the smaller depth. In rectangular panels, where the difference between the spans is appreciable, the drop panels at the wall in the short direction may be omitted. THICKNESS OF SLABS The thickness of the slab is governed, in flat slabs without drop panels, by the shearing and compression stresses at the column head. In flat slabs with drop panels, the only limiting stresses, as far as the thickness of the slab is concerned, are the diagonal tension stresses at the edge of the drop panel and the compression stresses in the center. The thickness of slab, determined by the stresses, may not be large enough to keep the deflection within desired limits. Since there are no reliable formulas for computing the deflection, most specifications impose a number of arbitrary and unnecessary limita- tions on the minimum thickness of the slab, such as that (1) the slab thickness shall not be less than 6 in., or that (2) it shall not be less than gz of the span for floor slabs nor less than zo of the span for roof slabs. It will be observed that the above limitations are general. They apply equally to flat slabs with and without drop panels. They apply also to all load conditions; the only distinction made is between floor slabs (irrespective of the design load) and roof slabs (also irrespect- ive of the design roof load). All these limitations originated when little was known about flat slab construction and are being carried along from specification to specification without investigation. The limitation of the thickness of slab to 6 in. originated when the four-way system was the only one in use and all flat slabs were built without drop panels. Under such conditions, it was not practicable to use thinner slabs and accommodate effectively the four layers of steel at the column. With the development of systems 326 DESIGN OF FLAT SLAB STRUCTURES using only two layers of steel, and also with the introduction of drop panels, the reason for this limitation has been removed. With this limitation, flat slabs cannot be used economically for small spans. There is no reason why, for small spans, a thickness of slab propor- tional to the span should not be used. The authors recommend that the limitation be revised to permit 4-in. thickness for slabs with drop panels and 44 in. for slabs without drop panels. The second limitation of the thickness of slab to a certain ratio of the span was made for the purpose of limiting deflection. It was based on the performance, in use, of flat slab structures, mostly built without drop panels. Since, in a floor consisting of nearly equal panels and with slabs of the same thickness throughout, the wall panels deflect most, the limitation was based on the performance of the wall panels. Assuming that the limitation is correct under such circumstances, its use would be justified for buildings in which the interior and exterior spans are about equal and the thickness of slab is uniform throughout the floor, as then the conditions in the wall panel would govern. In cases where the wall panels are smaller than the interior panels, it would not be correct to apply this limita- tion to the larger interior panels. The unreasonableness of the rule ‘s evident from the comparison of the conditions in Fig. 116, a and 6. Fig. 116a shows a building three panels wide, with exterior panels 18 ft. and interior panels 26 ft. Fig. 1166 shows a building two Fic. 116.—Section through Flat Slab Buildings. (See p. 326.) panels wide, each of the panels being 26 ft. According to the limita- tion under discussion, the minimum allowable thickness of the slab is the same for both cases, although it is plain that the slab in the second case would deflect much more than in the first case. Another objection to this rule is that it separates flat slab design into floors and roofs, irrespective of the loading to be carried. In many cases this leads to absurdities. For instance; in an office building, the design live load for the floors is 70 lb. per sq. ft. Actually, this load is seldom reached. The load on the roof, on the THICKNESS OF SLABS 327 other hand, may consist of cinder fill weighing about 60 Ib. per sq. ft., roofing weighing 10 lb. per sq. ft., and live load of 40 lb. per sq. ft. The total load, exclusive of the slab load, to be carried by the roof is 110 lb. per sq. ft., of which 70 lb. is permanently on the roof. Accord- ing to the rule under discussion for 20-ft. panels, the required thick- ness of the floor slab would be 7} in. for the design load of 70 Ib.; while for roof slab carrying 110 Ib. per sq. ft——of which 70 lb. is permanently on the slab—only 6 in. would be used. Authors’ Rules Limiting Thickness of Slab.—It is questionable whether any limitation of the thickness of the slab, other than that imposed by the stresses, is necessary. General rules, however, if made consistent, may be of assistance to designers who are not grounded in flat slab design. Therefore, the authors recommend the following rules: Two groups are to be considered: (1) slabs carrying, in addition to the dead load of the floor slab, loads of 100 Ib. per sq. ft. and under; (2) slabs carrying, in addition to the dead load of the slabs, loading of over 100 lb. per sq. ft. The minimum thickness of the slab Se drop panels, for the first group, should be 70 of the end span or {7s of the interior span; for the second group, 35 of the end span or =; of the interior span. For slabs with drop panels, designed as Peiieried under proper heading, the minimum thickness given above may be reduced by 3 of the thickness of the drop panel. Where the end spans are not monolithic with the wall columns, as in brick bearing jobs, the minimum thickness obtained above should be increased by 25 per cent. Thickness of Slab for Floors with Different Sizes of Panels.—If the panels are nearly uniform throughout the floor, the slab should be made of uniform thickness. The minimum thickness should then be based either on outside or inside span, whichever gives larger value. If the panels in one part of the floor are materially different from those in some other part, the thickness of the slab may be different for the different parts. The floor should be made level, and the difference in thickness made up by dropping the bottom surface of the slab. Where the different thicknesses of slab meet, the underside of the slab should be sloped gradually rather than stepped. 328 DESIGN OF FLAT SLAB STRUCTURES BENDING MOMENTS IN FLAT SLABS General.—As explained in the section on theory, a flat slab is subjected to bending in two directions. For instance, at the column head and in the center of the panel, the slab bends in radial and cir- —e—e— ZA te te: reas ~ Seat | Theoretical _7 Cantilever rie oe | ----> \ \ \ “eS e=—— / Slab Suspended along Slab Suspended along Two Edges \ _Slab Suspended on ~~~ Four Corners “~~~ bee @ mew oe oe we we wwe ew ew = = re ee ee eee see ——< ~— aaa swe sea anaes easererre= Suspended Portion Fic. 117.—Flat Slab Divided into Simple Parts. (See p. 328.) cumferential directions - while in the section between columns, the bending takes place in two directions at right angles to each other. The action will be better understood from Fig. 117, above, showing the flat slab divided into simple parts by cutting it along the points of inflection. BENDING MOMENTS IN FLAT SLABS 329 To simplify the computations, it has become established practice to resolve the bending moments to which the slab is subjected into two sets of bending moments acting at right angles to each other. Each set produces bending in one direction only. The two sets, combined, produce the actual deflection of the slab. Each set of bending moments, and the required amount of reinforcement for each, can be computed separately. The character of bending due to each set of bending moments, when considered separately, is the same as if a continuous uniformly loaded slab were supported along two opposite panel edges. This condition produces negative bending moments at the support and positive bending moments in the central portion of the slab. 1 - 1 Seation of Critical Neg. Bending Moment Column —<=>-- Strip eBending Moment Neg. Bending Moment tS) A - | Section of Max. Positive 3,- 1,, Section of Critical ah ee, (ee ee ht (& \ ol PNT Cex aS, 31 Sr mey! 9h ee), (@) 3,\ / ane ef coe Column ‘ OE. = Head Strip 3 - 3 Section of Critical Neg. Bending Moment Note: Solid lines indicate Critical Note: Solid lines indicate Strip in Section in one direction. Dash lines one direction. Dash lines indicate indicate Critical Section at right Strip at right angles. angles, : , Design Strip Sections of Critical Moments Fig. 118.—Sections of Critical Bending Moments and Design Strips. (See p. 329.) Sections of Maximum Bending Moments.—Sections of maximum bending moments are shown in Fig. 118, p. 329. As there are two sets of bending moments in a slab, there are also two sets of sections of maximum moments. Both sets are shown in Fig. 118. For the sake of clearness, Fig. 119 shows the sections of maximum moment for one set only. They are as follows: Section of Maximum Positive Bending Moment, coinciding with the center line of the panel. It is marked by 2-2 in Fig. 119, p. 330. Section of Critical Negative Bending Moment, coinciding with the edge of the panel and the edge of the column head. It is marked by 7-1 in Fig. 119, p. 330. | 330 DESIGN OF FLAT SLAB STRUCTURES Design Strips.—To simplify the computation of bending moments in any one direction, the panel is divided into strips running at right angles to the sections of maximum bending moments. (See Fig. 119.) The middle strip is concentric with the panel and is equal in width to one-half of the width of the panel. The column strips, placed one on each side of the middle strip, extend to the edge of the panel, so that the width of the two column strips is equal to the width of one middle strip. Similar strips can be drawn for the other set of bending moments. if 1-1= Section of Critical Negative rN Bending Moment 7 ee 2-2=| Section of Max. Positive Bending Moment oe ess Se ee |--_---—---_—--—- i Bending Moments act at right angles to Sections l= Length of span in the direction of Bending Moments l= Length of span at right angles to | Note that the designation of | and l,depends upon the direction of Bending Moment and not upon relative length of spans, Fic. 119.—Sections of Critical Moments and Design Strips in One Direction Only. (See p. 380.) Distribution of Maximum Bending Moments.—The distribution of bending moments over the sections of maximum moments is not uniform. The intensity of bending moments is largest near the column and decreases towards the middle of the section. Instead of using the varying distribution of bending moments, it 1s accurate enough to assume a uniform distribution over the section in each strip, but to make the intensity in the middle strip smaller than in BENDING MOMENTS IN INTERIOR PANELS IN FLAT SLABS 331 the column strip. The difference between the intensity of bending moment in the two strips is particularly large for negative bending moment. For the purpose of design, therefore, the total negative bending moment at each section is divided into two unequal parts. The larger part is assumed to act in the two-column head sections and the smaller in the middle section. A similar assumption is made regarding the distribution of the positive bending moment. BENDING MOMENTS IN INTERIOR PANELS IN FLAT SLABS The bending moments recommended by the authors, for use in computing design, agree substantially with those recommended by the 1924 Joint Committee. They are about two-thirds of the static bending moments. The reduction in bending moments is justified because the static bending moments do not take into account several factors which reduce tensile stresses in steel in flat slab con- struction. Let W = total live and dead load per panel, lb.; w = live and dead load, lb. per sq. ft.; 1 = length of panel, center to center of columns, in the direction of computed bending moment, ft.; 1, = width of panel, center to center of columns, ft.; c = diameter of effective column head, ft.; M = sum of positive and negative bending moments; M, = negative bending moment in two column strips; Mz = negative bending moment in middle strip; M3 = positive bending moment in two column strips; M4 = positive bending moment in middle strip. The formulas for bending moments given below apply to square and rectangular interior panels. For square panels, the bending moments in both directions are equal. For rectangular panels, the bending moments in the two directions are obtained by using, for / in the formula, the two dimensions of the panel, / and [;, successively. The sum of negative bending moment on section 1-1 and positive bending moment on section 2-2 is given by following formula. Sections are shown in Fig. 119, p. 330. Sum of Bending M ee, nterior Panel.— M = 0.09mi(1— 34) aed Lela toslegubesaanat (3) 1.oswi(1 — 4) ries iad wile mete all (35 ll oi 302 DESIGN OF FLAT SLAB STRUCTURES Deve 27e\" Values of {1 — 3] ane he Us (lors 37) may be taken directly from table below. 2c \2 2c \2 Values of (1 ——-—) and 1.08 (1 —-— ,- 31 31, a | C | Value of 1 nese EEE 0.1 |0.125| 0.15 |0.175| 0.20 | 0.225| 0.25 | 0.275 0,30 PAs . (: ie 0.870 | 0.841 | 0.810} 0.780 | 0.752 | 0.722 | 0.694 0.667 | 0.640 7) 2 1.03(1-25) |0.940 0.908 | 0.875 | 0.842 | 0.812 | 0.780 | 0.749 Oaeh 0.691 | The above sum on bending moments, M, may be distributed as given below: Negative Bending Moment on Section 1-1—I nterior Panels.— With Drop Panel Without Drop Panel Two Column Strips...... M, = — 0.54M M, = — 0.50M MiddlesStripayra te eee Mz = — 0.08M Mz = — 0.10M Positive Bending Moments Section 2-2—Interior Panels.— With Drop Panel Without Drop Panel Two Column Strips...... M3 = 0.23M M3 = 0.24M Middle: Stripsi.a de areas M, = 0.15M M, = 0.16M Permissible Variation in Distribution of Bending Moments.— A small variation in the distribution of bending moments, from that recommended above, is permissible. The positive bending moment in the column strips may be reduced by an amount not exceeding by 0.02M, provided that the positive bending moment in the middle strip is increased by same amount. The negative bending moment in the two column strips may be either increased or reduced by an amount not exceeding 0.03M. If the negative moment is increased, the other three moments, namely the two positive moments and the negative moment in the middle strip, may be reduced, the sum of reduction not to exceed the increase. If the negative moment, is made smaller, the other three bending moments must be increased proportionally, the sum of increase to be at least equal to the reduc- tion of the negative moment, WALL PANELS 033 Application of Formulas.—The formulas apply only to flat slabs extending in each direction at least over three panels, and only where the spans are equal or nearly equal. For bending moment formulas for unequal panels, as well as for flat slabs extending over two panels only, see Volume ITI. Use of Bending Moments.—The bending moments given above may be used ae for computing required areas of steel from the formula A, = —- or for computing stresses in steel from the formula a fae a where d is the effective depth of the construction at the section in consideration. Thus, in column strips, for negative bending moment the depth of the slab and drop panel, d1, should be used. For all other bending moments, the effective depth of slab alone should be used. For computing compressive stresses, the bending moments can- not be used directly, as it is necessary to make an allowance for factors which do not enter into the determination of the steel areas. For formulas for compression, see p. 341. WALL PANELS The design of wall panels is governed by the amount of restraint to which the slab is subjected at the wall. The restraint to be con- sidered is not only due to the wall column, but also to the wall beam, 1.e., the beam at the edge of the wall panel. In practice, the follow- oe cases may occur. Case (1). Wall panels with properly designed concrete wall columns and with wall beams capable of resisting torsion. Case (2). Wall panels with properly designed concrete wall columns, but without wall beams capable of resisting torsion. Case (3). Wall panels supported on brick bearing walls. The authors’ recommendations for tending moments in the three cases are given below. They differ from the requirements of the 1924 Joint Committee in that they differentiate between various condi- tions instead of making a general rule to cover all the dissimilar cases. Columns should have a ratio of rigidity of at least - = 1.5 4 Let, in addition to the notation given on p. 331, 3 = distance from center of column to outside of bracket at wall column. 334 DESIGN OF FLAT SLAB STRUCTURES Then ‘ we 0.09m71|1 us (3 i 7) ft.-Ib. af . eee lye Ci ; 1.081] 1 _ a(; 4- 3) ins: W is the total load on wall panel, but does not include the spandrel load. lis the span length at right angles to the wall. It 3 is equal to _ values of M in the above formula equal the values of M for interior panel. The moments given below are fractions of the moment repre- sented by the above equation. Case (1)—Slab Restrained by Wall Columns and by Spandrel Beams.—This design is particularly recommended for flat slabs carrying heavy loads. The wall beams should be nearly square in cross section and capable of withstanding torsional moment as on p. 336. Moments are in terms of M, from equation (7) above. —With Drop Panel —Without Drop Panel— Ratio to Ratio to Moment TEOTnSH Moment pe in interior in interior panel panel Negative Moments at First Interior Column Line: } Column Strip, M,.... 0.62M eats 0.58M L156 Middle Strip, Mz.... 0.12M 1.20 0.10M 1.20 Negative Moments at, Wall Column Column Strip, M1.... 0.42M 0.80 0.40M 0.80 Middle Strip, Me.... 0.08M 0.80 0.07M 0.80 Positive Bending Moment. Column Strip, M3.... 0.27M 1d 0.28M 1.18 Middle Strip, M,4.... 0.18M 1.20 0.19M 1320 Case (2)—Slab Restrained by Wall Columns Only, Spandrel Beam not Capable of Resisting Torsion.—This occurs when no wall beam is used, or when the thin concrete spandrel wall above the slab is utilized as a wall beam. Such design should be used only for light live loads. The recommended bending moments are the same as for Case (1), with the exception of the bending moments in the WALL PANELS middle strip, both positive and negative, which are larger. in terms of M, from equation (7), p. 334. With Drop Panel Moment Negative Moments at First Interior Column Line: Column Strip, M@1.... 0.62M Middle Strip, M2.... 0.138M Negative Moments at Wall Column Column Strip, M@,.... 0.42M Middle Strip, Mz.... 0.08M Positive Bending Moment. Column Strip, M3.... 0.27M Middle Strip, M4.... 0.20M Ratio to moment in interior panel 1.18 1.30 Moment 0.58 0.10M 0.4012 0.07M 0.28M 0.21M 335 They are —Without Drop Panel Ratio to moment in interior panel 1.18 1.30 Case (3)—Slab Supported on Brick Bearing Wall.—No reliable restraint at the wall. —With Drop Panel Moment Negative Moments at First Interior Column Line: Column Strip, M,.... 0.75M Middle Strip, Me.... 0.11M Negative Moments at the Wall : Column Strip, M,.... 0.20M Middle Strip, Mo.... 0.08M Positive Bending Moment. Column Strip, M3.... Middle eee. 2 4 0.32M 0.21M Ratio to moment in interior panel 1.40 1.40 1.40 1.40 Moment 0.70M 0.14 0.20M 0.08M 0.34M 0.22M The bending moments recommended, in terms of M, given by equation (7), p. 334, are: —Without Drop Panel Ratio to moment in interior panel 1.40 1.40 1.40 1.40 The negative bending moment reinforcement at wall is required to take care of any restraint caused by the brickwork above slab. Permissible Variations in Bending Moment.—The variations in bending moment for the wall panels are the same as recommended for interior panels on p. 332. 306 . DESIGN OF FLAT SLAB STRUCTURES Torsonial Moment in Spandrel Beam.—In Case (1) the spandrel beam should be able to resist a torsional moment at each end equal to one-half the moment M2 at the wall. For formula for torsional stresses see p. 95. Explanation of the Recommendations.—In wall panels, it has been observed that the loading produces not only appreciable bend- ing in the wall columns but also torsion or twisting in wall beams. The torsional resistance of the beams develops an equal amount of negative bending moment in the slab along the beam. This bending moment reduces the positive bending moment in the middle strip of the wall panels. If the wall beam is properly designed to resist the twisting moment, its resistance may be utilized by reducing the bending moments in the middle strip, as in Case (1). If no torsional resistance of wall beam exists, the bending moments in the middle strip must be increased, as in Case (2). In a number of cases, where flat slab has not been provided with proper spandrels, considerable trouble has been observed at the wall, where the brickwork above the slab or the thin concrete walls become distorted. Thin concrete spandrel beams have often bulged and cracked through torsion. Points of Inflection.—Points of inflection are points in the slab where the bending moments change from negative to positive. The bending moment at the points of inflection is, therefore, zero. For the purpose of determining the length of bars, the most unfavorable position of the points of inflection must be taken. For negative bending moment the most unfavorable condition occurs when all spans are loaded. For such case the distance of points of inflection from the edge of the column head may be assumed as equal to 0.211, where 1; = distance between the edges of column heads. ‘This location of points of inflection must be used in deter- mining the length of negative bending moment reinforcement. For positive bending moment the most unfavorable conditions occur when the adjoining spans are not loaded. For such condition the points of inflection are nearer the support and may be assumed equal to 0.1811. This location determines the length of positive moment reinforcement. FORMULAS FOR SLAB THICKNESS “The formulas for thickness of slab, given below, are not arbi- trary, but are based upon bending moments specified in previous paragraphs and upon allowable compressive stresses, as explained on FORMULAS FOR SLAB THICKNESS O37 p. 341. These formulas and the formulas for compression given on p. 341, are based on the same general formulas, which in one case were solved for the depth of slab and in the other for the unit stress. If a depth of slab obtained from the formulas below is adopted, it is not necessary to compute the stresses. Formulas are given for thickness of slab at the column head and in the center of the column strip. If no drop is used, the required thickness at the column head governs the thickness of the slab. For slabs with drop panel, the thickness must be found both at the column and in the center. Let M, = negative bending moment in two column strips (see pp. 332 to 335); Mz = positive bending moment in two column strips (see pp. 332 to 335). t; = thickness of slab and drop at the column; thickness of slab in center of panel; 1 = span length, center to center of columns, in direc- tion in which moments are considered; 1; = span length, center to center of columns, at right angles to 1; c = diameter of column head; width of drop panel, in direction at right angles to span, /; j,k, = constants for specified stresses in concrete and steel f. and f; (see p. 205 and table, p. 880); w = uniformly distributed dead and live load per sq. ft.; o I C;, C2, C3, C4, Cs, Ce = constants for different ratios ; and different unit stresses (see diagrams, p. 911). Basis of Formulas.—The formulas for thickness of slab are based on the same principles as formulas for depth of slab in other concrete construction. The bending moments M, on pages 332 to 335 are recommended only for computing reinforcement. They do not take into account the increased compression stresses caused by arch action, the compression which balances the tension resisted by concrete nor the effect of the size of column head on compression stresses. To provide for these in the formulas given below the bending moments M, were multiplied by a constant equal to ae (1 — i) To get the thickness of slab, 1.5 inches were added to the effective depth. 338 DESIGN OF FLAT SLAB STRUCTURES Thickness of Slab at Column Head as Determined by Negative Bending Moment.—The thickness of the slab at the columns, required by the compression stresses, should be computed from the following formulas. Flat Slab without Drop Panel. General Formula, as Determined by Negative Bending Moment, 2c M t= 0.69(1 —=s ‘) sacl. 37 SET e da ee Thickness ¢, will be in in. if M is in in-lb., f. in Ib. per sq. in., and I, in ft. Since, for interior panels, from formula on p. 332, 2 M, = 0.50M = 0.50| 1.08whi (1 — 34) | in.-lb., the formula for thickness becomes: Thickness of Slab for Interior Panels, 0.51 207 a — where 0.51 2c = 2280225). . il cea C 1 ( ) In exterior panels for slabs restrained at ends, the bending moment 2 ~ is M, = 0.58 | 10st (1 — : ‘) in.-lb. For freely supported slabs, the fraction 0.58 changes to 0.70. The thickness of slab required for the two conditions is Thickness of Slab for Exterior Panels. Slab Restrained at End. 0.56 2 6\ 77 = i= Sea — 35) Vy +1.5 = CalVw + 1.5 ROLL) where ; 0.56 24 — 1 St eae Be ° e e 7 ° * « e = » ° . e ae 3 4 | We) Thickness of Slab for Exterior Panels. Slab Freely Supported. Ben a | = t= $8139) Vw +1.5=C3livw+1.5 . (18) V fal seh or where " 0.61 Ql c= SE (1-54) ee eee FORMULAS FOR SLAB THICKNESS 339 Constants C1, C2, and C3 are given in diagram, p.911, for different ratios of ; and different unit stresses. The thickness, #, will be in in., if J is in ft. and w in lb. per sc. it. Flat Slab with Drop Panel. Thickness of slab and drop as determined by negative moment. General Formula for Thickness au Slab and Drop: . 2 c\ n = 0.50(1 ~ 39) 4/75 Th Sid baeaD Aint a tlh hi lt Soa Tay Thickness, ¢1, will be in in., if My is in in.-lb., b in ft., and f. in. lb. per Sq. in. Since, for interior panels, from formula on p. 332, 2 M, =0.54 M = 0.54] 1 .08whi?(1 ~ : ) | in.-lb., the formula for thickness becomes: Thickness of Slab and Drop for Interior Panels, per ee: ae et 31) w+ 1.5 = Calyfol + 1.5 (16) | Mien 8 t where ; 0.41 PAT C4 ae Sg 3 1) ° . ° ° ° ° . ° ° ° ° . (17) in exterior panels, for slabs restrained at ends, the fraction 0.54 in moment formula changes to 0.62, and for freely supported slab to0.75. The formula for thickness changes to: Thickness of Slab and Drop for Exterior Panels. End Slab Restrained. 0.41 fut foe = Fel - 35) $1.5 =Cslyfut+1.5 (18) 0.41 2 c\* Geet eee ibis erie JS geo ditasol 22 Ae eeh19 ; Jia 3 : ly) Thickness of Slab and Drop for Exterior Panels. End Slab Freely Supported. | Bree he ae 0.45 2¢ =— ee Py e ° ° e ° ° ° e e ° AL Go eg 31) oe where where 340 DESIGN OF FLAT SLAB STRUCTURES Constants C4, Cs, and Ce are given in diagram, p. 911, for different unit stresses and values of c. The thickness 1 is in in., if J, h, 0, and c are in feet and w in lb. per sq. it. Thickness of Slab as Determined by Positive Bending Moment. _The thickness of slab due to the positive bending moment should be based upon the largest value of the positive bending. This acts in the column strips and is given by formula for M3 on p. 334. The bending moments recommended on pp. 331 to 335 are for the purpose of computing the required amount of reinforcement. They do not ‘nelude the bending moment resisted by tension in concrete. Since the thickness of slab, as governed by compression, must be based upon total bending moment, the value of M3 used in computing thickness of slab is multiplied by a ratio 1.5. This represents fairly well the ratio between the total compression and the amount of the tensile stresses carried by reinforcement. General Formula for Thickness of Slab. M3 P= O.%nlaea tl... ee 0 any nF (22) Thickness will be in in., if Mo is in in.-lb. and J, in ft. For interior panels, M3 = 0.24) 1. 080h? (1 — - ‘) | and the formula for thickness becomes: Thickness of Slab for Interior Panels, 0.36 2c ad as ) where 0.36 2c = ——— 1 rae a Na = & val 3 ) oe For exterior columns, the fraction 0.24 in equation for M3 may . be taken as 0.29 for slab restrained at end, and as 0.34 for freely supported slab. The formulas for thickness of slab become: Thickness of Slab for Extervor Panels. Slab Restrained at End. 0.40 oe - i= deel 3 a ia ab ae 25) where : 0.40 2¢ C — ff baer Ji ceeee . . . ° . ° ° . . . . ‘ aa 3 ) ir COMPRESSION STRESSES IN CONCRETE IN FLAT SLABS 341 In the above formula, it was assumed that the size of the bracket, c1, is equal to one-half the diameter of the interior column head. If no bracket is used, the depth of slab must be increased by a ratio Lug bes yBel of 12° Soi Thickness of Slab for Exterior Panels. Slab Freely Supported at 0.43 A ee ee ee a Aan = SUA . i ae 3 51)! wtl=C/Vw+1 (27) where 0.43 le C ——— 1 eee * ‘ . ' a . ‘ F : A faut 23) see Constants C7, Cs, and Cy are given in diagram, p. 911, for different Pet : : ratios 7 and different unit stresses. | ~ Thickness ¢ is in in., if J is in ft. and w is in lb. per sq. ft. COMPRESSION STRESSES IN CONCRETE IN FLAT SLABS Not so much is known about the magnitude and distribution of compression stresses in flat slab construction as about tensile stresses in steel. The latter have been determined from extensometer read- ings in a comparatively large number of flat slab tests. Information on compression stresses from tests is much more meager. Since the modulus of elasticity of steel is known, it is possible in flat slab tests to get actual tensile stresses in steel by measuring deformations, and therefrom to get an idea of the magnitude of bending moment to be used in computing steel stresses. This, however, is not equally feasible in connection with compressive stresses, because the measured deformations are not a sure guide to the magnitude of the stresses in concrete. In the first place, the measured deformations of con- crete are not due to stresses alone, but consist of stress deformations and plastic deformations caused by a prolonged application of the stress. In the second place, the actual modulus of elasticity of the concrete, the correct value for which is necessary to convert deforma- tions into stresses, is usually unknown. While for steel the modulus of elasticity is practically constant, for concrete it varies not only for different concretes but also for different intensities of stresses in the same concrete. 342 DESIGN OF FLAT SLAB STRUCTURES Compression Stresses in Slab at Column Head.—Tests show that in the column head section the stresses are not uniformly dis- tributed along the whole length of the section used for computing maximum bending moments. The stresses are much larger at and near the column head, especially at the diagonal lines where com- pression failures at high loads first occur. The size of the column head influences the magnitude and distribution of the compressive stresses. For equal bending moments and equal length of section, the stresses will be smaller for larger column head. Tests indicate that the sum of measured compression stresses on a section of maximum bending moment is larger than the sum of measured tensile stresses along the same section. Theoretically, the two sums should be equal as they are produced by the same bending moment. The difference is due to arch action, which increases compression stresses and decreases tensile stresses, and also to the fact that a certain percentage of tensile stresses is resisted by con- crete, so that the sum of tensile stresses based on steel alone is not the total sum of tension on the section.. ; The bending moments recommended on p. 332 are mainly for the purpose of determining tension reinforcement. They are based upon observations of tensile stresses in steel. Therefore, it is plain that it is not possible to use them for computing compression stresses without some modification. For this reason, in the formulas given below for determining compression stresses, the bending moments are multiplied by appropriate constants. In flat slabs with drop panels, there is another element of uncer- tainty as to the width of the slab to be used in figuring compression stresses. The drop panel and the adjoining slab deform as a unit and have a common neutral axis. The extreme fiber in the drop, being much farther below the neutral axis, will shorten more than the extreme fiber of the slab. The compression stresses, therefore, in the extreme fiber of the drop are larger than for the slab just adjoin- ing the edge of the drop. or points some distance from the edge of the drop panel, the deformation of the extreme fibers in slabs increases gradually until it reaches the same magnitude as the deformation of the extreme fibers in the drop. For instance, in the Channon test ! with 20 ft. 3 in. square panels, 6 ft. 6 in. square drop panels, 8 in. 1 Test of a Flat Slab Floor of the New Channon Building, by H. F. Gonner- man and F. E. Richart: Proceedings of the American Concrete Institute, Sev- enth Convention. Vol. XVII, 1921, p. 182. COMPRESSION STRESSES IN CONCRETE IN FLAT SLABS 343 slab thickness, 12 in. thickness at drop, for maximum test load the measured deformation of extreme fibers at the edge of the drop was 0.00049, while the deformation of the extreme fibers of the adjoining slab was only 0.00020. At a distance of 6 in. from the drop, the deformation of the extreme fiber of the slab increased to 0.00033, and 14 in. from the drop, to 0.00053. This last deformation was even larger than the deformation of the drop. This indicates that while the effect of the slab at the column head next to the drop is small, it becomes considerable some distance from it. The difference between the stresses in the slab and in the drop will depend upon the ratio of the depth of the slab to the depth of the drop. For deep drops, the difference is greater and the distance is also greater from the edge of the drop to the point where the slab attains the same stresses as in the drop. With deep drops in the column head section, the effect of the slab outside the drop on compression stresses is negligible. In such cases, Formula (29), p. 349, may be used for computing compression stresses. This formula neglects the effect of the slab. With shallow drops, the effect of the slab outside of the drop may be considerable. In such cases, Formula (31), p. 344, will give better results. Formulas for Compression Stresses in Slab at Column Head.— Let M; = negative bending moment in column strips, in.-lb.; M, = tending moment resisted by concrete, in.-lb.; M, = bending moment resisted by compression reinforce- ment, in.-lb.; ! = span of panel, center to center of columns, in direc- tion for which bending moment was figured, ft.; 1; = span of panel, center to center of columns, perpendi- | cular to direction for which bending moment was figured, ft.; c = diameter of column head, ft.; b = width of drop panel measured at right angles to the span, ft.; d = effective depth of slab, in.; d, = effective depth of slab and drop, in.; A, = area of compression reinforcement, sq. in.; 4, k, n = constants for concrete. See p. 207. Cio, C11 = constants. See Table 17, p. 912. 344 DESIGN OF FLAT SLAB STRUCTURES Then Compression Stresses, Drop Only Considered Effective, C iz jkbd ° bd? 0.25(1 =n ong gk lb. per Sq: hw euntae) where Cio = (30) In the above formula 6 is in ft. and dj in in. Values of Cio for different © and jk are given in Table 17, p. 912. l Compression Stresses, Drop and Slab Considered Effective, 0. 25(1 2 1.25)\Mi lb. per sq.in.. (81) of 6D) Cefn) and Cio as given above. In the above formula J; is in ft. and d, in in. Values of C1; for different values of ; and a are given in diagram dy on p. 912. Compression Stresses, No Drop Panel Used, 0.125(1 ca 1.25) Mi Je = jklyd? = C105 98 a2 Ib=pereqeity a ae! Values of Cio for different ; and jk are given in diagram, p. 912. Lin ftsed7 anne In the above formulas, if My is in in. -lb., 1, 4, b, and c in ft. and d, in in., then f, is in lb. per sq. in. Values of jk.—In solving the equation for compression stresses, the values of jk must be assumed. For this purpose the values may be taken from table below. : Values of jk for Different Ratios of Steel p 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 jk 0.21 0.24 - 0.26, 0.28 0.80% 0:32 2g 0 ai Oise ates COMPRESSION STRESSES IN CONCRETE IN FLAT SLABS 345 Use of Compression Reinforcement.—Compression reinforce~ ment may be found useful in many instances where it is not desirable to increase the thickness of the slab. Since, with equal spans, the compression stresses at the first interior row of columns next to the wall are larger than at other interior columns, it may be economical in flat slabs without drop panels to use compression reinforcement there, instead of increasing the thickness of the whole slab. The thickness of the slab can then be determined from the moment at the interior columns and made uniform for the whole floor. Also, in many cases, odd panels are much larger than the rest of the slab. It may be more economical to use compression reinforcement in such panels, instead of changing the concrete dimensions. Formulas for Compression Reinforcement at the Column Head.— Since the maximum compression stresses act at the column head, especially in the diagonal direction, the reinforcement, to be most effective, should be placed there. Using notation on p. 343. The bending moment resisted by the compression reinforcement (see p. 234) is M, = Al(n — =“aa ~ a). Ca Sieh Meas When the concrete dimensions are fixed and it is necessary to introduce compression reinforcement for the purpose of keeping the compression stresses within working limits, the problem may be solved as follows: Compute moment of resistance of slab at the column head, M,, based on the allowable working stress in concrete. To get as large a value as possible, the moment of resistance is based on the formula which takes into account not only the compression resisted by the drop panel, but also that resisted by the slab outside the drop panel. This moment of resistance may be obtained by solving Formula (31) for M,. Therefore, Moment of Resistance of Slab at the Column Head, Ci 3 ed d 2 ce ° . e s e e es 4 em i (34) The values of Cio and C1; are given on p. (912). If J; is in ft., d; in in., f. in lb. per sq. in., M- is in in.-lb. The bending moment to be resisted by steel equals the actual moment, M,, at the column head minus the moment of resistance, M,, computed above. M. = 346 DESIGN OF FLAT SLAB STRUCTURES Thus, Moment to be Resisted by Compression Reinforcement, M, =M ~ M0 . . e and Required Area of Compression Reinforcement, i ps k (n—Dfdd — a)k—-a If M, is in in.-lb., f, in lb. per sq. in., and d in in., A’, is in sq. in. Compression Stresses for Positive Moments. ere compressive stresses due either to positive bending moment or to negative bend- ing moment in the middle strip may be computed from the formula below. Let, in addition to notation on p. 343, M,, = bending moment in the section under consideration, in.-lb. Compressive Stresses for Positive Moment, ul 5M, Je gklid2 Values of 1; are in ft. and those of d in in. Values of jk to be used in the above formula for different ratios of steel, p, may be taken from table on p. 344. tee (36) lb. per’sq: ine ae ee ae SHEARING STRESSES IN FLAT SLABS There is some difference of opinion as to the method of computing shearing stresses in flat slab construction and as to the allowable unit stresses at the various sections. The available test data on ultimate shearing resistance of flat slabs are meager. The tests on circumferential cantilevers and footings, which in some respects resemble the conditions in flat slab at the support, also give little information on diagonal tension. The test specimens either failed by tension (see p. 120) or the failure was complicated. The only reported tests on fat slabs carried to destruction are those conducted on the Western Newspaper Union Building ? during its demolition, and the tests at Purdue University on specially built panels designed according to the S. M. I. System and the Cor-plate System (see p. 99). None of these tests furnish any data on the ultimate resistance of concrete to diagonal tension, as all the slabs 2Test of a Flat Slab Floor of the Western Newspaper Union Building, by Arthur N. Talbot and Harrison F, Gonnerman. University of Illinois Bulletin No. 106. SHEARING STRESSES IN FLAT SLABS 347 failed by tension or tension in combination with compression. In the Purdue tests, cracks were observed at the column head, running diagonally downward from the tensile steel on the top of the slab toward the edge of the column head. The inclination of the cracks caused. them to be mistaken by some observers for diagonal tension cracks. Actually they were caused by the steel passing the elastic limit. The development of the tensile crack after the steel had passed the elastic limit was followed by a compression failure in concrete, a phenomenon observed in tensile failures of simple beams at points where there is no possibility of diagonal tension (see p. 22). It is‘generally agreed that the critical sections are at the column head and at the edge of the drop panel. Many codes and recom- mendations require that punching shear be figured at the edge of the column head and drop panel respectively. According to the Boston Code, the punching shear must be figured by the formula, v = ia’ and must not exceed 132 Ib. per sq. in. for 2 000-lb. concrete. The Chicago Code requires that the shear be figured by the same method, but the unit stress at the column is limited to 120 lb., and at the drop panel to 60 Ib. per sq. in. for 2 000-lb. concrete. The New York flat slab regulation specifies the same stresses as the Chicago Code, the difference being that the unit stresses are computed by dividing the total shear, V, by the area bd, instead of bjd. The modern tendency is to reduce the shearing unit stresses, especially at the column head. The authors recommend the following rule: Authors’ Recommendations for Shearing Stresses in Flat Slabs. —(1) Shearing unit stresses shall be computed at sections indicated in Fig. 120, p. 348. (a) At the column of critical shear the section is concentric with the column head and located from its edge a distance equal to Lio 14 in. (b) At the drop panel the section of critical shear is concentric with the drop panel and is located from its edge a distance equal to é— 1 in. } (2) Formula v = ia shall be used in computing the unit stresses where V is external shear in lb. at the section considered, 6 is the perimeter of the section in in., d, the effective depth in in. The resulting stress is in lb. per sq. in, 348 DESIGN OF FLAT SLAB STRUCTURES (3) The shearing unit stress shall not exceed 0.03f'., which for 1:2:4 conerete with f’, = 2.000 lb. per sq. in. equals 60 lb. per sq. In. This rule agrees substantially with 1924 Joint Committee Speci- fication, except that in formula for allowabie unit stresses the steel ratio, r, is omitted because there does not appear any justification for the same. Critical Section for Critical Section for nm Shear at Column--._. | Shear at Drop Panel BODE RD On SEN ir ikon Fhe a Fic. 120.—Section of Critical Shearing Stresses. (See p. 347.) Formulas and Constants for Computing Shearing Stresses.—The following formulas simplify the computation of shearing stresses. These may be used for square and rectangular panels, in. Let t; = thickness of slab and drop panel, in.; t = thickness of slab, in.; c = diameter of circular column head, ft. ; 2(t1 as Lees) ft.: 2. Radel b = side of square drop panel or long side of rectangular drop panel, ft.; > Otters Waco C3 0s Cre. Ae b, = short side of rectangular drop panel, ft. ; ba ipl le Le tre C4 > 91 12 ’ ap } 1 = length of long side of panel, center to center of column, tz 1, = length of short side of panel, center to center of column, tus; SHEARING STRESSES IN FLAT SLABS l ey ratio of short side to long side of panel; w = live and dead load per sq. ft.; v = shearing unit stress, lb. per sq. in.; d = effective depth, in. Then Shearing Unit Stress at the Column Head, C2 2 a nt 101) 1) Derisqy 1M: 3.147 X 12 3 or y= Craw lb. per sq. in., where C2 2 C2 2 m — 0.785() am — 0.785(5 ome Le ee oa 12 B3102 l l Note that / is in feet and d in inches. Shearing Unit Stress at Drop Panel, ee), | l 1 vy = —————- 74 vl Ib. per sq. in., Ae x 12." or y= Cis, lb. per sq. in., where (i) ast Ci3 = ———_—.. 42 Note that I is in feet and d in inches. 349 (38) (39) (40) (41) (42) (43) Values of C2 and C13 are given in table on p. 913 for different values of co and c3, respectively, and different ratios of panel sides, m. For square panels, m = 1. 390 DESIGN OF FLAT SLAB STRUCTURES When the drop panel is rectangular, the formula for shearing stresses changes to : y= Cis Ib. pet. sq. iD.,.. +. , ++» ytpip ene _where i + x “) Ci4 es = Sey aaa a aaa rer ee mere (45) 21(% oe 1) | No table is provided for C14. Note that 1 is in feet and d in inches. Diagonal Tension Reinforcement in Flat Slabs.—The critical section, as far as diagonal tension is concerned, usually is at the column head. Generally, it is advisable to select such dimensions for the slab that the shearing unit stresses do not exceed the allowable stresses for concrete alone. If it is necessary to reduce the thickness below that required by shearing strength of concrete alone, stirrups or other diagonal tension reinforcement may be introduced, in the same manner as in any other type of concrete construction. The diagonal tension reinforcement should be computed to resist the excess of shear, beyond that which can be resisted by concrete alone. The method of procedure is as follows: The stirrups will be placed concentrically with the column head. Compute shearing unit stresses at the critical section. From these subtract the allowable shearing unit stresses for concrete alone. Multiply the difference by the perimeter of the section, in inches, at which shear was computed. Assume radial distance between two concentric rows of stirrups. Multiply the shear along the perimeter just found by this spacing. This gives total shear to be resisted by stirrups in the assumed spacing. Select diameter of stirrups and find the tensile strength of one leg. Divide the total shear to be resisted by the tensile strength of one leg of stirrups. This gives the number of legs of stirrups required to be placed around the column head. Next, compute shearing unit stress of a section at a distance from the critical section equal to the radial spacing of the stirrups. With this shear, proceed as before. The stirrups must be anchored to the tensile reinforcement, which in this section is at the top of the slab, and must extend of the bottom of the slab, where they should be provided with hooks at the ends. REINFORCEMENT FOR FLAT SLABS oo BOND STRESSES Bond stresses should be computed for the negative reinforcement at the column head by formula, Bond Stresses, = Sai (See p. 262.) They should not be larger than the allowable unit bond stresses, which for plain bars are equal to 0.04f’. and for deformed bars, 0.05f'.. For 1:2:4 concrete the ultimate stress, f’, = 2.000 lb. per sq. in.; therefore, the allowable unit bond stress for plain bars is 80 lb. per sq. in. and for deformed bars 100 Ib. per sq. in. The following values should be used in the equation: W = total dead and live load on the panel; V = 0.35W for interior panel; 0.40W for exterior panel; xo = perimeter of all effective bars in two column strips used to satisfy the bending moment, /;; d = effective depth at column head. The values 0.385W and 0.40W, respectively used for external shear V, are equivalent to the load producing the bending moment resisted by the reinforcement, the perimeter of which is used in the equation. REINFORCEMENT FOR FLAT SLABS Required Areas of Steel.—After the bending moments for the various strips are computed and the thickness of slab and drop panel selected, the required areas of steel are determined from the formula _M jdfs 7 An approximate value of 7 = 7 may be used. For f; = 16000, the value of 7f; = 14 000, and for f; = 18 000, 7f; = 15 800 may be used. In the above formula, M is the bending moment for the various design strips and d the corresponding effective depth of slab from center of tension steel (which at column is in the top of slab) to surface of slab. In flat slabs with drop panels of the dimensions described on p. 323, the depth d at the column head equals the depth of the slab taken from the tension steel to the bottom surface of the drop, and the required area of negative bending moment reinforcement in (oe DESIGN OF FLAT SLAB STRUCTURES the column strips should be based on this depth. In computing the required area of steel at all other sections, the depth of slab without the drop should be used. When drop panels are omitted at wall columns, the reduced depth must be used in computing the required area of negative reinforcement in the column strips at the wall. To get the effective depth, Le., the distance from compression face of slab to center of tensile reinforcement, proper deduction should be made from the total depth. The distance to be deducted depends upon the number of layers of reinforcement. | The bars making up the effective reinforcement in any strip should be distributed over its whole width. They should be placed at right angles to the bending moment section and extended far enough on both sides, as explained in subsequent discussion. Con- versely, all bars cut at right angles by a bending moment section, if they extend far enough on both sides of it, as described under arrangements of bars, may be considered as effective at that section. The effect of bars in the horizontal plane, inclined at an angle other than right angle to a section of bending moment, as in the middle strip in a four-way system, may be considered as equal to the area of the bars multiplied by the sine of the angle of inclination. The same bars are also effective at a section at right angles to the above-mentioned section, and their effect is again equal to the area of bars times the sine of the angle of the bar with the second section. In square panels, the effect of diagonal bars is the same at both sections, while in rectangular panels the effect of diagonals is larger at the section parallel to the short side of the rectangle. Bars parallel to a section must not be considered as effective at that section. } The effectiveness of rings as used in the Smulski System, in resisting tensile stresses, 1s explained on p. 373. Obviously, the rules given above must be extended to fit the condition. Two-way System—In the two-way system, each design section cuts, in each design strip, only one band or group of bars running at right angles to the section. The area of the bars in each band should be made at least equal to the required area of steel in the strip in which the band is located. | In computing the effective areas for negative bending moment, not only the bent bars in the panel under consideration should be used, but also the bent bars extended from the adjoining panel, if the length of extension is sufficient to make them effective. REINFORCEMENT FOR FLAT SLABS BYa}s) Four-way System.—In the four-way system, the section of posi- tive moment in the column strips cuts the direct band of bars at right angles, and their area must be made at least equal to the required area of steel. In the middle strip, the section of positive bending moment cuts two diagonal bands at an angle. The effect of each band equals the area of steel in the band times the sine of the angle of inclination. Therefore, the area of bars in each band required to satisfy the bending moment may be obtained by dividing the total required area by twice the sine of the angle of inclination of the bars to the section. In a square panel, the angle of inclination is 45 degrees and the sine is0.71. If A,4 is the required area of positive moment reinforcement in the middle strip, the area of bars in each Ass i) A 54 ) 7, STE AP Bs Ba In the column strips, the effective negative reinforcement con- sists of the bent bars in the band perpendicular to the section and of the vertical components of the bent bars in the diagonal bands. The components, again, equal the areas of the bars times the sine of angle of inclination. Not only the bent bars from the panel under consider- ation but also the bent bars extended from the adjoining span may be considered as effective, if the length of extension is sufficient for this. If the bars from the adjoining panel extend only far enough to develop their strength, they should not be counted as effective in this panel. Smuilski (S. M. I.) System.—In the Smulski System, the section of positive moment in the column strip cuts the trussed bars and (in square panels) all rings in Unit A and two or three outside rings in Unit B. The positive moment section in the middle strip cuts the remain- ing rings of Unit B and the diagonal bars. The effect of the diagonal bars is equal to their areas multiplied by the sine of their angle of inclination to the section. The negative moment section in the column strips cuts the rings of Unit C, the bent portion of the trussed bars of Units A and B, and the radial bars. The rings and the bars at right angles to the section are considered as fully effective, while the effect of the diagonal bars and radials is equal to their areas multiplied by the sine of their angle of inclination to the section. In all cases, the required area of reinforcement should be dis- tributed between the various units and the various parts of each unit. — diagonal band equals 354 DESIGN OF FLAT SLAB STRUCTURES General Rules.—After the required amounts of steel at the sec- tions of maximum bending moments have been found from formula on p.. 351, the problem resolves itself into providing bars of such lengths, and bent in such a fashion, that not only the bending moments at critical sections but also those at intermediate sections are pro- vided for. In the region of the points of inflection, the positive and negative reinforcement should overlap, to take care of the movement of the point of inflection, due to partial loading. Tests show that, with columns rigid enough to withstand the bending moments due to unbalanced live load, the possible movement of the points of inflec- tion for different loading is small. If slender columns are used, the amount of positive moment reinforcement should not only be increased, but should also extend much nearer to the column than required with rigid columns. This means that, for this condition, the points of inflection for the maxi- mum positive bending moments are much nearer the column than for negative bending moment. Proper attention should be paid to bond stresses at the column head. There the external shear is large, and the rate of increase in tensile stresses is high, with the result that large bond stresses are produced. To keep the bond stresses within working limits, the use of bars of small diameter is advisable (see p. 262). Free ends of straight bars should extend beyond the points of | ‘nflection for a length equal at least to 20 diameters of the bar but not less than 12 in. to insure the bond needed to make them available as tensile reinforcement at the points of inflection. It is important to remember that, until the tensile strength of bar is developed by bond, it is not the area of the bar that determines its capacity for resisting stresses, but the length of imbedment, - measured from the point where the stresses are investigated, to the end of the bar. For instance, in a {-in. plain bar, the full tensile working stress is not attained until the length of imbedment’® reaches 8 ft.2in. Ina band of straight bars, therefore, contrary to common belief, the available area of steel is not constant throughout the bar. On the contrary, it is zero at the end and increases gradually as it is developed by bond. 3 Where values governed by working stresses are given, normal concrete test- ‘ing 2 000 Ib. per sq. in. at 28 days is assumed. REINFORCEMENT FOR FLAT SLABS 5915) When reinforcement consists of rings, it is not necessary to con- sider bond stresses except in determining the length of lap. When reinforcement consists of bands of bars, as in the two-way and four-way systems (see p. 358), most of the bars are of such length that they serve as positive and negative bending moment reinforce- ment. The arrangement of the reinforcement at the support is, therefore, dependent upon the reinforcement in the center of the span. In such cases, the number of bars required by the maximum positive bending moment is determined first; next, one of the schemes of bending, described below, is adapted. This fixes the number of bars which will be bent up and extended over the support. The area of the bent bars available for negative bending moment reinforcement is compared with the total required area of negative moment rein- forcement. If the bent bars are not sufficient, additional short bars at the support should be supplied. When reinforcement consists of rings and radials (see p. 367), the various units are designed more or less independently. Bending of Bars.—In band systems, a large proportion of the bars are bent up and extended over the support, as shown in Figs. 122 and 127. The best method is to bend the bars before they are placed on the forms, as this insures proper location and proper height of the bends. Another method, which is much less satisfactory, is to place straight bars on the form and then bend them by means of a tool popularly called a ‘‘ hicky.” Bars of small diameter can be easily bent in this way; bars 3 in. and larger should not be hickied. This bending method is much less desirable than bending steel before placing. Satisfactory results with the “hicky ”’ are obtained only when thorough supervision is available. The length of time required to place the steel is increased by the time it takes to hicky the bars, and the placing of concrete is thus delayed. When the bars are bent on the ground ahead of time, this delay is obviated. The hickying of bars is particularly dangerous in rush jobs, where the concreting very closely follows the placing of steel. Under such conditions, it is obvious that exact bending of steel would be an accident rather than a certainty. In early flat slab designs, no actual bending of bars was done. The bars were supported above the form at the column and were allowed to sag down or drape. This practice was not satisfactory, as the sagging, especially with larger bars, was gradual, so that the bars did not get down to the bottom of the slab for quite a distance. 356 DESIGN OF FLAT SLAB STRUCTURES Large sections, therefore, both at the top and at the bottom, were without effective reinforcement. As a result, cracks developed at the regions of sagging of the bars. Modern specifications prohibit draping of bars and require that bars be bent at proper definite places, and that both the positive and negative reinforcement be parallel to the surface of the concrete. 1924 Joint Committee Rules for Arrangement of Reinforcement in Two-Way and Four-Way Systems.—The 1924 Joint Committee provides the following rules for arrangement of steel in two-way and four-way systems: “The design shall include adequate provision for securing the reinforcement in place so as to take not only the critical moments, but the moments at intermediate sections. Provision shall be made for possible shifting of the point of inflection, by carrying all bars in rectangular or diagonal directions, each side of a section of critical moment, either positive or negative, to points at least twenty (20) diameters beyond the point of inflection as specified in Section 151. This recommendation is rational. It provides general principles which will give satisfactory results. Following this, however, in the same section requirements are specified for the two-way and four- way systems which depart from the sound policy of laying down principles of design upon which the details could be developed to suit the conditions. By prescribing the lengths of bars, this rule practically limits the design of flat slab two-way and four-way rein- forcement to one arrangement. Several very satisfactory arrange- ments, both in two-way and four-way systems, described later, would be prohibited by this rule in spite of the fact that they are perfectly sound and give good results. | Support for Flat Slab Reinforcement.—Flat slab reinforcement should be securely tied and supported, in such a way that no dis- placement of the bars can take place during concreting. The displacement of bars most to be cuarded against is in the vertical direction, since the effect of both the positive and negative reinforcement depends upon its distance from the compression face of the slab. This distance must be as great as shown on the plans, while on the other hand, the bars must not be brought so near the tension surface of the concrete as to destroy the fireproofing. With positive reinforcement, the problem is to keep the bars high enough above the forms to get proper fireproofing. The best method is to support the bars on metal chairs of proper height. There are REINFORCEMENT FOR FLAT SLABS 007 on the market, also, bar spacers which serve both as chairs and as spacers. Another method is to support the bars on small concrete blocks of proper height. To save the cost of such supports, in many cases, the bars are placed directly on the forms; then, during con- creting, they are lifted up by a special workman just after the con- crete is poured, and the concrete is allowed to flow under the bars. This method is not positive and should be prohibited, since the amount of concrete below the bars depends entirely upon the judgment and skill of the workman who lifts them. When pipes for electric wires are imbedded in concrete, they should be placed above and not below the reinforcement; otherwise, the reinforcement would be raised more than required by the plans. A crack is likely to form along the pipe. For negative reinforcement, supports are absolutely necessary. The bars must be kept the required distance above the form, since lowering of the bars reduces their effectiveness. In thin slabs, even small vertical misplacement may be dangerous. For instance, in an 8-in. slab without drop panel, with an effective depth of, say, 6¢ in., an accidental lowering of bars by 1 in. would reduce the strength of the slab by about 15 per cent. In band systems, the negative reinforcement may be kept above the forms by four heavy supporting bars, resting either on concrete blocks of proper height or on metal chairs. The supporting bars are placed a short distance from the points where the bars begin to bend down. ‘The number of blocks should be sufficient to enable the sup- porting bars to keep the steel rigidly in place during construction. The negative reinforcement in the middle strip should also be kept a proper distance above the forms. For this purpose, at least two heavy supporting bars, running at right angles to the band, should be used. These supporting bars should rest on concrete blocks or chairs. In the four-way and circumferential systems, the negative reinforcement in the middle strip consists of short. bars. These are often placed after the concrete is in place but before it has attained its initial set. .This method requires proper supervision. In designing proper supports for the bars, it should be kept in mind that the reinforcement is subjected to very hard usage during concreting. Very often, even well-placed reinforcement, with sup- ports strong enough to keep it in place, is bent completely out of shape by the workmen, when they are spading and placing concrete. This is particularly true in cases where concrete is spouted. The 358 DESIGN OF FLAT SLAB STRUCTURES adjusting of the spout requires a large number of men, who, occupied with their task, have no time to pay attention to the steel upon which they are walking. This emphasizes the absolute necessity of rigid arrangement of the negative moment reinforcement. Showing Reinforcement on Plans.—The best method of showing fat slab reinforcement on plans is to designate the various bands or units by letters, and give the reinforcement for each band or unit in a schedule. To explain the schedule, a typical exterior and interior panel should be prepared, clearly showing the arrangement of steel. The length and type of bending of various bars should be shown in sections. It is obviously insufficient to give the number of bars in a band required by the positive bending moment, without giving their length and clearly showing how the bars should be disposed, as then the arrangement of the steel at the support, which is the most heavily stressed section of the slab, is left to conjecture. This wrong method of showing flat slab reinforcement is unfortunately often found in general drawings submitted to the contractors for bids. No intelli- gent estimate of tonnage can be made on the basis of such plans. It leaves the field free to unsafe designs. FLAT SLAB SYSTEMS As far as arrangement of steel is concerned, flat slab construction may be divided into three eeneral groups, which, in the order of their appearance, are as follows: The Four-way System; The Two-way System; The Circumferential or Smulski System. The last-named system is also known under the abbreviated name of 8. M.1I. System. The three-way system may be considered as a modification of the four-way system. Four-WAyY SYSTEM The four-way system is distinguished by the use of four bands * of small bars at the column, running in four directions, namely, in two directions at right angles to each other—parallel to the center lines of the panel—and in two diagonal directions. While the Norcross patent was in force, the four-way system was exploited under 4A band isa layer of definite width of bars spaced at definite distances apart. FLAT SLAB SYSTEMS 309 several trade names. The best known were: Mushroom System, Cantilever Flat Slab System, Simplex System, and Barton Spider Web System. The Mushroom System was developed by C. A. P. Turner and used in a large number of buildings. At present it is used but little, if at all, because the features distinguishing it from ordinary four- way systems are either useless or harmful. The main reinforcement consisted of four bands of bars, running as in the modern four-way systems. At the columns, these bands rested near the top of the slab on a framework formed by “ elbow bars” and two rings placed on top of them. The “ elbow bars”’ were heavy bent bars with the vertical leg imbedded AA AAA in the column and the GFF RRRR or other leg extending into the slab, not horizon- tally, but diagonally 1 —— fais {BRO n aH downward. The band Cea o. of bars resting on the framework draped from the column. This left large sections of the slab, both at the top and LD, at the bottom, without aes: i effective reinforcement. The ‘ elbow bars ”’ and rings did not contrib- ute materially to the strength of the slab, Fic. 121—Barton Spider Web System. (See p. 359.) because, being located under the four bands of bars, they were too near the neutral axis. The Cantilever Flat Slab System and the Simplex System have practically the same arrangement of reinforcement. They differ only in the arrangement of chairs and spacing bars. The Barton Spider Web System has several patented features. The bottom reinforcement consists of four bands of straight short bars, which do not reach the column but extend on each side only a short distance beyond the points of inflection. The negative rein- forcement consists of two layers of bars running in two directions. This arrangement is shown in Fig. 121, p. 359, but is not recommended for use. 360 DESIGN OF FLAT SLAB STRUCTURES Arrangement of Four-way System in Interior Panel.—Several methods of arranging flat slab reinforcement, now in common use, will be illustrated. Any of the first three methods may be used. The fourth method, although used extensively, is not recommended, First Method—Each band, rectangular or diagonal, consists of long bent bars and short straight bars. The long bars are bent at both ends near the points of inflection, and the bent portions are carried, near the top of the slab, across the column head beyond the points of inflection for negative reinforce- ment of the adjoining spans. These bars serve as positive bending moment reinforcement in one span, and as negative bending moment reinforcement in the same span and in two adjoining spans. The short straight bars, placed at the bottom of the slab, serve as positive bending moment reinforcement only. They should extend, on both sides, at least 20 bar diameters, but not less than 12 inches, beyond the points of inflection for positive moment reinforcement as specified on page 336. In this method, after the required areas of positive and negative reinforcement have been computed, the number of bars in the rect- angular and diagonal bands are determined from the areas required by the positive bending moments. A sufficient number of bars are bent up, both in the rectangular and diagonal bands, to supply the required area of reinforcement in the column head section. Not only the bent bars in the span under consideration, but also the bars extending from the adjoining spans, should be considered as effective ‘n the column head section. The bars in each band that are not required for negative moment are carried straight. This method is illustrated in Fig. 122, folding page. | The negative reinforcement in the midsection consists of short straight bars. Second Method.—All bars in rectangular and diagonal bands are provided with one double bend, as shown in Fig. 123, folding page. They are of such length that one end extends at the bottom beyond the point of inflection and the other end is bent up to the top of the slab and extended across the column head, beyond the point of inflec- tion for negative reinforcement of the adjoining span. Each bar serves as positive reinforcement in one panel, and as negative rein- forcement in same panel and in the adjoining panel. The bars in . each band are so placed that a straight end alternates with a bent-up end. | . =e Sal ray ‘S$ y S| =x S Ss = : IGN TS \ SU eS WEE NOS RES SS SS 9" 9" a a > Band ¢ “| c ~-Points of 4h “i Points of _ + —_———— Inflection <= : / Inflection Ns Band A; odes mee” Band A by +4 20°Bar 20 bar 20 bar diam diam vi diam [| = | | =| Band B, m | Band 8 Notes: Distance of points of inflection from edge of column head for positive and negative moment is not identical. Distance of points of inflection from edge of column head equals: For negative moment reinforcement 0.2/;, and for positive moment reinforcement 0.181;, where 1; is the net span. Positive moment reinforcement to extend beyond points of inflection 20 bar diameters but not less than 12 in. Area of straight bars at the bottom should equal at least one-quarter of the total area of reinforcement in column strips. Fig. 122.—Four-way Reinforcement. First method. (See p. 360.) _Mnok ende of Do ™% a‘, le - ¥ hes ; \ | a a ot RS Se bs \ : i A ; : seme Ly yy PAI Scaiasoall ; o nee May EE — BA A pf ——— a a ie TS A a I NS I os MEE AS PS ECR OF SETI fC ee Ce Sar eee ree ~s EEE ANE LOSE ES POE AO LB Pen ae ay Aa Ss fk ID 8 ae. . ee —— oe ocr ae ee 8 OS as Tere a be — - we — 2 ; ber oy ;* ones Pe ees eee i Bane “Te : — pt ae SL o et a. a Lote a ee iden - bine — se Sree A i tie : Be tpn” - ae be "i # & P ome = tas sersaxers svitegon bas svitiadd tOT bead aimuloo to ogbs mow Pe bb | Jeomun weitegea: eH cabav ter asolyauteloks borg boron by Jere ton oft a pl orwdw ,1610 feeorertobtiet sooo Siicl are ota sueéhy weds Of, fai ah a 30. naan « hinogind for pusiliee Motewet row oe ot rome Weal a ln atic letat ods: le. Pod a HO este eons Satecerets | ¢ Pic. 1 (R06 bqastoay Realsdarantt \ wy! FLAT SLAB SYSTEMS d61 The number of bars in each band is determined for the positive bending moment. In the column head section, every other bar of each rectangular and diagonal band is bent up aid becomes effective as negative bending moment reinforcement.’ Also, every other bar from the adjoining span is extended into this span and becomes effective as negative bending moment reinforcement. The area of the effective negative bending moment reinforcement is therefore equal to one-half the sum of the reinforcement at both positive sec- tions in the span under consideration, plus one-half of the correspond- ing sum in the adjoining span. With proper depth of drop panel, this amount of steel may be sufficient to supply the required negative moment reinforcement in the column strips. If not, additional short bars should be provided. This happens for shallow drops or for slabs without drop panels. The short bars should extend, on both sides, 20 bar diameters,® but not less than 12 in., beyond the points of inflection for positive moment reinforcement as specified on page 336. The negative reinforcement in the midsection consists of short straight bars. Third Method.—All bars in rectangular bands are bent up and extended, at both ends, 9 in. beyond the points of inflection for negative moment reinforcement of the adjoining panels. (See folding Fig. 124.) The bars in the diagonal bands are straight and extend at each end, 20 bar diameters beyond the points of inflection for positive moment reinforcement as specified on page 336. With such an arrangement, the negative bending moment reinforcement at the column consists of two layers only. The area of the negative reinforcement. is equal to twice the area of the rectangular band steel. If this is not sufficient to supply the negative bending moment reinforcement, additional straight bars should be added. _ The negative moment in the midsection consists of short straight bars. Fourth Method.—In this method, all four bands consist of long bars, all of which are bent up in the region of the points of inflection and extend beyond the column center. Some of the bars are extended into the adjoining panel, beyond the points of inflection, and are considered as negative reinforcement there. Other bars are lapped at the column. For comparatively short panels, some of the bars of the rectangular bands are made two panels long. ’ See footnote, p. 354, 362 DESIGN OF FLAT SLAB STRUCTURES This is the earliest method of arranging flat slab reinforcement and was used in the Cantilever and Simplex Systems. It is in dis- favor now, because all the bars are bent at the points of inflection, thus leaving a section of the slab, where positive moments may occur, without any positive reinforcement. This method is not recommended. Cross Bars-—In each of the four methods, negative bending moment in the middle strip (as defined on p. 330) is provided for by means of short bars placed across the rectangular bands. Arrangement of Four-way System in Exterior Panel.—In exterior panels, the continuous bands are arranged in the same manner as in interior panels. In discontinuous bands, the straight bars com- posing the band are extended to the wall column or wall beam. This applies also to the straight ends adjacent to the wall in the Second Method. The bent bars at the wall are provided with hooks of sufficient size to develop the strength of the bars. The hooked ends are extended to within 2 or 3 in. of the outside face of the column or wall beam. This last is an important requirement but is frequently neglected. Usually, it will be found that at the wall column the bent bars alone are not sufficient to resist the negative bending moment. The required area of steel should be computed, and extra, short, hooked bars should be provided. The hooks should be placed in the wall beam, as near its outside face as possible, and these bars should extend into the slab beyond the point of inflection. The required number of additional bars will be particularly large when the drop panel is omitted at the wall column, in which case the required area of steel must be computed on the basis of the depth of the slab and not the depth of slab and drop panel. In the midsection, at the wall beam, short bars should be used near the top of the slab. They should be placed at right angles to the wall beam. The hooks should be placed not more than two inches from the outside face of the wall beam. Two-wAY SYSTEM In the two-way system, the reinforcement consists of bands of small bars running in two directions. This system was originated and patented by Frank F. Sinks, and developed commercially by the Condron Company, under the trade name of the Acme System. The question of patents is not definitely settled. In one instance, FLAT SLAB SYSTEMS 363 at least, the Sinks patent has been declared void by the court, but this is not necessarily a final decision. A typical arrangement of the Acme System is shown in Fig. 125, p. 363. Another variation of the two-way system is the so-called Cor- plate, developed by the Corrugated Bar Company. Some features 0.2t->10.21-~ : Al 2k rfenal Main Band | Extra Top Bars () , a e 4 « e a e « 4 a e e e e a ¢ 4 4 ‘ 4 e t 1 ~ = @ a a t] Cy q t e ¢ ‘ 6 a a i] i ‘ ' 1 ---(0.591—0.29¢)-----3 Si eee 0.211+0.29¢ \ --2 —=— ; ‘0.07 (I-¢) 0,07 (1-c}- i sriaatapeneny OMe) alate Fig. 125.—Acme System. (See D. 363.) of this arrangement are patented. This variation differs from ordi- nary two-way arrangements in that the steel in the column head section is not uniformly distributed through the band, but is more concentrated near the column than near the edges of the band. With bars of equal diameter, this is accomplished by making the 364 DESIGN OF FLAT SLAB STRUCTURES spacing of the bars smaller in the middle of the band and then increas- ing it toward the edges of the band. This is shown in Fig. 35, p. 101. Arrangement of Two-way System in Interior Panel.—Several methods of arranging reinforcement are now in common use. Three “+ Points of // Inflection — “9" Hook A & BO ‘ A&B S90 Bar ‘~-20 Bar _ 90 Bar Diameters Diameters Diameters Bending Details Notes: Distance of points of inflection from edge of column head for positive and negative moment is not identical. Distance of points of inflection from edge of column head equals: For negative moment reinforcement 0.2i;, and for positive moment reinforcement 0.1811, where /; is the net span. Positive moment reinforcement to extend beyond points of inflection 20 bar diameters but | not less than 12 in. Area of straight bars at the bottom should equal at least one-quarter of the total area of reinforcement in column strips. Fig. 126.—Two-way System. First Method. (See p. 364.) methods will be described and illustrated. Either of the first two methods may be used. The third method is described because it is sometimes used, but it is not recommended. First Method.—Each band consists of long bent bars and short straight bars. The long bars are bent up at both ends, and the bent FLAT SLAB SYSTEMS 365 portions are carried, near the top of the slab, across the column into the adjoining spans, extending 9 in. beyond the points of inflection for negative moment reinforcement. The points at which the bars are bent should coincide with the points of inflection. |} Points of _ a __Points of _ "Inflection <- 7 Inflection ~ li, / Z \.9" Hook} <; | “x A,& Bf” Bending Details Section Notes: Distance of points of inflection from edge of column head for positive and negative moment is not identical. Distance of points of inflection from edge of column head equals: For negative moment reinforcement 0.2/;, and for positive moment reinforcement 0.18l;, where J; is the next span. Positive moment reinforcement to extend beyond points of inflection 20 bar diameters but not less than 12 in. Fig. 127.—Two-way System. Second Method. (See p. 366.) The short straight bars should extend, at each end, 20 bar diam- eters, but not less than 12 in., beyond the points of inflection for 366 DESIGN OF FLAT SLAB STRUCTURES positive moment reinforcement as specified on page 336. This arrangement is shown in Fig. 126, p. 364. The number of bars in each band is determined from the positive bending moment. A sufficient number of bars should then be bent up to provide the required amount of steel for negative bending moment. With the arrangement recommended, the negative rein- forcement consists not only of the bent-up bars in the panel under consideration, but also of the bars extending from the adjoining span. It is good practice to have at least one-third of the bars in a band straight. If the negative moment reinforcement is not satisfied by the number of bent bars, additional straight bars should be used. These should extend, at each end, 9 in. beyond the points of inflec- tion for negative moment reinforcement. Second Method.—All bars in each band consist of bars straight at one end and are provided with a double bend at the other end. The straight portion serves as positive reinforcement and is of such length that it extends 20 bar diameters (minimum 12 in.) beyond the points of inflection for positive moment reinforcement. The point of bending of the bar coincides with the point of inflection. The bent-up portion of the bar is carried, near the top of the slab, into the adjoining panel, extending 9 in. beyond the points of inflection for negative moment reinforcement. Each bar serves as positive bending moment reinforcement at one section of maximum bending moment, and as negative reinforcement in the same panel and also in the adjoining panel. The bars in each band are so arranged that at each end the straight portions alternate with the bent-up portions. The total number of bars in each band is determined for the positive bending moment. The negative moment reinforcement will consist of one- half of the bars in a band in the panel under consideration, and one- half of the bars from the adjoining panel. Ordinarily, this is not sufficient in the column head section, and additional straight bars, extending, on each side, 9 in. beyond the points of inflection for negative moment reinforcement, should be used. This method is shown in Fig. 127, p. 365. Third Method.—In this method, all the bars in each band are bent up at both ends and carried near the top of the slab. Such an arrangement is not recommended, because it leaves a section of the slab, where positive bending moment may occur, without any rein- forcement. FLAT SLAB SYSTEMS 307 Arrangement of Two-way System in Exterior Panel.—In exterior panels, the continuous bands are arranged in the same manner as in interior panels. In discontinuous bands, the straight bars or the straight ends of bars should extend to the column or wall beam. The bent bars at the wall column should be provided with hooks, to develop their full strength. These should be extended to within 2 or 3 in. from the outside face of the column wall beam. This requirement is important. Usually, it will be found that, at the wall column, the bent bars alone are not sufficient to resist the negative bending moment. The required area of steel should be computed and extra, short, hooked’ bars should be provided. These should extend beyond the point of inflection. The required number of additional bars will be par- ticularly large when the drop panel is omitted at the wall column. CIRCUMFERENTIAL OR SMULSKI (S.M.I.) System © The Circumferential or 8.M.I. System was developed and patented by Edward Smulski. The reinforcement of typical interior and exterior panels is shown in Fig. 128, p. 369. A modification of the design is shown in Fig. 129, p. 370. . In general, the reinforcement consists of four types of units, which are usually designated by the letters A, B, C, and T. To distinguish between units of different panels, small letters are added. Thus, in end panels, the units become Ae, Be, Ce, Te, while at the wall Cw and Aw are used. When the steel in various floors is arranged differently, the floor is designated by a number placed before the name of the unit. Thus, 2A is Unit A in the second story, while RA is Unit A in the roof. Units A and B are placed near the bottom of the slab and serve as positive bending moment reinforcement, while Units C and T are near the top of the slab and serve as negative bending moment rein- forcement. . 6 Because of the business association of one of the authors, Mr. Smulski, with the Smulski Flat Slab System, we have refrained from comparing in the test the relative economy of the different Flat Slab Systems. Personal analyses of the theory made by the signer of this footnote show that, using the same principles of design and same bending moment, the Smulski System requires 20 to 24 per cent less reinforcement than the two-way or four-way systems. Tests of the system made by the signer of the footnote or witnessed by him prove the strength, reliability and permanence of the system.—Sanford E. Thompson. BOD. DESIGN OF FLAT SLAB STRUCTURES Bottom Reinforcement.—Units A are placed between columns. They consist of trussed bars and of rings. The number of rings depends upon the span and the load. For spans up to 22 ft., four rings are generally used. The ends of the rings are lapped to develop their full strength. The laps are staggered, so that not more than one ring is lapped in one section. The trussed bars are bent up at appropriate points and are car- ried, near the top of the slab, into the column head. In Fig. 128, they are extended into the adjoining panel and carried beyond the point of inflection. In the modification shown in Fig. 129, they are hooked on to the center ring within the column head. Units B are placed in the central part of the slab. They consist of rings and diagonal bars. The rings overlap those in Units A, thereby forming a continuous mat of steel extending over the whole slab. As evident from Figs. 128 and-129, the whole slab, with small exception of the area immediately around the column head which needs no such reinforcement, is provided with bottom rein- forcement capable of resisting stresses in any conceivable direction. The diagonal bars are bent up and carried, near top of the slab, into the column head, where they are hooked on the center ring. Units Aw, at the spandrel beam, consist of trussed bars and half rings that are provided with hooks at their ends. 7 Top Reinforcement.— Units C are placed at the column, near the top of the slab. They consist of rings and radial bars. | One of the rings is placed within the column head, while the remaining rings are placed outside the column head. The number of rings depends upon the size of the panel and the load. They are spliced to develop the full strength of the bar, and the splices are staggered. The rings form the top layer of steel. The radial bars are in the shape of a hairpin, with prongs of uneven length. The longer prong is placed near the top of the slab, directly under the rings. The hook of the hairpin engages the center ring, which is placed within the column head. ‘The free end of the longer prong extends beyond the points of inflection. The reinforcement at the column head, in addition to Unit C, consists of trussed bars from Units A and B. ae Units 7 are placed across the panel edge in the midsection, con- sisting of bars of a length equal to one-half of the panel length. Units Cw, at the wall panel, consist of radial bars similar to those of Unit C, and of half rings placed on the top of the radial bars. The half rings are provided at their ends with hooks of proper FLAT SLAB SYSTEMS 369 dimensions. To provide proper anchorage, the hooks are placed as near the outside edge of the wall beams as possible. Units Tw take the place of Units T at the wall. They are pro- _ vided on one end with a hook, which is placed as near the outside edge of the wall beam as possible. t hee, ML Ea Uprt pay} Hef > tte DEE Doel A cet EI 23 See Geteeiwe teen 20 = et XO < ie, OSS ASIEN SEES i Sag a ; tava AAS CDK PO Se ERO PEE TREY RESTS ee eee ane BS. 7 SST I ney, INSZ) SEK Nt : Unit T->\ 7 a M cory oti | CER DR CaO me Gan ee les SEEN i f ot Pea ge ne arg Seu tses “things a TN te Sea oe Bee? ” Us Bars. ‘Rings | “Radidis ea a ‘Radials Section M-M E-ed. Column Lc L.Colurnn b-feerab pari arecre aeci ee rf. 246% ! Trussed Bar Unit A Face of Wal/ ie Column ‘ v ~Center Ring wane nee ae += L S575 7 Panieianea aa £ a ¢ Trussed Sar Unit Ae tg fe Col In wall panels this end | fenters wall column “3 TH : / Sr Diagonal Span -----—------------ i ‘Center Ring Center Ringé. Trussed Bar Units8 \ a L. Column 4 --<—__—— | Fradials Unit C,Cw Fic. 128.—Typical Smulski Flat Slab Reinforeement for Interior and Exterior Panel. (See p. 367.) EXPLANATION OF ACTION OF THE SMULSKI SysTEM Since the reinforcement of the Smulski (S. M. I.) System is radically different from that of the Two-way and Four-way Systems, its action will bé more fully explained below. 370 DESIGN OF FLAT SLAB STRUCTURES The Smulski System takes advantage of the following prin- ciples: 1. Ring reinforcement prevents any deformation of the concrete enclosed by it. Any tendency of the concrete within the ring to spread or elongate in any direction is resisted by the steel ring. There- fore, any tensile stresses within the ring, irrespec- tive of their direction, are resisted by the steel ring. A properly lapped ring is a complete unit and does not depend upon bond to develop stresses in the bar, as is the case with straight bars. While the ring is not at all effec- tive outside of its circum- ference, it is fully effec- tive at all points within : it. Fic. 129.—Modified Arrangement of Smulski Flat. Because of this prop- Slab Reinforcement. (See p. 367.) erty, a ring is a valuable reinforcement where the direction of stresses is either unknown or changeable, as In @ flat slab, particularly at the column head. In test, to destruction, of the Western Newspaper Union Building,’ with flat slab floor, four-way system, the stresses in diagonal direction were much larger than in rectangular directions, also at ultimate load the stresses ‘n each band varied at an average from 18 000 lb. per sq. in. for out- side bars to 50 000 Ib. per sq. in. for center bars of the band. This means unequal utilization of steel, with the consequence that, to prevent excessive stresses in the bars carrying the largest stresses, a larger area of steel is required to resist the bending moment than if all bars were fully effective and equally stressed. ] VJ D © rt v4 , 6] ° on ey aes OCD aC) = is iat ia CSA x — O%: of esse sees eee wee PRE (ome == Saintes =a see eeteenee ones ON he) ‘Rings ‘Trussed Bars Lk / Radials Section Thru Column Head 7 Test of a Flat Slab Floor of the Western Newspaper Union Building, by Arthur N. Talbot and Harrison F. Gonnerman. Bulletin No. 106, Engineering Experiment Station, University of Illinois. FLAT SLAB SYSTEMS oll With ring reinforcement, all the stresses are resisted by all the rings at their full working value, irrespective of their position and intensity. Thus in Purdue test, for maximum load, the stresses in rings at center column varied from 33 300 to 41 900 Ib. per sq. in. while in radials the variation of maximum stress was from 41 100 to 44 700 Ib. (See p. 107.) Separating Flat Slab into Simple Parts.—It is a well-known fact that a continuous beam or slab can be replaced by simple parts without changing the stress conditions. Thus, the beam can be cut at the points of inflection, where the bending moment is zero. It is then changed into two cantilevers and a simple beam suspended from the cantilevers. The stresses and the bending moments would not be affected by this change. The same principle can be applied to flat slabs, which can be separated into following parts, as shown in Fig. 117, p. 328: Circular cantilevers at the column head; Slabs between columns; Slabs supported at four points subjected to stresses in all direc- tions. In designing the reinforcement, it is permissible to treat the separate parts independently and to provide in each of them a suffi- cient amount of steel to resist the particular moments to which they may be subjected. As the unit shear at the points of inflection is always low, not exceeding 40 lb. per sq. in., concrete is capable of taking care of the shearing stresses. Since it is not advisable to rely on concrete alone, the parts of the slab subjected to positive bending moment, and reinforced by Units A and B, are tied securely to the circular cantilever at the column head by the bent portions of the trussed bars and by the overlapping of Units A and C. The position of the points of inflection is variable for different positions of the live load. To provide for this and also to pre- vent secondary cracks due to temperature and shrinkage, the various units overlap, thereby tying the slab together and en- abling it to act as a whole if such action is required by any con- tingencies. Column Head Section.—At the column head the portion of the slab within the points of inflection acts like a circular cantilever, loaded uniformly over its area, and also along its circumference, by 312 DESIGN OF FLAT SLAB STRUCTURES the loads transferred to it from the rest of the slab. This portion is subjected to negative bending moments; i.e., the particles in the upper part of the slab elongate, while the particles in the lower part are compressed. After deflection, the cantilever assumes the shape of an umbrella. The negative bending moment at the column head is larger than the positive bending moment in the center of the slab. The amount of steel required there is therefore larger than in any other part of the slab. To reduce the amount of reinforcing steel required at the column head, the depth of the slab in this section is often increased by forming the so-called drop panel. The most unfavorable condition of loading for the column head section is that which occurs when all the spans surrounding the col- umn are loaded. In such a case, the shape of the cantilever will be as shown in Fig. 130, p. 378. Since, after deflection, any circle increases its radius as well as its circumference, the particles must elongate in radial as well as in circumferential direction and are therefore subjected to radial and circumferential stresses. The most effective tensile reinforcement consists of rings and radial bars. Compressive stresses act also in radial and circumferential directions. The compression acting radially composes the bulk of compressive stresses. Slabs between Columns.—The principal stresses in this part act mainly in one direction, which at first is parallel to the edge of the sanel and then gradually becomes inclined. - In addition, negative stresses due to cross bending, and also due to shrinkage and tem- perature changes, act across the principal positive stresses. The advantages of using rings in this part to resist the various stresses are as follows: (1) They intersect the lines of equal defiec- tion more nearly at right angles than straight reinforcement. (2) They bind the Units A and B, thereby preventing secondary cracks. (3) The rings in the two units supplement each other. (4) The arrangement 1s economical, as the rings cover the whole surface without waste of material. Central Part of Slab.—The central portion acts like a slab sup- ported at four corners and loaded with uniform load. The bending moment is positive, so that the top is in compression and the bottom in tension. The stresses act in all directions; the reinforcement consisting of rings, therefore, is fully effective. FLAT SLAB SYSTEMS 373 Action of Rings.—The rings resist all tensile stresses within them, irrespective of the direction in which they act. To understand the action of the rings, it is necessary to keep in mind that tensile stresses can not act without producing a corre- sponding lengthening of the materials. Also, it must be remembered that the rings are filled solidly with concrete. If the shape of ab&cd Arcs of Steel Ring before Deformation a,b, & c,d, Elongated arcs ab & cd after Deformation Points a oe oe eo Line of Inflection —_—— — __ son Lr Steel Ring “\\ also Points 4-3 Points 2a a roe a also Points 2- ]------ =-§teel Ring also Points 1 - 2-7 PSteel Ring also Points - a ae *—Points 4° ne aes eo = — ‘Neutral’ Axis 4-4'=Elongation of distance 0-4 Section A-A SectionA-A 2-2'= Elongation of distance 0-2 (equals 4- 4' plus Elongation of distance 2-4) Rings in Unit. C Rings in Unit B Fria. 130.—Action of Rings in Resisting Stresses. (See p. 373.) the concrete within the ring undergoes any changes due to tensile stresses, the tight-fitting rings must undergo ‘similar changes. When the slab is loaded and the concrete is elongated by stresses produced by the bending moments, it presses against the circumfer- ence of the enclosing ring and forces it to stretch proportionately to the magnitude of the stresses. Parts of the slab at the column and also in the center of the slab, subjected to stresses in all directions, are shown in Fig. 130, p. 373. AA is asection in any direction. When loaded, the slab is compressed near one surface and elongated near the other surface, so that in 374 DESIGN OF FLAT SLAB STRUCTURES Fig. 130 the distance 0-3 shortens by 3-3 ft, while 0-4 expands by 4-4 ft. The same is true of a section in any direction; therefore the circle 4-4 tends to assume the position 44 ft. Before assuming the new position, the concrete must stretch the ring by which it is enclosed and increase its radius and therefore its circumference. The concrete exerts a pressure along the circumference of the ring similar to the pressure of water In a reservoir. The steel ring, by its tensile resist- ance, partially prevents the movement, but it stretches to some extent, causing tensile stresses in the steel. Considering the second ring, it is evident that the movement of point 2 consists not only of the elongation of the distance 0-4 but also of the elongation of the distance 2-4. The outside ring, there- fore, shares the stresses with the inside ring. Any deformation of the concrete, irrespective of its direction, is taken up at once by all the rings placed outside of the place of deformation. All rings are effective in resisting stresses. . Forces Acting in All Directions.—Where the forces act in all directions, as in the center of a slab and at the column head, the ring stretches uniformly along its circumference. After deformation, the shape of the ring remains substantially circular and the stresses are uniform along its circumference. Forces Acting Principally in One Direction.—When the forces act principally in one direction, as in Unit A, the condition is similar to that of a solid disc of concrete with a tight-fitting steel ring around it, subjected to a force in one direction. Under the pressure of the enclosed concrete, the shape of the ring changes gradually into an oblong curve, with the concrete following and still pressing tightly on the ring. In this case the stresses in the ring are a maximum at the sections cut by a diameter perpendicular to the direction of the stress, and decrease to zero at points 90 degrees from the point of maximum stress. From the above, it is evident that the stresses due to the prin- cipal bending moment are small in the parts of rings of Unit A which are near the column head, so that they can resist stresses in diagonal direction in places where they run almost parallel to the diagonal bars. THREE-WAY SYSTEM The three-way system was developed by D. W. Morrow of Cleveland. It can be used only with a peculiar spacing of columns. OPENINGS IN FLAT SLAB BYES) The reinforcement then consists of three bands of small bars. The number of bars in each band may be determined for positivé bending moment. The arrangement of bars may be similar to that described in connection with the four-way system. The negative moment in the midsection should be provided for in the same manner as in the four-way system, and should consist of short straight bars. In Four -way System add at Column Diagonal bars to compensate for Diagonal-~_ Band omitted in this Panel. to carry Skylight Use Sufficient Amount of Reinforcement around opening ac) ied S = = = = D ~ a) > a) > is = , 8 ~~ ° ~~ compensate for omission of Strip add Short Bars on top to Middle Strip Bands. ‘2-2'Ra. 4 Ft. 0" Top —— at each corner ‘2 "Rd. 72" on centers Section Through Skylight Fra. 181.—Opening for Skylight. (See p. 376.) OPENINGS IN FLAT SLAB Proper design of the flat slab at openings requires considerable judgment. Openings up to one foot in diameter seldom require framing unless they occur at the columns. For large openings, it is often possible to strengthen the slab sufficiently to get along without 376 DESIGN OF FLAT SLAB STRUCTURES any beams. No general rule can be made to cover all cases. Inex- perienced designers are advised to provide beams at openings. Figure 131 shows a typical opening in the roof slab for skylight. With such an opening, the slab can be designed without any beams, the reinforcement on all four sides of the opening being made strong enough to carry the weight of the skylight. Sometimes, a curb is coe ees ee mee ee ee we a ae ew ws i ee ee Se A PS tet ee ee Fic. 132.—Opening for Staircase. (See p. 376.) provided, which may be built with the slab or separately after the slab is poured. If the curb is to be built separately, dowels should be provided in the slab, and when the curb is being built, the joint between old and new concrete should be bonded with neat cement paste to avoid leakage. ‘T’o prevent corner cracks, two -in. round bars should be placed diagonally at the top of the slab at each corner of the opening. Figure 132 represents another common type of opening, for a DESIGN OF BEAMS IN FLAT SLAB CONSTRUCTION BYi| staircase. A framing of beams is usually provided for such an opening. DESIGN OF BEAMS IN FLAT SLAB CONSTRUCTION When Beams are Necessary in Flat Slab Construction.—Some- times it is necessary to introduce interior beams in flat slab construc- tion, to carry for instance, heavy concentrated loads, or to frame an open‘ng. An illustration of the use of beams is given in Fig. 132, p. 376. Load Carried by Beam.—An interior beam in flat slab construc- tion should be designed for the load coming directly upon it, plus a flat slab load from a width of slab on each side of the beam equal! to one-fifth of the span at right angles to the beam. Often, in such cases, the beam is designed only for the load super- imposed upon it, and the flat slab, reinforced as if no beam were used, is assumed to carry the slab load. This is obviously wrong. The beam and slab cannot act separately, the behavior of the slab. being dependent upon that of the beam. Part of the slab on both sides of the beam acts as a T-flange for the beam, and, when the beam deflects, the slab deflects not as an independent slab, but as a flange of the T-beam. The position of the neutral axis is governed by the beam. The under side of the slab is in compression or, if not, the tensile stresses there are small. The effect of the slab reinforcement parallel to the beam (which would have been sufficient to carry the slab load if the beam were not there) is small; therefore, the load which would have been carried by the slab must be carried by the beam. Short bars must be placed at the top of the slab across the beam. Restraint of Interior Beams Framing into Columns.—Beams in flat slab construction are usually one panel long, and the tendency is to treat them as simply supported beams. This is not correct. The beams form a monolithic part of the flat slab construction. If they frame into the columns, they are restrained there to the same degree as the rest of the flat slab. When they frame into interior columns, they may be considered as fixed at both ends. When they frame into one interior and one exterior column, they may be con- sidered as fixed at the interior column and restrained at the wall column. i In Fig. 132, p. 376, for instance, beam B, is only one span long. If the rules for one-span beams were followed blindly, it would be 378 DESIGN OF FLAT SLAB STRUCTURES ee Peis at designed for oo in the center, with little or no provision for bending moment at the supports. This would be not only wasteful but also unsafe. From inspection of the construction, it is evident that the beam is restrained not only by the column but also by the flat slab on the other side of the column. The beam has a restraining effect on the slab and in turn, the slab restrains the beam. Bending Moments in Interior Beams Framing into Columns.— Interior beams framing into columns should be designed as follows: The span of the beam should be taken as the distance between the faces of the column and not between the edges of the column heads. — The following bending moments should be used. Let J; = net span, i.e., distance between faces of columns, ft.; w = total live and dead load, lb. per lin. ft. Then Interior Panels.— Negative Bending Moment at Columns, M = — +,wli? ft.-lb. or — wh? in-lb. . . . . (46) Positive Bending Moment in the Center, M = + sul)? ft-lb. or 0.75wl)? n-lb. . = . (47) Exterior Panels.— Negative Bending Moment at Intervor Column, M = — pywl:? ft-lb. or — 1.2wh? in-lb. . . - (48) Negative Bending Moment at Wall Column, M = — +,ul,? ft.-lb. or — 0.75wh;? in-lb. . . (49) Positive Bending Moment in the Center, M = + pls? ft-Tb. or wh? in-lb. - 2) a (a0) Interior Beams not Framing into Columns.—Interior beams that do not frame into columns such as beam Bg in Fig. 132, p. 376, are only partly restrained at the ends. The bending moments should be 2 wl assumed as 10 be taken as from center to center of the supporting beams. The negative reinforcement should not be hooked in the beam but should be extended into the slab. 2 in the center and - at the support. The span should WALL BEAMS 379 WALL BEAMS Wall beams are beams between exterior columns, used to carry the spandrels (i.e., the walls under the windows) or curtain walls. In flat slab construction, wall beams may be divided into two general types: (1) wall beams extending below the slab, and (2) wall beams extending above the slab. The various designs of the wall beams are shown in Figs. 133 to 137. Fic. 133.—Wall Beams below Slab. (See p. 379.) Wall Beams Below the Slab.—When wall beams extend below the flat slab, as in Fig. 133, they must be designed to carry not only the wall load but also a portion of the flat slab load. The reason for this is explained on p. 388, and is evident from Fig. 134, p. 381. The best practice, endorsed by the authors, requires that 20 per cent of the slab load in the wall panel be considered as carried by the wall beam. The slab load should be assumed as uniformly distributed over the whole length of the beam. For example, if the span of the slab at right angles to the beam is 20 ft., the dead and live load on a width of slab equal to 4 ft. should be considered as being carried by the beam. The authors recommend that the bending moments be based on the net span, i.e., the distance between the faces of the columns. The following coefficients are recommended: Let /; = net span, i.e., distance between the faces of column, ft.; w = dead and live load on wall beam, lb. per lin. ft. For interior spans the bending moments should be as follows: Negative bending moment, M = — ,uwli? ft.-lb. or wl,? in.-lb.; Positive bending moment, M = z;wl,? ft.-lb. or 0.75wl;? in.-lb. For end spans where the width of the corner column is less than one-tenth of the span, Negative bending moment, interior column, M = — xul,? ft.-lb. or — 1.2wl1? in.-lb. 380 DESIGN OF FLAT SLAB STRUCTURES Negative bending moment, corner columns, M = — 3,wl:2 ft-lb. or — 0.75wl,? in.-lb. ; Positive bending moment, M = },wli? ft.-lb. or wl)? in.-lb. For end spans where the end column is more than 14 times the depth of the spandrel and more than one-tenth of the span, the same bending moments may be used as for interior spans of the wall beam. At the end column, then, the same area of negative reinforcement must be used as at interior columns. For deep beams, the bending moment must be considered as resisted by the reinforcement in the beam only, because the tensile stresses in the reinforcement of the slab strip parallel to the beam are either non-existent or negligible. With shallow wall beams, the slab reinforcement parallel to the beam resists part of the bending moment in the wall beam. Usually, the effect of the slab reinforcement is small and is not worth con- sidering. If desirable to consider the slab reinforcement, it should be done as in the following example: Example 1.—If effective depth of beam isd = 15 in. effective depth of slab, d; = 10in.; k = 0.375. Then the depth of neutral axis for the beam is kd = 5.6 in. The depth of slab below neutral axis for the beam is 15 —5.6 = 9.4 in, while for the slab it is 10 — 5.6 = 4.4 in. If the stress in beam reinforcement is fs, the stress in slab reinforcement will be in proportion to its distance from the 4.4 neutral axis, or yn f; = 0.47f,. If As is the steel in the strip parallel to the beam, the tensile stresses resisted by it are0.47A sfs. To get the bending moment, this stress must be multiplied by the distance between the slab steel and the center of compression, which is 4.4 + 3kd =4.4 + 3 X 5.6 = 8.1 tinee lhe bending moment is M, = 0.47f, X 8.1 = 3. SAebe The same area of steel, if placed in the beam, would resist a bending moment M. =AcfsX34X15 =13.2Asf.. Comparing this with the above value, it is evident that the effectiveness of the slab steel is equal only to 0.27 of the effect- iveness of the same reinforcement placed in the beam. Requirements of Building Codes.—The requirement adopted by the authors, as to the amount of slab load carried by the beam, is incorporated in most modern codes. Some codes, however, still permit the wall beam to be designed for the wall load only. ‘This always gives unsafe results. The Code of the City of Boston requires that the wall beam be designed to carry the wall load and, in addi- tion, a slab load on a width of slab equal to 20 per cent of the clear WALL BEAMS 381 span of the spandrel beam (irrespective of the magnitude of the span of the slab at right angles to the beam). This requirement is safe in some instances and unsafe in others. In cases where the span of the slab at right angles to the wall is small, the requirement gives absurd results. 1924 Joint Committee Requirements—The 1924 Joint Committee distinguishes between a beam having (a) a depth greater than the depth of drop panel into which it frames, and one having (b) a depth equal to or smaller than the depth of drop panel. The beam (a) (deeper than drop panel) is required to be designed for all directly superimposed loads plus one-fourth of the distributed load for which the adjacent panel (or panels) is designed. Each column strip ‘prop ‘Slab Steel” ‘ Beam Steel ~ Pcie “Moment Arm i Sy for Slab Steel =8.1" Fig. 134.—Stresses in Beam and Slab Reinforcement at Wall Beam. (See p. 380.) adjacent to and parallel with the beam is required to be reinforced with one-half of the reinforcement which it would have required without the beam. The beam (b) (depth equal to or smaller than drop panel) is required to be designed for the load directly superim- posed upon it, while the flat slab strip parallel with it is to be designed to carry all the slab load. The requirement for beams (a) (deeper than drop panel) is logical and agrees substantially with the authors’ recommendation. The requirement for beams (b) is not correct, as clearly explained above and on p. 388. This requirement is based on the assumption of full working stresses in slab steel. From Fig. 134, it is evident that if the slab steel were stressed to full working stresses, the stress in beam steel would have to pass the elastic limit. Wall Beam below Slab with Concrete Curtain Wall above.—In many cases, the wall beam and curtain wall are built of concrete, 382 DESIGN OF FLAT SLAB STRUCTURES but, for reasons connected with construction, it may not be desired to build them at the same time or to make provision for the coopera- tion of the concrete wall beam below the slab with the concrete cur- tain wall above. In such cases, a definite joint is made between the curtain wall and the rest of the construction, at the top of the slab and at the columns. The concrete curtain wall is then built after the frame is completed. To insure weather tightness and to tie the cur- tain wall to the beam, dowels are provided in the beam and are then imbedded in the curtain wall. Also, recesses coated with asphalt are provided, in the column, for the walls, The curtain wall is rein- forced for temperature. The amount of this reinforcement will — depend upon the extent to which it is desired to avoid cracking. ' At least 0.25 of one per cent of longitudinal steel is recommended, particularly if the concrete in the wall is exposed. Window sills should be reinforced with two }-in. round bars. Satisfactory results have been obtained in some cases, in 8-in. curtain wall where 3_in. round bars were spaced 12 in. on centers; but the small saving in reinforcement does not compensate for the unsightly appearance of the temperature and shrinkage cracks which are liable to occur. The concrete curtain wall is not considered as a part of the beam. The method of design of the wall beam, in this case, differs from the previous case only in that the curtain wall does not need to be con- sidered as adding to the loading of the beam. Wall Beam above the Slab.—It is often desirable to bring the top of the window as close to the bottom of the slab as possible. In such cases, the wall beam is placed above the slab, as illustrated in Figs. 135 to 137. When no drop panel is used, the bottom of the wall is flush with the bottom of the slab. When a drop panel is used, the bottom of the beam is at the same level as the bottom of the drop panel. The design of the wall beam will depend upon the method of con- struction. Beam and Slab Built at the Same Time—The wall beam shown in Fig. 135 should always be built with the slab. The method of design will be the same for wall beams below the slab. In the design of wall beams shown in Fig. 136, the slab and the wall beam, for its full depth (including the column), are built at the same time. The beam is designed as fully continuous. The tem- perature steel in the beam is extended into the columns, to prevent cracks at the junction of the beam and the column. When beams WALL BEAMS 383 are deep, no bent bars are used, and the negative reinforcement at the column consists of short bars placed near the top of the beam. In such cases, the beam is tied to the columns so that it cannot con- tract freely. To prevent cracks due to shrinkage and temperature, at least 0.3 per cent of longitudinal steel is needed. This method, when the wall beam is properly designed and built, gives very satisfactory results. However, it is difficult to construct. The formwork for the wall beam is expensive, and difficult to keep in place during construction. It is difficult to prevent the concrete in the wall above the slab from running out into the slab, especially ~--- Sil ~---Brick 2" Diam. Bolt Continuous Angle____- Sor Brick Veneer Fic. 135.—Wall Beam above Slab. (See p. 382.) when the spandrel is so high that considerable pressure is developed. Rather dry mix should be used in such cases, and the concrete in the wall should be poured after the concrete in the slab has begun to stiffen very slightly. The same loads and moments may be assumed as in the case of beams extending below the slab. The bars in the band parallel to the wall may be assumed as effective in resisting the positive bend- ing moment in the beam, to the extent of 50 per cent of the required steel area in the beam. Stirrups should be used throughout the length of the beam. They act partly to suspend the slab load from the beam. Beam and Slab Built Separately.—In the design shown in Fig. 137 the portion of the beam above the slab is built after the frame is 384 DESIGN OF FLAT SLAB STRUCTURES completed. To get cooperation of the upper portion with the lower portion of the beam, sufficient stirrups are provided throughout the beam. Since no provision is made for resisting negative bending moment, the wall beam should be designed as simple supported, using the distance between faces of the column as the span. The design load should be taken as in previous cases. The reinforcement should be computed for three-quarters of the total effective depth of upper and lower portions of the beam. The fraction is used to provide for the possibility of imperfect cooperation between the two parts of the beam. With this type of design, the forms are often stripped before the upper part.of the beam is built. For this reason, the lower portion of the beam must be provided with re Steel Construction 0.21, +12 Negative Moment OTIC S0 eam aaa Reinforcement. : d. Temperatu Note: Column Bars are Spliced above. Construction Joint. Dowels from column below should extend proper length above Construction Joint. Construction Joint~. Fic. 136.—Concrete Spandrel Built Monolithic with Slab and Columns. (See p. 382.) sufficient reinforcement to carry safely the dead load of the slab without the assistance from the upper part. It is particularly important to provide negative bending moment reinforcement in the lower part of the beam, as shown in Fig. 137. The upper part of the beam should be provided with proper amount of temperature reinforcement. Since the beam is able to contract to some extent, the amount of temperature steel may be smaller than in previous case. The actual amount will depend upon conditions. ‘The authors recommend one-quarter of one per cent for temperature reinforcement. Recesses should be provided in the columns for the upper part of the beam. The joint should be water-proofed by using tar felt or similar material. Sometimes an attempt is made to tie the separately poured upper part of beam to the column by providing in the columns horizontal WALL BEAMS 385 dowels and longer horizontal bars near the top of the beam, and thus provide continuity for the upper part of the beam when built. This is not desirable for construction reasons. It is necessary to drill holes in the forms for reinforcement. These must be located with care so that the hole in one face is just opposite the hole in the other face. After the holes are drilled there is always some difficulty in placing the dowels in the forms. F inally, great difficulty is experienced in stripping the forms. The sides must be moved horizontally to clear the dowels. This movement cannot be con- veniently made on account of stirrups extending up from the slab. For these reasons, this method is not popular on the job and is not advocated. : Concrete Sill Cast Separately -~, §- : "Rd. Recess in Column., Horizontal Construction Joint Roughen Surface and coat with+ neat cement grout before pouring Spandrel. Ita. 187.—Concrete Spandrel Built Separately. (See p. 383.) Instead of providing horizontal dowels in the columns for the wall beams, the Aberthaw Construction Co. of Boston places hori- zontal pipe sleeves in the column forms at the proper height. After the column forms are stripped, steel bars of proper length, extending on both sides of the column, are inserted in the pipe. When imbedded in the wall beam, they serve as negative bending moment rein- forcement. GENERAL REMARKS It is impossible to design flat slab construction properly without a thorough understanding of the action of continuous structures and particularly of continuous flat slabs under varying conditions of loading. Rules and formulas given in preceding paragraphs are mostly for typical panels. In actual construction, odd panels will be found, the designing of which must be left to the judgment. of the designer. Sometimes, whole floors consist of an assortment of oe DESIGN OF FLAT SLAB STRUCTURES odd panels. In all such cases, the designer should be able to judge how the various parts of the floor will deflect when loaded. Follow- ing general formulas blindly may give unsafe results. Another perplexing point is the effect of beams on the action of the adjoining slab. This may be such as to require complete change in the amount and location of the reinforcement. Of importance it is to study the effect of openings on, flat slab action. Tensile Reinforcement Does Not Transfer Loads to Supports but Resists Stresses Due to Bending.—The designer should under- stand the function of reinforcement which is often misunderstood. There is a notion prevalent that bands of steel bars transfer the loads from the slab to the support. This is not the case. The function of the tensile reinforcement is to resist tension due to the bending moments, while the function of transferring the load from one section to another and then to the column is performed by concrete through its resistance to shear. While this is obvious and should not require comment, in order to correct the prevalent wrong notion, full explanation is here given. In the center, where the external shear is zero and therefore 1s no load to be carried, the tensile stresses in a band of bars is a maximum. These stresses decrease gradually until at the point of inflection they become zero. At the point of inflection, the effect of the reinforce- ment is practically negligible. While the stresses in the tensile reinforcement decrease from a maximum to zero, the load to be transferred to the column equal to external shear steadily increases from zero at the center until at the point of inflection, the load to be carried to support amounts to about three-quarters of the total load on the panel.. Thus, the effect of the steel is largest where there is no load to be transferred to the column and is nil at the point of inflection where three-quarters of the load on the panel need to be taken care of. This fact proves that the bands of bars do not perform the function of transferring the load towards the column. If they did, their stresses would increase, instead of decreasing with the increase of external shear. T he function of transferring the load from section to section is performed by shearing stresses in the concrete. In the center, where the load to be carried is zero, the shear is zero. With the increase in load to be transferred, the total shear increases in proportion until it reaches its maximum at the column. GENERAL REMARKS O87 Suspension Action Impossible in Flat Slab Construction.— Another misconception, fairly widespread specially among advocates of the four-way arrangement of steel, is that bands of bars act as sort of suspension bars. Nothing could be farther from the truth. Suspension action is always distinguished by the load being carried to the support entirely by tension stresses in the suspension bars. The stresses in the bars depend upon their sag. The horizontal component of the stresses equals the static bending moment divided by the area of bars and the amount of the sag. The suspension bars are under tensile stresses for their full length, the magnitude of which is such that their horizontal component is constant for the whole length of the suspension bar. A good example of suspension action is furnished by the suspension cables of a suspension bridge. The cables are anchored at each end and sag between the piers. They form the main means of carrying the load. No compression members are existing. The cables are free to elongate for their full length. No such action takes place in flat slab construction. The stresses. in each band of bars are a maximum in the center, then they decrease until they are zero at the point of inflection. Near the column, the stresses increase from zero at point of inflection to a maximum at the support. There is no continuous transference of tensile stresses as in suspension cable. The bands of bars may be continuous, but their action is not continuous. As a matter of fact, as far as tensile | stresses are concerned, the bars in flat slab could be cut at the point of inflection without affecting the construction, because the tensile stress there is zero. If a cable of a suspension bridge was cut, the bridge would collapse. A further difference is that for slabs of uniform thickness tke sum of tensile stresses at the support is much larger than the sum of tensile stresses in the center. In suspension structures, on the other hand, the horizontal component of the tensile stresses is constant at all sections. If bands of bars acted as suspension bars, the strength of the slab _ would be only a fraction of the strength flat slabs actually have. Flat slabs action is fully explained on page 328. Stresses Cannot Act without Corresponding Deformation.—It is of particular importance to remember also that stresses cannot exist without corresponding deformation of the materials and deflec- tion of the member. Neither can deformations of materials (other than shortening or lengthening of members free to move, due to 388 DESIGN OF FLAT SLAB STRUCTURES temperature) and deflection of members exist without stresses. Reinforcement cannot be effective in resisting tension when it is placed in such a way that it is not able to elongate. This is of importance in many Cases in connection with flat slab design, particularly in connection with the design of beams in flat slab construction. In Fig. 188, page 388, for instance, the bottom of the beam is some distance below the bottom of the slab. Some building codes permit the beam to be designed only strong enough to carry its own weight and the load directly superimposed upon it. The slab is assumed to act as if there was no beam, and the slab load is assumed as being carried from column to column by the strip of slab along the beam. ‘The tensile stresses in the slab are supposed to be resisted by the half-band of bars parallel to the ikial Neutral ~~ Stress in Slab reinforcement ‘near beam _-d7 Stress in beam reinforcement’ Fic. 138.—Action of Beam and Slab Reinforcement in Flat Slab Construction. (See p. 388.) beam. It should be clear that no such action can take place. The slab and the beam must deflect as a unit. The position of the neutral axis for both the beam and the slab is governed by the beam. The slab reinforcement will be near the neutral axis so that for quite a distance from the beam the amount of tensile stresses resisted by slab steel will be negligible. Before the slab bars can come into full action, the beam would have to fail. The design has sufficient amount of reinforcement, but it is placed in such a way that it cannot perform its function. The proper method of design- ing is to make the beam strong enough to carry not only the brick- work but in addition 20 per cent of the slab load. (See p. 379.) PROBLEMS IN DESIGNING FLAT SLAB CONSTRUCTION 389 PROBLEMS IN DESIGNING FLAT SLAB CONSTRUCTION Method of Designing.—After the spacing of columns has been decided upon, the design of flat slab floors or other plane structures may proceed as follows: 1. Assume dead load of slab based on assumption of its thickness. This weight added to the live load and to the weight of floor finish gives the total loading for the flat slab. 2. Select size of column head and dimensions of drop panel. (See p. ve 3. Determine restraint at wall column. (See p. 313.) 4. Compute required thickness of slab and drop panel, using formulas, pp. 8 to 28. If slab of uniform thickness throughout the floor is desired, the thickness should be computed for panels for which the expected bending moments are largest. 5. Compute shearing stresses at the column head and at the drop panel. (See p. 347.) 6. Compute bending moments Ef the various design sections for interior and exterior panels. (See pp. 332 to 335.) 7. Compute required areas of steel at the various design sections. (See p. 351.) 8. Select the flat slab system to be used and the desired arrange- ment of steel. (See pp. 360 to 369.) Select bars which will supply the required amount of positive and negative reinforcement in each strip. Decide upon the manner of bending the bars. Compute the areas of bent bars effective as negative moment reinforcement, and compare this area of reinforcement with the required area. If the bent bars are not sufficient, add short bars. 9. Check bond stresses at the column head. (See p. 351.) 10. Show on the plan, by sketches details of reinforcement for typical panels. Give a designating mark to each band of bars or reinforcing unit, and label the floor plan with these marks. In a schedule of reinforcement, give the number, length, and type of bars in each band or unit. Indicate clearly the length of the bars com- — posing each band or unit, and show bending sketches for bent bars. If the typical panel is to be used as a general guide for panels of different sizes, express the lengths of bars and the points of bending in terms of the span. 11. Show clearly the concrete dimensions, on the floor plan and in the schedule. 390 DESIGN OF FLAT SLAB STRUCTURES 12. Design wall beams. (See p. 379.) 13. Check ali columns supporting the flat slab, for strength and rigidity. Supply bending moment reinforcement in the exterior columns when required. (See pp. 305 to 319.) EXAMPLE OF FLAT SLAB DESIGN The use of the formulas and recommendations will be illustrated by the following example. Example 2.—Design an interior and exterior panel of a floor in a flat slab building. The dimensions of the panels are 20 by 22 ft., as shown in Fig. 139, p. 391. The design live load is 150 lb. per sq. ft. The floor finish consists of 13 in. of bonded granolithic finish (i.e., finish applied separately). Use following stresses in lb. per sq. in., fe = 800, fs = 16000, u = 100, v = 60, and ratio n = 15. Solution.—The method of solving the problem, as outlined on p. 389, will be followed. The reference numbers used in the example correspond to the numbers used there. 1. Assuming thickness of slab as 8 in., the design load is Live load ces): ce 6 6 Pe ee 150 lb. Slab Iodd ti eee ee ee eee 100 lb. 12 inifinish....90 eh eee 18 lb. Total . ota tae See aes w = 268 lb. per sq. ft. 29. The diameter of column head is made equal to 0.225 times the average 22 + 20 + x 0.225 = 4.72 ft. Assume 4 ft. 9 in. for the span. Therefore, c = diameter of column head. A rectangular drop panel will be used, with length of sides equal to 0.35 of the sides of the rectangular panel. The dimensions of the drop panel are 7 ft. by 7 ft. 6 in., with the long side parallel to the long side of the panel. At the wall column, a square bracket will be used, with a distance from center of column to edge, c: = 2 ft. 5in. Half drop panel will be used at the wall. 8 The slab is supported by concrete columns having the required rigidity. Concrete spandrels are also used. Therefore, the exterior panel may be classed as Case 1 (see p. 338). 4. It is desired to use a slab of uniform thickness throughout, and drop panels of the same depth. Therefore, the required thickness will be computed for exterior panels, which in this case require the maximum dimensions. Thickness of Slab and Drop Panel. (Use Formula (18), p. 339.) 4.75 The dimensions arel = 22 ft.,b = 7 ft., w = 268 lb. per sq. ft. For 5 Sey = ().22 and for specified stresses from diagram, p. 911, the constant C; = 0.0173. 1, = 0.0173 X 22 X V268 X 22 41.5 = 10.50 41.5 = 12 in. EXAMPLE OF FLAT SLAB DESIGN ool Thickness of Slab in the Center. (Use Formula (25), p. 340.) Using dimensions given above, and since for : = (0.22 and for specified stresses from table, p. 911, Cs = 0.0198, i = 0.0198 x 22V268 +1 =7.141=8.1. Use 8 in. slab. ak Be eerie 2 99'.0% ------- -—--- --»}e- --= ---- - + --- 22+. 0% ---------- SS | Column |. Middle strip___,| Twocolumn sirips..|_ _Midale strip ___,| Two column strips 38 strip short direction short direction short direction short direction at . t ES | Cx 2 ee = n Come Q <= H ‘SI i £3 Cy 2|.S eee = a Sirs 368 | ua - = E Sa S | : E Be | = : Sed S& 5 Ojon CT : 1D 2& a coo DAL P Ly ° § I oS = Ser oeeet SiS | & HOGI Pal ass oa Saat Ys oe | 3 S £& Se ae rial g ES SS Ea os eee ee Se. AU ae ee ee tl Srp a Be H See —th wes Ss TT a fois TH LS hae ad Dl Sangl Bo se] Bi | he Cine Sr a resell) ae et a Lae as nN ai) rH rarer eat as cs ee = = t | seuss A 1H eee aSSumaae 5 ae 4 aes ee re — fe c | Seeceiaes in SS | ae aoe! oS soneSauee = grey £21 0 Gem Ee; Se al i Tiara Er] VS fs eas s ieee Ht s\® Sat LH the samen a sus a : g E 3 S5n= & : 8 aa Bent Bars { ! ; Hooked Bars Band Cy ve Str Bars Band Co y 8 BandA&B l qs Sa ee k eASNOPIA OF 1 eraampre eS Gane? Nore. in plan all bars are shown only in bands Ay, Ar, 8 and Co - In other bands positions of outside bars only is indicated . Fic. 139.—Details of Flat Slab Design. (See p. 390.) 5. The points of critical shear are distant t:; — 13 = 12 — 14 = 10.5 from column head. The diameter of critical section 1s 10.5 e+ 2(t; — 14) = AER =i bite ero aD. The circumference is 78 X 3.14 = 245 in. The net area of panel outside the critical section is found as follows: (GN TEES a ee nr ree oe 20 <. 22:= 440 8a, ft. 676? Area inside of critical section, Te xX 3.14 = 33.3 Dem OU cee, 5s hi Ea Ma « = aw hee RG 406.7 sq. ft. 392 DESIGN OF FLAT SLAB STRUCTURES Total shear at critical section equal to net area times unit load, V = 406.7 X 268 = 109 000 lb. The weight of drop panel is neglected. The shearing stresses are, 109 000 48 5 Ib ¢ y = ———— _ = 48. . per sq. in. 245 X i X 10.5 Sha The same result may be obtained by using Formula (39), p. 349. Since 20 | = 22 ft, = 20, m = 5 = 0.91, o = 6.5 ft, - = 0.296, from table, p. 913, the constant Cy». = 0.85. Therefore l 22 : y= Ciw, = 0.85 X 268 X rer 47.7 lb. per sq. in. Since the computed stress is less than the allowable stress of 60 Ib. per sq. in.,: the thickness is satisfactory. 6. Bending Moment at Various Design Sections.— Interior Panel, W = 268 X 20 X 22 = 118 000 Ib. 4.75 : Long Direction, 1 = 22'ft., l = 20 ft., ; = = 0.22. M =1.08 X 118 000 x 22(1 — 2 X 0.22)? = 2 040 000 in.-lb. Negative Cae = — 0.54M = — 1100000 in-lb Moment \ M, = — 0.08M = —_ 164000 in.-lb. Positive 9, 3;= 0.23M = 472 000 in.-lb. Moment |Mi= 0.15M = 306 000 in.-lb. Ret d c’ Aa Short Direction, | = 20 ft., 11 = 22 ft., 7 Pog = 0.24. M = 1.08 X 118000 x 20(1 — 2 X 0.24)? = 1800 000 in.-lb. Negative (M, = —0.54M = — 972000 in--lb. | Moment \M, = —0.08M = — 144000 in.-lb. Positive a = 0.23M = 414 000 in.-lb. Moment |M,= 0.15M = 272 000 in.-lb. Exterior Panel, W same as for interior panel. Long Direction, M = same as for interior panel. First interior column line, Negative ie = — 0.62M = — 1 265 000 in.-ib. Moment | M. = —0.12M = — 245000 in.-lb. EXAMPLE OF FLAT SLAB DESIGN 393 Wail column line, Negative {Mi = —0.42M = — 857 000 in.-lb. Moment | M, = —0.08M = — 163 000 in.-lb. Positive en i 0.27M = 551 000 in.-lb. Moment | M, = 0.18M = 367 000 in.-lb. Short Direction. Bending moments same as for interior panel because the slab is continuous in this direction. 7. Required Areas of Reinforcement.—The required areas of reinforcement are obtained from formula As = For fs = 16000, 7f; = 14000. The jaf s values of effective depth, d, are as follows: At the column head, d = 12 — 1.5 = 10.5 in.; in the center of column strips, d = 8 — 1 = 7in.; and in the center of middle strips where two layers of steel are used, d = 8 — 1.25 = 6.75 in. Required Effective Required Number Location Bending Moment Area of and Diam- Depth Steel eter of Round Bars Interior Panel: Long Direction: Column strips M, = — 1100000 10.5 7.5 sq. in. 26—3"’ Middle strips M,= — 164000 7.0 12 7s, 1in. 6— 3” Column strips M; = 472 000 7.0 4.8 sq. in 16—3” Middle strips M,= 306 000 G2(D 88 p260s in 11— 3” Short Direction: Column strips M, = — 972000 10.5 .7 sq. in. 22—3/" Middle strips M,= — _ 144000 i200 .5-Sq. In. 5— 3" Column strips M;= 414 000 7.0 4.3 sq. in 14—3” Middle strips M,= 272 000 6.70 | 2.9-8q. in 10—3” Exterior Panel: Long Direction: First interior Pa a 1 265 000 10.5 8.7 sq. in. 29—3" Column line M, = 245 000 7.0 2.5 sq. in. g—3”’ Wall column -. = 857 000 | 10.5 5.9 sq. in. 20—3”" Column line M, = 163 000 7.0 1.7 sq. in. 6—3"" Column strip (M; = 551 000 aa 5.7 sq. in. 19— 3” Middle strip bps = 367 000 6.75 | 3.9sq.in. | 13—” ————— ae, 394 DESIGN OF FLAT SLAB STRUCTURES Reinforcement in short direction of exterior panel same as in short direction of interior panel. 8. The two-way system, with the arrangement shown in the first method, will be selected (p. 364). (The selection of this system is made for the purpose of illustration only and is not intended as a recommendation.) Column Strips.—In the column strip, two bars out of three will be bent up at both ends and carried across the column into the Udjoining span. In the long direction of the interior panel, where sixteen 3-in. rd. bars are used for positive bending moment, 10 bars will be bent up and 6 will be carried straight. The available reinforcement at the column will consist of 10 bent up bars plus 10 bars extended from the adjoining span, making 20 bars in all. Since 26 bars are required, 6 short bars must be added. In the short direction, 9 bars out of 14 will be bent up, making 18 bars available at the column. Since 22 bars are required, 4 short bars must be added to complete the required reinforcement. In the exterior panel, 12 bars will be bent up and 7 bars carried straight. The available reinforcement at the column head consists of 12 bent bars plus 10 bars extended from the interior panel, or 22 bars in all. Since 29 bars are required, 7 straight bars must be added. At the wall column, 20 bars are required and only 12 available. Therefore, 8 short bars, provided with a hook at the wall end, must be added. In the short direction, the same arrangement of steel should be used as in the interior panel. Middle Strips.—In the middle strip, one bar in three will be bent up at both ends and extended into the adjoining panel. The amount of negative reinforce- ment will then be equal to two-thirds of the amount of the positive reinforcement. 9. Bond stresses will be computed at the column head of interior panels, according to formula given on p. 351. V =0.35W = 0.35 X 118 000 = 41 300 lb. The perimeter of one §-in. rd. bar is 1.96 in., therefore the perimeter of all bars effective at the column is Zo = 1.96 X 22 = 48.2. 41 200 A382 KE IONS This is somewhat higher than the bond stress allowed for deformed. bars. To reduce the bond stress to 100 Ib., the number of bars would have to be raulti- plied by 194, or bars of smaller diameter used. 10. The details of reinforcement and the length of bars are shown in Fig. 139, p. 391. 11. Concrete dimensions are given in Fig. 139, p. 391. 12. The gross span of the wall beam is 20 ft. With wall columns 3 ft. wide, the net span becomes 17 ft. In this design, the wall beam will be placed under the slab. The load on the beam equals: U = 104 lb. per sq. in. Load from slab. 2, ...0.2 22 & 268 = 170 paper Spatid?tel jc.) sane 3 125 = 375 lb. per lin. ft. Glass "and Sash ise oe 7 eee aa 20 Ib. per lin. ft. Weight of beam below slab (assumed) 150 Ib. per lin. ft. 1 715 lb. per lin. ft. EXAMPLE OF FLAT SLAB DESIGN 395 The external shear equals 17.0 EE te ieee = 14 700 lb. The bending moment is computed as recommended on p. 379. Negative bending moment, M = 1 725 X 17.0? = 499 000 in.-!b. Positive bending moment, M = 0.75 X 1725 X 17.02 = 374 000 in.-lb. The smallest cross section for the beam, as determined by allowable shear, is 14 500 C0g Se 138.8. In. z X 120 i The depth of the beam is usually governed by the allowable distance from the top of the window to the bottom of the slab. It is usually advisable to make the beam as shallow as possible, so as to bring the window close to the under side of the slab. In this case, assume depth of beam h = 18 in. andd = 18 — 2 = 16. Minimum Width of Stem Governed by Shear.—Since, as computed before, minimum bd = 138, with d = 16, b = 438 = 8.7 in. It is desirable to use a larger value, b = 12, to get larger resistance to torsion. Areas of Steel.—Since d = 16 in., f, = 16000 lb. per sq. in., the areas of steel required by the previously computed bending moment are ' At the column, Pres. 405,0007) 7) ae 1 16.000. ak In the center, 372 000 re 1.7 5q. in. errata Cee Use four 3-in. rd. bars for positive bending moment reinforcement, the area of which is 4 X 0.44 = 1.76 sq. in. Bend two bars and extend them over the support. The available reinforcement at the column, then, equals 1.76 sq. in. Since the required area there is 2.2 sq. in., add one 3-in. rd. bar. The total area then is 1.76 + 0.44 = 2.2 sq. in. The straight bars will be extended into the column and will be available as compression reinforcement, with an area equal to 2x 0.44 = 0.88. Compression Stresses at the Support.—The bending moment at the support is 495000 in.-lb. The dimensions are 6 = 12 in., d=16in. The area of 252 tensile reinforcement is Asi = 2.2 sq. in., pi = Cen = 0.0115. The area ' ; 0.88 of compression reinforcement is A’; = 0.88 sq. in., p’ = eas = 0.0046. Since, with f, = 900 and f,; = 16 000, for a balanced design the ratio of rein- forcement is p = 0.0129, which is larger than the ratio used, the compression stresses are smaller than allowable and the design is satisfactory. Diagonal Tension Reinforcement.—Diagonal tension reinforcement may be computed as in ordinary design. However, the authors advocate the use of a larger number of stirrups near the ends of the beam than required by diagonal tension, because there they may assist the beam in resisting torsional stresses. 396 DESIGN OF FLAT SLAB STRUCTURES Joint Committee, 1924, Flat Slab Specifications.—The Joint Committee, 1924, flat slab specifications are reprinted as appendix in Vol. II of this treatise. CHIGAGO AND NEW YORK CITY FLAT SLAB REGULATIONS (In effect 1925) The regulations of the two cities are given below. Definitions. Chicago Code: ‘“ Flat slabs as understood by this ruling are reinforced concrete slabs, supported directly on reinforced columns with or without column heads at the top, the whole construction being hingeless and monolithic without any visible beams or girders. The construction may be such as to admit the use of hollow panels in the ceiling or smooth ceiling with depressed panels in the floor.” New York Code: ‘The rules governing the design of reinforced concrete flat slabs shall apply to such floors and roofs, consisting of three or more rows of slabs, without beams or girders, supported on columns, the construction being continuous over the columns and forming with them a monolithic structure. “Por structures having a width of less than three rows of slabs or in which irregular or special panels are used and for which the rules given below do not apply, the computations and the analysis shall, when so required, be filed with the superintendent of buildings.”’ Notation. ) 1 = distance center to center of columns of the side of square panel, or average distance of the long and short sides of rectangular panel; w = dead and live unit load, lb. per sq. ft.; W = total dead and live load on panel in |b. In New York the dead load should include weight of drop panel; W, = =total live load on panel under consideration. COLUMNS General Limitations for Interior and Exterior Columns. Chicago Code: ‘ The least dimension of any concrete column shall be not less than one-twelfth the panel length, nor one-twelfth the clear height of the column.” . New York Code: “ For columns supporting reinforced concrete flat slabs, the least dimension of any column shall be not less than one-fifteenth of the average span of any slabs supported by the columns; but in no case shall such least dimension of any interior column supporting a floor or roof be less than sixteen inches when round nor fourteen inches when square; nor shall the least dimension of any exterior column be less than fourteen inches.”’ Interior Columns. Chicago Code: ‘‘ The interior columns must be analyzed for the worst condi- tion of unbalanced loading. It is the intention of.this ruling to cover ordinary eases of eccentric loads on the columns by the above requirement. Where the | minimum size of column therein specified is found insufficient, however, the effect of the resulting bending moment shall be properly divided between the CHICAGO AND NEW YORK CITY FLAT SLAB REGULATION 397 adjoining slab and the columns above and below according to best principles of mechanics and the columns enlarged sufficiently to carry the load safely.” New York Code: “ Interior columns shall be designed for the bending moments developed by unequally loaded panels, eccentric loading or uneven spacing of columns. The bending moment resulting from unequally loaded panels shall be considered as 1/40W ,L, and shall be resisted by the columns immediately above and below the floor line under consideration in direct proportion to the values of their ratios of I/h. “Roof columns shall be designed to resist the total moment resulting from unequally loaded panels.” Wall Columns. Chicago Code: ‘‘ Wail columns in skeleton construction shall be designed to ~ . : WL WL resist a bending moment of "60. at floors and ay at roof. The amount of steel required for this moment shall be independent of that required to carry the direct load. It shall be placed as near the surfaces of the column as practicable on the tension sides, and the rods shall be continuous in crossing from one side to another. The length of rods below the base of the capital and above the floor line shall be sufficient to develop their strength through bond, but not - less than forty diameters, nor less than one-third the clear height between the floor line and the base of the column capital.’ (See Fig. 109, p. 318.) New York Code: ‘“ Wall columns shall be designed to resist bending in the same manner as interior columns, except that W shall be substituted for Wi in the formula for the moment. The moment so computed may be reduced by the counter moment of the weight of the structure which projects beyond the center line of the wall columns.”’ Slab Thickness.—The thickness of slab must not be smaller than 5/1 for floors and 751 for roofs. Absolute minimum is 6 in., except that New York allows 5 in. for roofs. Formulas for determining thickness of slab for the loading: Chicago f= VW for all cases New York ¢ = 0.024IV'w + 13 for slab without drop panels t = 0.021V w + 1 for slabs with drop panels. Effective Depth of Slab.—The effective depth of slab is the distance from the center of gravity of the effective reinforcement to the compression face of the slab. A protective covering for the reinforcement equal to 1 in., measured from outside face of the lowest layer of steel, shall be provided. Column Heads.—Effective diameter of column: head is as defined and de- scribed on page 319. Its diameter must not be less than 0.2251 for both codes. Drop Panels.—Drop panels are as defined and described on page 322. Thick- ness and width of the drop panel must be sufficient to resist shearing stresses at the column head and at the edge of drop panel and compression stresses at the column head. In square panels minimum length of drop must not be smaller than 3l. In rectangular panels New York Code permits square drop panels with length of side equal to } average span; while Chicago requires a rectangular drop with length in each direction equal to 4 the span in that direction. 398 DESIGN OF FLAT SLAB STRUCTURES At wall panels drop panels should extend at right angles to the wall from center of the column a distance equal to ¢l. Shearing Stresses.—The allowable shearing unit stresses for 2 000 lb. con- crete shall not exceed following values: At the perimeter of the column head 120 lb. per sq. in. and at the perimeter of the drop panel 60 lb. per sq. in. The V shearing stresses should be computed according to formula v = bid except that H V according to New York Code at the column head formula v = bd may be used. BENDING MOMENT COEFFICIENTS Design Strips and Bending Moment Sections.—The slab is divided into design strips as explained on page 330 and the sections of critical bending moment correspond to sections shown in Figs. 118 and 119. Chicago Code. Square or Nearly Square Interior Panels.—This rule applies to panels in which length does not exceed breadth by more than 5 per cent. oS ee a SS ES | Two-Way » Four-Way System System Column Strips Positive Bending Moment eoWl soWl Negative Bending Moment soWl soWl Middle Strip Positive Bending Moment zaaWl * aso Wl Negative Bending Moment za0Wl za0Wl * The reinforcement required by this moment is the area of steel in each of the two diagonal bands, and not the effective area at this section. The effective area of both diagonal bands, as customarily computed, is 42 per cent larger than the area required by this moment. Any other system of reinforcement in which the reinforcing bars are placed in circular, concentric rings and radial bars, or systems with steel rods arranged in any manner, whatsoever, shall comply with the requirements of either the two-way or the four-way system. Chicago Code: Wall Panels. Wall Panels in Skeleton Construction.—In wall panels supported by wall columns and wall beams, in strips at right angles to the wall increase positive bending moment by 25 per cent. . Wall Panels Supported on New Brick Walls.—Where wall panels are carried on new brick walls, thése shall be laid in Portland cement mortar and shall be stiffened with pilasters as follows: If a 16-inch wall is used, it shall have a 4-inch pilaster. If a 12-inch wall is used, it shall have an 8-inch pilaster. The length of pilasters shall be not less than the diameter of the column, nor less than one- eighth of the distance between pilasters. The pilasters shall be located opposite the columns as nearly as practicable, and shall be corbeled out 4 inches at the top, starting at the level of the base of the column capital. Not less than 8-inch bearing shall be provided for the slab, the full length of wall. The coefficients of bending moments required for these panels shall be the same as those for the interior panels except as: The positive bending moments at right angles to the wall shall be increased 50 per cent. CHICAGO AND NEW YORK CITY FLAT SLAB REGULATION 399 Wall Panels Supported on Old Brick Walls.—Where wall panels are supported on old brick walls there shall be columns with standard drops and capitals built against the wall which shall be tied to the same in an approved manner, and at least an 8-inch bearing provided for the slab, the full length. The coefficients should be same as specified for slabs on new brick walls. Where it is impossible to utilize the old brick wall in the manner described above, wall columns and wall beams should be provided as in skeleton construction and the panel treated as provided for skeleton construction. Chicago Code: Rectangular Panels. This applies to rectangular panels where length of panel exceeds breadth by more than 5 per cent but not more than 33.3 per cent. In columns strips of two- and four-way system use in each direction same bending moment as would be required by a square panel whose span length is equal to the length of the respective side of the rectangle. The amount of steel in the short direction, however, shall not be less than two-thirds of that required in the long direction. In the four-way system the amount of steel in the middle strip, positive and negative, shall be the same as that required for similar strip in a square panel whose length is equal to the mean of the long and the short side of the rectangular, panel. In the two-way system the amount of steel in the middle strip, positive and negative, running in short direction, shall be equal to that required for the same strip in a square panel whose length equals the long side of the rectangular panel. The amount of steel in the middle strip, long direction, positive and negative, . shall be equal to that required for the same strip in a square panel, whose length equals the short side of the rectangular panel. In no case shall the amount of steel in long direction be less than two-thirds of that in the short direction. New York Code: Square or Nearly Square Panels. Two-Way Four-Way System System Slab with Drop Panel Column Strips Positive Bending Moment gyW! Wl 100 Negative Bending Moment gyWl 35Wl Middle Strips Positive Bending Moment Wl TW! Negative Bending Moment zisW! zeal Slab without Drop Panel Column Strips Positive Bending Moment gsWl aeWl Negative Bending Moment 3gWl sgWl Middle Strips Positive Bending Moment zasWl Too! 133 0 Negative Bending Moment za5Wl za5WIl 400 DESIGN OF FLAT SLAB STRUCTURES New York Code: Rectangular Panel. Where the ratio of long span to short span is less than 1.1 the panel may be considered as square panel with panel length equal to average panel length. When the ratio of panel length is between 1.1 and 1.33 the bending moment coefficients specified for interior square panels shall be applied in the following manner: In two-way systems the negative moments and the positive moment at right angles to the long direction shall be determined ‘as for a square panel of a length equal to the greater dimension of the rectangular panel; and the corresponding moments on the sections at right angles to the short direction shall be determined as for a square panel of a length equal to the lesser dimension of the rectangular panel. In both cases the load W shall be taken as the load on the rectangular panel under consideration. ‘The amount of reinforcement in the short direction shall be not less than two-thirds of that in the long direction. In four-way systems, for the rectangular bands, the negative moment on the column head sections and the positive moment on the outer sections ahall be determined in the same manner as indicated for two-way systems. For the diagonal bands, the negative moments on the column head and midsections and the positive moment on the inner section shall be determined as for a square panel of a length equal to the average span of the rectangle. The load W shall be taken as the load on the rectangular panel under consid- eration. New York Code: Wall Panels. The negative moments at the first interior row of columns and the positive moments at the center of the exterior panels on moment sections parallel to the wall shall be increased 20 per cent over those specified above for interior panels. The negative moment-on-moment sections at the wall and parallel thereto shall be determined by the conditions of restraint, but the negative moment on the midsection shall never be considered less than 50 per cent and the negative mo- ment on the column head section never less than 80 per cent of the corresponding moments at the first interior row of columns. Panels without Drop Panel or Capitals or Both. Chicago Code: In square’ panels where no column capital or no depressions are used the sum total of positive and negative bending moments shall be equal to that computed by the following formula: Wl Cc c\3 M = —|]1.538 —4-4+ 4.18{- 8 | ie s(() | where M = numerical sum of positive and negative bending moments, regardless of algebraic signs; W = total live and dead load on the whole panel; 1 = length of side of a square panel, c. to c. of columns; c : : E es ratio of the radius of the column or column capital to panel length, /. This total bending moment shall be divided between the positive and the negative moments in the same proportion as in the typical square panels for ’ CHICAGO AND NEW YORK CITY FLAT SLAB REGULATION 401 two-way or four-way systems specified above for interior and wall panels, re- spectively. New York Code: The general rules apply to flat slabs with or without drop panels. Any difference in treatment of the two types of design is indicated in each case. No provision is made for flat slabs without column heads. The code states that for structures in which column capitals are omitted, special computations should be filed with the Superintendent of Buildings. Line of Imflection.—Both codes specify that in the design of reinforced con- crete flat slab construction, for the purpose of making calculations of the bending moments at sections other than the critical section, the line of inflection shall be considered as being located one-quarter the distance, center to center, of columns, rectangularly and diagonally, from center of columns for panels without drops. The New York Code further specifies that for panels without drops a fraction one-quarter should be changed to three-tenths. REINFORCEMENT Tension Reinforcement.—The required tension reinforcement should be M computed by formula A, = fj’ The effective depth d is the distance of the 3J center of gravity of all effective bars in the section under consideration from -the compression face of slab. The rules regarding the effective reinforcement are same as given on page 354, except that in the Chicago Code the area of positive reinforcement in the middle strips, as determined from the specified bending, is the area of bars in each diagonal band and not the effective area on the moment section in the strip. The effective area equals the area in both diagonal bands multiplied by the sine of the angle of inclination. This should be kept in mind when comparing the total amount of effective steel as required by the Chicago Code in four-way and two-way systems as well as in comparing with that required by other codes. ALLOWABLE STRESSES Allowable Stresses in Steel.—The allowable stress in steel in New York Code is f, = 16 000 lb. per sq. in. irrespective of the character of steel used. In Chicago Code the allowable stress for medium steel is f; = 18 000 lb. per sq. in. It should be noted that to allow for this difference in allowable steel stresses, the bending moment coefficients in the New York Code were made proportionally smaller, so that the areas of reinforcement as computed according to New York Code with f, = 16 000 are practically the same as obtained by using Chicago coefficients and stresses f; = 18 000. Compression Stresses.—Compression stresses in any section of the slab may be computed by ordinary beam formulas using the specified bending moment for the section under consideration and for width of beam, the width of the section. | In computing the compression stresses at the column head in slab with drop panels the width of beam in the formula should be taken as equal to the width of the drop panel. Chicago Code provides that if compression area at the column head is insuf- 402 DESIGN OF FLAT SLAB STRUCTURES ficient, it may be increased by introducing compression steel in the bottom of slab. The allowable unit stresses should not exceed the stresses allowed in beam design. At the column head same increase in stresses is allowed as in beams at the support. (Namely, 750 lb. per sq. in. for 2000 lb.concrete. ) Special Provisions Regarding Flat Slab Reinforcement. Splices in bars may be made wherever convenient, but preferably at points of minimum stress. The length of splice beyond the center point, in each direc- tion, shall not be less than forty diameters of the bars, nor less than two feet. The splicing of adjacent bars shall be avoided as far as possible. New York Code has this additional provision: ‘“ When the reinforcement is arranged in bands at least 50 per cent of the bars in any band shall be of a length not less than the distance center to center of columns measured rectangularly and diagonally; no bars used as positive reinforcement shall be of a length less han half the panel length plus forty bar diameters for cross bands, or less than seven-tenths of the panel length plus forty bar diameters for diagonal bands, and no bars used as negative reinforcement shall be of a length less than half the panel length. All reinforcement framing perpendicular to the wall in exterior panels shall extend to the outer edge of the panel and shall be hooked or other- wise anchored. ‘““ Adequate means shall be provided for properly maintaining all slab rein- . forcement in the position assumed by the computations.” Chicago Code provides as follows: . ‘‘ In order that the slab bars shall be main- tained in the position shown in the design during the work of pouring the slab, spacers and supports shall be provided satisfactory to the Commissioner of Buildings. All bars shall be secured in place at intersections by wire or other metal fastenings. In no case shall the spacing of the bars exceed nine inches. The steel to resist the negative moment in each middle strip shall extend one- quarter of the panel length beyond the center line of the columns in both directions. “Slab bars which are lapped over the column, the sectional area of both being included in the calculations for negative moment, shall extend not less than twenty-five one-hundredths of the panel length for cross-bands, and thirty- five one-hundredths of the panel length for diagonal bands, beyond the column center.” WALLS AND OPENINGS Chicago Code provides: ‘‘ Girders and beams shall be constructed under walls, around openings and to carry concentrated loads. ‘The spandrel beams or girders shall, in addition to their own weight and the weight of the spandrel wall, be assumed to carry 20 per cent of the wall panel load uniformly distributed upon them.” New York Code states: “In the design and construction of reinforced con- crete flat slabs, additional slab thickness, girders or beams shall be provided to carry any walls or concentrated loads in addition to the specified uniform live and dead loads. Such girders or beams shall be assumed to carry 20 per cent of the total live and dead panel load in addition to the wall load. Beams shall also be provided in case openings in the floor reduce the working strength of the slab below the prescribed carrying capacity.” CHAPTER VII CONCRETE AND REINFORCED CONCRETE COLUMNS This chapter gives formulas for plain concrete columns (p. 403), columns with longitudinal bars (p. 405), columns with spiral rein- forcement (p. 419), and columns reinforced with structural steel. In each case the authors’ recommendations for design, as well as the requirements of various building departments, are given. Details of design for columns with longitudinal steel (p. 412), for spiral columns (p. 432), and for columns with structural steel (p. 440) are presented and thoroughly discussed. | Particular attention is called to the section entitled ‘‘ Economies in Column Design,” p. 445. In this section, factors affecting the economy of reinforced concrete columns are discussed, and recom- mendations are given as to the most economical mix of concrete and the most economical proportions of steel to concrete. Reduction of live load to be used in column design is given on p. 452. Recommendations for columns subjected to bending are given on p. 458. The bending moments in columns carrying crane loads are given on p. 464. PLAIN CONCRETE COLUMNS AND PIERS Columns or piers, the unsupported length of which does not exceed four times the least lateral dimension, may be built of concrete without reinforcement. The load to be carried by the columns should preferably be centrally applied. If the load is eccentric, for rectangular or square columns the eccentricity must be less than one- sixth of the dimension of the column in the direction of the eccen- tricity. For larger eccentricity, the column must be reinforced. For central loads, the required area equals the total load divided by the allowable unit stress. This and other requirements may be expressed by the formulas below. 403 404. CONCRETE AND REINFORCED CONCRETE COLUMNS | Let P = total central load in Ib.; A = effective area of cross section of column in sq. 1n.; f, = compressive unit stress in concrete in lb. per sq. in. Then Total Load, P = Af,: seat 2 Required Effective Area, en Stress in Concrete, BP ey Ge In the above formulas the same units of area must be used for A as for fe. In the notation, it was assumed that the area, A, is in sq. in. and the stress, fe, in lb. per sq. in. Therefore, the load, P, is in Ib. If the area, A, is in sq. ft., and the load, P, in lb., the stress, fe, will be in lb. per sq. fis: Effective area is the area within fireproofing as discussed on p. 272. For piers imbedded in earth, such as pedestals above the footings, the total area may be considered as effective, for plain columns not exceeding in height four times the least lateral dimen- sion, as in such cases no fireproofing 1s required. Allowable unit stresses are same as given in table on p. 407. Formulas for eccentrically loaded columns or piers are given on Dory: REINFORCED CONCRETE COLUMNS Reinforced concrete columns may be divided according to the method of reinforcement into: Columns with vertical bars only; Columns with vertical bars and continuous closely spaced spiral; Columns with structural steel core; Columns with cast-iron core. The columns with structural steel and cast-iron cores may also be provided with spiral reinforcement. | Unsupported Length of Column.—The unsupported length of reinforced concrete columns shall be taken as: (a) In flat slab construction, the clear distance between the floor and under side of the capital; 1In European practice, P is in kilograms, f; in kilograms per square centi- meter, and, consequently, A in square centimeters. COLUMNS WITH LONGITUDINAL BARS 405 (b) In beam-and-slab construction, the clear distance between the floor and the under side of the shallowest beam framing into the column at the next higher floor level; | (c) In floor construction with beams in one direction only, the clear distance between floor slabs; (d) In columns supported laterally by struts or beams only, the clear distance between consecutive pairs (or groups) of struts or beams, provided that to be considered an adequate support, two such struts or beams shall meet the column at approximately the same level and the angle between the two planes formed by the axis of the column and the axis of each strut, respectively, is not less than 75 degrees nor more than 105 degrees. | When haunches are used at the junction of beams or struts with columns, the clear distance between supports may be considered as reduced by two-thirds of the depth of the haunch. The above recommendations agree with Joint Committee, 1924, Specifica- tions. : Longitudinal Bar COLUMNS WITH LONGITUDINAL BARS A typical design of a column with vertical steel only is shown in Fig. 140. The formulas given below have - been universally accepted for columns with longitudinal reinforcement. The Fic. 140.—Typical Design of various specifications and Building Column with Vertical Steel only. Codes differ only in the allowable unit erga hy | stresses and the allowable maximum and minimum amount of steel (see table on p. 409). Let P = total column load in lb.; A = effective area of cross section of column in sq. in.; A, = area of cross section of vertical steel in sq. 1n.; p =ratio of area of vertical steel to area of concrete =A,+A; f. = compressive unit stress in concrete, lb. per sq. in.; compressive unit stress in steel, lb. per sq. in.; f = average compressive unit stress, lb. per sq. in.; n = ratio of modulus of elasticity of steel to that of concrete. — | 406 CONCRETE AND REINFORCED CONCRETE COLUMNS General Formulas. Pee ee Aj,|) — ame Tp) tees (4) P= Af .where. f.=fi1+ @-— pee oe (5) Formulas to be Used in Designing Columns. Area of steel for assumed area of concrete, A, and specified stress, fc. P — Af. A, = ——> 7 in — Df. a Area of concrete and steel for assumed percentage of steel, p, and specified stress, fc. f=fll+ nm —Il)ph ee es iB A= LO ee ee (9) Age At Values of f for different percentages of steel are given on p. 916. Formulas to be Used in Reviewing Designs. Safe load for specified unit stress, f., and known dimensions of column. P=flA 4 Unit stress in concrete and steel for given column load and known dimensions of column. P Beker 6 Wares See Authors’ Recommendations for Columns with Longitudinal Steel Only. Effective Area.—Effective area of column is the area within fire- proofing, in columns exposed to danger from fire. ! In columns underground, also in reservoirs for water, full area may be taken as effective. In wall columns faced with 4-in. brick laid in Portland cement — mortar, the facing may be considered as fireproofing. COLUMNS WITH LONGITUDINAL BARS 407 Percentage of Steel—Minimum, 1 per cent; maximum, 6 per cent. Where, for architectural reasons, the concrete column is larger than required by stresses, the minimum percentage of steel should - be based on the area of concrete required by stresses and not by the area actually used. See Example 4, p. 410. Allowable Unit Stress, fe. ~ Ultimate Allowable Unit Ratio of Mix of Concrete Strength f’. Stress, fe Moduli of Lb. per Sq. In. Lb. per Sq. In. | Elasticity, n Vi et. 2 000 450 15 lime 13:3 2 500 570 12 ie? hk s.2 | 3 000 680 | 10 Limiting Length of Column.—Unit stresses in above table to be used in columns the unsupported length of which is equal to or smaller than forty times the least radius of gyration. Radius of gyration to be computed for concrete and steel. For longer columns reduce stresses as given on p. 434. Hoops or Ties.—One-quarter in. hoops or ties, spaced not more than 12 in. nor more than 16 diameters of the longitudinal bar. Fireproofing.—1% in. from outside face of steel. Joint Committee Specifications, 1924. Columns to be designed by Formulas (4) to (11). Effective Area.—Total area of column is considered as effective. Amount of Vertical Steel‘ The amount of longitudinal rein- forcement considered in the calculations shall be not more than 2 per cent, nor less than 0.5 per cent, of the total area of the column. The longitudinal reinforcement shall consist of not less than four (4) bars of minimum diameter of } in., placed with a clear distance from the face of the column of not less than 2 in.” , Allowable Unit Stress f, shall not exceed 0.20 f’.. Same limitation as to unsupported length of column as given above. | Lateral Ties. ‘‘ Lateral ties shall be not less than { in. in diameter, spaced not more than 8 in. apart.” | Fireproofing. (See p. 272.) 408 CONCRETE AND REINFORCED CONCRETE COLUMNS Requirements of Building Codes.—The table on p. 409 gives the requirements of the Building Codes of the cities of Boston, Cleveland, Chicago, New York, and Philadelphia, as they were in effect in the year 1925. Design of Columns with Longitudinal Bars Only.—In designing - columns with longitudinal bars only, the method of procedure is as follows: The load to be carried is computed first. The working stresses are then selected. Usually these are fixed by Building Codes; if not, the stresses recommended by the authors should be used. In a column there are two variables, namely, the size of the cross section and the amount of reinforcement. In determining the dimensions, therefore, it is necessary either to assume the size of the column and then compute the required area of steel from the formulas, or to assume the ratio of steel, p, and find the area of column and the amount of steel. The problem will resolve itself into one of the examples given below. | Example 1.—Given column load P = 400000 Ib. Design a square column of the smallest possible size, using the stresses allowed by the Boston Code. Solution.—Since it is desired to get a column of the smallest possible cross section, the richest mix of concrete (1:1: 2) and the largest percentage of steel (p = 0.04) will be used. The allowable stress for 1 : 1 : 2 concrete isf,; = 742 lb., R= 10. For p = 0.04, the average stress, fromf = fell + (n — 1)pl, 1s f = 742[1 + (10 — 1) x 0.04] = 1019 Ib. per sq. in. f = 392.5 sq. in. The side of the square equals 4/392.5 = 19.8 in. Adding 14 in. on each side for fireproofing, the size of column is 19.8 + 3 = 92.8in. Actually, a 23-in. square column should be used. The required area of steel is A = pA = 0.04 < 392)5"="15-7 “sq: in: sixteen 1-in. square bars may be used. Result: Column 23 in. square; Mix tiki tees Sixteen l-in. square bars. Example 2.—Given column load, P = 400 000 Ib. For architectural reasons, square columns are required. T he outside dimension of column must not exceed 25in. Find required mix of concrete and amount of steel, using Boston Code. Solution.—The richest mix of concrete will be used, as it gives the most eco- nomical design. From table on p. 409, for 1 : 1 : 2 concrete the allowable stress, according to Boston Code, is fe = 742 lb., n = 10. Fireproofing should be 13 in. thick on each side. 409 COLUMNS WITH LONGITUDINAL BARS ‘peay UUINIOD Jo 10390 IO UIBEq Jo U10}}0q 0} qBIs 100] Jo do} 9g} WO} 9oULySIP SI UUIN[OD Jo 4ySuez poysoddnsuy “4x04 UI S¥[NUIIOJ Aq poultaseyap o1¥ ‘049 ‘PROT 9[QEMOT[V [BIO], “YIOZ MON 3de0xe soryio [[e 10} BSugooidoeig urlyjIM wore st WOIyM ‘Bore aATJOIYS UO paseq SI [9e}S jo asvjusoIeg ‘Ifu sTenbo [e048 UI SSer4s JIU) :SALON “ul ‘bs G'°Q jO 18qQ Jod vole UmMuUITUTIU Sul YIM siveq F YSBo] IV -jooid “v6L UN vol cl e191 00% b: 6:1 ~[09 JO UOISUSUUTp JsvVoT]] “WIVIP GT | tT cI u& | SL {Urq} 009 SG: 41:1) 4 | Be | Brydpperyd » Gl : SE ne Oe he sd Ve Gish Pan F Ses oe Gl UOISUSUIp 4SBoT) “WIP GT | yt cl ui ZL | 1®4OL) 009 S240 Bi | By YOK MON ‘posn oq Jsnul [9038 Jo ‘ul ‘bs [| 4s¥o] VW Pert SCs see UrStp “Ur ST ‘ul £ uBvy} dJeT[euS uUBYy} 9IOUL CT OOF O00Z- P= Gal sieq ON ‘ur ‘bs #9 | you “req jo ral O8F | O0OF% |S: FL: 1 uUIN{OO JO BoIB 4JsBoT) “WIP CT ut GL uG OL ” O8¢ 006Z | 2: 1: 1) BF Se eae osBory() ‘sieq “pl tf Joos 4 - =. [eUIpnysuo] WINUTUTY) = ZT ut ra HOM SL. », | C09 | 0002 | F: 2:1) BI | %9 puepeas[y sivq jo Sul “WIBIP OT GI. |-Jyooid| G6F | 002 | Fi: 1 uevy} o10Ul Gl “og | 0&9 0082 |€: #1: T bes ou {21 ut ra utl OL jm) cL | OOSS | SZ: 1:1) VI | MH | °°’ “WOWsog sage JdJOULBIC, ¥! 07 eu wnt | wWnur suloedg ee qsvory 0} | Sut waly | pole -yg | “SN | -rarpy |-rxer | syIeutey ae poyiod §|-jooig| wu OAT} SIIND a ees ONT) ~ | OTE -9933q | ——__________—- ca aac ory SOL, a UOIsseIdUOD UI ssei}g | peurpnyaoy (CZ6I IB9X Ul OATIOOAH) SONID SNoWeA Jo Sepod Zurpiing jo s}uswosnbey 410 CONCRETE AND REINFORCED CONCRETE COLUMNS The effective side of the square column, obtained after deducting fireproofing, is 25 —- 2 X 1} =22in. The effective area is A = 22? = 484 sq. in. The load carried by concrete alone is Af, = 484 X 742 = 359 100 lb. The balance of the load to be carried by vertical bars is P — Af, = 400 000 — 359 100 = 40 900 lb. Since the allowable stress in steel is (n — 1)fp = 9 X 742 =6 680 lb., the required area of steel is P — Af, 40900 = = — = 6 Fs 12 C . (ne Ufee 6 680 oe s Fight 1-in. round bars, giving an area of 6.28 sq. in., may be used. Result: 25-in. square column; 1:1: 2 mix of concrete; Wight 1-in. round bars. Example 3.—Column load, P = 400 000 lb. Find dimensions of column and area of steel, using 1 : 1 : 2 mix of concrete when the amount of steel for economi- cal reasons is limited to 1 per cent, using Boston Code. Solution. —The allowable stress for 1 : 1 : 2 mix, according to Boston Code, is fo = 742 lb. per sq. in., n = 10. For 1 per cent of steel, the average stress is f =fedl + (n — 1p] = 742[1 + 9 X 0.01] = 809 lb. per sq. in. The required effective area of concrete then is 4 _P. _ 400.000 of oe BOS = 494 sq. in. Effective side of square column, d = V494 = 22.2 in. Required area of steel, A, = pA = 0.01 X 494 = 4.94 sq. in. Two 1-in. round bars plus six 7-in. round bars give an area of 5.18 sq. in. Since 14 in. fireproofing is required, side of column is d+2X1.5 = 22.243 = 25.2 mn. Actually the column side should be made 26 in. Column 26 in. square; 175) Bas Two 1-in. round bars plus six 3-in. round bars. Example 4.—The size of a rectangular wall column, as fixed by architectural requirements, is 40 by 16 in. The load is P = 150000 lb. Find the required mix of concrete and the amount of vertical steel, using Boston Code. Solution.—Allowing 14 in. on each side for fireproofing, the area of the effect- ive section is (40 — 3) X (16 — 3) = 481 sq. in. This section is larger than COLUMNS WITH LONGITUDINAL BARS 411 required to carry the load. Using 1 : 2: 4, mix, and one per cent of steel, the allowable average stress if f = f,[1 + (mn — 1)p] = 495 K 1.14 = 564 lb. per sq. in. The required area of concrete to carry a load of 150 000 lb. is P _ 150.000 i 064 The area of vertical steel will be based upon this required area of concrete. Therefore, = 266 sq. in. A, = 0.01 X 266 = 2.66 sq. in. Result: Use 1 : 2 : 4 mix, and six 3-in round bars. Reviewing of Column.—In reviewing columns already designed, two problems may present themselves. First, the dimensions of the column may be known, and the stresses may have to be computed for a given column load. Second, for known dimensions of column, it may be required to find the maximum load the column is able to carry. Often the reviewer finds that the design is unsatisfactory and he is called upon the change the dimensions. In such a case he may retain the size of column and increase or decrease the amount of steel, he may change the mix, or he may change the whole design. In any case, the methods given for the design are used. Example 5.—Dimensions and design of the panel are given. Live load is specified. From these the computed load on column is 300 000 lb. Dimensions of column are 24 in. square. Reinforcement consists of eight 1-in. round bars. Required fireproofing, 2 in.; mix of concrete, 1 : 13 : 3, n=12. Determine the stresses in concrete and steel. Solution.—With fireproofing of 2 in., the effective area of column is A= (24 — 2 X 2)? = 20? = 400 sq. in. The area of steel is A; = 8 X 0.785 = 6.28 sq.in. Hence, A + (n — 1)As = 400 + 11 X 6.28 = 469.1 sq. in. P The stress in concrete ore Formula. (11) oj. ene AL is f, = 300 000 630 1h . = . per sq. in. 469.1 on gees The stresses in steel are f, = 639 X 12 = 7670 lb. per sq. in. This example may also be solved by using Table 21 on p. 916. The value of p is computed, p= Hee = 0.0157. Then the average stress, f, is found, f an = 750 lb. per sq. in. For f = 750 and p = 0.0157, the stress, f;, is found from the table by interpolation. Result: fs = 639 lb. per sq. in. fs = 7670 lb. per sq. in. 412 CONCRETE AND REINFORCED CONCRETE COLUMNS DETAILS OF COLUMNS WITH. LONGITUDINAL STEEL ONLY General Requirements.—Steel bars used for column reinforce- ment must be straight. They must be placed in vertical position and be prevented from displacement by hoops or ties. In square and rectangular columns, the bars should be arranged symmetrically. They should preferably be placed along the perimeter of the effective area. Since vertical bars, if properly imbedded and placed sym- metrically, take their stress regardless of their location in the column, the effective area of the column (contrary to action in a spiral column) has no relation to the area of concrete located between the lines of steel. Thus, a requirement of 14 in. of fireproofing does not mean that the steel bars must be placed within 1} in. of the surface. On the contrary, the bars should be placed far enough away from the forms to give ample room for placing concrete. A minimum dis- tance of 2 in. is recommended. Sometimes, when a large number of bars is used and it is not possible to accommodate all of them along the perimeter, some bars are placed in the middle section of the column. This is not advisable, however, as the bars in the central part are hard to keep in place. They also interfere with the placing of the concrete. To get a symmetrical arrangement of reinforcement, an even number of bars should be used in a square or rectangular column. It is preferable that all the bars used in a column be of the same diameter. In many instances this is not possible without considerable waste of steel. Then the column reinforcement is made up of two groups of bars (each consisting of an even number of bars for square or rectangular columns). The bars in the two groups should vary in diameter by not more than one-eighth in. Thus, to get.an area of steel equal to 6.9 sq. in., five $-in. round bars plus five 1-in. round bars may be used. To use ten 1-in. round bars would mean a waste of 3.13 lb. of steel per lineal foot of column, which for a large number of columns may amount to a considerable tonnage. Seven 1-in. square bars give a close enough area and may be used in round columns. They cannot be used in square or rectangular columns, however, because it would not be possible to arrange them sym- metrically. There is one objection to using bars of different diameters in one column. Where experienced labor and good supervision are not available, there is danger that all small sizes may be used in one COLUMNS WITH LONGITUDINAL BARS 413 column and all large sizes in another. Under such conditions, bars of the same diameter should be used even if it means some waste of material. At least four bars should be used per column. In good practice, the bars should be at least 2 in. rd. to give sufficient lateral strength to the columns to withstand any bending which may come on them, either by unequal loading or by some accidental horizontal pressure. In long buildings, the columns in the end row of panels are sub- jected to considerable bending due to temperature changes. The building lengthens or shortens from the middle toward the ends, so that the top of the end columns must deflect by an amount equal to the total expansion or contraction of one-half of the length of the building. The bending stresses produced thereby may be large, particularly in top columns. To prevent cracks, additional steel should be used. When a large percentage of steel is used, it is advisable to use bars of large diameter so that the clear spacing along the perimeter between the bars is not smaller than 3 in. When the reinforce- ment of the column consists of very heavy bars, it is advisable, for practical reasons, to use at least four light bars. Their use is ex- plained under ‘‘ Assembling of Bars.” Lapping of Bars.—The stresses from the bars in the upper column should be transferred to the bars of the lower column by lapping of the bars. For this purpose, the bars from the lower column are extended into the upper column a sufficient distance to develop in them, by bond, the stresses carried by the bars of the upper column. (Lapping by extending the bars from the upper column into the lower column is not practicable, because the upper bars would have to be placed before the floor system below was poured and they could not be kept in position without considerable difficulty.) The length of the lap depends upon the stresses carried by the bars. For columns subjected to vertical loading only, the length of lap is determined by compression stresses in the bar. For columns subjected to bending, the length of lap may be fixed by tensile stresses. In wall columns, for instance, the full tensile strength of the bar should be developed by the lap. The length of lap may be taken from the table below, for the stresses to be developed and the diameter of the bar. The following simple rule may also be useful. For a bond stress of 100 lb. per sq. in. the required length of lap is equal to 23 diameters of the bar for each 414 CONCRETE AND REINFORCED CONCRETE COLUMNS 1 000 lb. of unit stresses in the bar. For a bond stress of 80 lb. per sq. in., the required length of lap is equal to 3g diameters of the bar for each 1 000 lb. of unit stress in the bar. Length of Lap for Different Stresses in Stee: and Diameters of Bar Length Apply Either to Tension or Compression Based on allowable bond unit stress of 80 Ib. for plain and 100 lb. for deformed bars Length of Lap in Inches Tne Siete Plain Bars Deformed Bars Steel, fs ng Diameter of Bars * Diameter of Bars * 2.) 8 | 8) 2) Dal 1g ee eee ee ee 6 000 O12 14 16 19S cr 8 Go) 11s et eee 7 000 11.| 141.16 |\19 | 22 1255 “Ol 1h eis tne 8 000 13 | 16 | 19 | 22 | 25 | 28 | 10 | 13 | 15 | 18 | 20 | 23° 9 000 14 | 18°] 21°] 25 | 28 | 82 | UP eee ee 20sec 10 000 16.| 20 |-283 |-27 | 8311-85 ) 134) 16" 19s 2a ease 11 000 17 | 21 | 26 | 30-| 34 | 39 | 14917 4 2is ees eeaer 12 000 19 | 23: | 28 | 338 .| 88 | 42° | 1b (19s 2a 5 S2ON eat 13 000 20 | 25 | 30 | 36 | 41 | 46 | 16 | 20 | 24 | 28 | 33 | 37 14 000 99 | 97 | 33:1 38 | 44 1.490 18 2226 Rae sanre 15 000 92 | 99 | 35 | 41 | 47 | 53 | 19 | 23) 285s a esses 16 000 25. | 31 138 | 44 | 50 1°56) 20) 25) BON eaon eres 18.000 28 | 35 | 42 | 49 | 56 | 63 | 23 | 28 | 34 | 39 | 45 | Sl 20 000 21 | 39 | 47 | 55 | 63 | 70 | 25 | 31 | 38 | 44 | 50 |.56 * Table may be used for round and square bars. If the number of bars used in the upper column is smaller than in the column below, some saving in steel may be made by extending upward only as many bars as there are in the upper column. The remaining bars may be stopped just below the top of the slab. Thus, if the fifth floor columns are reinforced with twelve {-in. round bars and the sixth floor column with eight 3-in. round bars, only eight Zin round bars need to be extended into the sixth floor, and the reinforcement of the fifth floor column may consist of eight long COLUMNS WITH LONGITUDINAL BARS 415 bars and four short bars. This method should be used only on jobs with good supervision and experienced labor, as otherwise confusion may arise and some of the columns may get a larger proportion of the short bars than others. When the number of bars in the upper'column is larger than in the lower column, the lower bars are not sufficient to develop all the upper bars. ‘Then, in addition to extending all bars into the upper column, it is necessary to use extra dowels equal in number to the difference between the number of upper bars and lower bars. These should be imbedded in the length required by bond in the lower columns and should extend the same length into the upper column. If the diameter of the lower column is larger than that of the column above, the bars, if continued straight, would come outside of the upper column. To be brought within the effective area of the upper column, the bars should be provided with a reverse bend, as shown in Fig. 141. This process is sometimes called “ goose- necking.” Care should be taken that the inclined portion of the bar is placed within the portion of the column restrained by the floor system, because if the bend were placed within the unrestrained part of the column there would be a tendency for the inclined bar to develop an unbalanced horizontal component which is harmful to the concrete. While this is elementary and obvious, it is not always appreciated by designers. In bars of small diameter, a small offset to bring the lower bars in line with the upper ones may be provided by “hickying.” Bars below # in. may be “‘ hickied ” after the bar is in place, but before concrete is poured in lower columns. Bars over ? in. in diameter cannot be readily bent when in place and should be bent before vac are placed in the form. An incorrect method of lapping the bars, often seen on con- struction, is shown in Fig. 142. The value of the bars as laps is lost, as the pressure from the column cannot readily be transferred to the inclined bars. Also, the bars may interfere with the flow of concrete during construction and pockets may form, thereby further weakening the columns. Butting Bars.—If the column is reinforced with longitudinal bars of larger diameter, say over 1 in., the stresses from the bars in the column above may be transferred to the bars in the column below by butting the bars. The ends of the bars must be milled to even 416 CONCRETE AND REINFORCED CONCRETE COLUMNS bearing. When in place, their ends are kept in position by tight- fitting pipe sleeves. From the construction standpoint, it is best to extend the bars from the column below about 4 in. above the rough slab. After the slab is poured, a pipe sleeve 8 in. long is placed over the bar and rested on the concrete. The bars for the column above are then placed in the pipe sleeve. If the number of bars in the column below is larger than in the column above, the extra bars may be stopped just below the top of the slab. If the number of bars to be butted in the upper column is larger than the number of available bars in the column below, dowels Fic. 141.—Correct Method of Splicing Fria. 142.—Incorrect Method of Lapping Column Bars. (See p. 415.) Column Bars. (See p. 415.) of proper length should be placed in the column below, against which the bars of the top column are butted. When the diameter of the upper column is smaller than that of the column below, the lower bars should be bent or goose-necked in the manner described in connection with lapped bars, so that the end of the lower bars will come directly under the bar above. Some designers slant the bars in the lower column for their full length to bring the ends in the proper position. T his is obviously wrong. © The butt joint is not capable of resisting tension. When the columns are subjected to bending, sufficient steel should extend from the lower column to the upper column to resist any possible tension. Either extra bending steel should be used, or else part of the column COLUMNS WITH LONGITUDINAL BARS 417 steel, sufficient to resist bending stresses, should consist of bars of small diameter and these developed by lapping. In cases where the bottom column has a larger number of bars than the top column, the extra bars not required for butting may be extended into column above to provide the required tensile steel. Dowels.—The stresses from the column bars are transferred to the foundation by means of short bars called dowels. Their total length equals double the length required for developing the stresses in column bars. One half of the dowel is imbedded in the founda- tion and the other half laps the column bars. The number and size of the dowels must ‘be same as that of the bars used for the columns above. If the depth of the foundation is smaller than the length of imbed- ment required for the dowels, their ends should be provided with a right angle hook, which should be placed above the footing rein- forcement. Foundation Plates.—Instead of using dowels, the stresses in the bars may be transmitted to the concrete by resting the bars on steel plates of sufficient dimensions to keep the bearing stresses on con- crete within working limits. In such a case, if the column needs to be anchored to the foundation, special anchor bars should be Bad veda provided. sai Column Ties—The bars 7? should be kept in position by column ties. They are usually made of }-in. round bars and spaced not more than 12 in. on centers (see requirements, p. 407). The bending details of ties for square and rectangular columns are shown in Fig. 143, p. 417. me. 143.—Bending Details of Ties for In determining total length of tie Square and Rectangular Columns. (See allowance should be made for the p. 417.) steel in corners. The wire should be lapped at a corner. This method is more satisfactory than lapping the wire between the corners, as the joint is more rigid and the steel bars are better kept in position. For corner columns, the ties should be arranged as shown in Fig. 144, p. 418. If Square, c=b Square = 404 I7t. TE CR = 2(b+e) +772 418. CONCRETE AND REINFORCED CONCRETE COLUMNS The ties for special columns should be designed with care, their purpose being kept in mind. Fig. 149 shows the ties for a special column. It consists of two parts, a rectangle and a stirrup. Elevation Section Fic. 144.—Ties for Corner Columns. (See p. 417.) Example of Column with Vertical Steel Only.—Figure 146, p. 419, iJlustrates a typical column with vertical steel only as used in the new buildings for the Massachusetts Institute of Technology.? All the details are clearly shown. Section Elevation Fic. 145.—Ties for Special Columns. (See p. 418.) Assembling Column Steel.—The column steel is usually assem- bled on horses. The skeleton, consisting of four corner bars and ties, 2Stone & Webster Co., Builders. Sanford E. Thompson, Consulting Engineer. SPIRAL COLUMNS is assembled first. Two corner bars are placed on horses, the proper dis- tance apart; the ties are then spaced and wired; then the other two corner bars are wired. If the weight of the total column reinforcement is not too great, all bars are wired before plac- ing; otherwise, the skeleton of the column reinforcement is dropped into place first and finally the remaining bars are placed and wired. When very heavy bars are used, say 1j-in. square bars weighing 5.3 lb. per lineal foot, the skeleton consisting of four bars, when assembled, would weigh several hundred pounds. It would have to be handled by a number of men, and there would be danger of damaging forms while placing the column steel. For this reason, when heavy bars are used for column rein- forcement, it is a good scheme to use, in making up the required area, four light bars, say #2-in. or {-in., to be used for the skeleton. A_ skeleton composed of such bars can be readily handled and at the same time is stiff enough to keep its shape until the rest of the bars are placed. SPIRAL COLUMNS Description of Spiral Columns.— Spiral columns are columns reinforced either with continuous closely spaced spiral reinforcement or of closely spaced separate hoops placed as near as possible to the outside face of the column. In practice continuous spiral is used in preference to separate hoops. Fae Se SS RS Gai iiake ‘ \ | fete | Roof 419 Slab 5-5 "# 5-6" 46" 5» About 9” =a Apart pt a) . ER vnh RM iRARhMMmEe ait hs Ges CEE igin nha oo re re \ t cuctunle tele te Iel lot? toRt Tol te tot ot oT oT SI ee pee | ame x ' © ~ << 20-3 $46" About 9” Lf ~/"@ J9-0" ee IECO: ay Lis -alpgke-F Sec. F-F SES hfe TA aes Ce ea C ro ae ice : on Se: 3175" AH See.C-C a RA si Fic. 146.—Typical Vertical Steel Only. Column with (See p. 418.) 420 CONCRETE AND REINFORCED CONCRETE COLUMNS The spiral reinforcement is based on the principle that concrete confined within the spiral is capable of withstanding very much | larger stresses than concrete free to spread sidewise. The function of spirals, therefore, is to confine the concrete. In performing this function the spirals are subjected to tensile stresses. Although spiral alone increases the strength of the column it is never used in practice without longitudinal reinforcement. There- apinaaay turn of Oowels--+ v1 of Spira 7 el eee Re ae BE at i Be A EE Ba Ie eB URNS BPS Oa __ ++} —_| ae ES Fire Proofing, OI v Fa ae ee aeee sane =e Length Of SPEC ea ar ean eed of Core oC ne eae 2 Ea BE has Gad Es We es Es eg | OJ op en Ae at Fic. 147.—Spiral Column. (See p. 420.) fore, for practical purposes a spiral column is a column reinforced with longitudinal reinforcement and with closely spaced continuous spiral placed near the outside face of the column. Such column is illustrated in Fig. 147, p. 420. There is much difference of opinion as to the method of computing the strength of columns with spiral reinforcement. ‘The reason for this divergence is discussed on p. 82, in connection with the test of columns. ‘The method recommended by the authors, as well as the methods required by the Joint Committee, and by the New SPIRAL COLUMNS 421 York, Chicago, Cleveland, par Des and Boston Building Codes, will be given. Let P = total column load in \b.; A = area of concrete core within spiral (the diameter of which equals the diameter of spiral measured to the center of wire) in sq. in.; A, = area of cross section of vertical steel in sq. in.; p = ratio of area of vertical steel to area of concrete core =A,+ A; pi = ratio of volume of spiral reinforcement to volume of concrete core; f. = compressive unit stress in concrete, lb. per sq. in.; f’. = ultimate compressive strength of concrete at 28 days, based on 6 X 12 in. or 8 X 16 in. cylinder tests, Ib. per sq. in.; f’. = compressive unit stress in steel, lb. per sq. in.; f, = allowable stress in wire composing spiral reinforcement (as used by New York Code). Authors’ Recommendation for Spiral Columns.—Formulas for Strength of Column.—For spiral columns designed according to rules given below the formula for strength is the same as for columns with vertical steel only. The effect of the spiral is expressed by allowing larger unit stresses. Formula for Total Load, peau Adil: A) fels, ww Ye eeueek es 1012) or, when A, = pA, feo T GRO ee (eee el eee Cae meer eres ES) c These formulas should be used in the same manner as explained for columns with vertical steel only (see p. 406). Formulas (4) to (11a) apply here also. Allowable Unit Stresses. The allowable stress in concrete on spiral columns is equal to fc = 0.35 f’,. The stress in steel equals n times the concrete stress. Spirals are not figured as adding directly to the strength of the col- umn. 422 CONCRETE AND REINFORCED CONCRETE COLUMNS The following stresses may be used: enn Ultimate Strength, Working Stress, Concrete Mix Se: it; n Lb. per Sq. In. Lb. per Sq. In. hod ye 3 000 | 1 050 10 1 ees es: 2 500 880 12 L224 2 000 700 15 0 8 a See eee Requirement as to the Amount of Reinforcement. Vertical reinforcement, not less than 1 per cent, nor more than 6 per cent of the effective area of column. 3 Spiral reinforcement, not less than 1 per cent of the volume of the concrete enclosed within the spiral. Pitch of spiral and other details as recommended under proper headings. Unsupported length must not exceed 40 times least radius of gyration. For larger lengths, use reduction of stresses given on p. 435. Fireproofing —Two inches for round and octagonal columns. For square columns with round spiral, minimum 14 in. Boston, Cleveland and Philadelphia Codes The formulas for spiral columns are the same as for columns without spiral. The amount of spiral reinforcement must not be less than 1 per cent of the volume of concrete within the spiral. For such columns the effect of the spiral is expressed in the allowable unit stress on concrete. P = Af. + (n — AS rr also when A, = pA, PL 4ytit @ = el Allowable unit stresses and requirements are given in table below. : 423 ‘IOJOUIVIP Ul ‘UL £ UY} SSO] JOU-SIVQ [BOTJAOA INO} 4SVOT 9B Oq [[VYS VIO, x “$1 wduInjoo JO Yysue] poysod -dnsun 0} Jo}oUr JO 9109 JO -BIp jo opis {sBa] JoJOUIvIpP 2 ‘UII We on/e, eMie| etuer eas) «ts JO O1}VI WINUIXBT] 4B | 10 (2 tel all aol ae oF ‘srs0eds 99144 syeads S IIqIQ‘saeoeds S [ volueyoow “2! aun 5 OMpoavy prnoys — |-[0o Jo yysue] 0} J J19J9W VIP ‘UI |Lo}JOUTvIPOpIs}No “UII eo SE RROP Un resis Or SN ieee at uS YU \xVo1 | %E | IevsJo UL |, FZ i “oF uuInjoo 1OU “UIBIP a JO YJSUV] OF 0100 _ 2 uvy} "ULI, 7A JO OFFBI UINUNIXBIT) TT | a10ur ON | %T |, YT YP qoue youd queo | wuinur | uNuUr uuinj[oy sur | -1urpy WINWIXB]Y | Jog | -UTPAL | -TXB] SHIVUIOY sg gee -joord | =~ : 0} 1dJIUIVIG, omy qo te. d {UOUI9D.IOFUTOY ghee: [BO1}10 A. GL OSL ZI 0068. Sl ae ase Si ‘erydyopeytyd GT O¢s 000 @ "* “puvypord;)y) CT OLL 00 G ol O86 OOS G OT CCoLT COE So ok ae a u0JSOG ‘ut'bs ‘ur'bs Jad "q’'T Jad *qry se aM de U ‘Ssal}Q jojJa1oUOD Jo SOND SUTYIO\, | YJsUaIIS FICE MOTT tee errr) sopog vIydepelyd pue ‘puvjeas[D ‘u0jsog ‘suumMyoD jerdg I0j sjusueimMbey 424 . CONCRETE AND REINFORCED CONCRETE COLUMNS New York Code Formula for Total Load on Spiral Columns, P= A4f,ttin— DAS. + 2Afe also P= Af[1+(a—Dp+ 23]. Serta From which, for known area of core, A, and percentage of spiral pi, Area of Vertical Steel, ie as A(f. ae 2pifs) A; = 18 in — if os For known area of core A and area of steel A,, Ratio of Spiral, Lee fA aE (n a LAs] (19) Pi ) Se (24a) where | f, = 300 + (0.10 + 4p)f’. lb. per sq. in,” aia Strength of Concrete, f’c.—The expected strength of concrete, f ee upon which the unit stress, fo, depends according to the report, depends not only upon the mix of concrete, but also upon the ‘‘slump ”’ of concrete during construction. (See Appendix, Vol. II.) Requirement as to Amount of Reinforcement. Vertical Reinforcement.—Not less than 1 per cent nor more than 6 per cent of the core. Minimum amount of steel to consist of at least six 4-in. bars. Spiral—Percentage of spiral not less than 1 of the percentage of vertical bars. : Pitch of spiral—Not more than § of the diameter of core, but not more than 3 in. Additional Requirements——Spiral must be held in place firmly and true to line by at least three spacer bars. Spiral reinforcement shall meet the requirements of the ‘‘ Tentative Specifications for Cold-drawn Steel Wire for Concrete Reinforcement ”’ of the Amer- ican Society for Testing Materials. Fireproofing —Two inches for round column or octagonal col- umns. For square columns with round spiral minimum 1% in. Use of Formulas for Spiral Columns. Boston, Cleveland and Philadelphia.—In these formulas for spiral columns, the percentage of spiral is fixed and does not enter into computations. The variables are the area of concrete core, A, and the amount of vertical bars, As = pA, as for columns with vertical pars only. The design of the column, therefore, proceeds in the same manner as for columns with vertical steel only, the difference being in the unit stresses. The problems in this connection and their solution are the same as given on pp. 408 to 411 for columns with vertical steel only. New York .Code.—In the New York City formula for spiral columns, there are three variables: the area of concrete core, A, the SPIRAL COLUMNS 427 percentage of spiral, P;, and the amount of vertical bars, A;. The problem of designing is best solved by means of tables given on p. 918. If no tables are available, the columns are designed by trial. The diameter of column is assumed first, then a ratio of spiral is adapted (see p. 432 for relation of p; to pitch of spiral, also tables on pp. 930 to 933); foe the amount of vertical steel is com- puted from formula : Cis a Fase | The problem may be also solved by assuming a size of column and ratio of vertical steel. Then the ratio of spiral is found from ney “Til. If the design is unsatisfactory or uneconomical, another assumption is made. Knowing spiral ratio p1, the pitch of spiral may be found by Formula (35), p. 432, for accepted diameter of wire. The pitch’ may also be found from tables on pp. 930 to 933. If it is desired to get the smallest possible column, in the equation’ P= Aj.|1 Sen 1)p -- air |, substitute for p and p, the largest _ allowable values (pi = 0.02 and p = 0.04). Then the smallest area is (26) (27) 12 sie (28) f.|1 + (n — 1)0.04 + 0.04 | If it is desired to use a special percentage of vertical bars and spiral, substitute the selected values of p and 7, in the above formula for P, and solve for A in the same manner as in the previous case. To find the stresses for given dimensions of column and given load, P. Values of A, p, and p; are known. Value of fs 18 also known as it depends upon the known strength of wire. Divide the load, P, by area, A, which gives the following relation: fi. + (n— lp + 2p | a in which all values except f. are known. To find the value of f, would require a solution of a second degree equation. To avoid this, a value for the ratio, Js may be assumed, and the equation : } q fr 428. CONCRETE AND REINFORCED CONCRETE COLUMNS solved for f.. The ratio of is then checked, using the value of f, just found. If it is near enough to the assumed ratio, the solution is correct; otherwise, repeat the performance with a revised ratio. Chicago Code.—The method of procedure with the Chicago formula for spiral column is the same as with the New York formula. The smallest column is obtained in the same manner. Also, the method — of procedure is the same if it is desired to use special percentages of steel. It must be remembered, however, that the percentage of vertical steel must not be less than that of spiral. To review a column for known values of A, p, p1, and for spiral column load, P. The stress f. is found in formula, f.= SS ye te ° All + (n — 1)p + 23npil in which all values, except f., are known. (29) Joint Committee Rule, 1924.—In the Joint Committee formulas for spiral columns there are three variables, namely A, p, and fz. The value of fe depends not only upon the quality of concrete, but also upon the percentage of steel. The ratio of spirals varies with the variation in the ratio of vertical steel. To design columns, it is necessary to assume percentages of steel, for which the stress, fe, is found from Formula (25). Finally, the area of column is found from P A lta ipo. (30) To get minimum size of column, assume the maximum value for p. The design is more involved if it is required to find reinforcement for an assumed size of column. For instance, if the area of column, A, is known, then — f+ — Del =9 (31) From this expression it is impossible to get directly the value of p, as the value of f, is not known, being dependent upon p. To solve the problem, two simultaneous equations must be solved, namely jl hie Del =p RNR ES and f, = 300 + (0.10 + 4p)f".. Renmirntons se ote? SPIRAL COLUMNS 429 : P In these equations, a and f’, are known and f, and p unknown. By eliminating the value of f., we get a second power equation. This, solved, gives the percentage of steel, p. To review a design for known values of P, A, and p. The stress f. is found from te ese Dol This value, however, does not give any information as to the safety of the column. To get this, it is necessary to substitute the values of fc and p in the formula f, = 300 + (0.10 + 4p)f’. and find the value of f’.. Then compare this ultimate strength with the ee ulti- inate strength of the concrete. From the above it is evident that working with Joint Committee formulas is complicated unless tables are available. (34) OCTAGONAL SPIRAL COLUMNS Octagonal columns with spiral reinforcement are sometimes used . because some architects prefer them to round columns on account of appearance and because it is easier to fit partitions to the flat surface of the column. In some instances, they are used because, through local conditions, they are more economical than round columns. As the spiral is made round, the effective area and the strength of an octagonal column are equal to those of a round column of a diam- eter equal to the short diameter of the octagon. The gross area of an octagonal column equals eee where d is the short diameter of the octagon, while the gross area of a circle is ee An octagonal 4 column requires, therefore, 5.4 per cent more concrete than a round column of the same strength. When comparing the cost of round and octagonal columns, in additon to the extra cost of materials, the question of formwork should be considered. In flat slab work, in addition to the column form, the form for the column head must be considered. The rela- tive cost varies with local conditions. SQUARE SPIRAL COLUMNS Tests prove that square spirals are not effective as such, since they are not capable of restraining the concrete laterally and prevent- 430 CONCRETE AND REINFORCED CONCRETE COLUMNS ing it from spreading under load. If it is necessary to use spiral rein- forcement in square columns, the spiral must be made round and the column designed in the same way as a round column of the same diameter. The concrete at the edges, which is outside of the spiral, is wasted. A square spiral column requires 27 per cent more concrete than a round spiral column of the same strength. Square columns with spiral reinforcement should be used only where for any reasons square columns are desirable or where round or octagonal column forms are not easily obtainable. The protective covering for the spiral in square columns varies +n thickness at different points in a square section. It is largest at the corners and smallest in the middle of the sides. This smallest thickness of the protective cover may be made } in. smaller than required for round columns, where the cover is of uniform thickness, but not less than 13 in. This reduces somewhat the excess of con- crete in square spiral columns. OBLONG SPIRAL COLUMNS Sometimes it is desirable to make the clear distance between the faces of the columns in one direction as small as possible, while Rounded Corners Rounded Corners Rounded Corners Square Corners Square Corners aos + _1 —T ee ‘ pee ee 8) es wees ES ea a oe oe co oe ee ee ig ec) ES a sos cn os a ee BS et ee ii ems SE = gem o oo oe ao pis Og pe rt _# =ae oo oe 8 tH ee iy eee ES: ron a oe Boek & oe pee ee ty oe ee ses fom rp Ht oo oe a oe Bee Pp} =e Pt =. =e et tts Bee Se 8g ee ae : a5 ww : . = gees GS Bs pee ee 7B Be: ea B oo oe =o a ees el Sg pee me Gee SS: Tae 8 coos aoe pee Sol ype ae oo: aan oon oe oo ee oo ge ese ae ee: i sam Se = tt — ‘es Cy eae 4——F rt a 6 c ‘Fic. 148.—Spirals in Oblong Columns. (See p. 431.) the clear distance in the direction at right angles is of minor impor- tance. In such a case, oblong columns may be used. Under some conditions, oblong columns may be spirally reinforced. DETAILS OF SPIRAL COLUMNS 431 The reinforcement of such oblong columns usually consists of two or more spirals. ‘The cross section may be either rectangular or oval, i.e., with rounded corners. To be economical, the relation of the small diameter to the large diameter of the column must be such as to permit the design shown on Fig. 148, a or b.. In the column shown in Fig. 148c, the concrete between the two spirals is outside of the effective areas. It does not contribute anything to the strength of the column, and is, therefore, wasted. The spiral oblong column actually consists of two spiral columns of small diameter, and should be designed as if the spiral columns were separated. It is obvious, however, that no extra concrete is required for protective covering between the spirals, as there the concrete 1s not exposed. The effective area in Figs. 148a and c equals the area of the two small spiral columns; while in Fig. 148b, where the spirals intersect, the actual area should be computed. The rules for spiral columns apply equally here. DETAILS OF SPIRAL COLUMNS Data for Designing Spirals.— Let pi = ratio of spiral = volume of spiral divided by volume of concrete core; A; = area of wire, square inches; A’, = equivalent area of vertical steel of same volume as ame | je d = diameter of column core, inches; s = pitch of spiral, inches; W = weight of spiral per foot of column height, feet. the spiral = pi Pitch of Spiral, s, for Known Ratio of Spiral, p: or Known Area A’s. 4A, = —-. Pee | aah hig he gene sane eink (35) VES 3.4 Ch...) RGR Pe eto De) Ratio of Spiral, p’, for Known Pitch s. 4A, Pi ds” (36) 432 CONCRETE AND REINFORCED CONCRETE COLUMNS Equivalent Area of Vertical Steel of Same Volume as the Spiral. Al, = 3.145 As. « er Weight of Spiral, W, per foot of Height of Column (not including spacers). W = 10.95 Ay. . ee POE Wire for Spiral—Spirals are usually made of cold-drawn wire having tensile strength of at least 80000 lb. per sq. in. The size of the wire is designated in the mill by gage numbers instead of by the diameter of the bar. The designers, however, use the diameter of the bar almost exclusively. The relation between the gage and the diameter of the bar is given in the table below: Standard Wire Used for Spirals Diameter of Actual Weight of Wi Gage Numbers Equivaient Diameter Area of Wire, ee Foot hh of Wire Round Bar, of Wire, sq. in. P lb oe in. in. : é 1 0.4900 0.189 0.636 : G 0.4305 0.145 0.490 : 3 0.3625 0.103 0.347 0 5 0.3065 0.074 0.250 3 1 0.2437 0.047 0.158 The spiral should be made of a continuous wire. If this is impos- sible, the wire must be spliced to develop the full strength of the wire by bond. It must be remembered that the spiral acts in tension and not in compression. Pitch of the Spiral.—The vertical distance, center to center, between two adjacent windings of the wire of the spiral is called the pitch of the spiral. The maximum pitch is always limited to a certain fraction of the core, and also, as an additional limitation, DETAILS OF SPIRAL COLUMNS 433 it must not exceed 3 in. The minimum pitch is seldom specified, but it is limited by the practical consideration that the clear dis- tance between the wires should be large enough to permit free flow of concrete from the core to the outside shell; otherwise, pockets might form at the wires, resulting in a porous fireproofing. The clear distance between the wires should not be less than 1 in., which fixes the minimum pitch for 3-in. spiral at 14 in., and for 3-in. spiral at 12 in. Spacers.—It is important, in order that the pitch of the wire be made uniform, that the spiral should be maintained in a vertical position and that it should be secure from displacement during pouring of the concrete. For this purpose, the spiral must be securely wired to the vertical bars. The best results are obtained by using mechanical spacers. The spacers consist of small angles, channels, or T-bars, one leg of which is notched at proper intervals to receive the spiral wire. At least two spacers should be used per column. Larger spirals may require three or four spacers. Usually, it is the best practice to have the spirals built up in the shop and deliver them to the job collapsed. In some cities (notably in New York), this has been against the regulation of the labor union. There the wire may come to the job coiled, but must be spaced and wired on the job. This increases the cost appreciably. In estimating the weight of spiral, the weight of the spacer must be added to the weight of the wire. The weight of spacers per foot depends upon the cross section of the spacer and also upon the diameter of the wire, heavier wire requiring heavier spacers. The weight per lineal foot of two T-section spacers for 7-IN. wire varies from 1.60 to 2.0 lb.; for 3;, and 2-in. wires from 2.0 to 3.0 lb.; for heavier wire 3.60 lb. Length of Spiral—By length of spiral is meant the vertical distance from the top to the bottom of spiral when erected. A good rule is to make the length of the spiral 6 in. shorter than the dis- tance from top of slab to top of slab. An extra turn of the spiral should be added at the top and bottom, for lap. With such design, the top of the spiral is well imbedded in the concrete, and at the same time there is no danger of any wires extending above the slab owing to inaccuracy in manufacture. It is true, as it is often argued, that the column is enlarged at the floor level by the floor system, especially in flat slab construction 424. CONCRETE AND REINFORCED CONCRETE COLUMNS where the column is enlarged at the top to form the column head. This may serve as an argument for stopping the spiral at some point within the column head. On the other hand, the spiral columns are almost always of rich mix, while in the floor and the column head 1: 2: 4 mix is used. The saving in the cost of spiral would be small in comparison with the advantage of having a continuous spiral column; therefore, the rule given above is recommended. Cost of Spiral.—The cost of spiral per ton is larger than the cost of bars per ton, to allow for the shopwork of bending and assembling. Also, the unit cost of drawn wire is larger than the unit cost of bars. The difference usually amounts to from $15 to $20 per ton. In figuring relative cost of spiral and vertical bars, it should be remembered that the length of vertical steel is greater than the length of spiral. The length of vertical bars equals the story height plus the lap of the bar while the length of spiral equals the story height minus 6 in. The cost of the spacers, if used, should be added to the cost of the spiral. Details of Vertical Bars.—Vertical bars should be extended above the top of the rough slab into the upper column, a sufficient amount to transmit by bond the compressive stresses from the column above to the column below. The remarks on p. 412 in con- - nection with vertical bars for columns with longitudinal steel, apply here also. LONG COLUMNS If the ratio of unsupported length of a column to the least side of smallest radius of gyration exceeds the values on which the formulas given in previous pages are based, the column becomes a long column and must be designed with reduced unit stresses. The formula given by the 1924 Joint Committee is recommended. Let P; = total safe load on long column; P = total safe load on short column (F not more than 40) ; 1 = unsupported length of column in inches; R = least radius of gyration of effective area in inches; f.1 = allowable unit stress on long column; f. = allowable unit stress on short column. STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE 485 Then ’ | R marie 2), sli doy etl hte (39) and R ie =f, Wea aa 120 suns ° abe . (40) It should be noticed that for : = 40, the expression in the brackets becomes unity. COLUMNS FOR FLAT SLAB CONSTRUCTION Special requirements for columns in flat slab construction are given in the chapter on “‘ Design of Flat Slab Structures.”’ STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE When small structural shapes are used and their area does not exceed 6 per cent of the effective cross section of the concrete, the column should be treated as a reinforced concrete column and designed according to the formulas given for columns with longitudinal bars, on p. 406. The angles should be fastened together with lattice work or stays, such as are used in structural steel design. The steel shapes should be covered with concrete of a minimum thickness of 2in. To prevent the separation of this fireproofing from the rest of the concrete, it is advisable to place outside of the angles 3 in. round ties spaced about 18 in. on centers. When the structural shapes are designed to resist the major part of the column load, the column becomes, in the strict sense, a struc- tural steel column imbedded in concrete. Its design differs mate- rially from that of the reinforced concrete column. The formulas for structural steel columns are not yet fully standardized and differ in building rules of different cities. Formulas recommended by the authors, and the requirements of several large cities, are given below. JOINT CoMMITTEE, 1924, RuLES ror STRUCTURAL STEEL AND Cast- IRON COLUMNS, INDORSED By AUTHORS The Joint Committee, 1924, provides rules for composite columns and structural steel columns. The allowable stresses for both types 436 CONCRETE AND REINFORCED CONCRETE COLUMNS are the same, the difference being in the requirement as to details. These rules are indorsed by the authors. Let f’. = ultimate compressive strength of conerete * at 28 days, lb. per sq. in. f, = compressive unit stress in metal core, lb. per sq. in.; A = area of concrete core, sq. in.; A, = area of structural steel, sq. in. ; P = total safe axial load, lb.; R = least radius of gyration of the steel or cast-iron section, in. h = height of column, in. Composite Columns.—Composite columns are columns consisting of structural steel or cast iron encased in a spirally reinforced con- crete core. : The following formulas should be used for design of composite columns: Safe Load on Composite Column, P =(A — A,)fo +pAgie pe where Allowable Unit Stress in Concrete, f, = 0.25f". ys eee (42) Allowable Compressive Unit Stress on Steel Section, | f, = 18.000 — 705, a Se ee ema an but shall not exceed 16 000 Ib. per sq. in. Allowable Unit Stress on Cast Iron, J, = 12.000 — 605, Me ren rn ee but shall not exceed 10 000 lb. per sq. in. The diameter of the cast-iron section shall not exceed one-half of the diameter of the core within the spiral. The spiral reinforce- ment shall be not less than 0.5 per cent of the volume of the core within the spiral and shall conform in quality, spacing, and other requirements to the provisions for spiral columns. 3 Ultimate strength based on tests of 6 by 12 in. or 8 by 16 in. cylinders. STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE 437 Ample section of concrete and continuity of reinforcement shall be provided at the junction with beams or girders. The area of the concrete between the spiral and the metal core shall be not less than that required to carry the total floor load of the story above, on the basis of a stress in the concrete of 0.35f’-, unless special brackets are arranged on the metal core to receive directly the beam or slab loads. Structural Steel Columns.—A structural steel column is a column consisting of a structural steel section which fully encases an area of concrete, and which is protected by an outside shell of concrete at least 3 in. thick. . Formulas for safe load and unit stresses are the same as for com- posite columns. The outside shell shall be reinforced by wire, mesh, ties, or spiral weighing not less than 0.2 lb. per sq. ft. of surface of the core, and with a maximum spacing of 6 in. between strands or hoops. Special brackets shall be used to receive the entire floor load at each story. The safe unit stress in steel columns, calculated by Formula (43), p. 436, shall not exceed 16,000 lb. per sq. in. RULES FOR STRUCTURAL STEEL or Various Burtpina CopEs 4 New York Code Regulation for Structural Steel and Concrete.— In columns of structural steel, thoroughly encased in concrete not less than 4 inches thick and reinforced with not less than one per cent of steel, the allowable load shall be 16 000 lb. per sq. in. on the structural steel, the percentage of reinforcement being the volume of the reinforcing steel divided by the volume of the concrete enclosed by the reinforcing steel. Not more than one-half of the reinforcing steel shall be placed vertically. The reinforcing steel shall not be placed nearer than one inch to the structural steel or to the outer surface of the concrete. The ratio of length to least radius of gyra- tion of structural steel section shall not exceed 120. Dea gee GOOG cli te ee Bis hts Oo eee This regulation differs from the Joint Committee Rules in one essential, namely, that no direct strength is attributed to the con- crete, the increase being taken care of by increased unit stress in steel above that allowed for ordinary structural steel columns. Philadelphia Regulation for Structural Steel and Concrete,— The Philadelphia ruling follows closely the New York rule, except * Building Codes are given as of 1925. As changes occur they will be made in subsequent imprints of this treatise. 4383 CONCRETE AND REINFORCED CONCRETE COLUMNS that the required sum of lateral longitudinal reinforcement outside of the shapes is to be 1 per cent of the concrete within the hoops, instead of within the steel section as required in the New York Code. Boston Rule.—No increase in strength is attributed to the ‘mbedment of structural steel in concrete. It is permissible, how- ever, to utilize the concrete casing to carry the load, for one or more stories between the brackets on structural steel. The rule reads: “ Reinforced concrete buildings may be supported by structural steel or cast-iron columns, fireproofed in first-class construction as provided elsewhere in this act. Brackets shall be provided to trans- mit the load from the floors to the column. Such columns shall be computed as follows: “ (qa) If the brackets are placed immediately below the floor, the structural steel or cast-iron columns shall be assumed to carry the load of all the floors above. “(b) If the brackets are placed immediately above a. floor, the structural steel or cast-iron columns shall be assumed to carry all the load above the brackets and the floor or floors below the brackets shall be carried on reinforced concrete encasing the metal, designed in accordance with the requirements of this act, to the next bracket below or to the foundation. In this case, however, the surrounding concrete shall be so separated from the steel or cast iron as to permit the separate action of both. “ Circular hollow steel or wrought-iron columns filled with con- crete shall be allowed to carry a load equal to the capacity of the metal casing plus the capacity of the concrete filling. The average unit stress in the casing shall be that specified elsewhere in this act for columns, and that in the concrete filling shall be in the same ratio to the unit stress in the casing which the modulus of elasticity of the concrete bears to that of the casing.” This rule is not logical, as it entirely neglects the effect of the concrete surrounding the steel, although it acknowledges its strength by allowing it to carry the floor loads when the concrete is separated from the steel section. Cleveland Code.—The Cleveland Code divides the structural steel columns into two classes: the columns with solid web, and the columns with open web. For solid-web columns, no allowance is made for the strengthening effect of the concrete, on the assumption that no bond can exist between large flat surfaces of the steel section and the concrete. They must be designed with the same stresses as ordinary structural steel columns. For open-web columns, the unit stresses for the steel are increased over those allowed for struc- STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE 439 tural columns. The rate of increase depends upon the type of col- umn and also upon the ratio of the steel area to the area of con- crete. The larger the percentage, the smaller is the allowable increase in unit stresses. This last provision is logical. The con- crete section has a definite strength and rigidity. When a structural column is imbedded therein, the total strengthening effect of the concrete, within limits used in practice, will be practically the same, irrespective of the magnitude of the area of the steel column. The effect per unit area of the steel section will be larger if the steel sec- tion forms a small part of the total section and diminishes with the increase of the section. The rule reads as follows: * Structural Steel Columns Encased in Concrete.— “(a) Solid Web Columns.—In columns of this type, the stcel shall be designed to take the total dead and live loads, and the con- crete shall be considered as fireproofing only. All columns of this type shall be wrapped with wire in such a manner as to securely hold the concrete in position. All loose scale or rust shall be removed before encasing the column in concrete. “(b) Open Web Columns.—In columns of this type, the steel shall be designed to take the full dead and live loads with the follow- ing unit stresses, with no added allowance for the concrete: “Gray columns and similar types, 20 000 — 300 X the percentage of‘steel. “ For four-angle columns, latticed four sides, 19 000 — 300 X the percentage of steel. “For four angles with latticed web, 17 500 — 300 X the percentage of steel. ‘“Open-web columns shall be wrapped with one-eighth (4) inch (or larger) wire at vertical intervals not greater than eight (8) inches. “The percentage of steel shall be based on the total area of the column, after deducting two (2) inches all around. “If the unsupported length of columns exceeds twelve (12 times the least outside dimension, the stresses shall be reduced as required for reinforced concrete columns. ‘In all columns under this section, positive connections shall be provided to transmit to the column steel the loads of all rod rein- forced beams and girders framing into the column flanges.” Chicago Requirements.—The Chicago Code provides for three combinations of structural steel with concrete. First.—‘‘Structural steel of box shape with lattice or batten plates filled with concrete. If no structural shape is less than 1 sq. in. in section and the spacing of the lacing or battens is not greater than 440 CONCRETE AND REINFORCED CONCRETE COLUMNS the least width of the columns, and the structural steel does not exceed 8 per cent of the area, then the column may be considered as a reinforced concrete column and designed by regular formulas in which the stress in concrete equals one-quarter (4) of the ultimate strength of concrete and the unit stress in steel equals the unit stress in concrete multiplied by ratio n.”’ Second.— For steel columns filled with, and encased in concrete extending at least three inches beyond the outer edge of the steel, where the steel is calculated to carry the entire live and dead load, the allowable stress per square inch shall be determined by the following formula: 18 000.--.70-2, 1b.,°. sea but shall not exceed 16 000 pounds.” | Third.—“ For steel columns filled with, but not encased in con- crete, the steel shall be calculated to carry the entire live and dead load. In this case, the above formula may be used, but the allowable stresses shall not exceed 14 000 pounds.” The above stresses should be compared with the following stresses allowed for structural steel columns not encased in concrete. 16 000 — 705 lb. persqy inst 2) a eaee eae In the above formula L = length of column in inches; R = least radius of gyration in inches. Deraits ofr DESIGN OF STEEL SECTION The details of the steel section depend largely upon conditions and upon the building department requirements. Naturally, when no allowance is made for the effect of the concrete surrounding the steel, the main object will be to get the best and cheapest steel sec- tion. On the other hand, when concrete is relied upon to resist stresses, it is desirable to get cooperation between the two materials, and a section must be selected which will provide best bond between the concrete and steel. The simplicity of the brackets and splices should also be taken into consideration, and a section selected which requires the least complicated brackets at the floor levels to transmit the floor load to the column. . Figure 149, a to f, inclusive, give the various sections in use. The cheapest section is likely to be the one requiring least shopwork and the one in which no material is lost for stay plates and lattice STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE 441 work. Such is the section shown in Fig. 149, e. The objection to it is that it cannot be counted upon to cooperate with the con- erete. In using this section, it is important to bind the concrete behind the flanges by means of ties, as otherwise the thin concrete may spall off. This section is accepted by the City of New York and by Philadelphia. In Cleveland, no increase of strength is allowed on such section. It can be used in connection with the Boston law, except that when concrete is depended upon to carry the floor load from floor to floor, the concrete behind the flanges cannot be con- sidered, as it is too thin. Y7 LLikifg lls - Tie or Spiral” _-Tie ~. Tie or Spiral Fig. 149.—Structural Steel Columns Encased in Concrere: (See p. 440.) The Gray Column, Fig. 149, b, is the best type of a concrete column. ‘The objection to it is the large cost of fabrication. The Star Shape Column, Fig. 149, d, has been used to some extent, as it is compact and fits well into a round column. It is par- ticularly adaptable under the Boston Law. It should not be used where considerable bending is expected unless bending moment is resisted by vertical bars. Splicing Structural Columns.—Splices should be designed in the same manner as for steel columns. No reliance should be placed on the concrete. The splices are best placed above the floor level. 442 CONCRETE AND REINFORCED CONCRETE COLUMNS The design of splices will depend upon whether the ends are machined or not. If the ends are not machined, to insure good bearing on the lower section, the splice must be strong enough to transfer the full load through the rivets to the lower column. ‘The number of rivets must be large enough to transfer the load from the column to the splice and then from the splice to the lower section. , If the ends are machined, as is generally the case, the load from the upper column is transferred directly by bearing to the section of the lower column. Only enough strength is required in the splice to keep the section securely in place during construction and to resist any bending moment to which the column may be subjected. In some types of columns, the concrete section may be counted upon to resist the bending moment, in connection with a proper amount of bars used to resist any possible tension. Steel Brackets.—The construction of steel brackets to transmit the floor load to the steel column depends upon the cross section of the steel column and also upon the construction of the floor. With a closed section, such as Fig. 149, d, ¢, or f, it is neces- sary to provide brackets at every floor for the load coming on the flange sides. In other sections, brackets may be placed only every two or three floors and concrete relied upon to transmit the load partly by bond to the steel section and partly to the brackets. In such cases, of course, the concrete must be strong enough to carry the floor loads as a column between the brackets. Under no circum- - stances should the brackets be omitted altogether and the bond relied upon to transfer the entire load. In flat slab construction, the brackets are usually placed within the column head. Angles properly riveted are sufficient. For heavy loads, stiffeners under the angles may be required. The angles should be placed in such a position that they will not inter- fere with the pouring of the concrete. This fixes their position away inside of the column head where the width of the column head form is much larger than the column, so that there is room for the concrete to flow into the column form. In concreting, care should be taken that the column is properly filled with concrete, especially when it 1s counted upon, directly or indirectly, to strengthen the column. In beam and girder construction, brackets should be supplied for every girder. These should be strong enough to transmit all the load to the steel section. When concrete is relied upon to transmit STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE 443 the load to the story below, it should be of proper dimensions and strength to carry the load as a column. Examples of brackets are shown in Figs. 150 and 151. Fig. 150.—Steel Brackets for Floor Beams. (See p. 442.) 4-33 ZZ pana RAS Fig. 151.—Steel Brackets for Flat Slab. (See p. 442.) Details of Base for the Steel Section.—The load from the steel section is transmitted to the concrete of the foundation by means of a base designed in the same manner as for structural steel columns, 444 CONCRETE AND REINFORCED CONCRETE COLUMNS The base for a structural column is either cast or built up of plates and angles. Where easily obtainable, a base of thick solid slabs of rolled steel may be used. The thickness of the plate may run up to 6 in. The problem in designing the base is to spread the load from the column on a sufficiently large area of the concrete foundation, so as not to exceed the allowable bearing stresses on concrete. After the area of the base is determined by dividing the column load by the allowable bearing stress, the base plate is strengthened, by angles in built-up bases, and in cast-iron bases by ribs of sufficient proportions to resist the stresses produced by the upward pressure of the foun- dation. The thickness of solid rolled plate is determined by the bending moment due to the upward reaction. It may be obtained from the following formula. | Let P = column load, lb.; a and b = dimensions of foundation plate, in.; c and d = outside dimensions of column, in.; t = thickness of plate. Then, for a stress in steel of 16 000 Ib. per sq. in., Thickness of Foundation Plate, (48) ata alti 3, P(b — d) t= 4] 213006 q 21 300a ~ Cast Iron wr Fig, 152.—Bases for Steel Columns. (See p. 443.) For square plates, both dimensions are equal. For rectangular plates, the larger of the two values should be accepted. ECONOMIES IN COLUMN DESIGN 445 The column must be bolted or riveted to the base. The bases must also be bolted to the foundation. If uplift is possible, the bolt must be strong enough to resist it. The bolts should be carried so far into the concrete, and anchored in such a way, that enough weight of concrete above such anchorage is available to counteract the uplift. Figure (a), (b), and (c) shows bases adaptable for steel colums. Cast-iron Core Imbedded in Con- crete.—A section of a column con- sisting of a cast-iron core imbedded in concrete is shown in Fig. 153. The cast-iron core is spliced by means of a pipe sleeve fitted on the outside Fic. 153.—Details of Column with of the core. Where the diameter of ©28t-iron Core. (See p. 445.) the core changes, special sleeve is provided. The ends of the core must be faced to even bearing on the lower core, _- Cast Iron . _Vertical Bars \---Spiral ECONOMIES IN COLUMN DESIGN Economical Dimensions and Concrete Proportions.—In a one- story structure, the problem is to select a column for which the cost of materials is a minimum. For structures several stories high, the additional question of formwork is introduced into the problem. The column load varies from story to story. If the cost of materials alone were considered, the greatest economy would be obtained by changing the size of the column at every story. However, the reduc- tion in column size requires remaking of column forms (except where they are easily adjustable) and also the lengthening of forms for beams and girders running into the column. Changing of column size is economical when the cost of these changes in formwork is smaller than the saving in materials resulting from the reduction in the size of the column. In many cases, the changes in formwork cost more than the saving in materials. No general rule can be established, since the relation between the cost of materials and that of labor varies greatly in different localities. As a general proposition, it is safe to say that the columns should be carried the same size for at least two stories. In many instances, the same size of column may 446 CONCRETE AND REINFORCED CONCRETE COLUMNS be maintained through several stories, by successively reducing the amount of reinforcement and changing the mix of concrete. The above suggestions do not apply to round columns, especially in flat slab construction. The column and column-head forms in such cases are usually made of sheet metal. As they are readily adjustable, no saving in cost is obtained by maintaining the size of the column unchanged. It should be remembered, however, that for standard forms the diameter of column is adjustable only for even inches; therefore, the diameter of column should always be in even inches. To get best results, it is desirable that the designer should become acquainted with cost data in the location in which the job is to be built. It is good practice to use the same mix of concrete for all columns on one floor. An exception may be made in case of wall columns, for all of which leaner mix may be used than in all interior columns. Confusion may result if some of the interior columns are made of one mix and the rest of some other mix. » Wall columns are usually rectangular in shape. To reduce form- work, it is advisable to maintain the same width of the column through its whole height and to effect the change in cross section by changing the thickness of the column. ‘This method reduces the cost of remak- ing formwork, as all remaking is confined to two sides; also the forms for spandrels are used without remaking. This suggestion does not hold in cases where, for architectural considerations, special designs of column are required. In such cases, the shape of the concrete column must conform to the shape of the finished pier. Where possible, the concrete column should extend to the edge of the window sash, and recesses should be provided in the concrete to ‘receive the sash. The width of the column should then be fixed exactly after the width of the sash is determined. Very often, this width is in fraction of an inch as it is obtained by subtracting from the span the fixed width of the sash. It is desirable to use columns of the same diameter throughout each floor of the building. When the loads of some odd columns in a floor are smaller than those of the typical column, but the difference is not appreciable, it may be advantageous to use not only the same concrete size, but also the same amount of steel as for the typical column, especially when it is difficult to distinguish the odd columns, by any peculiarity in their location, from typical columns. If a ECONOMIES IN COLUMN DESIGN 447 a smaller amount of steel is used for odd columns there is danger of interchanging the special reinforcement with the reinforcement of the typical column, to the detriment of the latter. Of course, this does not apply to special columns, such as columns in elevator walls and staircases, which, because of their peculiar locations, are not likely to be confused with the typical columns. Economical Mix of Concrete.—Under average conditions, the cost of 1 : 1 : 2 mix of concrete in place ° is about 14 per cent greater than that of the 1: 14:3 mix, while the strength of 1:1: 2 is 18 to 21 per cent greater than that of the 1: 15:3. The cost of the 1:1: 2 is 24 per cent greater than that of the 1 : 2: 4 mix, and the strength of the 1: 1: 2 is 45 to 50 per cent greater than that of the 1:2:4. The cost of the 1: 15:3 mix is about 8 per cent greater than that of the 1 : 2: 4 mix, and the strength is 20 to 27 per cent greater. : From comparison of the increase in cost of richer mixture with the increase in allowable stresses, it is evident that the percentage of increase in strength is much larger than the percentage of increase in cost. Therefore, the richer mix is more economical for columns than the leaner mix. To obtain best economy, therefore, a rich mix of concrete should be used, except in top columns where the size may be governed by the requirements as to minimum size and 1:2:4 concrete gives sufficient strength. For lower columns, - instead of increasing the size of the column, richer mix should be used. Economical Design of Columns with Longitudinal Reinforce- ment.—The cost of steel in a column is forty to fifty times larger than the cost of concrete of the same volume, depending upon the mix of concrete. The smaller value is for 1: 2:4 mix and the larger value for 1: 1:2 mix. The allowable stress on vertical bars is only from ten to fifteen times larger than the allowable stress on concrete. It follows that it is much cheaper to use concrete than steel to support the load. Therefore, if it is required to increase the carrying capacity of a column, it is cheaper, so far as the cost of materials is concerned to increase the size of the column than to add vertical steel. The most economical column, then, is one in which the minimum > The cost of concrete used in comparison includes cost of materials, cost of plant, and cost of mixing and placing. Cost of formwork is not included. Richer concrete requires smaller columns. In some cases, this reduces the cost of form- work, thus increasing the advantage of rich mix. 448 CONCRETE AND REINFORCED CONCRETE COLUMNS permissible amount of vertical steel is used. Some building codes allow 4 per cent of vertical bars as a minimum. ‘The authors do not recommend the use of a smaller amount of vertical steel than 1 per cent of the effective area of the column. Exception may be made to this rule in case of wall columns in which the column serves as a pilaster, and its area, as determined by architectural require- ments, is larger than required by the stresses. In such cases, the area of column required by the load for the specified stresses and for 1 per cent of steel should be computed and the area of steel to be used should be made equal to 1 per cent of the area of column as determined. (See example 4, p. 410.) The steel so obtained should be distributed over the whole section and the proper amount of ties used. If the column is also subjected to bending, proper bending reinforcement should be provided. | Economical Design of Spiral Columns.—The economical design of spiral columns will differ for the different types of formulas now in use. The economy of columns designed according to the formulas given on pp. 421 to 426 is discussed below. : Columns with Fixed Percentage of Spiral—Under this head come the columns designed according to the authors’ reeommenda- tions and 1916 Joint Committee recommendations and according to Boston, Cleveland, and Philadelphia Codes. The formulas used in design of spiral columns are the same as for columns with vertical steel only, with the exception that larger unit stresses are allowed. The relation between the ratio of cost of steel to cost of concrete and the ratio of unit stresses is same as for columns with vertical steel only. As in the case of columns with bars only, the most economical spiral column is one with minimum allowable amount of vertical steel. Rich mix of concrete also is more economical than lean mix, as the allowable strength increases more than the cost of concrete. Columns with Variable Percentage of Spirals.—Under this head come the rules of Chicago and New York. Chicago Code-—Formula (20), p. 425 [P=Af.-+ n—- Ase + 2inpiAf-] has three variables: area of concrete, percentage of vertical bars, and percentage of spirals. The effectiveness of vertical bars is the same as for columns with vertical bars only. They are, therefore, equally uneconomical means of increasing the strength in both cases. The spiral reinforcement in the formula is considered as two and one-half times as effective as vertical bars of equal volume. It is, therefore, a much more economical means of increasing the ECONOMIES IN COLUMN DESIGN A449 strength of the column than the vertical bars. Comparing with concrete, the increase in strength produced by the spiral is 2.5n _ times’as large as that produced by the same amount of concrete. This ratio equals 25 for rich concrete and n = 10, and 37.5 for lean concrete and n= 15. The ratio of the cost of spiral to that of concrete of equal volume is almost always larger than the ratio of increase in strength, consequently, it is cheaper to increase the strength of column by increasing its size than by increasing the per- centage of spiral, especially as an increase in the percentage of spiral reinforcement requires an equal increase in vertical steel. The most economical spiral column designed in accordance with the Chicago Code, then, is the one which has the smallest amount of spiral and vertical steel, but in which the concrete is of the richest mix. When the size of column is fixed, the cheapest means of increas- ing its strength is by an increase in the amount of spiral reinforce- ment and a proportional increase of vertical steel, rather than by an increase in the amount of vertical steel alone. New York Code——In Formula (16), p. 424, for-spiral columns ac- cording to this code [P = Af, + (n — 1) Asfe-+ 2p, Af,] the effect of the spiral reinforcement is sixty-seven times larger than that of 1: 11:3 concrete of the same volume and eighty times larger than that of 1: 2:4 concrete of the same volume.® Since the ratio of the cost of steel to the cost of concrete is usually smaller than the above values, it is evident that the spiral reinforcement is the most eco- nomical material in the make-up of the column so designed. The vertical steel, as in the previous cases, is the least economical material. The most economical column, then, is one in which the percentage of the spiral is the maximum allowable and the amount of vertical steel the least allowable. The richer mix is also more economical than the leaner mix for the reasons given in connection with col- umns with vertical bars. Although there is no requirement fixing the ratio of the percentage of vertical steel to that of the spiral, for good design it is advisable to use at least as large a percentage of vertical bars as of spirals. Relative Economy of Spiral Columns and Columns with Vertical Steel Only.—The relative economy of the two types of columns depends upon the formulas used in designing. As a general proposi- 2 6 This ratio is obtained from = using fs = 20 000 Ib. per sq. in. and te = 500 Cc and 600 Ib. per sq. in. respectively. 450 CONCRETE AND REINFORCED CONCRETE COLUMNS tion, spiral columns are more economical than columns with vertical steel only, with the exception of columns designed by the New York Code, as explained below. Boston Cleveland, and Philadelphia Codes.—In columns designed in accordance with these codes, spiral amounting to 1 per cent of the effective area is used, and the effect of the spiral is taken care of by an increase in the allowable unit stresses on concrete. Referring to tables on p. 409 and on p. 423, this increase amounts to more than 50 per cent of the stresses used for columns without spiral reinforce- ment. For instance, the allowable stress for 1: 2:4 concrete, according to the Boston Code, is 495 lb. per sq. in., while for spiral columns the allowable stress is 770 lb. per sq. in. The increase in strength amounts to 275 lb., which is 56 per cent of the lower value. In addi- tion, the unit stress in vertical steel is increased in the same ratio. To get the relative economy of a spiral column and of a column with vertical bars only, it is necessary to determine the increase in cost produced by the use of the spiral. This depends upon the ratio of the cost of spiral to the cost of concrete. Under present conditions (year 1925), the ratio of the cost of spiral to the cost of equal volume of concrete is 45 for 1 : 2: 4 concrete, A1.5 for 1: 14: 3 concrete, and! 36.for te: li 22 ‘contrete ier these unit costs, 1 per cent of spiral increases the cost of column by 36 to 45 per cent of the cost of the concrete within the effective area. Comparing the imcrease in allowable stresses (56 per cent) with the increase in cost (386 to 45 per cent), it-is evident that the increase in strength is much larger than the increase in cost; consequently, a spiral column is more economical than a column with vertical bars only. It may also be stated, on the basis of the above and of the discussion on p. 447, that the most economical column designed according, to the above codes is a spiral column of richest mix having the minimum allowable percentage of steel. New York Code.—According to the New York Code, the spiral reinforcement does not affect the unit stresses in concrete. It increases the strength of the column directly, the increase being expressed by the term Qi Afs (see p. 424). As explained on p. 449, the ratio of increase in strength effected by the spiral reinforcement is 7 Based on cost of spiral in place of 41 cents per lb. and on following costs of concrete per cu. ft. in place: 1:2:4 mix, 50 cents; 1: 14:3 mix, 54 cents; and 1:1: 2 mix, 62 cents. ECONOMIES IN COLUMN DESIGN 451 larger than the increase in cost. The deduction would be that spiral columns are more economical than columns with vertical bars only. This is not always true, however, because in computing the strength of a column with vertical bars only, the Code does not require any allowance for fireproofing, but permits the use of the total cross section of the column as effective area. In spiral columns, on the other hand, 2-in. fireproofing is required, so that for equal diam- eter of column the effective area of a spiral column is less than the effective area of a column with vertical bars only. The difference is particularly large in columns of small diameter. For this reason, spiral columns designed as explained above are not economical for small diameters of columns although they are economical for large diameters. Chicago Code.—According to the Chicago Code, spiral reinforce- ment increases the strength of column directly, as expressed by the term 23npiAf,, and indirectly by increasing the allowable unit stresses. The direct increase in strength due to the spiral equals twenty-five times the strength of an equal volume of concrete for 1: 1 : 2 mix; ov times for 1:13:38 mix; and 374 times for 1:2:4 mix. This means, for 1 per cent of spiral, an increase in the average strength in concrete over the whole effective area amounting to 25 to 373 per cent above the allowable stress on concrete for spiral columns, and 31 to 47 per cent above the allowable stress on columns with vertical steel only. The indirect increase in strength is caused by the larger allowable unit stress, f-, on spiral columns. While, for columns with vertical steel only, the allowable unit stress equals 3 of the ultimate compressive strength, the working stress in spiral columns amounts to } of the ultimate strength. This means an increase in working stresses, due to the spiral rein- forcement, of 25 per cent, not only for the concrete, but also for the vertical steel. When the direct and the indirect increases in strength, produced by the spiral, are added, the total increase in strength of a spiral column over a column with vertical steel only is found to be from 56 to 72 per cent, depending upon the mix of concrete. The increase in cost due to 1 per cent of spiral amounts to 36 to 45 per cent of the cost of concrete, as explained on p. 449. Comparing the increase in cost of concrete with the increase in strength of column, it is evident that the increase in strength is larger than the increase in cost. Therefore, a spiral column is more economical than a column with vertical steel only. 452 CONCRETE AND REINFORCED CONCRETE COLUMNS Economy of Structural Steel Columns Imbedded in Concrete.— The use of structural steel in columns is not economical. It should be restricted to cases where a reduction in size of column is required, beyond that possible with spiral columns of rich mix and with the largest. possible percentage of vertical steel. Usually, the struc- tural steel is used in the lowest floors only, while in upper floors concrete columns are substituted as soon as the reduction in load permits it. The objections to structural steel columns are as follows: Greater Cost of Structural Steel Columns as Compared with Rein- forced Concrete Columns.—Structural steel columns imbedded in concrete always cost more than reinforced concrete columns eco- nomically designed to carry the same load. ‘The increase in cost depenYls upon the specification used. It 1s, of course, much larger for specifications in which no allowance is made for the strengthening effect of the concrete. With the formulas recommended by the authors, the increase in cost is considerably reduced, as both steel and concrete are utilized in carrying the load. | Difficulties in Transferring the Floor Load and Tying the Floor to the Column.—These are particularly serious in the case of beam and girder construction, where it may be difficult to make the beams and girders continuous. In flat slab construction, the difficulties are considerably reduced, especially with systems of flat slab reinforce- ment in which a small number of bars, or none at all, cross the column (see p. 369). Difficulties in Erecting Steel Columns.—The steel cores are very heavy, usually weighing more than a ton. This offers a special problem in erection in a reinforced concrete job not equipped with facilities for lifting heavy members in the manner used in structural steel buildings. Therefore, it is either necessary to introduce special equipment on the job for handling the steel cores or to handle them at large expense without proper equipment. In either case, it is costly. To reduce the weight, it is advisable to limit the height of the steel cores to one story. REDUCTION OF LIVE LOAD IN BUILDINGS In designing columns, all the dead load and a certain proportion of the live load shall be used. By dead load is meant the weight of the floor system, floor finish, fixed partitions or walls, columns, REDUCTION OF LOVE LOAD IN BUILDINGS 453 and other fixed loads. Live load is the load that comes on the floor in the course of its use. It is generally recognized that a larger live load must be allowed for the floor system than for columns, because overloading of a small section of the building is much more likely to occur than simultaneous overloading of a number of floors carried by a column. The likelihood of the latter occurrence decreases with the increase in the number of floors to be supported. To allow for this, the reduction is small for the columns in the top floors and increases with the number of floors. Authors’ Recommendation.—The authors recommend that in all buildings the columns be designed for the total dead load, plus a proportion of total live load above the columns given in the table below. In this table are given proportions to be used for ordinary buildings and also larger proportions for warehouses and ReaNY. manufacturing buildings. Recommended Proportion of Live Load on Columns Proportion of Live Load on Columns Number of Floors Carried by Seine ney, Warehouses Buildings 1 1.00 1.00 2 0.85 0.90 3 0.80 0.87 4 0.75 0.84 5 0.70 0.81 6 0.65 0.78 7 0.60 0.75 8 0.55 . 0.75 9 0.50 0.75 10 0.50 0.75 For use of this table see Example 6 on page 456. Usually, no reduction is allowed for warehouses, the idea being that in a warehouse all floors are apt to be fully loaded at the same time. However, the chances of overloading a column in a ware- house are unquestionably much smaller than the chances of over- A454. CONCRETE AND REINFORCED CONCRETE COLUMNS loading a floor panel. In all warehouses, moreover, free aisles must be left for moving the goods. These two considerations convince the authors that some reduction is permissible for warehouses. The reduction recommended is smaller than for ordinary buildings. Boston Code.—The Boston Code makes the following provisions for reduction of live load: In all buildings except storage buildings, wholesale stores, and public garages, for all columns, girders, trusses, walls, piers, and foundations. Carrying one floor........ ashe kites wien ee No reduction. Carrying two floors..........+seeeeeeeees 25 per cent reduction. Carrying three floors..........++++++++++: 40 per cent reduction. ‘Carrying four floors2+). 7.325264 ao aaeae 50 per cent reduction. Carrying’ five floors mee .4 sae ote 55 per cent reduction. Carrying six floors or more........--.---- 60 per cent reduction. In public garages, for all flat slabs of over three hundred square feet area reinforced in more than one direction, and for all floor beams, girders and trusses carrying over three hundred square feet of floor, and for all columns, walls, piers, and foundations, twenty- five per cent reduction. New York Code.—New York allows a reduction of live loads only for buildings over five stories high. ‘The Code states: Loads on Vertical Supports—Every column, post, or other verti- cal support shall be of sufficient strength to bear safely the combined live and dead loads of such portions of each and every floor as depend upon it for support, except that in buildings more than five stories in height the live load on the floor next below the top floor may be assumed at ninety-five per cent of the allowable live load, on the next lower floor at ninety per cent, and on each succeeding lower floor at correspondingly decreasing percentages, provided that in no case shall less than fifty per cent of the allowable live load be assumed. Chicago Code.—The Chicago Code gives the following rules for reduction of live loads on columns: Walls, Piers and Columns—Dead and Live Loads.—(a) The full live load on roofs of all buildings shall be taken on walls, piers, and columns. REDUCTION OF LIVE LOAD IN BUILDINGS 455 (b) The walls, piers, and columns of all buildings shall be designed to carry the full dead loads and not less than the proportion of the live load given in the following table: Floor 17 | 16/15/14) 13) 12) 11/10 % 17 85] % 16 60! 85] % 15 75| 80| 85] % 14 70) 75| 80| 85! % 13 65| 70] 75] 80| 85 12 60| 65] 70} 75] 80 11 55| 60] 65] 70) 75 10 50} 55] 60| 65| 70 9 50! 50] 55) 60] 65 8 50] 50! 50! 55] 60 7 50| 50) 50) 50) 55 6 50} 50) 50) 50| 50 5 50) 50] 50] 50] 50 4 50| 50) 50) 50) 50 3 50) 50} 50] 50) 50 2 50) 50) 50] 50] 50 1 50) 50) 50} 50| 50 Zo 85 80 75 70 65 60 55 50 50 50 50 Zo 85 80 75 70 65 60 55 50 50 50 Griese ere Golo, pete Bee 2! ae SS) SS a ee SS To 85 80 75 70 65 60 55 50 50 Zo 85 80 75 70 65 60 55 50 To 85 80 75 70 65 60 55 To 85 80 75 70 65 60 Jo 85| % 80} 85} % 75| 80} 85 (c) The proportion of the live load on walls, piers, and columns on buildings more than seventeen stories in height shall be taken in same ratio as the above table. (d) The entire dead load and the percentage of live load on base- ment columns, piers, and walls shall be taken in determining the stress in foundations. (c) In addition to the entire dead loads, not less than the follow- ing proportion of the percentage of live load on the basement columns, piers, and walls shall be taken in determining the number of piles for pile foundations and the area of concrete caissons. Classes I and VII Classes IT, III and VI eee ete eee eee ee eee eee ee ee ee we eee ocee eet ee eee eee eee ee ee ee ee we amy Mv fanideV LEL! Aire Ml SE el el BA Pg In all foundations eccentric loading must be provided for. 75 per cent. 50 per cent. 25 per cent. 456 CONCRETE AND REINFORCED CONCRETE COLUMNS DESIGN OF COLUMN IN A BUILDING SEVERAL STORIES HIGH The procedure in designing columns in a building several stories high is given in the example below. Example 6.——Design an interior column in a building with ‘six stories and basement, where the conditions are as follows: Panel, 20 ft. square; Live loads, 1st and 2nd floors, 250 lb. per sq. ft.; 3rd to 6th floors, 150 lb. per sq. ft.; Roof, 40 lb. per sq. ft. Concrete dimensions are: 1Istand 2nd Floor 3rd to 6th Floor Roof Slabs ceca t pate 4 in. 7k in. 6 in. Drop panel,.....°. 4 in. og in. 3. in. Dimension of drop panel, 7 ft. 6 in. square. Column head, 4 ft. 6 in. Story heights, 12 ft. from floor level to floor level. Floor finish consists of one inch granolithic applied separately. Stresses recommended by the authors on pages 407 and 421 will be used. Reduction of live load recommended for warehouses on page 453 will be used. Solution.—The loads per panel in each floor will be computed first. Live load and dead load will be kept separately. Area of panel, 20 X 20 = 400 sq. ft. Ist and 2nd Floor 3rd to 6th Floor Roof Live load......... 250X400=100kips 150X400=60kips 40400=16 kips Sigh load i 2eue ce. 106 100 75 Finish vias sie 12 be 18 Witt ae eee ee 118X 400 =47 112 400 =45 93 X 400 = 37 Drop panel........ 3 2 2 Total dead load... . 50 kips 47 kips 39 kips Kips is an abbreviation of ‘‘ Kilo pounds ” and means thousands of pounds. Further work will be simplified by tabulating the values as shown in table, page 457. After the design load is computed the size of columns is adopted in accordance with suggestions contained in this chapter. The required amount of reinforcement is finally found. 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Bo 3 ue 2s s BS suolsuduII(] 939 19N0D | 5 2 Shee ah 8 Cogr ‘d ay) NOISAGC NWNIOO ALAAMONOD GAOAOANIAA AO AIA WVXA 458 CONCRETE AND REINFORCED CONCRETE COLUMNS The table may be simplified by eliminating the items ‘“‘ Columns and Column Head.” This could be done by adding to the dead load of the floor the average weight of the column and column head. In case where no live load reduction is permitted and no division of live and dead load is desired for designing footings, the live and dead load may be combined. The first column in the table, then, gives the design load. For wall columns the weight of the curtain wall may be added to the slab load if it is constant in all floors, or treated as a separate item. COLUMNS SUBJECTED TO BENDING Bending in Interior Columns.—For equal spans and uniform loading, there will be no bending in the interior columns for the dead load or for the_live load extending over all spans. Some bending moment will be developed when the panels on one side of the column are loaded and those on the other side are not loaded. Usually, no allowance is made for the bending stresses thus produced. The bending moment in a column due to unbalanced live load may be determined as given in Volume III. When the spans are not equal, bending moments will be developed +n the interior columns both by dead and live loads. The bending stresses produced by this bending may be considerable and should be properly provided for. . Bending moments should be computed by formulas given in Volume III. ‘These should be combined with the axial load in the manner explained in connection with wall columns. If necessary, tension reinforcement should be provided. This is required at the top of the column near the face outside the largest span, and at the bottom near the face inside the largest span. Wall Columns.—When wall columns are connected by reinforce- ment with the floor construction, negative bending moment is devel- oped in the floor construction at the column. This bending moment produces bending stresses in the wall column. The wall column, then, is subjected to the direct compressive stresses produced by the superimposed loads and to the bending moment transferred from the floor to the column.® 8 See ‘Design of Wall Columns and End Beams” by Edward Smulski, Journal American Concrete Institute, July, 1910. COLUMNS SUBJECTED TO BENDING 459 The magnitude of the bending moment at the wall column for ordinary conditions is given on p. 280. For special conditions, the rules given in Volume III should be used. In some cases, additonal bending moments in the wall column are developed by eccentrically applied spandrel loads or by the weight of the cornice. These bending moments may be of the same sign as the bending moment transferred from the floor construction, in which case they should be added to it; or they may be of opposite sign, in which case they should be subtracted from it. In the top floor, the bending moment from the roof must be resisted by the top column alone. In the lower floors, part of the bending moment is resisted by the column below and part by the column above. The proportion of bending moment resisted by each of the two columns depends upon their rigidity, as expressed by : and a where J and J; are moments of inertia and h and h, are heights of the respective columns. If these ratios are equal or nearly equal, each column may be assumed as resisting one-half of the bend- ing moment from the floor. If the difference between the ratios of rigidity is appreciable, the proportion of the bending moment, /, resisted by the column having a ratio of stiffness equal to h is 7 i 7 M. For the other column, 1 hy I, the bending moment equals 7 hi 7 M. The sum of the bending non moments is M. 7 It should be noticed that the moment in the column below the floor is negative, i.e., producing tension on the outside face of the column; while the moment above the floor is positive, producing tension at the inner face of the column. The deflection of the wall column is shown in Fig. 154, p. 460. The points of maximum tension are indicated there. Points of Application of Maximum Bending Moment.—The maximum negative bending moment may be considered as applied at the bottom of the beam or girder of the floor construction which produces the bending, and the maximum positive bending moment 460 CONCRETE AND REINFORCED CONCRETE COLUMNS directly above the floor. Recommendation for flat slabs is given on p. 315. If the respective bending moments are plotted at the points of application of maximum moments, the straight line connect- _~ Beam before Bending 4 Tension. SE Ie hn, BER, a oa ‘Tension ~~Beam after Bending Column before Bending “Column after Bending 4~<-Tension Tension, Tension rhe ee \ _...Tension | “~~. Beam after Bending “---Qolumn before Bending eee \ ~--Column after Bending --Tension Fic. 154.—Deflection of End Beams and Wall Columns. (See p. 459.) ing them gives the variation of bend- ing moments in the column. Steps in Design of Exterior Col- umns.—After the column loads and bending moments are determined, it is necessary to com- pute the maximum — compression stresses for the combination of the concentric vertical column load | and the bending moment, and to de- termine the amount of tensile reinforce- ment required to resist the tensile stresses for the most unfavorable combi- nation of the con- centric column load and the bending moment. The first step is to determine the required dimensions for the column for the concentric col- umn load alone. Next, the maxi- — mum compression stresses are computed for the combination of maximum concentric column load and the maximum bending COLUMNS SUBJECTED TO BENDING 461 moment. Hither formula on p. 177, or formulas on p. 182, may be used, depending upon the amount of tensile stresses developed for this condition of loading. The maximum compression stresses thus obtained should not exceed the maximum allowable stresses as recommended on p. 463. In computing the combined stresses, the full section, including the fireproofing, may be used, in the same manner as in beams and slabs at the support. If the stresses exceed the allowable unit stresses the section should be increased. The final step is to determine the required amount of tensile steel to resist the bending moment. For this purpose, the com- bination of the vertical column load and the bending moment which produces maximum tension should be used. This combination will be different from that used for determining maximum compression. There is a difference of opinion as to the method of determining the required amount of reinforcement to resist the bending moment in the column. The following method, developed by the authors, is recommended for use: For the top columns, where the vertical column load is small in comparison with the bending moment, the amount of tensile steel should be determined for the bending moment alone, the effect of the vertical load being disregarded. The amount of steel is deter- mined from the beam formula, A, = i where d is the distance from center of tensile steel to the face of the column. If the amount of steel required by the column design is not sufficient, additional bars should be used. These do not need to extend the full length of the column, as the bending moment decreases rapidly. For columns in lower floors, where the total dead load carried by the column is appreciable, the following method for computing tensile stresses in the column is recommended by the authors: Assume that the structure above the floor under consideration is loaded with dead load only, but that the floor under consideration is loaded with dead and live load. Under such conditions, provide sufficient tensile strength in the column to give a factor of safety of 2 in case of overloading of the floor ‘with the structure above not loaded). For this purpose, compute the dead load on the column above the floor under consideration. Add to it double the reaction due to live and dead load from the floor. A62 CONCRETE AND REINFORCED CONCRETE COLUMNS Compute the bending moment in the column produced by the dead and live load on the floor under consideration. | Multiply this moment by 2 and combine it with the column load computed above. Provide such an amount of steel in the column that the tensile stresses will not exceed double the allowable tensile stresses in steel. This method gives the same factor of safety against cracks in column as is used in the floor construction. This method will be understood from the following explanation: The column load above the floor under investigation is not affected by the loading condition in that floor. It may easily happen during the life of a structure, and even during construction of the structure, that the wall panels of one floor are overloaded, while no appreciable live load is placed on the floors above. With such loading, only the dead load above the floor can be counted upon as reducing the tensile stresses. This dead load is constant. It will not increase with the overloading of the slab. When double the design load is placed on the slab, the column load will be increased by the reaction of the slab, but the column load above the slab will not be changed. For this reason the constant dead load plus the reaction of the over- loaded slab is combined with double bending moment. Anchoring of Bending Steel in Wall Column.—The de in the column used to resist bending stresses should be properly anchored, else the bars may pull out instead of resisting tension. In the top of the top story columns, the outside bars should extend above the points of maximum bending moment a sufficient length to develop the strength of bar in tension. If this is not pos- sible by straight imbedment, the bars should be properly hooked at the top in the slab. The inside bars are compression reinforcement. In intermediate stories, the outside bars at the top of any column are sufficiently developed when they are extended into the column above, the proper distance from points of maximum moment to develop the strength of the bar in tension. The inside bars of any column serve as compression reinforcement below the slab and as tensile reinforcement above the slab. The length of imbedment of these bars in the column above, measured from top of slab, should be such as to develop the full tensile strength in the bar. In all above cases, an imbedment equal to 40 diameters of the bar for deformed bars, and 50 diameters for plain bars, may be con- sidered as ample. COLUMNS SUBJECTED TO BENDING 463 Allowable Unit Stresses for Members Subjected to Axial Com- pression and Bending.—Considerable difference of opinion exists as to the allowable compressive unit stresses in members subjected to axial compression and bending. The maximum stresses are neither pure compression stresses nor pure bending stresses, but a combination of the two types. (For explanation of difference between the two types of stresses, see p. 30.) The proportion of the bending stress to the compression stress entering into the make-up of the maximum stress varies. It is 50 per cent or less of the total stress for condition of loading where no tension is developed in the sec- tion, and more than 50 per cent in columns where considerable tension is developed. The allowable compressive unit stress in the two cases. should be different and should be fixed according to the proportion of the two types of stresses. The authors’ recommendation is based upon this principle. Authors’ Recommendations.—For members subjected to axial compression and bending, the maximum compressive unit stress should not exceed values given below. (a) When the whole section is under compression or when only negligible tension is developed, the maximum compressive unit stress should not exceed 1.4 times the allowable unit stress for axial compression. With allowable compression stress f, = 0.225 f’., the combined stress will amount to 0.315 fc. (b) When considerable tension is developed in the section, the maximum compressive unit stress should not exceed the allowable unit stress in bending, namely 0.4’. Joint Committee Specifications, 1924.—The recommendations of the Joint Committee, 1924, are as quoted below: “Reinforced concrete columns subject to bending stresses shall be treated as follows: , ‘“‘(q) With spiral reinforcement: The compressive unit stress on the concrete within the core area under combined axial load and bending shall not exceed by more than 20 per cent the value given for axial load by Formula (43). “(b) With lateral ties: Additional longitudinal reinforcement may be used if required, and the compressive unit stress on the concrete under combined axial load and bending may be increased to 0.30f’.. The total amount of reinforcement considered in the computations shall be not more than 4 per cent of the total area of the column. “Tension in the longitudinal reinforcement due to bending of the column shall not exceed 16 000 Ib. per sq. in.” 464. CONCRETE AND REINFORCED CONCRETE COLUMNS Cleveland, New York, and Philadelphia Codes.—Codes of the Cities of Cleveland, New York, and Philadelphia specify that the stresses due to direct stress and bending shall not exceed in the extreme fiber the allowable fiber stress in bending. Columns Carrying Crane Loads or Loads on Brackets.—Columns carrying crane loads or other loads on brackets are subjected to bending moments produced by the eccentric position of the load in. respect to the center line of the column. Let _ P = load on the bracket, Ib. ; a = distance of point of application of load to center of column, in.; b = depth of bracket measured from center of tension to center of compression; 1, 11, l2, = column dimensions, in. ; M = bending moment produced by load P, in.-lb.; Mi, Me, M3, M4 = bending moments in column at points indicated in corresponding figure; R = horizontal reaction, lb. Then, Bending Moment in the bracket Due to Load, P, M = Pa. sito. eee For cranes, the load, P, should be computed when the crane is in the most unfavorable position as far as the column in question is concerned. Sufficient allowance should be made for impact. The bending moments produced in the column by the bending moment, M, depend upon the end condition in the column, and also upon the position of the bracket on the column. The following conditions are possible: (1) Column fixed at the bottom and free at the top. The column is subjected to a bending moment equal to M = Pa for its whole length below the brackets. (See Fig. 155 (1).) (2) Column hinged at both ends. (See Fig. 155 (2).) Horizontal reaction, top and bottom, RPS COLUMNS SUBJECTED TO BENDING 465 Moment above bracket, M1 = Pils eee es ime Se (BL) Moment below bracket, a Mz = — Poh FRO ea Pei Ba, Te 2) Hinged 3 1 lg R& QING rie by Ve Pa ay nek oe ¥— ~Bracket Hao 1 i} 4 i} mos yt v 1 Hie ith ER ded Fixed Hinged (1) Column Fixed at Bottom (2) Column Hinged at Top and Bottom Fixed Fixed Fixed ( 3) Column Fixed at Bottom and Hinged at Top (4) Column Fixed at Top and Bottom Fic. 155.—Columns Carrying Loads on Brackets. (See p. 464.) (3) Column hinged at the top and fixed at the bottom: (See Fig. 155 (8)). Horizontal reaction, top and bottom, Rm aPaGl?— hb?) —bBle+ BD] . 24. « (63) 466 CONCRETE AND REINFORCED CONCRETE COLUMNS Moments, M,=0;..f we ee eS Mo = 1 Ble ft a (55) Ms = — Pa+ Ro + 0);. +. (56) M, = — Pa RL 0 (57) (4) Column fixed at top and bottom: (See Fig. 155 (4).) Horizontal reaction, top and bottom, R= BP eal Ile +b) +h? + (e+ d)71;. . - (68) Moments, My = — Pal? — Qh + db) +2 +5)"; (68) Mo = M,+ Rh... ee (60) M3; = M, + R(b +b) — Pa; WAM es GO (61) M, = PSP - Gh+HQ—-h)y+ a+ . (62) also M, I M, + Ri- Pa 2 My 2 Ri Simplified Formulas.—The above formulas may be simplified without appreciable error by assuming the depth of bracket, b, equal zero. The formulas change to: (1a) Column fixed at the bottom, free at the top. Formulas same as for case (1). (2a) Column hinged at both ends. (See Fig. 156 (2a).) Horizontal reaction, top and bottom, a eae po (64) Moment above bracket, M, = ki: Moment below bracket, Ms. = — Rh. (3a) Column hinged at the top and fixed at the bottom. (See Fig. 156 (3a).) Horizontal reaction, top and bottom, p= 1.5P7 3(1 +7) | 3 a COLUMNS SUBJECTED TO BENDING 467 Bending moments, Dee 0) beshioey srw ltl) a quueroyy (68) Mo = + Risploom) ols qo daneos, 21160) ee ah Rig. is eee 8070) YE ea i een rent A (4a) Column fixed at top and bottom. (See Fig. 156 (4a).) Horizontal reaction, top and bottom, a Lil R = 6P7 Bai ott Wire Mateus «i.e mist ate heteer « (72) Bending moments, Ms =— Pofi—4(2)4a(2)'].. . . @@ Mz=M,+ Rly; . ee eed Ete to t= NS - Rlg — Pas M,z=M,+ RI-— Pa = M34 Rly. Hinged Hinged R I } | ~ pero P 1 J Mey eae 7 ti M, as , | ‘aig H | pe vey , Hinged Fixed (2a) (3a) (4a) Itc. 156.—Columns Carrying Load on Brackets, b = 0. (See p. 466.) The case of a rigid frame with crane brackets is treated in Vol- ume ITI. . The first condition occurs: (a) when a post is firmly anchored to a heavy foundation; (b) when a column extends above the floor and is held sufficiently at the floor level, but has no support at the top capable of resisting a horizontal reaction; (c) when a post is imbedded in firm ground well tamped but is free at the top. In the last case, the stability of the post depends upon the resistance of the soil to compression. The post compresses the earth on the side A468 CONCRETE AND REINFORCED CONCRETE COLUMNS of the bracket on the top and on the opposite side at the bottom. The pressure on the earth exerted by the post can be found accu- rately enough from the formula 6M iS - ‘dh?’ U (77) where d = side of post; | h = depth of imbedment in ground. The second condition exists: (a) When the column is actually hinged on the top and bottom. Such cases practically never occur in concrete construction. (b) When the column rests on a small footing and carries a roof construction above, which is anchored to it sufficiently to resist any horizontal reaction, but the connection is not rigid enough to withstand bending. In such case the column is practically free to turn on both ends. The third condition takes place when a column is rigidly held by the floor system below and carries a roof construction anchored to it sufficiently to resist horizontal pull, but not rigidly connected. The same condition takes place when, with top free to turn, the col- umn is rigidly attached to a heavy foundation, or is imbedded in firm ground for such a distance that the resistance of the earth to com- pression holds the column firm. The pressure on the earth, due to the bending moment, can be computed as explained in connection with the first condition. The fourth condition takes place when the bottom is fixed as explained in connection with the third condition, and the top also is rigidly connected with a floor construction above. CHAPTER VIII FOUNDATIONS AND FOOTINGS Concrete excels as a material for foundations. Since the design of a foundation is governed not only by the weight of the super- structure, but also by the character of the supporting work or soil, this chapter includes a brief discussion of the method of determining the carrying capacity of soil and of the allowable soil pressures. The method of designing concrete is given in detail. The treatment of piles is discussed in Chapter IX. DETERMINATION OF CARRYING CAPACITY OF SOIL It is of prime importance for the safety of the structure that the foundation be designed with proper consideration for the conditions of the supporting soil. The character of the soil, therefore, must be carefully investigated. The most common methods are: wash borings, diamond drill borings, and test pits. Wash borings are made by driving a small pipe, up to 4 in. in diameter, and working a water jet inside of the pipe. The record of the material brought up to the surface gives a fair idea of the composition of the ground. The results, however, are not always sufficiently reliable, as the material is affected by the water. For more accurate investigation, Diamond drill borings are used. These are made by a sharp, hollow cutter which cuts through the soil and brings out a core of the mate- rial through which it passes. This gives reliable information on the conditions of the soil and the depth of the various layers. Borings, to be of any value, must be made by experienced operators. Test pits, as the term implies, consist of pits dug in the ground. ‘This method of exploration is too expensive where the character of the soil varies appreciably from point to point, as it requires a large number of test pits. It is restricted to shallow excavations. Test pits are often used in combination with borings, for more thorough investigation. After borings are made, the most important places are selected on the basis of the borings, for test pits. 469 470 FOUNDATIONS AND FOOTINGS Results of the borings should be shown on the foundation draw- ings, to serve as a guide for determining the size of the foundation. Many building codes require such a record as a part of the foundation plans. In addition to the borings, investigation should be made of the type of foundation used in the buildings in the immediate vicinity. In some cases, the information gained from the borings and from prior experience will limit the choice to one or two types of founda- tion. Often there is considerable latitude in the choice of foundation. This is the case when the layers of the materials composing the ground increase in hardness with the depth. For instance, if a layer of sand is underlaid by clay, which in turn rests on hardpan and rock, each of these layers could carry the foun- detion with an allowable unit pressure suitable for the particular soil. The question of depth is determined by relative economy. By going deeper, to a soil with large allowable soil pressure, the size and the cost of the footing would be decreased, but the cost of excava- tion would be increased. When there is danger of encountering water, it is cheaper to use larger footings on a higher layer than to go down to more solid ground. In many cases, however, the size and importance of the structure aré such that a foundation in unyield- ing soil is required and the footings are carried to hardpan or rock even when support could have been obtained higher at smaller cost. This is done because in compressible soils some settlement is unavoid- able, and even with carefully designed and properly proportioned - foundation it may be impossible to prevent uneven settlement. Sometimes it is found that a harder stratum of soil lies above much softer ground. In such a case, the foundation, if built on the upper layer, must be proportioned to the capacity of the underlying softer material. BEARING POWER OF SOILS As a general rule, a foundation should not be built on soil con- taining organic matter, such as loam, nor on filled or made ground. It should be built on firm undisturbed soil and should be of such dimensions that the allowable bearing on the soil is not exceeded. The sustaining power of soil depends upon its composition, the amount of water which it contains or is likely to receive, and the degree to which it is confined. Since the conditions vary in different localities, the allowable pressure on soils of the same name, as estab- BEARING POWER OF SOILS ATI lished by experience, varies to some extent. It is always advisable to be guided by the experience obtained in laying other foundations in the vicinity where the building is to be erected. The maximum allowable bearing values on soil, as specified by the Building Code of Boston, may be used as a guide in design- ing the foundation. These are selected because the nomenclature used in describing the soil is well defined, and the values have been arrived at as a result of broad engineering experience and numerous tests.! LTTE SS Ee gly Sa Re a 2 Ae een ee a 100 tons per sq. ft. UE EGR BE hs On ek ae re 10 tons per sq. ft. Gravel, compact sand, and hard yellow clay... 6 tons per sq. ft. Dry or wet sand of coarse or medium-sized grains, hard blue clay mixed or unmixed with sand, disintegrated ledge rock............. 5 tons per sq. ft. Medium stiff or plastic clay mixed or unmixed with sand, or fine-grained dry sand......... 4 tons per sq. ft. Fine-grained wet sand (confined)............. 3 tons per sq. ft. Soft clay protected against lateral displacement. 2 tons per sq. ft. Definitions —(a) Solid ledge: Naturally formed rock, such as the granites and others of similar hardness and soundness, normally requiring blasting for their removal. (b) Shale: Laminated slate or clay rocks, removable with more or less difficulty by picking. ) (c) Hardpan: A thoroughly cemented mixture of sand and peb- bles, or of sand, pebbles and clay, with or without a mixture of boulders, and difficult to remove by picking. (d) Gravel: A natural uncemented mixture of coarse or medium grained sand with a substantial amount of pebbles measuring one- fourth of an inch or more in diameter. (e) Sand (compact): Requiring picking for its removal. (f) Sand (loose): Requiring shoveling only. (g) Sand (medium grain): Individual grains readily distinguish- able by eye though not of pronounced size. (h) Sand (fine-grained): Individual grains distinguished by eye only with difficulty. (i) Hard clay: Requiring picking for its removal. 1Of particular interest are the investigations by Mr. Joseph R. Worcester, the results of which are published in Journal, Boston Society of Civil Engineers, January, 1914, p. 19. 472 FOUNDATIONS AND FOOTINGS (j) Disintegrated ledge rock: Residual deposits of decomposed ledge. . (k) Medium clay: Stiff and plastic but capable of being spaded. (1) Soft clay: Putty-like in consistency and changing shape readily under relatively slight pressure. If part of a structure rests upon unyielding soil, such as ledge rock or hardpan, and the rest on yielding soils, the allowable unit bearing value for the yielding soils should be reduced by from 50 to 30 per cent, the higher percentage to be used for soils with 2-ton capacity and the lower for soils with 6-ton capacity. The reason in obvious. Rock and hardpan are practically unyielding, and no settlement of the footings resting on them would take place. The footings resting on softer soils normally are expected to have some settlement. The structure resting partly on one and partly on the other material, therefore, would be subjected to unequal settlement. To avoid this, the normal settlement is reduced by reducing the allowable unit pressure for the yielding soil. GENERAL RULES OF DESIGN Excessive and Unequal Settlement Must be Guarded Against.— In designing foundations, the following facts must be borne in mind: (1) In yielding soils (everything softer than hardpan) some set- tlement of foundation is almost unavoidable. (2) The possibility of settlement should be reduced as much as possible by the use of conservative unit pressures. . (3) The foundation should be designed so that the settlement, if any, will be uniform, i.e., that all footings will settle approximately the same amount. The last requirement is of particular importance in reinforced concrete structures on account of the rigid connection between various parts of the structure: Uniform settlement, even of appre- ciable amount, does not produce any stresses in the members. com- posing the structure. When there is unequal settlement, however, i.e., when part of the structure settles more than the rest, even if the settlement is of small amount, stresses of a character and magnitude not anticipated by the design are produced in the various members of the frame. In extreme cases, the settlement of one portion may result in the failure of the structure. The first requirement will be satisfied by selection of the proper GENERAL RULES OF DESIGN 473 allowable unit pressure on the soil and of an area of footing large enough to prevent the load from producing a pressure exceeding this. To satisfy the requirement for uniform settlement, it is necessary to proportion the area of each footing strictly according to the load the column resting on it is expected to carry. Often, for the sake of sumplicity or appearance, columns carrying different loads are made of the same size, the size being determined for the largest column load. In designing footings, this is not permissible unless the dif- ference in loading is inappreciable. The use of the same area of footing for columns carrying appreciably different loads may mean un- equal settlement. This, of course, does not apply where the footings are carried to rock or hardpan and where no settlement is expected. As a general rule, a new structure should not rest anywhere on an old foundation unless both the old and the new structures rest on unyielding soil. When future extension of a structure is expected, the outside footings are sometimes made large enough to support the future column load. In non-compressible soils, this would be permissible. In yielding soils, it often causes unequal settlement and more or less harmful effect, since the footings with provision for future loads are larger in proportion than the adjoining footings, which are designed only for the load they are now expected to carry. The end row of columns, where provision is made for future exten- sion, would settle less than the rest of the structure; and cracks, of a size depending upon the amount of settlement, would be produced in the end panels. After the expected addition is built, another case of unequal settlement may occur, as the new footings may settle more than the old footings, which by that time may have settled to a standstill. Proportioning Area of Footing.—The area of the footing is obtained by dividing the load by the allowable unit pressure on the soil. There are several methods of accomplishing this. Footings Proportioned for Dead Plus Full Live Load—The most common method of determining the area of the footing is by dividing the full column load, plus the weight of the footing, by the allowable unit pressure. No distinction is made in such cases, between the effect of the dead load and that of the live load on the settlement of the foundation. Footings Proportioned for Dead Load Only.—In this method, a distinction is made between the effect of the dead load and that of the live load. It has been observed, in compressible soils, that the ATA FOUNDATIONS AND FOOTINGS rate of settlement is large at first and then decreases as the soil becomes compressed. In many cases, a large part of the settlement takes place while the structure Is in process of construction and when it carries only the dead load. The effect of the dead load is thus larger than the effect of the total dead and live load, because the live load does not come until some time after the completion of the building. In many buildings, full loading with live load is never attained. In ordinary structures, the ratio of the dead load to the total load is much larger for outside columns than for interior col- umns. If the area of footings is based on the total live and dead load, and most of the settlement is produced by the dead load only, it follows that the exterior columns will settle more than the interior columns. To prevent for this, some engineers and also some building codes (notably that of the City of New York) require that, if for different columns in abuilding the ratio of live load todead load varies, the areas of footings be computed on the basis of the dead load only, by dividing the dead load by a reduced soil pressure obtained in the manner given below. The loads on the foundation for the various columns are tabulated, and the dead and live loads are kept sep- arate. The column that has the largest ratio of live load to dead load is then selected. For this column, the area of the footing is computed by dividing the total dead and live load by the maximum allowable soil pressure. The reduced pressure to be used in pro- portioning the areas of the remaining footings is then found by dividing the dead load for this column by the area just determined. The areas of other footings are finally obtained by dividing the dead loads carried by the footing by this reduced pressure. The following example illustrates this method: | It is desired to determine the size of footings for Columns 1 to 3, carrying loads given in table ‘below, for a maximum allowable soil pressure of 6 000 Ib. per sq. ft. | ee Dead Load Ratio Calin (Including Weight Live Load, Total rissa (eat of Footing), Ib. Ib. ib Load 1 200 000 400 000 600 C00 2.00 Z 150 000 200 000 350 000 1.33 5) 220 000 200 000 420 000 0.91 ne as cece ——— GENERAL RULES OF DESIGN 475 In the table above, Column 1 has the largest ratio of live load to dead load; hence Column 1 is used for the determination of the re- duced soil pressure. The required area for this column, for a maxi- mum allowable soil pressure of 6 000 lb. per sq. ft., is 600 900 divided by 6 000, equals 100 sq. ft. The dead load of 200 000 lb., divided by the above area of 100 sq. ft. gives a unit pressure of 2 000 Ib. per sq. ft., which is used as the reduced pressure on the foundation. The areas for the remaining footings are found by dividing their dead loads by reduced this pressure of 2 000 lb. Footing Areas Solin B Pa Dead Load Divided’ by| Total Load Divided by j Reduced Pressure, | Maximum Allowable sq. ft. Pressure 1 200 000 100 100 2 150 000 75 58 3 220 000 110 70 In the table just above, the dead loads of Columns 1 to 3 and the areas of footings computed on the basis of dead load only, using the reduced pressure, are given. The areas of footings that would have been obtained if computed by dividing the total load by the maximum allowable unit pressure are also given, for comparison. After the areas are obtained, the stresses in the footings should be determined on the basis of the total dead and live load, but exclu- sive of the weight of the footing. Footings Proportioned for Dead Load Plus Fraction of Live Load.— Some engineers vary the above method by proportioning the area of the footings, not for the dead load only, but for the dead load plus a fraction of the live load (usually one-third to one-half). The pro- cedure is similar to that explained above. The area of the footing is found for the column with the largest ratio of live load to dead load, by dividing the total load by the maximum allowable pressure. Then the reduced unit pressure is computed by dividing the dead load, plus the selected fraction of live load, by the area. In deter- mining the areas of the remaining footings, the dead load, plus the same fraction of the live load, is divided by the reduced unit pressure, This method is recommended by the authors. 476 FOUNDATIONS AND FOOTINGS Footings Must be Concentric with the Column.—To get uniform distribution of column load on the soil, the footing must be built so that its center of gravity will coincide with the theoretical point of application of the downward load. For independent footings, under concentrically loaded columns, this means that the area of the footing must be symmetrical and its center of gravity must coin- cide with the center of gravity of the column. With a pile founda- tion, the center of gravity of the cluster of piles must coincide with the center of the column. If these conditions are not fulfilled, the pres- sure on the foundation will not be uniform, and overloading of certain portions, with consequent unequal settlement, will be induced. Eccentric Footings.—Footings for walls and columns on the property line are sometimes made with a projection on the inside only. The center of gravity of such footings does not coincide with the center of the column. ‘The pressure is eccentric, and the dis- tribution may be found as explained on p. 170. The maximum unit pressure is much larger than if the pressure were uniformly distributed. For an eccentricity equal to one-sixth of the width of the footing, the maximum pressure at outside of wall is double the average pres- sure and is zero at the other edge, Such design, obviously, is not permissible for columns carrying loads of any magnitude. It may be used only for walls carrying comparatively small loads, when they are kept from leaning outward by the pressure of the earth behind the wall. The active earth pressure, together with the passive resistance, may then be sufficient to offset the eccentricity. Attempts are sometimes made to offset the effect of eccentricity on the foundation by making the connection between the column and the footing rigid and designing the column strong enough to resist a bending moment equal to the column load multiplied by the distance of the center of gravity of the column from the center of the footing. However, no attempt is made anywhere to balance the bending moment supposedly developed in the column. Such pro- cedure is, obviously, insufficient. It is clear that it is not possible to develop bending in a column unless it is held rigidly somewhere, and unless the connection is strong enough to balance the same bend- ing moment. | For instance, the bending moment produced in the column may be transferred to the floor construction. The floor beam in the first floor would then have to be strong enough to resist this bending GENERAL RULES OF DESIGN 477 moment. This is not practicable for large column loads or large eccentricities, as it would require excessive strength in the beams. The combined footing, or cantilever footing, described on p. 522, offers a better solution. It is often stated, in defense of ents footings, that the col- umn load gradually spreads itself, so that by the time it reaches the ground it acts centrally on the footing. The unsoundness of this argument is evident. The column load cannot change its center of gravity by distributing itself to the foundation. Only an additional force could change its position. Even if the pressure should be uni- formly distributed on the foundation, there would still remain the unbalanced bending moment produced by the load acting on the top of the column and the upward reaction at the bottom acting eccen- trically in respect to the column load. This bending moment would turn, or tend to turn, the structure, even if the column and footing were perfectly rigid. Proper Design of Foundation at Property Line.— When the build- ing line coincides with the property line, the foundation must be placed within the building lines. In such cases, it is often impos- sible to build the footings for the wall columns concentric with the columns. To prevent eccentric. pressure on the foundation, the footing for the wall column may have to be combined with the interior footing, see Fig. 173, p. 5380, or the eccentricity taken care of by a strap beam extending from the wall column footing to the interior footing. ‘The design of several types of such footings is given on pp. 533 and 538. For combined footings, it should be remembered that the center of gravity of the loads must coincide with the center of gravity of the foundations. Sometimes, when the footings consist of caissons or caisson piles, it is necessary to place them some distance within the building line. Then the wall column, if placed at the building line, must be carried on a cantilever formed by extending the strap beam between the wall column and the interior column. Anchoring.—In structures such as chimneys, narrow buildings, towers, trestles, and viaducts, the foundation must be designed with due consideration of the lateral wind pressure. If there is any pos- sibility of uplift, the structure must be anchored to the foundation; and the weight of the foundation block, plus that of the earth above, must be sufficient to counteract the uplift with a proper factor of safety. 478 FOUNDATIONS AND FOOTINGS Minimum Depth of Foundations for Outside Columns.—Foun- dations for outside footings of a building, in localities affected by frost, must be carried below the frost line. In the vicinity of New York City, 3 to 4 ft. below ground level is considered ample. Footings on lines abutting adjoining property are often carried to a depth of, say, 9 to 10 ft. below established grade, in order that they may not be disturbed by possible excavations on the abutting property. According to law, in case of excavation on adjoining lot, the owner of the existing building is required to support his foundation if it does not extend to the depth below grade, usually 9 to 10 ft., established by law. Encroachment on Public Streets.—Before designing a founda- tion on the line of a building adjoining the street, it is necessary to investigate the local regulations. Some cities grant permission to encroach with the foundation under the street or sidewalk, so that it is possible to use concentric footings for wall columns. The distance of permissible encroachment is usually governed by the depth of foundation. In Chicago, for instance, permits are issued for founda- tions projecting into the street, according to the depth, for as much as 3 ft. at an 8-ft. level below grade, with the express understanding however, that any foundation less than 20 ft. below city datum shall be removed by the owner if it interferes with projected public works, such as sewers, subways, etc. New York, on the other hand, does not allow any encroachment with foundations, although it allows some encroachment with such parts of the building the removal of which would not affect. the stability of the building. Loads Used in Design of Foundation.—The size of foundation is computed by dividing the load by the allowable soil pressure, as explained on p. 473. The load used in the determination of the area of the foundation or footing is the total load coming on the column carried by the foundation, plus the dead load of the foundation and the basement wall, if any. The same reduction in live load is per- missible for the footings as is allowed in column design. (See p. 453.) In computing stresses in the foundation, due to bending, the dead load of the footings and walls must not be included, as they produce no bending moments or shears in the foundation. | How Stresses are Developed in Concrete Footings.—Footings are generally designed for the upward reactions of the soil or of the piles supporting the footing. Since the upward reactions are not GENERAL RULES OF DESIGN 479 active forces, the manner in which they produce stresses in the footing is not always understood. Obviously, the ground is not likely to rise in order to deflect the footings, and without deflection there can be no stresses. The following explanation will make the matter clear. The column load coming on the footing is not distributed evenly over the whole foundation. At first, a large portion of the load comes directly under the column. This tends to produce, and actually does produce, a small settlement of the footing directly under the column. If the column were not connected with the footing the settlement would continue until the ground compressed sufficiently to hold the load. Since the column is connected with the footing, the incipient down- ward movement of the column produces shear in the adjacent sec- tions and forces them down, although to a slightly less degree because of the bending produced. These sections in turn bring their neigh- boring sections into play, so that finally the whole footing presses on the earth and the whole footing settles. The amount of set- tlement is less at the edge than in the center, therefore, the footing assumes the shape of a curve of the same nature as the curve that would have been produced if the reactions of the ground were ac- tive pressures and were carried Position ~s. of Footing \ Ree A spOTt..) |. ¢ —-—------4|-—--- ---._--__. When a footing fails, it does not mean that the reactions of Original the soil move the footing upward, —geirement / A— ¥. Footing : j : , Cracks -’ ‘at Failure but that the projecting footing of Column : is not able to hold the column Fig. 157.—Failure of Footing. itself from settlement. The col- (See p. 479.) umn settles, while the projections of the footing, instead of following the column, break away from it, as is evident from Fig. 157, p. 479. Footings Must be not only Strong Enough but also Rigid.— From the foregoing discussion, itis evident that the footing not only must be strong enough to resist the upward reactions, but must be rigid enough to prevent the deflection under the reactions from becoming too great. It is obvious that if a projection of a footing under a loading equal to the reaction of the soil can deflect $ in., then, conversely, the column would have to settle 4 in. before the whole footing would be brought into play. With a flexible footing, 480 FOUNDATIONS AND FOOTINGS the pressure would not be uniformly distributed over the foundation, but would be a maximum under the column and a minimum at the edges. The difference between maximum and minimum pressures increases with the flexibility of the footing. PLAIN CONCRETE FOOTINGS The simplest type of footing consists of a plain concrete slab or block, the base of which is determined by the allowable unit pressure on the soil, and the depth determined by strength of concrete. For walls, for instance, the footing may consist of projections on both sides of the wall, as shown in Fig. 158a, p. 480. For a column, an independent footing, of a design shown in Fig. 158, b and c, may be Fig. 158.—Plain Concrete Footing. (See p. 480.) used. If the projections of the footing from the face of the column are small, the design in Fig. 158b may by used. Where the depth and the projections of the footing are large, the cost may be decreased by stepping it, as shown in Fig. 158d. The area of the base of a footing is determined by dividing the superimposed load by the allowable unit pressure on the soil. The depth of the projections is obtained by figuring them as cantilevers loaded by the reaction of the soil and supported by the column, REINFORCED CONCRETE FOOTINGS 481 The critical sections should be assumed at the face of the wall, column, or pedestal, and in stepped footings also at the face of each suc- cessive step. The ratio of the height of the projection to its length is governed by the allowable tensile stresses in plain concrete and the magnitude of the upward pressure. The table below gives the depth of the footing, h, expressed in terms of the length of the projection, l, for various unit pressures on soil and for an allowable tensile stress in concrete of 60 Ib. per sq. in., the unit stress recommended for use in foundations. In a stepped footing, the depths h, hi, and ho, corresponding to lengths of projections i, l; and lg, respectively, should be taken from the table. Required Depth h, of Plain Concrete Footing in Terms of Length of Projection, | Unit Pressure on Foundation, in Lb. per Sq. Ft. 2 000 | 3 000 | 4.000 | 5000 | 6000 | 7000} 8000 9000 | 10000 h | 0.81 | 1.00} 1.21 | Peay etapa 1.61 | 1.71 1.81 1.91 ! } The table assumes that the whole footing acts as a unit. This is the case only when the whole depth of the footing is poured in a continuous operation. REINFORCED CONCRETE FOOTINGS To distribute the column load over .a large area of the ground without carrying the foundation in successive steps to a considerable depth and using a large mass of concrete, the foundation may be built of reinforced concrete. Reinforced concrete footings utilize the compressive strength of the concrete and therefore are more economical than the I-beam type of design formerly used.” Reinforced concrete footings may be divided into five groups: (1) wall footings; (2) independent column footings of rectangular or square shape; (3) combined footings carrying more than one column; (4) cantilever or strap footings; (5) raft foundation. * See Second Edition of ‘‘Concrete, Plain and Reinforced,” page 643. 482 FOUNDATIONS AND FOOTINGS WaLuL FootinGs A wall footing, as a rule consists of a slab projecting the required distance on both sides of the wall as cantilevers. For small pro- jections, it is most economical to use plain concrete wall footing, in which. case the ratio of the length of the projection to the depth should be taken from table on p. 481. If the projections are large, it is cheaper to reinforce the footing. In figuring bending moments, each portion should be considered as a cantilever with the critical section at the face of the wall. The reinforcement, determined — from the bending moment in the usual fashion, consists of bars placed at the bottom of the footing at right angles to the wall. Special attention must be paid to bond stresses. The depth of the footing and the diameter of the bars must be such that the unit bond stress, based on the total external shear and determined by formulas given on p. 262, does not exceed the allowable unit stress. It is advantageous to use bars of small diameter. The use of deformed bars is advisable. Diagonal tension also must be considered (see p. 247). As a basis for figuring the diagonal tension, the shear is taken, figured at a distance from the wall face equal to the effective depth of the footing. It is preferable to design the wall footings of such dimen- sions as to avoid the use of diagonal tension reinforcement. In stepped footings, the steps must be of such depth that at no point of the footing the tensile stresses in steel, the bond stresses, or the diagonal tension in the concrete will exceed the allowable working unit stress. | Example of Reinforced Wall Footing. Example 1.—Design a footing for a concrete wall 10 feet high, 14 inches thick, carrying a concentric superimposed load of 26 500 lb. per lin. ft. The allowable pressure on the soil is 5 000 lb. per sq. ft. The allowable stresses are fc = 650, fs; = 16000, n = 15, u = 100 and v = 40. Solution. —The weight of wall is 10 x 44 X 150 = 1750 1b. per lin. ft. and the assumed weight of footing 950 lb. per lin. ft. Since the wall carries a load of 26 500 Ib. per lin. ft., the total load on foundation is 26 500 + 1 750 + 950 = 29 200 Ib. With an allowable unit soil pressure of 5 000 lb., the footing requires an area equal to 29 200 + 5 000 = 5.84 sq. ft. per lin. ft. of wall. A continuous footing 5 ft. 10 in. wide will be accepted. This gives a projection L =2.ft4 am on each side of the wall. If plain concrete were used, the required depth of the projection would be 3 ft. because, from table on p. 481, for unit pressure of 5 000 Ib., h = 1.31 = 1.3 X 2.33 = 3.08 ft. = = =_ - REINFORCED CONCRETE FOOTINGS 483 Depth of Reinforced Concrete Footing.—The total upward resisting pressure is 5 000 lb. per sq. ft. From this the weight of the footing may be deducted, leaving a net upward pressure of 4 840 lb. per sq. ft. The shear at the edge of the wall is 2.33 X 4 840 = 11 290 Ib. per lin. ft. The required depth for punching 11 290 12 X 120 = 7.85. This depth is less than required for other stresses. In wall footings, punching shear does not need to be considered. The cantilever is 2 ft. 4 in. long and loaded with 4 840 lb. per sq. ft. The bending moment, therefore, is M = 2.33 X 4840 X 14 = 158 000 in.-lb. per lin. foot. The required depth, using the slab formula on p. 208 (d = 0.028 4/M), d = 0.028+/158 000 = 11.13 in. This depth to steel is larger than required for punching shear and will be tried. Area of Steel.—Using f, = 16 000 lb. per sq. in., d = 11.13 in., 7 = 0.875, shear, for an allowable unit punching shear of 120 Ib. per sq. in., isd = the area of steel from formula is, A; = ——. fsgd 2 158 000 ~ 16000 X 0.875 X 11.13 8 = 1.01 sq. in. per lin. ft. | 3_in. round bars spaced 54 in. on center give an area of 1.01 sq. in. Bond Stresses.—The external shear at the edge of the wall is 11 290 lb. per lin. ft. With 3-in. round bars, 5} in. o.c., there are 2.28 bars per foot. The perimeter of a 4-in. round bar is 2.36. The perimeters of bars per foot equal Do = 2.28 X 2.36 = 5.38 sq. in. and the bond stress from formula u = =a oj 11 290 tea 660,875.06 11. 13 = 215 lb.-per sq. in. The bond stress is too large. To reduce it, any of the following methods may be used: (1) smaller bars may be used; (2) area of steel may be increased and same depth retained; (3) depth of footing may be increased and same area of steel retained; (4) bars may be provided with hooks (see p. 263). Any combination of the methods may be used. In this case, the first method will be tried. Five-eighth inch round bars 3{-in. on centers (3.70 bars per foot) will be substituted for {-in. round bars 5¢ in. 0.¢. This gives an area of steel per foot of wall 0.307 X 3.70 = 1.13 sq. in. The perimeter of a 8-in. round bar is 1.96 in. The perimeter of bars per foot is Yo = 3.70 X 1.96 = 7.25 and the bond stress 11 290 = = 160. 7.25 X 0.875 & IT.138 This is still too large. As it would not be practicable to use bars smaller than 3-in., since their spacing would be too small, it will be necessary to increase the effective depth in the ratio of +89, where 160 is the computed stress and 100 the allowable stress. The increased depth is 11.13 X +§§ = 17.81. To this must be added the protective cover. The total depth then.is 17.81 + 3.5 = 484 FOUNDATIONS AND FOOTINGS 21.31 in. Keeping the same reinforcement, with the increased depth, the unit bond stresses will be reduced to 100 lb. per sq. in. It is obvious that in such a case the full value of the steel in tension is not utilized, but is needed to satisfy the bond requirements. Diagonal Tension.— Diagonal tension will be figured at a section distant from the wall a length equal to the effective depth. (See Fig. 159.) This section is marked 1-1 in the figure. Since the upward unit pressure is 4 840 Ib. per sq. ft., the external shear there is V = 4840 X 0.83 = 4 020 lb. and the unit shearing 4 020 | 180,875 OO unit shearing stress and the depth is satisfactory. Final design. stress, 0 — 21.2 lb. This value is less than the allowable d= sin h = 214 it. 5_in. round bars 3% in. oO. ¢. Plan Peete eng His sag ve5 Ss : = D Ft 4" <= eae es a Ae Ror ce oe oe sd Ss Vesa ete AA eS A eras RPO Tee ET Stree a Sout? Bias Fe hile Gang Palen ee ¢ iE es ea ear Bitsy Lae ay aaa aay Aer ed SOT MES ae i ¥ ° i pr aie at on . * a2 ” , : ? . ° e in eee See pi gmias me SRS ern. 2 "Round, 5 Ft. 4” Ig. ee eR a he te a ees cn “Bowe on oan . Pea | rie ‘ae 27 --~ d=18¢-- S Gj cea ee Ree ( 3 9 Ais ~. i -----h Bites pei eb RYISTIAY ISR IRA, “< 5000 lbs per sq. ft. 1 —>| 3’k----------—-+----- 5 Ft, 4'------------------- 3 Geman penn nnn -- a= 5 Ft, 10 ------- ~~~ Section Fig. 159.—Details of Reinforced Concrete Wall Footing. (See p. 484.) Stepped and Sloped Wall Footing.—The diagonal tension com- puted above is much smaller than the allowable diagonal tension. The thickness of the footing may be reduced at the edges without exceeding the working stresses, either by stepping as in Fig. 160, or REINFORCED CONCRETE FOOTINGS 485 by sloping as in Fig. 161, p. 485. The principle of stepping and sloping is explained thoroughly in connection with independent footings. ROSSA SEC ee ‘ 37. Section Fiae. 160.—Stepped Wall Footing. (See p. 484.) Sometimes, instead of stepping, the footing is sloped as shown in Fig. 161, p. 485. The slope must be such that the stresses at the inter- mediate points are not excessive. gine SS ik Ue OR VLTENIIR ST, A STEWS d. bar 5 Ft. 4" Ig, gd on centers “= Section Fig. 161.—Sloped Wall Footing, (See p. 485.) INDEPENDENT CoLUMN FOooTINGS Independent column footings are of several types. The simplest type is shown in Fig. 167, p. 504. It consists of a rectangular or square slab of uniform thickness, reinforced in two or four directions. This type is economical only for small loads. 486 FOUNDATIONS AND FOOTINGS For larger loads, where the required depth is large, the stepped footing is more economical. ‘This design is shown in Fig. 168, p. 507. The sum of the depths of all steps is equal to the required depth for the footing. This footing is designed in the same fashion as the slab type, except that it is necessary to investigate the tensile shearing and bond stresses at each step. Pedestals.—The standard design of footings used by the authors and recommended for general use is provided with a pedestal which reaches from the top of the footing up to the bottom of the basement slab. (See Fig. 162.) The pedestal is of plain concrete and its cross section is sufficient to carry the column load with a unit stress of 450 lb. per sq. in. Since the pedestal is in the ground, no fireproofing e Top of Basement Floor -. SANS OE OM ATO ONS OV ey EE OB Not OB GSAt oar eF WHE Wisy UE) a Fic. 162.—Pedestals below Basement Floor. (See p. 486.) Sa Se Poa Wah Dare ie oh tenon |<-- Pedestal os : VTRY RYT AWN ANAM is necessary and its full cross section may be considered as effective. There are three reasons why the use of a pedestal is desirable. First, having a larger circumference than the column above, it decreases the depth of footing required for punching shear, also the bending moment. Second, it simplifies the construction when the top of the founda- tion is some distance below the basement slab, and especially when the tops of the footings for any reason are not on the same level. - In such a case, if basement columns were placed directly on the footings, their lengths would vary with the varying distance from the top of the footings to the level of the basement slab. The pedestals are made of such height that their tops reach to the level of the bottom of the basement slab, so that all the basement columns may be made of the same length. The column steel and column forms can then REINFORCED CONCRETE FOOTINGS 487 be made of the same length and can be ordered ahead of time without regard to the depth to which it will be necessary to carry the footings. Third, when the top of the footing is some distance below the basement, the use of the pedestal is economical because it replaces the expensive highly reinforced column by cheaper plain concrete. If the ratio of the height of the pedestal to its width exceeds the limit for plain concrete, the pedestal must either be reinforced or its width must be increased. The minimum height of the pedestal must be such that the sum of the depth of the footing and the min- imum height of the pedestal is not smaller than required by punching shear at the edge of the column. (For explanation of punching, see p. 492.) Dowels.—Bars, or dowels, of the same number and size as in the column above, should be imbedded in the pedestal or footing. When there is no chance of tensile stresses due to bending in the column, the dowels should extend 30 diameters into the pedestal (or footing) and the same length above the top of the pedestal, to be imbedded in the column. When considerable tension stresses are possible, the length of dowels should be governed by the tensile strength of bar. It is of no importance whether the column bars are placed immediately next to the dowels or whether they are placed -a few inches apart. The stress in either case is not transferred directly from-the bars to the dowels but through the concrete. If the column above is of rich mix and spirally reinforced, the stress transmitted to the pedestal at the area of contact may be larger than the allowable bearing stress for plain concrete. (For bearing stresses, see p. 270.) For this reason, it may be advisable to extend the spiral into the pedestal a sufficient distance to grad- ually transmit the high pressure from the column into the body of the pedestal. Some designers accomplish this object by placing several layers of lateral hoops in the upper part of the pedestal. , Stepped Footings.—The bending moments and shears in a footing are a maximum at the edge of the column and decrease toward the ends. Therefore, it is often economical to reduce the thickness of the footing by stepping. To maintain the same factor of safety at all points, the reduction in the depth of the footing should not be larger than the reduction in the stresses. The problem is compli- cated by the number of stresses to be considered, not all of which decrease in the same ratio. These stresses are: punching shear (if considered), bending moment, bond, and diagonal tension. A488 FOUNDATIONS AND FOOTINGS Figure 163 (a), (b), and (c), for the three ratios 0.2, 0.3 and 0.4 of width of pedestal (or column) to width of base, illustrates the rate of decrease of the various stresses and may serve as a basis for dimensioning the blocks in stepped footings. In this discussion, the footing is assumed to be provided with a pedestal. If the pedestal is absent, it should be considered as replaced by the column. ) In each of the three cases, the depths of footing required by punch- ing shear, by the bending moments, and by the bond stresses (the measure of which is the external shear) were computed at various points and plotted in the diagram. To get a direct comparison between the rates of decrease, the maximum values were accepted as a unit and the intermediate values as fractions of this unit. The variation of each item is shown by the curves starting at the edge of the pedestal from a common point. From the figures, it will be noted that the bending moments and the depths required by punching shear decrease in about the same ratio. The external shear, however, which is the measure of the bond stresses, decreases much more slowly. The depth required by diagonal tension is in the same yatio to the depth required by punching shear as the allowable shearing unit stress for plain con- | crete is to unit punching shear. However, diagonal tension does: not need to be considered nearer to the column edge than the depth of the footing (see p. 502). Therefore, it is seldom a determining item. Width of Blocks in Stepped Footings.—The curves showing the variation of stresses and bending moments may be used for the purpose of fixing the width of the blocks. The use of two blocks is recommended, as shown in Figs. 164 and 168. Also, it is assumed that a pedestal, as described on p. 486, isused. The thickness of the bottom block above the level of reinforcement is accepted as SIxX- tenths of the total effective depth, and the thickness of the top block as four-tenths of the total effective depth. The width of the top block depends upon whether the footing is designed so that the bond stresses are a maximum at the edge of the pedestal or at the edge of the top block. If the bond stresses at the edge of the pedestal are equal to the maximum. allowable working stresses, then the external shear curve governs the width of the top block and it must come outside of the shear curve, as indicated by the dash lines. REINFORCED CONCRETE FOOTINGS 489 BN . RS \ Ye i 1) =) a ay Values of eS c teat | | ‘ 0.1 | 0.15) 0.2 | 0.25} 0.3 | 0.35) 0.4 0.45 0.5|0.55| 0.6|0.65} 0.7 Qed 0.25} 0.23] 0.22) 0.20) 0.19 0.18) 0 16/0. 150. 14/0. 120. 11/0.09|0.08 0.15 | 0.26) 0.24] 0.23) 0.21) 0.20: 0.19) 0 17|0.16)0.14/0.13)0.11)0.10;0.08 0.20 | 0.27| 0.25! 0.24) 0.22 0.21! 0.19) 0.18/0.16/0.15|0. 13/0. 12)0. 10)0.09° 0.25 | 0.28] 0.26) 0.25} 0.23} 0.22) 0.20) 0 1910.17|0.15|0.14)0 12/0.11)0.09 0.30 | 0.29) 0.27| 0.26} 0.24, 0.23) 0.21) 0 19,0.18)0.16)0.14)/0 130.11)0.10 0.35 | 0.30) 0.29} 0.27) 0.25) 0.24) 0.22 0.20/0.18|0.17|0.15)0. 13,0. 120.10 0.40 | 0.31} 0.30] 0.28) 0.26}; 0.24) 0.23 0.2110.1910.17|0. 16/0. 14/0.12)0.10 0.45 | 0.33} 0.31) 0.29] 0.27) 0.25) 0.23 0. 22/0.20\0.18/0.16)0.14/0.13)0.11 0.50 | 0.34] 0.32] 0.30) 0.28) 0.26) 0.24 0.22'0.2010.19'0.17|0.15)0.13/0.11 0.55 | 0.35} 0.33} 0.31! 0.29) 0.27) 0.20, 0.23/0.21|0.19)0.17)/0 15|0.13)0.11 0.60 | 0.36] 0.34} 0.32} 0.30) 0.28) 0.26 0.2410 .22/0.20/0.16)0.16)0.14)0.12 0.65 | 0.37) 0.35| 0.33] 0.31) 0.29) 0.27 0 .25'0.23'0.20)0.18/0.16/0. 14/0. 12 0.70 | 0.38] 0.36] 0.34] 0.32) 0.30| 0.27 0 .25)0.23|0.21/0.19|0.17/0.15)0. 138 0.75 | 0.39] 0.37] 0.35) 0.33) 0.30) 0.28 0 .2610.24/0.22/0.20/0.17|0.15)0.18 | Diagonal Tension.—Tests indicate that in reinforced concrete footings, diagonal tension develops.at a distance from the face of the column equal to the effective depth of the footing. In figuring the V maximum diagonal tension, therefore, by the formula, v = bd (see p. 247), V should be taken as the upward pressure between the edge of the footing and a line concentric with the pedestal (or column) and located at a distance from the face of the pedestal (or column) equal to the effective depth of the footing. The value jd is the moment arm at the section considered, while 6 is the length of this concentric line. } For square footings, where c is the diameter of the column and d is the effective depth of the footing, the concentric line is a square with a side equal to (c + 2d). When the column is rectangular in section with sides c and ci, the concentric line for diagonal tension is a rectangle with sides equal to (¢ + 2d Jand (c; + 2d) respectively. For square footings, the intensity of diagonal tension is the same at all four sides of the concentric line. For rectangular footings, on the other hand, there may be considerable difference in intensity at REINFORCED CONCRETE FOOTINGS 503 the different sides of the rectangular concentric line. For this reason, it is advisable to compute diagonal tension for each side separately, using for V, in formula v = the external shear at one side of the Ms bjd’ concentric line, and for b the length of one side of the concentric line. The external shear, V, at each side may be computed by means of Formulas (14) and (16) on p. 501, in which case for = or : and a 1 the ratio of the sides of the concentric line to the sides of the base of footing should be substituted. The tables on pp. 495 and 502 may be used for finding the constants C,; and C75. The formulas for diagonal tension are: . For square footings, Car ee (e+ 2djd’ Pe sea Let Bes) Be, bow Ps (18) where C); is from the table on p. 495, corresponding to a value in the table vanes (¢ + 2d) a a For rectangular footings at line parallel to column side, c, ements 5F ” ~ (e+ 2d)jd’ ae where C;5 is taken from the table on p. 502 where the value . is equal c- 2d ci + 2d ay i to a and Si) + =H eS) = per ite eoe, ee Nee any UP Alay | ees | debe eee rally oe A GY, WY YE) /2 TAA SIRES AN ANY) we ee “WZ 4000 Ib. per sq. ft.'7 "7% “a || ts ps 7" bis " ise) a} \ Alot eees Equal Spaces ---7'- 11" ---~= vs -4f0 Fic. 167.—Details of Square Footing. (See p. 504.) 5.43 sq. ft., the area of footing outside the column is 100 — 5.43 = 94.57 sq. £6; The upward pressure on this area produces punching shear at the edge of the column, which for 1 : 2 : 4 concrete should not exceed 120 Ib. per sq. in. (See explanation of Punching Shear, p. 492.) The load producing punching shear may be determined by multiplying the total load on the footing (exclusive of the weight of the footing, which produces no shear) by the ratio of outside of column REINFORCED CONCRETE FOOTINGS 905 : ‘ 7 to gross area of the footing. It is, therefore, x 350 000 = 331 000 lb. (Same results may be obtained by multiplying the load, P, by constant from ; CoP oo table on p. 495, corresponding to - = Ch 0.23.) By dividing this load by a the perimeter of the column, or 112 inches, times the allowable unit shear, we get 331 000 OA hier: 112 x 120 mace Accept 25 inches as the effective depth of the footing. Find Diagonal Tension.—To determine whether this computed depth is suffi- cient, diagonal tension will be determined at a distance from column equal to - effective depth of footing, as explained on p. 502. The side of the square at which diagonal tension is computed is c + 2d = 28 + 2 X 25 = 78 in., or 6.5 ft. The BigsG ; ris ratio — to be used in finding the constant Cy: is ener 0.65, for which Cp, = a z(1 — 0.65?) = 0.144. (Value of Cy; may also be taken from table on page 495.) The value of diagonal tension from Formula (18), p. 503, is 0-144 X 350000 _ 4. = = - per 7 IMs ex eed et Hence, depth is satisfactory and no shear reinforcement is necessary. Find Bending Moment.—As explained on p. 497, the bending moment may be found by the use of the table on p. 497. The ratio of diameter of column to Find es 28 side of footing is OF 4008 DD table, by interpolation, Cy; = 0.0567. The bending moment, therefore, is from Formula (13), p. 497, = 0.2338, and the corresponding constant from the M = 0.0567 X 350000 X 10 X 12 = 2 380 000 in.-lb. Find Area of Steel.—The reinforcement will be placed in two directions, parallel to the sides of the footing. The area of steel in each band is found by dividing the bending moment determined above by the moment arm, jd, times M the allowable unit stress in steel. (Forme A,=- aj i IQs 2 380 000 JY SAGE as shekaliabanggc toe OE 7 25 x 16 000 Bivona Ss requiring twenty-three §-in. round bars. All these bars must be placed within the effective width, which is 28 + (2 * 25) +42 =8ft.3in.4 (See p. 499.) Add one bar at each side, making a total of twenty-five 3-in. round bars. 4Same result may be obtained from formula 3(a + c + 2d). (See p. 499.) 506 FOUNDATIONS AND FOOTINGS Find Bond Stress.—Bond stresses are determined by Formula (50), p. 262. The external shear outside the column is 331 000 Ib. as determined in computing 331 000 4 = 82750 lb. When shear producing punching is not known, the external shear ‘may be found from Formula (15), p. 501, using constant Cy. The number of effective bars per band is twenty-three 3-in. round bars, the perimeter of which is 23 X 1.96 = 45.1 in; therefore, the unit bond stress is sh 82750 See eit): punching shear. The external shear at one edge of column is V = U = 84 lb. per sq. in. This bond stress may be used for deformed bars, but is somewhat excessive for plain bars in 1 : 2 : 4 concrete (see p. 263). Hence, use deformed bars. The weight of the footing does not need to be considered in figuring bending moment, shear, diagonal tension, and bond stresses, because it is balanced by the upward reaction. It increases, however, the unit pressure on the soil; therefore, it was considered in determining the size of the base of the footing. SLOPED AND STEPPED FOOTINGS The design shown in the previous example is not economical for deep footings. In such cases, either a sloped or a stepped footing should be selected. The design of both types of footings will be illustrated in the following example. Example 3.—Design a footing for a 40-in. round column carrying a load P = 1000000 lb. when the allowable soil pressure is p = 6 000 Ib. per sq. ft. Solution.—Area of Footing—Assume weight of footing W = 50000 lb.; then total load P + W = 1000000 + 50 000 = 1 050 000 lb. P+wW_ ~ 1050000 _ = = (175 agit. 6 000 Use a = 13 ft. 3 in., making area of footing a? = 175.5 sq. it. Size of Pedestal.—(See p. 486.) For unit compressive stresses f, = 450 Ib. 1 000 000 450 The side of the square pedestal is c = 47 in. Use 48 in. square pedestal. Depth for Punching Shear at Edge of Column.—Area of footing is 175.5 314 X 3.33? Required area of footing is per sq. in., the required area of pedestal to carry load P is = 2222 8qi in. sq. ft. Area of column = Peg 8.7 sq. ft. Area outside the column ; . 166.8 ; is 175.5 —8.7 = 166.8 sq. ft. Ratio i755 = 0.95. Load to be used in com- puting depth is 0.95 P = 950 0001b. Since circumference of column is 40 X 3.14 = 125.6 in. and allowable shear v = 120 lb. per sq. in., _ 950 000 p9 25y69<) 120 == Go 10g REINFORCED CONCRETE FOOTINGS 507 This is the minimum depth from top of pedestal to plane of footing reinforce- ment. ‘The depth could also be found by using table on page 495, as is shown below. Depth for Punching Shear at Edge of Pedestal.—Since a = 13.25 ft. and Cc ? : side of pedestal, c = 4.0ft.,- = 0.3, and1 — (<) = 0.91. The load producing : a a ee o\2 punching is P : “a | = 0.91 P = 910000 Ib. For perimeter of pedestal, . 4c = 4 X 48 = 192 in., and v = 120 lb. per sq. in. 910 000 RCE eae aeons (ee 1 192 x 120 Zs allowing 33 in. from the bottom of footing to center of steel, the total depth of footing below pedestal h = 43 in. Same result for d; is obtained from table on p. 495. For es 0.3, Cr = 0.28. a u ee eg OO kag) ence => Tae) eS 3 ; eters 48 X 120 ae Top of Basement F x 4Ft.0 Sq. S Pedestal i ole y ay Top Block = i | = a I. Spgs tae CPLR og eer na ~ ap 4 N ' Bottom Block ; t rah UINNUIN WIN Gp CIN INN UNG SOON III SMINEIR SIN 44 - 5- Round bars, each way warn nn rn nnn eerie 13 Ft. 8" 8q----- ---------------—---? Fic. 168.—Stepped Footing. (See p. 507.) Shape of Footing.—Standard design shown in Fig. 164, p. 490, will be used. The dimensions a, c, d, and d; being known, the remaining values may be taken from table on p. 490. For Stepped Footings.—The bottom block is 0.6d: = 0.6 X 39.5 = 23.7 in.; c use 234 in. The top block is 39.5 — 23.5 = 16 in. For a 0.3, from table on p. 490, a1 = 0.5a or 6 ft. 8 in. For Sloped Footing.—For ae 0.3 from table, the value of az = 0.35a, or 4 ft. a 8 in. and d, = 0.2d = 8 in. 508 FOUNDATIONS AND FOOTINGS Diagonal Tension.—Compute diagonal tension at a distance equal to d; from face of pedestal. The side of the section isc + 2d, = 48 + 79 = 127 in. = 10 it. 7 in. The external shear to be used in computing diagonal tension may be 10.58 obtained by the use of table on p. 495. Call c = 10.58, a = 138.25, big yee a 13.25 = 0.799, for which, by exterpolation, Cr = 0.09. . Diagonal Tension for Stepped Footing.—The depth of a stepped footing at section considered is equal to the depth of the bottom block, or 23.5 in. Hence, unit shear from Formula (36), p. 247, 0.09 1 000 000 oe DA Ee eA int OS 197 SC D8 B Pe As this is smaller than the allowable shear, the design is satisfactory. Top of Basement Floor Slab ~ Pedestal ae Nis ~ ZS < A SS S 1) ” % “ ----99,58-—-- . Z ae ‘yT rial Footing (} i = i“ . ‘jad -----39,54---— TONG ANGINA RUNING NUANCE GINUNS OS 31-7 Round bars, each way Fic. 169.—Sloped Footing. (See p. 507.) Diagonal Tension for Sloped Footing.—The depth of a sloped footing at sec- tion considered is 17.7 in. Hence, unit shear 0.09 x 1 000 000 v= Wi X a et = 46.0 lb. per sq. in. With an allowable unit shear of 40 lb. per sq. in., this shear is too large. The slope of the footing should be changed so as to get, at the section considered, a 46.0 depth equal to 17.7 X ere 20.35 in. The revised slope is shown in Fig. 169, p. 508. Bending Moment.—Find bending moment at edge of pedestal. The side of ¢ 4 pedestal, c = 4 ft., side of square footing, a = 13/25 ft, Tate = = eae: = 048 Therefore, from table on p. 497, constant in formula M= C,,Pa® is Cy = 0.047 5 See p. 496. REINFORCED CONCRETE FOOTINGS 509 and the bending moment at edge of pedestal, M = 0.047 X 1000000 X 13.25 = 622 800 ft.-lb. or 7 474 000 in.-lb. Required area of steel for d; = 39.5 in. is : 7 474 000 ~ 16000 X 0.875 X 39.5 = 13.5 sq. in. s Thirty-one 3-in. round bars give an area of 13.70 sq. in. Compressive Stresses in Concrete.—Width of top block is 6 ft. 8 in., or 80 in.; 13.5 39.5 X 80 Since, with f; = 16 000 and n = 15, for an allowable stress of 650 lb. in concrete p = 0.0077, and the ratio actually used is only 0.0043, it is evident that in this case compression is considerably below the allowable unit stresses. Bond Stresses.—For sloped footings, with the slope accepted in design, the bond stresses should be investigated only at the edge of the pedestal where the external shear is the largest. For stepped footings, on the other hand, it is neces- sary to compute bond stresses also at the edge of the top block. Bond Stresses at Edge of Pedestal for Stepped and Sloped Footings.—The external shear at the edge of pedestal, for figuring bond, is one-quarter of the load for punching shear the depth, d = 39.5 in.; A, = 13.5 sq. in.; hence p = = 0.0043. 1 y= 10000 oor 5001 The depth is 39.5 in.; perimeter of thirty-one }-in. round bars is Zo = 31 X 2.35 = 72.8 in. Hence, the bond stresses, from formula wu = Sond’ Oj = 227 500 ~ 72.8 X 0.875 X 39.5 u = 90.5 lb. per sq. in. This bond stress is satisfactory for deformed bars. For plain bars, the stresses should be reduced as explained on p. 264. Bond Stresses at Edge of Top Block for Stepped Footing.—The width of the top block, as seen in Fig. 168, is 6 ft. 8 in., and the depth of footing at the edge is 23.5 in. The external shear, for figuring bond stresses according to Formula (14), p. 501, is ' 6.66 \? The same value would be obtained from formula V = CpP, where Cp = 0.18, from table on p. 495. The perimeter of bars is the same as computed above, or Do = 72.8 in. The bond stresses, computed as above, are 187 000 eee ee Seon in. “72.8 X 0.875 X 23.5 Path a 510 FOUNDATIONS AND FOOTINGS This shear is too large unless the bars are hooked at the ends. If forty-four 3-in. round bars are substituted for thirty-one 2-in. round bars, the perimeter changes to to = 441.96 = 86.2 185 000 Uu= = 104 lb. per sq. in. 86.2 X 0.875 X 23.5 This stress may be considered satisfactory for deformed bars. If further redue- tion of bond stresses is desired, it will be necessary to increase the width of the block. The use of still smaller bars would not be wise, as their spacing would be too small. (See also p. 264.) RECTANGULAR FOOTINGS Rectangular footings are often used for wall columns and for rectangular interior columns. If it is permissible to go outside of the building line, independent footings for wall columns will be found most economical. The wall columns usually being rectangular, the footing is also made rectangular. If not governed by other require- ments the shape of the footing should be made such that the length of the projections of the footing from the column edge on all sides are equal, since this requires the minimum amount of materials. The area of the footing is obtained first by dividing the total load on the foundation by the allowable unit pressure on the soil. The sides of the rectangle of the base of the footing, which should have equal projection from the edge of the column on all sides, may be obtained by trial or from the following formula. a=3e-a)+VA+i—«c)2, .. . (20) where a is one side of the rectangle, A is the area of base, and c and cy are sides of the column. The values may also be taken from table on tao | Design of Rectangular Footing —Compute the area of the footing by dividing total column load, plus Weel of footing, by the permis- sible unit pressure on the soil. Select sides of the rectangle which will give the required area. For most economical design, select such dimensions for the sides that the projections on all sides are about equal. Use table, p. 511. Compute depth of punching shear, as explained for square footings. (See explanation of punching shear, p. 492.) Check the depth for diagonal tension, as for square footing. 1 Length of Long Side, a, of Rectangular Footing Eacnemics (c— a)V A = c,; = short side of column. area of footing, c = long side of column, (c — ¢1)2, where a ? Bh 4 z 2 From Formula a long side of footing, A Long Side of Footing, a, in Feet REINFORCED CONCRETE FOOTINGS Values of c — c, in Inches 511 7 NNORHO OH OH DOH MDNDO Re) CH ODDROOHHANMMNOH HH oO Se I ee ee es | rs SserOnomMoaniwmowodnnwmrewnHwo oo = Cre KHORDRAOHHAHNANNMAOH SH = faa) See oe ee oe eB es | us DODMADHTAHAYTONSOH OS Q OnNRDAMOOHANNNAOH HS faa) Sen I cee I ee Be oe ee | : DHOHPOKRNDMODOARMDOOOL oO MOnMDARMRODHHANNMYONM HS ian) : 5 ee ee ee ce ee | < DOOMOCNKRAKRAOMDONMSO va) MOM ODOARMROOHHNANMMNOASH HSH N See OO coe ee I ce ee OO oe | p Peon own on OM D OM ONS Re) MOnRADADROOHHAHNANMHOOHH AN ee ee 2 SoOoOtFHnTOTOUMUMMDOWOMOKRHY + WMOnRWDDAROOHHNANANMNHHHSH N So ee oe ee ee | Si D199 MOrnMDMATHNHOHODOMDMOO AQ MOR DDAROHHHANNMNHOSH SH N Se OD ce oe BO re I oe ee ee | r: HTHNDONDDAMDOHOMOEMADOM oO MDOnnRORDRAOR HR HANNYOO HH N bn BD ce Oe ce oe BO ee ee | x MDOHOoOMMEMOMrRAoDOWOA ioe) MOnRRORAOHHHAHANANMHOHS coeeal Se ee Fe BO oe ee oe ee | z NANOCODHHONDACHDOOORNM Pe) WDOM~MKORDRAOOHHANNDMOH el Ss ee Oe ee I oe ee | - HHOr~rnmowonrnwootononon +H MOON ORMDAOOHHANNDOMH re bs cee ee eee oe oe eB ee | i SODSMANDORODOMRHDO aq MOOKnADADROOHHHANANMMN 5 ae Se I ce ee BO ee ce ee OO ee ee | ~ MPRDIMINDAHROOHONRHWDO SlHOOrnrnwmnananoon nt nna Ses OS ee A ce DO es A A oe ee | ie ODOrnRHHKhRMDAHOMKNOCOHHN oe HDOKRWADDAGADOHHNNMDHMN bere BD en I ee OO ee oe A cee I ee ee FI | ‘i KROMOONKRMARAOHWMOMLN Ro) HO OR DDAADOHHNANANDOD eo wesowsws we eS NS es re DOOMWINAMHONDHOOHANS rl HAMOOnRRARMRRMROOHHANANMN Seen I pe OO ce I es I ce BD ee BS | 2 MImMHHOHOHHOCOMOMK AD i HMDOKRKRDRBABOCOCHHHAHNANMNON Sn A oe A ee BD cee A ee FD ce FD ce oe (OD | AY So cmooackw S ooo So oH RS eo ee Ss oR eee Ss o rset SS oS SN es op 512 FOUNDATIONS AND FOOTINGS Compute moments in the long and short directions. Compute areas of steel, and select the bars so that the bond stresses do not exceed the allowable values. * Slope or step the footing as explained in connection with square footings. The width of the blocks may be the same as in square footings having a ratio of width of footing to width of column equal to the ratio of the rectangular footing in the direction considered. The blocks will also be rectangular. Example 4.—Design footing for column load P = 550 000 lb., column size c X c, = 48 X 24 in. Allowable soil pressure p = 6 000 lb. per sq. it Solution. —Assume dead load of footing W = 500001b. Total load P + W = 550 000 + 50000 = 600 000 Ib. Required area of footing is equal to (P + W) + p = 600 000 + 6 000 = 100 sq. ft. The sides of the column are c = 48 in., and c, = 24 in.; hence ¢ — ¢: = 24 in. Sides of a rectangle having an area of 100 sq. ft. and having equal projec- tions from the column on all sides may be found from table on p. 511. The long side is a = 11 ft., and consequently the short side, a, = 100 + 11 = 9.1 ft. or 9ft. lin. It will be exact enough for practical purposes to select a footing with a base of 9 ft. by 11 ft., having an area of 99 sq. ft. Depth for Punching Shear.—The ratio of the load producing punching to the 8 total load equals ( —~—*) =1—— =1-—0.08 =0.92. The load producing aay, 99 punching is 550 000 X 0.92 = 506 0001b. Perimeter of column bi = 2(48 + 24) —144in. Hence, depth for punching shear for working stress of r = 120 lb. per sq. in. is (irom formula d = 1200 /” 506 000 ‘ = ——— = 29 in 120 xX 144 Bending Moments.—Formula for moments is M =CpPa. (See p. 497 i nl Lita hoe eee ~ = eae a C1 ay = 9 ft. C1 = 2 ft ae eo 0.222 : a1 The constants Cy for the bending moments may be taken from table on p. 498. In the direction of the 11-ft. side, for : = 0.364 and Fy = 0.222," by interpolation Cy, = 0.037. In a similar manner, the constant toe the short side is found, using for : = 0,222 ana for z = 0.364. Ite 0.059. Bending Moments: Long cantilever side, M = 0.037 X 550 000 X aT 2 688 000 in.-lb. Short cantilever side, M = 0.059 X 550000 X 9 3 504 000 in.-lb. 224 000 ft.-lb. or I 292 000 ft.-lb. or REINFORCED CONCRETE FOOTINGS 013 Areas of Steel.—The area of steel, computed from formula A, = a is: A For long cantilever, A ENG 6.6 1 O cantilever, = = 6. Phat "0.875 X 29 X 16 000 a 3 504 000 For short cantilever, As: = = 8.6 sq. in. 0.875 X 29 X 16 000 The effective steel will consist of thirty-four 3-in. round bars 10 ft. 6 in. in | long direction and forty-four 3-in. round bars 8 ft. 6 in. in short direction. This should be placed within the effective width. Also, two additional bars on each side should be placed outside the effective width. Bond Stresses.—At edge of column, the shear on each side of the column to be used in figuring bond stresses may be found by determining the area of the trapezoid and computing the upward pressure acting on same. It may be found more easily by using the table on p. 502. Thus, for the shear along the long side of column, for which, as computed c é above for short cantilever, — = 0.364 and — = 0.222, we find a constant Cy; = a a1 0.27, and the shear V = 0.27 X 550000 = 148 000 lb. : Cc c : For the shear along the short side, interchange values of — and nie using for a a1 in the table 0.222 and for —! 0.364. The constant from the table then is 0.19, a Qa, and the shear V = 0.19 & 550 000 = 106 000 lb. The shear along the long side, which is 148 000 Ib., should be used in connec- tion with steel running at right angles to it, i.e., steel running in short direction. Circumference of forty-four 3-in. round bars is 44 X 1.57 = 69 in. Hence from formula u = ——, Zojd 148 000 = ——_—_______ = 8 Jb. . in. ** 69 X 0.875 X 29 eae This is satisfactory for deformed bars. For plain bars, the number of bars . would have to be increased in the proportion of the computed stress to the allow- able stress, or $4, or the bars provided with hooks at both ends. In this case it would be cheaper to use additional bars. Forty-six }-in. round bars will be required if plain bars are used. The shear along the short side of the rectangle is 106 000 Ib., and thirty-four 2-n. round bars are used perpendicularly to that side. The suin of the perimeters is Zo = 34 X 1.57 = 44 in. Using formula as above, 106 000 uU = 03 X 0.875 X 29 = 79 lb. per sq. in. This is satisfactory for plain and deformed bars. Sloped Footings.—The rules for sloping footings, given on p. 490 for square footings, may be applied here. Since the slope is outside of the shear curve, bond stresses are satisfactory. 514 FOUNDATIONS AND FOOTINGS Stepped Footings.—If a stepped footing is used, the dimensions of the blocks should be made according to table on p. 490. The sides of the step are 6 ft. 10 in. and 3 ft. 10 in. The sizes of blocks conform to the bending moment line, but not to the external shear line. The same amount of steel as’required at the edge of the column is satisfactory at the edge of the step for bending moment. It is, however, necessary to check the bond stresses, as they will be larger at the edge of the block than at the edge of the column. Bond Stresses at the Edge of Block.—The depth of the step is 17.5 in. The sides of the step are 6 ft. 10 in. and 3 ft. 10 in. The values of 2 and ” for the long a a 1 6.0 3.82 side are ane 0.55 and hae 0.43. The constant Cys for these values (either computed by Formula (16) or taken from table) is 0.22, and the shear V = 0.22 x 550000 = 120000 lb. Since twenty-nine }-in. round bars act perpendicularly to this section, from u = =~; Lojd 120 000 u = 58 xX 0.875 X 17.5 = 135 lb. per sq. in. This value for bond stresses is too large unless the bars are hooked. In the case under consideration, it is not feasible to reduce bond stresses by using smaller diameter bars, as it would require fifty-one $-in. square bars, the spacing for which would be too small. If hooked bars are not desired, either a larger number of 3-in. square bars must be used or the width of the step must be made larger, whichever proves to be cheaper. CorRNER CoLUMN FOOTINGS For architectural reasons, the corner columns are often made L-shaped. The footing must be made concentric with the center of pannn-=-=-Cly ae 4 Fic. 170.—Illustration of Corner Column. (See p. 515.) gravity of the column. ~ The center of gravity may be found from the following formula. REINFORCED CONCRETE FOOTINGS 515 Let a and a; = sides of the column, inches. c = the thickness of column, inches. e = perpendicular distance of the center of gravity from the base a inches. (See Fig. 170, p. 514.) Then ac + a;? — ¢? rare. ee ete er AG The perpendicular distance, ¢1, from the base, a1, may be found by ‘interchanging a with a, in the formula. _ The above formula may also be written in the following form, giving the distance from base a in terms of ay: C= Cre } PR rae oot ere, eS eR oe Brae, (22) where a1 + 2 =| a Cc — and — Qa, a1 c /a c sent) + a Values of Cy, = -— 2(1 fieek, & =) ai ay, Values of s a an Qa | OO OO | S|, | | SY NUL 0.2 0.4 | 0.433 | 0.451 | 0.468 | 0.484 0.6 | 0.385 | 0.403 | 0.419 | 0.434 | 0.450 | 0.464) 0.477) 0.489 0.8 | 0.350 | 0.367 | 0.383 | 0.399 | 0.414 | 0.428) 0.442) 0.454] 0.466 1.0 | 0.322 | 0.339 | 0.356 | 0.372 | 0.387 | 0.402) 0.416} 0.430) 0.443 1.2 | 0.300 | 0.317 | 0.334 | 0.350 | 0.366 | 0.382] 0.397] 0.411] 0.425 1.4 | 0.282 | 0.299 | 0.316 | 0.333 | 0.350 | 0.366] 0.381} 0.396] 0.411 1.6 | 0.266 | 0.284 | 0.302 | 0.319 | 0.336 | 0.352] 0.369] 0.384! 0.400 1.8 | 0.253 | 0.272 | 0.290 | 0.307 | 0.325 | 0.342} 0.358] 0.375] 0.390 2.0 | 0.242 | 0.261 | 0.279 | 0.297 | 0.315 | 0.332} 0.350] 0.366] 0.383 2.2 | 0.233 | 0.252 | 0.270 | 0.289 | 0.307 | 0.325} 0.342) 0.359] 0.377 2.4 | 0.225 | 0.244 | 0.263 | 0.281 | 0.300 | 0.318} 0.336] 0.353} 0.371 2.6 | 0.217 | 0.237 | 0.256 | 0.275 | 0.298 | 0.312} 0.330) 0.348] 0.366 2.8 | 0.211 | 0.230 | 0.250 | 0.269 | 0.288 | 0.307] 0.325! 0.344] 0.362 3.0 | 0.205 | 0.225 | 0.244 | 0.264 | 0.283 | 0.302) 0.321) 0.340] 0.358 | | 516 FOUNDATIONS AND FOOTINGS The design of a corner footing does not differ materially from that of any other independent footing. The punching shear is usually small, as the perimeter of the column is large in proportion to the load carried. The bending moments are determined in the usual way. INDEPENDENT FOOTINGS WITH PILES If piles are used to carry the column load, the size and shape of the footing will depend upon the spacing of the piles. If the column is round or square, the piles should be placed so as to make the footing as nearly square as possible. In many cases, it is not possible to space piles so as to obtain a square footing. The shape of footing for different numbers of piles is shown in Fig. 171, p. 517. For dif- ferent spacing, the shape of the footing will remain the same, but the dimensions of the sides will be altered proportionally. For concrete piles carrying large loads per pile, the spacing of the piles will depend upon circumstances, as discussed on p. 542. Design of Footing on Piles.—In the design of footings supported on piles, the upward reactions of the piles should be treated as con- centrated loads. In figuring punching shear, only the reaction of piles outside of the pedestal need be considered. Bending moments to be used in determining the steel areas are equal to the reactions of the piles multiplied by their distance from the edge of the pedestal. The piles producing bending moments in the direction considered are those within the trapezoid produced by the corner diagonals. If the lines of the trapezoid cut any piles, one- half of the pile so cut should be considered as acting in each of the two directions. | Placing of Reinforcement.—The reinforcement required by the bending moments should be distributed uniformly over the whole effective width of the footing, as in the case of footings resting on soil, because the object of the reinforcement is to make the whole footing act as a unit. Such an arrangement reduces the number of layers and also provides reinforcement for cross bending. Some designers treat each line of piles separately. They con- centrate the reiaforcement required by the bending moment pro- duced by each line of piles within a width across the line of piles somewhat larger than the diameter of the pile. Such arrangement is shown in Fig. 172, p. 518. 3 REINFORCED CONCRETE FOOTINGS 517 “| 2.6" t= > 2-6 4 1 Pile Footing 2 Pile Footing 4. Pile Footing § Pile Footing ("6 fares ey | ig 3 , ‘ , - 15 Pile Footing ait 42-2 % rid ‘ a ~ at 17 Pile Footing 18 Pile Footing 19 Pile Footing Fic. 171.—Shape of Footings for Different Arrangement of Piles. (See p. 516.) 518 FOUNDATIONS AND FOOTINGS This arrangement is wrong, for two reasons. First, there is no provision made for cross bending. It has been established that ina circumferential cantilever, such as an independent footing, circumferential stresses are developed in addition to radial stresses. The usual arrangement, where bars are placed in two directions, provides for these stresses even if they are not actually computed. In Fig. 172, there is no steel which could perform this function. The second objection is that, usually, | : to take care of bond stresses, a large num- Fic. 172.—Incorrect Arrange- ber of bars of small diameter must be ment of Reinforcement above used, and if these are concentrated in a Piles. (See p. 516.) small width, the spacing must be made very small. When, on the other hand, a small number of bars with larger diameter is used, the bond stresses are excessive, thus reducing the value of steel. Continuous WALL COLUMN FootiNnGs When it is not permissible to go outside the building line with the footing, independent footings cannot be used for the wall columns. In some such cases, it is possible to use a narrow footing of a width equal to the thickness of the basement column and extending the full distance between the columns. The basement wall may be used as part of the foundation. , Bending Moment Reinforcement.—The wall, as far as the reac- tion of the soil is concerned, acts as a beam extending between the columns. If the footing extends over several panels, the interior panels may be treated as continuous and the end panels as semi-con- tinuous. The bending moments recommended are somewhat differ- ent from those used for continuous beams. Let w = reaction of soil per lin. ft. of span in Ib.; 1, = net span in f{t.; | = span, center to center of column, in ft.; then Negative moment at center of interior span, wig wll; a! 12will, Fi : M= Tees in-lb. 2 yap (23) REINFORCED CONCRETE FOOTINGS 519 Positive moment at support, M, = “2 Tie eee suri cin. —lOsesoteety hs tow oe (DA) In exterior panels, increase the bending moment in the center and at the interior support by 20 per cent. It should be noted that the loads are acting upward. The eHatatt of the stresses and of steel, therefore, will be opposite to the position of stresses and of steel in ordinary beams. The design of such a footing is given in the example below. Example 5.—Design footings for wall columns spaced 20 ft. on centers, the load on which is P = 440000 Ib. The width of the column is 36 in. and the allowable soil pressure is p = 10 000 lb. Solution.—Assume weight of wall and footing W = 60000 lb. Total load on foundation is P + W = 440000 + 60 000 = 500000 lb. The required area P+W i 500 000 acne 10 000 Since the footing extends all the way between two adjoining columns, its length per column is equal to the spacing of columns, in this case 20 ft. The required width, therefore, is 50 + 20 = 2.5 ft. The wall column will be of the same thickness, to make the distribution of the load uniform over the foundation. Design of Wall.—The wall will be placed flush with the outside face of the column. The footing will be stepped inside. In this type of footing, the wall acts as a beam loaded by the upward reaction of the soil and supported at the columns. The reaction per lineal foot producing bending moment is obtained by dividing the column load by the length of the footing (i.e., the spacing of columns). The weight of wall produces pressure on the foundation, but does not produce any shear or bending moments in the wall. The reaction per lineal foot is 440 000 + 20 = 22 000 Ib. per lin. ft. of the footing, Shear and Diagonal Tension.—The thickness of the wall is governed by the requirement that the unit shearing stress shall not exceed the allowable unit stress. With columns 36 in. mice Hn net span is 17 ft. and the external shear at the edge of the column is V = 4,5 X 22 000 = 187000 lb. Since the effective depth of the wall is 11 ft., or 132 in., and the allowable unit shearing stress v = V 120 lb. per sq. in., the thickness of the wall, found from formula, b = Wd’ is a 187 000 a Reio0 ix 087s 182) Make wall 14 in. thick. Diagonal Tension Reinforcement.— Diagonal tension reinforcement consists of stirrups, the size and spacing of which may be determined as for a beam. Formula (40), p. 248, may be used. Also, tables on p. 899 will be found useful. 5b20 FOUNDATIONS AND FOOTINGS Projection of the Wall.—The base of the footing is wider than the wall; therefore, it must project outside the wall within the building. The projection of the base must be made strong enough to prevent it from breaking off. It should be designed as a cantilever and, if necessary, reinforced with bars placed near the bottom at right angles to the wall. If the projection is small, it should be made thick enough to withstand the bending stresses without reinforcement. The ratio between the depth and length of the projection may be taken from the table on p. 481. For large projections, it is more economical to use reinforcement. The steel must be placed near the bottom at right angles to the wall. It should run to the outside face of the wall and there be bent up and extended far enough into the wall to transfer to the wall the bending moment produced by the cantilever. The moment on the cantilever produces torsion in the wall. This is partly offset by the earth pressure against the wall. Bending Moment Reinforcement.—In the case under consideration, w = 22,000 Ib., 11 = 17 ft., andl = 20ft. The bending moments are: Negative bending moment in the center, ue a 12 X.22.000 X 17 X 20 id 16 Positive bending moment at the support, Mi = 22 000 X 17 XK 20 = 7 480 000 in.-lb. The required area of steel can be obtained from Formula (8a), p. 204, using effective depth of wall as d and j as). 5 620 000 $ X 132 * 16 000 = 5 620 090 in.-lb. Top reinforcement As = = 3.0sq.in. Use 4 — 1-in. rd. bars. ; 7 480 000 ; Bottom reincorcement at columns = = = 4:0 sq. ins aise z xX 182 & 16 000 4 — 1-in. sq. bars. Bond Stresses.—Shear at column edge, as computed previously, is V =187 000 lb. Perimeter of four one-inch square bars is to = 4 X 4 = 16 in. From formula, u = ——, Lojd Bi 187 000 ~ 16 x 0.87 x 182 u = 100 lb. per sq. in. If deformed bars are used, this is satisfactory. For plain bars, bond stresses must be reduced to 80 Ib. per sq. in. by using an increased number of smaller bars of the total required area, the perimeter of which equals 16 X 482 = 20 in. Nine j-in. round plain bars, having an area of 3.98 REINFORCED CONCRETE FOOTINGS o21 sq. in. and circumference of 21.3 in., could be used. The use of deformed bars, however, in this case is distinctly advantageous. Arrangement of Steel.—Since the wallis deep in proportion to the span, it is not advisable to bend any of the top reinforcement. The best method is to use straight bars at the top. Two one-inch round bars will extend the whole length of the footing and will be made 20 ft. long. Two other bars will extend 12 in. plus or minus beyond the points of inflection on each side. The distance between the points of inflection is equal to 2 of the net span of 17 ft., or 10 ft. 2 in.; there-. fore, the bars will be made 12 ft. 3 in. The negative steel placed at the bottom will extend 12 in. on each side beyond the + of the net span. Compressive Stress.—Since the ratio of the areas of steel to the area of con- crete is less than 0.0077, compressive stresses are smaller than the allowable values and need not be computed. Also, since p: is less than 0.0077, the value of j will exceed the value j used in the calculation. This discrepancy is on the side of safety, as it lowers the unit bond stress and the unit shear and reduces slightly the stress in the steel. Stresses in the Wall Due to Earth Pressure.—The design of the wall for earth pressure is discussed on p. 640. In the present case, the wall will be considered as supported at the top and bottom. The earth pressure equivalent to a fluid pressure of a 30-lb. fluid will be assumed. The maximum pressure at the bottom is 9 X 30 = 270 Ib. per sq. ft. The bending moment is found by Formula (4), p. 640. M =0.77 X 270 X 9? =16 800 in.-lb. per horizontal foot of wall. If the wall 14 in. thick is considered as plain concrete without reinforcement, the _tensile stress in concrete may be computed from the formula for homogeneous beams, f, = = It is _ 6 X 16 800 2149-9 Fe 19 142 Since, in addition to being supported on: top and bottom, the wall is also sup- ported at the columns, the actual stresses in the wall due to the earth pressure are much smaller than computed above, and would come below the 40-lb. tensile stress allowed for plain concrete. Therefore, no steel is required except some reinforcement for contraction due to temperature and shrinkage. This will be assumed as 3-In. round bars 18 in. on centers, both ways. In addition, the wall is reinforced with inch reinforcement. Projection of the Wall.—The footing is 30 in. wide and the wall 14 in. wide. With wall flush on one side, the projection is 16 in. The unit reaction of the soil (column load divided by area) is 8 800 lb. per sq. ft. Therefore, considering the projection as a cantilever, the bending moment at the edge of the wall is, M = 8800 x 18 x 8 = 93 900 in--Ib. per ft. The required depth, from Formula (11), p. 208, in which for f; = 16 000 Ib. per Sq. in., fe = 650 Ib. per sq. in., and n = 15, Ci = 0.028 is d = 0.028 V93 900 = 8.5 in. 522 FOUNDATIONS AND FOOTINGS The required area of steel, As = 0.0077 X 8.5 X 12 = 0.79. Allowing 33 in. below steel, the total depth of the projection is h = 8.5 + 3.5 =s12 in. COMBINED FooTINGS It is sometimes necessary to combine the footings of two or more columns. ‘This is required, for instance, when a wall column is placed at the building line and it is not permissible to project the footing of the column outside of the building line. An independent footing, in such a case, would be eccentrically loaded. To insure even distribution of the column load on the foundation, the footing for the wall column may be combined with the footing of the adjoin- ing interior column. | Combined footings are sometimes used for a row of columns, in cases where the soil is compressible and it is desirable to avoid inde- pendent settlement of any one column. Thus, in the new Massa- chusetts Institute of Technology buildings, due to the peculiarity of the ground, the footings of all wall columns were made continuous even when independent footings were feasible. Combined footings may also be used where the distance between adjoining columns is fairly small (as in corridor columns) and it is cheaper to build one combined footing for both columns than two independent footings. Pedestals.—The paragraph on Pedestals, p. 486, in connection with independent footings, applies equally to combined footings. General Requirements.—To insure uniform distribution of the column loads on the foundation, it is absolutely necessary that the center of gravity of the upward reaction of the soil should coincide with the center of gravity of the column loads. If the footing rests directly on the soil, then this requirement is fulfilled if the center of gravity of the area of the footing coincides with the center of gravity of the column loads. For pile foundations, the center of gravity of the piles (and not necessarily that of the footing) must coincide with the center of gravity of the column loads. Shape of the Base of Combined Footing.—The shape of the base of the footing will depend upon conditions. IH a combined footing is used for two unequal column loads, the base may be made in the shape of a trapezoid, the center of gravity of which coincides with the center of gravity of the column loads, REINFORCED CONCRETE FOOTINGS 023 The parallel Sided: are placed at right are to the line joining the two column centers, the longer side of the trapezoid being at the heavier column. It is more desirable, however, to use a base of footing rectangular in shape, as shown in Fig. 173, p. 530, as the design and construction of a rectangular footing is simpler than that ‘of a trapezoidal footing. The inequality of column loads is taken care of by extending the footing the required length beyond the heavier column. ‘The rectangular footing can be used only when the wall column load is equal to or smaller than the interior column load. With equal column loads, the combined footing should be sym- metrical when possible. Continuous footing for a row of equal col- umn loads should extend beyond the end columns, to provide sufficient bearing area on the soil for the same. The method of procedure in determining the base of facie is a8 follows: The required area of the footing is computed. The center of gravity of the column loads is found. This point is then con- ’ sidered as the center of gravity of the rectangle or the trapezoid (whichever shape is selected) and the dimensions of the base are so selected as to give the required area, as previously computed. If the combined footing is for a wall column and an interior column, the footing at the wall column is made flush with the wall. The distance from the center of gravity of the column loads to the outside face of the wall is, therefore, fixed. This fixes the length of the rectangle, because the total length must be equal to twice this fixed distance from the center of gravity of the column loads to the outside face of the wall. When the length has been found, the width of the footing is determined by dividing the required area of the base by the length of the rectangle. If the combined footing is for two interior columns, the length of the rectangle is not fixed. Any number of combinations of width and length of footing are possible. The arrangement is most eco- nomical if the length of the longer cantilever, measured from out- side face of column, is equal to 0.35 of the net span between the columns. After ae length is determined, the width of the footing is computed from the required area. Design of Combined Footing.—In most cases, oe bending moments and external shears in a combined footing may be found by statics. After they are determined, the footing is designed by formulas and principles given in the chapter on Reinforced Concrete. The combined footing may be a slab of uniform thickness (Fig. 524 FOUNDATIONS AND FOOTINGS 173). If the thickness required for punching shear at the columns is larger than the depth required for bending moments and diagonal tension, a pedestal of proper depth and width may be placed over the slab at the column. When the shearing stresses permit it, the cost of the footing may be reduced by eliminating some of the concrete in tension and using an inverted T-section for the cross section of the footing. The bottom flanges of the T-section act as cantilevers. If the tensile stresses exceed the allowable stresses for plain concrete, sufficient reinforcement should be used in the flanges of the T-section to resist the bending moment. The whole section must be built in a continuous operation. Bending Moments.—Combined Footings for Two Columns.—A combined footing for two columns consists of either one or two cantilevers and a center span between the two columns. The footings are subjected to the reaction of the soil, which acts upward; hence, the bending moments are of opposite sign to the bending moments in ordinary beams where the loads act downward. On the cantilever, the bending moments are positive, producing tension in the lower part of the footing. In the middle portion of the center span, the bending moments are negative, producing tension near the top of the footing, while at the columns they are positive, 1.e., of same character as for the cantilevers. Let M, = maximum bending moment at the column, Ms = maximum negative bending moment in middle of center span; w = upward soil pressure, lb. per lin. ft.; 1, = net span between columns in feet; l, = the length of cantilever in feet, I The maximum positive bending moments on the cantilevers, acting at the outside edge of the column or pedestal, are found by multiplying the upward soil pressure on each cantilever by the dis- tance from its center of gravity to the edge of the column or pedestal. Maximum Bending Moment for Cantilever. wl? . M, = rer ft.-lb. = 6wlo? in-Ib. . . . ~ (28) The positive bending moment at inside edge of the column (in the center span) is theoretically equal to the maximum bending moment in the adjoining cantilever. REINFORCED CONCRETE FOOTINGS 525 The magnitude of the negative bending moment in the middle portion of the center span may be obtained by combining the nega- tive bending moments for the span, considered as simply supported, with the positive bending moments at the columns, produced by the cantilevers. When the bending moments at the two columns are equal (or nearly equal), the maximum bending moment acts in the middle of the center span and is equal to the maximum bending moment for a simple span minus the bending moment at the column. Maximum Bending Moment in Middle of Center Span. Then wl? Mz = M; — — ft-lb. Bs Say ey gs ge (26) For inch-pounds, multiply by 12. For a large difference between the cantilever bending moments at the two columns, the maximum bending moment in the center span will act off center, and the above rule will not apply. In such case, the maximum bending moment can be obtained by plotting the pos- itive and negative bending moments and determining the largest value by scaling or analytically, from the principle that the maximum bending moment in the center span acts at the point of zero shear. If the column load at the right is P1, then the point of zero shear is at a distance equal ton! from the right edge of the footing. Let Is be the distance of this point from the inside edge of the column, then 2 My = My ~ 2 feb. | seaman ba The above formulas are based on the assumption that the bending moments at both sides of the column are the same. This assumption is correct if the effect of the rigid connection between the column and the footing is not considered. Actually, this effect is appreciable enough to affect the bending moments in the center span. For cantilevers smaller than 0.3 of the net span the bending moment at the columns in the center span is larger than the bending moment produced by the cantilever. On the other hand, for long canti- levers, part of the bending moment due to the cantilever is absorbed by the rigidity of the column, so that only one part of it is transferred to the center span. 526 FOUNDATIONS AND FOOTINGS To provide for this, the positive bending moment at the inside wl 2 ———— ANI aon edge of a column with cantilever should not be less than 16 Dy) wall column not less than ae Also, in the middle of the center span, sufficient reinforcement should be provided for a negative ~ wh? 20.’ smaller, as would happen for cantilevers longer than 0.38/;. bending moment even if the computed bending moment is Combined Footing for a Row of Columns.—A combined footing for a row of three or more columns may be treated as a continuous beam. With properly proportioned footings, there is smaller chance than in floor construction of unequal reactions on the various spans due to the variation of the live load, because there is small chance for the aggregate area of the floor tributary to one column to be much more lightly loaded by live load than the aggregate area of the adjoining columns. Therefore, not as much provision for the unbalanced load- ing need be made as is customary in continuous beams, and the bend- ing moment in the center of spans may be smaller than in continuous beams in a floor. It may be made more nearly equal to the true theoretical moment with all spans loaded. It is a matter of judgment just how nearly equal the two bending moments can be made. An inexperienced designer is advised to use the same increased bending moment coefficients as recommended for continuous beams. If the spans are unequal, special computations should be made. The tables and diagrams given in Volume III may be used. External Shear. Footings for Two Columns.—Shears for com- puting diagonal tension and bond stresses are determined by statics. The shear from the cantilever at the edge of the column equals the total upward load on the cantilever. The shear at the edge of the column in the middle span equals the shear from the center span, if considered as simply supported, plus the shear due to the bending moment on the cantilevers. When the cantilevers on both sides are equal, the shears due to cantilever action balance; hence there is no increase in shear. For unequal cantilevers, the external shear at the edge of the columns with the larger cantilever will be increased by a value equal to the difference between the bending moments on the two cantilevers about the center of the columns, divided by the distance between the column centers. At the other column, the shear will be decreased by the same amount, The same results may be obtained REINFORCED CONCRETE FOOTINGS 527 by subtracting at each end, from the column load, the upward reac- tion on the area between the edge of the footing and the inside edge of the column. External Shears for a Row of Columns.—The shears should be computed in the same manner as explained for continuous beams. Transverse Bending Moments.—If the width of the combined footing is larger than the width of the column, the upward pressure of the soil will produce a bending moment about an axis obtained by connecting the outside edges of the two columns. The part of the footing projecting beyond this axis acts as a cantilever, and the upward soil pressure produces a positive bending moment along the axis. This requires reinforcement placed near the bottom of the footing and running across the footing; this reinforeement may be either concentrated at the columns or spread along the whole width of the footing. While there is a preference for the former method, the latter method seems more logical. The transverse reinforce- ment is intended to prevent upward bending of the whole footing and not only of the portion near the column. The bending moment is produced all along the axis. It does not seem reasonable to resist it a great distance from the place where it was formed. The authors recommend, therefore, uniform spacing of the transverse bars over the whole footing. Diagonal Tension.—Diagonal tension stresses are likely to be large and should always be computed. Reinforcement should be used to provide for diagonal tension, when required by stresses. The external shears to be used may be computed as explained in the previous paragraph, under proper headings. The diagonal tension reinforcement is best provided for by means of stirrups, because, owing to the large depth of the footing relative to its length, it is seldom feasible to bend the longitudinal bars in such a way as to make them useful for shear and bending moment. Reinforcement for Combined Footing. Footing Carrying Two Columns.—The main reinforcement con- sists of longitudinal bars placed near the top of the footing in the center span and near the bottom at the two columns. The areas of steel are found from bending moments computed as explained on p. 524. At least 50 per cent of the steel in the center span should extend from column to column, The remaining bars may be bent down at 528 - FOUNDATIONS AND FOOTINGS proper places and carried near the lower face of the footing into the cantilever to serve as reinforcement for the center span at the column and also as cantilever reinforcement. Sometimes, the depth of the footing is too large in proportion to the length, so that the bend of the bars would be too steep. In such cases, the bars may remain straight and extend about a foot beyond the points of inflection. Bars should be investigated for bond stresses. The steel at the columns consists of bent bars carried from the middle span, short bars, or a combination of both. Bond stresses are always very high and should be investigated. Although the bending moment is the same at both sides of the column, the shear is much larger at the inside edge. ‘This may require a much larger amount of steel than that required by bending moment. Cross reinforcement should be placed as explained under proper heading. Bond Stresses.—Bond stresses in footings are always high and should be computed. In the middle span for the top steel, the shear at the point of inflection should be used in Formula (50), p. 262. For the bottom steel at the column, the bond stresses should be com- puted at both sides of the column. At the column, the bending moment is the same on both sides, so that the number of bars required by bending moment is the same. Usually, however, the external shear at the inside edge of the column is larger than the external shear at its outside edge. Since the bond stresses depend upon shear, they are larger at the inside edge. The explanation is that the rate of increase in bending moment at the inside edge of the column is much greater and the tensile stresses in steel are developed in a much shorter distance than in the cantilever. In such cases, to reduce the bond stresses, the amount of steel must be increased, also the steel must be anchored at its free end to permit large allowable bond stresses. The example below illustrates the design of a combined footing. Example of Combined Footing. Example 6.—Find the dimensions of a combined footing for a wall column with a cross section 38 in. by 24 in., carrying a load P; = 400 000 Ib. and an interior column 30 in. square with a load P2 = 580 000 Ib. The distance between column centers is 15 feet and the allowable unit pressure on soil p = 8 000 Ib. per sq. ft. The outside edge of the foundation must coincide with the outside edge of the wall column. REINFORCED CONCRETE FOOTINGS 529 Solution. Area of Footing.—Assume dead load of footing, W = 70000 lb., then total load on foundation, Pi + P2 + W = 400000 + 580000 + 70 000 = 1050 000 lb. For allowable unit pressure, p = 8 000 lb., the required area of Pe 1-050:000 footing is ee 131 sq. ft. Center of Gravity of Column Loads.—The distance of the center of gravity of the column loads from center of wall column is found by computing the moment of the column loads about the center of wall column and dividing it by the sum of the column loads. Dead load of footing needs no consideration in determining the center of gravity of the column loads. The column load Pz» is the only load producing a bending moment about the center of the wall column. This moment equals 15 X 580 000 ft.-lb. The sum of the column loads is 400 000 + 580 000 = 980 000 Ib.; hence, the distance from center of wall column to center of gravity 15 X 580 000 ee = 8.9 it, 980 000 Shape of Footing.—Use a rectangular footing with one side flush with the edge of the wall column. Its length will be determined from the requirement that the center of gravity of the rectangle should coincide with the center of gravity of the loads. The distance from the edge of the footing to the center of the wall column is 1 foot. Since the distance from center of wall column to center of gravity of the loads, as computed above, is 8.9 ft., the distance from the edge of the footing to the center of gravity equals 1 + 8.9 = 9.9 ft. The center of gravity of a rect- angle is in its middle. Therefore, 9.9 ft. found above, equals one-half of the length of the rectangle. Its full length, therefore, is 9.9 K 2 = 19.8 ft. Width of Footing.—Since required area of footing is 131 sq. ft., and its length 131 19.8 ft., the width equals Tee 6.6 ft. Use 6 ft. 9 in. Unit Pressure for External Shears and Bending Moments.—Bending moments and external shears are produced by the column loads only, exclusive of the dead load of the footing. 400 000 + 580 000 Unit pressure per sg. ft. is —79.8X6.75 = 7 340 lb. 400 000 + 580 000 Pressure per lineal foot of footing arta = 49 500 lb. Bending Moments.—As evident from Fig. 173, p. 530, the footing consists of the center span and of the cantilever beyond load P». Bending Moment of Cantilever—Length of cantilever to the edge of the out- side column is 2.55ft. Since pressure per lineal foot equals 49 500 lb., the bend- ing moment on the cantilever is total load multiplied by moment arm, or 2.55 : M, = 49 500 X 2.55 X = < 12 = 1930 000 in.-lb. This moment is positive, causing tension at the bottom face of the footing. 530 FOUNDATIONS AND FOOTINGS Bending Moment in Center Span.—At the inside face of the interior column, the bending moment will be either the same as produced by the cantilever or 1,2 eee ae ft. lb., whichever is larger (where w = pressure per lineal foot and 1; distance in feet between inner faces of columns). For w = 49 500 lb. h = wl 12 [15 — (1.0 + 1.25)] = 12.75 ft., the bending moment from formula Ae is 12.75? pies et IEE ius ~ = 503 000 ft.-lb., or 6 036 000 in.-lb. This moment is larger than the bending moment due to the cantilever and will be used in computing the required reinforcement at the column. Elevation "Ra, Stirrups 12 legs each , Plan Fic. 173.—Combined Footings. (See p. 529.) In the middle portion, the maximum bending moment will be computed at the point of zero shear. The distance of the point of zero shear from the outside edge of the wall column equals the wall column load divided by the pressure per 400 000 49 500 ft. from inside edge of column. The maximum negative bending moment is now found by statics. Disregarding the indefinite amount of restraint at the wall column, the negative moment, from Formula (27), p. 525, where M, = 0 and w = 49 500 lb. and J; = 6.1 ft., equals lineal foot. Itis — 8 1 ft. from outside face of column, or 8.1 — 2 = 6.1 6.1? M, = — 49500 X ih gant ak 921 000 ft.-Ib. or — 11 052 000 in.-lb. REINFORCED CONCRETE FOOTINGS 531 External Shears. Shear Due to Cantilever.—External shear due to the cantilever at the outside edge of interior column, is equal to the pressure per unit of length multiplied by - the length of the cantilever in feet. It acts downward and is therefore negative. Vi = 49 500 X 2.55 = 126 200 lb. Shears in Center Span.—The external shears at the inside edge of the columns will be computed by subtracting the upward pressures from the respective col- umn loads. External Shear at Inside Edge of Interior Column. V2 = 580 000 — 49 500 X 5.05 = 330 000 lb. External Shear at Inside Edge of Wall Column Vs; = 400 000 — 49 500 X 2 = 301 000 lb. Depth of Footing.—Depth of footing is determined either by bending moment or by diagonal tension, whichever gives the larger value. The depth should be checked for punching shear and, if required, a pedestal should be added on the top of the footing. Depth Determined by Bending Moment.—In this case, the largest bending moment is the maximum negative moment of the center span, M = 11 052 000 | M in.-lb. Using Formula (1), p. 204, d = C . ae in which width of footing b = 81 in. and for unit stresses f, = 650 lb. per sq. in., fs = 16 000 lb. per sq. in. and n = 15, C = 0.096 (see p. 880). Depth as required by bending moment 11 052 000 . d = 0.096 (ee = 5055 In. Diagonal Tension.—The largest external shear is at the inside edge of the interior column. For the depth found above, d = 35.5 in., and a width of foot- Vv ing equal to 6 ft. 9 in. or 81 in., the unit shear from formula v = bid? is J 330 000 — a BA 0.875 X 85.5 = 131 lb. per sq. in. This value is larger than the allowable working limit, which is 120 lb. per sq. in. The depth as determined for bending moment must be increased in the proportion of 130 to 120. The depth as required by diagonal tension is d = 35.5 X 434 = 38.7 in. Diagonal tension reinforcement must be provided for sections where v exceeds 40 Ib. per sq. in. Longitudinal Reinforcement.—Using depth of slab d = 38.7 in. as required by diagonal tension, the amount of steel at the various sections will be found from equation M maga 532 FOUNDATIONS AND FOOTINGS where j = 0.875 and f, = 16 000 lb. per sq. in. Center of Footing, M, = 11 052 000 in.-lb. A, = 19.5 sq. in: Twenty-six $-in. bars will be tried. Center Span at Interior Column, M = 6 036 000 in.-lb. Age=" 11 2-s05107 Twenty-six 3-in. round bars will be tried. The bars must be extended into the cantilever. Cantilever, M, = 1930 000 in.-lb. A; = 3.6 8q. In. This amount of steel is smaller than the amount extended from the other side of the column. Bond Stresses.—It is necessary to compute bond stresses in top and bottom reinforcement, as the footing may fail by slipping of the top or bottom bars. Bond Stresses in Top Reinforcement.—Since, theoretically, there is no bending moment at the wall column, the bond stresses in the top steel there will be com- puted as for simply supported beams (note that top steel in footings corresponds to bottom steel in ordinary beams). At the other end of the footing, the bond stresses are smaller, because there they have to be computed at the point of inflection where the external shear is smaller. The shear at the edge of wall column is V; = 301 000 lb. The bond stress with twenty-six {-in. square bars, having a total perimeter 20 = 26 X'3.D.= Olam from formula wu = Soja’ iS 0 301 000 08 1b : “= : meh. 91 X EX 38.5 Pe This is satisfactory for deformed bars. Bond Stresses in Bottom Steel—The amount of bottom steel is the same at both sides of the interior column. The external shear, however, is larger at the inside edge. The bond stresses will be figured for the larger shear. The perimeter of twenty-six 4-In. Sq. bars is 61.2 in. For Vz = 330000 lb. and depth 38.7 in., the bond stresses are 330.0000) 100 aa = = . per sd. 4 R198 SPO EST Be ae U To reduce the excessive bond stresses, use twenty-nine 5_in. sq. bars, the perim- eter of which is 29 X 2.5 = 72.5 in. 330.000 72.BI toe o0.t uU = 134: Further reduction of bond stresses by reducing the diameter of bars is not advis- able. Instead, the depth of the footing may be increased by providing a block << STRAP BEAMS TO CONNECT FOOTINGS 0933 at the column, and the bars will be extended beyond the point of inflection, in which case larger bond stresses are allowed. Transverse Reinforcement.—The projection of the footing beyond the edge of the interior column is 2 ft. 13 in., and beyond the edge of exterior column 1 ft. 93 in. The transverse bending moment will be figured for both projections. The upward pressure per square foot is 7 340 lb. Hence, transverse bending moments per foot of length of the footing are: een M; = 7340 X 2.12 X eee 198 000 in.-lb. 21.5 _M, = 7340 X 1.79 x oe 141 000 in.-lb. For d = 38.5in. f, = 16000 198 000 0.37 : Ss ae ae y Tepe Gin al . sq. 1nN. > 16 000 X $ X 38.5 ae Use 3-in. rd. bars, 10 in. on centers. 142 000 : A sq = 0.26 sq. in. ~ 16000 X 2 X 38.5 Use 3-in. rd. bars, 14 in. on centers. STRAP BEAMS TO CONNECT FOOTINGS Instead of using a combined footing for a wall column and an interior column, as in the previous discussion, it may be desirable to build both footings separately and to take care of the eccentricity on the wall footing by connecting it with the adjoining interior foot- ings by means of strap beam. ‘The strap beam may or may not be used to transmit the load to the foundation, depending upon the design. Its primary function is to resist the bending moment pro- duced by the eccentricity of the wall column load with respect to the wall footing reaction. Such an arrangement is particularly useful when pile foundation is used for both columns, and it is desirable to place the piles in clusters next to the columns; it is also useful when the footings for both columns consist of caisson piles. Bending Moments and Shear in Strap Beams.—For the purpose of determining bending moments and shears, the strap beam may be considered as a balanced double cantilever supported at the center of the wall footing. The short cantilever carries the wall column load, P,, and the long cantilever carries the balancing reaction, which is 534 FOUNDATIONS AND FOOTINGS provided by the interior column with which the long arm is con- nected. The conditions of loading are seen from Fig. 174, p. 534. To make the bending moments statically determinate, the strap beam is considered as free to rotate at the ends. With this assump- tion, from simple mechanics, the bending moment is zero at both ends and increases to a maximum at the point of support. _--For Free End of Long Arm P,=600,000 tb ~ ek Loading Diagram | 5-1" 8q. Bars Caisson---- 0 ee Design of Strap Fic. 174.—Loading and Moments in Straps. (See p. 534.) Let Pi = load on wall column, |lb.; 1 = distance between centers of the wall footing and the interior footing, ft.; e = distance between center of wall column and wall foot- re, tee R, = reaction on wall footing, lb.; | Rs = reaction on interior footing due to load P, lb.; M = bending moment. STRAP BEAMS TO CONNECT FOOTINGS 535 Then Maximum Bending Moment in Strap Beam. ese eae il, oe as es has) Reaction on Wall Footing. R, = P{1 +5) 1 Sar ous Sie a 20) Reaction on Interior Footing due ta Load P. Ro = — Pi, Ib. Mee eee ese La} Dowels Point of Zero Shear NV ‘- Stirrups Elevation Fic. 175.—Strap for Footing with Uniformly Distributed Reaction. (See p. 536.) It should be noted that the load for which the wall footing is designed is larger than the wall column load. The load on the interior footing is decreased by the reaction Ro. The bending moment in Formula (28) is based on the assump- tion that the reaction R, is concentrated in one point. This assump- tion gives the maximum possible bending moment. Actually, the reaction is distributed and.the bending moment is therefore smaller. Two cases may occur: (1) the reaction A; consists of uniformly 536 FOUNDATIONS AND FOOTINGS distributed upward pressure; (2) the reaction consists of several rows of piles. In both cases the maximum bending moment acts at the point of zero shear. For uniformly distributed reactions, the distance of the point of zero shear from edge of footing is found by dividing the column load P, by the reaction per lineal foot. Such a condition is shown in Fig. 175. Assume that the unit reaction of foundation is p and that P = pag; then ag is the location of the point of zero shear from edge of footing reactions. If a; is the distance between the load P and the point of zero shear, then M = —Pi(a—$). . Ree aS Lee Fic. 176.—Straps for Footings on Piles. (See p. 536.) For footings supported on piles, the location of maximum bending moment is at point of zero shear or point where shear changes sign. Usually this will be at the center of the last row or piles. If a is the distance from this point to the load Pi, qi is the rezstion of the first row of piles and its distance a2, gz the reaction of the second row of piles and its distance a3, then = — [Pya, — (qid2 + qoaa)].. . . - « (82) For a larger number of piles it may be necessary to use a3, a4, etc. Dead Load of Footing and Strap Beam.—The dead load of footing and strap beam should be added to the footing load, f1, in computing its size. In figuring the bending moment of the strap beam, however, the dead load of the strap may be disregarded. STRAP BEAMS TO CONNECT FOOTINGS 507 Design of the Strap Beam.—The wall footing is designed first, and its center fixed. This gives the value of e and l. With these, the bending moment is determined. The width of the strap beam is made the same as the width of the column. When the bending moment and the width are known, the depth is computed from Formula (1), p. 204 (a = ales) For the selected depth, the area of steel is computed from Pe jaf The depth required by the bending moment is usually larger than that required by diagonal tension. It should be noted that the external shear producing diagonal tension in the long cantilever is A, constant and is equal to P be Since the bending moment decreases toward the interior column, the depth of the strap beam may also be decreased. The minimum depth, of course, may not be smaller than required by diagonal tension produced by the shear Bir Reverse Bending Moment at Interior Column.—In the above discussion, it was assumed that the end of the cantilever at the interior column is free to turn. Since the strap beam must be connected either with the footing or with the column, this assump- tion is not correct. Owing to the rigidity of the connection of the strap beam with the interior column, some positive bending moment will be developed in the strap beam at the column. If the strap beam is fixed at the interior column, this bending moment is equal to one-half of the maximum bending moment, Pe. For intermediate conditions, intermediate values may be assumed. It is advisable to make generous provision for this bending moment. Provision for Unequal Settlement.—The two footings joined by a strap beam may settle unevenly. This may produce stresses of opposite character to the stresses for which the footing is designed. To provide for this, the strap beam should be reinforced on top and bottom for its whole length. The authors recommend the use of one-third the amount of the steel in the bottom. This recommen- dation is arbitrary. Anchoring Reinforcement at Wall Column.—In the short canti- lever, full tensile stresses in the kars must ke developed in the short 538 FOUNDATIONS AND FOOTINGS distance between the center of column and the center of footing. This distance usually is not sufficient for the purpose. The reinforce- ment must be extended to the edge of the wall column and then pro- vided with a hook sufficient to develop the tensile stresses in the bar. Without such provision the construction would fail by bond. Types of Strap Beams.—F'igures 174 to 176 illustrate several of the types of strap beams most commonly used. Example of Strap Beam Design.—In Fig. 174, p. 534, the columns rest on caissons. The data are as follows: P, = 600 000 lb.; P, = 1000 000 Ib.; 16 in. = 1.33 ft.; 20 ft., ® I I € 1.33 R, = Pa(i Bi | = 600 000 ( Ae) = 640 000 lb.; R, = Py = 40 000 Ib.; S T Pe = 9600 000 in.-lb. ; b= 44in. d= 45in. A, = 15.3 sq. in. The arrangement of reinforcement is evident from the figure. RAFT FOUNDATION When the allowable pressure on the soil is small, it may be neces- sary to spread the foundation over the whole area of the building. Such a foundation is called a raft foundation. It is often used in connection with piles, where they are driven in comparatively soft . strata and it is desirable to space them as far apart as possible so as not to overload the ground in which the piles have bearing. ‘The raft foundation is made either of flat slab construction or of beam and slab construction. | | Pressure on Foundation—As in the case of independent foot- ings, to prevent unequal settlement, it is necessary to design the foundation so that the unit pressure on the soil is uniform. Tor uniformly distributed floor loading, this is not difficult. If the area of the foundation is equal to the area of the building, then the load on the foundation will be equal to the sum of the loads on all floors. RAFT FOUNDATION 539 With uniformly distributed floor load, the foundation load will also be uniform. The difficulty begins when the floor loads are not uniform, or when some columns carry additional loads. (Columns carrying the tank, for instance.) At wall columns also, the loading is proportion- ately larger than at the interior columns because it consists not only of the floor loading, but also of the weight of the wall. To provide for the additional load, it is necessary to extend the foundation out- side of the building lines. The additional area to be provided is equal to the excess load on the wall column divided by the unit pres- sure on the foundation. When it is not possible to extend the foundation outside of the building, uneven pressure on the founda- tion will result. When this increased pressure does not exceed the maximum allowable pressure, the raft foundation may be used; but the construction between the wall columns and the first row of interior columns should be made strong enough not only to resist the upward pressures, but also to even up the settlement of the two columns. Uneven settlement of the wall column produces negative bending moment at the interior column and positive bending moment at the wall column. For large differences in unit pressure produced by the heavy columns, the design may be equalized by using piles under the heavy Arias If raft foundation is used in connection with piles, the TUN in column loading may be taken care of by using closer spacing of piles at the heavier loaded columns. Flat Slab Foundation.—The foundation may be of flat slab design, consisting of flat slab, drop panels, and column heads, as described in the chapter on Design of Flat Slab Structures, p. 319. The same formulas may be used as recommended for flat slab floors. In this case, the pressure on the flat slab acts upward and the column load downward. The loading is of opposite sign to the load- ing in a floor; therefore the bending moments are also of opposite sign. At the columns, the bending moments are positive (instead of negative), producing tensile stresses at the bottom; therefore the tensile steel must be placed at the bottom of the slab. In the central parts of the slab, the bending moments are negative (instead of positive, as in a floor slab) and the reinforcement must be placed near the top of the slab. To save form work, the drop panel may be placed below the slab. Instead of a column head of conical design, a cylindrical block of 540 FOUNDATIONS AND FOOTINGS concrete of a diameter equal to the diameter of the column head may be used. This is placed on the top of the slab. Cinder fill, equal in depth to the height of the column head, is provided on the top of the foundation slab. Fic. 177.—Details.of Flat Slab Foundations. (See p. 539.) Beam and Slab Foundation.—The design of the foundation may consist of a combination of beams, girders, and slabs. ‘The principles of floor design must be followed in determining the concrete dimen- sions and the amount of steel. As explained in the previous case, the loading acts in the opposite direction to the floor loading, so that the location of the steel will be opposite to that in the floor, RAFT FOUNDATION 541 The beams, or ribs, may be built either above or below the slab. When the ribs are above the slab, the beams are T-beams in the center and rectangular beams at the support. With ribs placed below the slab, the beams are T-beams at the columns and rectangu- lar beams in the center of the span. The design with beams above the slab is preferable, as better bearing on foundation is obtained. The space between the ribs is filled with cinders and the basement slab is placed on the top. It is necessary that the ribs be built monolithic with the slab. There is a tendency to pour the slab first, place the forms for the ribs on the slab, and finish pouring the ribs. This is absolutely wrong, as it is — impossible to make the ribs and the slab act as a unit when they are poured in this way. CHAPTER [IX PILES A foundation may be supported on piles when the ground, for a ereat depth, consists of layers of soft materials on which it is either impossible or undesirable to rest the structure. The piles may be driven either to solid bearing on rock or hardpan or far enough into the firm ground to develop, by frictional resistance, the allowable capacity of the pile. Piles may be of wood or concrete. The choice will depend upon economy and also upon the conditions affecting the permanency of the structure. For instance, when there is danger of their being attacked by teredo, or a likelihood that the level of low water will be materially lowered by future drainage, concrete piles should be given preference even at higher cost. The two kinds of piles will be treated separately. Spacing of Piles—The minimum spacing of piles depends upon two conditions. First, the piles must not be driven close enough together to compress the ground to an extent which would make impossible straight driving of adjacent piles. Second, the spacing must be such that the load transferred to the soil in which the pile has its bearing will not exceed the bearing value allowable on such soil. For instance, if the pile is driven into clay with an allowable unit bearing of 4 tons and the capacity of the pile is 20 tons, the area tributary to this pile must be at least 5 sq. ft., which requires a minimum spacing on centers of about 27 in. If piles are driven too close together the whole cluster with the surrounding earth may sink — as a unit. The minimum spacing on centers, as specified by most codes, is from 20 to 24 in. This spacing may be used if the ground is not compressible. New York City specifies a minimum spacing of 20 in., and a maximum of 36 in. In railroad work, closer spacing than 36 in. is not permitted. For use in building construction, under ordinary conditions, the authors recommend a spacing of piles of 30 in. on centers. | 542 WOOD PILES 543 Piles for Outside Bearing Walls.—All bearing walls should have at least two rows of piles. or buildings over three stories in height, if two rows of piles are used, the spacing of the rows should be at least 36 in. on centers. . WOOD PILES Wood piles are commonly used for foundations because of their low cost. They may be either of soft timber, such as pine, spruce, or hemlock, or of hard timber, such as oak, cedar, or hickory. The following are the requirements for wood piles: They shall be sound, close-grained, solid, and without injurious ring shakes or loose knots. All knots shall be trimmed close to the body and all bark peeled soon after cutting. Piles shall be of uniform taper; not less than 6 in. in diameter at the point, and not less than 10 in. at the butt, for piles up to 25 ft. in length; and not less than 12 in. at the butt for longer piles. ‘They shall be practically straight, a criterion of straightness being that a line drawn from the center of the butt to the center of the point shall be within the body of the pile. Short bends shall not be allowed. The carrying capacity of the wood pile depends upon the char- acter of the material into which it is driven. If the pile is driven to solid bearing on rock or hardpan, through firm material which is able to keep the pile from lateral displacement, its bearing capacity is equal to its strength as a short column. In such cases, the authors recommend 20 tons as the capacity of a pile. This capacity must be reduced when the pile is driven through material which, for any appreciable distance, does not offer resistance to lateral displace- ment of the pile. Such materials are water, mud, silt, peat, or fill. The pile should then be figured as a long column of a height equal to its unsupported length and of an average cross section within this length. If the pile is not driven to solid bearing, its capacity depends upon the frictional resistance in bearing soil. It transfers the load to the soil gradually by friction. The capacity then is best determined by a load test. Where this is impossible, the capacity of the pile, i1.e., the load to be used in design, is ascertained from its penetration under the last blows of the hammer. The most common formulas for the capacity of piles are the Wellington formulas, also known as the Engineering News formulas, given below. Their use is recom- mended by the authors. : 544 PILES P = capacity of pile, lb.; W = weight of hammer, lb.; h = fall of hammer, ft.; s = penetration of pile at last blow, in., or average pene- tration for several consecutive blows. 2Wh Then P = ear for pile driven with drop hammer, . . - (1) 2Wh : P= ea for pile driven with steam hammer... . (2) Modern specifications require that s be taken as the average penetration for a series of blows, usually five to ten, and further require that the pile should be driven until the successive blows produce approximately equal or uniformly diminishing penetra- tions. | When the piles are designed for a certain capacity, they are driven until the final penetration, s, becomes small enough to give the required capacity when computed by Formulas (1)° or (2) The weight of the drop hammer and the drop being known the max- imum allowable penetration is computed and the pile driven until the penetration is reduced to this value. For instance, if a pile is designed to carry 20 tons, the weight of the drop hammer is 4 000 lb. and the drop is 10 ft., the maximum allowable penetration giving the required capacity obtained from the formula is s = one inch. The pile is driven until the penetration is reduced to one inch. If it is impossible, with the length of pile at one’s disposal, to get small enough penetration at the last blows, the capacity of the pile should be reduced according to the value of obtained final penetra- tion and the proper number of piles added to give the desired resist- ance to the foundation. The size and strength of the footings should be increased if required by the larger number of piles. Cutting of Wood Piles. All wood piles must be cut below perma- nent low water, so that the entire pile will always be wet. In such a case, unless destroyed by teredo, the piles may be expected to last indefinitely. It is important to determine whether there is any likelihood of the lowering of the low water level through future drainage. If so, the cutting level must be established with this in mind. The pile must be cut ‘to sound wood which has not been injured by the hammer, as otherwise its value would be negligible. WOOD PILES 545 Driving Piles.—Piles should be driven under the direction of men experienced in this work. They should be driven straight, since the value of inclined piles for vertical loads is small. Some engineers require that piles driven to hard strata be provided with metal shoes of proper design. The depth to which piles are to be driven should be determined by borings, in order to prevent overdriving of piles, which is even more harmful than underdriving. An overdriven pile is one that has been forced farther down after the point has reached solid bearing. The result is that it buckles, breaks, or splits, at the end. In any case its value is either lost or greatly reduced. If possible, all the piles in a cluster should be driven to the same depth. Piles are driven by a drop hammer or a steam hammer. To prevent splitting of the tops, cushions are used. tee See 200.00 300.00 Estimated Fire insurance on building. None required. .....---+---+++: 235.20 14 cts. on $100 $19 117.50 $23 275.20 Yearly expense fireproof less than non-fireproofi. es Aare $4 157.70 The yearly saving of $4 157.70 capitalized at 6 per cent represents $69 295. Therefore, actual cost of concrete building is $119 705, in comparison with one of mill construction costing $168 000. Furthermore, there is a lower rate for fire insurance on the contents of the concrete building, which still further reduces the cost. Reinforced concrete has been used economically for dwelling houses, but only where cheap cottages can be built in groups of similar pattern. With this exception, wood is cheaper, on account of the high cost of forms. Precast blocks, requiring no forms, are simplest for this class of work, but unless the surfaces are tooled the appearance is apt to be monotonous. In estimating the labor where forms are used, allowance must be made for time lost in waiting for the con- crete to harden so that the forms can be raised. For this reason, a small gang of men should be employed—only enough to lay concrete to the height of one section of the forms per day. For cellar and foundation walls of all classes of buildings, including brick and frame, concrete is superseding rubble masonry except where rubble stone is taken from the excavation so as to be very cheap. Cement mortar plastered on to metal or wood lathing is used not only for outside walls, but in some cases for fire-resisting partitions in large buildings. ACTUAL COST OF REINFORCED CONCRETE BUILDINGS 567 ACTUAL COST OF REINFORCED CONCRETE BUILDINGS The tables presented on page 568 give the approximate average cost per square foot of floor area and per cubic foot of volume of plain rectangular reinforced concrete buildings of various sizes and heights. The costs include all details of construction, not only the concrete forms and reinforcement, but also windows, stairs, roof covering, and plumbing. Interior finish, which varies widely with the type of construction, is not included. The basis * of the tables is as follows: (1) Floor loads, 150 pounds per square foot. (2) Story heights: first floor on a 3-foot fill; other floors 12 feet from slab surface to slab surface. (3) Column spacing, 18 feet on centers. (4) Floor design: girders between columns in one direction; beams between columns in other direction with two intermediate beams. (5) Excavation and foundations. + Story Height — Outside Walls, Inside Walls per Linear Foot | per Linear Foot $5.45 45 $2.80 6.85 5.00 7.20 6.00 7.60 7.05 7.90 8.45 | Sha 6.15 3.75 (6) Filling under first floor: 3-foot fill at $1.00 per cubic yard in place. (7). Stairs: material and labor, $200 per flight per story. (8) Stairways and elevator towers: 2 stairways and 1 elevator tower for buildings up to 150 feet long. 2 stairways and 2 elevator towers for buildings up to 300 feet long. 3 stairways and 3 elevator towers for buildings over 300 feet long. (9) Floor finish: all floors of concrete with granolithic finish. (10) Walls: (a) Curtain walls between pilasters, 3 feet high and 8 inches thick; (b) Concrete walls for penthouses, 6 inches thick. Dimensions of pent- house are 10 feet by 10 feet; (c) Concrete walls around the elevator and stairway openings are taken 6 inches thick, the elevator opening being 10 by 20 feet and the * Values fixed with the advice of Morton C. Tuttle Co. They conform to 1925 prices. { Taken from paper presented before the New England Cotton Manufacturers’ Association, April, 1904, by Mr. Charles T. Main. Prices revised by Mr. Main to conform to prices prevail- ing about January, 1925. 568 BUILDING CONSTRUCTION Average Costs of Concrete Buildings per Square Foot of Floor Area (See p. 567) Costs include all items except interior finish ) ea a eee ae eee ae Cost in Dollars per Square Foot of Floor Area ee ee Width Feet Length of building, in feet Length of building, in feet 50 | 100 | 200 | 300 | 400 | 600 || 50 | 100 | 200 | 300 | 400 | 600 1-Story 2-Story 25 4.94; 3.86) 3.38 3. 08! 9.95) 2.91|| 4.83, 3.73; 3.27, 3.02 2.89! 2.74 50 | 3.52 3.02) 2.66] 2.40! 2.28/ 2.21 3.46 2.74| 2.43| 2.21| 2.13) 2.07 poe 8 91| 2.78) 2.43) 2.17 2.07| 2.00|| 3.04) 2.51) 2.17 2.02 1.921 1.83 100 3.04| 2.62]. 2.28] 2.07 1.92) 1.88|| 2.85 2.32) 2.04) 1.88 1-77 P71 150 2.93| 2.49} 2.17| 1.96 138)..1 77 2. 68| 2.19) 1.92) 1.75) 1.67 1.60 4-Story 6 to 10-Story 25 4.68| 3.54| 3.08| 2.89] 2.76| 2.64|| 4.68) 3.50| 3.06 2.85! 2.78} 2.64 50 3.25| 2.53] 2.26) 2.11) 2.05 1.96|| 3.23] 2.49| 2.24) 2.11] 2.05) 1.96 75 2.85) 2.28} 2.02) 1.90 1.83) 1,77} 2281) 2228) 2.02 1.88] 1.79] 1.75 100 2.64| 2.13} 1.88] 1.75 1.69| 1.64]| 2.62} 2.09) 1.86 1.73| .1.67| 1.62 150 2.49} 2.00] 1.77| 1.64 1.88] 1.52|| 2.45) 1.96) ders 1.62| 1.58) 1.52 First floor on fill. Slab supported on the ground. To use the table multiply the aggregate floor area (exclusive of the roof area) by the unit costs. Average Costs of Concrete Buildings per Cubic Foot of Volume (See p. 567) Costs include all items except interior finish Cost in Dollars per Cubic Foot of Volume Width Fee t Length of building, in feet | Length of building, in feet Ee ee eee 50 | 100 | 200 | 300 | 400 | 600 50 | 100 | 200 | 300 | 400 | 600 1-Story 2-Story pee ee ee eee 25 0.41110. 323/0.281|0.257/0.247/0. 243 0.403|/0.310 50 0. 293/0.251/0. 2210. 200)0.190)0. 183 0.289)0.228 0.167||0.253|0.209/0.183)0. 169)0.160\0. 152 100 0. 253/0.219/0.190/0. 173/0. 160)0. 156 0. 238/0.194/0.171|0. 156/0. 148)/0. 141 150 0. 245/0.207|0. 181/0. 162/0.152/0.148 0. 224/0.183/0.160|0. 145|0. 139]0. 133 4-Story 6 to 10-Story 25 10.390|0.295/0. 257/0.240/0. 230 0.219)|0.390/0. 291/0. 255)0. 236)0. 232|0.219 50 0.270|0.211)0.188|0.175|0.171)0. 162 0.270\0.207|0.186)0.175|0.171/0.162 | 0.190|0.169|0.156|0.150|0.145 0.173|0.154/0. 143]0.139]0.135 0.162\0.143\0. 1385/0. 13110. 127 0.272)0.251|0.240'0.228 0.202|0.186/0.177|0.173 75 0.26610. 232\0.202/0.181)0.173 75 0. 236/0.190,0. 169]0. 158)0.152/0.148 0.234 100 0. 21910.177|0. 156|0. 145\0. 141)0. 137 0:217 150 0.20710. 167|0. 148/0. 137/0.133|0. 127 0.205 Values are based on conditions outlined on page 567. The tables are taken from ‘‘ Concrete Costs”” by the same authors, converted to 1925 prices, and the values are made up from tables of unit times and costs given in the same book carefully checked by contractors’ esti- mates. For more complete details and for the unit values which are adapted to all conditions, see other tables and examples in “‘Concrete Costs.” Values are for symmetrical buildings. FLOOR LOADS 569 stairways 10 by 10 feet, these two being adjacent so that the one intermediate 10-foot wall serves for both openings; (d) For toilets, concrete walls 6 inches thick and 20 feet long, one wall for each 5 000 square feet of floor space. Walls 8 inches thick, including reinforcement and forms, 75¢ per square foot. Walls 6 inches thick, including reinforcement and forms, 68¢ per square foot. (11) Windows and doors: all openings for windows and doors, 75¢ per square foot. (12) Roof covering and flashing: five-ply tar and gravel roofing, 60¢ per square foot. . (13) Plumbing: two fixtures on each floor up to 5000 square feet of floor surface, and one additional fixture for each additional 5000 square feet, $150 per fixture. (14) Labor rates: carpenter labor, $1.10 per hour; steel labor, $1.10 per hour; and common labor, 65¢ per hour. (15) Concrete in place (including labor and materials): $12.00 per cubic yard, or 44¢ per cubic foot. (16) Form lumber: $45.00 per 1 000 feet B. M., delivered. (17) Steel for reinforcement: $55.00 per ton, delivered. For lighter loads than specified, the costs are slightly decreased, this decrease running up to 25 cents per square foot for a 75-pound load in a 10-story building. For a 300-pound load, the prices are increased from 12 cents for a 2-story building up to 25 cents per square foot for 10 stories. It must be remembered that the tables are based on rectangular, symmetrical buildings. Allowance must be made for irregular layouts, which increase materially the cost of form construction. , The variation in cost due to variation in spacing of columns is small. If columns are spaced 15 feet apart the cost is 6 per cent greater than where columns are spaced 25 feet apart. FLOOR LOADS Live Loads.—In designing any structure, it is necessary to deter- mine the design live load. This should be estimated from the investi- gation of the use for which the building is designed. The live load must not be smaller, however, than required by the building code of the city in which the building is erected. It should be borne in mind that the codes usually give the minimum values. If, from investigation, it should follow that the expected live load in the build- ing under consideration is larger than the live load required by the code, the larger live load should be used. 070 BUILDING CONSTRUCTION To illustrate good practice, the following requirements of the Boston Building Code are given: Live Load Required by Boston Building Code. Pounds per Class of Buildings Square Foot a ae ee ee Ts ts Armories, assembly halls, and gymnaSiUMS......---.eeeee eee eee 100 Fire houses: Apparatus floors.......--- Save, {ey hs a Se 150 Residence and stable floors. .....2.)5-- 5. 00 eee ee 50 Garages, private, not more than two ¢ars. ei Os a ee 75 Garages, public.) 0.0 0. PS 150 Grandstands ic. o. $0. oye be 6 i oo 100 Hotels, lodging houses, boarding houses, clubs, convents, hospitals, asylums and detention buildings: Public portions... 0h SEE a ee 100 Residence portions....... .¢.0. 2047 c <4 0 a> 7s se PDS Manufacturing, heavy s.. 20 se. oe oe Ne eens Fim se 250 Manufacturing, light. 2M O eee ee 125 Office buildings: First floor...sy.eecueeeeser sete nee nn 40s seco ee 125 ‘All other floors..«. iv.n0+ 2 s* 2h sree Been ates oe Vc Wud 75 Public buildings: Public portions). £105. oy 1200s. anaes oh «ate alec 100 Office portions: ...ag%s fneeiene oett ba aus FR. + Rs See stem 75 Residence buildings, including porches......-.++-+++++esrtees07? 50 Schools and colleges: Assembly halls... ¢22¢---4.- 0+ sess cee a) rr 100 Class rooms never to be used as assembly halls... 7720"). eee 50 Sidewalks: (or 8 000 pounds concentrated, whichever gives the larger moment or shear) .....-..+o+ 4209 es bimpee ns tote adel Rei 250 Stables, public or mercantile: Street entrance floors ......-+4te see 74-550 ei 150 — Feed roonit...,.1.c.00+seees seen eres et are +o oe tei 150 Carriage roomie i. ++ +l. 0s 5 Satna ie 2S 50 Stall TOO ..ntasct 4 oa byacpehee eee eee Perper 8 50 Stairs, corridors, and fire escapes from armories, assembly halls, and pymNasiumng.\ss..cres ++ ts feee ee ene: es ee aperae 100 Stairs, corridors, and fire escapes except from armories, assembly halls, and gyrimasiums. . ./))\75 .s yo) sigh se eee 75 Storage, heavy. slci+seaest osc stk: ae rh: oe eee te 250 Storage, light... 0.4). espa phe Sab or 125 Stores, retail... ce y oss ~== === -=20'-0"-=--- legative bending momen ba Ne Fat A-A oy ‘Moment curves ford seeing supported and conti beam (Parallel line Points where bars Dash lines denote limiting points for bending bars aia Limiting pos ot hie a ae, Sq. 70 ae — fut Teo Stirrup Support _ | Sli In 741A ain 2-159, 2620'Str | /st bar ly ‘ ‘ UO) wp ON imiting points ETRE SIAN icity ge 1S BERS td pe 8 pe 5 sy pg compression bars gt" i oke4 hen g> ' g 5-5"Sk#6 NL ‘Anchorage for Beam B3 Beam By. compression bars Ae ye- 15¢.T-O°SKES fl er rE NEES inte Pee ye ee 2=1"59. 25'-0"Str 1 3p 2'9 2640" 1~1'Sg. 29'-10"SK#1 = SLIPR BEAN SHS 1-1°Sg. 29'-10"Sh#2 “J-1"$ B5'-4" Skt Mes ap Boris 8-35 gf 5-5 Ok#8 32 g S'S Sk#8 In all beams same as Beam Ba Beam Bo_ shown in Bt Fig. 189.—Details of Design 35500 Max.v=——-———_- 4 yA 212000. sino ys eee en S a a am i ' . 14x 091x225 Min.y =—8£/39 _.- 40, BS cs 1/4" Edge of colurings on” “Tax o9ex 265 7100/4 is g Gent bars 1-1"Sq. each 63 60 Stirrups F "ha I. 3S | &3 .COre -Stirrups = £ of girder : ne { A \¢ of foam y 7 AS i edge of beam se: ee ee Papen STE Bent bars><-*> ~Stirrups Stirru, pss ——Bent bars= rem, Stirrups | ; y ‘lf 0. LS pephbe ued i= Net span fessiled eg: 9 ais J 5 iG r / laa hear diagram for beams Shear diagram for girder aces for bending down bars I" for Ist. bar or 2nd bar or grd bar P Bp for #04 bar | =. Ss eel C= apebrte a= cat €--Moment curve|at support bd Out Rivho sf a 4 Rririee wien elle bel 7 Pe 1-11" | 20" 15:4" | 204 7-.2"|7922"| 2-4" 34.9" |20"|11-8"| 20) 90"|193" | 2-4" | eet 7:5" |22"\ 1524" |224| 7:5" | 21d | 247” #4 |3524"|—| 929” [224 17:.3"122"1 9:3" lak” [227° | #5 [35:0 al2 2:9" |24"| 3:10"|24" 12*.9"\ 232 |2' 10" pri 0" —| 10-6" 24" 8.4” eceeae 2:10" eed fee 83/24/2110) 241 63" 2332470" edocs eee nent Curve for Girder 6; _ wa sketches for bars aight —s a i RN Ste Cont. Girder Positive B. moment g S Ss Posite B moment jor simply supported girder Bars from Ad- ( yoining Span i Bent Bars limiting points for 4 Kite? eee sia : ft. ire = ik aa ag ay ie Peo the aig" Limiting points. peony am i > OR te a Bg MY J-1"$ 230" Str. A, oti por UW [= 1'Sg 23:0" Str =o * compression T= 1° D 350" SKES ' oe 0" "SRO 4" oX = Dora ose G, Typical Panel. (See p. 579.) (To face page 579.) QUE Tae pI = 1 MWe 4 ry if} es < 44 \ Seb iif - Stabs oth, CF, es pags ele hid Labor! ist ot 4 ; ‘ Ry . ‘ x et Se ey. “ad , el ty sah? prt cee es Hitt > ta, & i) eS Te ane - Fea lige inte sttieh slit } & on oe | - — a i Gabi ious é Cad ok Sua . ~< -0 row t ee eon Bom F ‘ mans Wis a cS. vonrui BIR oes he ¢* "Eon SSE - a sit af soe Ee Jee Set ‘Sa. 25° Ob sat 1-1'Sa. 29-10 RAT NAA 1-1 Sia. 29-10" She Wales oak a B-2'¢ 5-3" GRPS : . ‘ bear Sy 5 DESIGN OF BEAM AND GIRDER FLOOR 579 Specified unit stresses in lb. per sq. ft. are as follows: f, = 800, fs = 16 000, n = 15, v = 40 without stirrups, » = 120 with stirrups, « = 80 for plain bars and 100 for deformed bars. At support of continuous beams, f, = 900. Solution.—The girder is placed in the short direction .and the beam in the long direction, as shown in Fig. 189, opposite p. 579. Constants.—The following constants, taken from table on p. 205, correspond to the assumed working stresses: Neutral axis and moment arm, k = 0.429,7 = 0.857. Constants for depth of rectangular beam and slab respectively, C = 0.083, C, = 0.024. Ratio of steel for balanced design, p = 0.0107. Bending Moments and Span Length.—Bending moments recommended on p. 278 will be used. Span lengths are recommended on p. 277. Slab. Loads: Live load, 200 lb. Slab load, 56 lb. (assumed) Granolithic finish, 12 lb. Total, 268 lb. per sq. ft. 20.0 Gross span oe ae 10 ft. O in. Assuming width of beam equal 1 ft., net span / = 10.0 — 1.0 = 9 ft. O in. Using formula M = ;5wl? ft.-lb. or wl? in.-lb. (p. 279). M = 268 X 92 = 21 700 in.-lb. Depth of slab from formula d = Ci+/M (p: 208) where C; = 0.024 d = 0.024+/21 700 = 3.5 in. Using #-in. concrete below steel, the distance to center of bars will be 1 in. and the total depth h = 3.5 + 1 = 4.5 in. Area of steel from formula A, = pbd. A, = 0.0107 X 3.5 = 0.088 sq. in. per inch. The spacing of 3-in. rd. bars is obtained by dividing area of one bar by 0.038. 0.196 It is —— =5.15in. Use ?-in. rd. bars 5 in. on centers. 0.038 The slab steel will consist of alternate straight and bent bars. The straight bar will extend over two bays and will be 20 ft. 6 in. long. The bent bars will serve as positive reinforcement in one bay, while the ends will be bent up and extended into the adjoining panel beyond points of inflection. (See plan.) Beam B,.—Beams B, extend between girders, therefore Formulas (66), p. 279, will be used. Span, center to center, 22 ft. Assuming width of girder equal 1 ft., net span, 22 —1 = 21ft. Spacing of beams, 10 feet. Loads: Slab load 268 «K 10 = 2680 lb. Beam, below slab, 300 Ib. (assumed) 2980 lb. per lin. ft. 580 BUILDING CONSTRUCTION For intermediate beams spanning between girders, use: M = +,ul? ft.-lb. or wl? in.-lb. for negative and positive bending moment. (See p. 279.) UM = wil? = 2980 X 21? = 1315 000 in--lb. External shear, V = 22 x 2980 = 31 300 lb. Breadth of Flange.—Breadtk. of slab equals either 16 times the thickness of slab plus the breadth of stem of beam, or 4 of span of beam, whichever is smaller. Assuming b’ = 12 in., b = 16 X 4.5 + 12 = 84 in. In this case, } of span of beam is smaller than the above value; therefore the breadth of flange will be made equal to + X 21 X 12 = 63 inches. Cross Section as Determined by the Shear. V = 31 300 lb. b’ (2 _ 5) = SS = 260 sq. in. (see p. 218) Depth Required tee t by Shear b’ 2 d 12 in. 21.6 23.85 14 in. 18.6 20.85 16 in. 16.2 18.45 Minimum Depth at Support.—In continuous beams, the beam at support isa rectangular one with steel in top and bottom. The depth should not be smaller than required at support. In this case, it is desired to bend up one-half of the reinforcement and use the other half as compression reinforcement. The avail- able amount of compression reinforcement, therefore, is equal to one-half of the tensile reinforcement, or p’ = 3p1.. Assume d = 0.1. Since the allowable stress iS 16 000 b tf. = 900, the ratio == — at support f, e ratio 15f, 15 X 900 = 1.18. In diagram for a = 0.1 on p. 908, using line a = 1.2, it is found that for pi: = 0.0182, p’ = 0.0091.* Cc This gives the required ratio between the compression and tension steel. The minimum depth, then, from Formula (26), p. 222, assuming b = 14 in. a= 1.054] 1 315 000 oye Aine 14 0.0182 X- 16.000 ames *In the diagram is drawn a straight line giving the required relation between the tensile steel and compression steel by connecting any two points for which p’ = 3p;. The intersection of this line with the line corresponding to eo = 1.2 gives the required ratio of tensile and compres- OSC sion reinforcement. For any other ratio a similar line may be drawn. Instead of the actual drawing of a line, the intersection may be obtained by using a straight edge. DESIGN OF BEAM AND GIRDER FLOOR 581 rM st Economical Depth of T-beam.—From Formula (18), p. 216, d = “f° 5 + 5: if the ratio of unit cost of steel to cost of concrete is r = 70. for b’ = 12 d = 21.7 + 2.25 = 23.95 b’ = 14 d = 20.0 + 2.25 = 22.25 b’ = 16 d = 18.7 + 2.25 = 20.95 Minimum Depth in the Center.—This depth is governed by the compression stresses in the center, where the beam is a T-beam. In this case the minimum depth is such that the neutral axis is within the a flange. Therefore ordinary beam formula, d = c,| >? is used. For fe = 800, fs = 16000, n = 15, C = 0.083. Since M = 1315 000 in.-lb. and b = 63 in., the required depth is 1315 000 nce vieseeg aa 1) 8 in, Depth Finally Selected—By inspecting the various depths computed above, the following values will be selected: b’ = 14in., d = 22.5in,, h = 22.5 + 2.5 = 25 in. This is the economical depth. It is larger than required at the support and some- what larger than required by shear. Area of Steel. t For d = 22.5 in., ¢ = 43 in., ade 0.2, the ratio of moment area is j = 0.91. The area of steel, then, is, from Formula (19), p. 216, 1315 000 As; = 0.91 XK 22.5 X 16 000 = 4.0 sq. in. . Use four 1-in. sq. bars with an area A; = 4 X 1 = 4.0 q. in. One-half of the bars will be carried straight and the other half will be bent up. The bars will be arranged in one layer. To distribute the steel properly, the bars will be bent up at two points at each end. The points of bending are taken from Fig. 100, p. 294. . Bond Stresses at Supports.—Since V = 31 300 lb. and the perimeter of bars is Zo = 4X 4 = 16 sq. in. j = 0.91, d = 22.5, the bond stresses are: 31 300 UU = 0.91 X 22.5 X 16 = 95 lb. This is satisfactory for deformed bars. Diagonal Tension Reinforcement.—Unit shear is plotted in Fig. 189, opp. p. 579. As it is desirable to utilize the bent bars for diagonal tension reinforcement, their location is marked on the shear diagram. As explained on p. 250, the bent bars may be considered as effective for a distance equal to d, provided their area is 582 BUILDING CONSTRUCTION sufficient to take the stresses. The area tributary to each group of bars is marked on the diagram. Section ab. In this section stirrups must be provided. Shear to be resisted 69 + 57 : ’ ; equals aa x 14 X 12 = 10600 lb. The area of } in. rd. stirrups with two legs equals 0.196 X 2 = 0.392 sq. in. and the available strength equals 10 600 392 X 16000 = 6300 lb. The required number of stirrups equals 6300 = 1.7. Use 2-4 in. rd. stirrups in section ab. Section be. This section extends 3d on each side of the bent bar, therefore, the bent bar is effective in this section. Shear to be resisted 57 -+ 39 a x 14 X 22 = 14800 Ib. Stress to be resisted by bent bar 0.7 X 14 800 = 10 400 Ib. Area of bent bar 1 — 1 in. sq. bar = 1.0 sq. in. Available strength of bent bar 1.0 X 16000 = 16000 lb. The available strength is larger than the stress to be resisted. Section cd. The bent bar effective in this section is the same as in section be. Since the shear is smaller than in the other section, the bent bar is sufficient to resist the stresses. Section de. ‘The shear in this section is small. Two 3 in. rd. stirrups will be used arbitrarily placed as shown in the diagram. Beams B).—The unit loading and gross span for beams B, are the same as for beam B,, but their net span is smaller and the condition at the support dif- ferent. Beams B, run into flexible girders and therefore must be considered as continuous beams designed by Formulas (63) to (68), p. 278. Beams Bz, on the other hand, since they run into stiff columns, may be designed as building frames by Formulas (69) to (74), p. 279. Net span of Beam B, is 1 = 22 — 2 = 20 ft. Unit load w = 2980 lb. per lin. ft. Negative bending moment (Formula (69), p. 279). M, = + wi? = — 2980 < 20? = 1 192 000 in.-lb. Positive bending moment (Formula (70), p. 279). M = 0.75 wil? = 0.75 & 2980 X 202 = 894 000 in.-lb. Shear V = 2980 X 2,2 = 29 800 lb. The bending moment for this beam is smaller than for beam B;. Usually it is desirable, for the sake of appearance, to use same size for beams B, as for B,. If not, the economical dimensions may be found as explained in connection with By. DESIGN OF BEAM AND GIRDER FLOOR 583 Using same dimensions as for B,; b= 14'm. a. ==,22,5 in. Py Rapes, Bis The areas of steel required by bending moments are: At support 1 192 000 a Seah 10.91 X 22.5 X 16 000 ana In the center of span ak 894 000 Bg ae eer 0162285446 000 iL. ture ten Use in the center 2-2” round straight bars = 1.20 sq. in. 2-1” round bent bars = 1.5/ sq. in. Toul 2.77 sq. in. This area is satisfactory. At the support the available reinforcement consists of two bent bars plus two bent bars extended from the adjoining span, making 4-1” round bars in all with an area of 4 X 0.785 = 3.14 sq. in. The available area of bent bars is smaller than the required area. The difference 3.64 — 3.14 = 0.50 sq. in. is supplied by a straight bar. As a result, reinforcement at the column consists of 4-1” round bent bars = 3.14 sq. in. Pay 4’ round straight bar = 0.60 sq. in. Total 3.74 sq. in. Check of Compresson Stresses at Support.—In this case the dimensions of the beam as well as the amount of tension and compression reinforcement are fixed. If the 2-2” round straight bars at the bottom of the beam are extended suf- ficiently into the columns to develop them by bond, they may be considered as compression reinforcement. The compresson stresses will be investigated b = 14 in. ye eae bien ld e225 =) sain: ; { 3.74 Tension steel A, = 3.74 sq. in. pi = 315 = 0.0119. 1.20 Compression steel A’; = 2 X 0.60 = 1.20 sq. in. p’ = ae 0.0038. The top bars are covered by concrete 1} in. thick. The distance from top to center of compression steel is = 0.089. ad ae + (7x4) =2in. t+ anda ity" 584 BUILDING CONSTRUCTION 15fe to p: = 0.0119 and p’ = 0.0038, which is ia = 1.52. When steel is stressed Using diagram on page 908 for a = 0.08, find the value of corresponding 16 000 in tension to f, = 16000 lb., the expression changes to = 1.52, and 15fe the corresponding stress in concrete is 16 000 gee cet cteten hm aye er FRSC Edn Pe a Stirrups.—Since the external shear of Beam Bz is practically equal to Bi, the same spacing of stirrups will be used for all beams. Girder G:.—Gross Span 20 ft. 0 in. Assuming that the minimum size of column is 18 in., the net span is 20.0 — 1.5 = 18.5 ft. Loading.—Concentrated load 2980 X 22 = 65 500 Ib. in center of span. Weight of girder (below slab) 300 lb. per lin. ft. Bending Moments. Concentrated loads, considering the span as simply supported. 65 500 _ 18.5 M = eae x Ys < 12 = 3 640 000 in.-lb. For continuous span, multiply by a ratio of 3- M = 3640 000 X 58; = 2 430 000 in.-lb. For uniform load, M = 300 X 18.35? 103 000 in.-Ib. Total 2 533 000 in.-lb. This bending moment will be used in the center and at the support. External Shear. 65 500 LABS = 32 750 lb., concentrated load 18.5 : ae’ V2 = 300 X re 2770 lb., uniformly distributed load Total V = 35 520 lb. Area required by shear, | t 35 520 v( ~ s) = ——— = 296 sq. in. DESIGN OF BEAM AND GIRDER FLOOR 5085 Minimum dimensions required by shear, b’ = 14 in. d — 21.6 in. b’ = 16 in. 1S 5 19.0 in. Minimum Depth at Support.—Since the ratios of steel in tension and com- pression for minimum depth for the girder are the same as for the beam Bi, namely, pi = 0.0182, p’ = 0.0091, the minimum depth at support = 1.05 4| 2 533 000 ee ieee 14.500. 018290 16 000) cs This depth will be accepted. It is larger than required by shear. Area of Steel in the Center of Span. t Ford = 26.2in.,¢ = 43 in; = 0.17, from table on p. 221, the ratioj = 0.92. The area of steel, then, is 2 533 000 Dg 0.92 XK 262 X 16 000 = 6.57. Use 6-1" rd. 4.71 sq. in. 3 straight 3 bent 2-1” sq. 2.00sq. in. 1 straight 1 bent Total 6.71 sq. in. The bars will be placed in two rows. They will be bent in two places. To find the points at which to bend the reinforcement, the bending moment diagram must be drawn, as explained in Vol. III. The variation of bending moments for concentrated loads is different from the variation for uniform loading, therefore, the diagrams on pp. 292 to 297 do not apply. Since the bending moment due to the dead load is small, the same variation of the total bending moment may be taken as for the concentrated load alone. The bending moment diagram and the bending diagram for reinforcement are shown in Fig. 189, opposite p. 579. Bond Stresses, V = 35 500 lb., jd = 0.92, d = 25.8. Zo = 6X 3.1442 X 4 = 26.8 in. Hence, 35 520 Ug eee oe = 65 Ib. 0.92 X 26.2 XK 26.8 Bond stresses are satisfactory. Diagonal Tension Reinforcement.—The shear diagram is drawn first. (See Tig. 189.) The location of the bent bars is then shown and the area tributary to each bent bar is marked off. The shear diagram is then divided into several sections. Each section is investigated as given below. 586 BUILDING CONSTRUCTION The shear diagram is divided into sections ab, be, cd, de, and ef. The stresses in each section will be computed and the required reinforcement provided. Section ab. In this section stirrups must be provided. Shear to be resisted 68 + 67 equals x 14 X 6 = 5700 Ib. Since the strength of 3 in. rd. U-Stirrup equals 0.392 X 16 000 = 6300 lb., one stirrup is sufficient. d Section be. This section extends 5 on both sides of the bent bar. The bent bar, therefore, is fully effective in this section. The diagonal tension to be : : 67 + 65 ‘ resisted is (v — v’)bs = oe < 14 xX 27 = 25000 lb. This produces stresses in bent bars in this section equal to A;f; = 0.7(v — v’)bs=0.7 XK 25 000 = 17500 lb. (See Formula (39), p. 248, also p. 155.) Since in this section two 1-in. rd. bars are bent with an area 0.785 X 2 = 1.57 sq. in. and a capacity 1.57 X 16.000 = 25 100 lb., the diagonal may be considered as taken care of. 65 + 63 65 +68 V4 x 27 = 24200 1b. Section cd. The stress to be resisted equals The stress produced in a bent bar equals 0.7 X 24 200 = 16900 lb. In this section 1-1 in. sq. bar is bent up with an area of 1 sq. in. and capacity of 16 000 lb. This is smaller than 16 900 lb., therefore, the bar is not sufficient to take care of the diagonal tension stresses. The difference 16 900 — 16 000 = 900 lb. Divided by 0.7 this gives the stresses to be provided for by stirrup 900 + 0.7 — 1290 lb. One } in. rd. U-stirrup in this section gives sufficient area. Section de. The stress to be resisted equals are xX 14 XK 27 = 23 300 Ib. In this section 1-1 in. rd. bar is bent up. Its capacity equals 0.785 X 16000 = 12600 lb. This divided by 0.7 and subtracted from the total stress to be resisted gives the stress to be taken care of by stirrups. 23 300 — (12 600 + 0.7) = 5300 lb. This stress may be added to the stresses in section ef. noe x 14X14" "1T 700 Tb. Adding to it 5300 lb. from the previous.section makes the total stresses 17 000 lb. This requires three 3 in. rd. stirrups. With the above figures as a guide stirrups are spaced as shown in the figure. One stirrup is placed on each side of the beam to take care of the load transmitted by the beam. Three additional stirrups are placed between the beam and the first bent up bar. The number of stirrups used near the column is larger than actually required. Section ef. The stress to be resisted equals END PANEL Slabs.—Slab in the end panel is of the same design as for interior panel. Beam B;.—The net span and loading is the same as for corresponding interior beam, namely Beam Bj. b 2d fu. w = 2980 lb. per sq. ft. FLAT SLAB FLOOR CONSTRUCTION 087 Since beams B; run into girders, they will be designed as continuous beams by means of Formulas (63) to (68). Negative Moment At Interior Column M, = — 1.2 wl? = — 1.2 X* 2980 & 212 = 1580 000 in.-lb. At Wall Column WM, = — 0.6 wi? = — 0.6 X 2980 X 212 = 790 000 in.-lb. Positive Moment is same as negative moment at interior column M = 1 580 000 in.-lb. Use same dimensions as for interior span Do 14. in. @ =. 22-5111. hp e250. ; M The required area of steel is found from As; = af, JIGS s In the center and at interior column 1 580 000 : 8 = 4.82 sq. in. ~ 0.91 X 22.5 X 16 000 At the wall column the moment is equal to one-half of the moment in the center, therefore, the required area of steel Asi a 7A I 2.42 sq. in. Use in the center 5-1” rd. bars = 3.92 sq. in. doi eeqs bar) )'="1),00 sq.in. Total 4.92 sq. in. Bend up 1-1” sq. bar and 2-1” rd. bars, which give an area of 1.0 + 1.57 = 2.57 sq. in. The bent bars extending from the adjoining interior span equal 2-1” sq. bars with an area of 2.0 sq. in. The total area, therefore, is 2.57 + 2.0 = 4.57 sq. in. Since the required area is 4.82 sq. in., the balance 4.82 — 4.57 = 0.25 sq. in. must be supplied by means of straight bar plaed on the top. In this case 1-3” rd. bar will be used. The bars are arranged in accordance with diagram on page 293. The reinforcement at the wall end of the beam is clearly shown in the eleva- tion of the beam. Stirrups.—Same stirrups will be used as for corresponding interior beam. FLAT SLAB FLOOR CONSTRUCTION Flat slab floor construction is used to a great extent in building construction. It is particularly adaptable to warehouses, industrial buildings, garages, and automobile service stations, where it is cheaper than any other type of floor construction, 088 BUILDING CONSTRUCTION In buildings with large spans and light live loads, flat slab con- struction may not prove economical, especially when column capitals are not permitted. For very long spans, over 35 ft., the dead load of flat slab may be too large. Advantages of Flat Slab Construction—Following are the advantages of flat slab construction over beam and girder con- struction. 1. Lower cost (see comparison of cost on p. 788). 2. Smaller depth of construction which results in saving in height of building. Under ordinary conditions the saving in height in a ten-story building over beam and girder construction is sufficient to add another story. This probably is of particular interest in cities, where the height of buildings is limited. 3. Flat ceilings give better distribution of light. 4. Due to absence of beams and corners in columns, flat slab construction resists fire better than any other type of concrete. 5. Greater efficiency of sprinklers, because there is no interference with the play of the stream of water. Design of Flat Slab Construction—Complete information regard-_ ing the design of flat slab construction is given in Chapter VI. LIGHT-WEIGHT FLOOR CONSTRUCTION Under this heading come floor constructions in which the solid concrete slab is replaced by closely spaced concrete ribs connected at the top by a thin concrete slab, usually called topping. The space between the joists below the topping is either filled by tiles, or is hollow. Such floors may be divided into the following groups, according to the method used for the elimination of the tension con- crete between the joists. 1. Hollow tile concrete floors. 2. Metal tile concrete floors. In some cases, the metal tile is replaced by removable wood forms, as in the Bransom System. ‘The design of this floor is the same as for metal tile. The forms are described in Volume II of this treatise. Use of the Light-weight Floors.—This type of construction may be found economical where light floor loads are carried on long spans, and where smooth ceilings are required. It is used to a great extent LIGHT-WEIGHT FLOOR CONSTRUCTION 589 - in schoolhouses, hospitals, and office buildings, in combination with either reinforced concrete or structural steel frame. It is not eco- nomical for heavy loads. Since the topping is thin, it should not be used where there is a possibility of concentrated loads. The light-weight construction is lighter than the solid concrete slab construction. This reduces the dead load on the girders and columns. The actual economy of the construction will depend upon the relative cost of the concrete and the tiles. In determining the relative economy of a solid slab floor and a metal tile floor, in cases where flat ceiling is required, it is necessary, to add to the cost of the tile floor the cost of the metal lath which is required in metal tile floors for suspended ceiling, also the difference between the cost of plas- tering on metal lath and on concrete. Comparison of Various Types of Light-weight Systems.—In deciding on the type of light-weight floor, the cost of the system, as well as the adaptability, should be considered. The hollow clay tile floor is better adapted for heavy loads than the metal tile floor, because the joists are spaced closer and therefore have larger shearing strength for the same depth. It is also more fireproof, as only the bottom of the joist is exposed to the fire. In metal tile floors, if no suspended ceiling is used, the narrow joist and the thin slab are exposed to the fire on all sides. The resistance of such floors to fire is much smaller than that of solid concrete floors. When the ceiling is not plastered, the appearance of hollow tile floor is not pleasing. Metal tile floors built with removable forms, may be used without plastering. REINFORCED CONCRETE HoLLow TILE FLOOR CONSTRUCTION For long spans, the weight of the floor construction. may be reduced by the use of a combination of hollow tile and reinforced concrete, in which hollow tile replaces the concrete in the tensile portion. The tile is lighter than the replaced concrete; therefore, the resulting construction weighs less than a solid concrete slab. The combination of hollow tile and concrete may be considered as intermediate between solid concrete floor and metal tile floor. It is considered by the authors as the best light-weight type. General Description.—The most common combination of hollow tile and concrete is shown in Fig. 190, page 591. ‘The construction consists of hollow tile, reinforced concrete joists poured between the 590 BUILDING CONSTRUCTION tiles, and concrete topping above the tiles, poured monolithic with the joists. The joists may run in one direction only, in which case the combination is called one-way hollow tile system (see p. 591), or they may run in two directions at right angles to each other, the construction in this case being called two-way hollow tile system (see p. 592). The joists are supported by reinforced concrete girders, by structural steel beams, or by walls. In the two-way system, sup- ports must be provided on four sides. The tile may be of burned clay, gypsum, or any other material adaptable for the purpose. Hollow Clay Tiles.—Burned clay tiles are most commonly used. Figure 191 illustrates a typical hollow floor tile made of fire clay and used in the one-way system. In plan it is 12 in. square and varies in depth from a minimum of 4 in. to a maximum of 15 in. The sur- faces of the tiles are usually scored, in order to bind the tile and the concrete more thoroughly and also to furnish a good plastering sur- face. When an all-tile ceiling is desired, special 1-in. tile slabs are placed between this under the joists. While this has the advantage of providing a uniform plastering surface, it increases the depth of construction by 1 in. The weight of the tile depends upon the material of which it is made and also upon the thickness of the walls of the tile. The weights given in the table below are for burned clay tiles as manu- factured by the National Fireproofing Co. Weight of Hollow Burned Clay Tile aa aan Le en ee Size of Tile Weight Size of Tile Weight in Inches per Tile in Inches per Tile D2 Wek 2 16 lb. | 8 X 12 X 12 32 Ib, 3X12 412 16 lb. . (Xe 36 Ib. 4 Xap s 18 lb. 10° * 125633 38 Ib. 5 xX 12x 12 QT oe 12 ie 42 lb. a Poul A i be 24 Ib. 15 X12 %12 50 Ib. TR ALRG 12 29 Ib. Se In the one-way arrangement, the clay tiles are placed with ribs parallel to the direction of the joists. When the tiles are sound and are properly placed, comparatively little concrete finds its way inside. 591 LIGHT-WEIGHT FLOOR CONSTRUCTION ee Se N/ ~ | | | | | | | | | | | | si | | | | | I | 4 Stag apart eo eae Section Through Joists ULE ee STs. =| S's = Ss }- 8 33 | Laika) aie) Section Through Girder (See p. 590.) Fic. 190,—Hollow Tile and Concrete Floors. 592 BUILDING CONSTRUCTION - When the tiles are broken or when the joints between the tiles are not properly sealed, considerable concrete may flow into the tiles, thereby increasing not only the weight of construction but also its cost. In the two-way arrangement, some designers use regular tiles, claiming that no considerable amount of concrete flows into the tiles through the open ends. The authors do not consider such designs desirable, because it is impossible to estimate the increased weight of slab and amount of extra concrete. Special tiles, for the purpose of sealing the open ends, are on the market. In the arrangement shown in Fig. 192, p. 593, each group between four joists consists of six tiles, forming a 24-in. square. With 4-in. joists, their spacing in both directions is 28 in. on centers. In another arrangement, the ends of the tiles are closed by means of channel tiles, the flanges of which are placed against the open ends of two adjoining tiles. The web of the channel is placed on the form, Fic. 191.—Typical Hollow Clay Floor and its width fixes the spacing Tile. (See p. 590.) of the tiles. The width of the joist is equal to the clear dis- tance between the flanges of the channel. Additional soffit tiles are used, and an all-tile ceiling is thus obtained. (See Fig. 201, p. 613.) Gypsum Tiles.—Hollow tiles may also be made of gypsum. The standard width of the gypsum tile is 20 in.; the spacing of 4-in. joists would therefore be 24 in. on centers. Gypsum tile is lighter than clay tile. An additional advantage is that it can be readily cut in the field. It has the disadvantage, as compared with clay tile, that it is much more brittle and therefore requires more careful handling. Joists.—The joists may be made of any width, by spacing the rows of tiles the desired distance apart. The minimum width of the concrete joists should be 4 in. This width has been adapted as standard by many users. A smaller width is sometimes used with burned clay tiles, but then the space is filled with cement mortar and not with concrete. LIGHT-WEIGHT FLOOR CONSTRUCTION 593 Topping.—The concrete topping over the tiles should be at least 2 in. thick, and should be poured in one operation with the joists Since the concrete topping rests on the tiles, no bending stresses can be developed in the concrete and no bending reinforce- ment is necessary. It is advisable, however, to use some shrinkage Fig. 192.—Hollow Clay Tiles for Two-way Combination Floor. (See p. 592.) and temperature steel to prevent cracking of the topping. The amount to be used can be smaller than that required in metal tile slabs, as the contraction of the concrete is reduced somewhat by the adherence of the concrete to the tile. In clay tile construction for light loads, the topping is sometimes omitted and the construction then consists of rows of tiles separated by rectangular concrete joists of the same depth as the tiles. Nat- 594 BUILDING CONSTRUCTION urally, the tiles must be strong enough to carry the load between the jOistseiee Design of Joists.—The floor loads are carried by the concrete joists, which, with the topping, form small T-beams. The design of the floor, therefore, resolves itself into the design of small T-beams. Each joist carries the dead load plus the live load of a strip of floor of a width equal to the spacing of the joists. The bending moments and shears are computed in the same manner as for regular concrete beams. The bending moment coefficients will depend upon the degree of restraint at the supports and upon the number of spans. The span to be used in design 1s discussed on p. 277. For simply supported floors, such as’ those in which the joists rest on brick walls, it is necessary to compute shearing stresses at the support, the amount of steel in the center, and the compression stresses in the center. The economical depth and the depth required by deflection are usually larger than the depth governed by the com- pression stresses. In continuous beams, in addition, the amount of steel and the compression stresses at the continuous support must be found. At the support, the topping, which in the center forms the flange of the T-beam, is in the tension zone, so that the joist becomes a rectangular beam. Usually, the compression stresses are high and compression steel is required. ~The depth required at support may be the govern- ing depth. , Shearing Stresses.—Shearing stresses involving diagonal tension should always be computed. If it is desired not to use any diagonal tension reinforcement, the unit stresses must not exceed the value allowed for plain concrete. Many building codes require that, in computing shear, only the width of the concrete joist shall be con- sidered as effective. At present, there is a tendency to make some allowance for the resistance of the clay tile to diagonal tension. The report of the Joint Committee of 1924 gives the following rule, which is endorsed by the authors: 3 “Shear in Beam and Tile Construction.—The width of the effective section for shear, as governing diagonal tension, shall be assumed as web plus one-half the thickness of the vertical webs of the concrete or clay tile in contact with the beam.” | Reinforcement of Joists——The amount of reinforcement is obtained from the bending moments, as recommended on p. 275. In selecting the number of bars, the bond stresses must be taken into LIGHT-WEIGHT FLOOR CONSTRUCTION 595 consideration. To keep the bond stresses within working limits, the use of bars of small diameter is recommended. Since the joists are narrow, the tendency is to use a small number of heavy bars. This produces excessive bond stresses. It is not advisable to use more than two bars in one row. If a larger number of bars are used, they should be placed in two rows, in which case proper allowance must be made in computations for the reduced moment arm. In continuous joists, it is advisable to use an even number of bars of nearly the same size. Bars forming about one-half of the total area of steel may be bent up and carried over the support. When the reinforcement of the joist consists of one bar only, it must be carried straight at the bottom for the full length of the joist. If it is not possible to divide the steel in half, the larger portion should be bent up. It is advisable to carry at least 40 per cent of the bottom steel straight to the support, and a larger amount is often required by compression stresses. The proper amount of steel should be provided at the support. The bent-up bars with the bars coming from the adjoining span are usually sufficient. If not, additional straight bars should be used. When no bars are bent up, the full area of steel required for negative bending moment must be provided by short bars. The bars should be bent up at such points that the center of the bend coincides with the point of inflection. They should be extended into the adjacent span a sufficient distance to develop the bar in tension for negative moment. ‘To serve this purpose, they must be extended at least 20 diameters beyond the points of inflection of that span. An attempt is often made to use the bent bars both as nega~- tive moment reinforcement and as diagonal tension reinforcement. The result is not satisfactory, as there is danger that neither of the stresses will be properly taken care of. At the end support, the proper amount of steel should be used to take care of any bending moments that may be developed. The magnitude of the negative bending moment to be taken care of will depend upon the amount of restraint. A minimum negative bending moment equal to ygwl* should be provided for. Points of Inflection.—The location of the points of inflection should be computed on the basis of the net span and not the gross span. The distance to the point of inflection should be measured from the edge of the support and may be accepted as 3 of the net 696 BUILDING CONSTRUCTION 1 span for interior spans and 45 for wall panels. When the bending moment is computed on the basis of the gross span, there is the tendency to consider the point of inflection as distant from the center of the support a distance equal to + of the gross span. For wide supports, this assumption fixes the point of inflection much nearer the support than it actually 1s, with the result that the negative bending moment is not properly provided for. While the assump- tion of gross span in figuring the bending moment is safe, it gives unsafe results in figuring the distance of the points of inflection. Spacing of Reinforcement.—The bars should be placed not nearer to the sides of the tile than ? in. This 1s required for bond and not for fireproofing, as the tile serves as a sufficient fire protec- tion. The bars should be kept a proper distance above the form. When two layers of bars are used, they should be separated by a separator bar one inch in diameter. In narrow joists, it is difficult to keep the bars in place without mechanical spacers and chairs. In many instances, it has been found after the concrete was removed that the bars were misplaced, in pouring concrete, to such an extent that there was no concrete between the bars. Diagonal Tension Reinforcement.—If the unit stresses exceed the allowable values for plain concrete, stirrups should be used. The size and spacing should be determined as explained on p. 247. The bent bars may be considered as resisting diagonal tension in the portion of the beam where they occur. Selection of Depth of Joist——In selecting the depth of joists, not only the stresses, but also the deflection must be considered. For a joist that is shallow in proportion to the span, the deflection may be excessive even if the stresses are within working limits. If the ceil- ings are plastered, the deflection should not exceed 345 of the span; if it is greater the plaster will crack. Economy of design should also be considered in selecting the proper depth of joists. The economical depth will depend upon the relative cost of concrete tile and steel. By increasing the depth of the joist, the cost of concrete and tile is increased; but the cost of steel is reduced. A depth may be considered economical when any increase in it would give an increase in cost of concrete and tile greater than the accompanying decrease in cost of steel. In many instances, the depth required for compression at the support is larger than the economical depth. At the support, the LIGHT-WEIGHT FLOOR CONSTRUCTION 597 joist is a rectangular beam with double reinforcement. The amount of compression steel may be assumed as one-half of the tension reinforcement, and the depth computed by means of the constants from Diagram, p. 908. The allowable compression stress at the support may be increased as explained on p. 282. Girders Supporting Joists.——The girders supporting the joists may be either of reinforced concrete or of steel. If reinforced concrete girders are used, the required flanges for the T-beams are provided by omitting the tiles adjacent to the edge of the girder. (See Fig. 190, p. 591.) Asa result, a solid slab is obtained on both sides of the girder, of the same depth as the depth of the joist. The flange so obtained may be deeper than required to resist the compression stresses. If this is the case, instead of omitting the tiles altogether, thinner tiles may be used. For instance, in a floor consisting of 10-in. tiles and 2-in. topping, by omitting tiles at the girders, tke thickness of the flange of the girder would be 12 in. In most cases, this is more than is required by the stresses. If 6-in. flange is required for the girder, instead of omitting the tiles at the girder, 6-in. tiles are substituted for the 10-in. tiles. As a result, the flange of the girder will be 6 in. thick. While this method saves some concrete, it may lead to confusion on the job. Support on Brickwork.—The joists should have a bearing on the wall of at least 6 in. The required area of bearing should be computed from the reaction of the joist and the allowable stresses on brickwork. Weight of Clay Tile Floors—In computing the dead load of the slab, the following weights may be assumed for burned clay tile with 4-in. concrete joist spaced 16 in. on centers and 2-in. topping. The weight of flooring and plaster should be added. If the thickness of topping is more than 2 in., the extra weight of concrete should be added. Thickness of tile, in... . 3 4 5 6 7 8 Diet LO ree Speen Amount of concrete per poettsin cu. tt....... 0.23 0.25 0.27 0.29 0.32 0.34 0.36 0.38 0.4 20.48 Average weight of tile and concrete, per sq. OS See Ae ee a ae 45% 5D. 255, 1000.65) 702 ome OmenOU. LOG Erection.—It is evident that for one-way arrangement of tiles the form boards need be used only under the joists. The width of the 598 BUILDING CONSTRUCTION boards must exceed the width of the joist by an amount sufficient to give support for the tiles. The tiles must be erected in a straight line and kept in place during concreting. Misplacement of the tiles would reduce the width of the joist and might have serious consequences. Tiles should be sprinkled with water before concrete is poured, as dry tile absorbs moisture from the concrete. If this is omitted, especially in dry weather, the topping will lose more water by absorp- tion than the joists, causing cracking and uneven setting of concrete. Since the joists are narrow, with a large amount of reinforcement, the large stones used as coarse aggregate would obstruct the flow of the concrete around the reinforcement and would form pockets. The maximum stone should therefore pass a {-in. sieve. The con- crete must be carefully spaded and worked around the reinforcement. The joists must be poured in one operation with the topping. Construction joints in the joist should be made in the center of the joist. Construction joints parallel to the joist should be made midway between the joists. It must be borne in mind that the topping acts as a flange for the joist, and nothing must be done which would interfere with this function. Example of Hollow Tile Floor.—Figure 193 gives the floor plan of a building in which hollow tile construction was used. The arrange- ment of panels and the sizes are given. ‘The computations for a typical panel are given below: Typical Design of Combination Hollow Tile and Reinforced Con- crete Floor.—The computations given below refer to the floor in the Ordnance Building shown in Fig. 193, p. 599.4 The floor is intended for office use. 2 Slab, Interior Span, 15 ft. 0 in—M = ae for positive and negative bending moments. Live:load 24.205. ae a eee 75 lb. per sq. ft. 9-in. cinder concrete fill and sleepers.. 15 T_in, wood flooring. ....--+ +++ +a 5 1-in. ceiling plaster. 4) --...+--.943 S Assume weight. of slaboe. 2!) eee 60 ‘Total loads... sepa eae w, = 160 lb. per sq. ft. 4 Arthur Wood, Architect, National Fireproofing Co., Engineers. LIGHT-WEIGHT FLOOR CONSTRUCTION 599 O"/ong perth orst on 4h a, 7* ef Seat a ere Sek aati ee ee ———- — = — Ss ff rrr 1-0" /ong per joist on top over Girder “ h. SS % > ~y se JZ He BLOM ate EES 610" ake > "x6"Tile mulhons filled with conerete .f and eee cay, = 59. Bar Bar rad 6 Reinforced concrete lintel 2-4 "Sa Plan of Triple Windows ras ie aie tee | Oe "Concrete top poured with the joists , 2” Cinder concrete Jill and, sleepers — = ab id ee xl2*hollow tile 7 Fs ‘Woo ay ee | po SE EL wake aX = wr" 1.000 al Bar we 8 ‘4 oy : 4-76" oe 3" -.t ie! pastas sIn Mullion 62+ nT oaks Section "A-A” -¢ Column Wall 8.—~ rk >| LS aoe on cae a ‘I sae Pe Pilaster 12"\~ yf section: C- ms as stirrups inverte Bh hc yal MCA 2 "Bars bent up These ae rups inverted. C73 j Bars pie alee || up ie Bal : End of bar froin adjacent span -- 3’-9"---> 4-5 Fae 7 Bars pilaster reinforcement fi = cells of tile with concrete down to Joundation jor solid pilaster End of bar from adjacent span Letail Girder "G” Alternate Courses ft ers C L. of Girder-- Typical Col Section Note:-/n alternate girders the# bars will be inside and thez’bars outside in order to permit bending a up of the f"bars in all girders. Cross-Section Girder "G Section’B-8” I'tg. 193.—Floor Construction in Ordnance Building, Washington, D. C. (See p. 598.) 600 BUILDING CONSTRUCTION With joists spaced 16 in. on centers, the unit load per joist, = 160°* 42 = 213 Ibsperlinait. Assume odulin unit stresses: f. = 650 Ib. per sq. in. fs = 16000 lb. per sq. in. v= 40 Ib. per sq. in. u= 100 lb. persq. in. (for deformed bars). If'n = 15,9 = ? (approx.). The joists may be considered as T-beams. Their fixed dimensions are: Width of flange. ...b = 16 in. Width of stem..... b’ = 5in. (4in. concrete + 1 in. for tile shells) Thickness of flange. . = 23 in. (2 in. concrete + 3 in. for tile shells) The maximum shear at edge of girder. Net span equals 15 — 1.5 = 18.5. y =e 140i The maximum bending moment, 213 x (15)2 x 12 12 Use 6-in. tile with 2-in. concrete top (weight = 60 lb. per sq. ft.) and, allowing 1} in. below center of steel, d = 6 in. The required area of steel is, Mire 48 000 f.jd 16000 X% X 63 ~ Use one 3-in. sq. bar in each joist (A, = 0.56 sq. in.). The area of steel was found for an assumed value of 7. The results will be checked by computing accurate values. Formulas on p. 134 are used. Check of Unit Stresses.—(Use formulas on p. 184.) QnAd+b2 (2X15 X 0.56 X 6.75) + 16 X (28)? M= = 48 000 in.-lb. Ane = 0.56 sq. in. kd = So > ax 15 X 050) Ce 2.22 in. < 28 in.; therefore the neutral axis is in the flange. hale 2222 = > = ae = 0.829, and 4=1—- 3k = 0.890. LIGHT-WEIGHT FLOOR CONSTRUCTION 601 Unit Stresses: M 48 000 Je = Figo ~ FX 0320 X 0.800 X 16 X (G75)? — 70 IP. Per sa. in M 48 000 : ee tra 06x08 X6.75 Per aa V 1 440 i = 48 lb. per sq. in. bid 5 X 0.890 X 6.75 Stirrups are necessary at ends of beam. Mes 1 600 weve ao < 0/890: X< 6.75 Bond stresses are satisfactory for deformed bars. Joist at Support——Assuming the negative bending moment over interior supports to be equal to the positive bending moment at the center of the span, the same amount of steel will be provided as in the center of the span. One {-in. square bar will be used at the top of each joist over the girder, with effective depth of 6% in. from the bottom of the slab. These continuity bars will be made 7 ft. long to reach the quarter-points of the span each side of the girder. The positive reinforcement will be carried straight through the girder into the adjacent span, for a distance sufficient to take care of com- pression stress in the steel equal to the maximum stress as deter- mined when the joist at the support is figured as a double reinforced beam. Ce Unit stresses due to negative moment at support. (See p. 139.) = 89 lb. per sq. in. 1+ 6.75 k = V2n(p + pa) + n2(p + p')? — n(p + p') = 0.418 _ WL — 3k) + 2p’n{k — a) — 2) _ 9 ga a k2 + 2n’n(k — a) ia ape 48 000 ad? 0.56 < 0.842'x'6.75 = 0.185 C= = 15 100 1b) per.sq.in. iF =f = 723 lb. per sq. in. (At support, a stress 15 per cent higher than 650 is permitted = 750 lb., per sq. in.) fe =fs vo ib = 6 050 lb. per sq. in, 602 BUILDING CONSTRUCTION Length of positive steel required beyond edge of support, as determined by bond for j-in. bar, fa 6050 x3 = 11.4 in., say 12 in. 4u «4. X 100 Exterior Span.—Use the same slab construction for the two exterior spans of 14ft.0in.; M = i We LU = 1385 lb. 2 ; yu = 218 x G4)" X 12 _'50 000 in.-Ib. 10 50 000 i. = +0 309 X 0.590 X16 x 6.756) ay 50 000 fe = DFO NC0BIO KOI ee 1 385 v= 5 xX 0.800 X 6.75 = 45.2 Ib. per sq. 1. Stirrups at ends are required. i“ 1491 ~ 3 xX 0.890 X 6.75 Bond stress satisfactory for deformed bars. U = 83 lb. per sq. in. STEEL TILE FLOORS Another type of light floor construction 1s the steel tile construc- tion. In this type the concrete in tension is eliminated by metal tiles of proper design. The resulting construction is shown in Fig. 194, p. 603. These steel tiles are often called “ metal pans,” or ““ tin pans. General Description.—Constructions of this type may be divided into one-way steel tile floors, in which the tiles run in one direction only, and two-way steel tile floors, in which the joists run in two directions at right angles to each other. In the one-way arrangement, each row of tiles consists of a number of tiles open on both ends, overlapping each other and provided at the ends with end caps. In the two-way arrangement, special tiles closed on four sides and on the top are used. They are often called domes. LIGHT-WEIGHT FLOOR CONSTRUCTION 603 The floor consists of joists from 4 to 6 in. wide, spaced about 26 in. on centers and connected by a thin concrete slab 2 to 3 in. ' \ Ss | | aoe | l | | Lo os | jie | | aad, =e. | 3) Lee 2 { th wee [ all | | Jp set Se ee 0 a en BOs PO eke al | | asian ed 3 A) Se OO ae re mew) Uy We oy ol on Oe ee Nea O ier | jets, Tie i." cS et ela il a 2 We er as ber " 7T | | [- ee sods = | | | atone | eee a nee ete i es SE See al [ oe “lieth aegeal E Darees, Leto | N See | | eee hel Vv i fee" | | hae ige cate rr | oo er ee Sa | | | S ] if) die, Seaeee x be | te a | | = ee v — oe a | | —e _— | | a Se ee eee eee | | Lois) 6 A Thy AN ror | ‘s S I | --—A bees ed ELISE 1 died RE eA gue The at Fl ow Lees ony ae Diwan Chaka Mice. to. 8 al ee r-- | ‘> | l ae Les €=-— 90" --- Section Through Joists Section Through Girder Fig. 194.—Typical Panel for Steel Tile Construction. (See p. 602.) thick, serving as a compression flange for the joists. The depth of joists below the topping varies from 4 to 14 in., depending upon the span and the load, 604 BUILDING CONSTRUCTION Steel Tiles.—Two types of metal tiles are on the market: the removable tiles, which act only as a form and are removed after the concrete has hardened; and the permanent tiles, which are left permanently in the construction. Both types are of the general design shown in Fig. 195 and differ only in the strength of the sheet metal from which they are stamped. The removable tiles are made from sheet metal of 14 to 22 gage. The heavier tiles are preferable, as they last longer and keep their shape better after rehandling. For tiles left. permanently in the construction, sheet metal of 24 to 26 gage is used. Permanent tiles must not be made too thin, or they will become distorted during construction. To stiffen the Holes about + Diam. 2 "depth of Corrugation 70 . 13- een a" \ Ff ~~ 37 abt. °EF3 * fe Red te Ok ak ak en 36” Covering Rad. = Width ; Pa Bae Se End View Jn a aa ; a aeceelts, Side View End View Tapered Steel Tile Frc. 195.—Details of Steel Tile. (See p. 604.) steel tiles and enable them to withstand the hard usage they receive during construction, they are provided across their tops and sides with corrugations formed by heavy dies during fabrication. The sides of the tiles are tapered and rounded at the junction of the slab and the joist. At the bottom, the tiles are provided with flanges one inch wide, which during construction rest on the forms and are lightly tacked thereto. Each row of tiles is closed at the ends by caps made of the same material and fitting the tiles. ‘The cap is shown in Fig. 195. Stock Dimensions of Steel Tiles. Standard Heights adopted by all manufacturers are 4, 6, 8, 10 and 12 in. Some manufacturers also carry 14-in. tiles in stock. LIGHT-WEIGHT FLOOR CONSTRUCTION 605 In other respects, the standards adopted by various manu- facturers differ to a great extent. Standard Width, measured at the bottom between the edges of the sides (not including the width of the one-inch flanges) varies from 20 in. to 25 in. This difference in width affects the spacing of the joists in the floor and should be taken into account when com- paring the strength of designs made by different firms. It is obvious that, with the same design for the joists, the floor will be stronger where the joists are spaced closer. For equal strength, the joists spaced farther apart should be thicker and have more reinforcement. In addition to tiles of standard widths, each manufacturer carries special tiles to be used in the last row near the edges of the panel, where the regular tile is too wide. Standard Lengths adopted by various manufacturers vary from 2 ft. to 4 ft. 6 in. Some manufacturers use two standard lengths. In one instance, they are 30 and 36 in. of covering length (the actual lengths are 13 in. greater to allow for lapping). By combining the proper number of tiles of each length and by selecting proper lapping, forms for any length of joist may be obtained. Tapered Tiles——In addition to tiles of constant cross section, tapered tiles are manufactured. These are used at the ends of the joists where widening of the joists is required. The tapered tiles at one end are of the same width as the regular tiles, and at the other end the width is reduced by about 5 in. ‘This taper is obtained in a length of 30 in. Often the tiles are tapered in height also. The purpose of the taper is to increase the width of the concrete joist at the support, where additional shearing area is required. In continuous joists, the increased width also relieves the large com- pression stresses at the support. Adjustable Metal Tiles.—Another type of metal tiles is shown in Fig. 196, p. 606. The differences between these and the tiles just described are as follows: The adjustable tiles are made of smooth, heavy sheet steel, No. 12 gage, instead of the lighter corrugated material used in other tiles. They are wider and longer. Their width is 31 in., against 20 to 25 in. used for the other tiles. The length is from 6 to 10 ft. The depth of tiles is adjustable, as is evident from Fig. 196. It should be noticed that the metal tiles may be removed without removing the form under the joist. The metal forms are shown in place in Fig. 197. The pleasing appearance of the ceiling in the photograph, Fig. 198, is due to the 606 BUILDING CONSTRUCTION use of smooth metal for the tiles and to the small number of joints. Both photographs were obtained from Mr. Wm. H. Gravell, engineer for the two jobs. | Method of Design.—The design of the floor system resolves itself into the design of the joists composing the slab and of the girders carrying the joists. For the purpose of design, the joists may be considered as small T-beams in which the topping serves as a flange. The following steps should be taken in designing: (1) Depth and width of the joist should be assumed for the pur- pose of determining the dead load. (2) Bending moments and 7 #12 Smooth Sheet Steel 60" wide 2 zy ~ ~) Ss Le _~ Bottom of = Joist Lengths of 6; 8' & 10° hake ne lea” Removable ------.___ Support for Centers Wood Shoring and Centering Fic. 196.—Adjustable Metal Tile. (See p. 605.) shears should be computed. For continuous joists, use formulas on p. 278. (3) Shearing unit stresses should be determined. (4) Required amount of steel and the compression stresses in the center should be found by means of the T-beam formulas. (5) For con- tinuous joists, compression stresses at the support should be com- puted by means of formulas for rectangular beams with steel in top and bottom. Proper selection of the depth of the joist requires judgment. The depth must be sufficient not only to resist stresses but also to prevent excessive deflection. The remarks made in connection with hollow tile floors apply equally here. If shearing unit stresses are larger than allowed for concrete, either stirrups should be used or the width of the joist should be LIGHT-WEIGHT FLOOR CONSTRUCTION 607 Fic. 198.—Ceiling in Penn Charter Gymnasium. (See p. 605.) 608 BUILDING CONSTRUCTION increased by the use of tapered tiles. The bent bars in continuous joists are seldom available for diagonal tension reinforcement, because, to be effective as negative bending moment reinforcement, the bent portion must be some distance from the points of maximum shear. In continuous joists, compression stresses at the support must be provided for. To reduce compression stresses in concrete, either compression steel may be used or the width of the joists may be increased by the use of tapered tiles. In this connection, refer to p. 594, where the same ‘subject is treated for hollow tile floors. Points of inflection should be computed as explained on p. 595. Reinforcement for the joists should consist of bars of small diameter, else the bond stresses will be excessive. The bars should be placed at least one inch in the clear from the edge of the joist. Not more than two bars should be placed in a joist 5 in. wide. Bars should be held securely in place. In all respects, the design should comply with the requirements eiven for hollow tile joists on p. 594. Design of Topping.—The topping, i.e., the thin concrete slab between the joists, serves two purposes: first, to carry the load placed thereon to the supporting joist; and second, to act as com- pression flange for the joists. For all practical purposes, the topping ‘s a concrete slab restrained at both supports (i.e., at joists), and carrying the load on a span equal to the distance between the edges of the joists plus the thickness of the topping. Since it is not prac- ticable to reinforce the thin slab for negative bending moment, to prevent cracking, the thickness of the slab should be such that the tensile stresses in the concrete, due to negative bending moment, will not exceed the safe value for plain concrete. (For 2 C00-lb. concrete this is 40 lb. per sq. in.) The minimum thickness of the topping should be 2 in., exclusive of the finish. For a width of core — of 20 in., this thickness is good for a live load up to 150 Ib. per sq. in. For larger live loads, the thickness of the topping should be increased. For larger spans, the safe load is proportionally smaller. In many designs, the bending stresses in the topping are disregarded, and the concrete is subjected to considerable tension. The same disregard is found in tables prepared by some manufacturers of steel tiles. For instance, in one table, a live load of 666 Ib. per sq. ft. is recom- mended for joist construction with a 9-in. topping, although this load would destroy the thin slab. The topping acts also as a flange for the joists, and its thickness must be sufficient to resist the compression stresses developed by LIGHT-WEIGHT FLOOR CONSTRUCTION 609 the bending moments in the joists. For deep joists of long span, the thickness of topping may have to be increased to keep the compres- sion stresses within working limits. From the above, it is evident that the topping is a vital part of the construction, and that it is of importance to prevent its separa- tion from the joist. Therefore, shrinkage stresses and temperature stresses must be resisted by reinforcement placed at right angles to the direction of the joist. This may consist either of small bars or of wire mesh. ‘The cross-sectional area of the steel parallel to the joist should not be less than 0.2 per cent of the cross section of the topping for cases where steel tiles are left in the construction and the ceiling is plastered. When the under side of the topping is exposed, at least 0.3 per cent of temperature steel should be used. In competitive designs, a small amount of temperature steel is used, such as {-in. bars 12 to 18 in. on centers. This practice pro- duces structures having a much lower factor of safety than required in other types of reinforced concrete construction. The small saving thus obtained is out of proportion to the harm done to the strength of the floor. When granolithic finish is used, it should not be considered as a part of the topping. Concentrated Loads.—The steel tile floor is not well adapted for concentrated moving loads. A concentrated load applied on the thin slab would overstress the concrete in bending. Another dis- advantage is that each joist would have to be designed to carry the full concentrated load, as the thin topping is not capable of bringing the adjoining joists into action. For this reason, steel tile floors should not be used for concentrated moving loads. When the position of a fixed concentrated load is known, the joist under the concentration should be strengthened. For instance, under partitions, double joists are often used. When the concen- trated load comes between the joists, the topping should be made thick enough to carry the load and should have the proper amount of reinforcement. If the position of the concentrated loads cannot be fixed in advance, or if there is a likelihood of its being shifted, it is best to use other types of construction better adapted for concen- trated loads. Construction.—Since the slab is thin and the ribs narrow, par-. ticular care should be given to the pouring of the concrete and the placing of the steel. 610 BUILDING CONSTRUCTION Topping Must be Poured Monolithic with the Joists Below.— The number of construction joints should be as small as possible. If it is necessary to place a construction joint at right angles to the joist, it should be placed in the center of the span. Construction joints parallel to the joist, when unavoidable, should be placed mid- way between the joists. To fill properly the spaces between the bars in the narrow joists, the concrete should be composed of coarse aggregate passing through 3_in, mesh. Large stones may bridge in the forms and cause pockets. It is particularly important to force the concrete below the rein- forcing bars in the joists. Puddling during placing of concrete is advisable. The steel tiles must be removed with care, so as not to injure the thin topping and the edges of the joists. | Examples of Steel Tile Floor Construction—An example of the use of the steel tile floor construction is shown in the designs of the hospital building on page 807. Another example is the design of the building for Barnes Foundation, shown on p. 630. Plaster on Rib Lath, Permanent” Lath with " Ri Metal Tile 3 : Rib. ~ Before removal After Plaster of Forms Applied Fra. 199.—Method of Forming Flat Ceiling when Permanent Metal Tiles are Used. (See p. 610.) Flat Ceiling. —In construction with metal tile floor, a flat ceiling is obtained by attaching metal lath to the concrete and plastering the lath with three coats of plaster. When permanent tiles are used, metal lath is laid on the top of the wood forms before the metal tiles are placed. (See Fig. 199.) The tiles and the reinforcement are then placed, and the lath securely . wired to the reinforcement of the joist. The concrete, when poured, engages the meshes of the metal lath; but this bond should not be relied upon to carry the plaster and should not be considered as replacing proper wiring of the lath to the reinforcement. After the forms are removed, the ceiling is ready for plastering. COMBINATION OF STRUCTURAL STEEL AND CONCRETE 611 The strength of lath for the ceiling should be sufficient to carry the plaster without undue deflection on a span equal to the spacing. The weight of the plaster, for ordinary three-coat work without ornamentation, amounts to 10 lb. per sq. ft. To avoid the use of furring channels, special types of lath with raised ribs are on the market. ‘They are sold under various trade names, such as Herring- bone Lath, Hy-Rib, Riplex, and others. Thir common property is that the expanded metal sheets are stiffened by rigid integral ribs running longitudinally with the sheets. The height of these ribs varies, depending upon the different uses to which they are put. For ceilings, the height of the ribs is usually 2 in. The thickness of material varies with the span between the joists. The gage of metal required for spans used in common practice is: for spans up to 24 in., No. 28 gage; for spans between 24 and 27 in., No. 26 gage; for spans up to 32 in., No. 24 gage. The rib of the metal lath should be placed against the support, and the mesh away from it. With this arrangement, the mesh offers a flat surface for plastering. The ribs are placed at right angles to the joists. Each rib should be anchored to the concrete. When removable tiles are used, the wire lath must be attached under the joists after the forms are removed. For this purpose, wire hangers, provided on one end with a hook, are installed in the concrete. After the metal tiles are in place, the wire hangers are pushed through special holes, which are provided for the purpose in the tiles, either on the top or at the sides, and placed so that the hook becomes inbedded in the concrete. After the metal tiles are removed, furring rods or channels, spaced proper distances apart, are suspended from the hangers and metal lath attached to them. With the rib lath described above, furring bars or channels may be dispensed with. If it is desired to have the ceiling some distance below the joists, suspended ceiling, as described on p. 613, should be used. COMBINATION OF STRUCTURAL STEEL AND CONCRETE A structural steel frame for beams, girders, and columns may be preferable to reinforced concrete for certain structures, such as city office buildings, simply for the reason that if the steel is fabricated in advance the buildings can be erected more rapidly although at higher cost. : 612 BUILDING CONSTRUCTION In structural steel buildings, slabs between beams are usually puilt of concrete. The design of a panel with steel beams and girders is shown in Fig. 200. Concrete slabs are reinforced in the direction perpendicular to the beams. The main slab reinforcement should be run over the top of the beams and near the upper surface of the slab so as to make the slabs continuous. Transverse temperature steel is used, as recommended in connection with concrete slabs. Often, wire fabric or expanded metal are used for slab reinforcement. Bars’ for Slab Directions Main Reinforcing below Bottom Flange Fic. 200.—Typical Panel with Steel Beams and Girders and Concrete Slab. (See p. 612.) Note: Main Reinforcement shown schematically by outside bars only. For typical arrange- ment see p. 210. The cross section through slab and steel beams is shown in Fig. 200a. In such construction, the steel beams and girders are designed separately as plain steel sections, as, due to lack of bond, no allowance can be made for strengthening effect of concrete. The slab is designed according to the principles of reinforced concrete. New York City, however, allows special empirical rules for design of cinder concrete slabs supported by steel beams. If the panel is square, or nearly so, the steel beams may be placed on four sides, as in Fig. 2006 above, and the slab is then reinforced by bars running in two directions, or by an adaptation of the Smulski System consisting of radials and cireles as shown in Fig. 128, p. 369. COMBINATION OF STRUCTURAL STEEL AND CONCRETE 613 The slab between beams also may consist of two-way tile and con- crete floor as shown in Fig. 201, p. 613. Floors may consist of a combination of steel girders and rein- forced concrete beams and slabs, in which case proper seats in the steel frame are provided for the concrete beams. The slab between steel beams is sometimes constructed in the form of a concrete arch. If not reinforced, the beam should be con- nected with tie-rods, to resist any possible unbalanced thrust. For arches with curved upper surfaces, a fill of cinders or a very lean con- crete is used for leveling. Fre. 201.—Details of Two-way Hollow Tile Floor Slab and Structural Steel Column Fireproofing. (See p. 613.) SUSPENDED CEILING The term “Suspended Ceiling” is used when the plaster ceiling is more than 6 in. below the supporting floor construction. Its construc- tion differs from the plaster ceiling directly attached to the floor con- struction, such as shown in Fig.201, p.613. Suspended ceilings are used: (1) To provide flat ceiling in case of beam and girder or joist construction in which case it is placed below the deepest beam. (2) To provide ornamental ceiling. (3) To lower a part of the ceiling. (4) To provide air space below the roof construction for preven- tion of condensation. 614 BUILDING CONSTRUCTION The suspended ceiling consists of plaster applied on metal lath which is suspended from the floor construction. The lath should be lapped both longitudinally and laterally so as to present a con- tinuous lathed surface. To get proper rigidity and strength the lath must be strengthened by furring as described below. All fur- ring and its supports and hangers must be strong enough to carry a load of at least 12 lb. per sq. ft. without undue deflection. The plaster will not crack for deflection up to sy of the span of the fur- ring. For ornamental ceilings, specially where false beams are used, the weight may exceed 12 Ib. per sq. ft., for which proper allowance should be made in design of the furring and hangers. The simplest design of suspended ceiling consists of lath, provided with longitudinal ribs or stiffeners, wired to furring channels, angles or flats placed on edge. Usually 3-in. furring channels spaced about 2 ft. on centers give satisfactory results. Each rib or stiffener of the lath must be wired at each intersection with the furring channel using a 16 gage galvanized or annealed wire. The furring channels are suspended from the floor construction by 7 gage wire hangers spaced about 2 ft. on centers along the furring channel. The wire hangers on one end are imbedded securely in the concrete of the floor construction before the concrete 1s deposited. The free end is securely wound around the furring channel. | To insert the hangers, holes are drilled in the forms proper distances apart through which part of the wire is made to project the required length outside of the form. The end to be imbedded in concrete is preferably looped around the reinforcement of the floor construction. When the lath is not stiffened, furring rods 2 in. diameter spaced 12 in. on centers should be placed across the furring channels and attached to them at each intersection. The lath is then wired to the furring bars. Where it is not desirable to use the large number of hangers required by the previously described method, following method may be used: The lath is attached to the 3-in. channels as in the previous case. These, however, instead of being suspended from the floor construction, are wired to from 13 to 2 in. channels spaced about 8 ft. on centers and suspended from the floor construction by 4-in. hangers. The spacing of hangers along the channel is about 4 ft. on centers. | Where wire hangers are not provided ahead of time, holes are COMBINATION OF STRUCTURAL STEEL AND CONCRETE 615 drilled in the concrete ceiling and expansion bolts are inserted. If the floor construction consists of hollow clay tiles, toggle bolts should be used to hold the hangers. Care should be taken that the load concentrated at each hanger is not large enough to break the tile. Suspended ceilings are shown in Fig. 202, p. 616. Simple ceiling as well as more complicated designs are shown. The construction of false beams is also shown. If more rigid construction is required, the wire hangers may be replaced by flat steel hangers which are ordinarily 1 in. wide and 1 or 38; in. thick. The hangers are provided with holes for 3-in, diameter bolts to be used for securing them to the furring channels. Special design of the supporting members must be made when the suspended ceiling is unusual in any respect either on account of the weight of the ornamentation or because the spacing of the supporting members is large. Ordinary principles of design should then be employed. | INSERTS FOR ATTACHING SHAFTING AND SPRINKLERS When it is desired to suspend shafting or sprinkler systems from the ceiling it is best to make provision for it by imbedding in concrete special inserts spaced proper distances apart. _ Inserts may be of the threaded type where the bolts are screwed into the threads in the body of the inserts. Such inserts may be used only where no great accuracy in placing is required. To take care of any misplacement of the inserts during construction, special adjustable slotted inserts are used, in which the bolt may be moved several inches and thereby adjustment can be made. In cases where it is not possible to fix the position of the shafting ahead of time, continuous inserts are used. ‘These may be of any desired length and may even extend the full length of the building. With sufficient number of parallel rows of continuous inserts, shaft- ing may be placed in any desired location. The inserts must be attached firmly enough to the forms, so that no misplacement.can take place during concreting. The simplest method is by nailing them to the forms with two or three nails. To facilitate this, many inserts have special projecting lugs for the nails. To save the labor of cutting the projecting nails after the forms are stripped, special means of attaching are developed. In some inserts a tight fitting wood block is placed within the slot of the insert. This 616 BUILDING CONSTRUCTION Furring Channel --y CaN VEZ Ceiling Suspended Below Beam from Concrete Slab Ceiling Suspended from Concrete Slab TUT TM PT OL SSL La SSS ——— TAIT I So SI TITS UTTT ATA I LL PL ELL Pista crete eet oe Hanoy =; VRP ee ey || ER SDE Ra FAT WIT EY PRI PT aa) aes A 574. .: Pa | Mea) CL MONLN ec eS » 3" ~Ixj§ Strap Iron False Beam and Cornice Suspended from Concrete Slab ‘Yaulted Ceiling Suspended "Section A-A- from Concrete Slab Fic. 202.—Examples of Suspended Ceilings. (See p. 613.) COMBINATION OF STRUCTURAL STEEL AND CONCRETE 617 block is provided with nails or screws which are fastened to the form. After the forms are stripped, the wood block is pulled out of the insert within the nail. There is considerable difficulty in attaching inserts to steel forms. An adjustable insert and a continuous insert are shown in Fig. 203, below. When no provision is made for attaching the shafting, small holes may be drilled in the ceiling for expansion bolts. Where considerable load is to be supported through-bolts may be required extending all through the slab. In such case sleeves of proper diameter should be imbedded in the concrete to receive the Fria. 203.—Inserts Used in Concrete Construction. (See p. 617.) bolts. To prevent concrete flowing into the sleeves during con- ereting, they may be filled with sand. Very often through-bolts for attaching shafting are placed across beams. The hole for these through-bolts is also obtained by imbedding a sleeve in the beam. The inserts should be of such design that they would not pull out from the concrete under the load of the suspending shafting or sprinkler systen. In the inserts shown in Fig. 203 the loop in the individual insert extends some distance into the concrete while in the continuous insert the bolts extend into the concrete and serve as anchors. Protecting Exposed Corners.—Exposed corners of concrete col- umns or curbs should be protected by steel angles of proper dimen- sions. These angles must be anchored into the concrete by means of bolts. There are specially constructed curb bars on the market which are provided with projecting anchors integral with the bars. 618 BUILDING CONSTRUCTION BASEMENT AND GROUND FLOOR SLABS This discussion applies to the basement floor slabs and to the first floor slabs in buildings without basement. The basement floor may rest directly on and be supported by the ground, in which case it may be built of plain concrete or reinforced with temperature steel only. If the ground is not firm enough to carry the slab load, reinforced concrete floor must be used. If the basement slab is subjected to upward water pressure, it must be designed for the upward pressure. Basement Floor Supported by the Ground.—The integrity of the slab resting upon the ground depends upon the firmness of the sub- grade and upon its uniformity. For this reason, the subgrade must be carefully prepared. All soft and spongy places must be removed, and the subgrade compacted by tamping. If the grade of the slab is above natural grade, the ground must be filled in with hard material. The use of soft material, trash, vegetable matter, and other perish- able material is prohibited. It is important to have the subgrade of uniform firmness, else the slab may settle unevenly and crack. If the subgrade is of sand or gravel, the basement slab may be laid directly upon it. A subgrade consisting of clay requires a subbase consisting of hard materials, such as gravel or clean, well-screened cinders. The subbase should be properly tamped and wetted before the concrete is laid. The wetting is helpful in compacting the subbase, but its main purpose is to prevent the absorption by the dry subgrade of water from the concrete. For this reason, the subbase or the subgrade, if a subbase is not used, should be wet at the time the concrete is placed upon its Plain Concrete Slab.—The basement slab may be built of plain concrete. To prevent cracking due to shrinkage and temperature changes, the floor should be divided into sections of not more than 100 sq. ft. To prevent buckling and heaving of the slab in case of expansion, it is advisable to provide elastic joints at the wall and at the column. In slabs subjected to heavy truck traffic, the joints should be rounded to prevent their being clipped off. Basement Slab with Temperature Steel.—If it is desired to limit the number of joints the slab should be reinforced with temperature reinforcement placed in two directions at right angles. Reinforce- ment should also be used in all cases where it is not possible to get firm and uniform subgrade. The reinforcement then prevents cracks due to BASEMENT AND GROUND FLOOR SLABS 619 bending, which occurs when a section of the subgrade yields more than the rest. The amount of reinforcement will depend upon the distance between expansion joints in the slab, upon the expected variation in temperature, and finally upon the load to be carried by the slab. The minimum reinforcement should consist of 2-in. round bars 12 in. on centers (or their equivalent). Wire mesh or expanded metal may be used to advantage. The reinforcement should be placed 2 in. above the bottom of the slab, and should be properly lapped. The reinforcement must not cross the expansion joints; otherwise, the slab would not be free to move at the porns and their purpose would be frustrated. Thickness of Basement Slab.— Basement slab should be at least 4 in. thick, including the wearing surface. Best results are obtained by using concrete of 1 : 2:4 mix for the base of the slab. The granolithic finish on the top should be of the same composition and should be laid in the same manner, as explained on p. 620 in connection with granolithic finish for rein- forced concrete work. In many instances, granolithic floor finish is not required, and the - basement slab is built for the full thickness of the same concrete. After the concrete is brought to the required grade, it is leveled off by a strike board and then floated and compacted with a wood float. Basement Floor Supported by Columns.—If it is not possible to get subgrade firm enough to support the load, the basement slab is supported on columns and designed in the same way as other sus- pended floors. The cost of the slab may be reduced by using intermediate short columns under the basement slab. For instance, in a 20 by 20 ft. span, intermediate columns placed in the center of the panel and halfway between the columns reduce the span of the slab to 10 ft. Of course, the use of intermediate short columns is economical only when foundation for them can be obtained without difficulty and at reasonable depth. Before deciding upon the type of construction, the saving due to reduction of span should be com- — pared with the cost of the additional column and foundation. Inter- mediate columns reduce the load on the main foundation. This should be considered when making comparison. Basement Slab Subjected to Water Pressure.—When the base- ment slab is subjected to upward water pressure, it should be rein- forced to withstand the difference between the upward water pres- 620 BUILDING CONSTRUCTION sure at high water and the weight of the slab. The live load in the slab cannot be considered as contributing toward the balancing of the water pressure, becausé it is not always in place. The bending moments due to water pressure are of opposite sign to those produced by downward load. Thus, at the column, the bending moment is positive and requires steel near the bottom, while between columns it is negative and requires steel near the top. The construction of the basement slab to resist the upward pres- sure will depend upon conditions, such as the amount of pressure and the distance of the footings from the level of the slab. If the footings are some distance below the slab, they cannot be used to support the basement slab. The construction then must be tied to the columns. Flat slab construction will be cheapest if the spans permit, as it requires no formwork. Tor long spans, beams and girders may have to be used. If the columns are reasonably near the surface, the beams and girders may be attached to the footings. They must be secured by reinforcement against uplift. Sometimes it is necessary to design the basement slab for upward water pressure as well as for downward live load. The downward load to be used in design is the difference between the total live and - dead load and the water pressure at low water. ; FLOOR SURFACES In reinforced concrete buildings the surface finish of the floors depends upon conditions and also upon the use to which the building is subjected. Obviously, for example, different finish will be required for a hotel than for a warehouse. Also different portions of the same structure may require different finish. GRANOLITHIC FINISH A common and durable floor finish for concrete and reinforced concrete floors is the granolithic finish. This consists of specially proportioned and specially applied concrete. ° Two mixtures for granolithic finish are now in common use, namely: 1. Mixture composed of one bag of Portland cement, and 2 cu. it. of fine aggregate such as very coarse sand. 2. Mixture composed of one bag of Portland cement, 1 cu. ft. of fine aggregate such as coarse sand and 1 cu. ft. of still coarser aggre- FLOOR SURFACES 621 gate consisting of clean, hard, crushed rock or pebbles. For still harder surface where appearance is of minor importance use one bag of cement, 4 cu. ft. coarse sand and 14 cu. ft. coarser aggregate. The size of the coarser aggregate should be such that it will pass when dry a screen having ?-in. openings and not more than 10 per cent will pass a screen having four meshes per lineal inch. The last named mixture gives harder floor finish and should be used where the floor is subjected to considerable wear, such as truck- ing. It is harder to travel and cannot be brought to so smooth a surface. It was designed originally by The Thompson and Lichtner Co. as a result of an extended series of tests with different composi- tions of material. Thickness of Granolithic Finish.—The thickness of granolithic finish depends upon conditions. It should not be less than { in. for integral finish, which is laid before the concrete of the structural slab is set, nor less than 1 in. for bonded finish, laid after the concrete slab has hardened. Where subjected to severe wear, an increase in thick- ness is advised. Thicknesses of 1 in. for integral and 13 in. for bonded are common practice. ' The “integral finish,” considered by itself is about one third cheaper than the “bonded finish,”’ due chiefly to the reduction in the amount of materials and saving in the cost of preparing base for “bonded finish.”? Another advantage of integral finish is, that there is no question of bond involved between the slab and the finish. In spite of these advantages the “bonded finish” is often preferred for the following reasons: The ‘integral finish”? must be applied before the concrete of the structural slab has set. Thus the finishing gang must work overtime in order to finish all the floor placed on that day. This adds to the cost especially where extra rates must be paid for overtime. Second, to avoid injury to the finish, the use of the floor for further construction work must be delayed. Third, when placing forms for upper floors it is not possible, even with proper precautions, to prevent entirely damage and scratching of the finish. Fourth, in inclement weather the finish is exposed to the action of the rain and frost, unless particular precautions are taken. F ifth, there is always danger that through carelessness or design, the “integral finish”? in part of the building at least will not be applied until the slab has hardened. All these items tend to reduce the economy of the “integral method.” | ? The “bonded finish” has these advantages. It is usually applied 622 BUILDING CONSTRUCTION after the building is completed and closed in. It may be applied at any time and at any desired speed. The fresh finish does not need to be exposed to use until it has hardened sufficiently. No special protection for “bonded finish” is required as the building is closed in. The most serious objection to “bonded finish’’ is, that if not properly bonded to the base, the granolithic may peel off. However, if applied in the manner recommended in the Specifications for Granolithic Finish in Volume IJ, satisfactory results will be obtained. It must be emphasized, that granolithic finish should be applied by skillful men, as poor workmanship, especially when coupled with poor selection of materials, is usually the cause of trouble with granolithic finish. Also special emphasis should be placed on the requirement in the specification just referred to that all laitance and loose material should be removed, that the slab be thoroughly roughened and wetted, and finally that the surface be coated with neat cement paste just before placing the finish. It is more economical to remove the laitance and roughen the slab before the concrete becomes thoroughly hard. Two other factors are of vital importance, namely: Selection of an aggregate free from dust or very fine particles which rise to the surface with the cement so as to produce a “‘ dusty ” surface and the avoidance of a wet, sloppy mix which also prevents a thoroughly hard, dense surface. _ | Advantages of Granolithic Finish.—Floors with granolithic finish are sanitary. They may be easily cleaned by a hose. To facilitate this the floors are pitched to drain through scuppers. The pitch is usually } in. to the foot. Granolithic finish is durable. If properly built it will resist severe wear, such as trucking for an indefinite period. Disadvantages of Granolithic Finish.—Often granolithic finish is objected to by the users on account of larger conductivity of heat. This objection is gradually disappearing for floors above ground as it was largely psychological. If a building is heated uniformly the floor slab also maintains a uniform temperature. The objection is often removed by oiling or painting the floors, thus relieving the cold appearing color of concrete. Granolithic is less resilient than wood. Work places for women required to stand all day should be covered with some material such as linoleum or rubber. For offices, also, a covering of this character is generally employed. 7 FLOOR SURFACES 623 Coloring of Granolithic Finish—Granolithic finish may be colored by mixing with the dry material for the concrete, in the entire finish, a proper amount, to be determined by trial, of mineral pigments. The materials should be mixed dry and thorough mixing is essential. As the admixture of coloring matter reduces somewhat the strength of the concrete, it should be limited usually to 5 per cent by weight of the cement. For description of coloring materials, see Volume II. Painting of Floors——Concrete floors may be painted to any desired color. If proper paint is used and is properly applied, it is as permanent as on wood floors. Naturally it wears out under foot traffic. Floors should not be painted where they are subjected to heavy wear. In such case the integral coloring is preferable. Directions for Laying Granolithic Finish —Complete directions for laying granolithic finish is given in Volume II of this treatise. TERRAZO FLOOR FINISH “Terrazo” is another type of cement finish. It consists of irregular marble or granite chips imbedded in cement mortar and ground to a smooth even surface.. The finish is usually 12 to 2} in. thick and is applied in two layers. The first layer, or binder course, consists of 1:3 cement and sand mortar and is about 1 to 1% in. thick. After the binder course is placed, tamped and leveled off, the finishing course is applied. This is composed of cement and stone chips, in proportions about 1 to 3 which are spread evenly over the whole surface and rolled. After this, additional chips are spread over the surface and rolled in. The finishing course is allowed to harden for about 24 hours. Then the surface is ground by rotary rubbing machine. Small surfaces may be rubbed by hand. Carborundum brick is used for rubbing both by machine and by hand. The color of terrazo finish depends upon the color of the chips. By proper selection of the material various color effects may be obtained. - Terrazo finish may be applied on green concrete slab or after the concrete has hardened. In the latter case, the base should be pre- pared as for granolithic finish by roughening, cleaning and wetting of the surface. Before the first course is placed the surface should be wet and then covered with neat cement paste as described for granolithic finish. Method of placing terrazo finish is fully described in Volume II of this treatise. 624 BUILDING CONSTRUCTION WOOD FLOOR FINISH Wood floor surfaces are often used for residences, apartment houses, offices and hospitals; less often for factories and seldom for warehouses. The top flooring usually consists of hardwood such as oak or maple. Beech and birch also are sometimes used especially in combination with maple, which they resemble in appearance and wearing qualities. Oak flooring is used in residences and more expensive buildings only. It may be either plain sawed or quartered. Plain sawed boards are obtained by sawing the whole log in parallel layers. Quartered oak is obtained by sawing the log parallel to the diagonal line in each quarter. Maple flooring, sometimes asl with other hardwood, is used in other types of buildings. Component Parts of Wood Flooring—Wood flooring consists of hardwood top boards, and wood sleepers and cinder concrete fill. Where heavy flooring is required, intermediate flooring, consisting of thin boards or heavy plank, is used. Dimensions of Hardwood Boards.—The boards may be either 3 or 4 in. wide. The 3-in. boards give better results than the 4-in. width, but the advantages gained are not worth the extra expense. The thickness of boards depends upon the use of the building. Flooring nominally 1 in. thick is most commonly used. Sometimes, when the flooring is exposed to hard usage, the nominal thickness is increased to 14 in. The actual finished thickness is about 7 in. smaller than the nominal dimensions. The boards are usually tongue and groved and blind nailed. Wood Sleepers.—The wood sleepers are usually spruce and their nominal cross section is 2 by 4 in. They are dressed on top and bottom and leveled on both edges so that the width on the top is reduced to 3 in. To prevent the sleepers from rotting when im- bedded in concrete they should be creosoted or kyanized. | Intermediate Flooring—Intermediate Flooring usually consists of 1 by 6 or 1 by 8 (nominal dimensions) spruce boards, surfaced on both sides. Heavy planks, ordinarily used for intermediate floors, are spruce. They are commonly 2 by 6 in. or 2 by 8 in. nominal dimensions sur- faced on both sides. The life of the floor is increased if the boards are creosoted or kyanized. | FLOOR. SURFACES 625 Description of Wood Flooring on Sleepers.—Wood flooring may be divided into one-layer flooring and two-layer flooring. One-layer Flooring.—One-layer flooring is satisfactory for light usage only. It consists of hardwood top nailed directly to 2 by 4 in. leveled sleepers. ‘These are spaced 16 in. on centers and are im- bedded in 2 in. thickness of cinder concrete. A ply of rosin-sized building paper is laid underneath the hard- wood flooring. Two-layer Flooring.—Two-layer flooring consists of hardwood top, intermediate creosoted softwood flooring placed diagonally and 2 by 4 in. leveled sleepers imbedded in 2 in. thickness of cinder concrete. ! | arallel to joist and H doh hea wine Fest op slab addz hd. ||\ | i i f 1 iBearitg Vaal zhdb"oc, 6"ac.,4°0"longs, y ey H H H \ \ t \ tent | t sige a fon ie = ae Pa ee ve) as | “| i I eh I i : : | Phe gt RH7" r G Hod feet) fale si nee & pata She _ a hy Typical Section Pe \Brich Bearing Wall~x, ——e=—e—! Frame. Plan of Second Floor Fic. 204.—Floor Plan and Details of Barnes Foundation Building. (See p. 629.) Paul P. Cret, Architect. William H. Gravell, Engineer. They should be laid in Portland cement mortar. The brickwork should be properly bonded. It is generally required that one course of headers must follow every five courses of stretchers. Face brick, REQUIRED THICKNESS OF BRICK WALLS .- 631 if properly bonded to the rest of the brickwork by means of headers, may be considered as part of the thickness of the wall. If the face brick is tied to the wall by metal ties only, it must not be considered as effective in carrying the load. Stresses in Brickwork.—The allowable stresses in brickwork depend upon the hardness of the brick and also upon the strength of the mortar. For supporting concrete construction, Portland cement mortar mixed in proportions as rich as one part cement to three parts sand (1 : 3) should be used for bearing walls. The following unit stresses may be used: Allowable Stresses, Ib. per sq. in. Hard paving brick material (crushing strength of single bricks PU meer Meee. ee ltt. ok ngs ca ewe ee chee es 350 Ib. rere UM eee nee, ie ac. Ue OLR AA, 250 lb. Hard common brick (crushing strength of single bricks 2300 lb. per Sopa meee eee Nt! Sd oi is Payee! vale Aa ke 2 200 lb. Bearing Stresses on Brickwork.— When the available area of the wall is not less than twice as large as the area of application of a con- centrated load, the allowable bearing stresses produced by the load may be equal to 14 times the stresses recommended above. For smaller ratios between the area of wall and the area of application of the load, the allowable bearing stress may be reduced propor- — tionally. Required Thickness of Brick Walls.—The minimum thickness of the brickwork for outside walls of industrial buildings, warehouses, hotels, office buildings, schools, and hospitals should be as given in the table below: Base- | 1st 2nd 3rd 4th 5th 6th Height of Building ment | Floor | Floor | Floor | Floor | Floor | Floor Re sey co ben oi © 12 12 PeOMILDTICS ee ee 16 12* 2 Brose BLOTIES is. Su. ess oe. 16 16 12* 12 Me tenes oii ce fe. el 20 20 16 16 12 BEPPOreL OPTICS, fhe) ia ws bis koe 24 20 20 16 16 16 SO i ares 24 20 20 20 16 16 16 * For schoolhouses use 16-in. wall. 632 WALL-BEARING CONSTRUCTION Limitations—The above table is valid for clear story heights up to 18 ft. For story heights from 18 ft. to 24 ft., 16-in. wall should be used instead of 12-in. For story heights from 24 ft. to 30 ft., 20-in. wall should be a minimum. No changes are required ‘n other items of the table. For walls 50 ft. long and under, the thickness of the wall may be decreased by 4 in. the minimum thick- ness, however, should remain 12 in. The length of the wall should be considered either as the total length or as the length between cross walls. | Walls used as enclosures for staircases, elevator shafts, or other shafts, when their length does not exceed 25 ft., may be built 12 in. thick for upper 50 ft. and 16 in. thick for the remainder of the height of the building. The thickness of the wall given in the table, or as modified by the above, should be considered as a minimum and should be increased if the stresses on brickwork exceed the allowable working stresses. Pilasters.—For heavy concentrated loads due to girders or trusses, pilasters, buttresses, or piers should be provided. The dimensions of these should_be such that’ the stresses on the brickwork are kept within working limits. In computing stresses, all dead load plus the proportion of live load recommended on p. 453 should be used. \ The wall between pilasters may be reduced in thickness by one- half of the difference between the thickness of pilaster and the required thickness of wall without pilasters. The reduced thickness of the wall should not be less than 12 in. for walls less than 30 ft. high and 16 in. for walls over 30 ft. When the thickness of the wall is reduced on account of the use of pilasters, their width must not be less than one-tenth of their spacing measured center to center. Interior Bearing Walls.—Interior bearing of brick walls used in concrete construction should not be less than 12 in. thick nor less than one-fifteenth of the clear story height. The thickness must be increased when required by stresses. In computing stresses in brickwork, the reduction in live load recommended on p. 453 may be used. Construction at the Wall.—When the ends of concrete beams rest on a’ brick wall, the bearing stresses should be investigated. If they are larger than the allowable stresses, as given on p. 631, the bearing area should be increased. This may be accomplished in several ways, as described below. CONSTRUCTION AT THE WALL 633 Steel bearing plates or steel shapes may be used to distribute the load on brickwork; the width of the beam may be enlarged; or, if this is not sufficient, projections may be provided on both sides of the beam. Such projections should be of the same depth as the beam, — to facilitate the laying of brickwork above the floor. The thickness of the bearing plates and the dimensions of the concrete projections may be determined by considering the part out- side the beam as loaded by the upward reaction of the brickwork and supported at the beam. It may be necessary to reinforce the concrete projections by bars placed near the bottom, parallel to the wall. A continuous distributing beam running along the wall may be used, especially in case of joist construction. This should preferably be of the same depth as the beams or joists. (See Fig. 204, p. 630.) A continuous distributing beam may be considered, for design- ing purposes, as a continuous beam loaded by the upward reactions, and supported by beams or joists. The reinforcement will therefore be required near the bottom at the supported beam, and near the top between the beams. Usually, however, continuous bars are placed near both top and bottom Where the distributing beam spans an opening, its action is reversed and it serves to carry the loading. Its dimensions will be governed, in this case, by the load and the span. If possible, the distributing beam should be of the same width as the wall, for the sake of simplifying the construction. When it is not desirable to expose the concrete, the width of the beam is made 41 in. smaller than the wall, allowing for brick veneer. The advantage of the distributing beam is that it ties the con- struction and in a measure prevents unequal settlement of the brick- work. It is always used in joist construction. The brick is stopped at the bottom of the joists, and for the next story is started on the top of the concrete. This gives a much neater job than if an attempt were made to brick in the spaces between the joists below the slab. The distributing beam should be reinforced with at least four j-in. round bars, two near the top and two near the bottom. Of course, special provision must be made over windows and doors, depending upon the span of the opening and the load to be carried. Provision for Negative Bending Moment at the Wall.—The brick- work placed above the concrete, especially when distributing beams are used, prevents free movement of the concrete construction. 634 WALL-BEARING CONSTRUCTION The amount of restraint so produced is problematic and depends upon the amount of brickwork and the load which it carries. It is larger in the lower stories and decreases toward the top. Therefore, it is not advisable to rely on it in computing the beams. ‘To prevent cracks, however, it is advisable to use a certain amount of reinforce- ment near the top of the beams and also along the distributing beams. One-half of the steel used at the first interior support will be found sufficient in ordinary cases of brick bearing jobs. This amount of steel corresponds to a bending moment for uniform loading equal to wi? Ye= 30° (See also p. 278.) CONCRETE WALLS Reinforced Concrete Bearing Walls.—Reinforced concrete may be used for bearing walls. Their thickness may be governed by the stresses, as in concrete columns. If built monolithic with the floor system, they may be exposed to bending and should be reinforced accordingly. ‘Temperature steel should be used. Special rein- forcement is required at the corners of the building to prevent the separation of the walls. Where a cross wall and a longitudinal wall are poured together, the horizontal bars of the cross wall should be extended into the longitudinal wall and hooked at the ends to pre-. vent a crack. If it is not desired to have a connection between the walls, they should be poured separately so that they may expand, contract, and settle independently. A recess should be provided in the longitudinal wall; a few dowels may also be used to prevent the opening of cracks between the walls, but their effect should not be sufficient to make the walls cooperate. Special reinforcement should be used above door and window openings. For wide openings, the necessary amount of reinforce- ment is found by considering the lintel as a beam. The reinforce- ment from the lintel should be run a sufficient distance into the wall to prevent a temperature and shrinkage crack at the junction. The thickness of the concrete bearing wall may be determined by stresses, the ratio of slenderness being taken into account. Some codes specify that the thickness of the conerete wall shall be three- fourths of the thickness required for brick wall. Reinforced Concrete Enclosing Walls.—The enclosing walls are sometimes built of reinforced concrete. When supported on the CONCRETE WALLS 635 framework, they do not need to be more than 6 in. thick, this thick- ness being sufficient to make the wall weather-proof. The walls should be anchored at all floors and reinforced with bars placed near their outside and inside faces; they should be made strong enough to resist a pressure of 30 lb. per sq. ft. applied either from the out- side or from the inside. This pressure represents the wind pressure and any accidental pressure that may be exerted from the inside. Temperature steel is usually sufficient to resist any stresses due to such causes. : Recesses for the walls should be provided in the columns; also bonds or dowels (1.e., short steel bars) above the floor beam and below the spandrel should be used. If walls are used above and below a beam, the same bars may be used for bonds above the floor and below the spandrel. The bonds may consist of 2-in. round bars spaced 24 in. on centers. These should be placed so that they come in the center of the wall. Temperature Reinforcement for Enclosing Concrete Walls.— When concrete walls consist of panels between columns which are built separately, the amount of reinforcement given below may be used. Spacing of ? in. Round Bars Thickness of at Each Face Wall Horizontal Bars Vertical Bars 6 in. 18 in. 24 in. 8 in. 14 in. 24 in. 10 in. pee Line 24 in. 12 in. 9 in. 24 in. If walls are monolithic with the columns, at least one-quarter of one per cent of horizontal reinforcement is required to distribute the cracks properly. These should lap at the columns. Reinforced Concrete Fireproof Partitions.—Reinforced concrete may be used for partitions, especially around staircases. The minimum thickness allowed by most building codes for reinforced concrete partitions is 33 in. Such partitions are considered fireproof and may be used for enclosing staircases, shafts, and elevator wells. The 636 WALL-BEARING CONSTRUCTION authors recommend in preference a thickness of at least 4 in., to give better space for pouring concrete around the reinforcement, with steel bars 2 in. round placed 12 in. on centers horizontally and 24 in. on centers vertically. The steel should be placed in the center of the wall. Special corner bars should be provided at the corners of the shaft or else the bars from one wall should be bent at right angles and extended into the other wall. The walls should be poured after the floor construction 1s completed. Bonds consisting of short bars spaced 2 ft. on centers should be provided above the floor and below the ceiling to tie the wall to the floor. Bars may be inserted into the pour hole. Pour holes spaced about 2 ft. on centers should be left in the slab above the wall to pour the concrete, although this is not necessary if the wall is under a beam, in which case holes may be left in the top of the wall forms and spouts arranged to receive the con- crete. ! Special reinforcement consisting of 4 or % bars should be provided at the top and sides of door and window openings. Short diagonal pars at the corners may also be used advantageously to prevent cracks. CHAPTER XII BASEMENT WALLS Basement walls, that is, walls below grade, in reinforced con- crete buildings or in steel buildings, may be thinner when made of reinforced concrete than when made of plain concrete. The prin- cipal reinforcement is supplied to resist stresses due to earth pressure. Some additional temperature and shrinkage reinforcement is needed, the amount depending upon the degree to which elimination of shrinkage cracks is desired, and also upon the design. More tem- perature steel is required for basement walls built monolithic with the column than for walls built separately between previously poured columns. Method of Construction.—The basement wall may be built either monolithic with the basement columns or separately from them. The latter method reduces the cost of wall forms. When the wall is built separately from the columns, proper recesses along the column should be provided. These recesses are formed by nailing, on the inside of the column form, a strip of wood of proper width and thick- ness. Dowels may be used to tie the wall and column. The joint between the column and the wall may be kept waterproof by insert- ing in the recess proper plastic waterproofing material. Thickness of Wall.—To be impervious to ground water, the basement wall should be at least 8 in. thick. Ordinarily, a 12-in. wall is used. This dimension may have to be increased if larger depth is required to resist earth pressure. In ornamental structures, the thickness of the masonry in ‘the first floor is sometimes greater than the required thickness of base- ment wall. If the difference is small, the greater thickness is adopted for the basement wall. If there is a large difference between the thickness of the masonry above and the basement wall, it may be economical to provide the basement wall at the top with a con- tinuous bracket to carry the masonry. (See Fig. 265, p. 754, and Fig. 290, p. 808.) 637 638 BASEMENT WALLS Earth Pressure.—For a basement wall below grade, the magnitude of earth pressure is determined in the same way as for retaining walls. The unit pressure will depend upon the character of the soil. Under ordinary conditions, earth pressure equal to the pressure of a fluid weighing 30 lb. per cubic foot 1s considered ample. When the wall abuts on a roadway or a railroad track, the moving load produces an additional pressure on the wall. ‘This can be taken care of by figuring a surcharge, as explained in connection with retaining walls (p. 838). For streets, a surcharge equal to 3 ft. of earth may be used. For railroad track, the surcharge should be from 6 to 10 ft. depending upon the distance from the track and the weight of the locomotive to be operated. Without surcharge, the pressure on basement wall varies accord- ing to a triangle. The pressure is zero at the ground level and a maximum at the level of the basement floor. There it equals the unit pressure multiplied by the depth in feet below the ground level. If the basement wall is 10 ft. deep and the unit pressure is 30 lb. per sq. ft., then the maximum pressure 10 ft. below ground is 10 X 30 = 300 lb. per sq. ft. } The effect of the surcharge is the same as if the ground level were raised by the amount of the surcharge. Thus, with 3 ft. sur- charge and a unit pressure of 20 lb. per sq. ft., the pressure at the ground level is not zero, but 3 xX 30 = 90 Ib. Ten feet below ground the pressure would be (10 + 3) X 30 = 390 lb. per sq. ft., ‘nstead of 300 Ib. for wall without surcharge. ‘The pressure, instead of being represented by a triangle as in the case without surcharge, is a trapezoid. Further explanation may be found in the chapter on Retaining Walls. Design of Basement Wall.—The direction of reinforcement to resist the earth pressure depends upon the manner in which the wall is supported. It may be considered as.a slab supported on the top and bottom, that is, on the first floor level by the floor construction -and on the basement floor level by the resistance of the basement floor and the passive resistance of the earth below the basement slab. Another method is to consider the wall as a slab, supported by the columns and spanning horizontally between them. The third method is to reinforce the wall in two directions. Wall Supported on Top and Bottom.—The first method, assuming the wall as a slab supported at top and bottom, is possible when there WALL SUPPORTED ON TOP AND BOTTOM 639 are no wide openings in the basement wall. At wide openings there is no connection between the wall and the top slab; hence, there is no support for the wall at the top. This method is economical when the spacing of the columns is more than 40 per cent larger than the depth of the wall. Its use is also advisable when the basement columns are slender and would have to be materially strengthened if any earth pressure were transferred to them. The main rein- forcement is placed vertically, near the inside face of the wall. Some horizontal steel is needed to prevent cracking due to tempera- ture and shrinkage. Under ordinary conditions, this would consist, for a 12-in. wall, of 2-in. round bars, 12 in. on center. Other thick- nesses Of wall may be reinforced proportionally. For a watertight wall, at least 0.3 of one per cent should be used. The temperature steel should be placed near the inside face. To prevent cracking of the wall at the column, due to bending moment produced by earth pressure, short horizontal bars should be placed at the column near the outside face. On the top of the wall, the earth pressure is transmitted and resisted by the floor construction. At the bottom, reliable support for the slab must be provided. When the basement slab is at least 6 in. thick and extends over the full width of the building, the base- ment wall may be considered as supported at the bottom by the basement slab. Ordinarily, it is not advisable to rely on the base- ment slab alone, because very often the earth behind the wall is filled in and earth pressure is exerted against it before the basement slab is poured. To avoid trouble due to this, the wall should be carried down into solid ground below the basement slab for a distance of at least 9 in. The passive earth pressure then resists the reaction of the wall. When the ground below the basement slab is not firm, it is not possible to rely on passive resistance of the ground for the horizontal support of the wall at the bottom. Unless the floor slab is designed to take the pressure from the wall there should be a beam extending from column to column and reinforced to transmit the load from the wall to the columns. Some designers provide a projection on the wall, on which the future basement slab will rest. If the slab is not reinforced and it rests on yield- ing ground, the projection is objectionable, because the slab after settling is likely to crack along the edges where it is held by the wall. 640 BASEMENT WALLS Bending Moments Due to Earth Pressure.—When the slab is supported on top and bottom, the bending moment and reaction may | be found from the following formulas: Triangular Variation of Pressure-—When the earth pressure varies as a triangle, Let w = maximum unit earth pressure, lb. per sq. ft.; h = height of wall, ft.; R, = reaction at the top; Rz = reaction at the bottom; M = maximum bending moment; x1 = distance of point of maximum bending moment from top. Then 1 wh Ry — 2 Lae . . . ° ° . ° e ° e . (1) 2 wh. Roan La oss hg M = 0.064wh? ft.-lb. . 2 on = 0.7 7wh? inlb» 4.) soe ee Maximum moment acts at a distance below the top equal to A = ah = 0. 58h. f “ A : A 2 c Cs 5 Trapezoid Variation.—When the earth pressure varies as a trape- zoid, in addition to the above notation, Let w, = unit earth pressure on top, lb. per sq. ft.; We = unit earth pressure on bottom, lb. per sq. ft. R, 2, 6 Ry = Mit Aes, _ , ORS be Seine a Minax = Se ft.-lb. (approx.). . . . (8) = 0.75 (w; + we2)h? in.-lb. (approx.) . . (9) The formulas for reactions are exact. The formula for bending moment is approximate and is based on the assumption that the total load is uniformly distributed. The formula for exact bending moment is complicated. WALL SLAB WITHOUT SURCHARGE 641 The distance between the point of maximum bending moment and the top may be obtained from the following formula: 1 pas oe 1+V2(2+o(¢—1)+1]h,. . (10) W2 where c = —. Wi When this value is found, actual maximum bending moment may be computed by ordinary rules of statics. Ordinarily, the approxi- mate Formula (8), p. 640, gives accurate enough results. Example 1.—Design basement wall for a building in which wall columns 3 ft. wide are spaced 20 ft. on centers. The distance from top of basement slab to top of first floor is 10 ft. and the top of the ground is 4 in. below the first floor. On one side of the building runs a railroad track producing an earth pressure equivalent to a surcharge of 6 ft. Solutton.—The wall, as shown in Fig. 205, p. 642, is supported at top and bottom. The theoretical support of the wall, at the top, may be assumed to be in the center of the slab, and at the bottom, 6 in. below the basement slab. The theoretical height of wall is h = 10 ft. 8 in. Wall Slab without Surcharge.—Taking the unit pressure at 30 lb. per sq. ft., the maximum pressure at the bottom is w = 30 X 10.67 = 320 Ib. per sq. ft. The pressure is represented by a triangle, for which the maximum bending moment, from formula M = 0.77 wh? (see p. 640). ‘M =0.77 X 320 X 10.66? = 28 000 in.-lb. per lin. ft. Assume 12 in. for thickness of wall; then d = 10.75 in. and the required M A; =---- amount of steel, from As jaf, 28 000 : As = (See = 0.19 sq. in. per lin. ft. 0.875 X 10.75 XK 16 000 1_in. round bars, 12 in. on centers, give an area of steel equal to 0.196 sq. in. per lin. ft. 0.19 ; : : The ratio of steel to concrete is p = ORT 0.0015. Since this ratio is smaller than 0.077, the compression stress in the concrete is below the allow- able value, and the thickness of wall is sufficient. : 10.66 The pressure transmitted to the ground at the bottom is 3 X 320 X re: 1140 Ib. per lin. ft. Since the wall is imbedded 12 in. in the ground below the basement floor, the horizontal unit pressure on the ground is 1140 lb. per sq. ft. This is satisfactory. 642 BASEMENT WALLS | I i] Noe el | SS | | Sia | ' Sic 27801 | | lo per sq. ft. ‘Y | eat pro 18 BES : a >| 8" —ec (ew ee we -—— — wo 10 Ft. 8*- - -e===- 10 Ft. 0 s{--7—-~--~-Span Length BOD Dot ae er With Surcharge oe wW = 320'1b.| per sq. ft. Without Surcharge fac: ------- ©---- 20 Ft, 0" ----------- SB age = With Surcharge Fra. 205.—Details of Basement Wall Supported on Top and Bottom. (See p. 641.) WALL SUPPORTED AT THE COLUMNS 643 Wall Slab on the Railroad Side.—The slab is subjected to a surcharge pres- sure of 6 ft. The earth pressures are represented by a trapezoid shown in Fig. 205. At the top, wi = 180 lb. per sq. ft., and at the bottom, w. = 500 Ib. per sq. ft. The bending moment due to a loading in the form of a trapezoid may be found by Formula (9), p. 640. M = 0.75(wi + w:)h? in.-lb. It is: M = 0.75(180 + 500)10.672 = 58 000 in.-lb. 58 000 Pepereeinens Bs TN bs A 2 San 8a in, 0.875 X 10.75 X 16.000 she 3-in. round bars, 94 in. on centers, give an area of 0.388 sq. in. 0.385 Mar : The ratio of steel, p = 12 < 10.75 = 0.003, indicates that the stresses in concrete are well within the allowable working limits. Wall Supported at the Columns.—When the basement wall is provided with wide window openings below the first floor level, or when bulkheads are provided in the slab, the wall cannot be con- sidered as supported at top and bottom, and it may be economical to span the wall between the columns. The bending moments produced by the earth pressure will depend upon whether the wall is continuous over several spans or is only one span long. The main reinforcement is horizontal, and is placed near the inside face of the wall. When the wall is continuous, negative rein- forcement at the column near the outside face should be provided. Some vertical temperature steel is commonly used. This may consist of 2-in. round bars spaced 18 in. on centers. Sometimes, additional bars of larger diameter are used to support the horizontal bars during construction. Bending Moment in Wall Slab.—If the ground is level, the earth pressure for any horizontal strip of wall is constant. The bending moments may be computed in the same way as for floor slabs. Since the intensity of earth pressure varies with the depth, the bending moments will be greatest for the lowest strip and will decrease toward the top. The thickness of the wall is usually con- stant for the whole height; therefore, with the decrease of bending moment, the amount of steel may be decreased. Providing for Negative Bending Moment.—The negative bending moment in the wall slab may be provided for by bending one-half of the horizontal bars and extending them to the adjacent span, as in continuous slabs. It is often difficult to handle long bent bars, 644 BASEMENT WALLS especially on account of the interference of column steel, behind which the ends of the bars would have to be placed. The construc- tion is simplified by keeping the positive reinforcement straight and by using, for negative reinforcement, short horizontal bars at the column near the outside face of the wall. End Spans of Wall.—In end spans, the amount of steel should be increased by 20 per cent. Bending Moment in Column.—The earth pressure transferred from the slab to the column produces bending in the columns. The earth pressure resisted by the column may be represented by a triangle or a trapezoid, and its magnitude equals the pressures on the wall multiplied by the span. In high buildings with heavy basement columns, the stresses in the column due to earth pressure are small, and do not have to be specially provided for. For heavy earth pres- sure and for light columns, the earth pressures may increase the stresses considerably. An inexperienced designer should compute the stresses in-columns due to earth pressure in all cases, until he develops sufficient judgment. The tensile stresses due to earth pressure in the center part of the column act at its inside face. When the column extends above the first floor, negative moments are developed at the junction of the column and the floor. The tensile stresses produced by this bending moment act at the outside face of the column. They should be added to the tensile stresses produced in the wall column by the bending moment transferred to the column from the floor construc- tion. Earth Pressure.—The variation of earth pressure along the ver- tical section of the wall is shown in Fig. 646. For a wall without surcharge, it varies as a triangle; and for a wall with surcharge, as a trapezoid. Along a longitudinal section of the wall, the pressure is constant. The wall is then considered as a slab loaded with uni- formly distributed load along its span. The magnitude of this uniform load increases with the depth. The load and the bending moments will be a maximum at the bottom and will decrease to a minimum at the top, in the same ratio as the pressure on a vertical section. Since the thickness of the wall is the same throughout, the amount of steel required to resist the earth pressure will vary accord- ing to a triangle or trapezoid. After the area of steel at the top and bottom has been computed, the amount of steel at intermediate points may be obtained by interpolation, or graphically as in Fig. EXAMPLE OF BASEMENT WALL 645 206. The varying amount of steel is provided by varying the spacing of the bars. When the spacing becomes too large, bars of smaller diameter should be used. Example of Basement Wall Supported by Column. Example 2.—Design basement wall, with columns 3 ft. wide and 2 ft. thick, spaced 20 ft. on centers. The distance from top of basement slab to top of first floor is 10 ft. Ground level is 4 in. below the level of the first floor. There is a continuous opening in the first floor slab, along the wall, for a bulkhead. On one side is a railroad track producing a pressure equivalent to a surcharge of 6 ft. (Compare the solution of this problem with that on p. 641.) Solution.—The wall cannot be supported at the top on the floor slab, because of the opening for the bulkhead. It will be considered as spanning between the columns. Span.—The distance, center to center, of columns is 20 ft. Since the column is 3 ft. wide, the net span is 17 ft. The design span, taken as the net span, equals 1; = 17ft. (See p. 277.) Wall without Surcharge——The maximum unit pressure at the bottom is w = 305 Ib. per sq. ft., as is evident from Fig. 206. The bending moment for continuous wall with 17-ft. span and 305 Ib. load, from formula M = wi? in.-lb. is, M = 305 X 17? = 88 100 in.-Ib. per lin. ft. of height. For a 12-in. wall, the effective depth is d = 10.5 in. The amount of steel: from formula A, = ——, is jdf g 53 88 100 ~ 10.5 X 0.875 X 16 000 = 0.60 sq. in. per foot. As -in. round bars, 6 in. on centers, will satisfy the requirement. The spacing of the bars for other sections of the wall is shown in Fig. 206. It may be noted that near the top of the wall, where the area of steel required by bending moment is small, a larger amount of steel is used to provide for tempera- ture and shrinkage. Walls with Surcharge.—The maximum pressure on the top is 180 lb. per sq. ft.; on the bottom, 485 Ib. per sq. ft. (See Fig. 206, p. 646.) The bending moments and the required amounts of steel will be computed at the top and bottom. These will be plotted and the amounts of steel at intermediate point determined. Top. Pressure 180 lb per sq ft. M = 180 X 17? = 52.000 in -lb. per lin ft. of height. A, = 0.354 sq. in. per ft. Use $-in. round bars, 12 in. on centers. LAs 20.60 Sq. In. at Depth x oe ET. cee ee ee ee eS ‘ z x Ss $ cite oa S io? ~ x : pee i eae K ' 19 if lor rs bs i: > 7 a 2 “be Ss zi < we tous a I 5 pal " | I 1) >| 1 ~ | Ss | ! \ S { | = \ | = 1 = f = eS 4 S 1 al = > aS ‘ ~~ = w N With Surcharge a 485 1b. per|sq. ft: Areas of Steel Section ToOOCUML PREECE SE ia peed a= =~ 20 Ft, 0" - = Fra. 206.—Detail of Basement Wall Supported at the Columns. (See p. 645.) BASEMENT WALLS FOR AREAWAYS 647 Bottom. Pressure 485 lb. per sq. ft. M = 485 X 17? = 140000 in.-lb. per lin. ft. of height. A, = 0.95 sq. in. per ft. Use 3-in. round bars, 4 in. on centers 0.95 ee OTS. PMI GR Se 12 Since the per cent of steel is smaller than the value corresponding to a compres- sion stress in concrete f, = 650 lb. per sq. in., the thickness of wall is satisfactory for compression. The required area of steel and the arrangement at intermediate points may be seen in Fig. 206, p. 646. Basement Walls for Areaways.—The design of walls for areaways depends upon conditions. Walls for small areaways, one panel wide, may be considered as supported on the cross walls, and reinforced with horizontal bars placed near the inside face. If the areaway extends over several spans, it is often possible to run struts at the top of the wall to the columns. The wall can then be designed as supported on top and bottom. Enough longitudinal steel must be used on the top to transmit the pressure to the struts. Such construction is shown in Fig. 265, p. 754. If struts are not permissible and it is not possible to support the top of the wall, it must be designed as a cantilever wall. The base must be made strong enough and rigidly connected with the wall. The principle of the design is then similar to that explained for retain- ing walls. _ The base may be placed either outside the wall or between the wall and the building. CHAPTER XIII ROOF CONSTRUCTION Loading.—Roofs should be designed for the dead load, the live load, and, in case of inclined roofs, for the wind pressure. The Dead Load consists of the weight of the roof construction, of the cinder fill or tile placed on the top of the roof, and of the roof covering. The unit weights of the various materials used in the construction of the roof are given in the table below: Dead Loads on Roof | Weight, Description Lb. per Sq. Ft Roof Covering: Five-ply felt and gravel. 2.....:... 2.50 +: >: 6 Four-ply felt, and grayel............9->/24 +) se eee ee 54 Tne «bcc bie cc din wb bus kin ove pom Web SOO ie cc ett ce ann een if Metal covering. oc sca wt. ce ee bee = nti oe 5 anne 13 to 2 Slate 8, to 2-in. thick. . 220.2 s 0 eb eet es 7.25 to 9.5 Insulating Materials: Cinders, per in. ..... 5. si ees se et eee oe oe #f Average weight of cinder fill............--++++seneeeeveseees | 70 Cork or lith insulator... 205 S200 2. 0 0 10 Tiles... cs «cc wu pins c unumrn :loppilcus Sgehensleets le (eto. ch Sains iain 12 to 25 Plastering on ceiling on concrete or tile.......----++++.++2255: 5 Suspended ceiling without ornamentation. .........+++-+++-++5 10 The Live Load for which roofs are ordinarily designed is intended to take care of the weight of snow, and also of the weight of men walking upon the roof. Most building codes require a live load of AO Ib. per sq. ft. for flat roofs or roofs with a pitch of less than 20 degrees with the horizontal. For roofs with a larger pitch than 648 DRAINAGE 649 20 degrees, the live load may be reduced to 30 lb. per sq. ft. In localities having severe winters, such as the northern part of the United States and Canada, the live load may have to be increased to 50 lb. per sq. ft. to take care of the additional weight of snow and ice. Wind Pressure.—Pitched roofs, when the pitch exceeds 20 degrees with the horizontal, should be designed for wind pressure, which should be assumed to act horizontally, with an intensity of 30 lb. per sq. ft. of vertical projection of the roof surface. In determining — stresses in the roof members, this horizontal pressure should be resolved into pressures normal and tangent to the roof. It is usually considered that, with a wind pressure acting on one side of the roof, the live load is applied only on the lee side, as it is scarcely possible that with wind force of 30 lb. per sq. ft. any snow would remain on the side exposed to the wind. Drainage.—A flat roof should be provided with a sufficient number of outlets to carry off the rain water. These should be properly distributed over the roof. To facilitate the running off of | water, it has been customary to pitch the tops of the roofs. Lately, the necessity of pitching the top is questioned, with much justifica- tion, as with properly distributed outlets, most of the water will run off without any pitch. Some water may remain on the roof until it dries out, but this does no harm to the roofing. The only advantage of pitching the roof, therefore, seems to be that in case of cracks in the roofing the leakage through the slab would be smaller with pitched roof than with flat roof. With reliable roofing, the chances of leakage are remote. Therefore, the authors recommend dead level roofs when high-grade roofing is used. Pitch of Roof——The customary pitch for gravel roofs is 2 in. per lineal foot. For smooth tile, this pitch may be reduced to # in. per foot. Pitching of the roof is accomplished either by pitching the whole roof construction or by building the roof slab horizontally and providing a sloped fill on the top of the roof slab. The method to be adopted will depend, first, upon architectural requirements (i.e., whether there is any objection from the architectural stand- point to an inclined ceiling), and second, upon the method of pre- vention of condensation. ‘Thus, if cinder fill is used on the top of the roof to prevent condensation, this could be utilized to give the required pitch. This also applies where light construction is placed above the roof to provide air space, 650 ROOF CONSTRUCTION _ The simplest method of pitching a roof is to raise one edge of the building and make one continuous pitch across its whole width. This method is particularly advantageous where the roof slab is pitched, as the building of the formwork is simpler than for more elaborate pitching. It may be used only in narrow buildings; otherwise, the difference in story height on two opposite sides would be large. For wider buildings, a ridge or a valley may be provided along the center of the building. With large roof areas, considerable thought must be given to the method of pitching. First, the number of outlets should be computed. These should then be placed in the most suitable positions, and the pitch in the roofs provided accord- ingly. Two or-three trials may be necessary to give the simplest » solution of the problem. It should be remembered that the down- spouts are best placed near columns, thereby restricting the number of possible positions for the outlets. Figure 207, p. 651, shows the location of the outlets and the method of pitching adopted in building. Roof Insulation.—Concrete roofs often require insulation to prevent condensation and loss of heat through radiation, in cold weather. Insulation of roofs is required in all buildings used for habitation, office buildings, hotels, hospitals, and schools. In ware- houses, and in some cases in factories, solid concrete roofs covered with tar and gravel roofing require no additional insulation. By condensation is meant the accumulation on the ceiling of moisture from the air, caused by a difference between the tempera- ture of the concrete roof and that of the air in the building. To each inside temperature and degree of humidity there corresponds a definite temperature of the roof at which condensation will occur. This temperature is called the dew point. The object of insulation is to prevent the outside temperature from reducing the tempera- ture of the inside surface of the roof to the dew point at the time when the conditions for condensation are most favorable, that is, when the difference between the inside and outside temperatures is greatest, and the humidity, or moisture content of the air, is also greatest. The condensation depends upon the following factors: (1) Greatest difference between outside and inside temperatures. Condensation increases with increase in difference. (2) Expected moisture content or humidity at the time of greatest difference in temperature. Condensation increases with increase of humidity. 651 ROOF INSULATION YIO/G D770) B1say gUIDIy /2az¢ asnoy ua fo uoljIa¢ (ogg ‘d aag’) “suIp[Ing ‘oD einqruimy oureg ur[g jooy—70Z “D1 Ilf 4apuly buimoys uoljoa¢ Mid 4apurg... f yA pasaagg saddoj sagnig xnvUuay)., = =—— ——— tL — ff a =f— HOM AUV|d-x 4/0 b0/4 No] it < SSS J00G JD Buspun7 yuDjd- gs e AOZD/IZUAA 49] Q we x0g OoY s waddo7 054! f0 [10J9q bs anaao a ssplg = Jauuny Fe’ 1/09 1addo7 ee 4add0),, Gi 4 _--- aga12U0] 49PUL) Gag fooy hey ysay £x2) UaaIIE As 4add07,9/— ea CV ee Rsluny fooy 1addg) 20 9/ 4Djj09 saddog-"*-uaai2g easy Ja 67 f| 652 ROOF CONSTRUCTION (3) Degree of conductivity of heat of the roof construction. For instance, for an 8-in. slab, the condensation will be smaller than for 9-in. topping in concrete joist construction. Humidity may be atmospheric or it may be produced by some process of manufacture. Atmospheric humidity can usually be taken care of without difficulty by proper insulation. Special humidity may require special ventilation in addition to insulation. In some cases, complete prevention of condensation is impossible. Methods of Roof Insulation.—There are several methods of roof insulation, and the choice between them will depend upon relative economy. | The methods are: (1) Providing dead air space between the ceiling and the roofing; (2) covering the roof slab with cinder fill; (3) covering the roof with insulators like fibrofelt, lith, or cork; (4) placing on top of the roof hollow tile either of clay or gypsum, (5) combination of tile with cinder fill or with fibrofelt. (The order in which these methods are arranged has no significance.) In determining the relative economy of various types of insula- tors, the dead load of the insulator also should be considered. For instance, the dead load of cinder fill amounts to 70 lb. per sq. ft., while the dead load of cork insulator or lith amounts to about 10 lb. The additional dead load requires stronger roof construction, also heavier columns. ‘The additional cost of construction must be added to the cost of the heavier insulator. Method of Providing Air Space.—The best method of providing dead air space between the roof and the ceiling is by a suspended ceiling, such as that described on p. 613. Air space of any dimen- sions may be thus obtained. A much less effective method, used in beam and girder and joist constructions, is by providing a plastered ceiling directly under the beams and joists, as described on p. 610. The airspace thus pro- vided is limited by the height of the beam or the joist. In addition, the joists are not insulated. A cheaper method of providing air space above the ceiling, used to a great extent in Canada, consists of a light wood framework built some distance above and supported by the concrete slab. Usually, this consists of 2 by 4 in. planks placed and anchored to the concrete slab, and vertical wooden studs resting on the planks and spaced about 4 ft. on centers. Stringers supported by the studs carry rafters upon which rests tongue and grooved planking covered with appro- INSULATORS 653 priate roofing material. Proper pitch can be easily obtained. An objection to this method is that the roof is not strictly fireproof. If the roofing is fire-resistant this is not a serious objection. A small air space may be obtained by placing wood strips of proper thickness, spaced from 12 to 18 in. apart, directly on the con- crete. On these rest the boards, and on top the roofing. Cinder Fill.—Cinder fill placed on the top of the roof slab is the most commonly used method of preventing condensation. The concrete slab is built level, and the desired pitch in the roof is pro- vided by varying the thickness of the cinder fill. To be effective as insulator, the fill in the valleys should be at least 6 in. thick. The slopes should be so arranged that the thickness of the fill at high points is not excessive. It is desirable to limit the average thickness of the fill to 12 in., in which case its average weight will be about 70 Ib. per sq. ft. The cinder fill is made of wet hard cinders properly tamped and covered with a mortar coat one inch thick. The roofing is placed on the top of this coat. To provide for expansion of the cinder fill, a one-inch joint, filled with asphalt or other bituminous material, is usually provided at the parapet. This is particularly important with brick parapets, because, if the joint were-not provided, the thrust due to expansion of the fill might injure the parapet, or, on the other hand, contraction cracks might allow penetration of water. Insulators.—Roofs may be insulated by placing, on the top of the concrete slab, insulators such as fibrofelt, lith, or cork board, and covering them with roofing. The fibrofelt comes in various thick- nesses up to 13 in. Lith and cork board come in sheets. Lith may be obtained in thicknesses up to 4 in. Both materials can be sawed like wood, to fit the roof. A coat of pitch is placed first, and the insulator is placed on the top and covered with another coat of pitch, upon which the roofing is applied in the regular manner. These materials show considerable resistance to loss of heat. Tile Insulators.—Lither clay or gypsum hollow tile may be used for roof insulation. When erected, the tiles are placed end to end so as to form a continuous air space. Gypsum tile offers larger resistance to loss of heat than clay tile, but under some conditions it deteriorates and loses its insulating properties, particularly when it is exposed to heat or when damp air can penetrate to it. When tiles are used as insulators by themselves, the roofing is 654 ROOF CONSTRUCTION placed directly on top of the tiles. Sometimes they are used in con- nection with cinder fill, fibrofelt, or cork. In such cases, the tiles are placed on the concrete and the additional insulator is placed on the top. Roof Covering.—Concrete roof slab may be made impervious to water without any special covering, by the use of properly propor- tioned concrete reinforced with sufficient temperature reinforcement and troweled to a dense and hard surface. Since this is not a positive insurance against leakage, in building construction the roofs are usually covered with waterproof roof covering. Concrete roofs are _...Flashing Form LQos Fic. 208.—Method of Application of Tar and Gravel Roof. (See p. 655.) used in reservoirs and structures of similar character where dampness of the under surface and occasional leakage are not objectionable. The roof covering materials may be divided into three classes: namely, roof coverings relying for waterproofing on coal tar or asphalt; roof coverings consisting of slate or tile; and metallic roof coverings, such as tin, copper, and corrugated iron. Tar and Gravel Roofing.—For substantially flat roofs, the tar and eravel roofing is used very extensively, as it gives a lasting roof surface at comparatively low cost. For roofs with large pitch, tar and gravel roofing cannot be used, as there is danger of the roofing running during hot weather. The maximum pitch for which this ROOF COVERING 655 roofing may be used depends upon the climate. In hot climates, it should not exceed one inch to the foot, and in cold climates three inches to the foot. The tar and gravel roofing consists of three materials, namely, pitch, felt, and gravel or slag. The pitch provides the waterproofing qualities; the felt serves to keep the pitch in place; while the gravel, which is imbedded in pitch on the top of the roof covering, serves as a protection for the roofing both against injuries by physical action, like walking and scraping, and also against the action of the direct rays of the sun. For satisfactory results, the materials and the workmanship must be of first-class quality. The method of application is as fol- lows: The pitch is spread ar first on the concrete; then | Fu Plies of Felt Ritchie. the required number of '(7¥ a ae, layers or plies of tarred \ CO aa! & al Af ee LA AO felt are placed. Thecearc —>E————————————— lapped in such a fashion that at each point of the roof there are the same number of layers. The : tarred felt must be covered Four heavy moppings of Pitch” with pitch, so that in no Fig. 209.—Section through Tar and Gravel place felt touches felt. The . Roofing. (See p. 655.) felt must be spread without wrinkles, as these destroy the waterproofing qualities and contribute very largely toward breaking of the felt. This method is illustrated in Fig. 208, p. 654. The relative thicknesses are evident from a large scale section through the roofing shown in Fig. 209, p. 655. The following specifications, universally known as “Barrett’s Specifications,” are recommended for use. First—Coat the concrete uniformly with Specification Pitch. Second—Over the entire surface, lay four plies of Specification Tarred Felt, lapping each sheet twenty-four and one-half inches over preceding one, mopping with Specification Pitch the full twenty- four and one-half inches on each sheet, so that in no place shall Felt touch Felt. Third—Over the entire surface, pour from a dipper a uniform coating of Specification Pitch, into which, while hot, imbed not less than four hundred pounds of- Gravel or three hundred pounds of 656 ROOF CONSTRUCTION Slag for each one hundred square feet. The Gravel or Slag shall be from one-quarter to five-eighths inch in size, dry, and free from dirt. General—The Felt shall be laid without wrinkles or buckles. Not less than two hundred pounds of Pitch shall be used for con- structing each one hundred square feet of completed roof, and the Pitch shall not be heated above 400 degrees Fahrenheit. Slate Shingle and Clay Tile.-—These materials are used on roofs with a pitch of more than 6 in. to the foot. Owing to the steepness of the roof, bituminous roofing cannot be used, as it would run. | The shingles and tiles are nailed to 1 by 2 in. strips, imbedded in concrete under each row of shingles. This covering 1s made water- proof by proper lapping of the tile at all joints. | Metallic Roofing.—This is usually expensive and is seldom used in connection with concrete roofs. It has been found by experience that the metal lasts much longer if it is placed on planks instead of ‘directly on concrete. | Parapets.—Buildings with flat roofs are usually provided with parapets extending above the roof. The purpose of the parapet is to give the building proper architectural proportions, to hide any drainage slopes, and finally, in non-fireproof buildings, to retard the spread of fire from the adjoining buildings. This last purpose need not be considered in reinforced concrete buildings. The height of the parapet will depend upon the architectural design. It may be uniform for the -whole building, or parts of it may be higher than the rest. The minimum height of parapets in business buildings is 3 ft. for outside walls and 18 in. for party walls. The parapet is usually built of reinforced concrete, brick, or tile faced with brick. ‘The selection of the material depends upon the type of the building and also upon economy. Concrete Parapets—Reinforced concrete parapets may have exposed concrete surfaces, or the surfaces may be faced with brick, stone, terra cotta, or any other suitable material. When exposed, the concrete surface may be plain or ornamental; but the ornamen- tation, if any, should be such as to require no elaborate formwork. ‘Concrete cornices are often used. They may either be pre-cast. or cast in place. When the shape of the cornice is simple, it may be built as a part of the parapet. For larger cornices with more intri- cate design, it is more economical to build the projecting parts sepa- rately from the supporting concrete parts. Proper provision, usually consisting of proper recesses ‘and steel anchors, should be PARAPETS 657 made to support the separately built parts. When the surface of the concrete is exposed, sufficient temperature reinforcement should be provided to prevent shrinkage and temperature cracks. When the parapet is faced with brick, an angle of proper dimen- sions should be used at the bottom to retain the face brick. When terra cotta is used, proper anchors should be provided for attaching terra cotta blocks. From a structural standpoint, the parapet may form a part of the supporting structure, i-e., it may form a beam supporting the roof load, in which case it should be designed to carry the loads coming upon it, according to the methods recommended for other concrete beams. In case the roof beam consists of a portion below the roof and a portion above the roof, it is best, from a structural standpoint, to have both parts built at one pouring. The design of the beam is then similar to that of any other beam. If, however, it is found economical to build the part above the roof separately, © special provision should be made to insure the cooperation of the two parts. This is usually accomplished by providing stirrups throughout the whole length of the beam, extending from the bottom portion to the top portion. With stirrups spaced closely enough, it is possible to obtain a beam that has the same strength as if it were poured at one time. In case of heavy loads, however, it may be advisable to increase the required amount of reinforcement, to provide for the possibility of incomplete cooperation of the two parts. : When a concrete parapet supports a heavy cornice projecting some distance outside of the building, it is subjected to bending moments produced by the weight of the projecting cornice. This bending moment equals the weight of the cornice multiplied by the distance between the center of gravity of the cornice and the center of the parapet. In some cases, it is possible to resist this bending moment by vertical reinforcement in the parapet, placed near its inside face. The amount of such reinforcement equals the bending moment divided by effective thickness of parapet. The maximum stress in this reinforcement acts at the junction of the parapet and the roof construction, and the reinforcement must be properly anchored in the roof construction. The roof members in which the bars are anchored must be strong enough to resist the bending moment transferred from the parapet. Sometimes, however, the parapet wall is not sufficiently strong and special provision, such as 658 ROOF CONSTRUCTION special brackets or extension of the columns above the roof, must be made to carry the cornice. When the parapet does not serve any structural purposes, i.e., when the beam below is sufficient to carry the roof load, it may be built separately. It then requires only temperature reinforcement and dowels joining the parapet to the roof. The thickness of the parapet should be at least 8 in. For very high parapets, this thick- ness may have to be increased to 12 in. At the parapet a cant, consisting of cinder concrete, should be built to prevent accumulation of water at the wall. A perspective drawing of a concrete parapet is shown in Fig. 210, p. 658. Better results are obtained by using special continuous reglets made of steel. These are tacked lightly to the inside of the forms for the parapet. Such Fic. 210.—Concrete Parapet. (See p. 658.) reglet is shown in Fig. 210, p. 658. Proper provision should be made, on the inside face of the para- pet, for attaching the flashing. For this purpose, there must be a longitudinal groove along the parapet, which may be easily obtained by nailing to the form a strip of wood of proper dimensions. Coping.—The coping for a reinforced concrete parapet may be built at the same time as the rest of the parapet, or it may be built separately. In the last case, special reinforcement consisting of three 3-in. bars should be used in coping. Brick Parapet.—A brick parapet should be at least 12 in. thick. It should be built of hard brick, laid in Portland cement mortar. If a cornice of any size is used in connection with a brick parapet, it should be anchored to the concrete frame rather than to the brick parapet. ‘To prevent any possibility of displacement of the parapet, either proper recess should be provided in the concrete or else steel dowels should be extended up from the concrete into the brick. To provide a slot in the brick for fastening the flashing in the connection, a special tile built with a groove of proper dimensions may be used to advantage. ROOF DESIGN 659 The parapet wall should be provided with a coping of stone, terra cotta, concrete, or cast iron. The inside face of the parapet should be coated with tar, so as to prevent moisture and frost from penetrating the masonry. Walks on Roof.—Where considerable walking is done on the reof, special walks should be provided to protect the roofing. A simple construction consists of 2 by 3 in. sleepers placed directly on the roof across the walk, with 1 by 2 in. slats, spaced 1 in. apart, laid over the sleepers. ‘Where the roof is used for a promenade, it should be surfaced with lasting material, such as vitrified tile. First the roofing is placed as usual; then cement mortar is placed on top, and the tile is imbedded in the mortar. Expansion joints should be provided in the mortar, to prevent injury to roofing. Flashing.—F lashing is provided to make the junction of the parapet and the roof waterproof and to prevent water from entering under the roofing at the parapet. There are several varieties of flashing in use. | Metal Flashing.—This may be of tin, galvanized iron, copper, or lead. It is usually nailed at the bottom to a nailing strip imbedded in the slab a proper distance from the parapet usually from 6 to 10 in. The vertical portion of the flashing is placed in a specially prepared groove or reglet in the parapet, the groove being filled with cement mortar after the flashing is in place. To prevent the mortar from coming out, the groove is tapered. The method of procedure is as follows: The felt is applied first, extending to the parapet. Then the metal is nailed on the top. Felt is then placed over the horizontal part of the flashing; and finally a coat of pitch and gravel is applied. Elastic Flashing.—Figure 211 shows the flashing advocated by the Barrett Company. No metal is used. The felt and pitch are brought to the edge of the parapet. ROOF DESIGN In buildings consisting of several stories and roof, all columns usually extend to the roof, so that the size of the panels in the roof is the same as the size of the panels in the floor. The difference between the floor and the roof, from a structural standpoint, is only in the magnitude of the loading. It is most economical to use for 660 ROOF CONSTRUCTION the roof a design of the same type as for the floor, so as to be able to use the formwork from the floors below. ‘Thus, if the floors are of flat slab design, the roof also will be a flat slab. If the floors consist of beams and girders, the roof will be of beam and girder design. The loading for which the roof 1s designed is usually smaller than the floor load; therefore, beams of smaller dimensions may be used in the roof. Before deciding, however, upon a change in the size of the concrete beams, the cost of the changes in formwork must be seriously considered. In many Cases, the saving in concrete - materials, obtained by using smaller concrete dimensions in the roof, ‘s more than offset by additional cost of the changes in formwork. A change in the amount of steel, instead of a change in the size of concrete members, will often accomplish the desired result more economically. If it is desired to change the dimension of beams, the depth of the beam, rather than its width, should be reduced. A reduction in width of the beam requires not only changes in the beam formwork, but also changes in the slab formwork, to provide for the greater distance in clear span between beams. The thickness of the slab can be reduced without affecting the formwork. It should be noticed that this automatically reduces the depth of the beam. Note that it is not the total depth of the beam, but the depth below the slab, which affects the formwork. In flat slab construction, a change in thickness of slab does not affect the formwork. Form- work needs to be changed only when the depth of the drop panel is reduced, and this change is very inexpensive. Sometimes it is desired to get larger spans in the top floor than in the lower floors, by omitting one or more rows of columns. In such cases, the entire design of the roof is changed. In beam and girder construction, the cross beams will be of a much longer span, requiring special concrete dimensions and, therefore, special form- work. The same spacing of the longitudinal beams, however, should be used as in the lower floors, so as to utilize at least a part of the formwork. When the lower floors are of flat slab design, the design of the roof must be altogether changed, and such arrangement of - beams used as will give greatest economy. Openings in Roof.—lIn buildings intended for manufacturing purposes, it is often desired to get additional light and ventilation in the top story by the use of skylights or monitors. With flat slab construction, the openings for skylights should be arranged so as to require the least amount of framing. If the sky- LONG SPAN ROOF CONSTRUCTION 661 light is placed in the central portion of the panel, it may be possible to get along without any framing. When framing is required, it consists of two beams at the edge of the skylight, supported by two girders running between columns at right angles to the beams. To make the connection of the skylight with the roof weather- proof, it is advisable to place the skylight some distance above the highest point of the roof at the skylight. This is accomplished by providing, around the opening above the roof slab, a concrete curb 6 in. thick and of proper height. The curb is usually built after.the slab is completed, and its height depends upon the height. of the cinder fill at the skylight. To join the roof and the curb, dowels are provided, extending from the roof into the curb, and a water-tight joint must be made by clearing the old surface and spreading upon it neat cement parts. Flashing is extended from the roof to the top of the curb, where it is nailed to wooden nailing blocks provided for this purpose. (See Figs. 211 and 212.) The nailing blocks should be imbedded in concrete before it is set. Metal apron set in place after Slashing has been nailed to top of curb Flashing run up over top of > curb and nailed to continuous _ Be _--Slate Surface Roofing creosoted wood nailing strip ~~ ~~ Three layers of roofing felt with alternate layers of Elastigum, ee run up the curb and out on we roofing felt at least 6" All plies of the Roofing felt are cut off at the angle of ~~~... curb and roof deck os “~--> Roofing... nr, y FE aR aS eg OSs gts aN Tn COs eye intel Daae piltiia May ES: ae RRNA Rane ante = ocr iS ie 3.9. a wdieita Fig. 211.—Flashing at Skylight, (See p. 661.) LONG SPAN ROOF CONSTRUCTION In long span roof construction, concrete is seldom as cheap in first cost as light steel truss construction, but it has the advantage of being fireproof and of requiring practically no expense for main- tenance. ‘The appearance is also in favor of concrete construction. The maintenance chrages for steel trusses are fairly high, as they need to be regularly repainted. Where steel trusses need to be fire- proof, concrete construction is more economical than steel con- struction. Several types of long span roof construction are in general use. These are simple girders, concrete trusses, rigid frames, and long 662 ROOF CONSTRUCTION span arches. Of interest also is the long span roof construction developed by Richard E. Schmidt. (See p. 671.) Simple Girder.—Simple girders have been used to quite an extent for long spans in garages and manufacturing establishments. They are usually supported by light concrete columns or by brick piers. The light concrete columns are seldom capable of offering appreciable restraint to the girders. Thero of construction may consist of girders 18 to 20 ft. apart supporting transverse beams spaced about 8 ft. and carrying a 2-in. concrete slab. Light weight concrete construction, as de- scribed on p. 602, may be substituted for the transverse beams and slab. To save formwork for intermediate beam the spacing of girders may be made smaller and slab spanned between them as If open joints occur between angle sills they should be sealed with Elastigum and capped with metal Angle bolted in place after flashing has been nailed at top of curb / Slate Surfaced Roofing Three layers of roofing felt with alternate layers of Elastigum, run 7 up the curb and out on roofing felt at least 6” Flashing nailed to creosoted-=__ wood nailing strip ae All plies of the roofing felt are cut off at the angle of-------- curb and roofing deck Fic. 212.—Flashing at Monitor. (See p. 661.) shown in Fig. 213, p. 663. The girders are usually of T-shape design. For long span girders to resist compression stresses, the required thickness of the flange may be larger than the 3-in. slab used in the roof, in which case the slab is thickened on both sides of the girder sufficiently to furnish the required compression area. To reduce the cost of the girder and particularly its dead load, it may be economical to make the thickness of the stem in the middle portion of the girder, where the external shear is small, only — large enough to accommodate the longitudinal reinforcement, and to increase it gradually towards the supports with the increase of the external shear. Before adapting such a design, it is necessary to investigate the cost of the concrete saved and the extra cost of the formwork. Since the dead load of the girder is reduced, there will be an additional saving in steel, due to the reduced bending moment, and also a saving in the size of supporting members. 663 LONG SPAN ROOF CONSTRUCTION ‘JoouIsuy Sur}[nsuop ‘TYS[NUIG pre apy ‘stoouIsuG “OO Suroomsuy uleyynog (‘299 “d 999) -wmr0yIpny MoTBS-UOSUTM IOJ JOpIN Jooy—"erg ‘py “1ozU99 UO “Ul G “J GT 0} “Ul F “4 CT UOI} pooeds siapitry ZY ee See Pe+9g-----------_- a----- “ (pa0yds) sing ‘bg #1. g. 4 ? (Paouds) sung “bg Fi-e--. (peoydsh avg bg Ly 1, WO ee SE | Sa¢ wy SSS3q°@s Nt SOS MNOS IMA ia NG. Pris - MMOH poke SS 7 = = = 29D, 0S ay 8G Jonbz g]-—-~ ~~ — -81 wz Sdg- p-- Pee ae al: é vol @'808- 6" 6 809-67 19“ A 664 ROOF CONSTRUCTION For the sake of appearance, it is advisable to provide in the center of the girder a camber of a depth at least equal to the total expected deflection of the girder. The objectionable appearance of -a sag is thus avoided. From the standpoint of design, long span girders do not offer any special difficulty. The bending moment is found from the formula for simple beams, the distance from center to center of the support being taken for the span. In computing the amount of steel and the compression in concrete, it may be advisable to use the formulas in which the compression in the stem of the T-beam is considered. T he simplified method given on p. 224 will be found of advantage in this connection. Particular attention is required in spacing the longitudinal rein- forcement. As the bars are more numerous than usual, it may be necessary to place them in several layers. The use of more than three layers is not advocated. Special provision should be made for sepa- rating the layers of bars. This may be accomplished by placing across the main reinforcement short pieces of one-inch square bars 3 to 4 ft.on centers. A proper proportion of bars should be bent up and utilized as diagonal tension reinforcement. Before bending the bars, the required amount of steel at intermediate points should be determined. ‘This requirement is particularly important in girders in which the restraint at the support is slight. (See Fig. 98, p. 292.) The points where reinforcement may be bent up will then be deter- mined by bending moment in simple spans, and they will be much nearer the support than is the case in continuous beams. The use of standards applicable to continuous beams may be disastrous. At least one-third of the bars should be carried straight to the end of the beam. Under no circumstances should the bars be stopped short at any intermediate point, as there is no means, in such cases, to develop the stresses in the short bar, without its slipping. When the ends of the girder are not free to move, short bars should be used on the top of the girder at the support, to insure against — cracks, even if the restraint is not counted upon. When a concrete girder rests on brickwork, the bearing stresses on the support should be investigated. If the weight of the girder pro- duces excessive bearing stresses, the bearing area should be increased. An example of long span girder construction is given in Fig. 213, p. 663, showing the roof girder used in the Winston-Salem Auditorium. The location of the girders is shown in Fig. 275, p. 781. BSVRSALEE ch ‘netic? Vek) * yaiyttes mes anys he PhS bes! ia Ab 2S yp ied.) 888-4 a stds) BIS, aS | era cats wh < a 2. Cr i ee wae nt : es . esd 73> %- . “ ead 2 A ’ ye 3. “t~} ~« me . = ‘> Sib raul mea Svar Zainl? AN CGR cam 8 | Sed \ ne SE, “OE AN ES % berks mr. are 2S) RR Ke 20, ae Yn ’ + ie 5 eae ia be [tae sane many a EP Vatar ee: ; Hid a I: te ; P pe iataeen 4B ae See ec eAG BOOS O45.°14.-9 ~--—-~- » ’ woe. pros B28 BF a a a8 - taratien of bars ard “done ai fr colvenn F |» s, j Py * oy GS a ee din ca coe, aA ON ‘eisieniht omandlpish “A sect FUABADTAB AAS HEE x Y4AM &. 051) (eee :: 2, -opliced tO 6. 5 " Sg RBI0" eer er a i 2-864 phere! at this reese 755 a . eh 7 hee went ee Yoneach face’ % a - §%---- Ree yu f 806 7 F/ Face | 25. 3 o- vid eres palate 62s s 1e2'9"" she OL Gabe 4 9" Qt ima Red stirrups 0350 | 3 "Spacer Bars §-0" ‘ ef & = oreo $0 Inher half beam} 860_. @ & 2 pee 2°01 similar spacing f-~! 752-752A Spliced. ~~ a ee eS hee. re ong 2) 1 oO" epost “4 Lae A = Si td Lond 2-2'RB. 1240 Ce a Rd Bar Oto eek” a ee Ba seme eet oil £ ok NS i sf ‘Rd Stirrups Asa SS Fr = 0351 12°28. | 2+ ‘ 1 ay 1 { { 1 t & | ' i P ' 5 1. < el Jiconnection, <5 **er a ev 0350 | Ss Asahi ee . _ 2 Si Se eke = 7 _-Reinforeement same 866-I-1"R2.8-3I'-0"™ * (865-1-1"R.B-40'-0" 8 3"Pd Stirru Sy as yor column on aise ee {pul TRE ae { a 035, gw: * es z opposite side, 864 -1-1"RB-45'-0% TRL. 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6S |} 865 -2-1"RB-40'0" U 6- x a 865-2 -1"RB-40"-0" 863 A-1-I"RB- d 864 -1-1"R.B-45°0" /5'-0" d 864 -1-1"RB- 45-0" 863 -1-1"RB-49°0" 862 A-I-I"RB- @ f 862 -1-1"RB- 40-0" ‘i F a i 2-1"R.B-13'-0" e 86! -2-1"RB-49'"-0" 867A 867 -4-1"8B-55°0" an fal B ON vi, Fig. 214.—Frame over Boiler Roof 2-86! 8B ia | ‘ /-862 K ge — oo G v i ee LI 755 -2-2'Rg-25'-08 ZAP adie aeeee : ; rte LAS EG -T5IA-2-2R “ 5 i 12'-0" 5-2-3 '28-25"0" 864 868) Gy, ir “4 i= '53-1-3 RB-35*0" et S's : ‘CO_ . 357 a ziae . 4-I°RBL6O'Lomp &s / { 2-/RB26-0 Long ‘3 oth B59 -/-/"RB- ranean CR NS > A --6-/"2B 26-0" Str ( i] . weg t 1 aoe ¥ 1 2-/2B.27°-6" Str /e4>| > / “LB. 28-0" Str 864-8644 Spliced N 8 Ge igs CL Oe Oe ed S 23°RB.12+0" 75/-A| ~864-2-1"RB-45'-0" a < - aes ~9" 859 4 Soliced cae Ely eae ae = -l"RB 1340" 859- (862-3624 Spliced 3 Rd. stirrup 0351 Note: Steed in right half except | where shown different is si- | milar to steel in left hat location of bars and hadi e-—“dowels asin column & inished Floor- ! I x 3 s ~~ S &p ra) at. me = ae : ‘a ra - x BY, g wre 8638 1-1"RB-13-0" PRB EEG TG. (Ae = tee 2 =F Ot — mn icy) ey 752 Hy ahd oer = ders !'RB27-9 Ig, i 186---3-- ————_____D 2 2-18 -/5'-0" ae Pe eet “4-867 Spliced fo 867A | S-'RB26-01g>— 1" B D4" 97 ony VR B.276" Long! # “— Section CC ( Showing bottorn stee/ of girder) 867A co 4-288 0" ms 36 37 38 39 40 Hey ) Wondensed Milk Co. (See p. 635.) (To face page 664.) Se ee 7) a ato ey . ) Tar re bhoes j : y i Sa? Gy, Sx hy pa DE BGS | Boe Whe | i ; 5 ee 7 & m haw mind ke newak of Z... | | % nla a bb Bees | \ i + aed é Is oe hii \ i ; , is - * re + a t : P 5 eran ote r | rT eee a mnie eamale eee : i 5 S| i th Ae, SAP AE re St a pisipemcinaatnl . ese = agrees bg & 26 : toon rarest mae pe tat aeratiaeranerendl oem ' maiboal e a Ved a3t . . J t i‘ Te oe Eee ' ; wes ; ES Ce ns am \ yew 3 wm 4 ” — ei ie ee” garde = » ern ect, Qi ya} 23 i spol bolts P.-... 54 ase & Sk 3 aes S88 G5 S&S SS aG/. 2 i SS 00791 norl~/ ~ 8/ ose Soe Les ‘ @ SU OGl= nt Toe! ZAQey ude wf u0-120-,a/~ [- b/ u0-,92-,al-1- £/ uO -9£¢ ~al-/- él 00-,82-_2-/-I/ wO& «0-07 -,ol-1-01 n0-,9%-,a/-Z2- 8 476 U8] Bazi ON YID a - —--— TF eee SS ———_— ————_——— — Scale for Reinforcement Diagram in Feet O123 456789 MIN 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Eee | ¥ Section EE ‘ Close stirrups after placing yoper DArs.~~. x _- Efbolts P---~.. }-----~--- ets = 8 Concrete lugsl in each truss must be set inplace when jpla- cing steel in truss and wired to same where possible. Special reinforcement at XX is the same as that shown at Y-Y opposite. & hae ee tt eset ae Wall Fratring” Mark No. Size &length Note: Place inverted stirrups | r ker 3 o] S ft ieee | k 5-/ - 797 5Q!-0" | =I “By cs Ea ai 1 S| z Me Baye, & ‘a en De : f-itaa-0" =a] + > = a Mark No, Size & Length ; += ) 3 ol o © | iu 29 - 2-12"10'0" eee = | ! “ y rs | ete | bea! FD Ee Uey ae doled Fak A Se | 20- — feo Section D-D S | ‘4 io, hs 3 -/0'-6'-0" ee | | ny .--Close stirrups 28 - 2-19" 1440" x I —- 2 7 ie & SN heh cease _————S aan shan pl a upper bars. : t eS ! £3 &, ‘adit Bars at X-X and Y-Y. } 2-B0Hor0" nn] kt a (These bars are placed after the | ! h /-9"'e ; red. I patt assembled units are hoisted and \ ! =, , : ! put inposition ) 1 22a- PP i ; | bLLorgrgn meee | te S fa 1 i f e+ EY “Bolts P through 1-02" ‘orm supporting steel eC ite fae Hider Plan of Roof. Scale for Reinforcement Diagram in Feet 2d, & 2ds,and removed 0123456789 10 1213 1415 16 17 18 19 20 21 22 23 24 25 26 27 26.2930 © en form ts set vertically 0 17 Ve. Section BB Hae bY Fia. 215.—Details of Foundry Building for Garfield Smith & Co. (See p. 665.) (To face page 668.) LONG SPAN ROOF CONSTRUCTION ~ 665 Long Span Roof Girders as Rigid Frame.—The cost of long span girders may be materially reduced, by building the girder mono- lithic with the supporting columns and designing the construction as a rigid frame. The use of rigid frames in America has not been as wide as the economy of this type of construction warrants. This is partly due to the fact that the design is somewhat complicated and that the building codes and standard specifications have no provision for such construction. The arbitrary bending moment requirements that are often imposed make proper use of rigid frames impossible. Design of Rigid Frames.—Rigid frames should be designed according to formulas given in the section on rigid frames in Volume III. These are just as reliable as the ordinary beam formulas, as has been proved not only by a number of successful constructions but also by numerous tests. Rigid Frame with Horizontal Girders.—Figure 214 shows a good example of a rigid frame design with horizontal girder. It has been used by Thompson & Binger in the construction of a boiler house for Hires Condensed Milk Co. in Morristown, N. Y. The span of the rigid frame in this design is 60 ft. and the height of the column 27 ft. 10in. The dimensions are shown in the figure. It should be noticed that the flange of the T-shape was obtained by thickening the slab on both sides of the girders. Of interest also may be the method of splicing of the bars, which was necessary because it was not possible to get bars more than 60 ft. long. The frame was considered as partly restrained at the bottom. This is an inter- mediate case between a frame hinged at the bottom and one rigidly fixed there. The assumption was justified by the fact that the columns were tied by reinforcement to substantial footings. Rigid Frame with Ridged Roof.—An interesting example of ridged roof construction may be found in Fig. 215, opp. p. 665, and also in the photographs, Fig. 216, p. 666, representing the foundry built for Garfield Smith & Company, by Villadsen Bros. The span of the frame is 70 ft., the height is 45 ft. The details of the design may be seen from the figure. The frames were designed as hinged at the bottom. Of interest also is Fig. 216, showing the method of erection of the reinforcement and of the formwork. The rein- forcement for the columns and the beam was designed so that it could be separated into three parts. Each part was assembled on the ground and placed in the formwork as a unit. After all three 666 ROOF CONSTRUCTION Fic. 216.—Views of Foundry Building for Garfield Smith & Co. LONG SPAN ROOF CONSTRUCTION 667 parts were in place, additional bars were placed, joining the three sections. The formwork was also divided into three sections. These were assembled on the ground and lifted into position by a derrick. Fig. 216 shows the central part of the form, with the rein- i 19n. 1912" mie l reoeeeeee 100A ee mle yeti iaglg BE A Oe |e Loe 7. ; ern Cc eee Das . ees gx te Vere BOTIMRTE, feat femoral eiste ECE {. APES? 88 2 ANDO ce oo oe a ER iS i | | ee ee zag ad 24 : | \ ] | 2913 va 195" 4 xe or = a E™ BG 13" oe" te ap 3" es BSXRK oye " °° a Hist OR er Section mt £RSS S Section Lm t PSSs8 a aR Fad 109 3; 1-143" ' H $5) 1-12 it tH 19, TSI" qu =) | rh a 1 igs ob BSA ETS 1 19,11. i Cranilme: Pale PF ait riick . 71 ' &$, 16 -¥-- Eb ery, 2 =f Pee 824 -->| 8 Fig. 217.—Hall Frame at Lausanne. (See p. 667.) forcement placed within, being hoisted into position. This ingen- ious method undoubtedly contributed largely to the economy of the construction. . Rigid Frame with Arched Roof.—A good example of this type of construction is seen in Fig. 217, representing a frame in a reinforced 668 ROOF CONSTRUCTION hall at Lausanne, Switzerland. This hall is rectangular in shape, 183 ft. 9 in. (56 meters) long, and 114 ft. 10 in. (35 meters) wide, ‘out to out. The construction consists of 8 frames spaced 26 ft. 3 in. (8 meters) on centers. Fach frame consists of an arch sup- ported on columns hinged at the bottom. The intrados is a three- centered curve, while the extrados is a segment of a circle, the radius of which is 82 ft. (254 meters). This makes the depth of the arch a minimum at the crown, where ++ ig 3 ft. 7 in. (1.10 meters) and a maximum at the springing line, where it is 8 ft. 3 in. (2.5 meters). The cross section of the frame is given in Fig. 217. It should be noted that the section is hollowed out, not so much to save material as to save dead load, thus reducing the bending columns and stresses. In America, owing to higher ratio of cost of labor to cost of material, this scheme may not be found economical. | The hinged effect at the bottom of the columns was obtained by building the foundation and the frame separately and employing a fan-shaped arrangement of dowels between the foundation and the frame. These dowels were effective in preventing any displacement of the frame, but were not capable of resisting any bending moments. Therefore, the frame can undergo angular movements due to changes of temperature and movements of foundation without producing any stresses. The following assumptions were made: snow load, 16.5 lb. per sq. ft. (80 kg/m?); wind load for calculation of the arch, 30.6 lb. per sq. ft. (150 kg/m?). A temperature variation of 15 degrees Centigrade was allowed. The shrinkage stresses were taken care of by assuming their effect to be equal to a fall of temperature of 20 degrees Centigrade. ConcRETE ARCHES FOR Lone Span ROOFS Reinforced concrete arches with tie-rods form an economical _ type of long span roof construction. Their use has not been as extensive as the advantages of the type would warrant, mainly because the design is somewhat complicated. The cost of form- work, which at first sight seems to be large on account of the curvature of the arch, need not be unduly expensive, because the ‘ntrados of the arch can be made up of a number of straight seg- ments. 1 La Technique Modern, Vol. 13, No. 1, Jan., 1921, pp. 12 to 14, LONG SPAN ROOF CONSTRUCTION 669 Method of Design.—The roof construction may consist of arch ribs placed across the building and spaced from 15 to 25 ft. on centers. The arch ribs support beams or joists running longitudinally, and these in turn support concrete slab. A monitor may be easily built on top of the arch. The arches may be constructed with the supporting columns, in which case the construction becomes a rigid frame with curved top; or they may be independent of the supports, in which case the horizontal thrust produced by the arch must be resisted by tie-rods and the arches may be considered as hinged at the supports. Formu- las for two-hinged arches, given in Vol. III, should be used. The concrete dimensions of the arch and the required reinforce- ment should be computed for a condition of load producing maxi- mum bending moments. Such a condition occurs when the live load is placed on one-half of the arch. The tie-rods, however, should be computed for the maximum thrust, which is produced by the live load extending over the whole span of the arch. In computing stresses, the effect of the length- ening of the tie-rods, due to the horizontal thrust, should be taken into account. The assumption in all arch designs is based on the span remaining constant. This naturally will not be true in the case under consideration, as the lengthening of the tie-rods will permit a slight displacement of the supports, correspondingly increas- ‘ing the span length. Wind pressure may also have to be considered in designing the arches, especially when the building is exposed and the rise is com- paratively large. The arch with tie-rods should be securely anchored to one support, but should be allowed to expand and move in the direction of the span length on the other support. If this motion is not provided for, the tie-rods will not become effective until the resistance of the support to sliding is overcome, because no stress can be taken by the tie-rods without corresponding increase in length. With such construction, no provision is required for temper- ature stresses. Example of Arch Roof.—Figure 218, p. 670, shows the design of -an arch 103 ft. 6 in. over all, with a rise of 17 ft. The spacing of arches in the building is 17 ft. center to center. The cross section of the arch, constant throughout, is 114 by 24 in. The reinforce- ment consists of three 7-in. round bars placed at top and three at bottom. These bars were properly lapped and also provided with 670 ROOF CONSTRUCTION Three-eighth-inch round stirrups were spaced e intrados of the arch To facili- hooks at the ends. uniformly the whole length of the arch. Th was nominally a segment of a circle with a 90-ft. radius. Roof at Balcony —_. =-=== = a4 3 ” als ..2'Rd Hangers, <> <-Tile with Turnbuck/4s “ R we oe ‘ h seonerete ’ “Beams ie rd ==! RSS | “” Hannan —bpean nn oman 10 Tz amnenn" au Concrete Post-->| ES pod 5 24 ohe- 24-3 2-2 PRdtofxtend Posto 24" over Support yeSky lights Concrete Beam EA Sees a 2"x10" 2 Gas Pipes-” \yis's 4 a7 Oe eac ra eam for Stub ods pl i7J ke— Section A-A iN Arch. Rib. ‘ pee i1g"x 24" Sor" 14> 3 we g ie Fenech I Z| >“ Bars fitch AL Stirrups { 2 "Tie od Top of Conerete. Roof Joist i eee 10'-0"----- x - as : : ty w" “ 4 6"x6 . o%jav || os ‘2 "ya" K 4"x: Ul ‘ Fig. 218.—Details of Arch Roof. (See p. 669.) tate the formwork, however, the intrados of the arch was made up of straight segments. This substitution is practically unnoticeable. A single 2}-in. tie-rod with turnbuckles resists the horizontal thrust. At each end of the tie-rod is a nut and bearing plate, as, is LONG SPAN ROOF CONSTRUCTION 671 evident from details, Fig. 218, p. 670. To prevent sagging, the tie-bars are supported by five 3-in. hangers. The roofing is carried by wood rafters 2 by 10 in., spaced 2 ft. on centers, seated in the ribs. The longitudinal skylights are sup- ported by concrete beams running longitudinally with the building and supported by the arches. The design was made by L. J. Mensch of Chicago.? Lone Span Roor Construction Long span roof construction of reinforced concrete has recently been developed and successfully used by Richard E. Schmidt, Garden & Martin, Architects and Engineers, of Chicago. A typical example of this construction is the armory designed to house the Illinois National Guard, Danville, Ill. The main features of this design are protected by patent. Figure 219, p. 672, and Fig. 220, p. 673, show details of the design and an interior view of this armory. The roof is supported by monolithically poured concrete trusses with a span of 85 ft. A monitor, with continuous steel sash both sides, is formed by canti- lever beams extending from the peak of the roof. Roof of monitor consists of a 3-in. solid slab suspended from girders at end of these cantilever beams, and formed on the curve of a catenary to avoid dead load bending stresses. Tie-rods are used as bottom chord for the trusses and between ends of cantilever beams, to avoid bending stresses in main truss members. Roof over gun shed, shown to the left of main truss, is of similar construction. Balconies for spectators are built between the main roof and the truss over gun shed, also at one end of building. Design of this balcony is shown by Fig. 219. CONCRETE ROOF TRUSSES Figure 221 shows a design of a concrete roof truss used in the Graumann Theater in Los Angeles. Other details of this structure are shown in Fig. 270, p. 764, and Fig. 273, p. 779. The following description is taken from Engineering News Record, July 5, 1923. The roof trusses are 126 ft. 6 in. in span, center to center of columns, 15 ft. 8 in. high; each truss has 61 tons of reinforcing steel and 231 cu. yd. of concrete, and carries a load of 750 tons. In the ? For full description, see Engineering News Record, Jan. 27, 1921, p. 167. CTL9 ‘a 909) “IIT ‘g]ptAuBd 7 S10ULLY Ul UOTJONASUOL) JOOY uedg Bu0'T Jo s[ivjod—'61z ‘OI ge - MeS MPyE spua pasdn ay a F a Ul Pd 9 “1 poy all PY, £1 pie Su BE| 90, 6-sall pyt PY, l- - A G/% 9 uf - 00, 2S fmGSi oy Spee jasdn spoy all py #E- t . Ta x i i 5 Pe apyonguny. 4 ee, pe Oe ee ee ee ee ed Dh w B “ gpua jasdn poy all PY tEA = PU yl-2- NX © =) © ca} 2B/ ssapaig sof wosboig busssofuray Seige S109 9018 70,9 S109 PY ie Se gronmunt 2B/ SDL Sy} says prpanag p 54m yim spua Jasan poy all PYFE PLTS S109 GO/S 29,8 S10G PY 2 PY Fe -€ ‘PY £/-2 Ey —-ubd? ajo/pala,u/- esas eae --unde puj----- 1 672 LONG SPAN ROOF CONSTRUCTION 673 building up of these trusses, the soffit of the truss, with its reversed decorative forms, was put in place first; the bottom chord steel was next laid out and fastened securely in place. Spacers, made of 1{ by 1{-in. bars bolted together, placed approximately 10 ft. centers, were used to hold this bottom chord steel in place. Cement blocks were used to carry the load. Care was taken in placing these blocks to see that they were staggered so as to allow for the placing of the concrete. The bottom chord consists of 14-in. square deformed bars, 65 ft. long; a 6-ft. lap was used at the splices with three #-in, U-clamps. Particular care was taken to see that these U-clamps Fra. 220.—Interior View of Armory at Danville, Ill. (See p. 671.) were securely in place and tight. The locations of these splices were all laid out so as to stagger with the splice in the bars adjacent to them. At the ends of the truss, the bottom chord bars were all hooked around the vertical column bars. Bars in the diagonal tension members were laid out so as to work in between each layer of the bottom chord. These bars were hooked on both ends with a cross bar running through the hooks, the top chord being rectangular in shape and reinforced as for a rectangular column. For discussion of the use and design of reinforced concrete trusses, see p. 769. | ROOF CONSTRUCTION 674 *qoopIYILV ‘yorjss1og UIMptT CTL19 d 299) ‘gojasuy sory Ul 1oyVoUL, s uUBUINBIL) IOJ SSNL]L, jooy—"12S “Pl en 103208 p4oyguiojzz0g Udi [01}40d fo u0lz2a§ joadhy p4oyg 10330g fo udjd [D!742d psoya doy fo udid . em a oe OOo Ge ame eae ae lg | SONI4I}S = pagihed yerisciariey ota on ub c §1.02G §/99 590g, [X 2 $40 /1 FX, ao ‘ by Y- ee ean ------ Eee ee slew GOCE DEO Sel OES Ferg mae tf OE OS ee ow i etoricn a ee e e i997» 2099, 9] SINIIAS F ~ je Pevees. HEE : d Pana tae rat ITKY 2 : SS Sarees eee _ += fe-—> = === 50799 sung asonbs F1-92 ae nOa0K SIDE DSi) 9S =a a >| SAWTOOTH ROOFS 675 SAWTOOTH ROOFS Where Used.—Sawtooth roofs are used in manufacturing plants where an abundance of light is required, but where it is necessary to avoid the glare due to direct sun rays and the consequent shadows. This is best accomplished by providing windows, facing north, in every or every other panel of the roof. The resulting construction is called sawtooth roof construction. Sawtooth roofs may be built with the windows vertical or inclined. The construction of the roof with vertical windows is the simpler, and also makes it easier to keep the windows clean. However, the vertical window does not provide as much light as an inclined window of the same size, and for this reason inclined lights are in more favor with architects. The maximum inclination with the horizontal is governed by the requirement that at no time of the year direct rays should be permitted to enter the room. This angle depends upon the latitude and may be expressed by the following formula: Let « = inclination of the window with horizontal ; a = latitude of the location. Then a = 113°30-—"a. In the vicinity of New York, a = 41° ; consequently, the angle of inclination is a = 72° 30’. The angle between the window sash and the inclined slab will depend upon the desired size of the window and the depth to window sash. If possible, the angle should be made 90 degrees as this facilitates formwork. Types of Sawtooth Roofs.—The sawtooth roof may consist of a frame, as shown in Fig. 222; or it may be provided with a horizontal tension member which resists the horizontal thrust produced by the inclined roof members. All intermediate beams should be parallel to the frame. Beams longitudinal with the skylight throw shadows | on the underside of the slab, thus reducing the effectiveness of the light. Details.—Special attention should be paid to details of the gutter, to make the windows weather-proof. The bottom of the window sash should be at least 18 in. above the gutter, on account of the possibility of snowdrifts. The gutter should be well drained. Down-spoutg should be placed at frequent intervals. A gangway lan 676 ROOF CONSTRUCTION x15 Beam, 2 ‘Sstirrups-— at “>> 2. : PN SIN 2 i Stirrups 7 _ Vv jp : ngs iwley Section A-A Sawtooth Frame framing into Columns Intermediate. Sawtooth Frame Supported by, Girder Se a Plan Roof. (See p. 675.) Fig. 222.—Details of Sawtooth PRE-CAST CONCRETE ROOF TRUSSES 677 at least 2 ft. wide should be provided to facilitate the cleaning of the gutters. To prevent ice from closing the down-spouts, the under side of the gutters should be heated. The details of roofing are shown in Fig. 223. State surfaced roofing nailed along top of curb “into creosoted wood nailing strip Pe Three layers of roofing felt with / alternate layers of Elastigum, run : / up the curb and out on felt in valley at least 6” Creosoted wood nailing strip~.. j-----Steep Roofing All plies of valley roofing___ j are cut off at the angle ~~ Roofing in valley nd carried up the slope Fig. 223.—Details of Roofing for Sawtooth Roof. (See p. 677.) PRE-CAST CONCRETE ROOF TRUSSES Pre-cast roof trusses were used for a shed of Pier 6, at Cristobal, Canal Zone. The shed is 945 ft. long and 159 ft. wide, and has a clearance under trusses of 25 ft. The arrangement of trusses is shown in Fig. 224, p. 678. The construction is three panels wide. The main trusses, placed across the building, are 39 ft. long and 7 ft. deep, and are spaced 45 ft. on centers. They are supported by columns. The longitudinal trusses, 45 ft. long, 4 ft. deep, and spaced approximately 10 ft. on centers, are carried by the main trusses. A 33-in. slab is placed on the longitudinal trusses. The depth of slab is small for the 10-ft. span and is possible only because there was no snow load to be provided for. The design live load was only 10 Ib. per sq. ft. The design of the trusses is evident from Fig. 225, p. 679. The design of the pre-cast columns is also shown. It is evident that the column on the top is provided with recesses to receive the cross trusses, as well as those longitudinal trusses which frame into the columns. All the columns and trusses were pre-cast, and completed before being placed in position. Locomotive cranes were used in erection. Interesting details of the method of pouring the pre-cast members and erecting them are given in Engineering News Record, June 24, 1920, p. 1232. 678 ROOF CONSTRUCTION AUXILIARY STRUCTURES ABOVE ROOF Pent Houses.—The enclosures surrounding the elevator machin- extend above the main roof. ery and the stair entran ce to the roof on the walls enclosin ¢ the pent house. The pent-house roof rests In many cases, a large part of the pent-house roof is staircase or elevator shaft. taken up by @ If no skylight to provide light for the A ------- 159+0"" Section 8-8 Gt eee 939'-9t-- OO Siti Girder Es | [==> | = -Back of Wall @L.of ERS ee b- Roof Plan, Sea End Fic. 224.—Partial Plan and Section of Pier 6 at Cristobal, Canal Zone. (See p. 677.) skylight 1s required (as in staircases well provided with windows) the roof consists of a solid concrete slab. The main roof construction must be made strong enough to carry the pent-house wall and roof. When the walls coincide with the outline of the opening, as is usually the case, the beams around the opening may be utilized to carry the pent house. Sometimes the location of the pent-house wall does not coincide with the location of the beams. Such designs should be avoided, because they require 679 PRE-CAST CONCRETE ROOF TRUSSES (229 ‘d aay’) ‘au07 [euey ‘Teqoysrig 42 ‘9 Jag ‘sossniy, fooy 4SBvd-91g Jo s[leyoq—ezz ‘pI A = fet Lae iAa it 1-7 A-M4 0-0 ae a ie 1 wUolaag eaetads U01}0a9 pyers Baas - - = .9 3 . Ff -ade py | ce _ aS } gh ; Ne uOl =102- s ME Pee) os F ec = ee NP ole eae Tne ie rye ies Sen ad pee ts! | d Z Fae “ef Ul =ado | gras Ss = [-L SAPs Urey ‘u0lz DAa/7 tee S . Bo er = --- 1 oy | - 2 =6 pag = Sear a | Bs i | ~y ef | 0K, | 1 yy 680 ROOF CONSTRUCTION SS ——————— Se ~-- aaa es: Cae 5 a 5 NJ Ls a-e'kd Eh: PTT Friis mpm . 's--4-2'Rd. Rods- 2 Top- N 3-2 Ral 16’ 16"Col. sacar cS 2 wir: 4,5 Rd. Rods------ (a 4 Hoops /2"0.¢. Col. Stee! fo be extenaen na a= = ——=———-— 70-9)" a ae a es ee ee eee 4 | | Va Le ow ome oe Len Se eee saan === 29-9 ------ =| Se ey a ee as 6" Slab 3'Rd-6"0.0.~ a 3 each way ; 51 BN FB a | oa Z panna 2019" -------- >| Plan of Tank Support Beam Schedule Fic. 226.—Water Tower. Clark Biscuit Co. at North Adams, Mass. (See p. 681.) Wm. Higginson, Architect. wi \ SH Lok a eg cael + ¥ ay ASO\" ‘ sy} had of Iris.ae = ’ a ve t ‘ y : Sonrn pes (A ye q 8 ‘ s 5 #3 : ? ani a ai ee eer “seal tint eosin ae anear ahs 599 = rae somarstte iow bene tenlnay are sie Sah a ae ae fag We A te ee es oe ee Ae ee oa ee ew Slee 2 ve $33 Se eer een ee ay eter ay gee See ee cer me me ee at ee ‘Yok rswot VMAS =. Sie ce ys wets ’ ee 4 Meni ten ine ej ste ssi i fe ns YS | R Framing Plan gt Tower Floor By --=9-08--——— ——}.9-0¢——-— >| | Q . < SS | | ! I | | | | | | | | ! | I | I | | | ! I ! ' | ) { | | | L aie yam 3 <=> FOUSuG 00,9 PY 8 —<———-— P1SuG Tower Roof Framing Fig. 228.—Details of Water Tower in I 4-2Rd 2-IRds. esr ris ie elt SOs ofS aBelbtrtebat sae Be: 0" qu aay Wikies, miei nei > -: BE Fy oes aa fet Ot | 4 2-W Ra. pape S PigkzO Ce) ns Eek: BT saggy ese axon 59 ge eno SHR RRR ATER A " Sep ee | se ha 6 0.C, ii me aes 4ip% eae | ze fal. Rd S3Rd) bd Ra Ul God he L2Rad, We 02, N 'Z 80.2 ih An U Teo. Ss L-—+£ S | A SI Rs, SL ive a 6"0.¢.Co/3.— “a ———-20+ 6'0.¢. Cols: > | Seetiona/ Elevation A-A Section 8-8 | Section C-C Designed for Bunte Bros. (See p. 681.) «fl face page 681.) ee ow a a TS a Te SEE SAE £ Do TEE EAE = od : gies, = fs An eee ee wor OD, i ye 4 Lav £00) 4 Pa L\ G8 syod vanlol, AUXILIARY STRUCTURES ABOVE ROOFS 681 special beam work for the roof, resulting in a considerable increase in the cost of formwork. Sometimes it is possible to reduce the cost by placing the beams supporting the pent house on the top of the roof slab, and utilizing them as a part of the pent-house wall. These beams can be built after the roof slab is completed. In some cases, it is desirable to construct concrete framework for the pent house and use the walls only as curtain walls. The pent- house roof is then supported by concrete wall beams. Water Tower.—When the water tank is placed above the roof, it may be supported on a concrete water tower. The first cost of a concrete tower is larger than that of a steel tower, but it is more sightly and requires no maintenance charge. Fie. 227.—General View of F actory Designed for Bunte Bros. (See p. 681.) In Fig. 226, p. 680, is shown a good design of a water tower used in the Clark Biscuit Co., building by Wm. Higginson, Architect. A good design of a water tower is shown in Fig. 227, p. 681, in the candy factory designed for Bunte Bros., by Richard Schmidt, Garden & Martin, Architects. Instead of detracting from the appear- ance, as would have been the case with an open water-tower, the tower in this design has been changed into an architectural feature. The details of the concrete design are shown in Fig. 228. CHAPTER XIV STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS Stairs serve two purposes: first, they provide a means of exit in case of fire; second, they are used for travel from floor to floor. In office buildings, tall hotels, warehouses, and factories, equipped with passenger elevators, the everyday use of the stairs is com- paratively small. In such structures, stairways are required mainly for exit in case of fire. This purpose governs their number, location, and design. Occasionally, however, there may be considerable local movement between two adjoining floors, and special stairways, conveniently located, may then be provided. These special stair- ways need not be enclosed and do not form a part of the fire exits. In private homes, schoolhouses, court houses, theaters, factories, and similar buildings without elevator service, the stairs serve mainly for everyday use- Their location is governed by convenience of movement from floor to floor, and their number may exceed that required for fire exits. “The stairs required as fire exits should be enclosed. Others do not need fireproof enclosures. ; Stairways are relatively expensive to construct. They also take up a large amount of space; hence, the tendency to use as far as possible. On the other hand, the proper number of stairs in proper locations may be of great value in the operation of the plant, or in the use of the building. No general rule is possible for cases where the stairs are designed for constant use. The conditions in each building should be thor- oughly studied before the location and number of stairs is decided upon. Rules for design of stairs as fire exits are more general. They will be treated below. Fire Exits.—There are two recognized means of exit from a build- ing in case of fire: first, horizontal exits through or around a fire wall; second, stair exits. Stair exits may be used alone or in com- bination with horizontal exits. 682 NUMBER OF EXITS REQUIRED 683 Horizontal Exits.—These are exits through a fire wall, a fire exit partition, or a fireproof wall separating two buildings. The exit may lead to another building or to another part of the same building. It should be provided with a self-closing fire door (1.e., a door nor- mally kept closed by some mechanical device) not less than 30 in. wide, the actual width depending upon the number of persons for whom the exit is intended. Horizontal exit also may be effected around a fire wall or a wall separating two buildings, by means of a balcony or an exterior bridge connecting two buildings or two parts of the same buildings. The balcony or the bridge should be at least 44 in. wide and constructed of incombustible material. It should be enclosed or provided with railings at least 4 ft. high. All doorways opening on the balcony or the bridge must be provided with self- closing fire doors. All window openings within 10 ft. of the balcony - must be protected by fire windows. The purpose of these pre- cautions is not only to protect adjoining buildings from fire, but also to prevent the entrance of smoke which would make the exit useless. There should be no obstruction on the balcony or the bridge. If the floors of the connected buildings are on different levels, gradients of not more than 1 ft. in 6 ft. should be provided. While the horizontal exit is considered more effective than the stair exit, its usefulness depends upon conditions over which the designer of the building usually has no control. For instance, the adjoining buildings may belong to another owner and may not be available for a horizontal exit. This prevents their being used as often as the stair exits. Stair Exits.—Stair exits may consist of enclosed interior stair- ways, smoke-proof towers, or outside stairways. Number of Exits Required.—The requirements as to the number of fire exits in a building differ to some extent in the building codes of the various cities. In some codes, such as that of Chicago, the number is based upon the area of the building; while in other codes it depends upon the number of persons occupying the various floors of the building. The latter method is the more logical of the two, because if the capacity of each exit is known or assumed, their number will be governed by the number of persons who are expected to use them. The building code, if applicable to the building, should be studied first. Where no code applies, the following suggestions may be adopted. 684 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS In all business buildings, at least two fire exits should be provided from every floor above the first, irrespective of the number of occu- pants of the floor. One of these should be a stair exit, while the other may be a horizontal exit. The exits should be arranged so that no part of any floor area shall be more than 100 ft. distant from a hori- zontal exit or an entrance to a staircase. Additional exits should be used if the number of occupants of any one floor served by the exits ‘exceeds the capacity of two exits. In all fireproof buildings that are more than three stories high, or occupied by more than fifty persons above the first floor, at least one of the stairways should be continuous and in fireproof enclosure. This should lead either to the street, alley, or open court, or to a fireproof corridor of proper width leading to the open air. In all buildings over 90 ft. in height, one of the stairways should be a smoke-proof stair tower. / Capacity of Stairways.—lt is a well-established principle that the capacity of the stairways as fire exits should be such that they may accommodate at one time all the persons engaged in the building. If horizontal exits are used, the number of persons to be taken care of by the stairways may be reduced proportionally. In determining the number and the width of stairways, the largest floor area served by them should be considered. The persons occupying such floor area should find accommodation in the staircase between their floor and the floor next below. The number of stairways, where the persons are distributed uniformly over each floor area, is determined by dividing the total number of persons per floor by the capacity of the stairways. Concentrated groups of persons or separated rooms may necessitate extra accommodation. In determining the capacity of the stairways, it is necessary to make assumptions based on judgment. The rules adopted by various building codes naturally vary and only designate minimum requirements. The New York Code allows per person on the stairs, a 22-in. width of stairs and one and one-half treads; and on the landing and +n the hall within the staircase, 33 sq. ft. of area. If, in addition, a horizontal exit is used, stairways need to be provided for only one- third of the persons occupying the floor under consideration. In an automatically sprinkled building, stairways for only one-half of the persons per floor need to be provided. When an automatically sprinkled floor is also provided with a horizontal exit, stairways ENCLOSED STAIRCASES 685 need to be provided for one-quarter of the number of persons occupy- ing the floor. In no case, however, should the number of fire exits per floor be smaller than two, one of which must be a stair exit. The National Board of Fire Underwriters recommends a some- what simpler rule. The capacity of the stairway per floor may be considered as equal to fourteen persons for each 22-in. width of the stairway, plus one person for each 3 sq. ft. of the floor area of the hallway within the staircase of the floor under consideration and of the area of the stair landing between the floor under consideration and the next floor below. If the building is provided throughout with automatic sprinklers, twenty-one persons may be allowed, in the above rule for each 22-in. width of stairways. Most of the modern codes agree as to the requirement that a width of 22 in. of stairs should be allowed for each person. The total width of the stairs must, therefore, be a multiple of 22 in., with a& minimum of 44 in. Enclosed Staircases.—In fireproof buildings, the staircases used as fire exits should be enclosed by fireproof partitions. The doorways leading to the staircase should be provided with self-closing fire doors. The unobstructed width of the stairs, landings, and hallways, within the staircase, should be at least 44 in. If a larger capacity is required, the stairs may be increased in width by increments of 22 in. This means that a minimum width of 22 in. is required per person. In a stairway 44 in. wide, two rows of persons can be accommodated. Tor three rows, the width must be 66 in., and for four rows 88 in. The open side of the stairs must be provided with substantial balustrades. Hand rails should be provided on both sides of the stairs, and are allowed to project into the minimum width, if such projection does not exceed 33 in. _ Stairs 88 in. wide should have an additional hand rail in the middle. The staircase should be free from any other possible obstruction. Winding stairs are absolutely prohibited in staircases serving as fire exits. At least one stairway should extend to the roof. The New York Code requires that if there are more than two staircases, at least two of them shall continue to the roof. An appropriate pent house built of brick or concrete should be pro- vided above the roof. The stairway should lead, in line of direct travel, to a street or open alley or to a hallway leading to a street. The hallway must be enclosed by fireproof partitions. The width of the hallway should 686 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS be sufficient to accommodate all who may use the stairways leading to it. It is important that balconies, used as horizontal exits, be kept free. In winter, all snow must be removed from them. Scuttles.—If no other provision is made for reaching the roof, scuttles should be provided, with a substantial iron ladder fixed in place. Scuttle opening should be at least 2 ft. by 3 ft. Smokeproof Stair Towers.—In tall buildings, one of the enclosed staircases is required to be a stair tower. The height of the building requiring this differs in different codes. The National Board of Fire Underwriters requires a smoke- proof tower or a horizontal exit for buildings over 90 ft. in height The New York Code requires smoke-proof towers for. buildings exceeding 85 ft. in height. In buildings provided with automatic sprinklers, this limit 1s raised to 125 ft. A smoke-proof tower is similar in design to an enclosed staircase, except that the doors leading from it do not open directly into the building but on to an open baleony. The door leading from the building also opens on the same balcony. Both doors are self- closing fire doors. Such an arrangement is seen in Fig. 229, p. 687. The advantage of this arrangement is obvious. The staircase, having no connection with the building, cannot be filled with smoke. LAYOUT OF STAIRS Arrangement of Staircases.—From a structural standpoint, it is advantageous to place the staircases in corner panels or wall panels, as such location offers the least interruption to the floor construction. also, the outside walls are then part of the enclosing walls of the staircase. When elevators are used, the staircase may be advantageously placed next to the elevator wall. Very often, it is possible to accom- modate the elevators and the staircase in the same floor panel, thus interfering less with the typical design of the floor construction. Some of the enclosing walls also serve for enclosing the elevator and the staircase. In many cases, the staircase and the elevator wall are placed outside of the area of the floor, as in Fig. 230. Although this arrange- 3 ment requires more walls, it simplifies the floor construction. SMOKEPROOF STAIR TOWERS 687 S N N u S90“ Interior of Building SSS s4$“s$3$“__o 7° ME QA ©a9» —_— Titi SSS SANS Outside Balcony Solid Floor Plan Elevation Smokeproof Tower with Outside Balcony Entrance j eee ae? : —7 Vestibule Y Opening to - aya ee my Ceiling Y | y Railing- ----}"] | g Floor Line Elevation Railing- “Vestibule Opening Plan Smokeproof Tower with Vestibule Entrance Fig. 229.—Smokeproof Stair Towers. (See p, 686.) 688 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS In factories, the stairs are placed in the service portion next to the lockers and lavatories. This section is usually separated from the working area of the floor. pers es Se aaa 60 Ft, 0 no nm ie AFL OG 2FL Sg SFLO" FLEE or 3FL.O" FLEE 4 FLO" oe - a ee SMa athe ee ——f DOr Ee Bot Ana's PA Dey D3 OS Sng id Tp Ry Oy ES) t A — =s oe Ha Sema WLLL LLZ LLL LLL LLL LL WELTY SS 7 / Door Guards: ue UN at ale ge ch Ft ax 8"X 10! Strut ie —8 Ft. 0 f = ASS -—-Tread ae [<1 —- Riser a Main ye =eaetg ~y Safety Reinforcement - Effective-” / Effective-~ Depth Thickness Details of Stairs Fig. 231.—Details of Treads and Risers. (See p. 688.) Treads and Risers.—The dimensions of the treads and risers will vary with different types of buildings. In commercial buildings, the width of tread, exclusive of the nosing, measured as shown in Fig. 231, p. 688, may vary from 93 to LAYOUT OF STAIRS 689 10 in. The latter figure should be used in all stairs serving as fire exits and also in all stairs that are much used. Secondary stairs, such as those leading to the basement, may have 9-in. treads. It : a well-established rule that the same width of tread should be used throughout the staircase. An exception is sometimes made for stairs leading to the second floor, which are used oftener than the rest, by giving these a wider tread and a smaller rise, as well as greater total width. The height of the riser, measured as shown in Fig. 231, p. 688, should be as near 7 in. as possible, and should not exceed 72 in. The height of the riser is obtained by dividing the distance from finished floor to finished floor by the desired number of risers; and the same height should be used for all risers, even if it happens to be an odd fraction. If the floor heights in a building vary, the heights of risers may have to vary, but this variation must be as small as possible. Special study should be made of the relation of tread to riser, for stairs in constant use, particularly in schoolhouses. If the width of the tread is increased, the height of he! riser should be correspond- ingly decreased. For ease of travel, as well as for appearance, the tread should be provided with a nosing, as shown in Fig. 231, p. 688. Slope of Subsidiary Factory Stairs—In addition to the regular stairs between floors, short stairways are frequently required in the operation of a factory. There is always a tendency to make these stairs too steep. Not only is this tendency observed in cases where it is necessary to avoid occupying too much space or to get in between machinery, but one often finds stairways, leading to a balcony or platform, so steep as to interfere with rapidity and-convenience of travel. This may cause serious loss of time, in case of accident or breakdown of the machinery reached by the stairs. Where space permits, short flights of stairs of this kind should have the same rise and tread as stairs between floors. If this is impracticable, one should adopt the rule, based on many years of practical experience, on the part of the authors, in factory operation and maintenance, that the slope of such subsidiary stairs be limited to 45 degrees. If this rule is made inflexible, some way will be found by the maintenance department to follow it. Stairs steeper than this are scarcely, if any, better than a permanent ladder, which occasion- ally must be put in. When ladders are unavoidable, spacing of rounds should be 12 in. 690 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS Runs and Platforms.—Runs or platforms, designed to give access to shafting or pulleys, or to facilitate the operation of machinery, should be built with at least 63 ft. head room to avoid the necessity of stooping. If obstructions make it impossible to allow this head room, short stairs should be built down and up again into the run, since it is less tiring and safer to walk up and down stairs than to stoop or bend. Intermediate Landing.—If the story height is more than 8 ft., the stairs should be provided with an intermediate landing. (Some codes allow single flight stairs for story heights up to 12 ft.) The width, i.e., the smallest dimension of the landing, should be the same as the width of the stairs. In straight runs of stairs (1e., when the stairs, the landing, and the next flight of stairs are in line), the horizontal distance between risers should be 44 in. Live Load.—All stairs, landings, and hallways should be designed for a live load of not less than 100 lb. per sq. ft. STRUCTURAL DESIGN OF STAIRS Two-flight Stairs.—Figure 232 shows the most common arrange- ment of stairs in industrial buildings. The stairs between two floors consist of two flights with an intermediate landing. ‘They may be designed by any one of the three methods described below. In each case, an inclined slab is used, with flat lower surface and with treads and risers formed on the top of the slab and integral with the slab. First Method.—The inclined stair and the landing slab are con- sidered as one slab, spanning between a beam at the floor level and a landing beam placed at the outside edge of the landing. (See Fig. 232, p. 692.) For designing purposes, the span of the slab is assumed to be equal to the horizontal distance between the two supporting beams. It is then designed as if it were a straight slab. The live load is taken at 100 lb. per sq. ft. of the horizontal pro- jection of the slab. The unit dead load of the stair portion equals the total weight of stairs divided by projected horizontal area. Ginee the stairs are usually poured separately from the rest of the building and are joined to the floor only by stair-dowels, the slab should be considered as simply supported, using M = jul?. The thickness of the slab is determined from Formula (11), p. 208. STRUCTURAL DESIGN OF STAIRS . 691 This thickness in the inclined portion is measured on a line at right, angles to the bottom surface, from foot of riser to bottom of slab. It is marked ¢ in Fig. 232. : The reinforcement is placed at. the bottom, longitudinally with the slab. Some of the bars are bent up and hooked at the landing beam, to prevent cracks due to negative moment. This is an important feature, though often neglected. Cross reinforcement should be used to prevent temperature cracks. As a rule, a 3 bar should be placed in each riser, and three or four bars of the same diameter in the landings. Landing Beam.—The landing beam, marked B3 in the figure, is designed to carry the reaction of the stairs and any wall load that may come on it. It may be suspended from the floor beam above, or supported by short piers resting on beams below. _ In wall-bearing jobs, where the outside walls of the staircase are bearing walls, the edge of the landing may be supported on the wall. No landing beam is thén required. | Second Method.—Beams are placed at both sides of the landing, as shown in Fig. 232, p. 692. The inclined stair slab is then designed as a slab supported by beams B1 and B2; the span is the horizontal distance between beam B1 and the center line of beam B2, and the loads are determined as in the first method. ‘The slab may be con- sidered as partly restrained at beam B2, which permits the use of 2 bending moment o Thickness of the slab in the inclined portion is as explained in the first method. The reinforcement is placed longitudinally at the bottom. Part of the steel may be bent up at beam B2 or special short bars, placed near the top surface of the slab, may be used. Cross reinforcement should be used as explained above. Landing Beams.—Landing beams should be designed as explained above, for the proper loads. They may be suspended, or supported by short columns. In wall-bearing jobs, beam B2 may be supported by the side walls. Beam B3 may be omitted, and the edge of the landing slab rested on the end wall. _ Third Method.—In this method, the landing beam B2 is retained, but B3 is omitted. The inclined slab is then supported by beams B1 and B2, while the landing slab acts as a cantilever of the inclined slab. The slab is designed as follows: The landing slab is considered as (02 STAIRWAYS, FIRE EXITS lardin ng beam’ supported by posts or wail or suspended from beam above. Plan Construction Joint Stair} Dowels Main-~ Reinforcement ‘. Stair} Dowels First Method Main ---~ Reinforcement Negative | -< Reinforcement : Mah) ven hes Reinforcement sy Joint and Recess” Stair Dowels AND ELEVATOR SHAFTS barti tions Balance of Panel Flat Slab Construction Plan- Alternate Design Stair / Dowels ; Main Reinforcement Negative Rein forcement Main Reinforcement ~ sy F Second Method rf fee encased ncrete Cantilever Third Method Fia. 932.—Structural Details of Stairs. (See p. 690.) r Reinforcement (. ‘Hooked at Bean) 7 a eS Sl — STRUCTURAL DESIGN OF STAIRS 693 a cantilever loaded by the dead and live loads on the landing and the wall load, if any. The load produces a negative bending moment in the slab, with a maximum at the edge of beam B2. The inclined stair slab is considered as a slab supported by beams B1 and B2. The span and the loading are determined in the same manner as for the first method. In figuring the bending moment, the moment produced by the cantilever should be taken into account, and two conditions must be considered: one with cantilever fully loaded, which produces maximum negative bending moment; and the other with cantilever loaded only by the dead load, which gives maximum positive moment. | Reinforcement and thickness should be determined for all sections of the slab, the thickness of the inclined slab, as in previous cases, being measured on a line at right angles to the bottom of the slab. The reinforcement should be arranged as follows: Some of the bars should run full length at the bottom and extend to the outside edges of the cantilever slab. The rest should be bent up at about one-fifth of the inclined span and extended near the top to the out- side edge of the cantilever slab. If necessary, additional short bars should be placed near the top to supply the required amount of steel in the cantilever. The top reinforcement should be properly supported by blocks or chairs. It must be remembered that the cantilever slab relies entirely for its strength upon the top reinforcement. Any mis- placement of this would materially reduce the capacity of the slab. Short rods should extend a sufficient distance from the cantilever into the inclined slab to develop full strength of the bar in tension. Landing Beam.—The landing beam must be made strong enough to carry all the load coming upon it. This will comprise the reac- tion of the inclined slab, consisting of one-half of the load upon it; the reaction of the cantilever, equal to all the dead and live load; and the reaction of the wall (if any). To these should be added the reaction produced by the cantilever, which equals the maximum bending moment on the cantilever divided by the horizontal span of the inclined slab. As in previous cases, the landing beam is either suspended or supported on short piers or side wall. Three-flight Stairs—Stairs are often constructed with three flights, as shown in Fig. 233, p. 694. The construction beeen two floors consists of three inclined slabs provided with two intermediate landings. 694 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS Y /\> A d 1/ ea ecule S 4h A 2. b -==0 Gs Site A Ad ji SIE ae eae aia ot =o NIG, LES ASSIA EGS Ep CLE ~— ] uw eer gran io oe i ite ~9.4"- * us K Plan of Second Floor Third Flor = Detail of Stair Tread and Nosing Solid Brick Wall,- On Mm « R : me w& Y aes —s ig ehh 1 Se t | | i second Floor § ee . Boy iat Dowels Sectionon Line A-A Section on Line B-B Frag. 233.—Stairs with Three Flights. (See p. 693.) be aL. ELEVATOR SHAFTS 695 In the typical design shown in Fig. 233, the stairs are supported by a floor beam and by the brick work at the landings. The two main sections are considered as slabs simply supported by the floor beam and by the wall. The reinforcement should extend the whole length of the slab to the edge. As shown in Fig. 233, in flights leading from the beam to first landing the bars are not continuous but are lapped at the juncture of the inclined slab and the landing. A continuous bar would have to be bent there in such a way that under tension there would have been a tendency of the bars straightening by pushing out the concrete in front of them. The third section may be considered as spanning between the walls. Sometimes the landings have no support on the sides. Then for designing purposes, the third section is considered as a simple slab supported by the landings. There is no definite point of support. The length of the span to be used in computations must therefore be determined by judgment. Ordinarily, this may be taken as the horizontal projection of the inclined slab plus 2 ft. The bend- Pate ing moment is found from formula = The reinforcement should be extended on the bottom from the outside edge of one landing _to the outside edge of the other landing, thus providing for any cross bending in the landing. The two main sections are considered as rant slabs supported by the beam on one end and by the wall on the other end. These _ slabs should be designed for the load on the slab plus the reaction of the third section. This reaction may be considered as distributed over the whole width of the landing. The reinforcement should extend the whole length of the slab to the edge of the landing. Cross bars should be provided in the treads, as in the other case. SPECIAL DESIGNS OF STAIRS Stairs may also be constructed by providing beams at the edges and then supplying a thin slab between the supporting beams. Such construction is seen in Fig. 204, p. 630. - ELEVATOR SHAFTS The elevators should be placed in groups and where possible near the staircase thereby reducing the number of places where the continuity of the floor is interrupted. Where staircase is placed 696 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS outside of the building, as shown in Fig. 280, p. 688, the elevator | shafts should also be so placed. Elevator shafts always require special framing. To simplify the construction, they should be located in such a way as to get the simplest framing possible. Sometimes the cost of framing may be reduced by introducing additional columns at the corners of the shaft. This is particularly advantageous in flat slab construction, because it avoids framing of beams into main columns which requires cutting of forms for column heads. The supplementary columns, of course, must be placed outside of the shaft. Sometimes the enclosing concrete walls of the shafts are made strong enough to act as bearing walls, thereby saving the cost of framing. The size of the shaft must be large enough to accommodate the elevator car with the clearances required by the manufacturer. The shafts must be built with care. All beams must be plumb under. each other, otherwise the clear space of the shaft is reduced. To prevent accidents, the gate sills should be provided with a nosing projecting into the shaft. Usually the sills are also provided with safety treads. Elevator Pit—A pit at least 3 ft. 6 in. deep must be provided below the lowest floor served by the elevator. For all elevators serving the basement, the bottom of the pit is below the basement slab. Sometimes elevators stop at the first floor, in which case the pit may be provided by depressing the first floor a proper distance. The pit must have the same clear dimensions as the balance of the shaft. ‘Therefore, no part of the footings or columns must project into the pit. Ordinarily the pits are supported by the ground. The sides and - the bottom slab are made 6 in. thick and are properly reinforced to resist earth pressure and also to prevent temperature cracks. Where the ground is not capable to support the construction the pit walls must be carried to the columns or supported on piles. Special anchorage is required when there is possibility of upward water — pressure. When built below water level, the pit should be waterproofed. Pent House.—The construction above the elevator shaft is called the pent house. This contains the supports from which the elevator is suspended and for modern elevators also the elevator machinery. The pent house, usually above the roof is properly enclosed and = ELEVATOR SHAFTS oe OS VEG So See SE pee (i ES LES Sa ia aoe SON ~------6+/9"-- ——— ee ee! Grating —----—,f/-'--+-- = <---- Section: oe ee en ae ---]/'-Q'> Se ee we wwe oe ee em rere Hee ---47-/Q"- = ewewe sore esses oes oc os = 65-9" === wee eme ne eee Be —]"-- mm Loe Plan -o---92tl|"=-- fe. ---------//'-'1------------- ae a ere er ere eo eee Section Fig. 234.—Details of Elevator Shaft. (See p. 698.) 697 698 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS covered with a sky-light. The size of the pent house depends upon the disposition of the machinery. Also there must be enough room in case of necessary repairs or replacements. The beams in the roof must be made strong enough to carry the weight of the pent house walls and roof and also of the elevator, counterweight and elevator machinery. These weights are usually heavy and must be well taken care of. To take care of the elevator loads 100% must be added to the actual loads for impact. Grating of proper strength must be provided below the machinery. Example of Elevator Shaft Details of elevator shaft are shown in Fig. 234, p. 697. This also shows the position of sheave beams and the reactions for which the supports must be figured. The loads given include an allowance for impact. ii ee a a CHAPTER XV STEEL WINDOW SASH Steel window sash is used in the majority of commercial rein- forced concrete buildings. The width of concrete pilasters is often determined by the width of the -window opening, which in turn is governed by the dimensions of the standard window sash. The dimensions of the recesses to be provided in the concrete are also fixed by the sash. Therefore, to do this work intelligently, the designer should familiarize himself with the design of the window sash, its standards, and the methods of erection of sash in the build- ings. | Standards of Window Sash.—The dimensions of sash are stand- © ardized as to design and dimensions, and sash of other dimensions than the standard can be obtained only at a considerably larger cost. Dimensions of windows therefore should be suited to standard sash. Steel Sections Used for Window Sash.—Window sash is built of specially rolled, solid, low-carbon steel sections. These are strong and rigid enough to withstand all the strains to which the sash is subjected in handling and in use. The sash frame, however, is not strong enough to carry any brickwork above the window— The following sections are used in window sash: Angle frame members, or outside rails compose the frame of a sash unit. Muntins are sections placed horizontally and vertically between the frame members. Weathering members are used at the ventilators to make the sash weather-tight. Mullions are sections used in connecting sash units to form a com- plete sash. | Composition of Sash Units.—A steel window sash consists of one or more sash units. The rectangular frame for each sash unit is built of four angle frame members. Between these, muntins are placed horizontally and vertically so as to divide the frame into the 699 700 STEEL WINDOW SASH required number of lights. A sash unit is often provided with a ventilator. Fig. 235, p. 701, shows a sash unit and the cross section of the various steel sections composing it, and also the construction of the ventilator. As evident from the figure, the ventilator is a specially con- structed small frame with the top and bottom of regular angle frame sections. The sides of the ventilator frame consist of muntins, to which, above the pivots, are attached steel weathering members. The top and bottom of the opening in the sash unit, to receive the ventilator frame, are built of angle frame members, and the sides consist of muntins with weathering members attached on the inside of the sash below the pivots. The hinges of the ventilator frame project beyond the plane of the sash; in some designs the hinges form a part of the weathering members. In other designs they are part of members that are double riveted to the ventilator and the fixed sash. The pivot passing through the hinges is placed 2 in. above the center of the ventilator opening, to insure proper balance. The ventilator is operated by either stay bars or chains. The stay bar is attached to the ventilator at the bottom and is provided with notches to permit any degree of opening, and when not in use it is locked to the frame by a catch. When the ventilator is provided with chains, a catch is attached to the bottom member of the venti- lator and a pulley to the top, while a clip is attached to the main frame to hold the chain. Composition of Window Sash—A complete window sash may consist of several sash units, all of which are of the same height but not necessarily of the same width. The units are connected by T-shaped mullions (see Fig. 235, p. 701), each mullion being bolted on both sides to the vertical frame members of two adjoining sash units. ‘To allow for adjustments in any direction, the holes in the mullion, which correspond to the holes in the frame member, are oblong in vertical direction, while those in the frame are oblong in horizontal direction. The mullion extends at the top to the edge of the opening, and at the bottom its flange extends to the bottom of the sash member, while its stem extends down 2 in. below the window opening. (See Fig. 237, p. 703.) Relation of Wall Openings to Sash Dimensions.—As will be noted from the details, the sash is secured in the window opening of a concrete building by grouting in a part of the flanges of the frame RELATION OF WALL OPENINGS TO SASH DIMENSIONS 701 members and the lower portions of the stem of the mullion. The size of the window opening, therefore, differs from the outside dimen- sions of the sash; but to avoid confusion, the sash is always desig- nated, not by its outside dimensions, but by nominal dimensions which correspond directly to the size of the window opening. In a window consisting of a single sash unit, the wall opening is identical with the nominal dimension of the sash unit. For windows con- sisting of multiple sash units, the width of the opening is equal to a uel \ ie 1 Ss Ma UL ie Sy" acta fe i S Se ad ee RO vox": . Se: % er Loma 90 C9 G w iS oN: LS oe: SS be c \ | ae Sere y Width of opening PD eect eet (For 14% 20" glass ) | When desired, stem of mullion can be turned in ~ if as shown by dotted lines ¢ | e sep Sash Unit----=2 i Ay ue Larimer Sash Umtt----- 2" \ Mullion Fig. 235.—Window Sash. (See p. 700.) the sum of the nominal widths of all the sash units, plus 2 in. for each mullion. In Fig. 235, a horizontal section through a sash illustrates the relation between the opening and the sash dimensions. The relation of the various parts is evident. The relations between the parts of the sash and the total dimen- sions are as follows: a = distance, center to center, of muntins = width or height of glass plus # in.; A = distance, center to center, of inside leg of frame members = a multiplied by number of lights. 702 STEEL WINDOW SASH Width of sash unit = A + $ in. Width of opening = sum of widths of all sash units, plus 2 in. multiplied by the number of mullions. Height of opening = A + % in. (in figuring A, a is height of glass plus 3 in.). Recesses for Receiving Sash.—In concrete and brick construc- tion, some of the frame members extend into the concrete or brick- work and are secured therein by grouting. In concrete, the necessary recesses or grooves are formed by nailing, to the formwork of the beams and columns, strips of wood of proper dimensions. (See Fig. 236b, below.) These strips are removed after the concrete has hardened, and form the required recesses in the concrete. In brick wall, the frame angles are inserted into a groove between bricks, as shown in Fig. 236a, below. For windows consisting of a single unit, 1 inch reveal is required at the jamb, as shown in Fig. 236¢. Ax< Width of Sash nr 5"| Width of Sash | and Opening and Opening \ gu pes: a. C Jamb for Brickwork dJamb for Concrete Jamb for Concrete Multi - Unit Openings , Multi - Unit Openings Single Unit Openings Fia. 236.—Details at Jambs. (Se2 p. 702.) Erection of Sash—The sash comes to the job in fully assembled units. Each unit is placed in position separately, the units next to the jamb being sct first. The top flange is moved into the top recess, and the side flanges of the frame are moved into the groove or recess of the jamb, and the remaining units are then set with their mullions. The mullions are first loosely bolted to two adjoining frames, the flange of the mullion always being placed outside of the frame. The stem of the mullion, in concrete construction, may be placed either outside or inside of the sash, as desired. | When sills are built in place, the brickwork or concrete of the spandrel is erected only to the bottom of the sill. The sash is then set and is supported at the bottom on blocks of proper, depth, which should be placed at the corners only, to prevent distortion of the GLAZING 703 sash. After all units are assembled, the window is lined up and plumbed. The mullion bolts are tightened and the sash is securely wedged against the jamb and the head. The sill is finished, and then all recesses are grouted. To keep the sash in place, wall ties are attached to it and concreted into the sill. (See Fig. 237, p. 703.) If stone or pre-cast concrete sills are used, they should be placed before the erection of the sash is begun. The sill should be provided with horizontal grooves for the stem of the mullions, and also with holes for expansion bolts for fastening the sash to the sill. The sash is erected from the inside. In setting, the top of the sash is first inserted into the groove or recess in the head. The sash is then Grout } Expansion ; Bolt eight of Sash hake " nd Opening Grout (ete ea _-Mullion a = és >S Q ~D>> <= as av Drs = eo eee ae - We pee on Solid Cut Stone or acs) el eee oe ear Precast— Poured / Precast Sills Poured in Place Sills in place Back of Sill Poured in Place Fig. 237.—Details at Sills. (See p. 703.) lifted sufficiently to clear the sill completely, and swing in position. Often, the sill is made of two parts, the outside part of stone, with the inside part cast in place. The stone sill is placed before the erection of the sash is begun. After that, the erection proceeds in the same manner as with cast-in-place sills. (See F ig. 237.) Painting.—Sash is always painted with one coat at the factory. After erection, the sash should receive two additional coats. Glazing.—The lights are inserted from the inside after the sash is erected. The glass is held in place by wire clips, four per glass. It should be carefully back-puttied in such fashion that the glass does not touch the steel. Putty is also applied on the inside. Special putty is used for steel sash. 704. STEEL WINDOW SASH Symmetrical Combinations of Sash oe a Glass Sizes i peers 18” 14x 20" Lights Height Lights Height ee pee eh Hite Number 2 3’ 13” 2 ram wales 9 3 4! 8” 3 or ae ec 4 6’ 23" 4 e109” | Wiahre 5 vie 83" 5 8/ 63” pening 6 OSS Me 6 LO see The over-all widths | The over-all widths shown below may be | shown below may be had in any of the|had in any of the above heights. above heights. ay PAM 3/ aes 3 4! PEYE 4! 102” 4 5/ aie 6’ 03” 5 6’ 67 Vf Go 6 g! 62” 9’ 103” g Nee Be 9 10°F Peo 12h eas” 10 10) 102” 12° 2G 10 11 lee 1S di 12) 41,” pO Re 12 1S Lip en ash 13 PESOS Ok i Ci dee add 13 14) bi Te Sy 14°, 1502235 LA 14 16g 0a 18 Gay 15 Wie wart 197) Vise 16 19° > 47" 23 ae” 18 pi Waar a ae Oe 19 phloem ee 24 ea 20 Number of Sash Units to Fill Opening PRoOARRWRWWWWWWNWNNHREE Number of Lights in Each Unit STANDARD SIZES OF SASH 705 STANDARD SIZES OF SASH Glass Sizes.—Standard glass sizes are 12 by 18 in. and 14 by 20 in. Glass can also be had in 10 by 16 in., and 16 by 22 in. sizes. The glass sizes used in ventilators are different from those used in the rest of the sash, since the ventilator frame is reduced one inch on each side. In ventilators consisting of four lights only, each light is reduced by one inch, both in width and in height. For ventilators of more than four lights, the outside lights are reduced as above, while the inside rows of lights are one inch less in height but the same width as standard glass. For fireproof buildings, wireglass should be used. Standard Sash.—Most manufacturers have standardized the dimensions of the sash. The dimensions of the window opening, therefore, should be such as to fit the dimensions of the standard sash. The standard sash is of two types: Warehouse Stock Sash and Standard Sash. Warehouse Stock Sash is carried in stock, assembled, at the warehouse. Standard Sash, while it consists of standard members, must be assembled after the order is received. There is ordinarily no difference in price between the two types, but since better deliveries can be had on Warehouse Stock Sash, it should be used in preference to the non-stock Standard Sash. Special sash can be manufactured, but its use should be avoided whenever possible, as the price is much higher and a much longer time is required for delivery. Design of Window Opening.—In designing the window opening, select first, the size of glass to be used, preferably from the two standards in general use, namely, 12 by 18 in. and 14 by 20 in. The larger glass gives the cheaper window. Estimate the approximate size of the opening, from the span and the desired width of pilaster. Decide tentatively on the number of lights and number of sash. Then compute the exact size of opening from formula given on p. 702. The number of ventilators then should be decided upon. The problem may be easily solved by referring to the table on p. 704, which gives combinations of different standard sash sizes. 706 STEEL WINDOW SASH UNDERWRITERS’ PIVOTED SASH Under some conditions, where it is especially important to make the window fireproof, Underwriters’ Sash is used. The Underwriters’ Sash differs from ordinary sash, mainly in that the glass is held on the inside, in addition to putty, by steel glazing angles 4 by 7 in., with the long leg bearing against glass and secured to the sash. The ventilators are provided with a chain con- taining one or two expansion clips at the top and a fusible link. When a fire occurs and the fusible link fuses, the ventilator closes and locks automatically. | CHAPTER XVI STRUCTURAL PLANS FOR BUILDINGS A set of plans for a building may be divided into several distinct parts. In the present discussion, the architectural and the struc- tural plans only are considered. The architectural plans consist of floor plans, sections, and ele- vations. They give an idea of how the complete structure will look. Usually, the structural features are only indicated. The dimensions of the structural members are given only when they affect the archi- tectural features or when, for any reason, they are limited as to size. The structural plans show the size and arrangement of the struc- tural members. The portions of the construction which do not con- tribute towards its strength, and which are built after the structural frame is erected, are usually omitted. Thus the piping,.the location of partitions, unless built of concrete—brickwork in skeleton struc- tures, and similar features are not shown on the structural plans. Inserts also are shown separately. Structural plans are of two classes: 1. General plans, forming a part of the architect’s set, 2. Detail working plans to be used in the field. GENERAL STRUCTURAL PLANS General structural plans are prepared before the contract is awarded and form a part of the contract drawings. The com- pleteness of these plans varies with the practice in the offices of the various engineers and architects. 1. Complete Structural Plans—Some firms prepare complete structural plans, giving the concrete dimensions and the required reinforcement for all members in the structure. 2. Typical Design Only—In many instances, the general struc- tural plans give only the design of typical members. The specifica- tions give the live loads and the allowable stresses. The actual 707 708 STRUCTURAL PLANS FOR BUILDINGS design of all members is left to the contractor’s engineer, with the understanding that the methods outlined in the design of typical members will be followed and that the plans are subject to the approval of the architect’s engineer. 2 Floor Construction not Designed—Often in flat slab construc- tion, the footings, columns and beams are either fully designed as in Case 1, or indicated by typical designs as in Case 2; while the selec- tion of the type of floor construction or system of flat slab reinforcing is left to the contractor, with the understanding that the selected design must be acceptable to the architect or his engineer. The live load, stresses, and bending moments are specified. 4. No Structural Plans Shown.—In a large number of instances (the majority of buildings), the architectural plans submitted for bids do not include any structural plans. The live loads are specified. Sometimes specifications also have requirements as to the stresses and bending moments. In cities having building codes, this is omitted and the specifications call for a design which would satisfy the requirements of the building code. The structural design is left to the contractor. | The method to be adopted in any particular case depends upon circumstances. Theoretically the first method, that of preparing complete struc- tural plans, should give best results. Actually, however, this method is recommended only where the organization preparing the general plans is composed of engineers well trained both in theory and field practice of reinforced concrete construction. | Under ordinary circumstances the engineers of architects’ offices cannot specialize in concrete construction, but must deal with all structural materials as necessity arises. While such experience tends to broad knowledge of structural engineering in general, it precludes the complete mastery of details of reinforced concrete. Therefore, either the second or the third method will give much more satisfactory results, especially when the working plans are checked. The method where no structural plans are shown, is not recom- mended unless the working plans are checked by a competent engi- neer employed by the architect or unless the working plans are pre- pared by an engineer selected by the architect. If the matter is left entirely to the contractor, the safety of the structure depends alto- gether upon him or upon the engineer engaged by him. It is obvious GENERAL STRUCTURAL PLANS 709 that this method either through intent or negligence may lead to unsafe designs. It is absolutely essential for the safety of the con- struction that the architect or owner employ an engineer in whom they have confidence, rather than to rely altogether upon the engi- neer of the material man or contractor. Often reliance is placed by architects upon the fact that the designs are checked and approved by the Building Department. In practice only the largest cities maintain engineers skilled in concrete construction. Even there mostly the checking is not complete but is confined to main features. In small cities the approval of plans is perfunctory. Purpose of General Structural Plans.—The purpose of the gen- eral structural plans is: 1. To enable the contractor to estimate the required amount of concrete, steel, and formwork. 2. ‘To serve as a positive guide in preparing the working drawings. These objects are not attained unless the plans and specifications show clearly what is wanted, particularly as regards the points upon which there is a possibility of a difference in opinion. If the engi- neer preparing the general plans does not make his intentions clear, the design, in these details, is left to the engineer preparing the working plans. . General Instructions.—Kach office should establish rules govern- ing the method of preparing structural drawings. To make the drawing clear, the lines having different meanings should be of different thickness and style. The following rules are suggested: Building line, dash and dot line, dashes about 2 in. long, medium thickness, diluted ink. Center lines of columns, light long dash and dot lines, diluted ink, dashes over one inch long. Dimension lines, light solid lines, diluted ink. Outline of concrete, solid black lines, medium thickness if visible, dash line of same thickness if invisible. Reinforcement, heavy solid lines in elevation. In section use solid circles for round bars, solid squares for square bars. Sections of concrete members should either be shown as concrete in the conventional way or shaded on the back of the tracing with a soft pencil. Care should be taken that the shading is not so heavy as to obscure any lettering nor lines of reinforcement. Objection may be raised to showing reinforcement by solid lines. It is usually invisible in the concrete, and dash lines would be more 710 STRUCTURAL PLANS FOR BUILDINGS logical. However, (1) dash lines require much more time to draw, and (2) in many cases when bars meet at an angle, dash lines do not make clear just where each bar goes. Solid lines, therefore, are preferable and should always be used. COMPLETE GENERAL STRUCTURAL PLANS Complete general structural plans consist of foundation plan and footing schedule; floor and roof framing plans with girder, beam, and slab schedules; and column schedules. The architectural plans usually show sufficient general sections and elevations, so that these are not repeated in structural plans. Special partial sections, however, will be found necessary. Numbering of Columns and Footings.—Two methods of number- ing columns and footings are now in general use. In one method, the panel pomts on the floor plan are numbered by consecutive numbers, 1, 2, 3, 4, etc. The numbers are then used to designate the columns and footings at these points, and so Cl, C2, C3, C4 would indicate columns 1, 2, 3,4 and FI, F2, F3, F4, etc., corre- sponding footings. These designations are used in column and footing schedules. This method is simple, but it has the drawback that it is often difficult to locate the columns on the plan, especially in a large or irregular building. Also, duplication of numbers is possible, especially when new columns are added after the plans are completed. In the second method, which is preferred by the authors, the mark of the column is fixed by its location. This is accomplished by the use of coordinates in the following manner. The main column lines that are horizontal on the plan are designated by small letters. The top column line is called a, and the others, in alpha- betical order, b, c, d, e, ete. The main column lines that are vertical on the plan are designated by numerals, 1, 2, 3, beginning at the left. The panel points, located at the intersections of the two sets of lines, are designated by the letter and the number belonging to the inter- secting lines. These designations are used for columns and footings. For instance, the panel point at the intersection of line b with line 3 is called b3, the column located there is called C—b3, and the footing F-b3. | Special column points located between the main column lines are designated by the addition of 4 to the number or a prime sign to COMPLETE GENERAL STRUCTURAL PLANS “11 the letter. Thus, panel point b3% is located on line b between lines 3 and 4. b’3 is located on line 3 between lines b and ec. This method of designation fixes the position in the building of every panel point. It is, therefore, preferable to the first method. Some contractors claim that the use of a letter and number is con- fusing in the field, but unquestionably the advantages overbalance any possible disadvantage. Numbering of Beams and Girders.—Typical beams and girders are designated by a number indicating the floor, a letter, B, G, or L, designating a beam, a girder, or a lintel respectively, and a second number indicating the distinctive beam or girder on the floor. Thus 2B3 would mean beam 3 on second floor. On one floor, there are | usually a large number of beams of the same span with the same concrete size and amount of steel. All these beams are given one common designating mark, which is marked on the floor plan in each place where the beam occurs. The dimensions of the beam are given in the schedule. It is advantageous to maintain the same designating marks on all floors, so that a beam 3B3 in the third floor will correspond to a beam 2B3 in the second floor and 5B3 in the fifth floor. Where the framing in the various floors is sufficiently different, this latter plan is not workable, and beams should be numbered on each floor without regard to numbers in other floors. Where the panel points are designated by coordinates, as explained in the previous paragraph, the odd beams and girders, which occur only in one place, may be designated by their location. .For a hori- zontal beam, the panel point at its left extremity is used; and for a vertical beam the panel point at its top end. Thus, a beam on the second floor, starting from the intersection of lines d, and 4, may be called 2B-d4. In a.building having vertical beams, the location may be still more definitely fixed by adding h for horizontal beams and v for vertical beam. Thus, 2B—d4v is a beam in the second floor, starting from panel point d4 and running vertically down. For intermediate beams, the prime sign and + may be used, as explained in connection with designation of columns. For illustration of this method, see Fig. 239, p. 714. This method of designating beams has been adopted as standard by several large contracting firms. It is of special advantage in the drafting room. If all beams are numbered consecutively, on the other hand, and in the process of design some odd beams are elimi- nated, the remaining beams must be renumbered or else the sequence 712 STRUCTURAL PLANS FOR BUILDINGS is destroyed. Also, beams are sometimes added after the schedule is finished, so that, for instance, beam B45 may be in the same panel with beam B3. This makes the checking difficult. FOOTING SCHEDULE (For Stepped Footing) st Sten | 2nd. Step], ]_Reinforcemert poner ¥ B edesta emarksS eres: C it — 6-6"| 420" | 5" B6'Sq 20-2'9 |22-¢'9 | Gd A NN NUNN Qa ae Stepped Footings Sloped Footings Fic. 238.—Method of Showing Independent Footings. (See p. 713.) Foundation Plan.—The foundation plan, usually drawn to a scale 1 in. = 1 ft., shows the location of the footings and basement walls. The elevation of the basement slab may also be shown on the foundation plans. This is especially desirable where parts of the basement are lower than the rest. When the condition of the ground is known, the elevation of the COMPLETE GENERAL STRUCTURAL PLANS 713 footings should be given. This helps in estimating the amount of excavation and the length of the pedestals. The outline of the footings may be shown on the plan. The rest of the information necessary for independent footings may be given in a schedule. A good form for a footing schedule is given below. ‘This in conjunction with the sketch, gives all information required for independent footings. It is evident that the same schedule may be used for square and rectangular footings. There is more difficulty in showing combined footings. Schedules will be found insufficient in this case. Details of combined footings should be shown in a sketch, which should give the concrete dimen- sions and the required steel. The reinforcement should be clearly shown by elevations and sections, in the position in which it is to be placed. This is important, as the design of combined footings is not sufficiently standardized. It is obvious that it would not be suffi- cient to give the number of bars only, without showing their location. If there are several combined footings of the same type, one sketch may be used for all. The various dimensions may be indicated by letters, and a schedule made up in the same manner as for independent footings. The length of the bars should be shown in the sketch in terms of the span. The determination of the actual length of bars may be left for working drawings. | The design of the wall should be shown by sections, which should be provided wherever there is a decided change in the height of the wall or the type of reinforcement. The reinforcement should be shown in such a way as to make the intentions of the designer clear. If horizontal bent bars extending into adjoining spans are used, one typical bending sketch at least should be shown; or a note should state clearly the design desired. Floor Plans.—Floor plans may be drawn to a scale 4 in. = 1 ft. and in some cases to a scale 7; in. = 1 ft. The larger scale is pre- ferred where the floor arrangement is at all complicated. Where the arrangement of beams does not vary from floor to floor, one floor plan may be used for all typical floors, even if the live loads are not the same. The difference in size of beams or in rein- forcement is taken care of by the schedules. For any floor in which the bulk of the construction is typical and special arrangement is restricted to a small portion of the floor, the typical floor plan may be used in conjunction with a partial plan of the special section. The partial plan should be properly tied to the 714 STRUCTURAL PLANS FOR BUILDINGS J " Hee By Ee opg2H 2BetH 2B4 284 OB4 86a > bio Pe oe) eo red i . am? | : Qe- kK = ‘ a ‘ - ls aC ' we o> -- ; Flat Slab Schedule /st. to 4th f/oor Slab &" I" Granolthie Jinish, bonded. Concrete 1.2/4 Column _ Head sie ~ 3" diameter; Drop panel 4"x 6-10'X6: 70” ile 10°-0"bia. | 172 9"Dia. nits leet aA a la Ne a ON ae ce a a Fy ae z0" x A124 017-8" 5) 72 120-61 1-8 226-101 =" ns| a see ses] || | dT | | cia ee 0 eee oo ic\/2| 7-4 |— 16 |77196"| — woz 36 | — wol2aieo| — wel | | | | TT Lg pilegt sah 9 2640"#3 dete 8" 65] 98127-74227" #5) Yo 33-61 2-77 #5) 439-31 2-75) 2 44-0} 1-8 S| G27 OWA Ye 4-512+1"| (FSSA ee MAUD: Bit Ne) ~ |Tep.|_Radials |Rings 30a 6-0"Dia. | 7*4"dia. | 8-8"Dia._| 10"0"Dia.| | | | | Lt WBS ence See Lap alice al La cos ae as ile Foti 1 fe fanged eaueiroleerapetr acre wiz eure Wiles coer a seared | tt |__| Cur | £25 GS i194 6-9"| — | iy izo— eo — pee er — we es eee lei = mre = pes Maes — medio at tI : SRD MER EE ie FS ee NOTE Sketches for bars not shown They are similar to those BIs> gale in ae ee Page 369. 7 “H Hf es Otss sera Te ” -0 —-19 aa “Thi No ——-19'-0 GY > i ed Fhe -19% Ot eo tig: pee 0"-—> Fig. 242.—Partial Plan sag and Sections f Section /-/ Section 2-2 —Brick Section #Slab building. Smulski System. fea oie! (See p. 719.) (To face page 719.) i ty . to Cg inal D2 to Dg ind. £,to£gind Col load in 000% Column Size Core S/ze oS S12 ; Ny la + 52 yy. A . 3 Vertical Bars Spirals or Ties .| 298 S a | 36°20" in | CoreSize [| 32%/6" ki Ver tica/Bars|~ 2" . Spirals or Ties ‘Core Size Vertical Bars Bas. Spirals or Tres S _| Dowels to Pedesta/ Top of | Footing 56" A Warrable Notes:_ Give thickness required for fireproofing Give length of spirals in respect to heights Give required lap for vertical bars A Give any other information reguired by special conditions Fig. 243.—Column schedule. (See p. 720.) center units by letter B, column head units by letter C, and bars across unit A by letter T. A number placed before the letter may be used to designate the story in which the unit is used, while a number following the letter serves to differentiate between varying units of the same type. (See Fig. 242, opp. p. 719.) 720 STRUCTURAL PLANS FOR BUILDINGS Column Schedule.—The column schedule should give concrete. dimensions, concrete mix, number and sizes of vertical bars, spacing, and size of column ties. For spiral column it should give diameter of core, size of wire, and pitch of spiral. The height of column also should be given in the schedule, as it assists the estimator in getting his steel tonnage and is an aid in preparing general working drawings. The load for which the column is designed should also be given. Norss should give all the other information not contained in the schedule. | The thickness of concrete for fireproofing should be clearly shown on a sketch or covered by a note. It is necessary to state how much lapping is required for column steel, and how high the spiral should be carried. A sketch of the column ties should be shown, particu- larly in cases of unusual columns. \ WORKING PLANS The difference between general structural plans and working drawings is evident from comparison of the purposes the two sets of plans are intended to serve. The eeneral structural plans are ‘ntended for making up estimates and as a guide for preparation of final working drawings. The working plans are intended for use in the field. They must be complete and exact. Dimensions of all concrete members must be given. Not only the number and size of reinforcing bars, but also their length must be given. Recesses and Dowels.—It is important to show on the working plans all recesses, dowels, and nailing blocks required in the concrete work. Recesses are required, for example, for window sash, while at the beams in staircases recesses and dowels should be shown. When concrete walls are used, dowels may be required in floor slabs to connect the wall to the slab, and also pour holes through which the wall concrete may be poured. Nailing blocks are required to nail the flashing and the doors, and sometimes at .the windows. The importance of providing recesses and nailing blocks is evident, as drilling holes in solid concrete adds unnecessarily to the cost. Some contractors do not show any recesses On the drawings, but rely upon the superintendent or carpenter foreman to make proper provisions. This practice is not satisfactory. Inserts.—Special plans should be made for the location of inserts. An attempt to show them on the floor plan, with other structural —— WORKING PLANS | 721 data, would result in confusion. Usually, details of typical plans are sufficient. | Building Lines.—To facilitate laying out the building in the field, the plans should be tied to a set of lines called “ building lines.’ These dre straight lines for each face of the building, which usually coincide with the main outside face of the building. For rectangular buildings, the building lines are rectangular; for polygonal buildings, they are polygonal, and so on. The building lines are established on the ground, and all measure- ments are made from them. All recesses in the wall, for instance, are measured from the building line. The outside faces of the con- crete for beams and columns are also located from the building line. The building lines should be shown on the floor plans and in sections through the wall and wall beams. In sections, the building line is indicated by a vertical line. The distance from outside face of concrete to this line should be given. For instance, in a brick veneered building the building line would coincide with face of brick. The concrete work is then 5 in. from the building line. Foundation and Footing Plan—The information regarding numbering columns and footings, on p. 710, applies to working plans as well as to general plans, and all should correspond. The foundation plan should have all the information required for the laying out, in the field, of all footings and basement walls. Footings may be shown in the manner described in connection with general plans (see p. 712). Outside footings should be tied to the building lines. Footing schedules should contain, in addition to the information given on general plans (see p. 712), the number of footings of each type and the number and length of bars in each footing. For com- bined footings, bending sketches, completely dimensioned, should be given for each type of bent bar. The position of all bars in the footing should be definitely fixed, in a clear manner, either by a detail drawing or a note. The column dowels should also be given in the footing schedule, rather than in the column schedule, because the steel is required at the time the footings are built. The necse- sary information as to the spacing of dowels should be given. In spiral columns, this is fixed by the diameter of the column core. The basement wall should be shown by sections and, if necessary, by elevations. Sections should be taken, not only for typical rein- 722 STRUCTURAL PLANS FOR BUILDINGS forcement, but also where the height, thickness, or amount of rein- forcement differs from the typical. The section should show height of wall, thickness, reinforcement, distance from building line, and all other information required in the field. Not only the size and the spacing of the bars should be given, but also their length and their number. The position of the first bar should be fixed. If bars are bent, a bending sketch should be given. The position of the bent bar should be shown in elevation. This is of particular importance where the bars are placed in an unsymmetrical position ; thus, in end panels, it is necessary to show how far into the column the bars should extend. Sometimes the different panels of the wall are given designating marks, and each panel is marked. A schedule, similar to that used for slabs, gives the required ‘nformation as to thickness of wall and amount of reinforcement. Elevations of the basement slab should be given. Sections should be given where the elevation of basement wall changes, showing the retaining wall. All pits should be clearly indicated, with dimensions and their locations tied either to a building line or a column line. Cross sections through pits should show depth, thickness of wall, and details of reinforcement. Sizes of all openings in the wall should be clearly given, and openings should be located with respect to column lines and floor elevations. Floor Plans.—Floor plans should be drawn to a scale } in. = 1 ft. If floor construction is complicated, a larger scale should be used. Floor plans show the position of beams, the slab reinforcement, and sections. Building lines should be shown on the floor plan. Where the center line of beams does not coincide with the center line of columns, the offset must be clearly shown. This is best accomplished by sections. Not only typical sections through the spandrel beam should be shown, but also sections in all places where the concrete dimensions or the distance from building line differ from the typical. All openings in the floor must be clearly dimensioned and should be tied to the adjoining center lines of columns or to building lines. When the floor plan is too complicated, especially if it shows slab reinforcement, the concrete dimensions and the details of the open- ings should be shown on a separate plan or a partial plan. This applies particularly to stair and elevator openings, which are usually WORKING PLANS 723 shown to }-in. scale in connection with stair details. Where recesses occur, they should be indicated in plan or in section. Beam Details.—All beams should be marked on the floor plan by the method explained on p. 711 in connection with general plans. Beam schedules should give complete information. A convenient schedule, more complete than the one suggested for general structural plans, gives the following data: number of beams; more complete information on concrete dimensions; not only the depth and width of the beam, but also the depth of the slab on each side. If the section of the beam is irregular, so that full information cannot be conveyed by the schedule, a cross section through the beam should be shown. | The information on reinforcement, also, is more complete than for general plans. Not only the number of bars but also their length is given in the schedules. To insure proper location of the bars in a beam, a detail of the beam with full reinforcement, including stirrups, should be given (see Fig. 240, p. 716). This is particularly necessary where the reinforcement is not symmetrical. In the end beams, this should show how far the hooked ends of the bars must extend into the column. Some designers give this information in connec- tion with bending sketches, by locating the ends of the bars with respect to the centerlines of columns (see Fig. 245). Marks of bent bars and stirrups (as noted below) should be noted in schedule. Straight Bars in a beam are not provided with any mark, as on the job all straight bars of equal size and length are stacked together, irrespective of their destination in the structure. This simplifies the work. | In listing bars, the straight bars are kept together, arranged by sizes beginning with the largest, and also grouped by length. Bent Bars should be given a designating mark, and each bar after it is bent should be tagged with its mark. Aluminum tags should be used and securely tied to the bars to prevent their loss. This is especially important where bent bars are shipped any distance and require repeated rehandling before they reach the job. Paper and cloth tags, sometimes used, are not satisfactory as they often get lost. This not only creates confusion on the job, but also is the cause of bars being used in the wrong position with great detriment to the construction. A bar without tag must be measured, and the bending sketch thus made compared with the bending schedules, to determine where the bar belengs. This is expensive in time and 724 STRUCTURAL PLANS FOR BUILDINGS money. Both the engineer responsible for the job and the con- tractor should insist upon secure tagging of bent bars. | Marking of Bent Bars.—Whatever method is used for marking bent bars, any possible duplication of numbers must be prevented. A simple method is to give the bent bars the mark of the beam. Thus, if a beam is marked, in the schedule, 2B3, the bent bars may be also marked 2B3. If more than one type of bent bar is used in a beam, the different types may be differentiated by adding a small letter to the beam mark. Thus, 2B3a, 2B3b, and 2Bé8e are three types of bent bars in beam 2B3. This system of marking is not adapted to stirrups, as often the same type of stirrup can be used for a number of beams of different mark. Marks U1, U2, etc., may be used for all stirrups of the same dimensions, irrespective of their location. Thus, if beams B1, B3, B5, and B7 have the same con- crete dimensions, they require the same type of stirrup and in each case the stirrup is marked Ul. Sometimes, to prevent confusion in marking, a number designating the floor is prefixed to the mark. In such cases, 2U1 is a stirrup for a beam in the second story. Another system of marking which is sometimes preferred, but which does not indicate location of bar and is thus less definite, employs figures only. , By this plan, the mark of each bar consists of three figures, such as 725. The first figure designates the size of the bar, representing the number of eighths of an inch in the diameter or side of the bar. Thus, a 4-in. bar will be represented by 100;. a 4-in. bar by 200; a 3_in. bar by 300; a g-in. bar by 400; a 5_in. bar by 500; a f-in. bar by 600; a g-in. bar by 700; a l-in. bar by 800; a 14-in. bar by 900, and so on. Bars of the same diameter but of different length and bending are differentiated by the last two numbers, which are con- secutive. ‘The first bent bar in the 4-in. diameter list will be 701, the next 702, and so on. This method was used in Fig. 215, opp. p. 664. Bending Sketches.—Every bent bar must have a detail sketch. It is not sufficient to show the general outline of the bar, as in general structural plans. All dimensions must be carefully worked out. The importance of this cannot be over emphasized. Poorly bent bars, even if detected before they are placed, must be rebent at the job at a higher cost than that of the original bending and often with vexing delays. The required dimensions are shown in Figs. 244 and 245. In double bent bars the important dimensions for bending are the height of the bend, measured from outside to outside of bar WORKING PLANS 725 (usually called “out to out ’’) and the horizontal length of the bend. The inclined length of the bent is not used in the field, but it must be computed to determine the length of the straight portions of the bar. If shown on bending sketch, it assists greatly in checking. Bending sketches are usually made separately from the beam schedules. In the beam schedule, the mark of the bar is given and the type of bending indicated by a small sketch, which is of assistance in locating the bar when needed in the field. In the steel schedule, the same mark is used. For convenience, bars of the same type of bending are placed on the same sheet. Thus, bending sketches are made for groups of double bent bars, groups of rings, and groups of stirrups. BENDING:.-SKETCHES For trussed bars, separate bending sketches are often drawn for each mark of bar. This method requires more work in the office, but it simplifies the work in the field. The foreman can give a workman a sketch representing each bar to be bent by him. Separate bending sketches may also be used for the purpose of indicating the position of the bars in the beam. Thus in Fig. 245, p. 726, the position of the ends of the bar is located in respect to the enter line of the columns. This is important where the position of the bar is not symmetrical. | Another method is to use one sketch for a number of bars, as shown in Fig. 244, p. 725. The sketch is drawn at the top of the schedule, and the dimensions for each bar are given in the schedule under the different headings. By proper marking of dimensions, one sketch can be used for all trussed bars listed on the sheet, with and without hooks. In Fig. 244 one sketch was used for four types of bent bars. This simplifies the work in the office. It is evident, 726 STRUCTURAL PLANS FOR BUILDINGS however, that the understanding of this schedule requires higher intelligence on the part of the workman than that of the other plan. — Stirrups may be shown either by making a drawing for each stir- rup or by making a table similar to the one suggested for bent bars. Slab Details in Beam and Slab Construction.—In many cases, all the information about slabs may be given on the floor plan. Thick- ness of slab, elevation, and type of finish should be marked directly on the plan. Reinforcement may also be shown on the plan unless the arrangement of panels is complicated, in which case a slab schedule must also be prepared. Sometimes the main reinforcement fecpoutside face of column | aN Tay Lees cae 18"! Out to Out a Le of column aye ear mee Te _ oe 20'- Ole = ee chose oh oat ae | Out to Out x 20" Fic. 245.—Separate Bending Sketches. (See p. 725.) is shown on the floor plan, while odd panels are covered by a schedule. Not only the spacing of the bars but also their number and length should be given. The position of the first bar in each section is located by a dimension from some fixed line. The remaining bars are located sufficiently by giving their spacing. Method of bending should be clearly shown. In many designs, bars of one panel length are used. Straight bars alternate with bent bars. The first straight and the first bent bars are located, and the rest are covered by giving the number of bars of each type and the spacing. Where bars extend over several panels, the bars of one type of bending alternate with bars of some other type of bending. Sometimes bars of the same f Peek os. « WORKING PLANS (27 type of bending are used but the ends are staggered. Each type of bar should be located as explained above. All these expedients are used to supply a sufficient amount of steel, not only in the center but also at the support, and with the smallest tonnage practicable. It is important to make the intention of the designer clear, since mis- placement of bars may give an excess of steel in one section and not enough steel in some other section. The more complicated the arrangement, the more clearly it must be shown. Position of the bars in the slab should be shown by a section indicating the type of bending required. When bars are to be bent kefore placing in the form, proper bending sketches should be pro- vided. The straight bars are then listed separately from the bent bars. Where the bars are “ hickied,” i.e., bent after they are placed on the form, the location of the points of bending should be given, and the height of the bend should be worked out and shown. In this case, all bars are listed as straight bars. Flat Slab Construction—Working drawings for flat slab con- struction should be worked out in the manner described on p. 717 in connection with general plans. The steel schedules, however, must be complete, and the length of all bars must be given. In the schedule for the Smulski System the length of each ring and the required length of lap must be given. If the bars are bent before placing, bending sketches must be prepared in the same manner as for beams. Different types of bars should be given an appropriate mark on the schedule. Bars should be tagged with an aluminum tag. If steel is bent on the forms, all bars should be listed as straight bars, care being taken to allow for the proper length required for the bend. Column Schedules.—Column schedules for working drawings differ from those described on p. 720 only in that the lengths of all bars are given. The number of ties and their bending sketches are also given. Where the column bars are bent at the ends on account of difference in size of the upper columns, bending sketches must be prepared. ; CHAPTER XVII ARCHITECTURAL TREATMENT OF EXTERIOR AND INTERIOR OF REINFORCED CONCRETE BUILDINGS! By Henry C. ROBBINS In this chapter the architectural treatment of the exterior and interior of reinforced concrete buildings is discussed. Illustrations of exterior and interior treatment, as well as details of construction of different types of buildings, are given. The development of the exterior treatment of reinforced concrete buildings has grown quite naturally with the increase in their use and with the growing knowledge of the architectural possibilities of this type of construction, which was originally developed for utilitarian purposes rather than for artistic merit. Beginning with the kind of building which frankly expressed its structural materials in piers, girders, and curtain walls of concrete, with liberal window openings, the concrete building has passed through various stages of architectural expression to its present state, in which, to all outward appearances, it may differ in no essential feature from other equally substantial forms of building construction. The clothing of the reinforced concrete building with brick, stone, or terra-cotta, or a combination of these materials with concrete, makes possible a variety of architectural treatments which in the details of construction present no more serious difficulties than are ordinarily met in the design of buildings of other types. The interior finish of concrete buildings may also be as simple or as elaborate as the requirements demand, without seriously .taxing the ingenuity of the designer as far as the practical execution,is concerned. It is intended in this chapter to describe and illustrate some of the accepted methods used in common practice to produce satis- factory results in both exterior and interior finish of such buildings. 1 The authors are indebted to Henry C. Robbins for this chapter, which has been especially prepared by him for this treatise. 728 a EXTERIOR WORK ¢29 EXTERIOR WORK For the purposes of this discussion, the exterior treatment of concrete buildings may be divided into: (1) Concrete exterior; (2) Concrete in combination with brick, tile, or other masonry ; (3) Brick exterior; (4) Stone or terra-cotta exterior. CoNCRETE EXTERIOR Simple Treatment of Concrete Exterior.—The simplest form of exterior treatment is found in the frank expression of the concrete - structural members with curtain and parapet walls of the same material, as illustrated in Fig. 246, p. 729. Buildings of this type are generally found in the warehouse or factory building class. Fic. 246.—Simplest Form of Factory Building: Concrete Piers, Concrete Spandrels, and Curtain Walls. (See p. 729.) Turner Construction Co., Builders. Details of concrete exterior are clearly shown in the part elevation, Fig. 247, and section, Fig. 248. As is evident from the illustrations, the exterior columns form the piers between the windows. The width of piers is governed, not only by the magnitude of the loads to ‘be supported, but by architectural requirements and by the desired 720 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS width of the windows. With steel sash, the window extends from pier to pier, therefore the width of piers is governed by the standard width of sash. When more than one window in a panel is required, different window units may be separated by concrete mullions. The piers are joined together by spandrel beams at the floor levels, acting as lintels of the windows, and also as supports for the floor construc- tion. Above the spandrel beams, and filling the space between the floor and the bottom of the windows, are concrete curtain walls. In Parapet Concer. Cap-------- x Flashing--..... Roofing me 3 Cinder Bire Sash Grouted in--" Sasi Grouted in~--. SIM Cast a Separate--"| 5] Curtain Wall.-- Reinforcement not Shown --- — sash Grouted Section ee ————— through = Wall Detail below Window Sl aS te Showing fe: —— _ Curtain ‘all -~ Plan of Pier between Sash Fic. 247.-—Details of Concrete Exterior. (See p. 729.) the simplest form, their tops may be simply beveled to receive the wood window frame. Very often, especially when the steel sash is used, a separate sill is cast on top of the curtain wall after the sash is in place. The roof spandrel beam may be continued above the roof level to form a parapet with a beveled top, or with a simple, square-edged projecting cap to give a finish to the wall. The simplest and cheapest arrangement of exterior wall treatment is that in which the face of the piers coincides with the face of the spandrels. This, however, presents a rather crude and monotonous appearance, EXTERIOR WORK 731 The monotony may be relieved at very slight additional expense by setting the spandrel beams back from the face of the wall columns. The curtain walls are then placed slightly back from the face of the spandrel beams, and the concrete window sill is made to project over the curtain wall. Such an arrangement is shown in Fig. 248, p. 731. The appearance is also improved by the introduction of stronger or more massive treatment of end panels. This is fre- quently justified by placing the stair towers in these locations and - Wood Sill Wood Sill- - Concrete Sill -- Reinforcement of — Concrete Sill Curtain Wall Not Shown Concrete __ __-Concrete Curtain Wall Curtain Wall eas Face of Column _.. Face of Column _-. Concrete / Spandrel Beam ». yi _-7 Anchorage Strip~.__ “for Wood Frame ~~~. ° ; fs = =o ee : Window Head Rea Ne - Window Head lial viet sd gis Concrete Curtain Wall Set Concrete Curtain Wall Brought Back From Face of Spandrel Out Near Face of Spandrel s Showing Concrete Sill Ouverhanging Inside to Take Wood Window Frame b Vic. 248.—Sections through Spandrel Beam. Concrete Exterior. (See p. 731.) changing the windows from the typical arrangement followed in the rest of the building to a smaller size with surrounding structural members of increased width. The reveals and projections obtained by the methods described above help to break the monotony of an otherwise flat surface by the casting of shadows, which, after all, are responsible for much of the effect in all architectural composition. The concrete curtain walls are shown in Fig. 248a and 6. The construction in Fig. 248b may be used where the thickness of the curtain wall is less than the thickness of the window frame, 732 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS The concrete curtain walls are usually built separately after the frame is constructed. ‘To prevent temperature cracks, they must be properly reinforced. (See p. 384.) Also, they must be tied to the building, and recesses or slots must be left in the columns to receive these walls as they are cast. The treatment of the surface of the concrete itself is discussed under a separate heading below. (See p. 734.) Elaborate Treatment of Concrete Exterior.—It is possible to carry the exterior design of concrete structures into quite elaborate detail, and very pleasing effects may thus be obtained. The means used for the purpose are: (1) more elaborate details of molded work; (2) special treatment of concrete surfaces; (3) combination of the two methods. The selection depends upon the purpose for which the building is to serve and upon the extra expense that can be devoted to orna- mentation. Pleasing effects may be obtained at a comparatively small cost, a combination of simple molded work with surface treat- ment of concrete. More Elaborate Detail of Molded Work.—The appearance of con- crete buildings may be improved by the addition of molded belt or string courses and cornices, and in more expensive buildings by the ‘ntroduction of special ornamentation. In selecting the design, the character of the material must be kept in mind. The molding and the design of the cornices should be such as present the least problem in form construction. Projecting members with square corners, simple molded forms such as quarter rounds or plain cove moldings with flat fillets, plain sinkages, and other similar forms are the easiest to produce and are best adapted for concrete work. The effect should be obtained by general proportions of the mass rather than by refinement of details. A simple but effective treatment of concrete exterior is shown in Fig. 249, p. 733, where the building is provided with simple cornice, pilaster caps, and bases. The entrance received rather more elab- orate treatment. Molded details may be poured with the structure, or they may be precast and placed in the construction in the same manner as other stone trim. When molded in place, the moldings can be poured simultaneously with the rest of the construction, or the supporting frame may be built first and the belt courses and cornices poured separately. EXTERIOR WORK 733 The latter method simplifies the formwork, as the forms for the projecting moldings are supported on the structure instead of on the elaborate support which would be necessary if the molding were poured simultaneously with the structure. In this way the molding ean be built truer to line, and better workmanship can be obtained. Also, special concrete may be employed. When belt courses are poured separately, proper recesses must be left in the supporting Fic. 249.—Building of the Frank E. Davis Fish Company, Gloucester, Mass. (See p. 732.) Densmore, LeClear & Robbins, Architects and Engineers. structure, while steel dowels, often called “‘ bonds,” should be pro- vided to tie the two parts of the structure. To prevent cracking, the belt course must be reinforced by separate horizontal bars. Wherever molded work is cast in place, it should be arranged, if possible, to bring the end of a day’s pouring at the molded section, thus concealing the line that is always in evidence, in the finished work of plain surfaces, where one day’s work ends and the next begins. Important decorative work is best done in concrete pre-cast in the form of “ cast stone ” and built into the main structure. See Fig. 250, 724 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS p. 734, in which the cartouches on either side of the entrance and the key-block were cast separately and inserted. Treatment of Concrete Surfaces.—Much can be done to produce an interesting exterior by the choice of surface treatment, which, Fic. 250.—Entrance Detail, Carter’s Ink Factory, Cambridge, Mass. (See p. 733.) Densmore, LeClear & Robbins, Architects and Engineers. if carefully handled, goes far toward making up for the lack of elaborate molded work. Exposed concrete may be finished in a variety of methods, ranging from the smoothing of the surface, primarily to eliminate form marks and to fill voids, to the picking or hammering of the surface to reveal an aggregate chosen particularly for its color. EXTERIOR WORK 735 A dense, smooth surface is desirable for concrete which is to be exposed to view and upon which no great amount of finishing work is to be expended. To produce such a surface, thorough spading of the mass as it is deposited is essential, and tapping the outside of the formwork with a hammer or mallet is helpful in further settling the mixture into place, thus forcing the heavy aggregate back from the surface’and bringing the mortar to the face. | : For all concrete work which is to be exposed to view, it is desir- able that the formwork be tight and smooth. It should be erected true to line and supported in a substantial and workmanlike manner. The materials should be as free as possible from defects which would show in the surface of the concrete. In the usual commercial work, however, it is impossible to eliminate all voids, irregularities, and fins or slight projecting ridges which are caused by the mortar oozing through the cracks in the formwork. ‘These defects must be corrected if a reasonably regular and unbroken surface is to be obtained. Surface Treatment by Rubbing.—Most satisfactory results are obtained at the least cost when the finishing is undertaken while the concrete is still green. The fins then can be smoothed down and the entire surface brought to an even finish, by rubbing with a wood or cork float, using a mortar composed of one part of cement to two parts of fairly coarse sand as an abrasive, the mortar at the same time filling the voids. With concrete which has set hard, it is obvious that more work is involved in producing a satisfactory finish. The fins must be removed by hammering, chipping, or rubbing with a carborundum stone, and the voids must be filled. After this has been done, a mortar composed of one part of cement to one part of screened sand is rubbed in, by means of a cork float manipulated in a rotary manner. The surface should be thoroughly saturated with water before the mortar is applied. After the mortar, applied as described above, has thoroughly hardened, the entire surface is rubbed with a carborundum stone while water is being applied with a brush, until a smooth, even surface, uniform in tone, is obtained. This method was used in finishing the building shown in Fig. 249, p. 733. It is essential that the mortar be thoroughly rubbed in during the first operation, so that no appreciable thickness of skim coat remains. 736 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS Otherwise, the coating will crack and peel off, if exposed to extremes of temperature. An effective variation in the treatment described above may be obtained by using white cement in combination with the fine sand, in place of ordinary gray cement. In this case, the second rubbing, after the mortar has set, may be dispensed with. When this is done, the thin surface coating should be kept wet as it sets. The so-called spatter or dash coats, so frequently employed in the finish of domestic stucco work, are not commonly used in reinforced concrete buildings of the commercial or industrial type, as they Fic. 251.—Example of Brush Finish. (See po 100) usually involve expense in the preparation and finishing of the surfaces to be treated, out of proportion to the results to be obtained. Brush Finish.—Green cement may be effectively treated also by washing with water to expose the aggregate, using a wire brush or a coarse, stiff fiber brush. Good judgment as to the proper degree of hardness of the concrete is required in using this method, since if the surface is too hard, the brush will not be effective in removing the mortar, while if it is too green, particles of aggregate will be dislodged, leaving a coarsely pock-marked surface. Ordinarily, this method of finishing must be undertaken in about twenty-four hours from the time of pouring. (See Fig. 251, p. 736.) EXTERIOR WORK 137 Bush Hammer Finish.—A more effective and also more costly finish than those described above is obtained by hammering or picking the surface to expose the aggregate. The bush hammer commonly used in stone masonry is a satisfactory tool for this purpose. The brushing method previously described approaches the hammered or picked treatment in result, but it lacks the sparkle of the latter because hammering not only exposes the aggregate, but cuts it as well. The condition of the concrete for a satisfactory hammered treat- ment must be the reverse of its condition for brushing; that is, it is necessary for it to be thoroughly set and hard, so that the aggregate is imbedded firmly enough not only to prevent dislodgment as the — - mortar is being cut back, but to withstand the actual cutting of the aggregate particles. This method presents a difficulty in the finishing at external corners, where it is practically impossible to prevent the cutting out of pieces of aggregate, thus leaving a ragged arris or corner. This may be obviated by first rubbing the surfaces at the corners to a smooth finish, as described above, and then hammer- ing up to a sufficient distance from the corners to leave a smooth draft line, of a width to be in scale with the hammered field. | (See Fig. 252, p. 738.) This treatment may be elaborated by choosing the surface aggregate for its color, but the common run of stone or gravel used throughout the work ordinarily will provide a satisfactory surface when exposed and cut. Coloring of Surface by Selection of Aggregates.—If a special color of aggregate is desired and is available only in small quantity or at an extra cost, it may be deposited in the outer surface only af the portions to be so treated. ‘This may be done by inserting a sheet- metal dam about 3 inches back of the inner face of the formwork, and depositing the special mixture between the dam and the form- work at the same time that the regular mixture is being poured in the balance of the member. The dam is raised gradually with the progress of deposition of the concrete. The spading is confined to the space above the bottom of the dam, else the two mixes may run together. It is not practicable to introduce this method in columns and piers, where the reinforcing steel, and particularly the column ties, complicate the handling of the dam; but in spandrel beams and walls, or in unreinforced members, it is quite feasible. Another method of introducing special concrete near the surface 738 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS of the building was used in the construction of the Hide and Leather Building in New York City After the stecl reinforcements for the spandrel had been fabricated as usual, expanded metal was wired to the bars, leaving a distance varying from 13 to 2 in. between the expanded metal and the outer Fic. 252.—Example of Hammered Surface Surrounded by Smooth Draft Lines. (See p. 737.) form. Into this outer space a special surface concrete was poured, consisting of one part of white Portland cement and two parts of a colored aggregate, the latter being composed of green stone chips, feldspar and quartz, ranging in size from 7 to 1 in. This surface concrete, which was hand-mixed and of dry consistency, was guided into the front space by a broad, flat chute, and tamped in place with 2Thompson & Binger, Architects and -Contractors.. EXTERIOR WORK 739 a special T-head as the latter being made by fastening a 1-in. flat, 6-in. long, to a 4-in. bar about 12 ft. in length. The balands of the beam, i.e., the space between the expanded metal and the rear form, was aired with the usual concrete. Pre- cautions were taken to pee the facing concrete pile up about 1 ft. higher than the structural concrete during the process, to prevent the common aggregate from flowing to the outside face. After the forms for the exterior surface had been removed and the necessary patching done, the concrete was permitted to season during the winter; in the early spring it was ready to receive its final finish. The two lower floors were bush-hammered by stone masons. The upper floors, sixteen in number, were surfaced by a carborundum _ grinding Pasieting. after which a colorless waterproofing was applied over the entire exterior. This treatment was given mainly to prevent the elements from attacking the surface of the concrete. CoNCRETE IN CoMBINATION wiTH Brick, TILE on OTHER Masonry This combination may be used in warehouse and factory’ con- struction, where the curtain walls are built of ordinary hard brick, and the cost approaches the cost of the simple all-concrete type. With proper treatment of concrete surfaces and use of more expensive filling materials, very ornamental effects may be obtained. Combination of Brick Curtain Wall with Concrete Piers and Spandrels.—The most common construction of brick and concrete is that in which the piers and spandrel beams are of concrete and the curtain wall of brick. Such construction is shown in Fig.. 253, p. 740, where the curtain walls and parapet are _of red brick and the rest of the fagade is carried out in concrete. ‘The section in Fig. 254 shows the construction more fully. Asin the case of the all concrete exterior, the size of the concrete pier is mainly governed by the required architectural proportions and the width of the window sash. The exposed concrete beam, which in the illustration rests entirely below the slab, carries the weight of the curtain wall and windows and also any floor load that may come upon it. The construction of the sill on the top of the brick curtain wall depends upon conditions and particularly upon the type of window. With steel window sash, concrete sills are commonly used, and are built after the sash is in place. } BUILDINGS CONCRETE B L TREATMENT OF ECTURA 740 ARCHIT ‘SIQQUISUT PUB SpOPIYOIY ‘SUIqqoY FD Ieo[HeT “etoursued, (‘6¢2 ‘d 20g) ‘“ssByy ‘UO}sog “SuIppIng uorueduor 8. YyNoA—'EgZ “OA EXTERIOR WORK 741 An interesting variation from the ordinary treatment of spandrel beam and curtain wall, when the beam is entirely below the level of the floor which it supports, is the “ upstanding ” spandrel, which involves a somewhat less efficient structural design for this particular member, but has the advantage of bringing the tops of the windows close to the ceiling. Figure 255, p.. 742, illustrates this form of construction and may be compared with Fig. 254, p. 741, the latter showing the usual dropped beam. The brickwork in the spandrels - may be plain or elaborate, the elabo- erick ration consisting of the use of differ- ent grades and colors of brick or different patterns. Brick or tile work is often intro- duced into concrete piers, walls, and spandrels in the form of pattern work, making possible most effective treat- ments in both color and design. The possibilities in the combination of these materials are limited only by the : ingenuity of the designer and the re- pig. 254 Seetion fProughic pare strictions of expense. It is, of course, drel Beam. Concrete and Brick obvious that wherever concrete ap- Exterior. (See p. 739.) pears it may be finished in any of the methods discussed under “‘ treatment of concrete surfaces.” The brickwork in the curtain wall may be replaced by ornamental terra cotta. Figure 256, p. 743, shows an interesting use of archi- tectural terra cotta in curtain walls and in the spaces over the arches just below the cornice. The spandrels are elaborated by the use of terra-cotta pediments at the third-story level. The terra-cotta work is green-glazed and gives a most striking appearance in combina- tion with the concrete, which is finished in white cement. Consideration of the general proportions of the mass and con- centration on a dominating feature bring about the same satisfactory results in architectural expression as may be obtained with the same study in any other type of construction. Figure 257, p. 744, is an excellent example of the possibilities of such study. While the main Wood Sill ~~ Masonry Sill - Brick -- Air Space : aan — ne ‘ a -Window Head 742 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS body of the building is given a touch of color in the brick spandrels and parapet wall, the tower is carried out entirely in concrete cast in place. Concrete Piers, Brick-faced Spandrel, and Brick Curtain Wall.— The concrete spandrel beam, which is exposed in the previously described type, may be cov- ered with brick. The wall construction then consists of exposed concrete columns with the balance, except for sFace of Column the windows, covered with brick. Figure 258, p. 745, illus- trates this method. In this case, as in all cases where masonry is to be placed against concrete as an ex- Anchorage Strip terior facing, a space of about For Wood Frame one inch should be allowed between the face of the con- crete and the back of the Fic. 255.—Upstanding Spandrel Beam Per- tse to allow for varia- mitting Window Head to Come up Near tions in the surface of the Ceiling. (See p. 741.) concrete member. The con- crete beam is then set back from the outside surface of the finished wall, a distance equal to the width of brick plus one inch. When masonry is to be used as a facing for concrete, it is cus- tomary to support the facing material on structural steel angles securely bolted to the concrete members. For ordinary cases, 3.in. bolts spaced 24 in. apart are used. This may have to be. increased where the angles carry heavy load. The bolts, with heads and washers, or with the ends bent for anchorage in the concrete, are inserted in the formwork, with a threaded end left projecting a suf- ficient distance to pass through the leg of the angle and receive a nut and washer on the outside. These bolts may be held tight, during pouring, by nuts, one just inside and one outside of the formwork. The steel angle supporting the brick facing is usually 6 in. deep in order to get the anchor bolt well up into spandrel beam. The spading of the concrete where the bolts occur must be done with care, so as not to displace the bolts or throw them out of line. as -- Wood Sill Masonry Sill -- 223 LLLLLL Z ih WLLL Air Space ---~- Concrete Spandrel- Ceiling“ -Window Head 743 EXTERIOR WORK CI#L “d 99g) A N ‘stoouIsun [OD 2 9U9014)-PooMyIo'T ‘19}80YO}IOg 6. out ‘SIOABY ofr] JO SUIp[Ng—'9¢z “SIT "Ad ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS ‘groping “OO Uoronrjstod JouINy, “WoTYIV ‘SOT[OM PP ploreyy (Tp, ‘d vag) ‘ssvyyy ‘uudT ‘A070 UOPYSIOID “WW “V—' LEZ “WA EXTERIOR WORK 745 Since the nature of concrete formwork is such that precision in the spacing of anchor bolts is somewhat difficult as compared with shop work in structural steel, it is well to allow for some slight varia- tion, particularly in the horizontal direction of the bolt spacings, by slotting the steel shape instead of punching it. (See Fig. 258, p. 745.) It is not practical to slot the angle vertically, but some allowance may be made for inaccuracy in vertical dimension, by making the slot slightly wider than the diameter of the bolt. This sell) - Wood Sill requires the use of r f Gis rer asonry Sill - Y) 3 a heavy washer un- LA |: Wi der the nut and a Yj thorough tightening puscpace-- Lop “Face of Column \ of the latter to pre- vent slipping under Brich-- the load of masonry. Backing of Cur- tain Walls.—It is Concrete eet obvious that the : backing -of curtain walls, described sieropnk above, may be of : the same material as for similar work in other types of fireproof construc- Frc. 258.—Spandrel Beam Faced with Brick, with Brick tion. The problem Curtain Wall. (See p. 742.) will vary with the character of the building and with the degree to which it is desired to avoid condensation. In most warehouses and _ fac- tories, a small amount of condensation is not harmful. In such cases the curtain walls are made from 6 to 8 in. thick, without any special precautions. In structures such as office buildings, schools, hospitals, and buildings for habitation, where moisture inside must be avoided as far as possible, the choice of the backing material and the method of laying it should be given prticular attention, to prevent the actual driving through of rain and also to eliminate condensation. Water troubles frequently appear even after the utmost care has been used in building, but the following methods may be used as being -) Sometimes Made of This Portion of Wall Is Single Course Brick -Fill Up With Mortar to Here Possible Irregularity in Face of Spandrel - Brick Facing -Angle Iron Support ——————— Anchorage Strip. et For Wood Frame \ Angle Slotted Window Head - For Anchor Bolt 746 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS good practice without involving great expense and as giving reasonably satisfactory results. If the wall is to be of brick, an air space of 1 or 2 in. in width should be provided between the outer and inner courses, and care should be taken that this space be free from mortar. A further precaution is to coat the inner surface of the outer coursing with a waterproofing material, this being done either as the wall is built up or after the outer coursing 1s built up to level and before the inner section is started. eieacgh er The outside face of the wall may be coated with a colorless waterproofing agent. The use of either of these methods of applied Face of Column waterproofing does not affect the conditions as re- Wood Sill-— Interlocking Tile Smooth Face 4 Y ZA “17 umd PZ282 ZZ A H b 4 Y H Z| ee 4 Ha Wa Yo Haan A zzz ef LOE Gl san 5 1 ; Hy hens teawarel TELIA Ha YLLEE: ze A ay 4 g 4 LLL H 0 HAE Y Crarzd W277 Z, Concrete _ N Spandrel N -Fill Up With gards condensation; but Serer SITE nas Mortar to Here the air space in the wall, - Possible if it is free from obstruc- Irregularity in : , Face of Spandrel _ . tions, provides reasonably Brick Facing effective insulation. The backing of the wall may be of hollow tile of the in- Sh LRU terlocking variety, which provides for bonding with Anchor -—~~_ Anchorage Strip - as for Wood Frame |.2° 37 a Fic. 259.—Spandrel Beam Faced with Brick, | . with Curtain Wall of Face Brick and Interlock- brickwork, or of the ace ing Tile. (See p. 746.) mon form of hollow tile bonded with brick or with metal ties. Figure 258, p. 742, shows a wall with brick backing, while Fig. 259, p. 746, shows a wall with backing of interlocking tile. For buildings where the interior faces of these curtain walls are not to be plastered, the methods of construction described above are within the limits of good practice, since the amount of moisture which will come through is not of serious moment and is, under ordinary circumstances, evaporated before it becomes troublesome. When walls are to be plastered, however, it is the opinion of the author that no method of construction or treatment is as satisfactory or positive as the furring of the inside of the wall to provide vertical — Anale Iron Support EXTERIOR WORK 747 air spaces. This may be done with ordinary wooden or stcel furring, upon which is placed the steel lathing and the plastering, or with tile furring blocks laid with the hollow spaces vertical. Brick EXTERIOR The next development beyond the partial use of masonry in connection with concrete is the complete clothing of the concrete skeleton structure with brick, in much the same manner as is prac- ticed in buildings of skeleton steel construction. Figure 260, p. 748, shows a group of buildings entirely covered with brick and with no indication as to the type of construction. This illustration was selected to show that concrete Wane may be used irrespective of the elaborateness of the architectural treat- ment. A less elaborate construction is shown in Fig. 262, p. 750. In the buildings shown in both illustrations, the outside treatment is independent of the size of concrete wall columns. These are, there- fore, designed only for the load, as their size is not governed by architectural considerations. ‘There the outside face of the columns must be in-one vertical plane, any reduction in size of the structural column is obtained by setting back the inside face in the different stories. If the exterior design calls for a buttressed column, the structural member may be reduced in thickness, as the column goes up, to correspond to the changes in the set-backs of the exterior face. In factory and warehouse construction, where the window occu- pies a large part of the wall, an arrangement shown in Fig. 261 may be used. ‘The window sash extends from concrete pier to concrete pier. The veneer is applied outside of the building only. The shape of the concrete pier is as shown in the figures. The recesses in the columns are easily obtained by using rectangular forms with blocking of proper dimensions inserted in the corners. The concrete wall beams ordinarily are set back just enough to allow a veneer of brick equal in thickness to one width of brick (usually 4 in.) plus one inch allowance for variations in the face of the concrete member. This may vary as necessary to meet the architectural treatment. If possible, the spandrel beam should be made deep enough to reach to the top of the window. ‘This saves the expense of extra lintels over the windows, ‘spooyIgo1y Buypnsuoyp ‘ugsatM “A eBSi0exy) pus SoOTPM PPM PlolPH ‘saspling pues szyooyryory “OD F SOA PP “WW yyouUusyy (LE, d aag) “sseyy ‘oulpyoorg ‘1oueyy AV Voply—'09G “OMA RHSR CSR pay ai he Y SSE IN EXTERIOR WORK 749 2 Masonry Ze aa ge Size of nak proportional. to overhang beyond column Elevation Fill up with mortar yp Supporting angle sl Ls to Aca! there Concrete spandre, Angle slotted jar anchor wale ES Injface of spandre/ Y 4 ‘ EZ, LA VY Spa a, ‘Masonry Ties attached to rl 4 Section A-A Detail Elevation Details _of Brick Exteror Masonry Ties” ain Mas asonry Ties \ Brick Facing Details of _Facing_of Columns with Brick Fig. 261.—Partial Elevation and Sections of Brick Veneer Building. (See p. 747.) TREATMENT OF CONCRETE BUILDINGS JRAL eget 50 ARCHITE 7 ‘s10PORIYUOD “OD Suyooidoity peyuy “qooqyyory “sog “gq 981004) (2$2°d ag) “AN ‘OveyNg Ul oseIeyH JopyeVIG—ZIZ “OI a oe ee eee ee ee ee EXTERIOR WORK | 751 Support of Veneer.—The support of brick facing or any other such architectural treatment on the face of the structural members is accomplished by angle supports, in the same manner as described above in the discussion of masonry spandrel walls. (See p. 742.) The angle supports should extend along the spandrel beam and, in case of wall columns, across the face of the columns also. If span- drels are flush with wall columns, continuous angles may be used. When the spandrels are set back, separate short angles should be used across the column. The spacing of bolts and the method of securing them to the forms is the same as given on p. 742. The steel supports for the facing should be bolted onto the structural members at each story. The angle on the spandrel is usually 6 in. deep to get the anchor bolt well into spandrel beam, while the angle on the column face can be 4 in. deep unless stronger angle is required by the load. Ties for Facing.—In addition to the angle support, metal ties should be used for anchorage of the masonry to the concrete. These should be placed vertically, not more than six courses of brick apart. When the spandrel beam is no more than 20 in. deep, no ties are required. Horizontally the eae may be 24 in. In the simplest forms, the ties consist of 4-in. wire, preferably brass or copper, imbedded partly in the concrete and extending from it a sufficient length to reach well into the brick masonry. In constructing, holes are drilled in the form, the required distance apart, and the wire inserted to be buried in the concrete. The part of the wire to be imbedded in brick extends outside of the form. To avoid the expense of drilling holes, ties of special design may be used. With these, the part intended to serve as anchorage in masonry is tacked to the formwork, while. the part to be imbedded in concrete is bent back at right angles to the formwork. After the concrete is poured, this part of the tie becomes buried in the concrete mass. When the forms are stripped, the ties will detach themselves from the forms, but will of course remain anchored in the concrete. The ties should be of sufficient length and of such flexibility that the portion left outside the concrete surface can be bent to meet the coursing in the masonry and be laid into the joints as the facing is built up. If this method is used, it is not necessary accurately to predict the masonry jointing or to place the ties exactly at joint lines. 752 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS Fic. 263.—Building of the Salada Tea Co., Boston, Mass. (See p. 753.) Densmore, LeClear & Robbins, Architects and Engineers. EXTERIOR WORK 753 STONE OR TERRA CoTtTa EXTERIOR The use of stone or architectural terra cotta for the facing of reinforced concrete structures ordinarily presents no more difficult problems than are met in similar buildings where steel forms the structural members. 7 y TILT TTT ARLE 5 | = agen KXAN \ Architectural -—-[>4 Terra Cotta N | / Sa G oF hice . ‘ SL we. <4 rahe we fs ASS A gm Noeeere | Architectural —_ Terra Cotta = Ly ZZN jp 4 Y Anchor ---}A =. Ly) : =, Y Showing Architectural Terra Cotta Facing Showing Stone Facing Fic. 264.—Sections Showing Facing with Architectural Terra Cotta and Stone. (See p. 755.) Figure 263, p. 752, shows a concrete building, the lower stories of which are faced with limestone and the upper story with architectural terra cotta. In general, the manner of support and anchorage of the stone or the terra cotta is similar to the methods described above in the discussion of brick-faced buildings. In the building illus- trated, the first two stories of stonework rest on the foundation wall, 754 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS (‘egy “d 29g) ‘aspliquiey) ‘KZojouyoa J, JO 9yn}T}sUT S}JOSNUOVSSB JA] ‘sjIno0D IOUT; puw io[eyy, uo sapvor J [eyuoureuIQ—'Cgz “DTT rae = oer ace? re EP hey > ULUNIOD Ams 9 iY) S parsofulay fo aon = S ‘ f Q Se or (Ih ae lh acre ce SEIT as fe hr N fo -& Re ae | Ss SoS SS ql , : 7 igs Ne Ss © © e te 3S “ Xu8E 120 eR H as we Se BS) S| te w% iv] oF woes S BE Seo Sa} 3S ES eae 3 S SSS oy ee ae a 38 ae Soe a8 Sores CE eS = XN i y < ie ns = at SO = Cee tac 2 =f = Rs) is} oO Re) ; “ x he S & oe a cS) ‘s S SS INTERIOR FINISH 759 while the remainder of the facing is carried on angle supports at each floor level. Particular attention must be given to the anchorage of members having sufficient projection beyond the normal face of the wall to introduce possibilities of overturning. The use of heavy ties an- chored in the concrete and turned into the stone, or of angle shapes secured to the structure and overlapping the top of the stone at the back, are methods frequently used; but each case is worthy of special study to determine the most satisfactory method. See Fig. 264, p. 753, for an illustration of facing with stone and archi- tectural terra cotta. Sections through ornamental facades on major and minor courts of the Massachusetts Institute of Technology buildings,? are shown in Fig. 265, p. 754. Attention is called to the method of facing of the concrete and also to the different methods used for anchoring the stonework. In some cases eye bolts were set in concrete spandrel beams spaced about 4 ft. 6 in. on centers, through which 1 inch base were placed. Anchors were carried from each stone and fastened around these bars. In other cases anchor bolts were anchored in the concrete and carried up between the joints of the limestone. Each anchor bolt was provided on the top with a plate and nut which hold the stones in place. INTERIOR FINISH The finish of concrete which is exposed to view in the interior of buildings is open to as great a variety of treatments as is the case in exterior work. Much has been done in the way of unusual and interesting finished surfaces; but in the great majority of con- crete buildings the architectural treatment of exposed interior con- crete has not received the attention that has been given to exterior work. ‘This is true because, except for the simplest possible handling of the material, other mediums in common use are as available and as readily applied as in any other type of building construction. For purposes of discussion, the interior finish of concrete buildings may be divided into the following classes: (1) Simple finish. (2) Special finish. (3) Concrete painted. (4) Plaster finish. 3 Stone and Webster, Builders, Sanford E. Thompson, Consulting Engineer. 756 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS Simple Finish.—Buildings in which the concrete structure is allowed to show on the inside are generally of the utilitarian type, in which good practice demands no more than a relatively clean, smooth surface. There are usually two kinds of defects in the concrete as it comes from the forms, which require attention to produce a satis- factory finish. These defects are the “ fins,” left projecting from the face where the mortar has oozed through the joints in the form-_ work, and the “ voids,” caused by a lack of homogeneity in the mix- ture and due usually to improper consistency of the mix or to insul- ficient spading as the concrete is deposited. In ordinary practice, it is usually sufficient to smooth the projec- tions by hammering and to fill the voids by rubbing in a rich cement mortar, using a wood or cork float or carborundum stone according to the degree of fineness desired. For a really smooth surface, the methods described in the discussion of exterior treatment may be used. | If the work is to be painted, it is essential, in filling the voids and smoothing, that no appreciable thickness of skim coat of mortar be left over the face of the concrete work, as such a coat is likely to peel under paint and flake off. The choice of materials for formwork and the quality of work- manship used in placing it have an important bearing on the elimina- tion of labor in finishing. For all work except that which is intended to be left in a rough state, the wood for forms, against which the concrete is to be poured, should be planed to bring the boards to an even thickness and to eliminate the marks of the grain to prevent ridges, which are sure to develop when boards of even slightly differing thicknesses are used. The fitting together of the forms should be done with care, to elim- imate as far as possible the cracks through which the mortar can flow. Knot-holes should be covered with tin or other suitable thin material, and where the finished surface is not to receive plaster the forms should be oiled. | Steel formwork also requires care in handling and placing, if a workmanlike finish is to be obtained. : Figure 266, p. 757, shows a ceiling surface resulting from the use of ordinary wood formwork and a column and column head cast in steel forms. Improvement in the manufacture and setting of steel formwork has advanced to the stage where concrete as it is left after the forms INTERIOR. FINISH 757 have been removed may approach the appearance of a plastered surface; but this is accomplished only with unusual care and special forms, and consequently with added expense. Special Finish—For more elaborate work than is commonly expected in the average concrete building, many devices may be used to produce attractive and unusual results. The forms may be left rough, purposely showing the grain of the wood as prominently as possible, to be treated later with paint to imitate a wooden surface. Fic. 266.—Factory Interior, Salada Tea Building, Boston, Mass. (See p. 756.) Plaster casts may be introduced into the forms, particularly into the sides and bottoms of beams, to give a desired pattern or design. In this case, relatively fine aggregate must be used for the concrete which is to come next to the cast. The wooden formwork may be chosen for kind and quality, and may be left in place, solidly anchored into the concrete, when the mixture is poured. The possibilities of interesting architectural effects, to be pro- duced in ways similar to those mentioned above, are many and will suggest themselves to the resourceful designer, not only as to methods to be used but how and when to use them. 758 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS Painted Concrete.—Painting of interior concrete surfaces may be done in the same manner and with the same materials as would be used on ordinary plaster surfaces, except that it is advisable to precede the painting with a non-burning sizing coat which is especi- ally prepared for use on concrete surfaces and which can be obtained from most paint manufacturers. The object of such a sizing coat is to prevent chemical action between the concrete and the paint, which, if not provided’ against, causes unsightly spotting of the painted surface. It must not be assumed that the use of paint will hide the form marks or cover up other irregularities. In fact, such treatment, particularly if a glossy finish coat is used, will often make the faults in the surface appear more marked than if left unfinished. Painting does, however, give a surface which is more easily cleaned than if the concrete were left bare, and permits color treatments superior in appearance to the untreated concrete. | Either lead and oil paint or lithopone paint is satisfactory for use on concrete surfaces. Ordinary cold-water paint is also used extensively in industrial work for ceilings, and for walls above the height at which the paint is likely to be rubbed off, that is, 6 or 7 ft. above the floor. 3 In all painted work, the concrete should first be cleaned of any oil which may remain on the surface, due to the oiling of forms. Plaster Finish.—The use of plastering as a finished surface is as much an accepted form in concrete construction as it is in any other type of building construction, but the problem of obtaining a satis- factory bond between plaster and concrete continues to be an impor- tant one. Concrete which is to be plastered should come from the forms as porous as sound construction will permit, and the use of oil or grease on the formwork should be avoided. The plastering of concrete ceilings requires unusual care, in order to avoid bond failure with the consequent dropping of sections of plaster finish. It is probable that the only method of securing positive results in this direction is that of furring and wire lathing the surface as the foundation for the plaster work. This is an expensive treatment, and common practice does not require its adoption except in the most important cases. There are on the market preparations in the form of specially prepared plaster which may be applied to the concrete surface as a foundation for the finished plastering, and which provide a reason- INTERIOR FINISH 759 ably satisfactory bond. An accepted practice for ceiling work is to prepare for the first or bonding coat by roughening the smooth portions of the concrete surface, by picking or hammering, and to make sure that if oil has been used on the forms it is thoroughly cleaned off. The amount of roughening varies widely, ranging from a surface hammered practically all over to hammered or picked spots of an area the size of the palm of the hand and spaced 3 to 6 in. Fic. 267.—Main Office in Salada Tea Building, Boston, Mass. (See p. 760.) apart. It is obvious that the more thoroughly the roughening is done, the less is the likelihood of bond failure. It is of the greatest importance in connection with the plastering of ceiling surfaces that the plaster be applied in as thin coats as possible, and in no case should more than two coats be used. The plastering of vertical surfaces does not present as serious difficulties as are met in ceiling work, average concrete being a satisfactory base, particularly if the special plaster bonding coat mentioned above is used. Rough formwork, brushing the concrete while it is still green, or other similar methods may be used, where practical, to 760 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS produce vertical surfaces suitable for plastering without the necessity of special treatment. Ornamental plaster work may be carried out on concrete members by the same method as commonly ased in connection with other forms of construction. It is important, however, that no molded work be done on horizontal surfaces without furring or other suitable anchorage. at An interesting example of the interior treatment of a reinforced concrete building is shown in Fig. 267, p. 759, and this may be con- trasted with a view of the interior of another story of the same building, Fig. 266, p. 757, where the construction is frankly exposed _-7 Profile of Structural. = oe Concrete Cap aa. |fecAanger en 502 gree NS ee: see eal STS A OE A BODE | TT 5 Finish Ceiling Line? On ee ei) rR: SSS », : : ~ E ENS ly _Seagliola ------ -—l yg (Imitation Marble) if VW { of Salada Tea Building. (See p. 760.) to view with no further treatment than painting. In this particular building, the construction is of the flat slab type with the usual flaring column heads. ey The room shown in Fig. 267, p. 759, is in the first story of the building and is used as an office. The ceiling was furred down about one foot, concealing the sprinkler piping, and at the same time bringing the ceiling level down to a point on the flaring head of the column where the diameter of the structural head is such that a correctly proportioned column capital of cast plaster could be formed around it. (See Fig. 268, p. 760.) The columns themselves are encased in scagliola or imitation marble, with molded bases of the same material and plinths of dark marble. The molded ceiling, INTERIOR FINISH 761 carried on channel iron furring, is hung from the bottom of the structural slab of the second floor, the anchorage being provided by hangers imbedded in concrete or sockets for holding threaded rods, installed in place on the formwork and incorporated in the slab when it was poured. For details of suspended ceiling, see p. 616. 3 A much simpler treatment of a circular column head is shown in Fic. 269.—Office Building Interior, Columns, Walls and Ceilings Plastered. (See p. 761.) Fig. 269, p. 761. The plaster molding at the top cf the capital in this case is run without furring, as it approximates the controur of the simple molding of the steel form. It is obvious that, in beam and girder construction, the use of cornices, capitals, moldings, etc., is relatively simple, provided that such furring is used as is ordinarily employed under liké conditions in other types of fireproof construction, CHAPTER XVIII CONCRETE IN CONSTRUCTION OF THEATERS AND AUDI- TORIUMS Reinforced concrete is well adapted for theater construction. It has been used for this purpose in a large number of cases, in suc- cessful competition with steel construction. In localities distant from large steel fabricating plants, it 1s particularly economical, since the materials for reinforced concrete theater construction are more easily obtainable than structural steel members. It is also useful for baleony cantilevers in combination with steel trusses, whereby the difficult connections between the steel cantilevers and | trusses are avoided. | | The theater, for the sake of discussion, may be divided into the following parts: orchestra floor, balconies, proscenium, and roof. A general idea of a design of concrete theater may be obtained from | Fig. 270, showing a section through Grauman’s Theater Building in Los Angeles and Fig. 275 showing cross section through Winston- Salem auditorium. , | | 3 ORCHESTRA FLOOR The space under the orchestra floor is often entirely unexcavated or excavated only in part. When the ground is solid enough, the concrete floor for the unexcavated portion may rest directly upon it and be treated in same fashion as the ground floor in ordinary build- ings. To prevent cracking due to contraction and shrinkage, some reinforcement, consisting of small bars or mesh, is advisable. The orchestra floor in the Winston-Salem auditorium, shown in Fig. 275, p. 781, rests directly on the ground. If the space under the floor is excavated, or if it is not desirable to rest the slab on the ground, the construction of the orchestra floor consists of one-story columns supporting a reinforced concrete floor system. It differs from ordinary building construction in that the 762 BALCONY DESIGN 763 slab is level only near the stage, and then slopes up. In many cases the slope is gradual at first and then more pronounced. The longi- tudinal cross section is then a curve with a horizontal tangent at the origin. When the rows of seats are parallel to.the stage, all cross sections are alike. The only difference between such construction and ordinary building construction is that the columns in successive rows will increase in height and the beams or girders longitudinal with the building will be inclined, while the members across the building will be level. Ordinarily, however, the seats in each row are placed on a curve which is either a segment of a circle or, more often, a three-centered curve. Since all seats in a row must be on the same level, the orches- tra is not a plane but a warped surface. This complicates the floor construction. The best solution is to place inclined girders longi- tudinally with the building, and then arrange the beams tangentially to the curve of the seats. All beams in a row will then be on the same level. Unless fixed by the requirements of the space below the orchestra floor, the spacing of the columns carrying the orchestra floor will depend upon economic considerations, and particularly upon the cost of foundation and length of columns. If the cost of foundation for each column is small and the columns short, they may be spaced more closely. For difficult foundation work, larger spacing of columns will be found economical. Usually, a spacing of 18 to 20 ft. will be found satisfactory. Although beam and girder construction is generally used for orchestra floors, flat slab construction in many cases is more economi- cal. Either type of floor is designed according to the principles discussed on pp. 303 and 575. The cross section through Grauman’s theater, Fig. 270, p. 764, shows a design of supported orchestra floor. In this case the construction is of the beam and girder type with girders running longitudinally with the building. The steps for the seats may either be made integral with the floor slab or be built separately of cinder concrete on top of the structural slab. BALCONY DESIGN The most difficult problem in theater construction is to design the balconies without using intermediate columns in the orchestra. This is usually effected by the introduction of one or more fulcrum CONSTRUCTION OF THEATERS AND AUDITORIUMS 764 400/4 Y3L OCU Fe *qooUTYOIV eee UIMp oT (egy ‘d aag) “lovBoyL § ueunedy ae UOTIEY SSOIN— "OLS “OL go —--- Ago | : a) | | le Al Zi call a | rere eet 8 ez ole Id D1789Y040 = 49D SI 4100/4 43990°7 Se auIUuDZZay pel nig Las tines 17 ¢ 692 1 Su aa oa Me a YP ‘ — ssndl aa” ae a wuNiuadsodd [[-L --- — — — — Fe és 968 doa, sassnif fooy ee a 400/] 439 oie aaa 89 01 1D ee Se es ee SS a poe ph pete i ea ie 9PE [7 buldod Ou LS OLTEE IT IAS JO FOL aa FULCRUM GIRDERS 765 girders running across the theater and serving as supports for the inclined cross beams carrying the balcony slab. As it is impossible to place the fulcrum girders at the edge of the balcony, the inclined cross beams must be cantilevered out. often to a considerable length. FULCRUM GIRDERS Arrangement of Fulcrum Girders—The arrangement and the number of fulerum girders depend upon the width of the theater. The simplest arrangement is shown in Fig. 271 a, p. 766. One fulcrum girder is placed as near the edge of the balcony as possible. It is supported on columns concealed in the wall, or, if this 1s possible without obstructing the view, a few feet from the wall. The latter arrangement is practicable, for example, when columns can be con- cealed in boxes or placed at the edge of outside aisles. Full-span fulerum girders are particularly adaptable for small spans. They have been used, however, for spans as large as 126 ft. The balcony seats are usually placed on a curve. With a fulcrum girder placed across the theater, as in the first arrangement, the cantilever arms at the sides may be too long. They may be reduced by using side diverging girders, one on each side; one end of each resting on a column placed in the side wall and the other end on the main fulerum girder. (See Fig. 274, p. 780.) Where the width of the theater is large, it may be economical to use the more complicated arrangement of fulcrum girders shown in Fig. 271 b, p. 766. Here two side girders are supported by four columns, and a middle girder spans between the side girders. This has the advantage of requiring spans not larger than 50 or 60 per cent of the width of the theater. It is economical to make the span of the side girders smaller than that of the middle girder. The aggre- vate leneth of the three girders in this arrangement is greater than the length of the one girder in the arrangement previously described, but their size is much smaller. The required depth for the middle eirder is much smaller than would be required for a girder spanning the whole width of the theater, and it can be placed, therefore, nearer the edge of the balcony and thus reduce materially the canti- lever arm of the balcony beams. Position of the Fulcrum Girder.—Several considerations affect the position of the fulcrum girder in relation to the edge of the balcony. The first is the depth required for the girder. The available depth 766 CONSTRUCTION OF THEATERS AND AUDITORIUMS ‘3 the distance between the top of the balcony floor and the line fixed by the minimum headroom for the seats below. At the edge of the balcony, this available depth is made as small as possible, and usually equals only a few inches. Since the balcony seats are stepped up, the available depth increases rapidly with the distance from the balcony edge. As far as the economical design of the girder alone is concerned, it would be placed where the available depth is equal to the economical depth for the span of the girder. The second consideration is the design of the balcony cantilevers carried by the fulcrum girders. The length of the cantilever arms 4-—-t#--- 1-7-4 \ | i / I 4 [Te al? ‘ fe \ z Balcony Cantilevers Ni Circular birder b Fic. 271.—Arrangement of Fulcrum Girders. (See p. 765.) obviously depends upon the distance between the fulerum girder and the balcony edge. As explained in Vol. II, the cost of the canti- lever is a minimum if the length of the cantilever arm is about equal to one-half of the anchor arm. It will be found, in most cases, that the two considerations are conflicting, and the problem is solved by considering the economy of the construction as a whole. Usually, it will be found economical, for the construction as a whole, to place the girder at a distance from the edge where the available depth equals the minimum depth for which a girder of the required span can be designed. The sacrifice in the economy of design of the girder will be more than offset by the economy in cantilever beams. FULCRUM GIRDERS 767 Shape of Fulcrum Girders.—The fulcrum girders may be either T-beams, trusses, or arches. The design of each type is discussed below under the proper heading. Fulcrum Girders as T-beams.—The slab in the balconies is thin, and cannot serve as a flange for the girder. If a rectangular beam has not sufficient compression area, a special compression T-flange must be provided and the fulcrum girder built as shown in Hign27 7, p. 783. If possible, the flange should be large enough to make compression steel unnecessary. For allowable width of the flange see p. 218. The flange should be as symmetrical as possible about the center line of the girder. In most cases, the width of flange on the balcony side is limited by the shape of the balcony, but on the opposite side any width is available. For this reason some designers use a flange appreciably wider on one side than on the other. The authors do not approve of this, because in unsymmetrical beams the neutral axis is inclined and stresses in steel are not uniform. Spe- cial formulas would have to beused for unsymmetrical beams. In the design on p. 783, the flange was made symmetrical. The total width was fixed by the available width on the balcony side. This width was not sufficient, and therefore compression steel was used. The web of the T-beam is fixed by the shear and also by the required width for the placing of the large number of bars. Some- times it is economical to make the web wider near the supports, where the shear is a maximum. Also, the dead load of the beam may be reduced by paneling the web in the central portion of the beam. In designing the girder, especially where the flanges are thin in comparison with the depth of the girders, it is advisable to use the formulas in which the compression stresses in the concrete below the flange are taken into account. A simple exact method is given on p. 224. The fulcrum girders are almost always of single spans. Since the columns supporting them have a much smaller moment of inertia, it is not possible to restrain the ends of the girders to any extent. Therefore, they should be considered as simply supported, but top reinforcement at the support should be provided in sufficient amount to resist any negative bending that may be developed. Special attention should be paid to the design of reinforcement. To accommodate the large number of bars, an unusually large number of layers are required. To develop full stresses in a large number of 768 CONSTRUCTION OF THEATERS AND AUDITORIUMS bars placed in several layers with a comparatively small amount of concrete between the layers, the rules given below must be strictly followed: First-—The bond stresses must be kept within working limits, while at the same time, to reduce the number of bars, their diameters should be made as large as permitted by bond stresses. Second.—The bars in each row must be kept in place by positive means. The rows of bars should be separated by short bars of proper diameter placed across the main reinforcement between the layers, and spaced about 2 or 4 ft. on centers. The diameter of these cross bars must be equal to the diameter of the largest bar, but must | not be less than one inch. Third.—From 50 to 60 per cent of the number of bars should be bent up and carried to the top of the beam. ‘The points of bending should be determined from the bending moment and well distributed. “Some of the bent bars should be extended along the top to the end of the beam. All of the bars should be provided with hooks at their ends. The bending of the bars serves two purposes: (1), the bent portions act as diagonal tension reinforcement; and (2), by anchoring the trussed bars in the compression zone the development of their strength is insured. Since the bent bars are distributed through the beam, a large mass of concrete assists in developing the tensile stresses in the bars. The number of the straight bars depending upon bond alone is materially reduced. All straight bars should be provided with hooks at the ends of the beam. Fourth.—Stirrups should be used throughout the whole length of the beam, and kept in place by bars placed in the top of the beam. 3 Fifth.—The T-flange of the girder should be provided with cross bars having a total cross-sectional area of at least 0.2 per cent of the section of the flange. These should be supported and kept in place by longitudinal bars placed in the flanges. Sizth—Under no circumstances should the longitudinal rein- forcement be reduced simply by stopping bars at points of reduced bending moment. Designs with a number of straight bars shorter than the span have been used, but cannot be condemned too strongly. It should be evident that stresses in the short bars cannot be induced without slipping of the bars (see p. 289). This would be doubtful practice even in small beams with one or two layers of bars. In FULCRUM GIRDERS 769 large beams, no amount of saving in steel that may result from this arrangement is sufficient to warrant the bad practice. Seventh.—If there is any possibility of negative moment, top reinforcement, properly anchored at ends, should be used even if no advantage is taken of the restraint in computing the positive moment. Fighth—Columns supporting the girders should be designed to take care of any bending moment. Fulcrum Girders as Trusses.—For large spans, the amount of concrete may be reduced by using concrete trusses instead of T-beams. fectangular Truss.—The trusses may be obtained simply by providing rectangular openings throughout the girder. The truss would then consist of chords and vertical members, except at the ends, where solid web may be used for the distance required to provide for shear. ‘This type of construction is statically indetermi- nate. ‘The chordsand the vertical members are subjected not only to direct force but also to bending. The bending stresses in these mem- bers depend upon the external shear and are larger near the end than in the middle portion of the truss. Method of design of trusses is given in Volume III. | Triangular Truss—To avoid the large moments in the members due to shear, concrete trusses are often designed with diagonals. _ The truss consists of triangles, and the direct stresses may be com- puted in the same manner as in steel trusses. In addition to direct stresses due to'truss action, all bending stresses due to the dead load of the members, and to any load placed’ between the panel points, should be computed. In computing bending moments, the members may be considered as continuous or fixed beams supported at the panel points. The stresses due to the bending moments should be combined with direct stresses as explained on p. 187. Because of the rigidity of the joints, secondary bending stresses will develop. These, however, are no larger than in a steel truss under similar conditions. In the shallow steel trusses commonly used in theater construction, the rigidity of joints, due to heavy gusset plates, is about the same as in concrete construction. In both cases, proper allowance ‘should be made for secondary stresses. Reinforcement in the truss members should be placed so that it is able to resist tension due to bending moments to which the members may be subjected. In designing the top and bottom chords, it must be remembered 770 CONSTRUCTION OF THEATERS AND AUDITORIUMS that the computed stresses will not be uniformly distributed over the whole sectional area of the member. T he depth of the chords is large in comparison with the theoretical depth of the truss. After deflection, the top chord will shorten more at the top edge than at the bottom edge, although for chords of small depth this difference in deformation is small. In theater construction, owing to the large depth of the chords, the difference in shortening of the two edges is appreciable. The resulting variation in stresses is proportional to the variation in the deformations. The member must then be designed for an average unit stress smaller than the maximum allow- able stress. ‘The same principle applies to the stresses in the bottom chord. Allowable Compression Stresses in Trusses.—Since the compres- sion members in trusses are subjected to direct compression (similar to column stresses), the same unit stresses allowed for columns should be used, and not the stresses allowed for extreme fibers in flexure. However, when bending moments are considered in addi- tion to direct stresses, the formulas on p. 170 should be used and the allowable combined stress may be accepted as explained on p. 463. Design of Tension Members in Trusses.—In tension members, all the tensile stresses are resisted by the reinforcement, and the area of concrete is made only sufficient to properly cover the bars. The same spacing of bars should be used as recommended on p. 273. 7 While a small number of large bars is easier to handle and makes a compact tension member, its use is recommended only in cases where the tension member extends the full length of the truss, where any splicing is accomplished by mechanical means, such as turnbuckles, and finally where the stresses are transferred to the concrete at the ends by nuts and anchor plates. When the tensile stresses in bars are developed by bond, a large number of small bars is advocated. In addition to direct tension, bending, due to dead load, and second- ary bending stresses must be considered. In some cases, bending moments are so large that the concrete may assume the duties of resisting compression stresses produced by bending. Formulas given on p. 191 should be used. Usually, however, the bending stresses are not sufficient to change the tension, due to direct stresses, to compression. The proper design of the details and connections between members is of great importance. Bars should be spliced only at panel points. To avoid crowding of bars, as few splices as possible should be used. FULCRUM GIRDERS 771 The bars should extend over several panels and should be arranged so that only a few bars are spliced at any one panel point. At the joints, the bars of one member must extend far enough into: the adjoining member (or the intersection of adjoining members) to develop the full strength of the bar, taking into consideration, of course, the character of the stresses to which the bar is subjected. A bar effective in one panel of the bottom chord, for instance, should extend beyond the panel point into the adjoining panel for a length equal to 40 diameters of the bar for deformed bars, and to 50 diam- eters for plain bars. The solid concrete at the joint, which cor- responds to the gusset plates in steel construction, must be of suffi- cient size to enable the concrete there to transmit the stresses from any one member to the other members in the joint. Each joint, in addition to the reinforcement of the members extending into it, should be provided with special horizontal and vertical reinforcement in the shape of hoops or stirrups. These serve to take care of any secondary stresses and preserve the integ- rity of the joint. End Panel Point.—At the end panel points, proper provision should be made for diagonal tension due to the external shear. The structure may fail by forming a diagonal crack between the com- pression and tension members. To prevent this, the vertical area of concrete at the junction of the two members must be sufficient to keep the unit shearing stresses, computed as in ordinary beam design, from exceeding the maximum allowable working values. In addi- tion, the authors recommend that the minimum horizontal cross section of the joint should be equal to the reaction divided by 100 lb. _In this distance, the cross section of vertical stirrups should be equal to the reaction divided by 16000 lb. Stirrups must be looped around the tensile bars at the bottom and around the compression bars at the top. All tensile stresses are resisted by the reinforcement. The area of concrete is made only large enough to cover the bars. The tensile reinforcement in the bottom chord, if it consists of large bars, should be provided with a screw and nut at the end and an anchor plate of sufficient size to transfer all the stresses from the bar in bearing on the concrete. Reinforcement consisting of small bars should be provided with hooks at the ends. If possible, the tensile bars should hook up into the compression member. If not, they should hook around the reinforcement of the compression member. 772 CONSTRUCTION OF THEATERS AND AUDITORIUMS Relative Economy of T-Beam and Truss.—The relative economy of the T-beam and the truss for fulcrum girders depends upon the cost of formwork. The truss has much less concrete and steel, but the formwork is complicated and the placing of concrete and steel more difficult. In many instances, the truss without diagonals, | (that is, with rectangular openings) even if it is necessary to use solid web at the ends, may prove more economical than the lighter truss with diagonals because of cheaper formwork. In cases where a decided saving cannot be demonstrated, T-beam design should be used in preference to the truss as it 1s easier to build. Fulcrum Girders as Arches.—Arch construction is well adapted for fulerum girders for large spans. As it is not feasible to make the support unyielding enough to take the horizontal thrust, the arch must be provided with tie-bars of sufficient area to resist the thrust. If properly designed, the cost of the arch is smaller than the cost of the truss under similar conditions, mainly on account of the simplified formwork. For designing purposes, the arch should be considered as hinged at the supports, which makes the horizontal thrust the only statically indeterminate quantity. The cantilever beams may be placed between the top of the arch and the bottom level of the tie-bars. Tension bars of sufficient area to suspend the cantilever from the arch must then be used. The method of design is discussed in Volume III. Special Concrete for Fulcrum Girder.—To save dead load, it may be economical to use richer mix of concrete for fulerum girders. A mix of 1:13:23 or even 1: 1:2 may prove economical, especially when, with the leaner mix, compression reinforcement would be required. BALCONY CANTILEVERS The balcony floor slab is carried by beams, which in turn rest on the fulerum girders. As already explained, the fulcrum girders must be placed some distance back from the edge of the balcony, and therefore the balcony beams must be cantilevered out beyond the girder. Reinforced concrete is well adapted for the balcony cantilevers and is being used for this purpose, not only in reinforced concrete theaters but also in connection with structural steel fulerum girders. BALCONY CANTILEVERS 173 As a general proposition, the cantilever beam carrying the balcony consists at one end of the cantilever arm, resting on the fulcrum girder, and at the other end of span the anchor arm, anchored to the support to prevent uplift. In modern theaters, the length of the balcony may be considerable. With only one fulcrum girder, the length of the cantilever arm may exceed 30 ft. If possible, the cantilever arm should not be longer than the anchor span. In no case should the uplift caused by the load on the cantilever arm (multiplied by proper factor of safety) with the anchor span unloaded, exceed the available downward reaction at the point of anchorage. Moments and External Shear in Cantilever.—The beams must be designed for three assumptions as to the disposition of the live load, each of which produces maximum stresses in some part of the con- struction. The assumptions are as follows: 1. The cantilever arm is loaded, the anchor span not loaded. 2. The cantilever arm is not loaded, the anchor span loaded. 3. Both arms are loaded. The first condition, where only the cantilever arm is loaded, produces maximum stresses in the cantilever arm and also maximum negative bending moments in the anchor span. The uplift is also a maximum for this condition. The second condition, where the anchor arm ant is loaded, produces maximum positive moments in the anchor span. The third condition, where both arms are loaded, produces maxi- mum shears in the anchor span and also maximum reactions on the fulcrum girder. The moments and shears for the live load should be combined with those for the dead load. For some conditions of loading, the stresses produced by the live load are of opposite sign to those pro- duced by the dead load, and if the two are added, the dead load stresses may balance partly or fully the live load stresses. Under such conditions the combination of live load stresses and dead load stresses by a simple addition, is not permissible, but the live load stresses must be multiplied by the required factor of safety. When the live load is doubled, the stresses produced by it are doubled also. The dead load stresses, however, remain the same, so that the reduc- tion in the live load stresses due to the dead load stresses is propor- tionally only half as large for double the live load. The following example explains the matter more fully. Assume 774. CONSTRUCTION OF THEATERS AND AUDITORIUMS that the anchor span is 20 ft. long, cantilever arm 12 ft., uniformly distributed dead load 1 000 lb. per lin. ft., uniformly distributed live load 1 400 lb. per lin. ft. Dead Load Reactions: Due to load on cantilever arm: At fulcrum girder, lett. 5. eee 12 000 Ib. downward At fulcrum girder, right.......... 3 600 lb. downward At. end SUuDpOTt ses ace oe .. 8600 lb. upward Due to load on anchor span: At fulcrum girder, right.........-. 10 000 Ib. downward At'end SupDORisie- = saa -~. 10000 Ib. downward "The sum of the two reactions gives: Reaction due to dead load: At fulerum girder, left.........7¢5: 12 000 lb. downward At fulcrum girder, right, 10 000 + 3 600 = 13 600 lb. downward At end support.. .10 000 — 3600 = 6 400 lb. downward No uplift due to dead load. Live Load Reactions, Cantilever Arm Only Loaded: At fulerum girder, right. .......... 14000 lb. downward At fulcrum girder, left...2 72-1. eee 5 040 lb. downward Atend'support: 25.20. 73 5 040 lb. upward Uplift due to live load, 5 040 lb. Combining the uplift due to the live load, equal to 5 040 Ib., with the downward reaction due to the dead load, equal to 6 400 lb., it will be found that no uplift will occur for this condition of loading, and therefore no anchorage would seem to be necessary. But in such a case, practically no factor of safety against uplift would exist. If, now, the live load on the cantilever is doubled, the uplift due to the live load will also be doubled, or 2 X 5 040 = 10 080 lb. The downward reaction due to the dead load will remain the same. The net uplift, therefore, for double live load on the cantilever, will amount to 10 080 — 6 400 = 3 680 lb. mtr | To get a factor of safety of 2 against uplift, it is necessary to anchor the structure strongly enough to resist an upward pull of 3 680 Ib., even if for the combination of single live and dead load a downward reaction is obtained. BALCONY CANTILEVERS 115 The same applies to the stresses due to negative bending moment in the anchor span. To provide a proper factor of safety, it may be necessary to provide negative moment reinforcement in places where the combination of one live load and dead load give positive moment. This difference between ordinary beam design and cantilever design must be constantly borne in mind. To combine dead and live load moments when they are of oppo- site sign, the procedure is as follows: The moments due to live load are multiplied by the desired factor of safety. ‘The moments due to dead load are subtracted. The difference is then divided by the same factor of safety and the dimensions of the member determined in the usual fashion. In many parts, a reversal of stress may take place; that is, for some loading the stress will be tension, and for other loading, com- pression. The frequency of the reversal is not sufficient to require any reduction in unit stresses. | Depth of the Cantilever Beam.—In most cases, for designing purposes, the cantilever beam is considered as a rectangular beam. The required depth of the cantilever beam is governed either by shear or by the negative bending moment, both of which are a maxi- mum at the fulerum girder. In the cantilever arm, because of the slope of the balcony, the beam decreases in depth, and this decrease may be larger than the decrease in bending moments and shears. Therefore, it is not suffi- cient to determine the depth and the amount of steel at the support; it is also necessary to investigate the stresses at intermediate points and design the sections to correspond. In the anchor span, the shears and bending moments are largest at the fulcrum girder. The depth of this girder, therefore, is made the same as for the cantilever arm at this girder, and the beam is then tapered toward the other support. The taper may be uniform, or a fairly steep hunch may be used at the support followed by a gentle taper toward the end of the beam. The depth of the cantilever beam must be the same on both sides of the fulerum girder, so that the compression stresses on one side may be resisted by compression stresses on the other side of the girder. If the depth were different, torsion would be developed in the girder. In some designs, the moment produced by the cantilever arm is resisted on the other side of the girder by a tie beam on the top and a strut at the bottom. It is obvious that in such designs 776 CONSTRUCTION OF THEATERS AND AUDITORIUMS no provision is made for the shear produced by the bending moment, and as a result heavy secondary stresses are developed in both members. Reinforcement for Cantilever Beams.—The reinforcement deter- mined for the negative moment in the cantilever beam at the fulcrum girder must be extended on both sides. In the cantilever arm, at least one-third of the bars should extend on the top for the full length of the arm; the rest may be bent down to resist diagonal tension. In determining the points where it is permissible to bend the reinforcement, it is necessary to remember that, while the bend- ing moments decrease with the distance from the support, the depth of the cantilever decreases. For steep balconies, the decrease in depth may be almost as rapid as the decrease in bending moments, -n which case no reduction in negative steel is permissible and all bars must extend on the top for the full length of the cantilever arm. In the anchor span, some of the top steel from the cantilever arm should be carried on the top, for the full length of the anchor span, because, for the most unfavorable position of the load, negative moment will be found throughout the span. This happens when the stresses due to live load on the cantilever arm only are multiplied by the factor of safety and are combined with the dead load stresses. The rest of the bars may be bent down, when permitted by the moments, and extended at the bottom for use as positive moment reinforcement. Sometimes, the cantilever is of such shape that it 1s difficult to extend the bars along the top from end to end without difficult bending. In such cases, bars may be spliced, provided ereat care is used to break joints. Some steel should be used at the bottom of the cantilever span, even if not required for compression. Positive bending moment reinforcement in the anchor span should be treated as in ordinary beams. Cantilevers in Steel Construction.—Concrete balcony cantilevers may be used with economy in connection with structural steel ful- crum girders. In all cases, the girders are deep enough to allow sufficient depth for an economical concrete member. The economy effected by the use of concrete instead of steel consists not only in the saving in the cost of the member itself, but also in the simplifying of the connections between the girder and the cantilever. The design of a cantilever supported by a steel girder does not differ to any extent from that of a cantilever supported by a rein- forced concrete girder. The only interesting part is the method of Ted BALCONY CANTILEVERS (S22 ‘da aay) "ssnay, wans[N YY 79019 Aq poytoddng JeAsTI}UB_D 9JO1IDUOD) PIdIOJUlEY—’Z/)Z ‘OLA _ Spox Ss Spa rsa Ae ee eee Te OO a care aa a wuoji0g pub Jo/ SUD Ul SININZS$ Z iia ae ~---4---- 7783 CONSTRUCTION OF THEATERS AND AUDITORIUMS supporting the cantilever on the steel truss. In Fig. 272 is given acan- tilever used in the theater auditorium of the State-Lake Building in Chicago.! The part of the supporting girder is also shown, as well as the detail of the seat for the cantilever. The concrete poured between the gusset plates and resting on the bottom chord, serves as a seat for the cantilevers. Cantilever Trusses for Balconies.—Ordinarily, cantilevers are rectangular beams. For large spans, trusses have been used, as shown in Fig. 278, p. ¢#9.- BALCONY FLOOR CONSTRUCTION The top of the balcony is stepped. The most common construc- tion consists of vertical ribs, spanning from cantilever to cantilever and forming the risers of the steps, and a thin slab spanned between the ribs. The thickness of the slab is usually 3 in. The ribs are about 5 in. thick, and their depth is governed by the required height of riser for the step. The ribs must be designed as beams, each carrying half of the load on the two adjoining slabs. The slabs should be reinforced with bars extending into the ribs. Although negative moment will be developed in the slab, it is not practicable to bend up the slab steel. Wire mesh is sometimes used as slab reinforcement. The top of the cantilever extends to the top of the floor and is also stepped. Another method sometimes followed is to use an inclined slab between the cantilever beams (or inclined slab and joist construction) and then build the steps of lean cinder concrete built on the top. In this way the formwork is simplified, but the construction may prove heavier and more expensive than in the previous case. ROOF CONSTRUCTION The roof construction may consist either of simple girders or of trusses. Arches with ties may also be used. The long span roof construction is discussed in Chapter XIII. THEATER AT WINSTON-SALEM, N. C. Examples of Theater Construction—A good illustration of the adaptability of reinforced concrete in theater construction. is shown 1 For description, see Engineering News-Record, Vol. 83, July 24, 1919, p. 178. 2 For full description, see Engineering News-Record, July 5, 1923. kde: BALCONY CANTILEVERS ‘PITY “Wolssiog UMpy (‘BLL ‘d aag)) ‘sejesuy sory ur Jozvoy J, s,uBUINeI) Jo AMOD[eg I0J SsnIy, IOAITUBO—"E)z ‘DIT — ety ees eiighoe: rssog bs 2-2, sings 2-91 Poe Ree f a“ 7 ¢ g 6a Ba tl iN Sapsl9 Auorjo i -----U/Dag ee ey u9Ghi---=-=3 -=-,,0-,6-—3 wO>L- “us <------,0-, 41 ie _-z--/[Off 8J8-40U09, 2 “9 = HE 7d 780 CONSTRUCTION. OF THEATERS AND AUDITORIUMS Proscenium Beam 1 ee Say otic iS ay Se First Balcony Framing Plan Winston-Salem, N. C., —First Balcony Framing Plan. Fia. 274 (See p. 784.) Auditorium. 781 THEATER AT WINSTON-SALEM WNMOPIPNY “OD *N ‘UreTeg-uoYSUT AA YsINoIy} WOTaG ssorg—eyz ° (UMOYS Siubag /apUBAS) Y-y LoK.0aS /DUIpny/bLO7 E[---- --—— ‘ 3U/7 YsoMapIs 1d 960497909 100) / pualliasDg ne oi6d BSA La > Me lFi ‘ ‘ ri 4 > | ’ ‘ i” ' - ~707€L Elz suipag fo puz ~ p40ddnS 0 J 4% Le# 10) U0 Jog09 | ~~ —~ 4 ESS E0F- Whe ¢--70 en g, 7 6 “ BSOAAMMAAAASSSSANSSSS PTT TTT] ODE See OZEN TS EDS, pets Fespuods sng 7204 Ul a/Do3 CONSTRUCTION OF THEATERS AND AUDITORIUMS 782 Fs oe ae “a -iqoulsug Burypnsuod ‘Tys[NUIG PABMPH ‘g190uISUG OD SuLeoulsuG, UIOYINOS (PRL d aag) “wnTTOypPNYy Coy NT (UeTBg-UOysUT MM — “JOAOTIUBL) Kuooeg ysatq Jo spreyeqd—'9Le “OM fiuoojng 38414 yGnosyy u0!2298 Sth Fo ee a ee ee ee eam me FD erie ape ae? Fn ei oa a a ee at te a decals wie? es a 20 ¢L py aS) q = ae 2 5 9 5 ae S ts CO & S = et 8 Mine 3 ls S we ae ale = @ Be ane o 4 x SS nr S ” ‘ —— Ss ~ = - LS S See oD (ats ~— YW -” i i) oO ale 2 5 A ~ a ~4 | ‘ . 3S % a . = I~ Gels tf : N — g = es 784 CONSTRUCTION OF THEATERS AND AUDITORIUMS in Fig. 274 to 277, taken from the design of the theater in Winston- Salem, N. C., designed by W. C. Northup, Architect. The structural plans were prepared by The Southern Engineering Go., with Mr. Edward Smulski as Consulting Engineer. The plan of the first balcony is shown in Fig. 974. The cross section through the theater is shown in Fig. 275. Fig. 276 gives reinforcement in the first balcomy cantilevers. Fig. 977 shows the design of the fulerum girder. CHAPTER XIX REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS In this chapter, dealing with the different types of buildings in which reinforced concrete is commonly used, the economy of con- struction is briefly discussed and the design live loads are given. Recommendations are made as to proper spacing of columns and proper type of floor construction. Floor finish is also discussed. The following types of buildings are considered: warehouses, p. 785; cold-storage warehouses, p. 792 ; manufacturing buildings, p. 792; buildings used for the manufacture or storage of automo- biles, p. 795; office buildings, p. 803; hotels, apartment houses, and hospitals, p. 805; schools, p. 808. WAREHOUSE CONSTRUCTION Reinforced concrete is well adapted for warehouse construction, and a large majority of fireproof warehouses are built of this material. Type of Floor Construction.—The majority of reinforced concrete warehouses are of flat slab construction. Beam and girder con- struction or joist construction is rarely as economical as the flat slab for this type of buildings. Joist construction should not be used for heavier live load than 125 Ib. per sq. {t., because of the possibility of heavy load concentrations for which joist construction is not adapted. An idea of the relative economy of several types of floor construc- tion may be obtained from the table on p. 786, taken from a paper by A. F. Klein.! This table gives costs of three types of floor con- struction for a warehouse consisting of eight stories and basement, with 20 by 20-ft. panels designed for a live load of 250 lb. per sq. ft. The following unit costs were used: * Analysis of Cost of Types of Fireproof Construction. By Arthur F. Klein. Journal Western Soc. of Engineers. Vol. XXIX, No. 7, July, 1924. 785 936 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS eL 1g = “37 "bs rod 4s09) ZOL 0% = “35 7D tod 4sod EL $50 PH = 3800 THIOL $G' 1S = ‘bs rod 4sor) 9zI' 0$ = “HT rod 4s0¢) ¥Z SPS S$ = 3809 THIOL Sr 13 = Pol 0$ = 26-668 S$ = 3800 TBIO.L "yy ‘bs “yy NO Jod 4s079) rod 4s0d 00° Zz 20°0 | 34 bs 009 € \00 Ze 10°0 | 33 PS 009 € \00 ZSe WC, CSN es ae ystuly JUoULE| OL 6tF Oe) ‘sq] F668 |20' SEG 670 0 ‘SdT 086 OT\28°8Z9 |670°0 "SCT PON Cll cee ee BUTOLOJUIOY oo rT | 00'9T| “#46 p0cer | 0081). a6 oo'ser |odset) “a6 se AS ck SEN ay - SULIO Of LIS ‘PO | 33 N12 F6P = 00 SSG HO |9F NOGLE 66° 8&G FPO | M9 SFG seseeseessemgrU0r) FHS: 1 :SUUINIOD) c,'0z¢ —s« [S800 ‘Sq] GPS OT\ZE'SL6 — [8¥0'0 ‘sq] 6TE OZ|SP'996 [8700 SOE GE | OC. 2 oe ee [2078 SuIo1OsUIOyY 00 06 00 OT; ‘sed 6 OL 986 1¢°0 | 93 ‘bs OSE Z@ ITP STS yey toap be eGs Tey (spoueg doiq ‘S1epaty) . |pue sued ‘slopixy) SULIOM OFZ. | YEO 14? ‘bs 909 & [OO OTT T | 180 | ¥ ‘bs 009 & (00 FFOT | 62°0 | “5 ‘he O00 G3 ee ees (sqv[g) SULIOY Ze OCT 1$| PH OF] “HF “NY 829 @ 08 6&E I$ | FF 0$| “3 M2 SPO & ez GPS 1$| PHOS) HF MOTSSS [oO gjyo1vU0D 7: Z: T :LOO[ ee Mua po ee ee 4soc 4sor) 4soc) 4sor yug Ayxyuene) 4son ug Ayyuent) 4sod ont AyyUeNny) a Mae eel Laks ee es So eS Se uorydiioseq IOO[ IOO],J JOpITY) JOO] J Jopiry pus qejg I1I—08Z “St—*e adkyp|pue weog—61Z “SI1—Z adh ilqvig AeM-Z—8L6 “SLA—T edX J, ‘qyStoy A104S8 BOO “FF OT “UT ‘bs sod “q] 0GZ ura “a “WV Ad peol ory “35 0% 44 “45:02 SPPUPd -juoUTESt PUB SOLIO}S g “OSNOTETE MA WoOHINI}SUOD OOTY JO sodXy, 20IG] YM Supping Jo 4S07) jo uostredui0) WAREHOUSE CONSTRUCTION 787 aleankeata 110" Lis ead Girders 10° x Ig" syed Typical Floors a a0: xed” -2-Way Slab and Girder Floor for 8-Story Warehouse. (See p. 786.) Fig. 278.—Type 1. eceqent Fish S/abs Beams Girders SEN IO MAL Reo! Proof a Typical Floors 6" 20"x 24" 20"x 28" x ‘ ! | ! ! ! ! Fig. 279.—Type 2. Beam and Girder Floor for 8-Story Warehouse. (See p. 786.) Cement dele c R S| Roof drop panels 727" xP T'x3e, : S| Typ. Fl drop panels 7-7"x7*7'x45 SS = us Co/umn Heads 4-6” diameter ©o wy 1 Roof Slabs 6" yn Typical Floor 83 Fic. 280.—Type 3. Flat Slab Floor for 8-Story Warehouse. (See p. 786.) 788 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS Cost of Concrete—Cement, $2.40 per barrel net. Crushed stone and torpedo sand, $2.74 per cu. yd. delivered. On this basis, cost of 1 :2:4of concrete in place is $12.50 per cu. yd., or 44 cents per cu. ft. ; and of 1: 1 : 2 concrete $15.00 per cu. yd., or 54 cents per cu. ft. Lumber for formwork figured at $47.00 per thousand feet, board measure. Reinforced Steel—Cost of steel based on $60.00 per ton, base, with a standard increase in price, for sizes below base. Following additional costs were used: Per ton Trucking: .. 2... - #1 eel th Sines = one $3.00 Standard bending of steel.........-.+-+++++-> 6.00 Bending stirrups. ./...5. .- =>". pe 18.00 Shop details. .....-.4- +--+ = =e a 3.00 Spirals... +a cen eines o> 2 90 . 00 Placing of steel. 2... . 1417 eee 20.00 The cost of formwork is based on the assumption that forms for 9} floors will be provided, and therefore that they will be re-used four times. , Values given in the table are total estimated costs of an interior section of a building equal in area to 20 by 20-ft. panel and 8 stories high. The total costs are as follows: Cost per section Cost per of 8 stories. sq. ft. Flab slab design. ...+ +. «25+ e-0+ +++ += nie $4 054.13 $1.13 Square panel with four hears Aer ac. 5 ce 5 320.02 1.48 Beam and girder design. .....--+++-+++es000 5 645.24 1.54 The cost of flat-slab design is much lower than that of beam and slab designs. The difference in cost between the two-beam and girder types is small. It must be noted that the above figures cannot be used for esti- mating the cost of the frame of an entire building as the cost of exterior panels will be larger, owing to heavier construction of floors, additional cost of spandrels, and also additional columns. In figuring the cost of an interior panel, the cost of one interior column only needs to be included, while for an exterior panel it is necessary to use one exterior column plus one-half of an interior column. In a corner panel, extra cost of corner column must be considered. WAREHOUSE CONSTRUCTION 789 Spacing of Columns.—Column layout should be governed pri- marily by the requirements of the occupants. If no other consider- ation governs, the most commonly used and most economical spacing varies from 18 to 20 ft. The panels, preferably, should be made Square or nearly square. Where the shape of the building can be made to suit, its length and width should be multiples of the accepted square panel. The end panels may be made somewhat smaller than the interior panels, as the moment coefficients in end panels are larger than in interior panels. With smaller span, it is possible to maintain the same dimensions in exterior and interior panels. When the dimensions of the building are fixed by the size of the plot, the choice of the panel sizes is restricted. For instance, if the outside dimensions are 100 ft. by 120 ft., 20 by 20-ft. panels follow almost automatically. The 100-ft. length, divided into four parts instead of five, would give a panel length of 25 f{t.; divided into six parts, it would give a panel length of 16 ft. 8 in. The possible panel sizes in a 100 by 120-ft. building are 16 ft. 8 in. by 17 ft. 1 in., 20 by 20 ft., and 25 by 24 ft. The smallest panel may be cheaper than the 20 by 20-ft. panel, but the building will not be as useful. The 25 by 24-ft. panels are too expensive. Ordinarily therefore, 20 by 20-ft. panels would be selected. If the building is not rectangular, the main column lines should be laid out at right angles to each other and the skewed portion composed of odd panels. This rule does not apply when the building lines form arhomb or rhomboid. In sucha case, the opposite sides are parallel, but the intersecting sides are not at right angles. The column lines then may be placed parallel to the sides of the building. Design Live Load.—The magnitude of the design live load depends upon the uses for which the warehouse is designed. No warehouse should be designed for a smaller live load than 125 Ib. per sq. ft., a live load permitted for furniture storage. This load should never be used if there is any possibility of a change in the ownership of the warehouse. For general storage,a minimum live load of 150 Ib. per sq. ft. is recommended. In warehouses several stories high, a good plan is to design the first floor for 250 to 300 lb. per sq. ft., the next floor or two for 200 lb. per sq. ft., and the balance for 150 Ib. per sq. ft. ‘The rentability of such a warehouse is much superior to that of one in which all floors are designed for a small live load. In determining the live load for a warehouse which is to be rented, it should be kept in mind that the adoption of too small live loads 790. REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS may restrict the use of the warehouse considerably, and may make ‘t less rentable, so that the small economy in construction may result in actual loss. The difference in cost between a warehouse designed for 150 lb. and one designed for 250 lb., at prices prevailing in 1925, is only 10 cents per sq. ft. This means an increase of about 8 per cent in cost of the structural frame, but only 3.5 per cent of the cost of the complete structure. | In large cities, the capacity of a floor is posted. Although in many cases this is only of theoretical value, still there is a possibility of the law being enforced. A warehouse that is constantly overloaded will develop cracks which, even if not dangerous, are unsightly and limit the market value of the building. The difference in cost between a building with light live load and one with heavy live load is negligible when the larger usefulness of the latter structure is taken into con- sideration. Loading Platform.— Where possible, provision should be mide for railroad deliveries of goods, by the addition of a covered loading platform extending the full length of the track. The width of the loading platform may have to be made large enough to afford temporary storage for the goods. The loading platform should be designed for a minimum of 300 Ib. per sq. ft. Where ground is available, the side tracks are run outside of the building and the platform projects from the building. It is usually covered with a permanent canopy. In U-shaped buildings, the tracks may be placed in the court. | | In cities where the ground is very valuable, it is not possible to waste the space which would be required for tracks outside the build- ing, and the tracks are therefore run into the building. A straight track, running parallel to a column line, can easily be accommodated in one panel width. The curve required to connect the side track with the main track should be placed preferably in the court or out- side the building. If this curve is placed within the building, it is necessary to offset some of the columns in such a way as to give the required clearance for the train. Sometimes the position of the offset columns is maintained in all floors. Usually, however, it is cheaper, and gives a better job, to maintain uniform spacing of the columns above the second floor. The columns above the track are then carried by heavy girders placed below the second floor. Such girders will usually be of rather large dimensions. If possible, they should be built of reinforced concrete, even if the cost of the materials WAREHOUSE CONSTRUCTION 791 is somewhat larger than for steel construction, because the reinforced concrete construction is simpler and therefore results in a smaller total cost. If the headroom is not large enough for a concrete girder, steel girders must be used. In such cases the supporting columns below the girders also should be of structural steel. Sometimes the depth of girders to carry the offset columns is made equal to the story height and the girders are placed between the second and third floors. The center line of the tracks should be at least 8 ft. from the face of the building. The top of the loading platforms should be 4 ft. above the top of the rails. Loading Platform for Trucks.—Each warehouse of any size must have provisions for loading merchandise on trucks and unloading it from them. The truck-loading platform is usually placed inside, in order to make the loading operations independent of weather con- ditions. It is often necessary to omit some columns in the driveway to facilitate the operation of trucks. This is expensive, as the upper columns must be carried on heavy girders. The level of the driveway should be 3 ft. 8 in. below the floor level, so as to bring the platform on the level with the truck floor. Elevators.—In modern warehouses, elevators are used to move goods up and down, and their location, number, and size are therefore of particular importance. Ifit isimpossible to ascertain ahead of time the number of elevators required, special provision should be made for installing additional ones in the future. This is done by providing the same framing as used in the elevator shaft, and covering the opening with a temporary removable slab. This expedient is often resorted to in cases where provision is made for additional stories. Until the upper floors are added, the space may be used for storage. With the increase in number of stories, the temporary slab is removed and additional elevators installed to provide for the increase in traffic. : Floor Finish for Warehouses.—Granolithic finish is commonly used in warehouses. When there is much heavy trucking and the main travel of trucks is along a well-defined lane, special hard finish is provided for the truck lanes. 792 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS COLD-STORAGE WAREHOUSES The foregoing discussion of warehouse construction applies equally to cold-storage warehouses. ‘The difference in construction is only due to the use of insulators in the cold-storage warehouses. The insulator usually consists of lith or cork board 4 to 8 in. thick, depending upon the degree of insulation required. The floors are insulated by laying the insulator on the concrete in hot asphalt. A 3-in. concrete slab is placed on the top of the insulator. To be effective, the insulation must be continuous. Not only the floors but also the columns must be insulated. The exterior columns usually consist of two sections, one carrying the floor load and the other the brickwork. A space is left between the two columns, sufficient for the placing of the insulator. For details of Insulation see Engineering New-Record, July 24, 1924, p. 140. Live Load for Cold Storage.—Live load depends upon the weight of goods to be stored and may vary from 150 to 250 lb. per sq. ft. Floors carrying refrigerator machinery and tanks should be designed for special loads. The live load for floor sections carrying freezing tanks should be 350 lb. per sq ft. To the load should be added the weight of the refrigerating coils suspended from the ceiling. | Type of Floor Construction.—Flat slab is the most commonly used floor construction for cold-storage warehouses. The state- ments made in regard to spacing of columns and other matters, in connection with ordinary warehouses, apply with equal force to cold-storage warehouses. MANUFACTURING BUILDIN GS Manufacturing buildings are of two types: One-story buildings, and multi-story buildings. One-story buildings are generally used where the process of manufacture requires breadth, length, and height. The roof spans in such buildings are large. While reinforced concrete 1s often used in one-story construc- tions (see Garfield Foundry, p. 666), steel truss construction will be cheaper in first cost. The additional cost of upkeep, however, may more than balance the difference in first cost. : Fireproof multi-story buildings are almost universally built of reinforced concrete. Spacing of Columns.—In designing a factory, a machinery layout of the floors should first be made, with due consideration of economy MANUFACTURING BUILDINGS 793 of construction of the building. If no special arrangement of columns is required by the process of manufacturing, their spacing is governed entirely by economy of construction. While 18 to 20-ft. panels are usually most economical, larger spans may prove much more satis- factory from the standpoint of usefulness of the building. The addi- tional cost of long spans may be more than offset by greater freedom in the arrangement of the machinery. For buildings with special machinery, such as printing presses, the spans are governed by the dimensions of the panels. Thus, in the Youth’s Companion Building,” the floor plan for which is shown in Fig. 281, p. 794, the spans were governed by the arrangement of the presses. In shoe factories, where the process of manufacture progresses longitudinally with the building, construction two spans wide is con- sidered advantageous. Thus, in the Ideal Shoe Factory,? shown in Fig. 282, the spacing of the columns across the building is 20 ft. _ From the standpoint of cost of the structure, it is advisable to use equal spans throughout the building. However, to fit special require- ments, the dimensions of the spans may have to be varied. Thus, in the Clark Biscuit Co. building, the spacing of the columns across the building was not equal, as it was adjusted to suit the require- ments of the manufacturing process. Type of Construction.—For single-story buildings, the roof con- struction may be of any of the long-span roof types described on pp. 661 to 679. For multi-story buildings, flat slab construction is most com- monly used. In cases where the spacing of columns is not adaptable to flat slab construction, beam and girder construction may be used for heavy loads and light weight construction for light loads. Design Live Loads.—Design live loads depend upon the purpose for which the building is intended. Light manufacturing buildings may be designed for 100 to 125 lb. per sq. ft. Ordinary manufac- turing requires a live load of 150 lb. per sq. ft. Heavy manufacturing buildings may have to be designed for 200 to 400 Ib. per sq. ft. Por- tions of the building used for storage should be designed accordingly. Full live load should be used in designing beams and slabs, also joists in light-weight floors. For girders carrying an area over 300 2 Densmore, LeClear & Robbins, Architects, Sanford E. Thompson, Consult- ing Engineer. 3 Mowll & Rand, Architects, S.M.I. Engineering Co., Engineers, 794. REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS (26. ‘d ag’) “uo sOg ‘Zurpting uotuvdwm0g s,yynoK “Ue 1OOLY IST q7— Tks “Pld $40} DNA/T sabuassDdd wooy busyjidp 67> ot SADE E @ oie @ @ pe agnya ow) @ @ & @ @ ubiseag qoig io}4 fo $1 400]4 Of oer sOhaae g “ ssozona/7 jy blasly PO4S 2M A He er re 7 CRN Sig panes x cn: oh: on: oa: a: 3 oa: $ 5 5 AY i eal Te hchhehel CPEEEBOIEOEE. a Gf ~ EL RQ’ f. 224 NA SSS wn cane nna ean n a nen WONG Bn nn gener RASTA i cme ee ee ee re ee ea ee EE Oe a Lee ee oe AUTOMOBILE BUILDINGS AND GARAGES 795 sq. ft., as well as for flat slab construction, a reduction in live load of 15 per cent is permissible. Loading Platforms.—See corresponding section on previous Bee under ‘‘ Warehouse Construction.” Floor Finish.— Because of the large variety of structures embraced in this section, no general recommendation as to floor finish is possible. Obviously, heavy manufacturing buildings require different treat- ment from light manufacturing buildings. In buildings with con- siderable trucking special floor finish may be required. aa, La ip ae mene en | Nee ical au I-08 et "2 : 2 Ste v 4 SF ~ & ; & 1 Flat Slab Floor ‘Construction iC f ins re) fo) D fo) ) ° . ° 3 | K--20:07>+ 20:0'> & Paes S we ee Giyp Rae | . i lL k 17-0 tl aly a , ee re ee te aah aha 80: 93°-------£-- 4 2:Il5 s ifs Q $ eae aaa tg S S 4 % 1 | ! iy | ! tof oy y fh | Fig. 282.—Ideal Shoe Factory, Brockton, Mass. Oye p. 793.) Granolithic finish seems to be the best all-round type suitable not’ only for ordinary use but also for trucking. However, the oper- atives sometimes object to it, and for this reason hardwood floor is often used. In some processes, the slight dusting which occurs in new granolithic floors is objectionable. This dusting will occur even with the best granolithic, unless it has a ground surface. For further discussion of floor finish, see p. 620. AUTOMOBILE BUILDINGS AND GARAGES Concrete is used to a great extent in the construction of buildings for use in connection with automobiles. These may be divided into buildings for the manufacture of automobiles, which are similar in 796 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS design to other manufacturing buildings, and buildings for the storage of cars, such as garages and service stations. The garages may be designed either as commercial garages, where the automobiles are handled within the building by garage men, OF as private garages where the owner drives his cars in and out. Such garages are built by hotels and department stores for their patrons. The difference sn the two cases in the arrangement of the columns may be consider- able. Buildings for service stations and storage will be discussed below. Live Load.—All garages and service stations should be designed for a live load of 150 lb. per sq. ft. This live load should be increased if the garage is likely to be used for storage of heavy trucks. Full live load should be used in the design of beam and girder or joist construction, for slabs, beams, and joists. For the design of the cirders, however, when the area carried by the girder exceeds 400 sq. ft., the live load may be reduced by 25 per cent. The same reduc- tion may be made in design of flat slab construction. The reason for the suggested reduction 1s obvious. While it is possible to place a wheel load on beams and slabs so as to vet the effect of the full live load, it is hardly possible to load a large area sufficiently to produce the effect of a uniformly distributed load of 150 lb. per sq. ft. Size of Cars and Trucks.—The spacing of columns and the width of aisles depends upon the overall dimensions of the cars. These are given below. Size of Passenger Cars.—In the table below, the various makes of passenger cars are divided into five groups, and the average dimen- sions for each group are given. Overall Dimensions of Passenger Cars By re Width | Height Ford, Star, Overland, Chevrolet. £0 eee = 11 ft. 7in.| 5 ft. 5 in. | 6 ft. 4m. Durant, Essex, Buick, Studebaker, Maxwell, Dodge, Hupmobile, Dort......---++---> 13 ft. 6 in.| 5ft. 5 in. | 6 ft. 4m. Willys-Knight, Jewett, Franklin, Buick, Oldsmobile, Lexington, H. C. Stutz, . Hudson. cere ae ee ee ree 14 ft. 6 in.| 5 ft. 8 in. | 6 ft. 6 im. Nash, Apperson, Jordan, Paige, Marmon, Studebaker, Haynes. ...-----+-+s+ 7070? 15 ft. Gin. | Dito meO eee in. Cadillac, McFarlan, Pierce Arrow, Packard. | 16 ft. 6 in. | 5 ft. 8 in. | 6 ft. 7 in. TE In each group, cars are arranged according to size, AUTOMOBILE BUILDINGS AND GARAGES (ey From the above table, it is evident that an average space of 6 ft. 6 in. by 15 ft. is sufficient for most makes of passenger cars. The width of the storage aisles, from center to center of columns, there- fore, may be made from 13 ft. 6 in. to 15 ft. size of Trucks.—The dimensions of trucks vary more widely than the dimensions of passenger cars. average overall dimensions for trucks of different capacities. Overall Dimensions of Trucks In the table below are given a a a ea ee Capacity in Tons Length Width Height 5 to 7 15 ft. 5 ft. 6 in. ¢ ft, 10 in. 1 to 13 18 tt. 5 ft. 9 in. 8 ft. 6 in. 2 18 ft. 6 in. 5 ft. 9 in. 9 ft. 3 in. 3 20 ft. 8 in. 6 ft. 6 in. 10 ft. Oin. 5 21 ft. O in. 7 ft. 6 in. 11 ft. 9. 0 ae tee me SASSY Sth | spn Aa/7 | 409. \ SSAA y- uoroags SS ISSNISNSSS J-7 Uoj0ag pUaLLIasDg 2D //D\f 3}a40U09 a0 90 fo 3 985-7 Sop /O79 aur fo7 SS i 8 OH Po . . p40 /4a2uNED gd Y/ = *@© yoo} 9 ayoug SSSVQ“¢s SSS SAAN 135 SNANAAY ’ ’ oy SS SS jhayonig: Ss 800 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS to prevent mud guards from touching them. The view of both ends of the ramps should be unobstructed, else there would be danger of collision between the cars coming in and those going out. ,4-Ply Barret Specification Roof /4-Ply Barret Speci if tation Roof fa Concrete’ Girder fill alt-blazed. Terra Cotta... Pest Ste pee ill tobrade 4 Brichy ‘2 ‘inh 'e bray , Coping “3 1g oe Brick yf 2 Finish-. sey a ieee GORE eis ha" 0" Ji | = oo level I] rr: A y 19s i oer ere _Level IO 0 ag sie Te M0" Architectural "72.7 he Goud ath, w) ae Le i Terra Cotta’ ear ' ip ena} = 8 ee ety 1a" i ee | meg 2: 26 high ba: i eee Powe ; £6 -- 7 “ps a ae = pe: Concr 13/0" |i ® gas pe he petal eas Curb: = i! | L-3 TU ral -"T oF tau | ae eri 2 ees H-0" ata bt Cinders 5 tank, aie L-2 rade s : I-33 | aF IL-7 |i y MiCement Finish UE ta tone 56" Ppt a . ey * Th .19- fies Domaine fo ee ee as 22-6 Fae 21-3 "appl I sx, Wee : , " \ : | " Soe eee * a ees 4 ea x ’ ee etd ees " H ee © i ae 1 fess es ae < 9 Ke Ngo te sez epete-t seepage se iepe [Sta Steel od tie Te ASP Let Pie ote ad ed Sey OSHS ee Be ue Curb. Rak Neo ee oe ea ea) BiB Curb, i i! " i ee Tee Sar : oop. ee nn © Wee se i ea a pom ai “af WL J 0x66 -! yw oo i lh ak \ Wise F Lt Ram p Ramp | Kal. Door ré«8 Curb | 3 Se ae Ramp rl th | mi: . a ; a aw. Dolan Den tp ee: a Pee Ry | % 6. PipeRail~ | Tike you be ee J Ls) aie ae) we ee Fea “Plaster PS i it Y2 Common Level 11. I 3: ~ | Brick 8 a 2 ' 7% : Set r " 8 8 r le } 25 Cement Finish over Entire Area; eee Thickness. ‘Grade foDrains , Dm sete Ep ee fed Phd et Se PS Sh 4 i! di Hy iF f f f 4 _ es is i Ne ee — =e — b= a te ‘ SIN Peay 5 892647" repens tenn. > (2 Fats | woth 2 2 aes be 2S ae aun mania LEZ IIE ‘maa 3k (9.3 e- 226" ake - 22-6 “4 ke 2216" He 8.6 TE DO Ad, Plea isk | moe 18) 198" nanan names enc tansts 49'S Fig. 284.—Details of Statler Garage in Buffalo. (See p. 801.) Geo. B. Post & Son, Architects. United Fireproofing Co. Contractors. A curved inclined driveway, as used in the six-story Washington Garage + in Chicago, is shown in Fig. 283, p. 799. The average” ‘Holabird & Roche, Architects. For further description see Engineering News-Record, July 24, 1919, p. 188, . AUTOMOBILE BUILDINGS AND GARAGES 801 inclination of the driveway is 11 degrees, with 13 degrees at the inner curve and 9 degrees at the outer curve. The width of the driveway between curbs is 10.ft. 3 in. on the tangents and is increased to 11 ft. on the turns and reduced to 8 ft. at the middle of the curve. D’Hume Ramp System.—With ordinary height between floors, the ramp must be very long and therefore must occupy too much of the garage space. This is remedied in the D’Hume Ramp System, by constructing the garage in two parts so arranged that the story in one part falls halfway between the stories in the other parts. With this arrangement it is necessary to build ramps only large 8 Concrete Cap V.- 12" Brickwork Steel Sash /4 420" rine level /4 level 13 Pipe Rail... 8x8 je I nr We level 8 2a Fig. 285.—Section through D’Hume Ramp. (See p. 801.) enough for one-half of the story-height. A design containing ‘D’Hume ramps is shown in Fig. 284, representing a floor plan and cross section of the Statler garage in Buffalo. Photograph of the exterior of this building is shown in Fig. 262, p. 750. It may be of interest to mention that in this building slag was used as coarse agere- gate for concrete. Section through the ramp is shown in Fig. 285, p. 801. . Circular Ramps.—Another solution of the same problem is employed in the Eliot Sheet Garage in Boston, where a continuous circular ramp is provided. This arrangement is less satisfactory than the previous one, as it occupies too much space. 802 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS ‘ggouIsuy ‘Od Sutszs0ulsug “J ‘IN ‘S ‘Se yory ‘suivojg 32 paedoyg (‘208 ‘d aag) “sseyA[ “WOJSOg Ul UON}BIS ZdIALIG UlP{UVIY— "98S “Ol J0opy PIE OF a pune YSDpl s4iDye aaiqu0y” Bi d 0) 100 rs; | © ‘ 4 & Y2fIG Of 409) J . © © a yy, apn, p sussnyo) ybly 9 B ain p punoso SgINg) ap2Jouo) gsn) 94049U0') oe 1009 Pd/2D ULL. aids] 100] UO 7,9 \ ‘ suowips0d DL ,P- Jof saauan youg Ny i OFFICE BUILDINGS 803 Typical Arrangement of Columns.—Figure 286, p. 802, shows the arrangement of columns as used in Franklin Service Station in Boston, designed by Shepard & Stearns, Architects, 8. M. I. En- gineering Co., Consulting Engineers. The center span is 30 ft. wide, while the end span, on both sides, is 21 ft. 4 in. wide. The longi- tudinal spacing of columns is 27 ft. 93 in. Construction of Garages.—Flat slab is the most widely used floor construction for garages. It is used for spacings of columns up to 30 it. In cases where the spacing of columns is not well adapted to flat slab construction, beam and girder or light-weight floor construction may be used. The beam and girder or clay tile construction is prefer- able to the joist construction, because the thin slab between the joists is not well adapted for carrying heavy concentrated loads, unless provided with special reinforcement. Also, the thin slabs and narrow exposed joists would suffer considerably in case of a fire. Floor Finish.—Granolithic floor finish is practically always used in buildings intended to house automobiles. OFFICE BUILDINGS Reinforced concrete is being used in office buildings to an increas- ing extent. Ordinarily, concrete is cheaper than fireproofed struc- tural steel, particularly when flat ceilings are required. Limit of Height for Concrete Buildings.—Contrary to popular opinion, there is no limit to the possible height for concrete buildings, except that imposed in each particular case by considerations of economy. While not long ago a height of twelve stories was consid- ered as the practical limit, even by engineers engaged in concrete construction, in recent years a number of buildings of considerably greater height have been erected. Usually, there is an objection to tall concrete buildings on account of the large size of the columns required in the lower floors. This objection is easily removed by sub- stituting, in these floors, structural steel columns encased in con- crete. The Medical Arts Building 5 at Dallas, Texas, is a notable exam- ple of a tall concrete building. It is 19 stories and 255 ft. high above basement floor. °C. E. Barglebaugh, Architect. For description see Concrete Frame and Exterior for High Tower Building, Engineering News-Record, April 5, 1923, p. 610. 804 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS The Hide and Leather Building, in New York,® is 18 stories, 215 ft. above the curb. In both of these instances, the concrete frame proved cheaper than structural steel. Shape of Building.—The most important problems in the design of office buildings are those of light and ventilation. In modern office buildings, the courts are arranged so as to limit the width of the building to 75 ft. A typical floor plan is shown in Fig. 287. The passenger elevators are placed in one central location instead of being divided into groups. J pox Fig. 287.—Typical Floor Plan for Office Building. (See p. 804.) Theoretically, in long buildings, two groups of elevators, one at each end of the building, would serve the purpose better. However, offices very often occupy large areas, extending the full width of the building. In such cases it is not possible to make all offices accessible from both groups of elevators, and confusion results, especially for visitors. Spacing of Columns.—In old office buildings, columns were placed along the sides of the central corridor. The cross section then con- sisted of large end panels and a small center panel. Such an arrange- 6 Designed and built by Thompson & Binger Co., Raoul Gautier, Chief Engi- neer. HOTELS, APARTMENT HOUSES, AND HOSPITALS 805 ment was satisfactory for small offices or subdivided offices. Now, with the advent of large, open office spaces, the office often occupies the full width of the building. Equal spacing of columns is therefore much more advantageous. This also results in cheaper construction, as the large spans required in the first case are replaced by smaller spans. If necessary, the space can be subdivided into smaller offices without difficulty. The corridor may be placed either in the center, dividing the space into offices of equal depth, or off center. Loading.—A live load of 70 lb. per sq. ft. is usually sufficient for the upper floors of an office building. The first floor is usually designed for 150 lb. per sq. ft. In some locations it is advisable to design the office building for 120 lb. per sq. ft., to make it available for light manufacturing. Dead load should include a proper allowance for partitions. This _ is usually 25 lb. per sq. ft. Weight of plaster should also be included in the dead load. In light-weight floors, special joists should be provided for heavy partitions running in the direction of the joists. To provide for par- titions by increasing the dead load is obviously insufficient. The partition load is carried by one joist only, as the thin topping is not sufficient to transfer the load to the adjoining joists. Types of Construction.—Light-weight floors, such as clay tile floors and metal tile floors with plaster ceiling, are considered the most economical for office buildings. Flat-slab construction is now gaining in favor, especially for office buildings designed for heavier loads. The original objection of architects to flared heads is gradually disappearing, especially since, with increased knowledge, the size of column heads and their shape may be made to suit architectural requirements. The interior of a flat slab office building is seen in Fig. 269, p. 761. Finish.—Hardwood finish in offices and terrazzo or granolithic finish in corridors have been used very extensively. In recent years, linoleum, or rubber parquetry, is growing in favor. Also, grano- lithic finish, usually painted, is often used, but objection is made to it on account of hardness, noise, and conductivity of heat. HOTELS, APARTMENT HOUSES, AND HOSPITALS Reinforced concrete is well adapted for the construction of hotels and apartment houses, In hotels, a corridor is usually placed in the 806 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS middle and the rooms on both sides of it. In old hotels, the columns were placed in the corridor wall, so that the width of the building was divided into two large spans at the ends with a short span in the middle. In modern hotels, each room has a bathroom or closets placed next to the corridor. This makes it possible to place the columns in the corners of the bathroom, farther away from the cor- ridor, so that the end spans become practically equal in length to the center span. The cost of construction of the equal spans is smaller than that of the spans used in the old arrangement. The longitudinal spacing of the columns is governed by the size of the rooms. Similar construction is possible in apartment houses. In hospitals, it 1s usually necessary to place the columns in the corridor wall. Design Loads.—Hotels, apartment houses, and hospitals should be designed for a live load of 50 Ib. per sq. ft. for private rooms and 100 lb. per sq. ft. for corridors and public rooms. In computing the dead load, proper allowance should be made for partitions. The weight of plaster and flooring should also be added to the dead load. These items are often neglected, with the result that the design is of inadequate strength. Type of Floor Construction.—The economical type of floor con- struction naturally will vary with conditions.. Usually, light-weight construction, as described on pp. 588 to 611, will prove satisfactory. Flat slab construction is not used to any great extent, because of the objection to the flared heads. However, for spans up {O1LS- ih, it is possible to build flat slabs economically without flared heads. In many cases, the rooms can be arranged so as to permit even spacing of columns not more than 18 ft. apart, which permits the use of flat slab construction (without flared heads), at a cost much lower than that of light-weight slab. This is not possible where spans over 18 ft. are required in the bulk of the structure. Floor Finish.—The most popular floor finish for this type of build- ing is hardwood flooring, particularly in the living portions. Ter- razo is generally used for corridors, kitchens, service portions and bathrooms. | In recent years, linoleum and similar coverings are gaining in favor on account of their lower cost and great durability. With linoleum, there is also a gain in thickness of about 2 in. Objections are made to the use of granolithic finish for buildings of this type. The hard appearance, which constitutes one of the HOTELS, APARTMENT HOUSES, AND HOSPITALS 807 pee ~ < Clinie Laboratory Mm _-4-Surgical Students mil oat lass ates arto! Coat Room. |_| | <0 2 Special Clinic | i * Laboratory~ 7 e & = Internes-~.. Solarium F iq. 288.—Layout of Rooms in Hospital for State of Wisconsin. (See p. 808.) North Win 7 laced _voists. “21 _y_ EEA re 3t of \4 SSS II, Lnteys 7-I"Rods in top /4-I"Rods in bottom Section A-A re ge! “UPS Detail at X-Y & Rod, 1749" 2 Rod, 22-3" eee eee | 5 2 Rod,4' 35 Joists | Dumb waiter shaft 9 S3 joist, Interior ¢ of --7, Et SP ous GF tiers columns building il j BiJoists . | 34 Rows of tile | loggia , iia? y Wall 4 Ay tee/ columns | Les Oz columns Beam Ib JUUL in loagia os ae Section of Column X? — S-Lintel Z Lintels Z 2" Rod! Spe gn Section o other Lintels Linte/] Z Fra. 289.—Design of Floor Construction for Hospital for State of Wisconsin. (See p. 808.) 808 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS objections, can be remedied by painting the floors a warm color. Design of Hospital—A typical design of a hospital for the State of Wisconsin” is shown in Figs. 288 and 289, p. 807. The arrange- Third Floor Gr. 61.60 ee =e ee ee ee First Floor Gr. 29.60 Fic. 290.—Typical Cross Section Showing Classrooms Opening on Corridor. (See p. 810.) Massachusetts Institute of Technology, Cambridge, Mass. ment of rooms is evident from Fig. 288, while the construction details. are shown in Fig. 289. SCHOOL BUILDINGS Reinforced concrete is used in the construction of school buildings, either for the whole frame or for floors only. Live Load.—The following live load may be safely used in the design of the various parts of the school building: 7Structure and Facilities of a Modern Hospital. By Arthur Peabody, Engineering News-Record, April 10, 1924, p. 608. SCHOOL BUILDINGS 809 Live load lb. per sq. ft. Bee ey ne a ae 50 Pea ANC SUAITS: oo. ee one ee ke ns 100 Assembly rooms and auditoriums....... 100 Oo) ToL! ES Sel lead a aaa a 100 ee i i gs bec ot an 100 Wall-bearing vs. Skeleton Construction.—School buildings are seldom over three stories high; therefore, wall-bearing construction is used to a great extent. Recent investigation show that skeleton con- struction may be built at the same cost and sometimes cheaper. It has the additional advantage that the concrete structure may be erected without any interference from the brickwork. Type of Floor Construction.—The type of floor construction depends upon the size of the room and the purpose for which it serves. Class Rooms.—The average class rooms are 32 ft. long and 22 ft. wide. They are arranged either on both sides of the corridor or on one side only. In wall-bearing construction, the walls along the corridor may be built as bearing walls. The floor construction then spans from wall to wall. The cost of the construction may be reduced by omitting the interior bearing walls and substituting columns placed along the corridor, with longitudinal beams to carry the floor as well as the corridor partitions. The class rooms must be built without any interior columns. Flat ceiling is also often required. In class rooms beam and slab construction may be used for the floor, the beams spanning across the building, which is the short direc- tion of the rectangle. The spacing of the beams should be about 8 ft., which permits the use of 3}-in. slab. When flat ceilings are required, a suspended ceiling, consisting of steel rib lath suspended at the beam and at one or several intermediate lines, should be used. Where hollow tiles or steel tiles are available, light-weight con- struction, as described on p. 588, may be used for the floor. The joist may span in the short direction only, or a two-way system may be used, with joists running in two directions at right angles to each other. . Sometimes a solid concrete slab, reinforced in two directions, may be used. The hollow tile and solid concrete ceiling may be plastered directly 810 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS below the construction with two-coat plaster. Metal tile construc- tion requires metal lath underneath with three-coat plaster. Special Rooms.—In ‘ndustrial training schools, the sections occupied by workrooms, shops, and drawing-rooms are much larger than the class rooms, but there is no objection to intermediate col- umns. Columns may be spaced with due regard to the economy of the structure. Flat slab construction is often used in special rooms. Gymnasiums and Auditoriums.—Rooms serving for gymnasiums ‘and auditoriums are usually of considerable length and width, with- out any intermediate columns. 3 In some cases, an attempt Is made to combine the gymnasium and auditorium. This is usually not successful, because the floor of a gymnasium must be level and therefore is not well adapted for auditorium purposes. Gymnasium and auditorium are often built side by side, so that the gymnasium or part of it may serve as a stage for the auditorium. In such case, means are provided for shutting off the whole or a part of the gymnasium so that both rooms may be used separately, if desired. Usually, no additional story is placed above either of these rooms. The roof construction is ordinarily carried by steel trusses. While these are cheaper as to first cost than concrete members, they require larger cost, of maintenance. If a permanent roof is required, it may be built as explained in connection with long-span roof construction (p. 661). 7 The galleries and the ‘nclined floor in the auditorium may be designed as explained in connection with theater construction. Building for Massachusetts Institute of Technology.—Figure 290, p. 808, shows typical framing plan for Buildings 1, 3 and 5 of the Massachusetts Institute of Technology The beams were arranged across the building. In portions of the building where classrooms are placed on both sides of the corridor, the columns are placed along the corridor. Fig. 290 gives the widths of the classrooms and of the corridor. Stairs.—As all the up-and-down travel in school buildings is by means of the stairs, their number and location must be carefully worked out. The stairs must also serve as fire exits and must be 8 Designed and built by Stone & Webster Engineering Corporation; William W. Bosworth, Architect. The concrete design and construction was under the supervision of Sanford E, Thompson, Consulting Engineer, SCHOOL BUILDINGS 811 properly protected. As a general rule, each school building should have at least two staircases, located at opposite ends of the corridor. A larger number may be required. The width of stairs is usually 4 ft. 6 in. The rise of steps varies, according to the age of the pupils, from 6} to 7 in., and the tread from 104 to 11 in. Floor Finish.—The floor finish in class rooms, auditorium, and gymnasium is usually of hardwood. Linoleum and rubberized par- queting also give satisfactory results. For corridors, lavatories, and other service portions, granolithic finish or terrazzo may be used. CHAPTER XX REINFORCED CONCRETE CHIMNEYS FORMULAS FOR REINFORCED CONCRETE CHIMNEY AND HOLLOW CIRCULAR BEAM DESIGNS REINFORCED concrete chimneys may be regarded as vertical can- tilever beams supported at the base. The loadings to be provided for are (1) the weight of the chimney and (2) the wind pressure. Although the design is somewhat complicated by the fact that the beam is circular and hollow, the treatment is nearly identical with that of ordinary rectangular beams. In fact, the analysis which follows is based upon the several fundamental assumptions adopted in reinforced concrete beam design with only one additional assump- tion, viz.: that, since the concrete is usually thin as compared with the diameter of the chimney, no appreciable error 1s involved in assuming all material as concentrated on the mean circumference of the shell. An analysis for shear 1s given on p. 819. An example of chimney design and review is given on p. 829. Although specially devised for a chimney, the formulas are appli- cable to any hollow beam. The principles involved in the demonstration of the thickness of steel and concrete are taken by permission from the analysis by Messrs. C. Percy Taylor, Charles Glenday, and Oscar Faber.! NOTATION W = weight of the chimney above the section under consideration, lb. ; M = bending moment about that section due to wind or other cause, | in.-lb.; P = total compression in concrete, Ibs T = total tension in steel, lb.; n= i — ratio of modulus of elasticity of steel to that of concrete ; 1 Engineering (London), Mar. 13, 1908. | 812 DERIVATION OF CHIMNEY FORMULAS 813 fc = maximum compression unit stress in concrete (measured at the mean circumference), lb. per sq. in.; fs = maximum tevsion in the steel, lb. ate Sq. in; D = mean diameter of shell, in.; r = mean radius of shell, in.; ¢ = total thickness of shell, in.; te = thickness of concrete ailyei in. ts = thickness of an imaginary steel shell of mean radius 7, and having a cross-sectional area equivalent to the total area of reinforcing bars, in.; total area of vertical reinforcing bargin the section under consideration, sq. in.; k = ratio of distance of neutral axis, from mean circumference on compression side, to diameter D; j, 2, Cp and Cp = constants for any given value of k. (Tables 1 and 2, p. 828); jD = distance between center of compression and center of tension; 2D = distance from center of compression to center of force due to weight. = T DERIVATION OF CHIMNEY. FoRMULAS Referring to Fig. 291, if f; is the maximum. intensity of stress in the concrete at the mean circumference on the compression side, then the intensity of compression in the steel at that point is nf,. Since f, is the maximum intensity of stress in the steel at the mean circumference on the tension side, then the variation of the stress in the steel, across the section cd, is represented by the straight line ab, which cuts the line cd at e, thus locating the neutral axis or the line of zero stress. A con- stant value having been assumed kf > for the modulus of elasticity of the Fy. 291.—Resisting Forces in a concrete in compression, it therefore Reinforced Chimney. (See p. 813.) follows that, at any point of a given section, the stress in either the concrete or the steel is directly pro- portional to the distance between that point and the neutral axis. LLM Neutral _ Center _ ont dit ea bae 814 REINFORCED CONCRETE CHIM NEYS Calling kD the distance between the neutral axis and the mean circumference on compression side, as shown in Fig. 291, we have by similar triangles kD __h D) topahe ete whence 1 a i 1 Nc By this formula, the position of the neutral axis may be determined for any combination of fe, fs, and n. The formula for k is the same as for rectangular beam. If now, as shown in Fig. 292, represents half the angle subtended at the center by the portion in compression, we have cos a = (1 — 2k) from which, for any given value of k, cos becomes known, as well asa and sina. Thus, having located the neutral axis for any given etnias combinations of f.,/,and , and ra ie 7 ~Centerof bearing in mind that the stress ae las at any point of the shell is pro- portional to the distance be- tween that point and _ the neutral axis, we can now de- termine the total force on the ei compression side, the total Tension force on the tension side, and also the location of the center of compression and the center of tension. Considering a small radial element subtending an angle dé, as shown in Fig. 292, we have in this element, since the length of an arc is its radius times the angle, Center of f. Compression --_f-/f., Concrete in---%¥X\1 Compression Steel in -~ Compression Fic. 292.—Distribution of Stresses in a Reinforced Concrete Chimney. (See p. 814.) area of concrete = t.7rdé, area of steel =Hrdu: The distance from the element to the neutral axis is r(cos 6 — cos a), while the distance from the neutral axis to the point of extreme stress, DERIVATION OF CHIMNEY FORMULAS 815 fc, is r(1 — cosa). Therefore, the intensity of stress on this ele- mental area is EAGOS =: C08 cr) in the concrete, r{(1’— cos a) te and r(cos 8 — cosa) . te Perce) in the steel. Assuming these intensities at the mean circumference to represent the average for the entire element, we have the total force on the elemental area (concrete and steel) cs f.r(cos 6 — cos a) dP = (t, + nt,)rdé Milne. The total compression force, P, acting across two alpha is, therefore, “fer(cos 8 — cos a) P= (to + nt,)2 , (1 — cos a) dé. Integrating this expression, we obtain 2 : Piss Werth + Ms) eos gy cin Qa — a cos a). Since any given position of the neutral axis determines a, as shown above, this equation may take the form LP = 10rtiee nt iB ieiehanaadn i avons leah gn in which Cp is a constant for a given position of the neutral axis. (See Table 1, p. 828.) When the magnitude of P has been determined, its location, with respect to the neutral axis, may best be found by taking its moment about that axis and dividing by P, thus giving the distance from the neutral axis to the center of compression 1,, as shown in Fig. 292. As before, the compressive force on an elemental area is oe f.r(cos 6 — cos a) The distance of this force from the neutral axis being r(cos 6 — cos a), we have as its moment about that axis fr?(cos 86 — cos a)? dM, = (t, + nt.)rdé race awe >) 816 REINFORCED CONCRETE CHIMNEYS while the moment of the total compressive force P is ate + niga f r( GOS Bare cos a)? f.r(cos 8 — Cos a)" 7, (1 = cos "(1 0os poe = (t, + nts) ee afer | (cos? 6d cos a) | Jo — Zeosa f cos 6d0 + cos? af as}. es Te (2) P (sin a — a COS @) Following a similar method of procedure, it is possible to deter- mine the total tension and the location of the center of tension. In accordance with our assumption that the concrete does not resist any tensile stress, it 1s evident that in considering the forces on the tension side of the section we are concerned merely with the steel. On the tension side, a small element, therefore, has an area = t.rdé. The intensity of stress on this element, being proportional to its distance from the neutral axis, 1s r(cos 8 + cos a) — Js r(1 + cos a) while ve total tension on the small element is (cos @ + cos a) (1 + cos @) The total force, 7’, on the tension side of the section is, therefore, (x — a) T =2 ip dip W008 23 ane a) 49 0 av Soar el ot (1 + cos a) Integrating, we have T = Farley (sin a + (t — @) COs @). + cos a) DERIVATION OF CHIMNEY FORMULAS 817 Since, as before, any given position of the neutral axis determines a, this equation may take the form OSCE ERS OREM SRE at in which Cy is a constant for a given position of the neutral axis (see Table 1, p. 828). By a method similar to that used in considering the force on the compression side, we may write the moment, about the neutral axis, of the force on a small element on the tension side as r(cos 6 + cos a)? (L+ cosa) ’ while the moment of the total tensile force, 7, about this axis is (7 — a) . vs ab: r(cos 6 + cos a) ea af SO ayaa eyede! dM r= t.rd of. Integrating, we have 2 : AD cy eon anes ids = 2 3 ae My = 1,7 Iq A he a) cos? a + 3 sina cosa + 3(r — a)]. Dividing M, by T, we have as the distance of the center of tension from the neutral axis | ( (7 — a) cos? a + 2 sina cosa + 1(r — a) ) - (sina + (x — a) cosa) ls = eet (4) From Formulas (2) and (4), it is evident that the distance be- tween the total force in compression and the total force in tension (i.e., 11 + 2) may, for any given po- sition of the neutral axis, be expressed as a constant times the diameter D. Thus, 1; + lz = jD, as shown in Fig. 293. Likewise, as shown in Fig. 293, zD may represent the distance from the center of compression to the center of the chimney, z also being a constant for any given position of the neutral axis. In a chimney, the tensile and Fie. 293.—External and Internal compressive stresses which we have Forces Acting upon a Chimney. been considering are produced by (See p. 817.) a combination of wind pressure and the weight of the chimney. Thus, on any horizontal section, cd, as shown in Fig. 293, the forces external to that section are: <—__. Wind 818 REINFORCED CONCRETE CHIMNEYS the horizontal pressure of the wind, causing a moment M about the section, and a central vertical load W representing the weight of that portion of the chimney above the section under consideration. These forces are resisted, and held in equilibrium, by the forces P and T, which represent the compressive and tensile stresses in the concrete and steel. The system of forces, as shown in Fig. 293, must be in equilibrium. Hence, taking moments about the force P, we may write TID = M — WeD: But YA OO Ba Therefore CrfrtgD = M — W2D. Whence aes M — WzD : Cn Dae The total area of steel, A, = 277ts. Therefore ad In(M — W2zD) fA are CfgD GAT haga Se ity sae From Table I, p. 828, it may be seen that the constant 7 changes but slightly for a considerable variation in the position of the neutral axis. Taking * = § for all cases, equation (5) may be _ 8(M — WeD) As = =o37 7 ae (6) While this formula is not exact, the error involved is inappreciable for almost any case, so that formula (6) may always be used in- stead of formula (5). Applying now the condition that the summation of all vertical forces must be zero, we have P—-T=W. Substituting values of P and T as previously found, the equation becomes | Crfar(te + nts) — Cafert, = W. Transposing and solving for ¢., we obtain ba W a (Cof — Cofen)rts te G pf c¥ DERIVATION OF CHIMNEY FORMULAS 819 The total thickness of the shell is t=t,+4,, gk W oh (Cofs eT Cpfn)rt, Cpf r For convenience in use, after A, has been determined by the whence = ee formula given above, by substituting r = = and t, = ab this formula 1. ; for t may best be written QW + (Crh — Caf.n)At t= A, Meee at +. ype ° . ° . (7) In view of the fact that Formulas (5), (6) and (7) contain the constants z, J, Cr and Cp, which, as has been shown, are dependent for their value solely upon the location of the neutral axis, it is evi- dent that, for any specific values of f., f,, and n, which in turn will determine the position of the neutral axis, the expressions for A, and ¢ will admit of a further simplification. For given values of Jc, fs and n, the necessary thickness of shell and area of reinforcement may be expressed merely in terms of the moment of the wind, M, the weight, W, and the mean diameter, D. The expressions, as given, however, seem best adapted to general use, and when sup- plemented by the tables given on page 828, are rendered quite sim- ple of solution for specific values. In Table 2, p. 828, are given values of &, the location of the neutral axis, for various combinations of f,, f, and n; while Table 1, p. 828, gives the corresponding values of the constants C p, Cr, 2 and j for various positions of the neutral axis. Shear, or Diagonal Tension—When the necessary thickness of shell and vertical reinforcement have been determined, the size and spacing of the circular steel hoops must be considered. The external forces produce shear and diagonal tension, which may be analyzed similarly to like stresses in rectangular beams, and the reinforcement necessary to resist the diagonal tension, which is a function of the vertical tension, may be determined. Usually, this reinforcement is not so great as that which it is advisable to insert for the proper distribution of temperature stresses, but nevertheless it should be determined, in order to be sure that it is sufficient in quantity. The concrete should never be relied upon to carry any tension or 820 REINFORCED CONCRETE CHIMNEYS vertical shear, because the expansion from the heat may cause vertical cracks in the concrete. These need not be considered dangerous, if sufficient horizontal reinforcement is provided, any more than the vertical cracks in a brick or tile chimney. Considering the stresses due to vertical shear, it may be easily shown that at any horizontal section of a chimney the vertical shear per inch of height is the total horizontal shear on that section divided by the distance between centers of tension and compression, jD. With this as a basis, there may be developed a formula for practical use in determining the necessary area and spacing of horizontal steel hoops at any given section. Thus, Let i; = height of chimney above section under consideration, ft. ; F = effective wind pressure against chimney, lb. per sq. {t.; f, = allowable tensile stress in steel hoops, Ib. per sq. in.; D = mean diameter of shell in. ; po = ratio of area of steel hoop to area of concrete. The total external shear on any horizontal section of a chimney is equal to . D V — jah while the maximum longitudinal shear per inch of height is from beam formula ‘ Die Sy: _DAF_ hk 1X4) = 95 GD aoe Having seen that for all positions of the neutral axis 7 remains practically constant, and giving Jj an average value of, say, 0.783, we obtain, as the expression for the maximum vertical shear per inch of height, v X 2t = 0.106hF, while the shear or diagonal tension in one foot of height is Stresses to be resisted by hoops = 12 x 0.106/F. The area of steel in one foot of height of chimney will be 12¢po, and the stress the hoops in this height are capable of sustaining on | their two sections is Available strength of hoops = 2 X 12tpofs. | SLIM CHIMNEYS 821 Equating the stress to be resisted with the available strength, we have | 12 X 0.106h,F = 2 x 12tnof,, whence ehh ied WGA TS art This ratio of steel is for shear or diagonal tension only. To provide for temperature stresses, or rather to distribute the strains so as to prevent the localization of cracks, an additional amount of horizontal steel is needed. This may be provided for arbitrarily, by assuming 0.25 per cent steel, or rather 0.0025 for temperature stress in addition to the steel for shear. Expressing this as a formula for ratio of steel, we have h,F The use of small rods, spaced 6 to 10 in. apart, except in the upper part of the stack where the spacing may be greater, is advised. The spacing of hoops in many of the chimneys already built has been 18 to 36 in., but as such chimneys have frequently cracked quite seriously, more recent designs have called for 8- or 9-in. spacing throughout the entire stack. | Note on Slim Chimneys.—Since, in designing a chimney, the selection of certain allowable working stresses in the concrete and in the steel will fix the position of the neutral axis, it is evident that the ratio of these working stresses limits the compressive area of the section. Hence, for a very high chimney in which there is a large compression in the lower sections, it is possible that the selection of an ordinary working stress in the steel of 14.000 or 16 000 Ib. per Sq. in., together with the customary working stress in the concrete of, say, 500 Ib. per sq. in., would locate the neutral axis so near the compression side of the section as to make it impossible to obtain sufficient compression area to withstand the compressive forces with- out exceeding the allowable unit stress in the concrete. If, therefore, the thickness of shell as computed from Formula (7), p. 819, should work out materially larger than the assumed thickness, recomputation should be made on the basis of a smaller working stress in the steel, thus changing the position of the neutral axis so as to allow a larger proportion of the section to carry com- pression. In such a case it may be necessary to make a series of 822 REINFORCED CONCRETE CHIM NEYS trials with different working stresses in the steel until the computed thickness checks with the assumed thickness. In high chimneys of small diameter, it may be impossible to utilize a working stress in the steel greater than 7 000 or 8000 Ib. per sq. in. DESIGN OF HOLLOW CIRCULAR BEAMS The analysis of a hollow circular reinforced concrete beam, whose thickness, compared with its diameter, is small, is similar in principle to that of a chimney. In this case the weight of the member acts in the same direction as the external forces, so that in Formulas (6) and (7), W, the weight in the axial direction, is zero. The forces of compression, P, and tension, 7’, are equal. The area of steel and the thickness of shell are therefore obtained from Formulas (6) and (7), pp. 818 and 819, by making W = 0. DESIGN OF REINFORCED CONCRETE CHIMNEYS High factory chimneys of reinforced concrete are being built in this country and abroad. ‘The cost, especially of those over 100 ft. high, is usually much less than that of a brick chimney. If designed and built upon the same principles and by the same methods which have proved essential in other types of reinforced concrete construc- tion, they can be depended upon to give permanent satisfaction. Reports? on a large number of chimneys have shown that con- crete is unaffected by the heat from an ordinary steam boiler plant. — The temperature in such chimneys seldom exceeds 700° Fahr., while 400° to 500° Fahr. is more usual. Experimental tests also indicate that concrete is not appreciably injured by temperatures of 600° to 700° Fahr.° To provide for extremes, it is advisable, however, to build an independent inner shell of concrete or firebrick for at least a portion of the height. Concrete should not be used for a chimney in con- nection with special high temperature furnaces. Since concrete and steel have substantially the same coefficient of expansion * there is no danger of heat causing a separation of the reinforcement from the concrete. 2A special investigation of reinforced concrete chimneys was made . by Sanford E. Thompson in 1907 for the Association of American Portland Cement Manufacturers. Many of the points here discussed are summarized from the report, which is printed as Bulletin No. 18 of the Association. — ? 3 Tests of Metals, U.S. A. 4 See Volume IT of this treatise.. DESIGN OF REINFORCED CONCRETE CHIMNEYS 823 The expansive effect of heat is a more serious question. Stresses are set up in the shell of any masonry chimney, because of the hot interior and cold exterior surfaces. A concrete chimney, however, has thinner walls, and the stress is therefore less than in one of brick or tile; the concrete chimney is also better reinforced. Provision for temperature stresses are discussed in the paragraphs on design, which follow. Construction.—A reinforced concrete chimney, because of its height and shape, is more difficult to construct than many other kinds of concrete construction, and it therefore should be handled by experienced builders. It is essential in chimney construction that the materials be very carefully selected. The sand, as well as the cement, should be tested by determining the actual tensile strength of mortar made from it. The stone preferably should be of the nature of a hard traprock, 3-In. maximum size. Proportions 1 : 2:3 have been found to give good results. A dry mix should not be used, since insufficient water will produce a porous concrete which does not adhere to the steel. The consistency must be wet enough to quake and form a jelly-like mass when lightly rammed, so as to properly imbed and bond the reinforcement. No exterior plastering should be permitted, because it is likely to check and scale. The steel should be round or deformed bars of good quality. Bars with flat surfaces, like T-bars, are inferior because the flat surfaces give a poor bond and the angles make the placing of the concrete difficult. Deformed bars of small size, quite closely spaced, are especially good for the horizontal steel to dis- tribute the temperature stresses, and high-carbon steel of first-class quality also has advantages for the horizontal reinforcement. Example.—The design of a chimney built in Brooklyn, N. Y., in 1907, is illustrated in Fig. 294, p. 824. Design of Reinforced Concrete Chimneys.—A reinforced con- crete chimney consists primarily of a concrete shell with vertical steel bars disbributed uniformly along the circumference of the chimney. The shell must be of proper thickness and the steel bars sufficient in size and number to withstand the stresses due to the weight of the chimney and to the action of the wind. A chimney of this type differs essentially from one of brick, in that the diameter at the base is so small as compared to the height that the chimney would over- turn under a heavy wind were it not for the vertical bars of steel which serve as anchors and hold it on the windward side. REINFORCED CONCRETE CHIMNEYS 824 Section C-D eee Aree a ee a a a ae peer ee oh ee ren ce adh) © Q $23/1NO IV ind) ict n bent ered Senne Deeion cea o8 Ua AT nO-19F pe aes 0-08) 2 20706 = 7) 10nCe Section AB (See p. 823.) S Fic. 294.—Design of Chimney of the Edison Electric Illuminating Co. INS --16 -----28'-"---- Brooklyn, ; > t= So a ee re = gee Oe .. -—— x S4o0g 4a, gf | S4I2G 424 001 la 72 910) GA ool alt beat OS: 710~06 n0-OL n070L 10701 nG TOL DESIGN OF REINFORCED CONCRETE CHIMNEYS 825 Wind, in blowing against a chimney, causes compression on the side opposite the wind and tension on the side against which the wind is acting. This compression is resisted by the concrete and steel on the leeward side, while the tension or pull is taken by the steel on the windward side. In addition to the vertical reinforcement, a reinforced concrete chimney should be provided with horizontal hoops of steel, the object ‘of which is to stiffen the vertical steel, to distribute cracks in the con- crete due to a difference in temperature between the interior and exterior, and to resist the diagonal tension. In designing a reinforced concrete chimney, the problem, then, is primarily to determine at various horizontal sections the necessary thickness of the concrete shell and the required amount of vertical reinforcement, so that the allowable working stresses in the concrete and in the steel shall not be exceeded under the action of the forces to which the structure may be subjected. The complete analysis and development of the most useful formulas are given in the first part of this chapter. The final formulas to be used in design are repro- duced below. The problem of the determination of stresses due to the difference in temperature between the interior and the exterior of the shell involves many uncertainties. The heat tends to expand the inner surfaces, producing tension in the outside surface of the shell and compression in the interior surface. Although the distribution of the stress is not clearly known, the variation of the heat through the shell not being uniform, tentative computations indicate high stresses, so that it is a question whether vertical temperature cracks can be entirely prevented any more than they can be prevented in brick or tile chimneys. The function of the horizontal steel may therefore be to distribute these cracks and to resist the vertical shear or diagonal tension. This horizontal steel should be distributed, therefore, by using small diameter bars closely spaced rather than large bars spaced further apart. Because of the possibility of vertical tem- perature cracks, the concrete should never be relied upon to carry tension or vertical shear, and the amount of horizontal reinforcement to resist this may be obtained in a fashion similar to the determina- tion of vertical stirrups in a beam. The analysis for the shearing stresses is given on page 819, and the final formula is presented below together with suggestions for adapting the horizontal reinforcement to temperature stresses. 826 REINFORCED CONCRETE CHIMNEYS The amount of vertical reinforcement, the thickness of the shell, and the percentage of horizontal reinforcement may be obtained from the following formulas. Let W = weight of the chimney above the section under considera- ton, 1b.) M = moment of the wind about that section, in.-lb.; f, = maximum tension in the steel, lb. per sq. ak f. = maximum compression in the concrete ineadived at the mean circumference), lb. per sq. in.; s n= E, = ratio of modulus of elasticity of steel to that of con- crete; D = mean diameter of shell (i.e., diameter of center of ring), in.; r = mean radius of shell, in.; t = total thickness of shell, in. ; A, = total area of vertical reinforcing bars in the section under consideration, sq. in.; k = ratio of distance of rauriat axis, from mean circumference on compression side, to the mean diameter D; 2, Cp, Cr = constants for any given value of k, Tables 1 and 2, p. 828 ; po = ratio of cross-sectional area of steel hoop to vertical sec- tional area of concrete; h, = height of chimney above section under consideration, ft.; F = effective wind pressure against chimney, lb. per sq. it.; Then Area of vertical bars BOE Wa2d) 3 Ale rae Crf.D e e e e e e e e e e (9) Thickness of shell A, 2W + (Crfs — Cefn) — a eee ge we A “aa Cif.D xD Ratio of steel hoops hiP Po = agrgey + 0.0025: 1 (11) Formulas (9), (10), and (11) correspond to Formulas (6), (7), and (8), pp. 818 to 821. DESIGN OF REINFORCED CONCRETE CHIMNEYS 827 In the formula for po, the first term gives the ratio of steel to resist vertical shear or diagonal tension, and the second term is an arbitrary ratio designed to distribute the temperature strains. To best distribute the temperature strains, a maximum spacing of 6 in. to 10 in. is recommended for the horizontal bars. In the formulas, the terms z, Cp and Cy are constants, the values of which are fixed for any given position of the neutral axis. By means of Tables 1 and 2 (p. 828), these constants may be easily and quickly determined, so that the solution of Formulas (1) and (2) is rendered quite simple after the diameter and height of the chimney have been selected and the bending moments due to the wind have been computed for the various sections considered. _ The thickness of shell must be assumed in Formula (1) in order to deter- mine the average diameter, D, and to compute the weight, W. A new computation may be made to correct this, if necessary. For economical distribution of concrete and steel, computation must be made for several sections in the height. It is not advisable to make the thickness of exterior shell less than 5 in. at any place, but the ~ number of steel rods may be gradually reduced toward the top. Summary of Essentials in Design and Construction—In the investigation ° referred to, the essential requirements are sum- marized as follows: | (1) Design the foundations according to the best engineering practice. (2) Compute the dimensions and reinforcement in the chimney with conservative units of stress, providing a factor of safety in the concrete of not less than 4 or 5. (3) Provide enough vertical steel to take all of the pull without exceeding 14 000 or at most 16 000 Ib. per sq. in. (4) Provide enough horizontal or circular steel to take all the vertical shear and to resist the tendency to expansion due to the interior heat. (5) Distribute the horizontal steel by numerous small rods, in preference to larger rods spaced farther apart. (6) Specially reinforce sections at which the thickness in the wall of the chimney is changed, or which are liable to marked changes of temperature. (7) Select first-class materials and thoroughly test them before and during the progress of the work. > See footnote, p. 822. 828 Table 1. Values of Constants Cp, Cr, Neutral Axis, (i.e., for various values of k) (5), (6) and (7), pages 818 and 819, and (9), (10), and stance of neutral axis from mean circumference diameter D. Value of k to suit the condition For use with equations (11), page 826. & is ratio of di on compression side to the mean REINFORCED CONCRETE CHIMNEYS of the problem is obtained from Table 2, below. i) is) <> On oO 100 .150 200 250 300 390 400 450 500 .590 .600 Table 2. Location of Neutral Axis for Vario 0 0 DONNER err e Cp .600 852 .049 .218 .310 .510 .640 765 .884 .000 113 224 Cr Z 3.008 ' 0.490 2.887 0.480 Dat h a 0.469 2.661 0.459 2.551 . 0.448 2.442 0.438 2.333 0.427 2.224 0.416 2.113 0.404 2.000 0.393 1.884 0.381 1.765 0.369 J 0.760 0.766 0.771 0.776 O.779 0.781 0.783 0.784 0.785 0.786 0.785 0.784 Stress, f,, Tensile Stress, /s and Ratio of Moduli, n. (See p. 827.) Maximum Tensile Stress in Steel, fs ee k Ratio of Depth of Neutral Axis to Depth of Steel Below Most Compressed Surface of Beam m = 10 eee as Maximum compressive stress in concrete, fc n=i12 ————_—_ Piet) ie) z and j for Different Positions of the a us Combinations of Compressive Maximum compressive Maximum compressive stress in concrete, fc stress in concrete, fc 300 | 400 | 500 | 600 | 700 | 300 | 400 | 500 600 | 700 -310| .375| .428] .474| .512) .360). 285] .348] .400| .444|.483) 334]. 264| .324| .375] .418] .456) .310]. 246] .304| . 353] .395] .433) . 290). 231] .285| .334| .375} .412| . 272). 217| .270| .316| .356} .392) .257/. -204| .255| .300| .340] .375| .243). 198] . 242] . 286] .324|. 360) . 231). 184] .231| .272| .310| .344) .220). 175] .220| . 261] .298] .330} .210). 166] .210] . 250) . 285} .318) . 200). 160] . 201] . 240) . 275] .306) .192). 152] .194| .231| .264| . 296) . 184). DESIGN OF REINFORCED CONCRETE CHIMNEYS 829 (8) Mix the concrete thoroughly and provide enough water to produce a quaking concrete. (9) Bond the layers of concrete together. (10) Place the steel accurately. (11) Place the concrete around the steel carefully, ramming itso thoroughly that it will slush against the steel and adhere at every point. (12) Keep the forms rigid. The fulfillment of these requirements will increase the cost of the structure; but if the recommendations are followed, there should be no difficulty in erecting concrete chimneys which will give thorough satisfaction and will endure. In connection with reinforced concrete chimneys, the problems which arise are of two general kinds: (1) A problem in design, involving the determination of the neces- sary thickness of shell and required amount of reinforcement at the various sections of a chimney of given height and diameter. (2) A problem in the review or investigation of a chimney of given height and diameter having a certain thickness of shell and a given amount of reinforcement, to determine the stresses in the concrete and the steel under the action of certain forces. The application of the foregoing formulas to such problems and the use of the accompanying tables may best be illustrated by the following numerical examples. EXAMPLES OF CHIMNEY DESIGN Example 1.—Given a chimney with height above section considered, 110 ft.; mean diameter at section considered, 10 ft.; allowable pressure in concrete (f,), 500 Ib. per sq. in.; allowable tension in steel (fs), 14 000 lb. per sq. in.; ratio of moduli n, 15; wind pressure (on normal plane) 50 lb. per sq. ft., weight of con- crete taken as 150 lb. per cu. ft. What is the necessary thickness of shell and amount of reinforcement at the given section? Solution.—As in all chimney designs, it is necessary here to make a trial assumption of the thickness of shell in order to estimate the weight. Suppose Wwe assume a 6-in. shell for the entire height above the section. Assuming that a wind pressure of 50 Ib. per sq. ft. on a normal plane corresponds to 8 of 50 Ib. or 30 lb. per sq. ft. on the projected diameter of a cylindrical surface, we have the bending moment due to the wind, M = [10.5 X 110 X 30] X +32 X 12 = 22 869 000 in.-lb. and the total weight of the chimney above the section, W = 3.1416 X 10X 0.5X 110 X 150 = 259 180 lb. 830 REINFORCED CONCRETE: CHIMNEYS f. = 500, fs = 14000, and n = 15, Table 1 gives k = .349. For ce I 349, Table 2 gives Cp = 1.687, Cr = 2 335, 2 eae, Substituting in equation (1), _ 8(22 869 000 — 259 180 427 120) i 2.335 X 14000 X 120 = 19.6. =* 5 Therefore, 19.6 sq. in. of steel are required. If 3-in. round rods are selected, 45 of them will be required. Substituting in equation (2), we have 19. 2> 259 180 + [(2.335 X 14.000) — (1.637 X 500 X 15)l5 77a 1.637 X 500 X 120 19.6 3.1416 X 120 = 6.6 in. Therefore a 6.6-in. shell would be used. In general, the values of A, and ¢ as thus obtained should be readjusted by computing W on the basis of the computed thickness of shell. In the case at hand, however, the original assumption of a 6-in. thickness corresponds, for all practical purposes, to the computed thickness of 6.6 in., so that recomputation is, in this case, unnecessary. If the walls of the chimney taper in thickness, the value of W must be altered accordingly.® When the required thickness of shell and amount of vertical reinforcement have been determined, there remains the question of the necessary horizontal or ‘ circular reinforcement. Substituting in Formula (3) for fs, say, 14.000 lb., we © have 110 X 30 Po = 18.8 14000 X 6.6 + 0.0025 = 0.0044. Area of steel, As = 6.6 X 12 X 0.0044 = 0.35 sq. in. Thus 3-in. round rods — should be spaced 6{ in. on centers. In a similar manner, any other section of the chimney may be proportioned. REVIEW OF A CHIMNEY Example 2.—Given a chimney with height above section considered, 90 ft.3 @ , mean diameter at section considered, 8 ft.; thickness of shell at section con- — sidered, 6 in.; vertical steel at section considered, sixty 3-in. round rods; wind pressure (on normal plane, 50 lb. per sq. ft.); weight of concrete taken as 150 lb. per sq. ft.; ratio of moduli, n, 15. the section under consideration? 6 In relatively high chimneys steel cannot be stressed to 14 000 lbs. per sq. in. (see p. 821.) a ee a ee q 4 What are the maximum stresses in the concrete and in the vertical steel at ; DESIGN OF REINFORCED CONCRETE CHIMNEYS 831 Solution.—A problem of this kind must necessarily be solved by a method of successive trials, since the position of the neutral axis is not-known. The location of the neutral axis is determined by the values of f,, fs and n, two of which, in this case, are unknown. The method of procedure, therefore, is to assume outright a trial position of the neutral axis, select the constants accord- ingly, substitute in equations (1) and (2) and solve them for f, and foe Then see if the position of the neutral axis, as fixed by these values of Ts ana Ts and the given n, is the same as the position assumed at the start. If the two positions agree, then f, and f,.as found are the actual stresses; if not, a new position of the neutral axis must be assumed, new constants selected, and new values of f; and f, computed from equations (1) and (2). Thus a series of trials must be made until the location of the neutral axis as assumed is consistent with the computed values of f, and f, together with the given n. In this problem, assuming 30 Ib. pressure on the projected area, we have the bending moment due to the wind, 90 M = [8.5 X 90 X 30} x ms < 12 = 12 393 000 in. lb. and the total weight of the chimney above the section, W = 3.1416 X 8 X 0.5 X 90 X 150 = 169 646 lb. A, =60 X .3068 = 18.41 sq. in. Now suppose we assume the neutral axis at, say, k = .400 For k = .400, table on p. 828, gives Cp = 1.765, Cr = 2.224, 2 = .416. Substituting in equation (1) we have 8(12 393 000 - 169 646 x .416 x 96) 18.41 = 2.224 X fs X 96 whence f, = 11 400. Substituting in equation (2) we have, | 18.41 3.1416 ee lord 1765 f,0 96 3.1416 X 96 2 X 169 646 + (2.224 x 11 400 — 1.765.f, 15) whence f, = 416. Now fs = 11 400, f, = 416, and r = 15 gives k = .354 which does not cor- respond with our original assumption of z = .400. Evidently the true & must lie somewhere between the assumed and determined values, hence if we now assume, say, & = .375 and recompute, we obtain fs; = 11000 and f, = 435, the values of which together with n = 15 gives k = .371 which checks fairly well with the assumption of k = .375. For all practical purposes we may therefore say that the maximum stress in the steel is 11 000 Ib. per square inch, while the maximum stress in the concrete is 435 lb. per square inch. The results indicate that both the thickness of shell and the amount of steel are greater than are necessary for safe stresses. CHAPTER XXI RETAINING WALLS For walls designed to resist the pressure of earth or water, con- crete is generally superseding other classes of masonry. In most localities, its cost is less than that of rubble masonry. Its adaptabil- ity for thin walls and for certain classes of face work often makes it a suitable substitute, in complicated designs, for first-class masonry, with a consequent large saving in cost. In combination with steel, its possibilities for special designs are almost unlimited. Water-tightness, often an essential element for this class of struc- tures, receives general treatment in Vol. II. Portland cement con- crete may be made water-tight more readily than stone masonry laid in mortar of similar proportions to the cement and sand in the concrete, since large voids or stone pockets in the concrete are more easily prevented than the “ rat-holes ”’ so frequently found in the © bedding of stones in mortar. Moreover, with careful selection of — aggregates and skill in laying, thinner, impermeable walls may be built of concrete—strengthened with steel reinforcement—than is possible with stone masonry. Reinforced concrete retaining walls cannot be designed by “rule of thumb,”’ and therefore a careful consideration of the forces acting — on them, and of the stresses in the concrete, is presented in this | chapter. Since.the earth pressure is the controlling factor, it will be necessary to introduce a practical discussion of this before taking up the details of the design and examples of the two principal types. Relative Economy of Plain and Reinforced Concrete Retaining — Walls.—Retaining walls to support the pressure of earth may be ~ designed: (1) of gravity section with plain concrete or stone masonry; (2) of thin reinforced concrete section with spreading base or footing. Reinforced concrete retaining walls are almost always more economical than gravity sections of either plain concrete or masonry. Sa. WEIGHT OF EARTH 833 In the gravity section, the materials cannot be fully utilized because the section must be made heavy enough to prevent overturning by its own weight. Counterforts or buttresses are of comparatively little advantage because, in stone masonry, or even in plain concrete, ‘the wall is likely to break away from them. In reinforced concrete retaining walls, on the other hand, a part of the sustained material can be used to prevent overturning, and the section need only be made strong enough to withstand the moments and shears due to the earth pressure. Since the wall is lighter, exerts less pressure on the soil, and may be made, if necessary, with a very broad base, the special foundations or piling that are often necessary for a gravity wall may frequently be avoided. Reinforced concrete, properly designed, can be depended upon as absolutely reliable. The economy of a reinforced concrete wall, over a wall of gravity section of either stone masonry or plain concrete, is obvious because of the saving in material. The cost of forms is practically the same for reinforced designs as for gravity section. Mr. J. I. Oberlander reports! that twenty-three bids submitted on alternate designs of gravity and T-shaped reinforced concrete sections showed the average total cost per linear foot for the gravity section to be about one-third greater than that for the reinforced section. The unit price for the concrete in the reinforced section, however, was about 20 per cent greater than that for the gravity section. | | WEIGHT OF EARTH In the calculation of retaining walls, and many other structures, the weight of earth in place is a prime factor. The weights of dry material, based upon experiments by the authors, are represented in the following table. Most of the figures for. weights of earth give the weights per cubic foot after excavation in a loose or a compacted condition. In the authors’ experiments, the excavation was meas- ured, so that the weights represent the material in place. As fills will eventually assume much the same characteristics as earth in original excavation, the figures may be employed for either natural earth or filled material. The weight of earth containing water varies with the character of the material and with the conditions. Gravel containing ordinary moisture weighs about 2 per cent more than 1 Engineering and Contracting, May 19, 1915, p. 457. 834 RETAINING WALLS dry gravel; and sand may weigh from 3 to 10 per cent more,” depend- ing upon its fineness, since the finest sands absorb the moist water. Wet muck weighs about 75 lb. per cu. ft: These percentages assume that the bank is provided with natural drainage; if the earth is literally filled with water which cannot run off its weight will be increased by a quantity of water nearly equal in volume to the voids in the material, which vary, with the character of the material, from 20 to 50 per cent of the bulk of the earth in the bank. Many of the values appear high, but they are the result of care- ful tests. Average Weight of Ordinary Earth Before Excavation — Lb. per Cu. Ft Cand oO eee 105 CGravelina J) 110 igs 2 135 Gravelly clay ..0si.1> 5-00 e) Se = ae 130 TiO aT ce ed’s wid eomeael « Card nln a eal ee eee 90 Hardpan «...-..+ 9+ ¢17 9 oar oiae 2 een 130 Dry muck 7 uo... ers se AQ Drainage of Retaining Walls.—The drainage of retaining walls is a highly important matter, and lack of drainage may cause either complete or partial failure. A common plan is to lay a drain of tile or of broken stone along the back of the base, opening at the ends of the wall or discharging through weep-holes. The Delaware, Lackawanna & Western R.R.® builds 4-in. tiles through the wall at frequent intervals along the footing, with a right- angle elbow, turned up, on the inner side. A chimney of loose rubble, about 2} by 3 ft., runs from each weeper up to the top of the wall. The Rock Island Railroad,* in some track elevation work, laid a line of tile drain along the back of the wall and carried the water through the abutment by a weeper running to a storm sewer. Mr. Lindenthal’s high walls (see p. 873) were drained through weep-holes, placed every 10 ft., through each wall. From each weeper one 4 by 4-ft. dry rubble chimney was built up back of the wall to the surface. 2 This larger weight applies to moist sand in place. Loose, moist sand on the other hand, as stated in Vol. II, weighs less than dry sand. ; 3 Engineering Record, Jan. 3, 1914, p. 29. 4 Engineering News, April 8, 1915, p. 670. EARTH PRESSURE 835 Another good method of drainage consists of a layer of loose stone over the entire back, with weep-holes placed at intervals. Frost.—The depth of foundation must be sufficient to prevent heaving of the material in front of the wall, and to protect it from frost. A depth of 3 ft. may be given as a minimum, while 4 or 5 ft. is necessary in temperate or very cold climates. Even with the base safely below frost level, special precautions are sometimes necessary to prevent heaving by frost-grip on the side of the wall or abutment. Such a case, cited by Edward H. Rigby,5 was encountered in China, where frost gripped the side of bridge abutments to a depth of 5 ft. and lifted them, railroad, girder bridges, and abutments, clear off the pile foundations. Piers and abutments with sloping faces were lifted as much as those with vertical faces. Computations showed that the average lifting power of the frost was 1 000 lb. per sq. ft. of exposed surface, and that the remedy was to design the piers and bridges to overcome this force by dead weight. EARTH PRESSURE The principal force governing the dimensions of a retaining wall is the earth pressure. The magnitude of the earth pressure depends upon the character of the soil, its moisture content, the slope of the earth back of the wall, and also the inclination of the back surface of the wall. Rankine’s Theory.—The magnitude of the earth pressure is affected by so many variables, that it is incapable of being solved by an exact method applying to all conditions. The best theory thus far advanced is Rankine’s, based on the assumption that the earth is composed of granular particles without cohesion, held together only by friction developed between them. Direction of Earth Pressure.—The unit earth pressures—also the total earth pressure—are assumed to act in a direction parallel to the slope of the ground (for level ground or aestheley sloping up). This is shown in Fig. 295, p. 836. Distribution of Earth Pressure.—The unit earth pressure is assumed to increase proportionally with the depth below the ground level. The variation of the pressure may be represented by a tri- angle, as shown in Fig. 295. The unit pressure is zero at the ground level and a maximum at the bottom of the wall. 5 Engineering News, March 5, 1908, p. 260. 836 RETAINING WALLS Let P = resultant earth pressure in lb. on a vertical surface for a length of wall equal to one ft.; h = total height of wall, ft. ; h; = depth below top of wall of any point, ft.; w = weight of earth, lb. per cu. for | 6 = angle of inclination of earth behind the wall; ® = angle of internal friction of the earth; C, = constant depending upon 6 and ¢. (See table on p. 837.) ——---—--—-},—-—----—- 4. Mioep Be Cp wh les ecVerres wey We Value of Cp 1s different in the two cases Frc. 295.—Distribution of Earth Pressure. (See p. 835.) Unit Pressures on Vertical Surface.— Unit Pressure at Any Depth, hz, below Ground Level, _| Lape Shea cos 6 — Vcos? 6 — cos? ¢ x= w cose i : cos 8 Vcos? 6 — cos? () calling Ciel ans 5 008 5 — Vos? 6 — cos? ¢ : cos 6 + V cos? 6 — cos? ¢ a = Cytthe! » + +o ee ee Maximum Unit Pressure at Bottom, p = Cwh...> ig ee ee : 3G Values of C, are given in table, p. 837. Special Cases of Earth Pressure.—The above formula is general. Below are given formulas for special cases. When ground is level, 6 = 9, and the constant C, changes to 1 —sin¢ oar 1+ sin ¢ EARTH PRESSURE 837 When ground slopes at angle equal to angle of repose (6 = ¢), the constant becomes Cz = cos 6. Total Pressure.—Since the pressure is represented by a triangle, the total pressure is obtained by multiplying the maximum unit pressure at the bottom by one-half of the height. The point of application of the pressure is at one-third of the height from the bottom. Thus, Total Pressure, Tei Ri, ae eo LS te 1, (A) Values of C, may be taken from table below. Values of Constant Cp. cos 6 — V cos? 6 — cos? } cos 6 + Vcos? 6 — cos? b Cp = cos 6 Values of Constant Cp Slope with Horizontal Angle ; f Pato | 1to1s | 1to2 | 1t023| 1to3 | 1to4 | Level | Friction ! ? Corresponding Angle of Slope 6 | fo ego a0 e206, 40’ 121° 50’ | 18° 30% | 14° 0 0 p 55° 0.18 0.13 0.12 0.11 0.11 0.10 0.10 | 0.57 50° 0.29 0.18 0.16 0.15 0.14 0.14 0.13 | 0.64 Tee hee... 9 0.26 0.22 0.20 0.19 0.18 OPDD. WOT ae 0.36 0.29 0.26 0). 24 0.23 O22 O00 SOOM shane. 0.58 0.38 0.33 0.31 0.29 0727 |) 0782 SoU) sie Meet pit 6 1b ire, Svs xi ay 0.54 0.44 0.40 0.37 0733 |, 0°87 EET VAS EE Ss Per 0.60 0.52 0.46 0.40 | 0.91 RMIT Aee Ma is ras to) ec lckia wey os 0.72 0.58 0.49 | 0.94 Nortr.—If the angle of internal friction of the earth is unknown, the following average values may be used: Coal, shingle and broken stone, 50°; earth, 35°; clay, 30°; sand dry, 30°; sand moist, 35°; sand wet, 20°. Surcharge.— When the earth behind the wall is loaded in any way, as for example, when a highway or a railroad track runs along the 838 RETAINING WALLS wall, or when the embankment is used for the storage of material, this loading causes additional pressure on the wall, which should be provided for by replacing the weight of the load by an equiva- lent surcharge of earth. The height of this surcharge, hi, is the load per square foot divided by the weight of a cubic foot of earth. Thus, a load of 500 Ib. per sq. ft. is equivalent to a surcharge of 5 ft., if the earth weighs 100 Ib. per eu. It, | The earth pressure on a re- taining wall with surcharge is Fia. 296.—Earth Pressure with Sur- represented by a triangle with the chatge, | (See P8338.) zero point at the top of surcharge. The maximum pressure at the bottom is p = C,w(h + *1), where hi is the distance from level eround to top of surcharge. (See Fig. 296, p, 838.) The pressure resisted by the wall is represented by a trapezoid, the top of which is C,wh; and the bottom C,w(h + hi). The total pressure equals the area of the trapezoid. In addition to notation on p. 836, Let h, = distance from level of ground to top of surcharge. Total Pressure with Surcharge, ahr si ma te eres Bk ae Chath) P-= iC w(h + 2h Height of Point of Application of Pressure above Bottom, he Bhi h = FL, 3 oe Ree tay x Wall with Inclined Back.—The earth pressure, #, on an inclined plane ab (Fig. 297) is the resultant of P, the horizontal pressure on the vertical plane ac, and W, the weight of the prism of earth abc, and acts at one-third the height from the bottom. Resistance of Wall to Sliding—The horizontal component o the earth pressure causes the tendency of the wall to slide on its base. This force is opposed mainly by the friction between the base of the wall and the earth, which is equal to the vertical pressure multiplied DESIGN OF RETAINING WALLS OF GRAVITY SECTION 839 by the tangent of the angle of friction between concrete and earth. Let . F = total frictional resistance of the base; Wi + W2 = weight of concrete and earth above the base; N = vertical component of earth pressure; y = angle of friction between earth and concrete, Then F = (Wi+ W2+ N) tan y. Woke Site) LeWi eat) 26) Ne ts ah) If the friction is larger than the horizontal component of earth pressure multiplied by a factor of safety, the wall is secure against sliding. This rule may also be expressed as follows:® If the angle which the resultant pres- sure makes with the vertical is smaller than the angle of friction, the wall is safe against sliding. . The resistance to sliding is increased by the passive pressure of the earth in front of the wall, and particularly in front of the toe. Also, the cohesion of the earth back of the wall assists the resistance to some extent. In many cases, the resistance to sliding is in- 16: 297-— Harth ‘ ees Pressure on Inclined creased by a vertical downward projection of the Back of Wall. base or a key, which may be placed in the middle (See p. 838.) of the base or at its heel end. Passive Earth Pressure.—The passive pressure of a mass of earth, that is, its resistance to movement or compression, is many times as great as the active pressure. Because of the possibility of shrinkage of earth fill in front of the wall, away from the wall, the passive pressure of the earth in front of the wall cannot be counted on to counteract active pressure due to the earth behind the wall DESIGN OF RETAINING WALLS OF GRAVITY SECTION The thickness of gravity retaining walls is frequently determined by rule-of-thumb, but this is an unsatisfactory procedure. On work of any importance, much more economical results are obtained by special designs, governed by the character of the foundation soil and the earthbacking. Partial failures—tipping forward, cracking, and 6 The horizontal component equals (Wi + W2-+ N) multiplied by the tangent of the angle of the resultant with the vertical. If this angle is smaller than y, its tangent will be smaller; therefore, the horizontal force is smaller than friction. 840 RETAINING WALLS sinking—are prevalent among retaining walls. In one case, a heavy gravity wall failed under the weight and impact of the backfilling dropped through a distance of 30 or AO ft. from a large drag line scraper bucket. Where the foundation is so poor that such action is possible. the line of pressure should pass through the base, well within its middle third. Uneven settlement is then less likely to take place; and in any event the line of pressure has more opportunity to shift, without causing either uplift or excessive pressure on the base, than if the line passed through the forward edge of the middle third.’ The methods of determining the line of pressure and of computing pressure on foundation are similar to those discussed in connection with reinforced walls. If empirical rules are to be used, one that is easily remembered is to make the base three-eighths of the height, for walls without sur- charge. Another is to make the base at least as thick as would be necessary if the wall were to be subjected to water pressure under a head two-thirds the height of the wall. A table of empirical values, adopted by Mr. Trautwine for thickness of base of masonry walls to resist earth pressure, is given below. Thickness of Retaining Walls of Gravity Section with Earth Surcharge By Joun C. TRAUTWINE. (See p. 840.) Ratio of Thickness of Ratio of Thickness of Height of Earth to | Base as Ratio to | Height of Earth to | Ease as Ratio to Height of Wall Height of Wall Height of Wall Height of Wall 1 0.35 2 0.58 a | 0.42 275 0.60 1,2 0.46 3 0.62 1.3 0.49 4 0.63 1.4 0.51 6 0.64 1.5 0.52 9 0.65 eae) 0.54 14 0.66 hare 0.55 25 1.8 0.56 or more 0.68 ee ee Walls designed by these empirical methods are unsafe under unusual pressures, such as quicksands, and detailed analyses must 7 Certain failures of this type are discussed by Charles K. Mohler in Engineer- ing News, Oct. 18, 1910, p. 384. 8 Suggested by Engineering News, Sept. 26, 1912, p. 593. DESIGN OF REINFORCED CONCRETE RETAINING WALLS 841 be made. On the other hand, the designs obtained by empirical methods are in many cases unnecessarily conservative. The height of the wall is assumed to be measured above the sur- face of the firm ground in front of it. The batter of the face of a retaining wall is customarily limited to 13 inches to the foot, and the back is usually vertical. This fixes the width on top. The values in the table may be employed for long walls of con- crete with no reinforcement. In many cases, because of the mono- lithic character of concrete, a ratio of thickness to height from 10 to 20 per cent less may be adopted with safety, if the character of the filling back of the wall precludes excessive pressure, and if the base is slightly spread. For more accurate determinations of gravity sections, the principles which follow, relating to reinforced designs, are applicable. When two walls enclose a narrow fill, they may be tied together by rods, as discussed on p. 872, and thinner sections used. Similarly, the ordinary single wall may be anchored to the ground behind it. DESIGN OF REINFORCED CONCRETE RETAINING WALLS General Principles.—A reinforced concrete retaining wall may fail as a whole, if it settles excessively or slides on its base. Its failure may also be caused by the failure of its component parts. For instance, in a cantilever wall, the vertical slab may break away from the foundation, or the toe cantilever may crack or shear off. To prevent the failure of the wall as a whole, the base must be properly proportioned. In a properly proportioned retaining wall, (1) the pressures on the foundation must be within the allowable working limits, and must be properly distributed over the base; and (2) there must be sufficient resistance to sliding of the wall. To prevent failure of the component parts, each part must be designed according to the accepted rules, to resist the stresses to which it is subjected. Also, the parts must be properly connected: Earth Pressure to be Used in Design.—The base of a retaining wall is usually several feet below the level of the lower ground, in order to protect it from frost and to reach solid ground. There is, therefore, a certain amount of earth in front of the wall. In fixing the amount of earth pressure to which a wall is subjected, the question arises whether it would not be permissible to utilize the passive 842 RETAINING WALLS pressure of the earth in front of the wall to resist a part of the earth pressure. This is not permissible. The earth in front of the wall is usually a fill, which, until it becomes compacted, has very little passive resistance. Furthermore, the fill may shrink away from the wall. Finally, the fill may not always be in place at the time the earth pressure acts. A good rule is to design the wall for the amount of earth pressure acting above the top of the slab at the end of the base. (See Fig. 298 below.) Total Earth Pressure : i y Pressure / on upright Pressure on upright ‘ Total } ressure\ ¢ { ' { \ H ; | ’ | ' - \ wm : { U t 1 | Resistance | Resistance / | of this earth of this earth unreliable y | unreliable \ | NTN | FARTING : | ’ \ ! ; 5 { \ nie: ‘ ! @ Level Ground Reinforcement not shown. _(b) Sloped Ground Fic. 298.—Earth Pressure on Reinforced Concrete Retaining Walls. (See p. 842.) For walls supporting earth with a sloped surface, the height of the wall to be used in the formulas for earth pressure is the distance from the top of the slab at the end of the base to the top surface, measured on a vertical line. This distance 1s marked h; in Fig. 298. The width of base should be designed for this earth pressure. The amount of earth pressure to be used in designing various parts of the wall is discussed under the proper headings. Factor of Safety for Retaining Walls.—In’ all self-contained retaining walls, the earth pressure is resisted by the vertical loads acting on the retaining wall. It is a common characteristic of rein- forced concrete retaining walls that they utilize not only their dead load, but also the weight of earth resting on their base, to counteract DESIGN OF REINFORCED CONCRETE RETAINING WALLS 843 the earth pressure. The amount of available weight of earth depends upon the width of the base back of the wall. The distribution of the pressure on the foundation depends upon the length of the toe, 1.e., the part of the base in front of the wall. For any accepted dimensions of the base, the available vertical loads are fixed. To get the pressure on the foundation, it is necessary to decide upon the proper assumption as to the earth pressure. The combina- tion of the expected earth pressure, found from Formula (4), p. 837, with the vertical loads, would give the pressure on the founda- tion for working conditions only. This, however, does not solve the problem completely. The retaining wall should be investigated for accidental earth pressure, the magnitude of which is equal to the expected earth pressure multiplied by a selected factor of safety. The object is to make the wall stable not only for working conditions, but also to insure a definite factor of safety, just as in other structures. Ordi- narily, structures are subjected to one kind of loading. After the stresses at working loads have been computed, the stresses at increased loads may be computed by simple multiplication. If proper working stresses are adopted, the problem of the factor of safety is automatic- ally solved (see p. 125). Retaining walls, on the other hand, are subjected to two kinds of loading, largely independent of each other. The pressure on the foundation is a resultant of the earth pressure and the weight of the wall and of the earth above the base. The earth pressure may increase without necessarily increasing the vertical loads. For twice the earth pressure, the pressure on the foundation will not be doubled, but instead its distribution will be altogether changed. The point of application of the resultant on the foundation for increased earth pressure will be moved towards the toe, as is evident from Fig. 303, p. 857. The difference between the maximum and minimum pressures will be larger. The authors recommend that all walls shculd have at least a factor of safety of 2, as far as pressure on foundation is concerned. Allowable Unit Pressures on Foundation and Allowable Distri- bution.—The proper selection of a base requires some Judgment, as it is impossible to make rules applicable to all conditions. In no case, of course, should the pressures on the foundation for working conditions exceed safe bearing values on soil, as given on p. 471. If it is impossible to support the wall on soil within a reasonable depth below the surface, piles may have to be used, 844 RETAINING WALLS For yielding soil, the pressure on foundation for working condi- tions should be distributed as uniformly as possible. Therefore, for expected earth pressure, the resultant — should be well within the middle third, but for earth pressure multiplied by the factor r = 2, the resultant should strike the base at the outside edge of the middle K L\ RaVEAW : Rath 7 a be <8 2 third. The factor r is in this case ! equal to the factor of safety. For un- TENTS yielding soils, such as hard pan or rock, the resultant for working loads may be allowed to intersect the base at the edge of middle third. In this case factor ,y = 1. For earth pressure times factor of safety of two the resultant would be outside the middle third and there would } be theoretically, tension (or rather uplift) ‘Pressure on foundation gn the outer edge, but the pressures at the Fic. 299.—Proportions of T- ¢o¢ would still remain within safe limits seen Segre BAY and therefore the wall would not fail. For intermediate conditions of the supporting soil, intermediate values of r (between 1 and 2) should be accepted for which the result- ant is allowed to reach the edge of middle third. The ratio r must not be confused with the factor of safety. Formulas for Width of Base——After the magnitude of the earth pressure is decided upon and the ratio, 7, is selected, the approximate width of base may be computed from following formula: “Let b = width of base in feet, ft.; . h.= height of wall from bottom to top of ground, for level eround; distance from bottom to top of surcharge, for wall with surcharge, feet; h, = distance from base to top of ground measured on a vertical erected at outside edge of base, for wall with sloped ground, feet; distance of point of application of forces R and N from center of base, ft. m = ratio of length of toe (measured to outside face of wall) to width of base. Fig. 299); : selected ratio of earth pressure, for which the resultant ‘s allowed to reach outside edge of middle third, to expected earth pressure. © II 3 II DESIGN OF REINFORCED CONCRETE RETAINING WALLS 845 Cy = constant for determining earth pressure, p. 837; Cy = constant for determining width of base, p. 845. Width of Base (approx.), when resultant strikes edge of middle third, b= 0.944) h for | . 4 ae Gre EE hs or level ground, (8) os rC, cos 6 b = 0.94 ee 1 Si a for sloped ground, . . (9) or b = Ch for level and sloped ground, where C; is a constant from the table below, depending upon the ratio of toe to base, m, the constant of earth pressure, Cy, from _table on p. 837, and the selected ratio r, for which the resultant should strike the edge of middle third. For sloped walls, the value of C, should be taken from the table, that value being used which corresponds to the value of C, Pte icd by the cosine of the angle of slope. VALUES OF CONSTANTS C;, Teen od if ae Sa Gk Semana Values of rCp for Level Ground and rCp cos 6 for Sloped Ground Value of m | | | 0.05)0.10/0.20/0.30/0.40|0.50\0.60:'0.70/0.80/0.90/ 1.0] 1.2}1.4]1.6]1.8]/ 2.0 0 0.21)0.30/0.42/0.52'0.59/0.67)0.72/0.79/0.84/0.89)0.94/1.03!1.11/1.18]1.2611.32 0.05 0. 20/0. 29|0.40/0.49/0.57/0.64|0.69/0.75/0.80!0.85/0.90'0.99/1.06/1.13/1.20/1.27 0.10 0.19|0.28)0.39}0.48/0.55/0.62/0.67/0.73/0.77/0.82|0.87'0.96]1.02/1.09]1.16)1.22 0.15 0.19|0.27/0.38/0.47|/0.53/0.60|0.65/0.71/0.75)0.80,0.85/0.93/1.00)/1.07/1.13/1.19 0.20 0.18!0.27/0.37/0.46/0.52/0.59|0.64/0.70/0.74/0.79/0.83/0.91/0.98/1.05}1.11]1.17 0.25 0.18/0.26/0.37|0.45/0. 52/0. 58)0.63/0.69/0.73/0.78)0.82/0.90/0.97/1.03)1.10/1.16 0.30 0.18'0.26/0.37|0.45)0.51/0. 58/0.63/0.68/0.72/0.77/0.8110.90/0.96]1._03}1.09}1.15 Ueao 0.1810. 26!0.3710.45/0.51/0.58)/0.63/0 68/0.72/0.77/0.81/0.89|0.96/1.02}1.09]}1.15 0.40 0 18/0. 2610.37 0.45/0.52 ee ai 0.73)0.78'0.82/0,90/0.96/1.93/1.10}1.15 | | | If it is desirable to get uniform pressure on the foundation, the resultant should strike the base in the middle. For such a condition, the formula for width of base becomes: S46 RETAINING WALLS Width of Base (approx.), when resultant strikes in the middle C ae Ram bY nat b 0.55) aryl for level oTOuNid) sind tan (nb)? in.-lb. 11 The vertical component of the pressure, represented by abcd in the triangle of pressure, acts on the heel cantilever, as shown in Fig. 302. The vertical component of the earth pressure at the heel is Cpwh sin 6, and at the edge of upright slab Cpwh, sin 6. To get vertical components for unit of length of base, — these values must be multiplied by tan 6. Thus, at the heel, the vertical com- ponent per lineal foot is Cpwh, sin 6 tan 6, and at the wall Cpwhz sin 6 tan 6 CANTILEVER WALLS 855 The formula for upward moment is the sameas for level ground. = tpl — n) + np] ft.-lb. | ere ge OA 2[p2(38 — n) + npi|(nb)? in.-lb. | Resultant Moment to be Used in Design (Acting Downward), en) 2 Nk vss ce vw we se, (OD) For small angles, 5, the value of M3 is negligible. Moe or a 4 ana Seng 2 1 i bic ' 1 ae ' ee 5h : 3: 1 C, d, cd Cpwh x eae ; 4 y 1 ' 1 aS ' uy eS r) ' is ecad ) 4 L} : ! bio a & | (abe,d,= t * || Earth Pressure cael Fs ~==~12 on Heel Cantilever h. RY pe en = 5 ae: b || c,d ‘ab’= Vertical ; tS : qs Component 4 , cae ‘ c CP . i} ) ' pON a "4 aa BHC a ea nt Stele es ale b ie Se == ee Be s gg Ps: Upward p, Upward Pressures, -- a on Heel Conte on Toe Cantilever eS ~np, +(1-n) Py 1-m) Pt mp, Fia. 302.—Forces Acting on Base Cantilevers of T-wall. (See p. 854.) ) Temperature Reinforcement.—To prevent open vertical cracks in the wall, which are unsightly, the wall, even if well reinforced horizontally, should be built in sections not more than 60 ft. long. Grooved lock-joints should be provided at the ends of the sections. The amount of longitudinal temperature reinforcement necessary will depend upon the degree to which elimination of cracks is desired. It should not be less than 0.2 per cent of the volume of the concrete. For water-tight walls, this should usually be increased to 0.4 per cent. Two-thirds of the temperature reinforcement may be placed near the outside face, and one-third near the inside face of the wall. 856 RETAINING WALLS Adjoining bars should be properly lapped, to develop their full strength by bond. Deformed bars are more effective than plain bars in distributing the fine cracks that may occur. EXAMPLE OF T-SHAPED RETAINING WALL Example 1.—Design a retaining wall, 12 ft. above ground, to support a sand fill, which slopes upward at a rate of 1 to 2, making an angle with horizontal of 26° 30’.. Weight of sand is 100 lb. per cu. ft., and its angle of internal friction 35°. Usel:2:4 concrete and mild steel with ration = 15 and working stresses in Ib. per sq. in., fc = 650, fs = 16.000, v = 40, wu = 80, for plain bars, and u = 100 for deformed bars. It is desirable that the resultant should strike the base at the edge of the middle third for not less than 14 times the expected earth pressure. Thus, ratio Teli Be Solution.—If the base is imbedded 4 ft. below the surface, to protect it from frost, the total height of wall from base to top is 16 ft. Proportions of Wall.—The depth of wall is 16 ft. A ratio of length of toe to base, m = 0.3, will be assumed. For an angle of repose ¢ = 35° and an angle of slope 6 = 26° 30’, the earth pressure constant, from the table on p. R37 18 C, = 0.38. The value of rCy cos 6 = 1.5 X 0.38 X 0.894 = 0.51. For this value and from m = 0.3, the constant from table on p. 845 is C, = 0.58, and the length of base b = Cyh = 0.58 X 16 = 9.3 ft. Use 9 ft. 4 in. From this the length of toe, mb = 0.38 X 9.3 = 2.8 ft) “Use 2 ft. 10 in. Check of Assumed Base Width.—The assumed proportions will now be laid out to scale as shown in Fig. 303. To facilitate computations, the concrete slabs are represented by rectangles. The thickness of the upright wall and of the base is assumed as 12 in. The weight is now computed. Taking unit weight of concrete at 150 lb. and of sand as 100 Ib. per cu. ft., Distance from Heel to -Moment of Weight per Foot of Length ' Weight Center of Gravity oak Elen! Slab W,=15.0 1.00 X150 = 2 250 lb. 5 ft. 6 in. +6 in. =6.00 ft. 13 500 ft.-lb. 9.33 Base We= 9.33 1.00 X150 = 1400 |b. aie Pee ft. 6 550 ft.-lb. 15.0+17.7 5.517.7+2X15.0 Ve= §5.5———— X100= 9 000 lb. ee Se tt ek Oe BET Ae Sorry 3° 17 7-Eib0 Total weight.occ ones sys os os W = 12 650 lb. Total moment... ...----- 44 150 ft.-lb. Distance from heel to center of gravity of all loads, equal to total moment divided by total weight, is 44156 = 3.5 ft. The earth pressure acting parallel to slope is represented by a triangle. It is. assumed that the pressure on the face of the base is resisted by the passive pres- 857 EXAMPLE OF T-SHAPED RETAINING WALLS : NY 6 4 006 GLISJ m0 J02H2A —-g, nog z= 4 M4 M \ TReee Vomee M 107s i Fe Be ROLES ! ' S 2 Se oe $3 \a\s ; : 3a \O e, —& iret \ 94 068 NY ay rurynsoy fo 91 000 6="M—~S_ : ---- 5 Ft, 44------ a 91 €6 eS : ' i 35 4 xy t i] t ~ : ek | oR | S| aS ' I Se) 22 al | : Spa) Net 23 Ea | = Ww ' L ~ © | 1 ‘ n ‘ 1 ' ‘ 1 ' 1 i] ' I \ ! ‘QI LGES Se > 2% uN iS ® ag D Fae Ss Sw Q 2150 =m™ ™ 1 OLS & Pressure on Foundation for Expected Earth Pressure | Pressure on Foundation for Earth Pressure X Yr (See p. 856.) Fig. 303.—General Dimension of Wall. 858 RETAINING WALLS sure at the toe. The total earth pressure, from Formula (4), p. 837, where C, 018318 =x 0.38 X 100 X 17.72 = 6000 Ib. This applies at one- third of the height from bottom. The components of the earth pressure are: H = P cos 6 = 6000 X 0.894 = 5 360 lb. V = Psin 6 = 6000 X 0.446 = 2 680 lb. Plot the resultant downward force, W, at a distance of 3.58 ft. from the heel, and also the earth pressure, P, in proper position. Combine the two forces as shown in the figure. The resultant, R, strikes the base well within the middle third. Combine again the force, W, with rP, i.e., for earth pressure multiplied by ratio r = 1.5. For this condition, the resultant strikes the base at the edge of the middle third. Pressure on Foundation.—The pressure on the foundation is obtained from Formula (14), p. 847. The vertical component is 15 330 lb. and the eccentricity, ; 9 ft. 4 in. 5 e = 5 ft. 41in. Eas = Sin. = 0.67 ft. 15 250 6 X 15 330 X 0.67 = = 1:645 +) 712 1b, a tte Hasne PP oe 9.33? + 712 Ib. per sq. From this, pi = 2357 lb. per sq. ft.; p2 = 933 Ib. per sq. ft. For the other condition, the pressure acts praetically at the edge of the middle third. Therefore, the maximum pressure at the toe equals twice the average pressure. —7 2 pS es 3 5 TO) b. e oq. nN. Graphical method of determining pressure on foundation for known average pressure and location of force, is shown in Fig. 303, p. 856. Resistance to Sliding —The horizontal component of the earth pressure is 5 380 = 0.352 15 330 — is much smaller than the tangent of the angie of friction, hence the design is satisfactory. (See p. 839.) Toe Cantilever.—For explanation, see p. 852. The maximum bending moment, from Formula (19), p. 852, when m = 0.3; mb = 2.82 ft py ="2 257 1b a 933 Ib. M = 2|2 357(3 — 0.8) + 0.3 X 933]2.827 = 106 000 in.-Ib. H H = 5380 lb. The normal resultant is N = 15 330. Ratio N ae The maximum external shear, from Formula (21), p. 852, 2 357(2 — 0.38) +0.3 X 933 = ) f 0.8 Ae" 2.82 = 6050 Ib.) y. 2 EXAMPLE OF T-SHAPED RETAINING WALL 859 The depth of slab for bending moment, from formula d = C,;V M, where C;, = 0.028 (Table 2, p. 880). d = 0.028V106 000 = 9.12 in. As = 0.077bd = 0.0077 X 12 X 9.12 = 0.84 8q. in. 3 bars, 4’ on centers has an area of 0.921 sq. in. Bond Stresses now will be investigated. The perimeter of a 2-in. rd. bar is o = 1.96. Since there are 3 bars per lineal foot, the value of So = 3 X 1.96 = 5.88. The unit bond stress is V 6 050 ee eee 196 Ib: Lojd 5.88 X 0.875 X 9.3 The bond stress is excessive. To reduce it, increase either the area of steel or the depth. In this case, the depth will be increased in the ratio of the computed to the allowable bond stress. d, = 9.12 X 128 = 11.5 in. Use a total depth of 15 in., which gives about 3 in. protection for the steel. The bars must be extended 40 diameters beyond the junction of the toe and the wall, to develop their full strength. For 3-in. bars, this amounts to 2 ft. 1 in. Shear, 6 050 . = = 48 0 Ib. +1h; * ~ 11.6 X 0.875 X 12.2 AG tan Tt Upright Slab.—For explanation, see p. 850. As evident from Fig. 305, p. 862, the height of upright slab above the junction of slab and toe is hy = 14 ft. 9 in. Since Cp = 0.38 and w = 100 Ib. per sq. ft., 6 = 26° 30’, cos 6 = 0.894. The bending moment, from Formula (18), p. 851, M = 2 X 0.38 X 0.894 * 100 X 14.753 = 218 000 in.-lb. Thickness of slab, from formula d = C iv M is d = 0.028V 218 000 = 13.1 Use h = 15 in. With 13 in. protection outside of bars, the distance to the 0.7 : center of steel is 1.5 + “e ='] 875 in. Area of steel, from formula A, = pbd, where p = 0.0077, is A; = 0.0077 X 12 X 138.1 = 1.21 sq. in. This is satisfied by 3-in. rd. bars placed 4.37 in. on centers. Use 44-in. spacing at the bottom. At the top, where the bending moment is zero, the minimum practicable depth may be used. This is considered to be from 8 to 12 in. In this case, 10 in. depth is selected at the top and the slab made to slope uniformly to the - Maximum depth. 860 RETAINING WALLS If it is desired to make some of the vertical bars shorter than the rest, bending moments at intermediate points should be found, and the required areas of reinforcement computed. Since the moments vary with third power of the height, the values at intermediate points may be obtained by multiplying the maximum moment by third power of the ratio of the height of the point to the total height of upright slab. Effective M Height hz | Ratio, iis 's) See al Depth, Aye fad h k nch-pounds eo fsJ nches eq. In. Pit Os 1 o 3 400 9.35 0.03 ih = 188 4 $ 27 200 10.6 0.18 8h = 11.05 A 27 §2 000 11.85 0.56 he = 14.75 1 1 218 000 13.1 121 ; ee ee For maximum bending moment use 3.in. rd. bars spaced 4% in. on centers. The length of bars may be obtained by drawing a free-hand sketch on cross- section paper, aS shown in Fig. 304. The height is laid out to a convenient scale, for instance, one division for one foot. At each elevation, the required area of steel is laid out on a horizontal line, the divisions being used again as a measure. The points connected give a curve representing the required area of steel at any height. The reinforcement is laid out as used. In the sketch, an assumption was made that one bar in six would be carried to the top, making the top spacing of bars 25} in. Next, the spacing of bars is reduced to 123 in. by using two bars in each 253 in. Then four bars are used in 254 in., and finally six bars, making the bottom spacing 44 in. The areas of bars are reduced to effective areas per lineal foot of wall, and plotted. ‘Thus, one 3.in. rd. bar in 253 in. gives 0.207 sq. in. per lineal foot, two bars 0.41 sq. in. per lineal foot. The intersection of the vertical lines representing the areas of bars, with the curve representing the required areas of reinforcement, gives the theoretical point where the bars become unnecessary. The bars should be extended 40 diameters beyond this theoretical — point. In important work, hooks should be provided at ends to prevent slip of bar. Heel Cantilevers.—For explanation, see Pp. 953. Since nb = 5 ft. 6 in, n= a = 0.59, p: = 2 357 |b., po = 933 lb., A = 16 ft. Oin., ki = 17 ft. 9 ing h, = 15 ft., sin 6 = 0.446, tan 5 = 0.499, the maximum bending moments are — as follows: Bending moment due to weight of earth and slab (Formula (32), p. 854). M, = 2.2 X 100(2 X 17.75 + 16.0 + 2)5.5% = 357 500 inl. Bending moment due to vertical component of earth pressure, Formula (34), p. 855. M, = 2 X 100(2 X 17.75 + 15.0)0.38 X 0.446 xX 0.499 x 5.52 = 25 900 in.-lb. - EXAMPLE OF T-SHAPED RETAINING WALL 861 The maximum bending moment due to upward pressure, from Formula (33), p. 854, is M; = 2(933(3 — 0.59) + 0.59 X 2 357]5.52 = 221 000 in.-lb. The sum of the bending moment is M = My, + Mz — M; = 357 500 + 25 900 — 221 000 = 162 400 in.-lb. EE 1 Ee : aC In06 1000 wah mE : Ce : RE | i-He -5 il ae ee raphe aan d dnd used | to oh A BEE vealFoot | | | ole oe Wl HHH ‘ai use| V-/4t6{ |_| [) meen Pees SCH E Ee Bert heh at wl ETA = el Fia. 304.—Points of Stopping at of Vertical Bars in Upright Slab. (See p. 860.) Thickness of slab, computed as before, d = 0.028V 162 400 = 11.3, ee) 3 te k= 13-28. Usa 132 in A, = 0.0077 X 12 X 11.3 = 1.04 sq, in. Use 3-in. rd. bars, 5 in. on centers, with an area of 1.06 sq. in. Temperature Reinforcement.—In the upright slab, temperature and shrinkage reinforcement will consist of horizontal bars placed near both faces of the slab. The total amount will be assumed as equal to 0.2 per cent of the total concrete area. This requires 12 X 10 X 0.002 = 0.24 sq. in. of steel per foot of height at the top of the wall, and 12 X 15 X 0.002 = 0.36 sq. in. per foot of height at 862 RETAINING WALLS the bottom. Of this steel, about two-thirds will be placed near the outside face and one-third near the inside face, as shown in Fig. 305. To support the bars near the outside face, either spacers or vertical bars should be used. Design of L-walls.—L-walls consist of two instead of three cantilevers, as the toe cantilever is omitted. The method of design S —— += | 1 eg ! 1 \ | | \ { = ns ie Eh eed is <|~~s. All longitudinal bars & © " . | | <4 > a 3 rd, continuous = ea ~ with 40 diam. splice. Z ' ) N = GC ie be ? { | Is | aan ae i \ Be ! 1 I Arad a | i a | MM Seve igi 2 3 Me: cc 8 : FS Pa wig | oD t ee } : + j ~ ; oO w i bas vy Ea = | ro \ vo ee i | ! \ : | Yi) Y Bis | i 3] i] Ss 1 2. [aig S t I | = ! Py “8 t bey 2: * I 5 iad 4"on 2 Ft. 7 { ws 8 centers | a"chorage Construction 1-1 Ft. 3” Joint -2 Ft, 10'->}< E><——- - 5 ft, 3 —--- ------ == — 9 Ft, ¢’-------- Fic. 305.—Details of T-wall designed in Example. (See p. 862.) of the upright slab and the heel cantilevers is the same, as explained in connection with T-beams. ‘The arrangement of reinforcement 1s shown in Fig. 306, p. 863. Erection of Cantilever Walls.—In designing retaining walls, the — method of erection should be considered. Ordinarily, it is desirable to erect the base slab first, before the forms for the upright slab are WALL WITH COUNTERFORTS 863 erected. The forms for the upright slab are then supported on the base. To avoid the necessity of bracing forms on the bottom, a small section of the wall, say, 6 in. high, is built with the me base. The wall eo UE forms then are bolt- ate ed to this stump. 3 The wall rein- Al forcement extends OOM TG sA d ; Reinforcement into the foundation slab. There would as be considerable diffi- eI) _--- Main Tension Reinforcement culty in keeping the 3: HOB OIRO Lo long vertical bars es in position during Re: pouring of the foun- ae dation slab. To . soe avoid this, dowels = aN a are inserted instead OF riction bq] Dowels of full-length bars. Joint --- 7 PE One-half of each ; dowel is imbedded ae in the base and the nome le other half extends Fic. 306.—L-shaped Retaining Walls. (See p. 862.) above the construc- AYR tion joint to lap the tension reinforcement of the upright slab. The length of dowels is such that each dowel is capable of developing, by bond, full strength of bar below and above the joint. To transfer shear at the construction joint, a key should be used as shown in Figs. 305 and 306. ,Main Tension Reinforcement for Base Cantilever BS a= 28. Deis Ceo Slot ia ans WALL WITH COUNTERFORTS To reduce the amount of steel and concrete in the upright slab and the toe cantilever, a design is often used with a counterfort between the two cantilevers, as shown in Fig. 307, p. 864. Relative Economy of Cantilever Walls and Walls with Counter- forts.— Whether the T-section of reinforced wall or the wall with counterforts is the more economical depends upon certain conditions. 864 RETAINING WALLS The principal condition is the height of the wall, but the intensity of the earth pressure and the relative cost of concrete and steel and forms also enter into the consideration. The T- section is preferable where skilled labor is scarce, as the construction is sim- pler and the placing of steel easier. The form st construction in the coun- 8 terforted wall is consid- < erably more expensive. Comparative studies of the two types indicate that the counterfort type is scarcely ever economi- cal when the height is less than 18 feet. For lower walls, it will be found that the \ additional cost of form- Fig. 307.—Isometric View of Wall with Counter- work for the counterfort forts. (See p. 863.) more than balances the saving in material. Design of Wall with Counterfort.—General proportions of the base — will be the same as for a cantilever wall of the same height and sub- | jected to same earth pressure. The proportions should be checked, in the same manner as shown in the example for cantilever walls, p. 857. Stresses Acting on a Wall with Counterforts.—An idea of the distribution of stresses in a wall with counterforts may be had from Fig. 308, which shows the possible causes of failure of such wall. Figure (a) shows a failure caused by cracking of the counterfort. This occurs when the amount of tension reinforcement near the out- side face of the counterfort is too small. Figure (b) shows a failure caused by the separation of the upright wall from the counterfort, which is caused by insufficient anchorage of the wallto the counterfort. Since the upright slab is reinforced with horizontal bars, it cannot act as a vertical cantilever, as in T-wall. Separation of the upright wall from counterfort, therefore, means failure of wall. Upright Slab-.. --Counterfort ace WALL WITH COUNTERFORTS 865 Figure (c) shows failure of the upright slab produced by earth pressure. ‘This is caused by insufficient amount of tension reinforce- ment. An additional type of failure may occur by rupture of the toe can- tilever. This is not illustrated. It is similar to that shown in Fig. 301, p. 850, in connection with failures of T-wall. At the junction of the wall and the base, the slab is not free to defect horizontally, as it is held by the base. A vertical bending moment will develop there, which should be resisted by short vertical bars placed near the outside face of the wall. Direction of \ Direction of | Earth Pressure -\ Earth Pressure \ ce we — i, Sy Ete. ee as Horizontal Section through Wall Failure of Upright Slab \ Crack in ae —~Counterfort ge 4 ane si ae D.. Failure of Counterfort Upright Slab Separated from Counterfort (a) (b) (c) Fig. 308.—Failure of Component Parts of Wall with Counterforts. (See p. 864.) Upright Slab in Wall with Counterfort.—The upright slab is a con- tinuous slab loaded by earth pressure and supported by the counter- forts. The principal reinforcement, therefore, is horizontal. In designing, consider horizontal strips of wall, say, one foot wide. The loading on each strip will be uniform and equal to the intensity of earth pressure acting at the height of the strip. The bending mo- 2 ment should be figured by formula M = oe for interior panels, and 2 ay we 10 for end panels. 866 RETAINING WALLS The earth pressures at the various strips varies from a maximum at the bottom of the slab to zero at the top. The thickness of slab is computed for the maximum pressure at the bottom of the slab. The slab may be made of uniform thickness. However, if the differ- ence between the maximum thickness and the practical minimum is large, the wall may be sloped. When the thickness of the slab has been decided upon, the amount of reinforcement required at different heights should be computed. Since the unit pressure varies according to a straight line from zero at the top to a maximum at the bottom, the bending moments also will vary in the same manner. If the heights are plotted on a vertical line and the moments at the bottom on a horizontal line, the triangle will represent moments at various points. These may be scaled. Since the wall is a continuous slab, reinforcement should be pro- vided at the supports. The reinforcement in the middle of the span will be placed near the outside face of the wall, and the reinforcement at the counterforts near the inside face of the wall. Toe Cantilever of Wall with Counterfort—The method of design of the toe cantilever is the same as for cantilever walls. The formula given on p. 852 may be used for bending moments and shear. If the length of the toe cantilever is appreciable, some saving may be effected by using cantilever ribs spaced at the same intervals as the counterfort. The base slab may then be considered as spanning between the ribs. Horizontal Slab Back of Wall.—The horizontal slab forming the base of the wall may be considered as though it were composed of continuous strips parallel to the wall, supported by the counterforts. Each strip is uniformly loaded, but the loading of the various strips of the slab varies from a maximum at the edge of the base to a minimum at the wall. For level ground, the loading of each strip is composed of the weight of the strip and of the earth above it, minus the upward pressure at the foundation. For inclined ground, the downward load consists, in addition to the weight of earth and slab, of the vertical component of the earth pressure. If 6 is the angle of slope, C, the earth pressure constant, hy the distance from top of slab to top of ground at the end of base in feet, and hz the height of wall above base in feet, then the intensity of the vertical component of earth pressure equals: WALL WITH COUNTERFORTS 867 Vertical Component of Earth Pressure at End of Base, Nmax = C,pwh, sin 6 tan 6 Ib. per sq. ft.. . . (86) Vertical Component of Earth Pressure at Wall, N = C,whg sin 6 tan 6 Ib. persq. ft.. . . (87) The loading is represented by a trapezoid, from which the load at any point may be computed. Counterforts.—Counterforts are really upright cantilever T-beams held in position by the base slab and loaded by the earth pressure transferred to the counterfort by the vertical slab. A portion of the vertical slab may be considered as a flange of the T-beam. The amount of earth pressure on the cantilever may be determined as follows: | Let ho = height of cantilever above base slab, ft.; s = spacing of cantilevers, ft.; C, = constant from table on p. 837, for earth pressure; 6 = angle of inclination of earth pressure; w = height of earth per cu. ft. Then Total Pressure upon Counterfort Acting Horizontally, eee oS ID Pe ee eS (88) Maximum Bending Moments in Counterfort, M = 1C, cos 6 whg?s ft.-lb., or 2C> cos 6 whe3s in.-lb. (39) For level ground 6 = 0, consequently cos 6 = 1. The maximum bending moment and the total pressure act at the junction of the counterfort with the base slab. The pressure and bending moment at any other point may be found by substituting the proper value for he. The amount of steel to resist bending moment may be found in the same manner as for an ordinary T-beam. The bars must be securely anchored at the bottom by bending them back into the base slab. The bending moment decreases rapidly. Since the cantilever slopes, the available depth of beam also decreases, but in a lesser degree. It will be found that it is possible to reduce the amount of tensile reinforcement by making some of the bars shorter than the rest. To find where the bars may be stopped, plot the required amount of steel and the available amount of steel. Carry the bar at least 40 diameters beyond the point where, theoretically, 868 RETAINING WALLS its area can be dispensed with. Because of the importance of the counterfort, it is advisable to provide hooks at the ends of all bars except the longest. Horizontal stirrups should be provided, not only to resist diagonal tension in the counterfort, but also to tie the wall to the counterfort. (See failure of wall, Fig.308(0).) The stirrups must be looped around the tensile reinforcement in the counterfort and extend through the counterfort as far as possible into the slab, where the ends should be provided with hooks. The counterfort should also be tied to the base by means of vertical stirrups, the area of which should be sul- ficient to transfer, by tension in steel, all the downward load from the base to the counterfort. Spacing of Counterforts.—The quantity of concrete and steel in a wall depends upon the spacing of counterforts. By reducing the spacing of the counterforts, the thickness of the wall and the base slab are reduced, hence the amount of concrete and steel is reduced. But, at the same time, the number of counterforts is increased, which means increase in formwork. The most economical spacing of counterforts is the one for which the total cost of materials and formwork is a minimum. Ordinarily, 12 in. is considered a practical minimum thickness for the vertical longitudinal slab at the bottom. Investigation shows that the most economical spacing of counterforts is the one for which the required thickness of vertical slab is equal to this minimum thick- ness. Considering 12 in. as the minimum thickness of vertical slab at the bottom, and using the following stresses in lb. per sq. in., fe = 650, fs = 16 000, the spacing of counterforts and the corresponding unit earth pressures at the bottom are given in the table below: SS ee 8 ft. 0 in./8 ft. 6 in. Spacing in Feet.......-. ft. .OUne i tbegoe te 9 ft. Oin.|9 ft. 6 in.|10ft. Oin. Unit pressure, lb. persq. ft.) 2 000 lb.| 1 860 lb. 1 640 Ib |1 450 Ib. |1 300 Ib. |1 160 lb. |1 050 lb. ee ee Thus, when the vertical slab is 24 ft. high above the base slab, weight of earth w = 100 lb., and the constant C, = 0.40, the pres- sure at the bottom is p = 0.40 X 100 X 24 = 960 Ib. per sq. ft: For this pressure, the economical spacing of counterfort is 10 ft. For the same conditions, except that the ground surface, instead of being level, is inclined 1 to 23 with the horizontal, constant SPECIAL DESIGNS OF RETAINING WALLS 869 Cy = 0.60, the pressure at the bottom is 1440 Ib.; consequently, the economical spacing will be 8 ft. 6 in. instead of 10 ft. From the above, it is evident that, as would be expected, the economical spacing of counterforts depends not only upon its height but also upon the intensity of earth pressure. Thickness of Counterfort.—The thickness of counterfort should not be smaller than required by diagonal tension. The external shear to be used in determining the thickness is the total pressure from Formula (88), p. 867. The counterfort must be wide enough to accommodate all rein- forcement without crowding, preferably in not more than two layers. Erection of Counterforted Wall—The counterfort should be poured with the vertical wall, in one continuous operation. No construction joints should be made in the counterfort. In the wall, construction joints should be vertical and placed midway between counterforts. If the base cantilever is poured separately from the counterfort, concrete keys should be provided at the juncture of base and coun- terfort to transfer shear from the counterfort to the base. Design of Wall with Counterforts.—Figure 309, p. 870, shows a typical design of a wall with counterfort, worked out in accordance with the rules previously given. SPECIAL DESIGNS OF RETAINING WALL Special designs have been worked out with considerable ingenuity, where local conditions require departure from the standard sections. In railroad construction, it is often necessary to build retaining walls where every inch of available room up to the edge of the right-of-way is valuable, and where no trespassing on adjoining property is per- missible. Cellular Wall.—The simplest solution, in case of a railroad fill, is to use the L-shaped wall with or without counterfort. In cases where such a wall produces excessive pressure on the toe, the cel- lular type of wall, shown in Fig. 310, p. 871, may be used. It con- sists of a continuous base a, two parallel longitudinal walls b and 6y, and tie wallsc. The space between the longitudinal walls is filled with earth. The earth pressure is resisted by the weight of concrete and the weight of fill between the longitudinal walls. The pressure on the 870 RETAINING A Iternate Arrangement 0 Slab, Steel rf we 7 Temperature Bars 5 All Horizontal Bars sf au ‘ " N *) 5 ; SO. eb aaa ; = Sr ee : mlb y : mie Ses Le . wey PV itros be ad [et ee a ee ' Lol Gee ‘ ae Ch RR ES : aS) WN 21 a ei \: re) As ee ee ee a Sa oe BS S| | | : %:=S ae 8 ia La eS BEC ii Rls Rec SOS aa Be! SS ; a ara eae BBQLS, me iano | vase SS Cross Section Rear Elevation Fic. 309.—Details of Wall with Counterforts. (See p. 869.) SPECIAL DESIGNS OF RETAINING WALL 871 foundation is obtained by combining the horizontal earth pressure with the weight of the concrete and the fill. All possible conditions of loading must be considered. The line of pressure may be obtained graphically, by the method explained on p. 856 in connection with cantilever walls. The distribution of pressure on the foundation depends upon the location of the point of intersection of the line of pressure with the base. It may be computed as explained on p. 849. is eA ati eave FESR SNR ; er ys 2 Fg? ' 4 AeA b ' Be NAV ee Ne | ee ee Bea f I 3% VA pel 3 GaN ( efarth Cn oe be Le SN A) ge Earth 7 é oe oF eNO: Pressure Te EES CASE ee Sa XS ee ! SF SN Wee ere rr Ne AWE ZA FEB Me On LSS SAI RRNESE GI] el \N) P \ oN : wes ‘: i aye CIN oes Eye: oat Nita St —~ 5 . < Vain oa 3 MW A; Witte ZAG S ; WIDE NACA +f Ih eo TS : WARY VARTA AN 76' --=---- 9 Ft, 101--=----S4 ee — 16" = =a gt ~—------ ih Oe Reinforcement not shown. Fic. 310.—Cellular Wall. (See p. 869.) For designing purposes, the longitudinal walls are considered as slabs supported by the tie walls and loaded by the earth pressure produced by the fill and the surcharge caused by the train load. The reinforcement consists of longitudinal bars placed near the out- side face of the wall, between the tie walls, and near the inside face at the tie walls. Vertical reinforcement should also be used, to prevent shrinkage and temperature cracks. The cross walls should be reinforced with horizontal bars of sufficient area to tie the longi- tudinal walls and prevent their separation under the pressure of the 872 RETAINING WALLS earth in the cells. The base should be designed to resist the differ- ence between the weight of earth and the upward pressure of the soil. They may be considered as supported by the cross walls. Lacher Design.!2—Where the cellular type produces much pres- sure on the toe, the Lacher design may be used. This consists of a is -------- 8 Ft,- 64 ------- Z fg! Gree 2 j E ied ae = NOE nae =) fra 08 mea RITE? aN MN oS pe | | ee : D | Pst | f is Vetch age ls Dag Ee | fs (ane: 1 Ft.- 6" Walls n | \ | | 4 a Spaced 12 Ft. on centers me’ c cs 2 | Y | es “(| | s \ ie I . { Ui : : y ou ' sie ! ) RU). : ne ' tj r " a < i t vs <3 I = SS ah b =p W Reinforcement not shown. Fiq. 311.—Lacher Design of Retaining Wall. (See p. 872.) continuous base, a, longitudinal wall, 6, and cross walls, c, spaced 12 ft. on centers, and a longitudinal beam, d. The track is sup-_ ported by pre-cast concrete slabs resting upon a beam, d, and a wall, b. The cost of the Lacher design, as estimated by Mr. Prior, 12 J, H. Prior, in Engineering and Contracting, May 10, 1911, p. 530. SPECIAL DESIGN OF RETAINING WALL 873 was larger than the cost of the cellular design. The Lacher design has the advantage, however, of producing lower pressure at the toe. Wall with Tie Rods.—An interesting wall is that designed by Gustave Lindenthal for the New York Connecting Railroad. Here two longitudinal walls, 65 ft. high, enclose a railroad fill, nearly 60 ft. wide, carrying a surcharge loading of four tracks (EZ 60 loading) and 100 per cent impact. The longitudinal walls are connected by transverse walls about 50 ft. apart. The earth pressure produced by the earth between the longitudinal walls is resisted by 21-in. round tie-rods, spaced 10 ft. apart vertically and horizontally.18 The ends of the tie-bars are threaded and provided with nuts. The pressure is transferred to the concrete by means of longitudinal and vertical 8-in., 163-lb. channels, which are placed near the outside face of the walls. The bolts at the ends of the tie-rods bear against the anchoring channels. While the earth filling was being placed. the tie-rods were surrounded with concrete to form an 8-in. square sec- tion, for protection against rusting. 18'The computations are discussed in Engineering News, May 6, 1915, p. 886. CHAPTER XXII TABLES AND DIAGRAMS This chapter gives tables and diagrams required in design. For convenience of reference the tables are grouped according to the subject matter. The list of tables in the body of the book is also given. General PAGE Table 1.—Working Unit Stresses... 1.1... eeeee cee ecceeererccess 879 Constants for Rectangular Beams and Slabs. Table 2.—Constants for Rectangular Beams and Slabs, n = 15...... 880 Table 3.—Constants for Rectangular Beams and Slabs, n = 12...... 881 Diagram 1.—Constants hk and p for Different Stresses f- and fs...... 882 Diagram 2.—Value of k, j, and i for Different Ratios of Steel........ 883 : Explanation of Use of Diagrams given on Pages 882 and 883. Design of Rectangular Beams Table 4.—End Spans of Continuous Rectangular Beams............ 884 Table 5.—Interior Spans of Continuous Rectangular Beams.......... 885 Design of Slabs Table 6.—Thickness of Slab and Reinforcement Required for Given Live Load. f, = 650, fs = 16000, n.= 15...............++-+5; 886 Table 7.—Simply Supported Slabs, fe = *SB0ORCSSs Table 8.—End Spans Continuous Slabs, | fs = 16,000 22889 Table 9.—Interior Spans Continuous Slabs, n= 15. 890 Table 11.—End Spans Continuous Slabs, Table 12.—Interior Spans Continuous Slabs, 874 Table 10.—Simply Supported Slabs, So = (3005 2 SBE TABLES AND DIAGRAMS 875 Design of T-Beams PAGE Diagram 3.—Values of Maximum Steel Ratios p» for Different Ratios 894 Diagram 4.—Values of Constants Cq for Different { Ratios “ fs = 16000 894 Diagram 5.—Values of Maximum Steel Ratios p, for Different Ratios 895 Diagram 6.—Values of Constants Cg for Different Ratios | f, = 18000 805 Table 13.—Values of k, j, and Cy for Different eee, 896 Explanation of Use of Table and Diagrams given on Page 897. Design of Stirrups | ee IN OPCS ITIUIDS «eee ase §. dogoenigvie wink ected oc larrenee tS cal) 899 Table 15.—Number of U-stirrups in Uniformly Loaded Beam........ 900 Table 16.—Spacing of U-stirrups in Uniformly Loaded Beam........ 901 Explanation of Use of Tables given on Page 902. Beams with Steel in Top and Bottom '( fe = 650, 750, 800, 900 Diagram 7.—Ratios of p’ for Given 1, } f; = 16 ef | 904 Jo =_800, a 1000 Diagram 8.—Ratios of p’ for Given pi, } f; = 16 oe | 905 fc = 650, a 800, 900 Diagram 9.—Ratios of p’ for Given pi, } f, = 18 ne 906 f. = 800, "500. 1000 Diagram 10.—Ratios of p’ for Given 71; Jee 18 es 907 Diagram 11.—Ratios of p’ and p; for Different at and a = 0.02 to 0.12 908 Diagram 12.—Ratios of p’ and p; for Different a anda = 0.14 to 0.24 909 Diagram 13.—Values of 7 for Different Steel Ratios p.............. 910 Explanation of Use of Diagrams given on Page 903. - Flat Slab Design Diagram 14.—Constants in Formulas for Thickness of Slab at Column.. 911 Diagram 15.—Constants in Formulas for Thickness of Slab in Middle EER OTIS acs Seip Ra PME RE Meld PORTE CHA toe 911 876 TABLES AND DIAGRAMS PAGE Table 17.—Flat Slab Constants Co and C1: for Computing Compres- sion in Concrete: ¢. ues pas cn essen + « ein aie skenetteianenete aea 912 Table 18.—Flat Slab Constant C12 for Computing Shearing Stresses at Column Head ix. ss ste se ne a 94.007) > Rape ne 913 Table 19.—Flat Slab Constant Ci; for Computing Shearing Stresses at Edge of Drop Panel.........--.++ + == #0 geen i 914 Column Design } Table 20.—Properties of Column Sections, Areas, Volumes, etc...... 915 Table 21.—Average Stresses Unit f for Different Values of f-and p.. 916 Table 22.—Average Stresses f: in Spiral Column, New York Code.... 918 Table 23.—Average Stresses f2 in Spiral Column, Chicago Code...... 919 Diagram 16.—Total Loads on Round Columns for Different Stresses f. 919 Diagram 17.—Total Loads on Square Columns for Different Stresses f. 919 Table 24.—Safe Loads, Square Columns with Vertical Reinforcement, f, = 450. (1:2: 4 concrete) ........ 6c eee eee eee ee eee eee. 920 Table 25.—Safe Loads, Square Columns with Vertical Reinforcement, f, = 570. (1:14:38 concrete) ........ eee e eee eee eens 921 Table 26.—Safe Loads, Square Columns with Vertical Reinforcement, . f, = 680. (1:1: 2 concrete)................-- +s esses eee 922 Table 27.—Safe Loads, Round Columns with Vertical Reinforcement only, fo = 450. (1: 2:4 comorete)..-. 0/100 tse 923 Table 28.—Safe Loads, Round Columns with Vertical Reinforcement only, fe = 570. (1: 13:38 concrete)........-..--.++2+-.20-:. 924 Table 29.—Safe Loads, Round Columns with Vertical Reinforcement only, fe = 680. (1:1: 2 concrete).......... 6... sense eee 925 Table 30.—Safe Loads, Round Columns with Vertical Reinforcement and 1 per cent Spiral, fe = 700. (1:2: 4 concrete)........... 926 Table 31.—Safe Loads, Round Columns with Vertical Reinforcement and 1 Per Cent Spiral, f. = 890. (1: 14: 3 concrete) ....... 0 ae Table 32.—Safe Loads, Round Columns with Vertical Reinforcement and 1 Per Cent Spiral Reinforcement, f. = 1060. (1:1: 2 con- crete). 2600 ee a bee cece ne ei Oe 2 os eae 928 Table 33.—Column Spiral Forming 1 Per Cent of Core............. 929 Table 34.—Values of pA in Spiral Columns, §-in. wire............- 931 Table 35.—Values of 1A in Spiral Columns, 7-in. wire............. 932 Table 36.—Values of pA in Spiral Columns, 3-in. wire...........-- . 933 Explanation of Tables 34 to 36 given on Page 930. Members Subjected to Direct Load and Bending N Diagram 18.—Constants C, in Formula f= Coa, for Za. hw. see 934 : . N Diagram 19.—Constants C, in Formula de Coup for 2a = O:9R. =... 935 ‘ : N Diagram 20.—Constants C, in Formula f. = Ce—- for 2a = 0.8h 936 Choo Ae ee TABLES AND DIAGRAMS 877 PAGE Diagram 21.—Constants C; in Formula f, = Cm COC SLOT Ayo 937 Diagram 22.—Values of k for Different Eccentricities, \ Both Sides 938 Diagram 23.—Values of Ca, ) Reinforced 939 Table 37.—Ratio of “e for Different Steel Ratios p............... 940 Diagram 24.—Values of k for Different Eccentricities, | Tension Side 940 Diagram 25.—Values of C4, J Reinforced 941 General Data Ty cou. ogy SRS SSIS gail eel ae iowa a 942 Table 39.—Areas of Groups of Bars of Uniform Size................ 943 Table 40.—Area of Square Bars in Slabs for Different Spacing....... 944 Table 41.—Area of Round Bars in Slabs for Different Spacing....... 944 Slee -ELOMCrUes OL GCCUONS. .. 0. secede ett cee eteeeeree 945 TABLES IN THE BODY OF THE BOOK Following tables are given in the body of the book. Properties of Reinforcement Areas, Weights and Perimeters of Round and Square Bars.......... 12 Wire Uecderateitrercementse. MEPS) oS ae’ PUI LS ace, 15 Properucs a: Mxpanded Metal) 2. oiy en. TE a a. 16 Properties of Welded Wire Fabric.....:.....-.... 057200000 be. ea ee. 17 Properties'of Triangle)Mesh Fabricy).. 0...) oo 18 Bond Stresses Length of Lap for Different Stresses in Steel and Diameters of Bars... 414 Spacing of Bars in a Beam Width of Beam Required by Different Number of Round and Square ISR oe ale a eee Nn ne OR ee ee 270 Distribution of Concentrated Load on Wide Slab Effective Width of Slab for Concentrated Load.................... 70 Beam and Slab Design Constants in Beam and Slab Design for Selected Unit Stresses. ...... 205 Distribution of Load on Slab Reinforced in Two Directions......... 212 Columns Stresses Recommended by the Authors for Columns with Longitudinal SPST See, IR ae eet Oe ae: SE ae 7 CP eee ete Semen 407 Stresses Recommended by the Authors for Spiral Columns.......... 422 Requirements of Building Codes of Various Cities Columns with Weoevocimal tteck ONY oes: skeen ew sles 8 a 8 ee ae eae ria ae 409 878 TABLES AND DIAGRAMS Requirements of Boston, Cleveland and Philadelphia Codes for Spiral Columns. 2. 0. Prt Standard Wire Used for.Spirals. ......--+.+5-> => s)he =o eee Recommended Proportion of Live Load on Columns......-.---+---- Footing Tables Required Depth of Plain Concrete Footing in Terms of Length of Pro- FOCtION «esse ene cet ce terete ees 070 0m Muelle ae age cee cae Dimensions of Independent Stepped and Sloped Square Footings... a Constants C;, for Square Footing in Formulas for Shearing Stresses. . . Constants C,, for Rectangular Footing in Formulas for Punching Shear. Constants C,, for Square Footing in Formulas for Bending Moments. . Constants C,, for Rectangular Footing in Formula for Bending Moment Constants C;, for Rectangular Footing in Formula for Diagonal Tension Economical Length of Long Side, a, of Rectangular Footing <2. iad Constants C,, in Formula for Center of Gravity of Corner Column... . Chimney Tables = Values of Constants Cp, Cr, z,7 for Different Positions of Neutral Axis. Location of Neutral Axis. . 4: 4<- 1...) .4> «+5 gp esis ee ee Retaining Wall Tables Average Weight of Ordinary Earth before Excavation. .../:1. 9, 22%. Values of Constants Cp in Formula for Earth Pressure.........-----. Constants Cp in Formula for Length of Base.......-----+++-++++-> Spacing of Counterforts............+ +1: Aven Res ote Se Sena Thickness of Retaining Wall of Gravity Section.........-+++++++++- Miscellaneous Tables Weight of Hollow Burned Clay Tile. .......--+++++++seereteee ress Weight of Clay Tile Floors.......-----+1++ +++ 55: apse siete ede Required Thickness of Brick Wall.......-----+-+-+-++eesssss css Dead Loads on Roofs... ee oe oa Live Loads Required for Different Buildings..........---. Fe iii nee Dead Loads of Floor Finish, Plaster, Partitions, etc........-..-.---- Overall Dimensions of Passenger Cars... .....--- ++ se ee eee eters Overall Dimensions of Trucks. ......00eeessnenrcneene renter nt os PAGE 423 TABLES AND DIAGRAMS 879 Table 1.—Working Unit Stresses. Allowable Working Stresses. Nota-| Percent-| For Kind of Stress. tion, |_28¢ of | 2 000 Remarks. Crushing} Lb. Strength|Concrete, at 28 | Lb. per Dave. | mq. in. (1) (2) (3) Ae (5) BCE ed Perea a) City og fs is a 2 a5 t T Length of pier Exh) COMPLOSSION 6. cca ene ese te 2ar0 450 not to exceed 4 diameters Vertical steel 1 to | Length at a Columns (as Mee A ae if 2265 450 pee ys described BUS Oe on page} Vertical steel 1 to | omen a hl 561) 4% and spirals d40least Oe Sie aa Hah alee 700 P aah gaits radius of gy- | ration moiniaryee st, .. o fe 40 800 Use in beam for- Compression mulas in extreme } In continuous beams Der s. .} adjacent to the| BUupporiuss.. > 2). te 45 900 Shear (punching shear). ..........|./.... 6 120 Shear (as (Beams without web measure of | reinforcement....| v 2 40 diagonal ( Stresses above tension... | Beams with web re- 40 Ib. must be | inforcement...... v 6 120 provided for with web re- inforcement Head { Plain: barsiviies. 4: u 4 80 bap vic a Deformed bars.....} u 5 to 6 |100to120 ( Structural grade....| fs 16 000 |Ib. per Steel in ten- sq. | in. MOU a fees. Intermediate and high carbon steel.| fs 18 000 |Ib. per Stja} ins. es ee ed 8 ol ee * Strengths at 28 days for different proportions of concrete are given on p. 205. See Volume II for discussion of strength of concrete. + See Formulas (56), p. 271 and (57), p. 272. 880 TABLES AND DIAGRAMS Table 2.—Constants for Rectangular Beams and Slabs To be used in formulas for Depth of Beam, d = oy ord = Na and Depth of Slab, d = C1\/M; in formulas for Moment of Resistance, M M or M = Rbd?; in formula for Area of Reinforcement, A, = pbd, or Aa Ratio of Moduli of Steel to Concrete, n Working | Working Ratio Strength | Strength | Depth of of of Neutral Steel, |Concrete,} Axis to Lb. per | Lb. per | Depth of Sq. In. Sq. In. Steel. fs te k 14 000 500 0.349 550 0.372 600 0.392 650 0.409 700 0.428 750 0.446 800 0.462 850 0.477 16 000 500 0.319 550 0.339 600 0.358 650 0.378 700 0.397 750 0.414 800 0.429 850 0.444 900 0.458 18 000 500 0.294 550 0.314 600 0.333 650 0.351 700 0.369 750 0.385 800 0.400 850 0.415 900 0.429 20 000 500 0.272 550 0.292 600 0.311 650 0.328 700 0.344 750 0.359 800 0.374 850 0.389 900 0.403 Ratio of Moment Arm to Depth of ee Lies j Ratio Area of Steel to Beam Above Steel. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 _ bd? jdfs = 15: Constants. For Beams. For Slabs Cc R C1 0.114 CEL 0.0329 0.106 89.4 0.0306 0.099 102.1 0.0286 0.093 lib.2 0.0268 0.088 128.6 0.0254 0.083 142.2 0.0240 0.080 156.2 0.0231 0.077 170.4 0.0222 0.118 Tio 0.0341 0.110 83.0 0.0318 0.103 95.0 0.0297 0.096 107.5 0.0277 0.091 120.4 0.0263 0.086 13320 0.0248 0.083 146.9 0.0240 0.079 160.6 0.0228 0.075 174.5 0.0216 0.123 66.3 0.0355 0.113 77.4 0.0326 0.106 : 88.9 0.0306 0.099 100.8 0.0286 0.094 113.1 0.0271 0.089 126.2 0.0257 0.085 138.7 0.0245 0.081 151.9 0.0231 0.077 165.4 0.0222 0.127 62.0 0.0367 0.118 To CG 0.0341 0.109 83.5 0.0315 0.103 94.9 0.0297 0.097 106.7 0.0280 0.092 118.8 0.0266 0.087 131.3 0.0251 0.083 144.0 0.0240 0.080 157.0 0.0231 TABLES AND DIAGRAMS 881 Table 3.—Constants for Rectangular Beams and Slabs To be used in formulas for Depth of Beam, d = cy ord. = ee and Rb’ SE AGE : bd? Depth of Slab, d = Ci/M ; in formulas for Moment of Resistance, M = Ca” : [ M or M = Rbd?; in formula for Area of Reinforcement, A, = pbd, or A, = cafe Jas Ratio of Moduli of Steel to Concrete, n = 12. Working | Working Ratio Ratio of Ratio Strength | Strength | Depth of Moment | Area of b Constants. of of Neutral | Arm to | Steel to Steel, |Concrete,| Axis to epth o Beam Lb. per | Lb. per | Depth of Steel Above For Beams. For Sq. In. Sq. In. Steel. (1 aa 4 Steel. Slabs. 3 fs te k J p Cc R Cy 14 000 700 0.375 0.875 0.0094 0.093 115.6 0.0268 750 0.391 0.870 0.0105 0.088 129.2 0.0254 800 0.407 0.864 0.0116 0.084 141.6 0.0242 850 0.422 0.860 0.0128 0.081 152.4 0.0234 16 000 700 0.344 0.885 0.0075 0.097 106.7 0.0280 750 0.360 0.880 0.0085 0.092 118.8 0.0266 800 0.375 0.875 0.0094 0.087 i3ivo 0.0251 850 0.389 0.870 0.0103 0.083 144.0 0.0240 900 0.403 0.866 0.0113 0.080 157.0 0.0231 1 000 0.429 0.857 0.0134 0.074 184.0 0.0214 18 000 700 0.318 0.894 0.0062 0.100 100.0 0.0289 750 0.333 0.889 0.0070 0.094. kee VL 0.0271 800 0.348 0.884 0.0077 0.090 123.4 0.0260 850 0.362 0.879 0.0085 0.086 130.1 0.0248 900 0.375 0.875 0.0094 0.082 148.8 0.0237 1 000 0.400 0.867 0.0111 0.076 173.0 0.0219 20 000 700 0.296 0.901 0.0052 0.103 94.3 0.0297 750 0.310 0.897 0.0058 0.098 104.1 0.0283 800 0.324 0.892 0.0065 0.093 115.6 0.0268 850 0.338 0.887 0.0072 0.088 129.2 0.0254 900 0.351 0.883 0.0079 0.085 138.3 0.0245 1 000 0.375 0.875 0.0094 0 1 0.0226 .078 164. Use of Tables 2 and 3.—The constants in Tables 3 and 4 may be used for determining concrete dimensions and the required amount of steel for rectangular beams and slabs for known bending moment M, and known stresses f., fs, and n. Care should be taken to use the table for proper value of n. For 1:2: 4 con- crete n = 15 is commonly used. For 1:13:3 concrete n = 12 and1:1:2 concrete n = 10 are common. In formulas for depth, if M is in inch-pounds and } is in inches, the depth will be also in inches. With M in foot-pounds, b should be in feet, and d will also be in feet. In both cases the constants remain the same. To change bending moment from foot-pounds to inch-pounds multiply the foot-pounds by 12. 770 BRneSaeene ee TEE | Poo ECE oan SceReeSaeeaEe 760 nee 150 a4 aon Coo 140 Poth rH i 730 HH “ i & 110 ~ er a6 Z “ PATA TCO AA ~ ay Coageerae 3 90 Te a A > Be enue oo ae ae 2 7. Pot LA ZL 80 aia av arenaMe’s ava sua8 Aer 70 Ad ARP a nae Zari | vi Ai Tt Boe Ratio of Reinforcement p Dracram 1.—Values of R and p for Rectangular Beams and Slabs. For Different Values of f,; and fe. Use of Diagram 1.—Following problems may be solved by means of this diagram: 1. Determine values of R and the ratio of steel, p, for any given stresses f- and fs. In solving this problem find point of intersection of the curve corresponding to the stress f, with the curve corresponding to the stress f,. Vertical line down from this point gives the required ratio 7, while a horizontal line gives the value of constant f. For intermediate values, interpolate. 2. Find stresses f,; and f, for known concrete dimensions b and d, known area of steel A, and given bending moment M. M A Se In solving this problem find R from R pies and p from p = 7 Find in the diagram the point of intersection of a vertical line corresponding to the determined value of p with a horizontal line corresponding to the value of R. This point by interpolation between the f, curves gives the stress f, and between the f, curves the stress fs. 882 TABLES AND DIAGRAMS 883 Values of jandk g wae Values of Pp DiacraM 2.—Values of J, k, and” for Different Ratios of Steel p. c Use of Diagram 2.—From this diagram the ratios 7 and k may be obtained directly for any ratio of steel p. This diagram may also be used for computing stresses in a beam with known dimensions b and d known amount of steel As and given bending moment M. As : In solving this problem compute value of p from p = bd Find from diagram the corresponding value of 7 and . Compute stress in steel from formula c Ss ere in which all values are known. 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AC a pe NI NAC a aaa Fag Fs aS RR IE 08 0 OOT g OIL | Set | 99T | GOS | LHS | BOE | BLE | COP (Si cee he eee ee G10 OOT 8 18 | GOL | S&T | LOT | 90% | ES% | FTE TQEUIEORS CIE |a- ak ee ee ee 69 0 v6 $d 19 | 98 | SOL | SET | 89T | 602 | 09 | 9ZE gs SGT od RASS EG SO ROS SES MRE v9 0 18 L og |¢9 | 78 | LOT | S&T | OAT | &Ta | 696 ZPecialpre Ose cee 69 0 18 £9 ze | cr | 09 | 62 | ZOT | O&T | 9OT | ate Vir iGGeLLop tere |s tao) o | en 6g 0 GZ 9 ST. | 6Z. | Tr 129. | Ste (586 | Sal. 1 Sut WiC ike cele 1OUs amt | os 870 69 4g et |ez | 2e | Ig | 69 | 26 | Zt | O9T | cle qsz | 168 | GSS | 968 | &V 0 G9 g TL Oz. | Te hee OO | oe eat Ser I1z | 26Z | 81h | 4¢9 | TIOT}| 2.0 9g iP el | oo. | 2e | &F | G4 | POT | SPT | S02 L6@ | OSY | TéZ cé 0 og v ot | gz | er | 89 | 16 | S&L | 26T | G0 | 867 LZ 0 wW a5 6 ST 1S 1054 | 925 | L1Te | SB ie OTe 1c 0 LS & 400,7 orenbg sed spunog Ul ‘pvory eATT 9F8S ‘ut “bg "aT ‘ul es SS ee TIPLM JO oo 7 Jd qs Jo 4 Spo eT SIE Sie vl) eo Ble ole: eee OLE 0 8 js 9 G v Pee ‘a 1431 MA qejg Jo jo vary yO ssouxort, oo,J Ul ‘7 yysueT uedg ‘ut ‘bs sod “gq, 000 Z = 7f yj}Suerjs uoIsso1duro09 o}etury[N YIM 91910U09 IO} OS/) “1-4 = Pp ‘uydep satjoegy ST = U ‘000 8ST = *f ‘008 = °f ‘Sasset}s 31 /) a = jy ‘jueuoUr suIpued G ‘suedg puq—sqzg snonurju0j— TT 1481 893 TABLES AND DIAGRAMS are | 27¢ | veg | ope | gag | eeotl coz | ter | e2¢ | evo | gee | eee occ ose | sep | zie | zoo | ziz.| 2g ccc ee oe eco | ese | ocr | pee | ogo | ogg (occ 16z | cee | zor | per | 299 | ezodccc cc bec G67 Il 862-1. 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Tal eyne We) el are ed wv) fen of fete oe 6 as aw (ope ier oe. es a = Rae apt is. see ee eee ee oe eee H Ke) So S19 Sa SOS |S OOS Sirs rt ‘ut ‘bg THPEM fo yoo 10d “SV 2248 JO Voly Uw ‘000 8T = “f ‘008 = °f ‘sesserys 41U) ‘suvds 101I9}U[—SqvIS snonunu0j— ZT s1quL 20L lM ° { YYsuei4s UOTsse1dw0d 0} BUITYIN YIM 940.1900 IO} 98) = py ‘juoWOW SuIpusg a1 qBIg jo JUSTO MA qa, Saal) ria min canis | I CNH SHaD DOOnKKO q ion d dine Ssouyory J, 894 TABLES AND DIAGRAMS DIAGRAMS FOR T-BEAMS. OL rat 1 0.015 S & x Hoes 0.010 © g = 2 S 0.005 0.0055 0 0 0.05 OI02 S078 S 0. <5 dbs Values of 4 i wih DracraM 3.—Values of Maximum Steel Ratio pm for Different Ratios 7 Governed by Compression Stresses. (See p. 216.) 1000 & 900 =, u 800 se : a . Se SS) ae 702 Face BEES == Seeee o pie dene : SSenseecses > % os cS =e a = = 3 3 600 = = % s : 8 = UTA TT aan HAN ANN anand tt LH a “alo 015 oS "930 035 Values of # For Example, see p. 896. t , Dracram 4.—Values of Constants, C, for Different Ratios 7 In Formula for M Minimum Depth, d = ~~. (Seep. 216.) Cqbt 895 TABLES AND DIAGRAMS DIAGRAMS FOR T-BEAMS “qd fo sanjo, 0.35 0.30 025 Ge Sata Taal 020 O15 Values of AWE BesA\\ge 010 maar VA @ YT AW be | | IAN 0.05 HEE “4 fo sanjo/ t - DiaGcram 5.—Values of Maximum Steel Ratio p», for Different Ratios (See p. 216.) Governed by Compression Stresses. P5-- fo sano, S S s) S S Ss) S S S S S S Ss S S Ss iS D a ~ © % -s “o i | iH | | | | | 4 TT i / LY {|| 1 LTT VALLETTA LETTE TTT AT TTT BA AT PERERORURDGRD GEDA GR ORER) CRROG) ARRGR) /URRRORY 400) NED AUUED RR eT Ra Re ASURURRUEARD? AURLY-ANRUAY UOLARAY ABRY (ORY (ALA HTT A LEECH OEE TTT UE EEE IL HTT NATTA HH HEHE AULT QO META HTT ATT TTT Se 9 AL anil il TAT HTN ATT 9 PA I CTT ATTTTT TPT ATT SATA PORREDY CCUURRRURRGL NY ZORUERD) (01 SAT MYT A AUT ATT ASA SAO FLT TTT a HELL ATTA LOA OAT] ATTA TTT TT HALT ATT TTT TT HTT ATA HTT AT HECELCLL AA TEECt REHEAT HERE TTA HITT TTA HATTA ITAA nnd 19% LATTA LT Vaal 41 AUT | S S S S S S S Ss S S SS S S D 9 ~~ ey Ly vt ~ Pp fo sanjno, 0.20 025 Values of 4 For Example, see p. 896. In Formula for t 7 ? Diacram 6.—Values of Constants, Ca, for Different Ratios (See p. 216.) Hake Cor Minimum Depth, d 896 TABLES AND DIAGRAMS Table 13.—Use to Find Stresses f, and f, in T-Beams. (See p. 898.) As . Values of k, 7, and C'r, for Different Values of a (See pp. 224 and 898.) Ratio nAs bt it ooo Coco coco cooe oooo oooo wo bo Ratio of Moduli n Value of k. Use to find 7 and CT below. « 0.10 . 136 .208 . 269 .321 .345 . 367 387 406 .424 441 457 472 (aayes) “CeKola) (aioleny haleayer Me) A On ) Ratios of Thickness of Flange to Depth of Beam, 4 O12 420514 SOM GeO as 0.145} 0.154] 0.164 0:217| 0.225) 0.233] 0.242 0.277| 0.285] 0.292} 0.300 0.329] 0.336] 0.343] 0.350 0.352] 0.359] 0.365} 0.372 0.373] 0.3880] 0.387] 0.393 0.393] 0.400} 0.406) 0.413 0.412| 0.419] 0.425) 0.431 0.430} 0.436] 0.442] 0.448 0.447) 0.453} 0.459] 0.465 0.463] 0.469) 0.474] 0.480 0.478; 0.483) 0.489] 0.494 Values of 7. 0.20 . 250 . 308 357 379 .400 .419 .437 .455 .471 .486 . 500 ooo ,ocoo ooo oo 0.22 . 258 .315 . 364 386 407 .426 .444 .461 .476 .491 . 006 coo _ Sooo coco oo 0.24 |0.26 | 0.28 | 0.30 | 0.32 . 26710. .323)0. .37110. .393)0. .413]0. .432)0. .450]0. .467/0. .482/0. .497)0. .511/0. ooo ooo, occ oo Use to find fs in Formula fs = 275)0.283 331/0.338]0.346]0 379|0.386]0.393 0. 400|0.407 |0.414/0 420|0 .427 |0.433)0. 439|0.445/0.452|0 456|0.462/0.469|0 473/0.479/0.485)0. 488|0.494/0.500/0 503]0.509|0.514/0. 517/0.522)0.528)0. M Asjd’ Ratios of Thickness of Flange to Depth of Beam, 4 sake .914)0. .912)0. .910)0. .908)0. .907)0. .905/0. .904/0. .903)0. .902/0. .901)0. .900)0. .899)0. .899/0. 250410. .897)0. .896/0. .896)0. .895)0. oEeo coco ,oCcoo coco CoCo (See p. 304 400 421 440 .458 ATS 491 . 506 520 533 —— 913 909 906/0.907 904/0.904 902/0.901|0.900 900|0 .899 |0.897 8938/0. 896 |0.894/0. 897/0.895/0.892/0. 896|0.893|0.890)0. 894|0.891/0.888)0. 893|0.890/0. 8860. 892/0.889/0.884)0. 891|0.887 /0.883)0. 890|0.885|0.880)0. 889 |0.883|0.880)|0. 888/0.881|0.879|0. 887|0.879|0.878)0. 886/0.877 |0.870)0. 224). 0.20 | 0.22 | 0.24 | 0.26 | 0.28 | 0.380 | 0.32 0.22 0.028) 0.032) 0.035 OPTS Fs 20RISS OSLG Owls 0.948) 0.940) 0.937) 0.933) 0.930 0.948; 0.940} 0.935] 0.931] 0.927 0.947} 0.939| 0.934} 0.929) 0.925) 0.924 0.947] 0.938) 0.933} 0.928] 0.923) 0.921 0.946! 0.938] 0.933] 0.927} 0.922] 0.919 0.946] 0.937) 0.932} 0.926} 0.920) 0.917 0.946] 0.937) 0.931} 0.925) 0.919] 0.916 0.946] 0.936] 0.931) 0.924) 0.918) 0.914 0.945) 0.936] 0.930! 0.924} 0.917) 0.913 0.945} 0.936] 0.930} 0.923) 0.916) 0.912 0.945] 0.935| 0.929) 0.922) 0.915) 0.911 0.945} 0.935) 0.929} 0.922) 0.915) 0.910 0.945) 0.935] 0:929|} 0.921] 0.914] 0.909 0.944} 0.935) 0.928) 0.921} 0.913) 0.908 0.944] 0.934] 0.928) 0.920} 0.913| 0.907 0.944] 0.934} 0.928} 0.920) 0.912) 0.907 0.944} 0.934] 0.927) 0.920} 0.912) 0.906 0.944) 0.934] 0.927) 0.919} 0.911| 0.905 0.944] 0.933; 0.927} 0.919} 0.911} 0.904 0.943} 0.933!) 0.926} 0.918) 0.910} 0.903 0.943! 0.932) 0.926) 0.918) 0.910) 0.903 0.943} 0.932) 0.925}) 0.917} 0.909) 0.902 0.943) 0.931) 0.925! 0.916!) 0.909 on Values of CT in Formula fe = CTrfs. Values of & 0.24 0.26 0.28 0.30 0.32 0.34 0.36 O38 (P0540 e0e 42 0.019} 0.021} 0.023) 0 026) 0.029) 0.031) 0.034] 0.038] 0.041] 0.044] 0. 0.024} 0.026; 0.029] 0.032) 0.036) 0.039} 0.043} 0.047| 0.051] 0.055] O. | 0.039| 0.043) 0.047 OO 0.056} 0.061} 0.067] 0. 048 060 073 x or TABLES AND DIAGRAMS 897 USE OF DIAGRAMS AND TABLES FOR T-BEAMS Use of Diagrams 3 and 5.—These diagrams give maximum ratio of steel, Pm, Which can be used in a T-beam without exceeding the allowable compression stresses in concrete. Also they may be used to check a design for compression stresses. Example.—Given M = 1800000 in. -lb., f. = 650, fs = 16000, n = = 15, d=18in.,?=4im.,6=58 in. Find Peneired area of steel and see hee: gies stresses are satisfactory. M Solution.—Area of steel is found from A, = fja where 7 is assumed as 0.89. sJ 1 800 000 7202 Aes MM alence dhe tation of ateciiia pce *~ 16 000 x 0.89 X 18 OE TS AR ne Be ea a eR t 4 0.0067. For Car 0.222, and the stresses, f, =650, f; = 16000 from Diagram 3 the ratio is pm = 0.0053. This is larger than the computed value of p, consequently the compression stresses are larger than the allowable. By locating p = 0.0067 in the same ae it is found that the compres- sion stresses are about 770 lb. per sq. in. Therefore the depth, d = 18 in. cannot be used unless the compression falas is made thicker or compression steel is used. Ordinarily it is cheapest to increase the depth of beam. Use of Diagrams 4 and 6.—These diagrams give the minimum depth of T-Beam as governed by compression stresses for any given dimensions b and t, given bending moment M and specified unit stresses f, and fe. Example.—Find minimum depth of T-beam as governed by compression stresses fe, when M = 1 500 000 in. lb., t = 4 in., b = 58 in., and the stresses are fc = 650, fs = 16 000, and n = 15. M Solution.—Minimum depth will be found from formula d = Gwe in which d all values except Cg are known. Assume : = 0.24. (This value may cor- respond to minimum depth required by shearing stresses.) From the diagram 4, corresponding to specified stress f,, and assumed “ the constant is Cg = 386. Minimum depth therefore is 1 500 000 = ————__ = 16.7 in. B86 X58 X40 898 TABLES AND DIAGRAMS Use of Table 13.—Rule for use of Table 13: Find k, use k to determine 7 and thus f; in formula A, = Use k also to find Cr and hence fe. jaf s Example.—Given M = 1 900 000 in. Ib. resisted by a T-beam with the fol- lowing dimensions: 6 = 58 in., t = 4 in., d = 22 in. A, = 6.2 sq. in. Find stresses f, and fs, assuming m = 15. nAy - 15°%-6 bt §=58X4 Find in Table 13, corresponding to the above ratios, k = 0.35. With this value t 4 Solution.—Compute the ratios = 0.4 and 7 ap =. 0.18. t : of k and on 0.18, find from the second part of the table 7 = 0.920, and, by inter- 1 900 000 polation, Cr = 0.036. Stresses in steel therefore fs = 0.920 X 22 X 6.2 Fe 15 200 lb. per sq. in., and f, = Crfs = 0.036 X 15 200 = 547 Ib. per sq. in. BASIS FOR TABLE 14 Asf: . 4 J , in which 010 Table 14 is based on the general formulas = s = spacing of stirrups, in inches; A, = area of two legs of stirrups, in square inches} f, = tension in stirrups = 16 000 lb. per sq. in.; v, = unit shear at the point under consideration to be resisted by stirrups, pounds per square inch. V = average external shear at the point under consideration in pounds; »y’ = unit shear to be resisted by concrete, pounds per square inch. b = width of rectangular beam or width of stem of T-Beam in inches. I The value », depends upon the method of distribution of diagonal tension between the stirrups and the concrete. For Method 1, where concrete is assumed to resist a definite amount of ’ Vv : diayonal tension, 1. =7—> —v’. For this the above formula for spacing bid changes to s = a which is identical with Formula (40), p. 249. a —v'b For Method 2, where concrete is assumed to resist one-third of diagonal tension and the stirrup the balance, 01 = > ae This substituted in the above 3 jd ; . 1.5Asfs ° . . . ® eoneral formula gives s = v which is identical with Formula (44). jd Values of v: may be taken from a shearing stress diagram prepared as explained on p. 252, TABLES AND DIAGRAMS 899 Table 14.—Spacing of Stirrups. (See p. 902.) Spacing of Stirrups, in Inches Unit Shear to be Resisted Width of Beam }, or Stem b’, in Inches by Stirrups, | . Je a AUR aaa haa | 8 | 10 | 12 | 14 | 16 | is | 20 | 22 24 \ | + in. rd. U-Stirrup 10 26 20 16 j 13 11 10a) 59 8 fe 6 20 Loo LU 8 6 5) 5) 4 4 3 3 30 9 6 9) a uy 3 3 2 2 2 40 6 +) 4 3 3 2 2 2 2 2 50 > 4 3 2 2 2 Pl ae 60 4 3 3 2 2 2 80 3 2 2 2 100 3 2 120 2 2 140 2 | % in. rd. U-Stirrup 10 59 44 35 29 25 22 20 18 lor 15 20 29 22 18 15 13 Lies 10 9 8 t 30 19 15 12 10 8 7 6 6 5) 5 40 15 11 9 ie 6 5 5 4 4 4 50 11 8 fi 5 5 4 4 3 3 3 60 10 r) 6 5) 4 4 3 3 3 2 80 7 o 4 4 3 3 2 2 2 Z 100 6 4 3 3 2 2 2 2 120 5 4 3 2 2 2 2 140 4 3 2 2 2 SS a 3 in. rd. U-Stirrup 10 Peete 00-1 ba 4G? 1 BO Pra) SI 28 | 26 20 B21 7-00 fh iol 2G. dees 0 hay 16 14 | 13 30 BO 26 Pee 21 17 15 13 11 10 9 9 40 26 | 20 16 13 11 10 2 8 (s 6 50 20 15 12 10 8 és 6 6 5 5 60 Li 13 10 S 7 6 6 5 5) 4 80 13 10 8 6 5 5 4 4 3 3 100 10 8 6 5) 4 4 3 3 3 120 9 6 5 4 4 3 3 3 140 7 5 t 4 3 3 10 1338 |100 | 80 | 67 | 57 | 50 | 44 | 40 | 36 | 33 20 Gree) tO 33: We Sy t 25° 1225). 201g 17 30 44. | 33 | 26 | 22 19 16 15 13 12 11 40 a Pa ee A ae 14 12 11 10 9 8 50 25 19 15 12 11 9 8 7 7 6 60 22 if 13 1B! 9 8 7 7 6 5 80 17 12 10 8 7 6 5) 5 4 4 100 13 10 8 7 6 5 4 4 4 3 120 it 8 fs 5 5 4 4 3 3 3 140 9 ri 6 5 4 3 3 3 For explanation, see pp. 252 and 898. For example see p. 902. 900 TABLES AND DIAGRAMS Table 15.—Number of U-Stirrups in Uniformly Loaded Beam. (See p. 902.) Number of Stirrups per Beam is 2Ns = Got 2N, = number of stirrups per entire beam; 1 = span of beam in feet; b = breadth of beam in inches (in T-beam, breadth of stem); v = shearing unit stress in beam at support in pounds per square inch; ~ »’ wi : eee VLLNLL_| at | NI S N 007 Values of P, F800, f,= 16000, n=I5 9, = ratio of tension steel; iy NS SS IN So eS = 5 5 5-8) 8 Sesvare so:.S SS Ss 3S 3°95 0 Values of Pp, fF. = 900, 752 16.000, a= 15 ratio of compression steel; a = ratio of depth of compression steel to depth of tension steel. Diacram 7.—Ratio of Compression Steel, p’ for Given Ratio of Tension Steel p1. (See p. 903.) Stresses in Concrete, fe Stress in Steel, ts and n 650, 750, 800, and 900; 16 000; 15. 904 I 006 005 0.05 a | 1 004 tH PN NV] bee nl ITN 004 Hill Hint UT TT ane ni I nt of p! i ‘ : N Unt TH a a NS iN 003 Ul Sau N N N ! NT] ut 4 NN N ‘ : [ 9 STC LY LR LF NCD CE A uy i N Values Ny WO II we 4 fee Ue Ls LP CL | 002 sui Ny} : NI N Ny NN Q07 | 0.010 Ore 0014 \H 0016 0078 0020 0.022 Wt 0.024 0026 0028 Values of p, Values of p, Fe= 800, fs = (6,000, = 12 Se = 00, 4s = 16,000, n= 12 Q03 of Q02 Va/ues 00) \—— Oe? PD WIALL Values of p, Se=/000, F,= 16 000,17 =12 p: = ratio of tension steel; p' = ratio of compression steel; a = ratio of depth of compression steel to depth of tension steel. Dracram 8.—Ratio of Compression Steel p’ for Given Ratio of Tension Steel p:. (See p. 903.) Stresses in Concrete, fc = 800, 900, 1 000; Stress in steel, fs = 16000; and nm = 12. eee & ~ aes) (aS Ket 2 005 ee ee — aa aes Oe > =. : > aoa Ge Se oM, Vale ara ae oom + (ran SS GRAS SSE RE Bee AF aa AS et kt — Q04 PF rf, US 003 ee oy po a as rT TF % ee ee ee aa aan fe} 003 Sy oe a ee he} eT a Re = —— t= 10? ft aoe Eee eS oats Geeta Een SS CS REA een PE Se er oe Ae ae pt Se ESS ae 002 pt ey RES SE ee Ses ne comes ee ee ee ee Q07 eet ame ne 00) Se a i a eo ed a Re eas iF iT ee a a a See as aS 0 aS! eS ane tee 0 ELEN EE: ELE | ee FSG ee @DBDON WY SH & NY ODD SON wW SH DON KYO DWH So GDRVNRrRWRPTHANAWNAAD Serr tT evrnwnvWaun NY oe ogoouogeeao ogeogogaogaeese 8 Ss S S&S SseossssgsgsssSss SS8sssssegssqg Valves o Values oO i 1 Tc = 650, fs = 18,000, n= 15 Te = 750, Js = 18 000,07 = 15 OO ee ee eee (6 Q04 es SS a Ye a Ee ee eee ree Sa ae oases Sk RE ed Sh Me Be a Sara ee GS ee SD pf tf 004 pene we Pees, ase ey Ses Pe Sea 003 eae a a Ea ees a ey," Be SSeS SS Gey GE es RI ae hs ie BS eS = ao Sa ee Es Va ar ccsenee enn] Se aaa| ean Tees ee ee HI LS ~~ Rei roe 8 EE ES ue 0.03 oT AES a ee ° eo ee ee Be } {|_| _}_4 Se BE nn ne Ee Ss ORAS Ol Cae Bee Q02 a a SS IP OGG OS AES Ease Sy OPN eS OLN LE ee me EES TIAL LAL LA eee ry a a A A = ey v pA et HOOD i I ee 2 002-4] [GY eT nae EET Ss PML eee tt h eae ES ) Z 4 Se es ee [eee if 4 Pea as “ Se ce ee ae es 6 a ee =n eed j Sia ie ae ee A eS RW 0000 C07 BEE Ee A a ea ttt tt _ BE B90 75 8 A ES 00/7 a TNT a a ae Se jared Baer Bes Q0/ (ES /SU) 07h ES Ge es a) CE EG liSaeeses LK es SS SS as tee 3 WLLL a ae Gas PS DE een pees [| ee Rw Se ee Ws eee eed Ee RS SSS ee | a ge | AY aes Ee Ree es Bere Se Pee as eee Grae es BS foe gah aM aes Oe ea ee a SS inane Bec ek od 0 an SUS eee ON ee See SSssgsesss 8 os & = 8.8 8.5 -coeeees ssscososoosoo SSS os Ss S$ SS -o SS) sac 7S Values of -p, Values of pP, Te = 800, Js = 18 000, 0=15 Jc = 400, fs = 18 000,77 = 15 p: = ratio of tension steel; p’ = ratio of compression steel; a = ratio of depth of compression steel to depth of tension steel. Diagram 9.—Ratio of Compression Steel p’ for Given Ratio of Tension Steel pi. (See p. 903.) Stresses in Concrete, f, = 650, 750, 800, 900; Stress in Steel, fs = 18 000; and n = 15. 0.07 006 Q05 Qo of p’ Q03 Va/ues O62 Ee i = = Po aa a 4 OFT asa, e, =o a [| Set BE SIF a OF LF ASS. t-- fi Hh 4 Ci BF, soos e: oma a, Po t+— 5 vows A 4 ve LA, 22 am = 007 v 7 == STS NSW we So SOUS DO WD © SQ Ss 88 co FOr FON NS USOmS © Ss Se eS NO NN a Serr pe So OT BON CNN AG nD SSS 97S OSS Oo S S SS: S SSSSSECSCSECSSS SSSSSSEESSSE Values of P, Vatues of P, Sc = 800, fs = /8 000, n= I2 fc = 900,7; = 18,000, n=12 Q04 Q03 CE Hs Q02 Values 001 - Be P1 = ratio of tension steel; Va/ues F-= 1000, fr = 18,000, n= 12 of P, Pp 4 = ratio of compression steel; a = ratio of depth of compression steel to depth of tension steel. DraGcraM 10.—Ratio of Compression Steel p’ for Given Ratio of Tension Steel p:. (See p. 903.) Stresses in Concrete, fc Stress in Steel, and fs 800, 900, 1 000; 18 000; 12. 907 908 TABLES AND DIAGRAMS saa uea'iy TTT TAL VA HY] A La Al / fi | ( AA, aA r | V 7407 a0 002 AL 0.002 CI .007 010 Qdls a 02025 2030 0035 0040 .007 ae 2015 0. ma 2025 0.030 0035 0.040 (SUG NOE a-0.04 Ss U] 0.030 Hy Tern my .mney aunaay co Mh ra ro anny 408 ty ACTA Po ELE ,Aanny ane 1 Ht! ATO Coe Aun” nnne 4o HL ra A es A 557400008 0.040 Fr Ol im ae A | f VY wi efi ‘alan l ew, a Ap dan sues 0.030 el 7 71] WA ) v, (AV [IAT I wad | N = [| A as S x Ss S a A | NE Ch Katin oj Cornrpression Stee/ p J ad 4 fo 0.050 5 TA A ney VY fifi Vy UAT A | iE HA, VY VIA VT PARA XL Yar: | | A AK Me pad AL Cae eae os tq ian ooe 0.002 007 0010 Q015 0020 aoes 0030 0035 a040 .007 0.010 001 ie > S Q2 2025 0030 0.035 0040 a=0.1/2 Ratio of Tension Stee/ p, a:Ratio of Depth of Steel in Compression to Depth of Steel in Tension Dracram 11.—Relation between Tensile and Compressive Steel in Beams with Steel in Top and Bottom. (See pp. 238 and 903.) TABLES AND DIAGRAMS 909 0050 (TT Lt Alf aie 0020 Seat 55.057 dann bY eat LAT et LAT | a 0002 a0 .Gn87.4074 0002 & Vit 1007 0010 Q015 0020 0.025 Q030 0035 2040 007 0010 2015 0020 0025 0030 0035 0040 a=0/4 a=0/6 ; 0050 0.050 CTT TAT) 2930 poo oo O60 A000 An) Gy Gey Oey Anne ar UariAF an ty Pian HY CO HA B) (70 AB 007 S08 G08 SER nee’ A PA 7 BA) a7 A645 4NbY Suan a “au odo Hef ATT oo DO 4 rl S00 AS i sn ann y, ry PY PVA ot bP I] au aie any aERy. 7; 7} ) HANAFI) Acerca cous V28 0.030 1H BY, etd 4 0030 4 10 aw Ny SERED CAERUD SERB pa PTY PAT TT] rT AULT ATTA TAT 181940407 G00 .00"9/4008 pin Paicae 0020 FA fr! go20H SEGEe wl aReae : AHH f a Rath s 0070 WHE 4 0070 ReaaR LA Paul 2002 Ae HH Z BECEEH EHH HEH 007 0.010 0015 A020 0025 0030 0035 a040 .007 2010 0.015 0020 aes 0030 2035 0040 a=018 Ga. 0.050 nEEEAS TOMO nan/08 CC i smnuen/ ann aang , 55/000007,0008 7 GUSnny dene 0040 np48 ARGO) 400888 1040 ay 47a 4 C f) a wos Ht i Ho ay 4a6 ue aan 1m 0407000 408 a y, 0.020 FI 0.020 anP74 An An Ya 767 07 G00'4020 488 HAT ATTA aT HEA Hy 0.010 Fee 0.070 TET PLA ee I PLL EY TT edy Gay aan TT ALITY SUScROnG8 AvGV Ann 4058 Hc 2 ALEC 2002 0.00. -007 0010 0015 0020 0025 0030 0035 0040 907 Q010 Q0IS 0020 0025 0030 2035 0040 a= 022 a=0.24 Ratio of Tension Steel p, a=FRatio of Depth of Steel in Compression to Depth of Stee/ in Tension Diagram 12.—Relation between Tensile and Compressive Steel in Beams with Steel in Top and Bottom. (See pp. 238 and 903.) TABLES AND DIAGRAMS 910 on hy oe gia 6 a 0.20 0.22, 0.24 G fo sanjog Js nfo Values of Ratio of Depth of Steel in Compression to Depth of Steel in Tension. a= —Values of j for Beams with Steel in Top and Bottom. (See p. 240.) DracRamM 13 TABLES AND DIAGRAMS 911 USE FOR FLAT SLAB DESIGN 003 aa Constants C4 Cs Ce 0.025 7 S S No Constants Cy Co C3 a Sold lines Jor {,=16000 lb per Sgn. Dashlines for f,.=18000 Ib. per Sa. in. 0.07 S Sey Values of :: of Flat Slab at Column. 0024 cor LLL RN SERGE TERRE RRRSSe! Me sla Le Hebd ET AEP eee bh a] f Solid Ines for f =16000 /b per Sq. in Dash lines for 7,=18000/t per Sq.in. 0.04 O02 Ml Values of r Diagram 14.—Constants C1, C2, C3 and C4, Cs, Cs in Formulas for Thickness (See pp. 338 and 339.) 7 0.078 nei 0.076 Constants C 0.0714 0.072 I = CEE CSS Sct Mica back helt >) (LESSER eee fon) - rol = N N ON Values of i Diacram 15.—Constants C;, C3; and Cy in Formulas for Thickness of Slab in Middle of Panel. (See pp. 340 and 341.) 912 TABLES AND DIAGRAMS Table 17.—Flat Slab Constants Cio and C,; for Computing Compression in Concrete. Values of Cio Cc 0.28(1 _ 1.28) Cy = 7 in Formulas (29), (31), and (82), p. 344. a a c i 0.1 0.195 | 0,15:120 Avo) aOeee 0.225 | 0.25 | 0.275 | C10 0.647 | 0.625 | 0.603 0.581 | 0.559 | 0.537 | 0.515 0.493 | Values of Cu b 1 bY fa \4 d : Cu = i +. ( — *) (“) (2.55 ~ 1.5) ji Formula (31), p. 344. ee ES fl .tttttC» | = Af,. (See p. 424.) fe = 600, fs = 20000, n = 12. Values of fi eS eee Ratio of Ratio of Vertical Steel p. Spiral 3 | HY 0.01 0.015 | 0.020 | 0.025 | 0.030 | 0.035 | 0.040 fe = 500 J, = 20 000 n = 12 TN. al ee ee 0.01 970 1 005 1 040 1075 1110 1 145 1 180 0.0125 1 070 1105 1 140 1175 1 210 1 245 1 280 0.15 1170 1 205 1 240 1 275 1 310 1 345 1 380 0.175 1 270 1 305 1 340 1375 1 410 1 445 1 480 0.20 1 370 1 405 1 440 1 475 1510 1 545 1 580 fc = 600 fs = 20 000 n = 12 mix = bilgo3 Sepa eu ERI cck ec 0.01 1 066 1 099 1182 1 165 1 198 1 231 1 264 0.0125 1 166 1199 1 232 1 265 1 298 1331 1 364 0.015 1 266 1 299 1332 1 365 1398 1431 1 464 0.0175 1 366 1 399 1 432 1 465 1 498 1531 1 564 0.020 1 466 1 499 1 532 1 565 1 598 1631 1 664 Note: For spiral design recommended by the authors average stress f may be formed from T able 21, p. 916, for appropriate values of f,, and tables on pp. 926 to 928 may be used for design. Pp. 929 to 934 give data applicable to all spiral columns. : ok ree Ht fi Ht In Ur “th NE PAR eer er ttt Me ‘i Aye il) STP i} ‘il i aN pat | ih ‘nai a Pit ia ~~ secadligert my TRERADE « at, bE 1a] ke +t wee’ 3 aa] 3 ‘TT sth +45 . if A a? q r 17) | > — ao ~ te — a ee — a | fit on i}. gi Sie nity i ie | thi | e 2 ; Hil a ‘i é i Ne 4344 ne 1 i Witt 4 Ri ERE Hilt H ia ii rH Hert a | Huta UL » +o for teg—h fed | HHALANSHILNNEREL ee hit mike Titahe ti 4-4 4nd ra HT | riti ‘ee . i 7 %. . e+ oe ane Ht rf QF) NENT TN | i} ab,’ s * mh t o.. } i H! THUAN ie BY CREED, CER, NEN TTING HE tt BUTI) HUINSTLIN IL | PATTEMPLATE NN te TiN NTT titi | NUNN Min SINUS HUTS < VIN NUTNT ING HTTTTTTT TT | HH ass th UENUINGI NG INGA f Mth SS NON NNT (3 i | Lin bs adhes pee : ; en seinpenaiee tient 4 | RN PRICE § “Or, a vi tit ; wt } e AN " i il Ve We NN oe in Le uu ANH " it ih il Dl es Ni oy indfh diosa} Lato! —-.iNF rensnina 11 noes baenti (To face page 679.) to ea es age ee ee , STs SS \ OTS SS Bae: eee ee ER ees Cae ee mer ty eres ee ey ee Ee oR PRRs (Si. RGR) fn, RPS, REG, ED eam. Ges SE, NEL REG, a oa ie Bi ee Ce RI “SR REE yee: | EN SS, ESS ee T Column Loae Based on Formula .—Total Loads on Round ‘| es A ee eee ee ies ee Ol ae eee Bey oe ee eee oe a Sa, Tee 's, SE Sats. SE a. Wi ee ee, ee TS, Gres, “i ST Se RRR TS SERS ns prs Sieg gig eS | ee Ge CE SS \o ee Ey ‘i a A ee SE HW, AE hc | Se Bs rs ee Sees 1 ee Ses eS Se ee a a it eae ee Te, Se ik a eee Se, WE. ES EA) S| ee ee ae" ee | ed Ee SE ee ee ee et So A ee ee a ee ee ee i a es ia ee el ee ee ee ee ee 00 9 ive) a * 6 Ee ee ES ae. OS, WS RE 8, E.R ee" | a ee ee ee 4 iso ae ST S te Le eee ee ee ee eee a Oe ee ee ee ee ee ee te Lee ee ee ee ee ee SS Ss ca a a = aa Yee ee ee ee el ae a eee ee ee on a ee ae ee ee a REIS - Sale ee rae ae oe ~ GES Ea a a a TS RN I, te, “Ge SG, HE. Gat IR, WSS SE es SS ee - e Ieee eae aes pe ot de I 5 a SS as Oe ek ES a, a Ee, Ea G ea WS | RE Pe ee eS Ee GS 6 ee eS Pe ee (ia, Sa mee, ea Re Se ea ene. Gea Sf Rae ee RS eS eet a a aS RN eet S aa ‘a, ‘ee BOR idle. -O) Aneesh RS ei SS, SS». SS oh SP ee De Be eee 5 Se aa om ae < Te eo Se SEE BOE Le St SORE Bee Sse 1 72 RO GS a, SE, Gee on BS oo a ee ee Ee 4 ae (i Se tas bh Gh SSeS Se OE Se Wg), OE Ee ee) ee Re Ge 00 G Q — | A ee Se eS a ee aoe Si “nod ee eens ie a. Es SS = iS ts ER ORT Pant ee for MR oe ee ies 1 en ee DO ES SR eS RE ee, as Gh GHG aire Pee 1 er 5 a Bes Tee eG ee ——f SS Be he SSR 2), AE es - e eee S in ceo ee Sn a oe 0OOr CS ae ee a ee ame. ~ (ie, A. US, i, GR ER ee ee ee : 1 GREE, \L_3S. IEE, le ee ee ee ES s $2. SE ROE. Se, CS ae 2 ee ee Ee Re 1 SS (TSS BD, Th OD als LS RE a (Sk I ke, Gag ee Pe ee I ee Be .o SS SL ON ee ee ee ee CAS See SS a Se ee ee ee ee Se ce, RE RT ii. ee, a2 a ES OlORS | eee eee ee ee SI ei tT be SG ML. SR, SE ENS, he GE ee ee eee =~ fees Se, SS SS Rly, a Gi, Sl SS ee eee SEE SIMs GS |. Ee, I, Se. el SG a, ET ee ee 1 EES ae RR CR, NU, SR sy ON OE Gere = eS ES TSE ae ee eae ee eR a i Sh ee ee SS a ES En, ee ee a eS Se Nee See eee eee en ee Se | 00 é as SS ee es Ra a A i SEE FE RE Be RS ES, ES, Ee = a eae ee See ee a Ce ys RN Se , G, , , SR eet Eee es oe es ce a Se ie, Sa eS ee See Lip aeaeee ee er ee ee Se ee ae ee ae a | rae (ae es SS Sa ae a ee ES SE ee eS, Dy RE es aa Soars | (wenn ieee) Ge Da ee 2 Re GE MRS Be Ee ES ES EME Sate ESS TS w= CT | ee i Si Sh. Ss Gees aaa GN SS SET] eae) (GEE 8 (eae TR te a SSS SSS SSS SS ae es ets Re FS a he as ess Sa, Se a ee Sead ee ee —— =| SS SS Se - eS ees a a oe : PE ORR LEAT aa =] —— SERS Cae ace Som aume bors sees —- S La) sayauy ui ‘uunjop punoy So 4ajaluvig anijraaff7 Ze Z ‘i De : Satis in Inches S 2 1000 Pounds h Af. (See p. 406.) Fffective Diameter of Round Column. / he 1800 ™ limns for Different Average Stresses, f. wpe i a ¥ a : Seer leeiee eee PORES OTT ‘termcairaes OTS ER ombdinel sr A ee eee . omar ne EF a= i . erieeeene Wemremnte: Sane oreer) Sateen oo Te ‘ ‘ ‘ em t ~—- Pic a he Ss ee aga react : atte pe OD eianaag vhs ‘> ro | : ee ee en See ee ae — tum ~ oo mage aa AE! A SE A AT AE ST CRE LE SA ANE LAG LS A A PLS AS A AS SA AT AAS AAS TAS MIRE NNN ANNU == - E meee. Saree eke nar Som ome TT ————————————————— or 08\ ‘ J ‘ = "% : en 4 - | > a a od =) tae v : we A en 1 tn wn is: i ee kee — ge wa) oo os % nee oa pe “ 2 ve re we “ a Fi Cy : ¢ e : r i< e} es : * £ - é * » - ° " a 7~ 1 ai 2 + 4 seu, iv ‘ geereeulog> Seana AO revs eusyost Al ‘ Bs ‘ ey y re = i LF ea a ee ih TUT Hit Pet hy PM ett . i | Lal ie BERT hia* eR, ut a0 He! ERE PUTT Eee 4H lh ih DRTNG aTN| iit taal fi ate PEELE EERE Ora PAN NE ae TNT HN un Re re Ud ss 4 i hee HSH To ae HINA TINT inst TTT TLELLETTTTLEE a STRUTS SUINGUINTTTET TT sti AUTTTNUTTTNTT | i tt WE aS TST mur Nitti HFS ia i A HE oe iil mn UT TI EN Hit iH HA lite! Until Hida UTRTUN ie ub ‘S ‘ANU haut Hn WNL THEE ieee a PMT NSN NG NG! UNH PINTS NUAONCONINUA LICE i mI | erent ths nN ONS SUNN TL TYEE Poa ia Ay Uh ANS TN ASE i i fe HT SEENON HU NUT) HA Hi THE | me Hi Tt tt HLH | RLU ET Mit qi} CT {SVL ‘AT AT {AACA VUE EEACUUT TALES ESCA SS (ime 3 1900 teat ¢ face pave WIR) Sas, YE Teese ot A Rois “Covi XE q Ac we Tia IMG QI 3 a Css » Try =) oe sic te Side es Sguare Column, tn Inches 40 TTT HATTA Me (MNURTUAE cA 38 IMAL ATTA TTT HE | y | UTES ACT TT A | | IAA ALIA A all il iin A LN Ppa HUUATAUUATEUUNAYUOHUAAENUMGOENAD VED AQUA He “Alli PETTITTE TUTTE ATT ATT HY ese so KN NTT A A DNATA TTT TET ATTTIL HAVANT CATT oi oo LTTE VATTTEATTTAI enT dt Dal : HNHENUARRV HAART TOPO 2 TT ATAU ata Halll! AA AA TAL ee A A fl, og HHMI MATT ATA Ue ill at Ae eT PATEL SVS HL ALT | og HLT A A Ane CTT acon / LA YAS Loon oa : i , HHT i hfe Vaan HAT i / eH Wat CUCU | es CHU iy Lie OUTTA TTT TT uaa IVA OTT ht A Lhhy WAN ral HUTTE DAMN AAMC FoA ANNTNUNAARAL AHEAD HOUTA oes We ‘a EUCLA ALLEL TT TTA Wy, Uy tal ATE ie} Ht HH | Hy ATT TT =, Hs WW PeOUUTEEERTTTTAETUUUUUTUARUREEERESOTEUUUUOTERUOTATUAT EEUU TTEGHAE|) vo AAT LE o AMMA 2. § 3 si % Sb Mp 9 Se S ~N OO’ Column Load Diagram 17.—Total Loads on Squari Based on Formula AM PRL nN i ty cal HU i B MAL rail ie vail HI CTE TE Tt eat ieee ec ll ll ie ail sect AANANNNUAN UL He aL eA Hat TTT Fal A cl U0 esOOCTUNEETETTUNEEETELAAOGEERATTAAEET ATT ANS iN ATT ree wt is ec TTT ee alll Ye cMO0EUUUAATAAEUAUAAATEQBEUUUOATASOOAUUUUAAGEDOAU THEE ie TTT HAI HAT Ve ll = rl PORTE eet (ETAT LPT IES Tt = ms 5 TRS, Pee Paes re Bea “il duties Weick sed aati” Za Pewee HLA ATATALTETHELA A ETLALAT HELL AY THT EE Gas meee Macon is Gee ieee. Wipemmee RE CEs wen os | Wildes ER Se, Se LY +4} ir er EE NRT 2 NET OI a HHA AEE PUTTU TIPE UDINTENEECUHESEOUOUUTOUEUURAONEUAHAEENTOUAT UN OTE CTT TTT ff cc on PUTETRETATETTEUEREEEREUAAAALE ACMA TTT TTT HL a att A ee ih S > S S S S S S > N <> ~ ~ 1/000 Pounds imns for Different Average Stresses f. Af. (See p. 406.) (To face page 919.) Effective Side of Square Column, in Inches ys od Ai fr jonas AA: im wr ? Grete we atte: § mais . 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LI O'8 |96I |9°9 |€8t |O'9 |2Z2T jeg jOZT |9°F lFOT Jo'y lzcT lee jogr |z'% lert foe lzet let lost SI 91 89 |49t j2°9- (OSt |I'S jISt ig°h |SPl 40°F HOFL sige “vet 19% leer 4.449) 2 4)00° 138.6516 18.33 30 706.9 6.90 | 5.91 5.17} 4.60] 4.14] 3.77 | 3.45 31 754.8 riot) 6211 5.34 | 4.75} 4.28 | 3.89] 3.56 32 804 .2 7.36 | 6.30 5.51 4.90 | 4.42 | 4.02 | 3.68 33 855.3 weconte 0.00, |) 6,695 |= 5:06) 42 56.1524215. 1) 13.80 34 BOTs O gee, 19 6269) )) 5286 |) 5521 4.69 | 4.28} 3.91 30 962.1 &.00 |r 6.89 6.03 | 5.36] 4.83 | 4.40 | 4.03 36 1017.9 Brae], 7.09.1) 6.20) |) 5.5205 74.9% |» 4,52 10.414 37 1075.2 8.51 7 2041) G2a8—| ace Ode teaok hola 4h 65. de 26 38 1134.1 &. (4) 7.49 |. 6.54 | 5.82 | 95.251. 4.78 | 4.37 39 1194.6 BO ie 4.68. 1, 6.42 1e.5.987)45.38) |. 4.90 (9.4248 40 1256.6 9.20 7.88 6.89 |r,6.13 5.52 | 5.03 | 4.60 Values above upper heavy line are for more than 2 per cent of spiral. Values below lower heavy line are for less than 1 per cent of spiral. To Find Ratio of Spiral p;.— Divide value from table by area of concrete core. To Find Weight of Spiral.—Multiply the corresponding value of p,A by 3.4 and by the length of spiral plus twice the pitch (to allow for extra turns on top and bottom). Add weight of spacers. (See p. 433.) To Find Load Carried by Spiral. Chicago Code.—Multiply value of p,A from table above by 24nf,., which for 1: 13 : 3 concrete equals 18 000 Ib. and for 1: 1 : 2 concrete equals 18 125 lb. New York Code.—Multiply value of A from table above by 40 000 lb. 932 TABLES AND DIAGRAMS Table 35.—Values of pA in Spiral Columns. Where pi = Ratio of Spiral and A = Area of Concrete Core These values are equivalent areas of vertical steel having same volume as the volume of spiral q,;-in. Wire Diameter| Area Pitch of 75 wire. oO of Core Core d A ae) PRES 3.0 10 18.2 3.14 ere Leon 11 G5.) 3.45 1.88 iek2 12 bo 3.76 : 2.06 1.88 13 or 4.08 oO. ; Deo 2.04 14 153.9 4.39 oe oe 2.40 220 15 176.7 Aa. os 2. 25 22300 16 BOToL 5.02 2. 3: 2.74 Dee 17 227 .0 5.oo 3) on 2.92 2.66 18 254 5.65 a aie 3.08 2.82 19 283.5 5.96 4. on 3.26 2.98 20 o14.2 Gs20 ae ~A 3.45 3.14 21 346.4 6.59 4. 43 3.60 3.29 oe 380.1 6.90 Le 4. aokt 5.40 23 415.5 ziS BX 4. 3.94 3.60 24 452.4 fixe oe Lie 4.12 3.76 25 490.9 ve a Dd. 4.29 3.92 26 530.9 8. Gx ae. 4.45 4.08 27 572.6 8.4 6. Bi 4.63 Ae ze 28 615.8 on 6. 5. 4.80 4.39 29 660.5 9. 6. 6. 4.97 4.55 30 706.9 9. vf. 6. 5.14 re a 31 754.8 9. 79 6. 5.31 | 4.86 a2 804.2 LO 76. 6. 5.49 5.02 33 855.3 10. re: 6. 5.66 5.18 34 907.9 10. ce ve. 5.83 bbe a5 962.1 10. Se fh §.00 5.49 36 1017.9 celle 8.4 “i 6.17 | 5.65 aire 1075.2 1 8. Te 6.34 5.81 38 1134.1 Le 8. ; 6.51 5.96 39 1194.6 12? 9. 6.69 6.11 40 1256.6 123 9. 6.86 6.28 Values above upper heavy line are for more than 2 per cent of spiral. Values below lower heavy line are for less than 1 per cent of spiral. To Find Ratio of Spiral p1.—Divide value from table by area of concrete core. Weight of Spiral—Multiply the corresponding value of piA by 3.4 and by the length of spiral plus twice the pitch (to allow for extra turns on top and bottom). Add weight of spacers. (See p. 433.) ' Load Carried by Spiral. Chicago Code.—Multiply value of piA from table above by 23nfc, which for 1: 14:3 concrete equals 18 000 lb. and for 1: 1 : 3 concrete, 18 125 Ib. New York Code—Multiply value of piA from table above by 40 000 lb. TABLES AND DIAGRAMS 933 Table 36.—Values of p,A in Spiral Columns. Where p; = Ratio of Spiral and A = Area of Concrete Core These values are equivalent areas of vertical steel having same volume as the volume of spiral 3-in. Wire Diameter; Area Pitch of 4 Wire. of of Core Core d A A EES: 15 . 225 | 25 pares) 320 10 78.2 4.10 ook 5, Da 2.46 Zeek Nee oe OF 11 95.0 4.51 3.86 of 3.00 Zeit 2.46 QoDb 12 Toe! 4.92 4°21 3. o.26 2.95 2.69 2.46 13 T3a227 DOG Ae DOsinre © BO ey Pea | 2.66 14 153.9 OTA |. 4:91 4. oLo2 3.44 Del ot 287 15 L7G 27 6.15 ne 20 4. 4.09 Hess 0 Ov oUsbw o. OF 16 201.1 Cesar h D262 4. Ae 3.94 SOS) § oLeo 17. 227.0 CAO he nb OF a 4.64 Daa ers eeell 3.48 18 2545 foot .0. 32 ie 4.91 4.43 4.03 4-35-69 19 283.5 TO 667 py B19 4.67 | 4.261] 3.89 20 Deo ie ae 200). «7.02 Gp l4e15 5.46 4.92 4 48 | 4.10 21 346.4 8.61 Or 6.45 mee aL dabieeac 14) lene) 22 380.1 9. re 6.75 6.01 se Bl 4.93 | 4.51 23 AID. 9. 8. PACOC Oe: a OSOO um eOrto th ol 1 24 452.4 9. 8. (ead DO Te. On 908. D088. --. 4,92 25 490 .9 10. 8. TOTS Ose 6.15 DL Out 7O. 12 26 530.9 10266° | 9213 7.98 TALOUMOROTE tp. 82 bh plod 27 572.6 11.07 | 9.48 | 8.29 CO Leen 04 OL 05: PMS 03 28 615.8 | 11.48 | 9.838 8.60 | 7.64 | 6.89 6.27 5.74 29 660.5 Hoo LOLS. lee 8: 90° 27,92 als 6.50 8 5.94 30 706.9 12730-10253 9.21 8. fie GEO PG ale 31 154.8 12-71. 110288 |. 9.52 8. is OF04 a Gaoo a2 804.2 Hee 2a th ON 82) he Se (eG TL) 6756 33 855.3 boot mle bs. LO. 13) fy 9. 8. aoe G2aP 34 907 .9 13.94 | 11.98 | 10.44 | 9. 8. 7.62 6.97 BO 962.1 Peewee be LO ica fee 8. Peo Ue ks 36 1017.9 14-46 | 12.63 | 11.05 0735) e138 805 S06 1° F238 3d 107522 toot? | 12-98-4011 .386 0-10-1009 104" 8.29. 1-77.59 38 1134.1 15.58 | 13.34 | 11.66 7 10.37 | 9.35 8.51 7.79 39 1194.6 15.99 | 13.69 | 11.97 | 10.65 9.59 S24) 799 40 1256.6 16.40 | 14.04 | 12.28 | 10.92 9.84 | 8.96 | 8.20 Values above upper heavy line are for more than 2 per cent of spiral. Values below lower heavy line are for less than 1 per cent of spiral. To Find Ratio of Spiral p:.—Divide value from table by area of concrete core. To Find Weight of Spiral—Multiply the corresponding value of p,A by 3.4 and by the length of spiral plus twice the pitch (to allow for extra turns on top and bottom). Add weight of spacers. (See p. 433.) To Find Load Carried by Spiral. Chicago Code.—Multiply value of »:A from table above by 23n¢, which for 1: 11 : 3 concrete equals 18 000 lb. and for 1: 1: 2 concrete, 18 125 lb. New York Code.—Multiply value of pi1A from table above by 40 000 lb. TABLES AND DIAGRAMS ETNIES WET NT NUNN il i \ i PASTE UTNE NEAT AACE TTT nt NININININ Hi i Re cat TTT : = = 4 tN NI nth CTUANTAUATNTNIN ATA CTT TTT = e INIT Ht COSTA TTA Hl TEDSTER AANA NAAT og -o AAA N 900 A NACNDANSNANNAARAANNTI NNNINNAANNASNMMNTTIN UNNI AAANAATTTT N SUMAN ASAT 200 NINTANANRANNANA Att | SAINT 9 2) S) NS © ly v > \ Le Ss vy ce) ~ © lo en JO sen] Df (See p. 177.) (See p. 177.) ants C, for Members Subjected to Direct Compression and Flexure. Use in Formula for Max. Compression Stress, Te D1aGRAM 18.—Const 935 RAMS 1 q AND DIAC TABLES TINTINT TEINTENTEN q N qn NT HHH TURININIK HATA NUUNTINENTENT NTN = = == = 2 = = = = = = 4 = ace : eee eS eee ee = = : a | i = | i Hl a 30 20 Values FeO hee 15 10 of Eccentriety " h=1.2d. n= 15 THE Values of k, ratio of depth of neutral axis to depth of steel below compressed surface of beam As aj ak | Both Sides Reinforced. p = bd where A, is Reinforcement at Both Sides. DracraMm 22.—Direct Stress and Flexure. Ratio of Depth of Neutral Axis, k, for Different Eccentricities. Part of Section in Tension. (See p. 184.) TABLES AND DIAGRAMS 939 0.350 TIV TN NAT iil \ \ vA \' JAA ANN i Hill a) wi NIU oS Sree NTS ah Beaette, esata | Mae STS SIS 9.250 LA aS “s ES é PTS = i Min 4 a os i Tegel nist a HUT S | N TH Hd 0.200 ee = pe 7 9220 ATR eet ULI TPE a “ sti mee © } unas < in & 0.750 8 h=1.2d. n= 15 A Both Sides Reinforced. p = re where A, is Reinforcement at Both Sides. DracraM 23.—Direct Stress and Flexure. Constant Cz in Formula (125), p. 182. Part of Section in Tension. (See p. 184.) 940 TABLES AND DIAGRAMS Table 37.—Ratio of “ for Different Steel Ratios p. (See p. 186.) Steel Ratios p ~19.002/0.003/0.004/0. 005/0. 006/0. 007|0 .008\0.009)0.010)0.011)0.012 0.565/0.570/0.576/0.581)/0. 586/0. 590)0 .595|0 . 600/0 .605/0 .609)0.613 Ne TT el lll! a | \\ cn | NN UREA 0.7 NACE NN NAT a i NIN === wpe? Bi Sah SN TAI. > col AGASSI & s ‘mht a iS AWN : S oH NAS SRNTTTTHHTL S 06 lil HLA WINKS un SR 0.4.0 : HRN eh RSS|! 05 nadine te SSS STHUAUVUTAUNAUUAAL NaS SHilillins << o TMT UTM rs i it 0.2, agit 0.2 ie ae d=09h. n= 15 As Tension Side only Reinforced. p = ao DiacraAm 24.—Direct Stress and Flexure. Ratio of Depth of Neutral Axis, k, for Different Eccentricities. Part of Section in Tension. (See p. 187.) TABLES AND DIAGRAMS 941 TL. pea TTI ETAT tl LL : | Ht men : HNN ie 4 °| a 0.20°1° iN |x a.19~ | 2 lI : a NT 0.18 > ~Im | th Pi Ul) 0.171 3S ti ot Clea s cya tt 0.15 — S ae iM ut : . 0.3 04 O05 Values of k d=09h. n= 15 As Tension Side only Reinforced. p = bd Diacram 25.—Direct Stress and Flexure. Constants C, in Formula (133), p. 187. Part of Section in Tension. (See p. 187.) 942 TABLES AND DIAGRAMS Table 38.—Areas, Weights, and Circumferences of Square and Round Bars. One cubic foot of steel weighs 490 Ib. | b % Square Round bs Square Round = rs Bars. Bars. 5 S Bars Bars. pail Ae Beles Balan | pe nes 4 teeipese mlb 5 po a ia aa| ". | Ba | se | oo | Be ee] | eee as| we “8 a0) ae oh ats) wa a ny =a ik. és| ¢ | B2| ¢ | 88] Re Ss] ¢ | Be] Fg | Be | Be ‘a+ © DH © HO | OF |g+ © ‘Oy o oO | (OR = < |e aS) Boe oh) oe 4 |5 = rf) 2 | 4.000; 13.60} 3.142) 6.283] 10.68 7; | 0.004) 0.013) 0.003} 0.196) 0.010 275| 4.254) 14.46) 3.341] 6.480) 11.36 : 0.016] 0.053] 0.012} 0.393) 0.042|| 24 | 4.516} 15.35] 3.547) 6.676) 12.06 3; | 0.035} 0.119) 0.028) 0.589) 0.094 23;| 4.785) 16.27| 3.758] 6.872) 12.78 a 0.063! 0.212) 0.049] 0.785] 0.167|| 2% | 5.063) 17.22) 3.976) 7.069) 13.52 5 | 0.098] 0.333] 0.077} 0.982) 0.261 275| 5.348] 18.19) 4.200] 7.265) 14.28 = 0.141] 0.478) 0.110] 1.178] 0.375|| 22 | 5.641] 19.18) 4.430) 7.461) 15.07 z= | 0.191} 0.651} 0.150) 1.374) 0.511 275| 5.941] 20.20] 4.666] 7.658, 15.86 S 0.250} 0.850] 0.196] 1.571] 0.667|| 24 | 6.250) 21.25) 4.909] 7.854) 16.69 2; | 0.316} 1.076) 0.249) 1.767) 0.845 2%;| 6.566] 22.33) 5.157] 8.050) 17.53 & 0.391] 1.328! 0.307] 1.964] 1.043]| 22 | 6.891) 23.43] 5.412) 8.247) 18.40 11 | 0.473] 1.608) 0.371} 2.160) 1.262 244) 7.223] 24.56] 5.673] 8.443} 19.29 3 0.563 1.913] 0.442} 2.356] 1.502]| 22 | 7.563) 25.00) 5.940) 8.639) 20.20 13 | (0.660) 2.245] 0.519} 2.553) 1.763 213) 7.910] 26.90) 6.213] 8.836; 21.12 t 0.766! 2.603! 0.601] 2.749] 2.044]) 22 | 8.266] 28.10} 6.492) 9.032) 22.07 15 | 0.879) 2.989) 0.690) 2.945 2.347|| 248) 8.629) 29.34) 6.777) 9.228) 23.04 1 | 1.000! 3.400] 0.785] 3.142] 2.670]| 3 | 9.000) 30.60) 7.069} 9.425; 24.03 13; | 1.129) 3.838) 0.887} 3.338) 3.014 37s| 9.379] 31.89] 7.366) 9.621) 25.04 13 1.266] 4.3031 0.994! 3.534] 3.379]| 34 | 9.766) 33.20} 7.670) 9.813] 26.08 133 | 1.410] 4.795} 1.108) 3.731) 3.766 33;|10.160| 34.55} 7.980|10.014) 27.13 13 1.5631 5.3121 1.227] 3.927} 4.173)) 3% |10.563} 35.92) 8.296/10.210) 28.20 135 | 1.723) 5.857| 1.353} 4.123) 4.600 375|10.973) 37.31) 8.618)10.407| 29.30 ie 1.891| 6.428] 1.485] 4.320) 5.049]) 3% [11.391] 38.73) 8.946)10.603) 30.42 175 | 2.066) 7.026) 1.623} 4.516) 5.518 375|11.816| 40.18) 9.281)10.799) 31.56 13 2.250] 7.650| 1.767|} 4.712) 6.008]| 34 |12.250) 41.65} 9.621|10.996} 32.71 1% | 2.441] 8.301] 1.918} 4.909} 6.520 37%5|12:691} 43.14) 9.968}11.192) 33.90 Le 2.641] 8.978] 2.074) 5.105] 7.051|} 33 |18.141) 44.68/10.321]11.388) 35.09 134 | 2.848) 9.682] 2.237] 5.301 7.604)| 344/138.598| 46.24/10.680/11.585) 36.31 12 3 063) 10.41] 2.405] 5.498] 8.178]| 32 |14.063) 47.82)11.045)11.781) 37.56 123 | 3.285) 11.17] 2.580} 5.694 8.773)| 342/14.535) 49.42)11.416/11.977| 38.81 1i 3 5161 11.95] 2.761) 5.891] 9.388} 3% |15.016) 51.05)11.793)12.174| 40.10 128 | 3.754| 12.76} 2.948) 6.087/10.020 343|15.504| 52.71)/12.177|12.370| 41.40 TABLES AND DIAGRAMS Table 39.—Areas of Groups of Bars of Uniform Size. of Bar in Inches Diameter [ fe sie [= O\eo | Pa on ale olen 5° ole oer ois takes wee le lca wl oj: fh fox tale | ale | + Sh 1s oon 5° re coo 38 i a aie oe Olan al alo BO cole mI cos pm top oooo ooo °o Number of Bars. 943 cS HS oO Oo. 6 We} eo) HEH © an er) bee ee {2 31 .49 .70 .96 .25 . 58 .95 . 36 81 . 30 .83 .40 .00) .33 .81 .45/11.34/13. Rounp Bars 0.29) 0. 7.36) 8. 8.91/10. .84/10.60|12. 34 . 54 TT .05 37 74 15 .60 .09 .63 21 .83 .50 6.96 59 37114. Ke OOO 9): 39)11. . 39 .61 .88 . 20 .57 -99 .46 NSE 97 53 3 .15) 4 .81) 5. .52) 6 .28) 7. 7.95| 8. 82/11. 88/13. 14}15. Square Bars bo i) e wWNnNndD = 9.38/10. 25/13 .50/15. 44 .68 .98 34 15 ~22 .73 .31 94 62 .36 15 .00 ,86/10. 94/12. 23/15. 75|18. se - CO © wWwhdo bs Ke oO O° wWwnNDhND tl i) Porno re, .85 ll a) x - © © 10. 12. 15. Ow bd = row Nd onna «J J eH OO aS co eB who d& =] =] oe kK © 1Ou) .00 TABLES AND DIAGRAMS 944 Table 40.—Area of Square Bars in Slabs for Different Spacing. Area of Reinforcement in Square Inches wnAOK oe TAA HS rr sooooo lead 2) OR ee Oa Se ne | remo oO Or ky a8 rl GD 6} XH 1D ae | o-oo oC oo Rae ec da a a se DOM aMAAN © 2 | AAA HDS ee | coooooo 2.18] 2.00] 1.71] 1.50] 1.33] 1.20) 1.00 oor oO Don ooo So ~~ ono orRao oo ° 4 MDoONk ~roOon oonn MHAaKRKRATHTHRWMN ee HNANMHOKR OAM a SOS Oe. OC 6.0, 0 4 = Sener | esa ee a ee ye oe ee ee ee ratontrRe nt OM A — BAHT DOCWAHMI et | oo oS oC 6 Cm mo ge eh eae a ea at Ae a ree es re 4 HHoHNMMDORANMNMMOE’ (te) ae ee eS eto ava er at ER Re ey FS ; ero 6 ODO mete RQ ee SS ee a MHANMDANMONDNMHEHNA eal nit gf ANAMAMNOHDONHOD oy H Ge S16 S.C Cnt eri ° ier Area) Ue See. ee &6 ei eee oe a ha HAANHORANNNWONH s ml doccoodooonnANAN fl el ae 2k 2s ie es BE ee ee a We ae be beg eae we i ANAMMOWDONMNNNOM SO Ri) oododonnnnaan Ne eae ee ee ee ee ed OOaNKRMOMOUREADANOYSO jg /mAWSRanWSaRES ae | eoeooooonnAANND 5 eis Si eG baa a rk Pe HOMDODOOOHAMON HT wg | ARNSDOMOCANSO Mj) odocoonannaANaD 5 it le 5 ee ee (pete ate eee aS 2 NMMWORONNMNANDOWS eS | COOCOOnR THA ANANMMH ESI ee a a Ds a a ORRNONRKRORRAO wg | R*SRABHARASAL H | COOCOOnFRHTAANMOA SH HH Seesteycr athe 2 eas eS, Ee ee Go Oe ee ae if MMOHMAMNDM AION OS SSS Scot A to lae st aS Dia- meter (In.) wo ko (eo mo ojo wo is wf e9|0 +2 mice ofS un|oo ae old 2 Pale) “es So Di- men- sion (In.) Table 41.—Area of Round Bars in Slabs for Different Spacing. Area of Reinforcement in Square Inches Spacing of Bars. 12 In 9 In. .57| 1.35] 1.18] 1.05] 0.94] 0.78 0.60} 0.54] 0.50] 0.43] 0.37] 0.33] 0.30} 0.25 0.74| 0.67] 0.61] 0.53) 0.46] 0.41] 0.37) 0.31 0.89] 0.81] 0.74] 0.64] 0.56} 0.49) 0.45) 0.37 1.06] 0.96] 0.88] 0.76] 0.66] 0.59] 0.53) 0.44 1.24! 1.13] 1.04] 0.89] 0.78] 0.69] 0.62} 0.52 1.44] 1.31] 1.20] 1.03] 0.90] 0.80] 0.72} 0.60 1.66] 1.51) 1.38] 1.18] 1.03] 0.92] 0.83] 0.69 0.47} 0.43] 0.39) 0.34| 0.29] 0.26] 0.24) 0.20 3.14) 2.69) 2.36] 2.09) 1.88) 1.71] 1 0.26] 0.24] 0.22] 0.19] 0.17) 0.15) 0.138 .51] 0.45] 0.40) 0.36] 0.33] 0.30] 0.26) 0.23) 0.20} 0.18} 0.15 0.18} 0.17} 0.15} 0.13 5 In. Ced2 8 In BK 2 In. 0.20 0.52 0.66 0.82 1,18 1.38 1.84 4 In. 0.23 0.75 0.92 1932 2.07 1 2 In 0.36] 0.31] 0.26 0.66} 0.53] 0.44) 0.38) 0.33) 0.29 1.19) 0.99).0.85 1.47} 1.23] 1.05 2.20) 1.78) L482 27) L105 99 2.65 3 ; In. ‘ PEP AW ML aA OL dle rent Sno lie a Ole2eod sl 0.72} 0.60} O 1.18] 0.94] 0.78] 0.67] 0.59 1.49 1.84 ae 2 In 0.29} 0.23} 0.20] 0.17) 0.15) 0.13 3.11] 2.48] 2.07] 1.78] 1.56 3.61] 2.88] 2.40) 2.06] 1.80) 1.60 4.14 4.71 2 In. 0.46 0.90 wc io wo rio eo wie rit anf erico n{ ries ofS s2\c0 TIS cin SIE eleo Sie Section ees Aa YI), * RECTANGULAR CELL }-—— b1- | ae 4 HOLLOW CIRCLE | For reinforced concrete sections the reinforcement may be considered as replaced by having an area equal to n times the area of reinforcement and placed at the same (See p. 173.) concrete, distance from gravity axis. * Applicable only to homogeneous (not reinforced) beams. TABLES AND DIAGRAMS Table 42.—Properties of Sections. Moment of Area A Inertia I bh3 bh — 12 b2 a 12 2 Ue 12 BH-—bh ds (BH3— bh’) Area of flange A1hi2+Asho? -+area of web 12 Ay Ao(hi the)? =Ayt4. $O 7 4(A,+Ae) ; rt ur 7 a (r4— 114) wr%- 17) ars Distance of Neutral Axis from Most Strained Fiber * h %Y b 2 3bv/2 H a ho Athg— Agh oo da) h Ayjhg— Ach to = - --—_———_ a 2 2CAy Aa) f r 945 Radius of Gyration, Nf iA 0.289d 0.2896 0.2896 | BH3— bh3 ‘12(BH—bh; wv INDEX A Abrams, D. A., 52, 53, 55, 57 Acme flat slab system, 362 . Adjustable metal tiles, 605 Aggregates, choice of, 6 coarse, 3 coloring surface by, 737 effect on modulus of elasticity, 201 fine, 2 floor surfaces, 621 grading, 3 selection, 1 size, 273 Air space below roof, 652 Alden Park Manor, 748 Anchored ends, working stresses, 263 Anchoring bars, 265 chimneys, etc., 477 column reinforcement, 269 compression reinforcement, 268 footing reinforcement, 537 hooks, 163, 269 plate, 163 problems of, 267 retaining wall reinforcement, 853 steel in wall columns, 317, 462 straight imbedment, 163 web reinforcement, 244 Angle of bent reinforcement, 291 column heads for flat slabs, 320 Apartment de Luxe, metal tiles for, 607 houses, 805 Arch roof, details, 676 Arches. See also Vol. III. fulerum girders as, 772 Area, column sections, 915 reinforcing bars, Table 12, 942, 943, 944 Armory at Danville, 673 Assembling column steel, 418 Assumptions, rectangular beam theory, 126 Auditoriums, concrete, 762, 810 Automobiles, dimensions of, 796 Auxiliary structures above roof, 678 Axial compression and bending, col- umns, 463 B Bach, C., 24, 38, 49, 50, 52, 58, 59, 60, 62, 65, 93, 95 Backing of curtain walls, 745 Balanced design, 203 rectangular beams, 132 Balcony, cantilevers, 772 cantilever trusses, 778, 782 design, 763 floors, 778 framing, 780 fulerum girder for, 765 Barrett’s Specifications, tar and gravel roofs, 655 Bars. See also Reinforcement aluminum identification tags, 723 spacers, 211 table of sizes, 12, 942, 943, 944 Barton spider-web reinforcement, 359 Base for structural steel columns, 443 Basement slabs, 618 supported by columns, 619 thickness, 619 under water pressure, 619 Basement walls, 637, 638 waterproofing of, 637 Batter, retaining walls, 841 Beam, design tables, 880 947 948 Beam and girder schedules, 716 and girder type floors, 575 Beam and slab, 383 foundation, 540 Beams. See also Wall beams See also T-beams balanced design, 1382, 203 bending moments to use, 275 compression below flange, 137 constants for rectangular, 205, 880, 881 continuous at support, 281 depth, formula, rectangular, 131, 204 design constants, 205 details on plans, 723 diagonal tension in, 147 formulas for, 247 reinforcement for, 248, 252 tests, 38 . failure by compression, 50 by tension, 35 false, 615 formulas, design, rectangular beams, 204 steel top and bottom, 234 T-beam, 216 unsymmetrical T-beams, 142 wedge-shaped beams, 140 review, rectangular beams, 207 T-beams, 224 haunch, 284, 405 hollow circular, 822 neutral axis, rectangular, 130, 207 steel top and bottom, 139 T-beam, 134, 220 numbering, 711 reinforcing for, 41 resisting forces, 129 restraint of, 275 in flat slab, 377 safe loads, table, 884 schedules for, 715 shearing stresses, 143, 150 spacing of tension bars, 273 spandrel, 334, 459, 746 steel in top and bottom, theory, 137 simplified formulas, 234 diagrams, 904 to 909 INDEX Beams, straight-line theory, 127 and actual conditions, 142 stresses, actual and computed, 24 test of continuous, 64 rectangular, 20 steel top and bottom, 48 T-beam, 35 theory of rectangular beams, 124 steel top and bottom, 137. T-beam, 133 unsymmetrical T-beams, 142 wedge-shaped beams, 140 unbalanced design, 132 wall, 379, 578 width required by reinforcement, Table, 274 -Bearing stresses, 270 brickwork, 631 Bend, angle of reinforcement, 291-297 test, 8 Bending, columns subjected to, 458 combined with, direct tension, 189 direct compression, 187 thrust, 187 Bending bars, point of, 287 slabs, 355 beam reinforcement, points of, 292- 297 diagrams, for beam reinforcement, 292-297 Bending moments, beam design, 275 building frames, 279 coefficients, from test of continuous beams, 65 combined footings, 524 continuous beams, 277 diagrams for determining point of bending of bars, 292-297 exterior columns, 314 flat slabs, 328 See also Flat Slabs interior panels, 331 wall panels, 333 footings, 495 permissible variations in flat slabs, 335° strap beams, 533 — Bending sketches, 724 INDEX Bent bars, marking of, 724 Bent-up bars as web reinforcement: area and spacing, 156, 249 example, 582, 587 formulas for, 156, 249 limiting spacing, 250 Bergstrom, Edwin, 764, 779 Bignell pre-cast pile, 552 Bond, allowable unit stresses, 263 anchorage in beam reinforcement, effect on, 26 deformed bars, 53, 263 design formulas for, 262, 268 distribution of stresses, 57 surface of bar, 53 on temperature cracks, 299 elements of, 51 function of external shear, 261 green concrete, 56 hook in beam, effect on, 263 influence of age, 54 of mix, 54 length of lap for, table, 414 load, effect of position on, 58 mix, effect of, 59 necessary to develop tensile strength of bar, 354 plain bars, 53, 263 problems of, 264 proportions of concrete, effect, 59 ratio length of imbedment to diam- eter bar, 52 resistance, 52, 260 shape of bar, effect, 53 steel to concrete, explanation, 260 stirrups, 40 strength, 50 stress, requirements, 132 stresses, 10, 56, 151, 263 flat slabs, 351 footings, 499, 500, 528 fulerum girders, 768 importance of computation, 261 small vs. large bars, 262 unit stresses, allowable, 268 Bonded finish for floors, 621 Borings, results of, 469 Boston Building Code, 347 949 Boston Building Code, bearing values of soils, 471 columns, 408, 421, 488, 448, 450 live loads, 454, 570 structural steel and umns, 438 width of flange in T-beam, 218 conerete col- ' Bracket loads on columns, 464 Bransom System, floors, 588 Brick, exterior, 747 Brick parapets, 658 piers, strength, 77 stresses on, 78, 631 veneer building, 749 wall and concrete frame, 739 wall thickness, 31 Brickwork, bearing walls, 335, 629 Bridges. See Vol. III British building rules, 202 Brush, finish, 736 hammer finish, 737 Buckling. of columns, 81, 87 Building Codes. See Code in Ques- tion Building construction, 564 Building, frames, formulas, 279 lines, 721 plans, 707 tests on completed, 113 Bunte Bros., factory, 681 Bureau of Standards, 77 Butting bars, 415 Buttresses, retaining walls, 833 ae: Ca, values of, 939 Cp, values of, 941 Ca, values of, 894, 895 C., values of, 934 Cn, for square footings, 495 _ Cp, for rectangular footings, 495 Cy3, for square footings, 497 Cys, for rectangular footing, 498 C;;, for rectangular footing, 502 Cp, values for earth pressure, $37 Cr, table of values, 896 Ci to Cy, diagram, 911 950 C0 and Ci, table, 912 C12, table, 913 C13, table, 914 Caisson pile, 560 Caissons, 477 Cantilever balcony truss, Grauman’s Theater, 779 vs. ordinary beam, 775 Cantilever retaining walls, 849 Cantilevers, design of, 772 Capacity of stairways, 684 Carrying capacity, soil, 469 wood piles, 542 Carter’s Ink Co. Factory, 734 Cast-in-place piles, 554, 558, 559 Cast-iron core, concrete columns, 445 Cast stone, 733 Ceilings, ornamental, 614 suspended, 19, 613 Cellular walls, 869, 871 Cement, 2 flash set, 2 mixing, 4 protection of reinforcement, 9 ramming, 2 slump test, 4 testing, 2 water in, 4 Chairs, metal, 356 Cheapest exterior treatment, 730 Chenoweth pre-cast pile, 551 Chicago Building Code, 218, 322, 347 columns, 396, 408, 421, 425, 439, 448, 451 exits, 683 flat slabs, 312, 317, 396 live loads, 454 Chimney, Edison Electric Ill. Co., Brooklyn, 824 Chimneys, concrete, 812 constants for neutral axis, 828 construction, 827 design, 822, 827 essentials, 827 mix of concrete for, 823 temperatures, 822 thickness concrete, 819 Cinder fill, roofs, 653 INDEX Cinder fill, foundations, 540 Circular beams, 822 plain sections, stresses in, 172 reinforced section, 179 Circular ramps, 801 Circumferential flat slab reinforce- ment, 117, 367 Clark Biscuit Co. Building, 793 Clearances, elevator shaft, 696 Cleveland Building Code, 218 columns, 408, 438, 448, 450, 464 Coefficient of contraction, concrete, 300 Cold weather, laying in, 299 | Cold-storage, live loads, 792 warehouses, 792 Coloring of surfaces, 737 Columns. See also Exterior Columns. See also Interior Columns. See also Spiral Columns. See also Wall Columns Column caps, 760 Columns, 403 action under test, 86 allowable unit stresses, axial compression and bending, 463 composite columns, 436 plain concrete columns, 879 spiral columns, 421 vertical bars only, 406 axial compression and bending, 463 bars, 412 bars, lapping of, 413 butting, 415 in spiral column, 434 bearing stresses, 271 bending in, 458 bracket loads, 464 buckling, 81, 87 building codes for, 408, 409, 422, 437, 450 cast-iron core, 445 comparison of cost, 445 composite, 436 concrete, effect of richness, 76 concrete vs. brick, 77 core. See Effective area cost of concrete vs. forms, 445 crane loads, 464 INDEX 951 Columns, deformations, plain steel, 89 | Columns, reinforcement, influence of, 80 design example, 456 dowels, 417 eccentric load on, 164 economy of design, 445 economy spiral vs. vertical steel only, 449 effective area, 406 exterior. See Exterior columns factor of safety, 79 failure, 76, 79 fireproofing of, 272, 407 flat slab construction, 305 formulas, general, 158 for design, long columns, 435 plain concrete, 404 spiral columns, 421, 422, 424, 425, 426 vertical steel only, 406 heads for flat slab, 319 angle of, 319; 320 slab thickness at, 338 stresses at, 199 hinged effect at bottom, 668 interior for flat slabs, 305, 310, 312 lapping of reinforcement, 413 limitation of size for flat slabs, 307 limiting length, 407 live load on, reduction of, 453 load on footings, 479 long, formulas for, 434 tests on, 93 mix, influence of, 80 modulus of elasticity, 76 ’ numbering, on floor plan, 710 oblong spiral, 431 octagonal spiral, 429 octagonal vs. round, 429 orchestra floor, 763 plain, test of, 81 point of maximum bending moment, 459 Poisson’s ratio, 76 proportions of concrete for, 446 ratio of rigidity in flat slab, 459 reduction of live load, 452 reinforcement, anchorage of, 269, 413 | disposition of, 414 uniform size of, 412 repeated loads, test under, 77 review of design, 411 round with 1 per cent spiral, tables of loads, 926, 927, 928 round with vertical reinforcement, tables of loads, 923, 924, 925 schedule for, 719 spacing in garages, 797 spiral design of, 419, 421, 422, 430, 448 test of, 82, 89 splicing structural steel, 441 splitting effect of bars, 79 square columns, table, 920, 921, 922 square spiral not effective, 430 test of, 88 steel, assembling of, 418 strength, formula for, 160, 421 stresses in spiral, 422 with vertical steel only, 205 structural steel in concrete, 435 supporting basement, 619 tensile steel, amount, 461 test of, 76 ties, 417 total loads on, 919 unsupported length, 405 ‘vertical and spiral reinforcement, _ test of, 84 vertically reinforced, test of, 78 yielding of, 108 Combined footings, 522 bending moments, 524 external shear, 526 for interior columns, 523 shape of base, 522 unequal loads on, 522 Comparative costs of columns, 445 floors, 786 Comparison of floor surfaces, 627 Composite columns, 436 piles, 555 Compression, below flange in T-beam, 137 due to eccentricity, 191 failure of beam by, 30, 32 952 Compression, reinforcement for beams, 50 anchorage of, 268 diagrammatic solution, 237 formulas for design, 234 flat slabs, 345 influence on deflection, 63 problems in design, 232 ratio of, 236, 283 supports, 282, 283 T-beams, 225, 230 stress in concrete at 223 stresses in flat slabs, 341 Concentrated loading, stirrups for, 254 Concrete, definition of. See item in question Concrete buildings, cost, 567 height, 803 shape of, 804 types, 572 Concrete chimneys, 812 coefficient of contraction, 300 Condensation under roofs, 649 Condron Company, 362 Considére, 22 Consistency of concrete, 26 Constants. See also C4, ete. beam and slab design, 205, 880 Continuous beams: design, 285 details, 281 diagrams for points of bending bars in, 293-297 equal span, formulas for, 278 minimum depth, 222 points of inflection, 289 span of, 277 tension in steel at supports, 288 tests, 64, 115 unequal spans, 280 unsymmetrical loading, 280 vs. building frames, 277 web reinforcement for, 157 Continuous slabs, formulas for, 209, 279 arrangement of reinforcement, 210 wall column footings, 518 supports, INDEX Coping, 658 Corner column footings, 514 Corrugated Bar Co., 99, 105, 363 Cost, columns, 445 columns and footings, 577 reinforced concrete buildings, 567 spiral reinforcement, 434 various floors, 786 types of construction, 564 Counterforts, 833, 863, 867 Coverings for roofs, 654 Cracks, 20, 66 cantilever slab, 119 diagonal tension, 40, 147, 153, 157 percentage reinforcement, 300 shrinkage, 300 spacing of, 300 temperature, 299 Crane loads on columns, 464 Cret, Paul P., 629 Cross bars, 37 fulcrum girders, 768 in slabs, 211 over girders, 211 Cross-section Massachusetts Inst. Tech., 808 theater auditorium, 781 Cubes, strength of, 50 Cummings pre-cast pile, 552 Curb bars, 617 Curbs for garage drives, 798 Curtain wall backing, 745 Curtain walls, 572 D Dash coat finish, 736 Dead load, balcony cantilevers, 774 on floors, 571 roofs, 648 é Deere and Webber Building, test on, 113 Deflection, beams, 61, 115 : compressive steel, influence on, 63 end beams and wall columns, 460 flange width, influence on, 62 slabs, 69, 103, 198, 326 ‘steel, influence of, 61 theoretical and actual, 65 INDEX Deformations, concrete, 72, 341 in reinforcement, 28 of columns, plain steel, 89 plastic in slab, 109 relations to stress, 387 spiral colums, 82 steel, 72, 117 ‘Deformed bars, purpose of, 10 Delaware, Lackawanna and Western R. R., 834 Densmore, Le Clear and Robbins, 733, 734, 740, 752, 759 Depth of beam increase by haunch, 284 beams, formula for, 131 cantilevers, 775 footings, 493 foundation, 470 Designation of structural members on plans, 710 Detail of column cap, 760 Detailing methods, 716 D’Hume Ramp System for garages, 801 Diagram, solving stress by, 176 Diagonal tension: allowable unit stress, 244 assumptions for, 153 beam and tile floors, 596 bent bars for, 248, 251 chimneys, 819 combined footings, 527 distribution of between concrete and steel, 245, 246, 247 element of beam under, 152 failure by, 46, 147 footings, 482, 502 for moving loads, 258 formulas for, 148, 247 reinforcement in flat slab for, 350 reinforcement for, 247 requirements for web design, 218 stirrups for, 247, 252 stresses, 241 test, 35, 38, 48 uniformly distributed load, 156 Diamond drill borings, 469 Direct stress and bending, 169. See also Thrust and Bending mo- ment 953 Direct stress and bending, allowable stress, 463 circular section, 179 formulas for design: total section in compression, 177 part section in tension, 180, 182, 186 negative thrust and moment, 167, 189 plain section, 169 reinforced section, 173 sign of bending moment and position of thrust, 166 steel in tensile face only, 189 tension, 162 Distributing beams, 633 Dowels, for columns, 417 on plans, 720 pedestals, 487 retaining walls, 863 Drainage, retaining walls, 834 roofs, 649 Drop panels for flat slabs, 322 advisability of, 322 at wall columns, 324 dimensions, 323 shear stress at, 914 1) Earth, weight of, 833 Earth, pressure, basement walls, 638, 644 distribution of, 835 passive, 839 Rankine’s theory, 835 retaining walls, 841, 842 special cases, 836 surcharge, 837 variation of, 640 Eastern Concrete Construction Co., 60 Eccentric footings, 476 Eccentric loading, 164 See Direct Stress and Bending See Thrust and Bending Moment Eccentric thrust, relation to bending moment, 165 stress caused by, 171 954 Eccentricity, effect of, 177 limiting, 178 Economical, arrangement floor beams, 575 depth, T-beams, 220 design, column reinforcement, 447 rectangular beams, 213 mix for columns, 447 Economy, cantilever vs. counterfort walls, 863 columns, spiral vs. vertical steel only, 449 concrete construction, 564 concrete vs. rubble masonry, 832 deformed bars, 10 floor construction, 806 foundation depth, 470 plain vs. reinforced walls, 832 reinforced concrete, 123 reinforced concrete footings, 481 reinforcing steel, 9 rich mix, 213 structural steel columns, 452 T-beams vs. truss, 772 various types floor, 785 Edison Electric Illuminating Co. Chim- ney, 824 Effective area, columns, 406 section for thrust and bending, 179 width slabs, 72 Elastic flashing, 659 joints, ground slabs, 618 limit, 50 Eleannes reinforcing bars, 14 Elevator pit, 696 provision for additional, 791 shafts, 682, 695 Elevators, warehouse, 791 Eliot St. Garage, Boston, 801 Emperger, 93 Engineering News pile formula, 543, 559 Exits required in buildings, 683 Expanded metal reinforcement, 15, 18 Expansion joints in roofs, 653 Exterior columns, bending moment on, 314 and concrete INDEX Exterior columns, design of, 312, 460 rigidity, 313 Exterior treatment buildings, 728 F fc, allowable unit stress, 879. fs, allowable unit stress, 879 Faber, Oscar, 812 Facing with terra cotta, 753 Factor of safety, 30 columns, 79, 80 definition of, 125 reinforcing steel, 8 retaining walls, 842 Factory, floors, 795 Failure, at footing, 479 cantilever walls, 850 green concrete, 46 retaining walls, 841 torsion, 95 wall with counterforts, 865 False beams, 615 Fillets for T-beams, 146 Finish, brush, 736 exterior, 728 hammer, 737 interior, 755 floor surfaces, 620 Fire exits, 682 protection, reinforcement in beams, 272 Fireproof partitions, concrete, 635 Fireproofed columns, 88, 93, 407 Fireproofing, 422 First baleony fulcrum girder, 783 Flange, compression below, 137 T-beam, compression neglected, 133 T-beams, 36 width, influence deflection, 62 Flanges, shearing in, 37 Flash set of cement, 2 Flashing, 659 Flat ceiling, 610 slab design, 911 Flat slab, action of, 198 action of rings, 373 INDEX Flat slab, advantages of, 304 beams, 377 bending moments, 328, 331, 332, 334 Chicago Code, 398 New York Code, 399 bond stresses, 351 Chicago regulation, 396 columns for, 305 bending moments in exterior, 314 exterior, limitation of, 312 interior, limitation of, 307 column heads, 319 author’s rules, 319 Chicago Code, 397 New York Code, 397 compression at column head, 342 in center, 346 Chicago Code, 401 New York Code, 401 compression reinforcement, 345 description of, 304 design details, 360 design strips, 330 diagonal tension reinforcement, 350 drop panels, 322 authors’ rules, 322 Chicago Code, 397 New York Code, 397 example of design, 390 formulas for area of steel, 351 formulas for bending moments, 331 interior panel, 332 wall panels, 334 formulas for compression stresses, 343 at column head, 344 in center, 346 interior columns, 305 authors’ recommendation, 307 Chicago Code, 396 New York Code, 396 maximum bending moment, tribution of, 330 method of design, 389 New York regulation, 397 openings, in, 375 points of inflection, 199 problems in design, 389 dis- 955 Flat slab, sections of maximum bend- ing moments, 329 separated in simple parts, 371 shearing stresses, 346 critical sections for, 348 formulas for, 349 Smulski system, 369, 376 test of, 104 steel, working unit stresses, 351 Chicago Code, 401 New York Code, 401 stresses at column head, 199 stresses in central portion, 200 suspended action impossible, 387 systems: Acme system, 363 Barton’s spider web system, 359 four-way system, 358 Smulski system, 367 three-way system, 374 two-way system, 362 theory, 194 thickness of slab, 325 authors’ rules, 327 Chicago Code, 397 New York Code, 397 thickness of slab at column head, formula, 338 thickness of clab in center, formula, 340 two-way system, 364, 365 test of, 99 wall beams, 379 above slab, 381 below slab, 379 wall columns, 312 authors’ recommendation, 312 Chicago Code, 397 New York Code, 397 Flexure, 164, 173 Floor, girders supporting joists, 597 joists, 592 loads, 569 plans, 708, 713, 722, 804 surfaces, 620, 627 Floors, balcony, 778 beam and girder design, 575 example, 578 956 INDEX Floors, concentrated loads on light- | Footings, diagonal tension in, 502 weight, 609 concrete and tile, 597, 689 construction of light-weight, 609 cracking of topping, 594 dead loads on, 571 economical arrangement of beams, 575 economy, relative of different types, 786 factory, 795 finish, 805, 806 flat slab construction. slab garages, 795 granolithic finish for, 622 hardwood finish for, 625 hollow tile, 590 hospitals, 807 light-weight, 588 painting of, 623 school buildings, 808 steel tile, 602 structural steel and concrete, 613 tile, 590 two-way tiling, 592 types, 574, 787 warehouse, 791 wood block, 626 wood finish, 624 Footing plans, 721 Footings, 469, 481 See also combined footings area of, 473 bending moments on, 495 bond stresses in, 499 column loads on, 479 combined, 522 concentric, 476 constants for formulas, 495, 497, 498, 502, 515 continuous wall column, 518 corner column, 514 critical sections of plain concrete, 481 of reinforced footing, 496 for diagonal tension, 502 dead loads on, 536 depth of plain table, 481 See Flat eccentric, 476 external shear in, 500 failure of, 479 independent column, 485 independent stepped, 490 piles under, 516 plain concrete, 480 projection of wall, 520 proportions of independent, 491 punching ‘shear on, 492 rectangular, 510 schedules, 721 sloped, 484, 491, 506 sloped vs. stepped, 491 stepped, 484, 487, 490, 506 strap beams for, 533, 536 stresses, variation of, in independent, 489 types, 481 wall, 482 Formulas. See subject notation for, 128 Foundations, 469 at property line, 477 beam and slab, 540 design loads for, 478 encroachment on public streets, 478 flat slab, 539 outside columns, 478 plans, 712, 721 plates for, 417 raft, 538 steel cylinder, 562 Foundry for Garfield, Smith & Co., 666 Four-way flat slab details, 718 system reinforcement, 358 Framing plan, 714 Frank E. Davis Fish Co., building, 733 Franklin Service Station, 802 Freytag, 87, 88 Frictional resistance of piles, 543 Frost, power of, 835 Fulcrum girders for 769 Furring, 614 inside walls, 746 theaters, 765, INDEX G Garages, 795 column spacing, 797 inclined driveway, 799 live loads, 796 Garfield & Smith Co., 665 German building rules, 202 Giant pre-cast pile, 552 Girders, numbering on plans, 711 schedules, 715 supporting joists, 597 Glass, sizes of, 705 Glazing, 703 ' Glenday, Charles, 812 Goldbeck, A. T., 69 Goosenecking, 415 Gow, Charles R., 547 Graf, O., 95 Granolithic floor finish, 609, 620, 623 Graphical spacing of stirrups, 252 Grauman’s Theater, 671, 674, 764, 779 Gravell, William H., 629 Gravity walls, design of, 839 Gymnasiums, 810 Gypsum tiles, 592 H Hammered surface, 738 Hardwood floors, 625 Hatt, W. K., 99, 104, 112 Haunches, 284, 405 Havemeyer bars, 13 Heel cantilever walls, 852 Height of concrete buildings, 803 Herringbone lath, 611 Hickey for bending bars, 355, 415, 727 Hide and Leather Building, 738, 804 Higginson, Wm., 681 Hinged effect, bottom of columns, 668 Hollow circular beams, 822 tile, 589, 590, 598, 809 Homogeneous beams, stresses-in, 241 Hooks in reinforcement, 59, 60, 63, 266, 269 Hoops. See Spiral reinforcement chimney reinforcement, 819, 825 strength, 820 957 Hospitals, 805 design, 808 State of Wisconsin, 807, 808 Hotels, 805 Howard, J. E., 80 Hughes, C. A., 97 Humidity, 652 Humphrey, Richard L., 25, 27, 62 Hy-rib, 19, 611 I Ideal Shoe Factory, 793 Identification tags on bars, 723 Imbedment, formulas for, 267 length, 267, 462 Imitation marble, 760 Inclined drive, garage, 799 stirrups, 243 Independent footings with piles, 516 Inertia, moment of, 174, 281 Inflection, coefficient of, 65 points of, 110, 289 Inserts for shafting, 615 Insulation of roofs, 650 Interior, cclumns, 305, 310, 312 finish, 755 office building, 761 panels, bending moment in, 331 J j, formulas, rectangular beams and slabs, 130, 207 steel top and bottom, 240 T-beam, 134, 221 j, rectangular beams and slabs, tables, 205, 880, 881 j, steel top and bottom, diagram, 910 4, T-beam, table, 221, 897 Jetting, pre-cast piles, 551 Joint Committee, year 1916, 246, 448 year 1924, 161, 245, 246, 282, 324, 331, 333, 337, 405 1924 Full Report. See Vol. IT Joist, depth, 596 loads from slabs, 74 Joists for floors, 594 958 K k, diagrams, direct stress and flexure, 938, 940 rectangular beams, 883 k, formula, direct stress and flexure, 183, 198 rectangular beam, 130, 207 steel in top and bottom, 139 T-beam, 134, 137 k, tables rectangular beams and slabs, 205, 880, 881 T-beam, table, 897 Kahn, Arthur, 803 Kahn trussed reinforcing bars, 13 Kellog, Harold Field, 748 Kenneth, M. de Vose and Co., 748 Klein, Arthur F., 785 L L-shaped retaining walls, 849 Lacher design, walls, 872 Laitance, 5 Lap, plan bar, 121 Lapping of bars, length, 414 Life Savers Inc. Building, 743 Limestone, 272 Lindenthal, Gustave, 834, 873 Linoleum floor finish, 805 Lintel beam, 103 Live load, balcony cantilevers, 774 beams, table of, 884 building codes, 454 buildings, 452, 570, 805 cold storage, 792 columns, 453 floors, 569 garages, 797 reduction for columns, 452 authors’ recommendation, 453 Boston Code, 454 Chicago Code, 454 New York Code, 454 reduction for large areas, 454 school floors, 808 Load transference, flat slabs, 386 Loading, concrete beams, 20, 884 INDEX Loading, concrete chimneys, 812 eccentric, 164 effect on neutral axis, 27 effect on reinforcement, 27 floors, 569 for foundation design, 478 office floors, 805 platforms, 790 roofs, 648 slabs, 69, 102, 886 slab stress at maximum, 109 slabs to joists, 74 tests of buildings under, 113 Lockwood, Greene Company, 743 Long columns, 434 Long column span roof construction, 661, 671 Longitudinal reinforcement. See also Vertical reinforcement. for shear, 146 fulcrum girders, 768 steel, columns with, 405 Lord, A. R., 88, 113 Losse, L. H., 25,27, 62 M MacMillan, A. B., 626, 627 Manufacturing buildings, 792 Marking of bent bars, 724 Mass. Institute of Tech., facades, 754 framing plan, 714 section, 808, 810 Maurer, EK. R., 220 Medical Arts Building, Dallas, 803 Melan arch, 122 . Mensch, L. J., 671 Metal flashing, 659 lath, 18 roofing, 656 Mills, Charles M., 299 Minimum depth T-beam, 219 Mix of concrete, basement slabs, 619 concrete chimneys, 823 concrete columns, economy of, 447 for use, 4 fulcrum girders, 772 granolithic finish, 620 INDEX Mix of concrete, influence of, 24, 80, 83, 87 pre-cast piles, 550 stress in steel and concrete, due to, 87 Modulus of elasticity, 31, 80 columns, 76 concrete, 38, 202 effect of, if constant, 127 for beams, 127 variation with stresses, 142 rupture, plain concrete beam, 48 Moisture protection of steel, 273 Moment arm, formulas, beam with compression steel, 240 rectangular beams, 130, 207 T-beam, 134, 221 tables and diagrams, beams with compression steel, 910 rectangular beams, 205, 880, 881, 883 T-beams, 897 Moment coefficients, 102 Moment of inertia, effect on bending moment, 281 for different sections, table, 945 Moment of resistance, rectangular beam, 130 T-beam, 136 — Monotype reinforcing bars, 14 Morris, C. T., 70, 74 Morrow, D. C., 374 Moving loads, diagona: tension for, 258 Mullions, 699 Muntins, 699 Mushroom system of reinforcement, 359 N- National Board of Fire Underwriters, 685 National Fireproofing Company, 590 Negative thrust, 165 Neutral axis, formulas, direct stress and flexure, 183, 193 rectangular beam, 130, 207 steel in top and bottom, 139 959 Neutral axis, formulas, T-beam, 134, 137, 224 tables and diagrams, direct stress and flexure, 938, 940 rectangular beams, 205, 880,881, 883 T-beam, 897 tests, 27, 35 New Channon Building, test, 342 New York Building Code, 160, 218, 322 ender concrete, 612 columns, 396, 408, 421, 424, 437, 448 450, 464 flat slab regulations, 347, 396 live loads, 454 panels, 399, 400 piles, 544 smoke-proof stair towers, 686 spiral columns, 426, 918 stairways, 685 steel cylinder foundation, 562 New York Connecting R. R., 873 Nichols, John R., 196 Nielson, N. S8., 196 Norcross system of reinforcement, 358 Northup, W. C., 784 Notation, 128 Numbering of columns, 710 of girders, 711 O Oberlander, J. T., 833 Office buildings, 803 floor plan, 804 interior, 761 Office, Salada Tea Co., 759 Openings, roof, 660 slabs, 375 Orchestra floor, 763 Organic matter, effect, 2 Ornamental ceilings, 614 facades, 754 Outside columns, foundations for, 478 ie p, values for beam and slabs 882 p’, ratio for given pi, 904 960 pa, values in spiral columns, 931 Paine Furniture Company’s roof, 651 Painted concrete, 758 Painting window sash, 703 Panels. See also Interior panels. See also Wall panels. . building codes for, 400 sizes, warehouses, 789 steel tile construction, 603 wall vs. interior, 104 Parabolic distribution stress, 30, 143 Parallel bars, spacing, 273 Parapets, 656 Passive earth pressure, 839 Pedestals, 486 pile, 557 Peerless pile, 557 Penetration of piles, 543 Penn Charter Gymnasium ceiling, 607 - Pent house for elevator shaft, 696 Pent houses, 678 Philadelphia Building Code, 218 columns, 408, 421, 437, 448, 450, 464 Pier 6, roof truss, Cristobal, 677 Piers, strength of brick, 77 Pilasters, 632 Piles, 477, 542 See also Cast-in-place piles See also Pre-cast piles Bignell pre-cast, 552 brooming, 545 caisson, 560 capacity, cast-in-place, 558 cast-in-place, 554 Chenoweth pre-cast, 551 composite, 555 concrete, 546 Cummings pre-cast, 552 cutting of wood, 544 driving, 545 driving pre-cast piles, 550 Engineering News formulas, 543 frictional resistance, 543 Giant pre-cast, 552 Gilbert pre-cast, 552 inclined, 545 independent footings, 516 INDEX Piles, outside bearing walls, 543 pedestal, 557 Peerless cast, 557 penetration, 543 pre-cast, 546 raft foundation, 539 Raymond cast, 554 retaining walls, 843 Simplex cast, 555 spacing, 542 stresses in pre-cast, 548 Wellington formulas, 543 wooden, 543 Pipes imbedded in concrete, 357 Pitch of roofs, 649 of spiral reinforcement, 431 Plain concrete, columns, 403 factory, 729 footings, 480. section, direct stress and bending on, 169 general, formula for stress on, 169 Plan, Pier 6, roof truss, Cristobal, 678 Plans, building, 707 flat slab building, 718 floors, 713 foundations, 712 working, 720 Plaster casts, 757 finish, 758 Plastic deformation of concrete, 34. Plate anchorage, 163 Platforms for access, 690 Plinth, 322 Point of bending reinforcement, 287 Points of inflection, 110 beam and tile floors, 595 continuous beam, 289 effect of reinforcement on, 386 flat slabs, 119, 336 Poisson’s ratio, 76, 121, 197, 199 Post, George B. & Son, 750, 800 Pre-cast piles, 546, 547, 548, 550 roof trusses, 677 Pressure on raft foundation, 538 wall foundations, 841 Pressures on retaining walls, 836 Prior, J. H., 872 INDEX Probst, T. E., 64, 67 Properties of sections, 945 Proportions for concrete, 4 Protective covering, reinforcement, 272 Pull-out tests, 51 Punching shear, 492 R R, values of, for beam and slab, table, 205, 880, 881, 882 diagram, 883 r, effect of cost of formwork on, 221 Raft foundation, 538 Railroad sidings for warehouses, 791 Ramps for garages, 798 . circular, 799 D’ Hume, 801 Ramp Building Corporation, 797 Rankine’s theory, earth pressure, 835 ‘Raymond pile, 554 Recesses on plans, 720 Rectangular bands, 88 Rectangular beams, see also Beams approximate formulas, 207 balanced design, 202 compression reinforcement, 232 economical design, 213 formulas for design, 203 review, 206 plain sections, stress formula, 170 safe loads, table, 884 stress distribution, 180 theory, 127 table of constants, 880 Reduction of live load on columns, 452 for large areas, 453 Reglets, 658 Reinforced concrete bearing walls, 634 definition, 122 Reinforced concrete, definition, 122 Reinforcement, 7 allowable stresses, 8, 879 anchorage of. See Anchorage. angle of bend, 291 bearing stresses, 54 bending bars, 355 961 Reinforcement, bending diagrams for, 291 bond stresses, 10 _ circumferential vs. band, 117 Clinton wire fabric 17 compression, 50 corrosion of, 9 definition of, 7 deformation in, 28, 72 deformed bars, 10 diagrams to determine points of bending, 292-297 Elcannes bars, 14 expanded metal, 15 factor of safety, 8 failure of, 22 foor plan drawing, 726 four-way flat slab system, 353 fulcrum girders, 767 grade steel, 8 hard steel, 9 Havemeyer bars, 13 identification tags for, 723 intermediate grade, 9 Kahn new rib bars, .13 Kahn trussed bars, 13 lapping of bars, 413, length of inbedment in columns, 462 lap, 414 spiral, 433 metal lath, 18 minimum, for columns, 413 for economy, 136 moisture protection for, 273 Monotype bars, 14. necessity of securing in place, 356 neutral axis, influence on, 27 over doors, ete., 634 per cent, influence of, 25 pitch. of spiral, 432 plain vs. deformed, 10 point of bending in beams and slabs, 209, 287, 290 properties of bars, 12 properties of bars, steel, 7 protection of, 9, 272 ratio in slabs, 211 T-beams, 894 962 Reinforcement, rib lath, 17 rings, action of, 373 shapes, 10 shock, protection from, 10 shrinkage stresses, 298 slabs over girders, 211 Smulski flat slab system, 367 spacing for aggregates, 273 splicing of, 61, 265, 267 steel, brittleness of, 10 — structural shapes for columns, 88, 44] supports, 209 tables of sizes, 942 temperature, in walls, 855 temperature stress, 298 temperature under, 9 three-way flat slab system, 374 torsion, resistance to, 94 triangle mesh, 17 two layers, 30 two-way, flat slab system, 352 vertical spacing, 274 weight of spiral, 433 wire, 14, 17 yield point, significance of, 202 Relative cost of buildings of different materials, 564 Repeated loads on columns, 77 Reservoirs. See Vol. III Restraint of ends of beam, 275 slabs at wall, 333 Retaining walls, 832 cantilever, 849 cellular, 869 drainage, 834 economy of types, 832 factor of safety, 842 failure, 841 L-shaped, 849 line of pressure on, 840 piles under, 843 pressure on foundation, 846, 848 sliding of, 838 T-shaped, 849 temperature reinforcement, 855 upright slab, 850 Rib lath, 19 INDEX Rich mix, economy in rectangular beams, 214 “ economy in slabs, 213 Rigby, Edward H., 835 Rigid frame, with arched roof, 667 roofs, 665 Ring reinforcement, 370 Rings, effectiveness as reinforcement, 352, 373 splicing, 121 stress distribution in, 110 Riplex, 611 Risers, stairs, 688 Road construction, 12 Roadbed, distribution of load on, 75 Roads and Pavements. See Vol. II Robbins, Henry C., 728 Rock Island R. R., 834 Roof, air space below, 652 arched with rigid frame, 667 cinder fill for, 653 concrete arches for, 668 condensation under, 649 construction, 648 coverings, classes, 654 dead loads, 648 design, 659 drainage of, 649 expansion joints in, 653 foundry for Garfield Smith & Com- pany, 666 girder, Winston-Salem Auditorium, 663 insulation of, 650 long span, 661, 668, 671 openings in, 660 Paine Furniture Company, 651 pitch of, 649 rigid frame, 665 sawtooth, 675 theater, 778 trusses, concrete, 671 Grauman’s theater, 674 pre-cast, 678 walks on, 659 wind pressure on, 649 Rubble masonry, 832 Runs and platforms, 690 INDEX S Safe loads, rectangular beams, 884 slabs, 886 Sagar, W. L., 97 Salada Tea Company Building, 752, 757, 759 Sand, organic matter in, 2 testing, 3 Sash, dimensions, 700 standard sizes, 705 Underwriters’ pivoted, 706 Sawtooth roofs, 675 Scagliola, 760 Scheit, H., 61, 64 Schmidt, Garden & Martin, 671, 681 School buildings, 808 Scuttles to roof, 686 Sea water, concrete in, 6 Sections, properties of, 945 Separate bending sketches for bars, 726 Settlement, foundations, 472 unequal, 472 Shafting inserts in ceiling, 615 Shapes, reinforcement, 10 Shear, beams, 143, 244 cantilevers, 773 chimneys, 817, 819 column heads, 913 definition of, 143 diagram, moving loads, 258 drop panels, 914 external, magnitude of, 144 footings, 500 formula for stress, 148 horizontal, 144, 150 longitudinal, cause of, 144- longitudinal T-beam, 144 magnitude, solution for, 145 strap beams, 533 stress, 148, 148, 151, 244, 346 vertical, 56, 144, 150 Shepard & Stearns, 803 Shock on concrete, 10 Shrinkage stresses, reinforcement for, 298 Sidewalks. See Vol. II Simplex pile, 555 963 Simplex system reinforcement, 359 Sinks, Frank F., 362 Skeleton concrete buildings, 572 Skylight openings, 660 Slabs. See also Basement slabs See also Flat slabs area round bars, 945 square bars, 945 arrangement of reinforcement for, 210 basement, 618 bending moment in continuous, 209, 278 bending of reinforcing bars for, 210 breaking load for wide, 72 cantilever, tests on, 116 constants, table of, 880, 881 for selected stresses, 205 continuous, 208 cross reinforcement of, 211 over girders, 211 design, 205, 208 constants for, 205, 880, 881 tables, 886, 888-893 details and schedule, 716 on floor plan, 726 distribution of load, square or ob- long, 212 eccentric loads on, 70 economy of rich mix, 213 effective width for concentrated load, 69, 70, 72 example, 579 formulas for design, 208 reinforcement, arrangement of, 210 formula for, 211 over girders, 211 safe loads on tables, 886, 888-893 span of, 209, 277 steel, points of bending, 209 ratio, 211 stresses in, 131 structural steel beams, 611 stresses in wide, under trated load, 69 Slabs, supported by four beams, 211 temperature steel for, 618 tests, 99, 116 concen-= 964 INDEX Slabs, thickness, formula for, 208 Spiral columns, formulas, Boston, two-way hollow tile and structural Cleveland, and Philadelphia steel, 613 Codes 4222) Slate shingles, 656 Slater, W. A., 31, 113 Sleepers, 624 Sliding of retaining walls, 838 Slip of tension bars, 51 Sloped footings, 490 Smoke-proof stairways, 686. Smulski, Edward, 99, 110, 111, 112, 116, 224, 233, 367, 784 Smulski flat slab tests, 104 Smulski flat slab system, action of, 369 flat slab sections, 719 reinforcement, 353, 367 _ schedule, 727 Soil, bearing power of, 470 carrying capacity of, 469 definitions of varieties, 471 effect of organic matter on bearing power, 470 unequal settlement, 472 Southern Engineering Company, 784 Spacers for reinforcement, 275, 357 bar, 211 spiral columns, 433 Spacing, area bars for various, 945 columns, 793, 797, 804 reinforcement. See type in question Spalling, 91 Span length, 277 Spandrel, beam faced with brick, 746 beams, 334 loads, 459 torsion, 336 Spatter coat, 736 Special sash, 705 Spiral columns, 419 See also Columns description, 419 design, 431 details, 482 economical design, 448 equivalent area vertical steel, 432 columns, example, 456 formulas, authors’ recommendation, 421 Chicago Code, 425 New York Code, 424 oblong, 430 octagonal, 429 ratio length to diameter core, 422 relative economy, 449 square, with round spiral, 429 square spiral not effective, 430 tests on, 82 Spiral reinforcement, columns, 160, 419 cost, 434 length, 433 pitch, 431 relation of pitch spiral, 431 weight, 432 Spitzer, 79, 93 Splices in reinforcement, 61 Splicing, rings, 121 staggering of, 163 structural columns, 441 Spouting of concrete, 357 Sprinklers, 577 Stairs, 682, 810, 811 capacity, 684 enclosed, 685 intermediate landings, 690 landing beam, 691 layout of, 686 methods of design, 690 nosing, 688, 689, 694 reinforcement, 693 slope of, 689 special design, 630 three-flight, 693 treads and risers, 688 two-flight, 690 winding, 685 Stairways, smoke proof, 686 Standard sizes of sash, 705 Standards for window sash, 699 Statical limitations, flat slab theory, 194 Statler Garage, 750, 800, 801 Steel. See Reinforcement bearing plates, 633 and ratio of INDEX 965 Steel, brackets, 442 columns, 314 cylinder foundations, 562 ratio modulus to that of concrete, 201 tile, dimensions, 604 tile floors, 602 window sash, 699 Stepped footings, 487 Stirrups, 36, 38, 96, 242, 251 action of, 153 area of vertical, 154, 248 counterforts, 868 design tables, 899 diagonal tension in, 153 formulas for design, 247, 248 fulerum girders, 768 graphical spacing, 252 held by top bars, 290 inclined, 243 limiting spacing, 156, 250 ) number of, for uniformly distributed loading, table, 900 spacing depends upon type of loading, 253 spacing for vertical, 154, 247 | for inclined, 156, 249 for uniformly distributed load, table, 901 spacing table, 899 support for, 290 theory, 147 types of, 243 Stone & Webster, 755 Stone exterior, 753 Straight-line formula, 127, 142 theory, 123 Strap beams in footings, bending at interior columns, 537 connecting footings, 533 loading, 534 pile foundation, 536 types, 538 Strength, brick piers, 77 Stresses, actual and computed, 24, 125, 128 allowable unit, table, 879 allowable, in concrete at supports, 223 Stresses, assumptions of distribution in rectangular beam, 143 bearing in hooks, 163 direct and flexure. See stress and bending eccentric thrust, caused by. See Direct stress and bending effect of moduli on, 202 interdependence between steel and concrete, 206 parabolic distribution of, 30, 143 relation to deformation, 387 rings, distribution in, 110 steel and concrete compared, 87 unit, 247 variation, 127, 143 wall with counterforts, 865 Structural steel, and concrete balcony cantilevers, 776 and concrete columns, difficulties of, 452 and concrete columns, economy, 452 and concrete floors, 611 Structural steel columns, base for, 443 brackets for, 443 encased in concrete, details of, 440, 435 Support of veneer, 751 Supports, tension in steel at, 288 working stresses at, 282 Surcharge, 638, 837 Surfaces, coloring of, 737 floor, 620 rubbing of concrete, 735 treatment of concrete, 734 Suspended ceilings, 19, 6138, 652 Salada Tea Building, 760 Suspension action of flat slabs, 387 Symbols, table of, 128 Direct T T-beams, 35 area of steel in, 135, 221 compression below flange, 133 simplified formula, 224 compression flange, 116, 217 reinforcement, 225, 230 966 INDEX T-beams, depth at center governed | Tensile strength of concrete, 21 by compression, 219 design, 215 design diagrams, 894, 895 tables, 897 details of design, 223 diagram of steel ratio, 894 economical depth, 220 exact vs. approximate formulas for, 221 failure by tension, 35 fillets, effect on strength, 146 flange, effective width of, 36 formulas for design, 216 simplified, | compression flange, 224 simplified irregular flange, 228 unsymmetrical flange, 142 formulas, derivation of, 133 fulerum girders, 767 how obtained, 215 longitudinal shear in, 144 minimum depth, 135, 220 minimum depth at support, con- tinuous beam, 222 moment arm for, 221 moment of resistance, 136 neutral axis for, 133 table, 897 shearing stresses in flanges, 37 stresses in, 224 test of, 35 theory of, 133 vs. truss, 772 width of flange, 36, 217 T-shaped retaining walls, 849 Tags on reinforcing bars, 723 Talbot, Arthur N., 28, 35, 47, 76, 80, 83, 88, 89, 93, 118, 148 Taper of piles, 542 Tapered tile, 605 Tar and gravel roofs, 654 Taylor, C. Percy, 812 Temperature in chimneys, 822 Temperature reinforcement, ment walls, 635 retaining walls, 855 stresses, reinforcement for, 298 below base- Tension, and bending combined on rectangular section, 191 center of, in chimneys, 816 concrete beam in, 140 diagonal, 147 direct, 162 entire section, subjected to direct, 190 members, composition of, 162 thickness of, 164 trusses, 770 Tension reinforcement of columns, 461 of slab across beams, 146 © steel, center of span, 287 Tension, failure by, of T-beam, 35 Teredo, 542 Terra cotta exterior, 753 Terrazzo floors, 623, 805 Tests, 20 beams with compressive reinforce- ment, 50 bend test of bars, 8 bond of concrete and steel, 51 cement, 2 columns with vertical steel, 78 spiral reinforcement, 82 ¥ structural steel, 88 continuous beams, 115 cracks in beams, 115 Deere & Webber Building, 113 diagonal tension, 38 flat slab construction, 98 long columns, 93 octagonal flat slab, 116 plain concrete columns, 75 pull-out of bars, 51 sand, 3 ¢ slabs, 71, 100, 117, 120 slump, 4 Smulski system, 104 T-beams, 36 Turner-Carter Building, 115 value of, for structures, 113 Wenalden Building, 114 Test pits for ground exploration, 469 Theater balcony truss, 779 Theaters, concrete construction, 762 INDEX Theory, flat slab, 194 least work, 313 rectangular beams, 126 reinforced concrete, 122 Thompson & Binger, 665, 738 Thompson, Sanford E., 116, 621, 755 Three-flight stairs, 693 Three-way system of reinforcement, 374 Thrust and bending moment, 164 allowable unit stresses, 463 character of stresses, 463 circular sections, plain, 172 reinforced, 179 columns under, 458 design of, 460 condition of loading for maximum stress, 187 diagrams for design: part of section in tension, 938, 939 tension side reinforced, 940, 941 whole section under compression, 934-936 effective section, 179 general formula, 169 negative thrust, 190 part of section under compression, 191 whole section under tension, 190 plain section, 170 relation between position of eccentric thrust and sign of bending moment, 166 reinforced section, general formulas, 173 circular, 179 * one surface intension, 180 rectangular, 175 reinforcement in tension face only, 185 sign of, governed by bending mo- ment, 165 Tie rods, long arched roofs, 669 retaining walls, 872 Ties for facing, 751 sag in, 163 Tile floor, construction, 589 assumptions, 967 Tile floor, weight, 590, 597 Tile insulators, 653 Tiles, adjustable metal, 605 gypsum 592 hollow clay, 590 metal, 611 removable, 611 tapered, 605 Toe-cantilever walls, 852 Top bars to hold stirrups, 290 Topping for hollow tile floors, 593 for metal tile floors, 608 pouring, 610 Toronto tests, 97 Torsion, cause of, 94 failure by, 95 lintel beam, 104 resistance of concrete in, 94 spandrel beams, 336 stresses due to, 94 ultimate stress, 96 Traprock, 272 Trautwine, John C., 840 Treads, stairs, 686 Treatment of concrete surfaces, 734 Triangle mesh, 17 Truck loads, distribution on road bed, 75 Truck-loading platforms, 791 Trucks, dimensions of, 797 Trusses, 769 Turneaure, F. E., 22, 220 Turner,’ C: P? A;, 359 Turner-Carter Building, test on, 115 Turner Construction Company, 729, 744 Two-flight stairs, 690 Two-way system reinforcement, 362 flat slab tests, 99 Types of concrete buildings, 785 floor construction, 785, 787 U Unbalanced design, rectangular beams, 132, 203 Underwriters’ pivoted sash, 706 908 Unequal settlement, footings, 537 raft foundation, 539 Uniform loading, stirrups for, 253 Uniformly distributed load, diagonal tension with, 156 United Fireproofing Company, 760, 800 Universal Portland Cement Company, 565 V Villadsen Brothers, 665 Variations in bending moment, 335 Vertical bars for columns, 434 Vertical shearing, measure of diagonal tension, 241 spacing, reinforcement, 274 Voids due to reinforcement, 273 W Walks on roofs, 659 Wall beams, 379, 382 general remarks, 385 Wall bearing construction, 572, 629 vs. Skeleton construction, 809 column footing, continuous, 518 columns, anchorage of steel, 462 bending of, 458 deflection of, 460 footings, 482 foundations, 841 panels in flat slab, 333 panels in flat slab, New York Code, 400 Chicago Code, 400 restraint of flat slabs, 333 tie rods, 872 Walls. See also Basement walls See also Retaining walls basement, 637 supported at columns, 643 supported top and bottom, 638 concrete, 634 waterproofing of, 637 Warehouse, column spacing construction, 785 elevators, 791 floor finish, 791 INDEX Warehouse, live loads, 453, 789 panel sizes, 789 wash borings, 469 Wawrzyniok, O., 61 Washington Garage, inclined drive- way, 799, 800 Water jet, pile driving, 545 Water pressure on basement, 619 Waterproofing walls, 746 Watertight work, 6, 98 Water tower, Clark Biscuit Company, 680 Watertown Arsenal Tests, 77, 80 Wayss, 87, 88 Weather tightness, curtain wall, 382 Web area of T-beam, 218 determined by diagonal tension, 218 resistance, effect of age on, 47 Web reinforcement, 242 action of, 152 anchorage of, 244 area and spacing, 248 cantilevers, 158 i continuous beams, 157 description of, 243 design of, 251 proportion of diagonal tension resisted by, 153, 245 tests, 48 theory, 147 usefulness of, 157, 242 Web resistance, effect of horizontal steel on, 47 Wedge-shaped beams, 140 Weight, clay tile floors, 597 earth, 833 of bars, table, 942 Welded wire fabric, 17 Wellington formulas for piles, 543 Wenalden Building Test, 114 Westergaard, H. M., 196 Western Newspaper Union Building, 370 destruction test, 346 Width of flange, T-beams, 217 Wind, chimney loads, 812, 825 pressure on roofs, 649, 669 INDEX 969 Winding stairs, 685 Working stresses, table of, 879 Window opening, 705 sash, 699, 701, 703 | Y inston-Sal Auditori 663, 762 : : ; Bees a8 7a0) 78 eas 3 : ’ | Yield pont effect of reinforcement on, Wire reinforcement, 14, 432 Arena Wiren, George R., 748 5 is re fin 93 Withey, M. O., 48, 58, 59, 79, 83, 91, 93 ee a ee Young, C. R., 97 Wood, block floors, 626 3 ee Pie floors, 624, 805 Youths’ Companion Building, 740, 793, piles, 543 794 cutting top of, 544 4 sleepers for floors, 624 Working plans, 721 Zipprode, R. R., 31 eceer reeee eee cece Oe CG reoree seers eeere * eoreo peeer e2 er © R : C3 i) r enecage & Ser * ath a ata “at 2 Date Due ne por RPP ica bare pet aS serene, ra ae t Bis jiter nt 7 \ ¢ y tons pene ES — as x