OK8INDERS iitli First Sirt-c N JOSC, CAL. ^, G, :D|VlliH, ^, G. SlVlllt^ Safe Building, Safe BaiLDiNG A TREATISE GIVING IN THE SIMPLEST FORMS POSSIBLE THE PRACTICAL AND THEORETICAL RULES AND FORMUL/E USED IN THE CON- STRUCTION OF BUILDINGS BY LOUIS DE COPPET BERG, F. A. I. A. ASSOCIATE AMERICAN SOCIETY OF CIVIL ENGINEERS VOLUME ONE SECOND EDITION, REVISED BOSTON TICKNOR AND COMPANY 2U Cremont .Street 1890 Copyright, iS86, 1S87, 18SS, and 1889, by TICKNOR AND COMPANY. All rizhts reserved. S. J. Parkhill & Co., Printers, Boston. To MY FRIEND AND PARTNER, J. CLEVELAND CADY, M. A., F. A. L A., THIS BOOK IS AFFECTIONATELY DEDICATED BY THE WRITER. 201605'; •^ TABLE OF CONTENTS. PAGE LIST OF TABLES "*'"' ^"^ LIST OF FORMULAE INTRODUCTORY REMARKS IX-XU 1 CHAPTER I. Strength of Materials 2-84 CHAPTER 11. Foundations 85-95 CHAPTER III. Cellar AND Retaining Walls 96-116 CHAPTER IV. Walls AND Piers 117-153 CHAPTER V. Arches 154-179 CHAPTER VI. Floor-Beams and Girders 180-240 CHAPTER VII. GRArmcAL Analysis of Transverse Strains. . . . 241-271 LIST OF TABLES. Table I. — Distance of extreme fibres, i, r, a, p-, for different sections Table II. — Value of n in compression formula Table III. — 'Wrinkling strains .... Table IV. — Strength and weights of materials Table V. — Strengths and weights of stones, bricks, and cements Table A''I. — Weights of other materials . Table VII. — Bending moments, shearing and deflection Table VIII. — Strength of 1" wooden beams Table IX. — When to calculate rupture or deflection Table X. — Angles of friction of soils, etc. Table XI. — Values of p and slope for retaining walls Table XV. — Iron I-beam girders .... Table XVI. — A-'alue of y in formula for unbraced beams Table XVII. — Continuous girders Table XVIII. — Trussed beams Table XII. — Wooden floor-beams Table XIII. — Wooden girders Table XIV. — Iron I beams for floors Table XIX. — Explaining Tables XX to XXV Table XX. — List of iron and steel I beams page 12-21 23 26 37^2 43-45 46 58,59 62 63 09 101 199 205 218,219 220, 221 at end of text. VIU LIST OF TABLES. Table XXI. — List of iron and steel channels . . at end of text. Table XXII. — List of iron and steel even-legged angles, " Table XXIII. — List of iron and steel uneven-legged an- gles " Table XXIV. — List of iron and steel tees ... " Table XXV. — List of iron and steel decks, lialf-decks, and zees .-..., " LIST OF FORMULA. NO. OF PAGE FORMULA. 1. — Fundamental formula — Stress and Strain . . . . • 21 22 2. — Compression, snort columns 04 3 " long columns • * " 05 4..— Wrinkling strains 28 5. —Lateral flexure 30 6. —Tension . 32 7.— Shearmg, across gram 30 8.— " along grain 9__ " at left reaction "'' 10. " at right reaction "^^ ll._Vertical shearing strain, any point of beam . . • . • 3i ^^ (( " " " cantilever ... 35 35 13. — Longitudinal shearing 14. —Amount of left reaction (single load) ^"^ 15. _ " right . . , . ■ 1(3___ " left " ■ (two loads) ** 17.— " right .... 18. — Transverse strength, uniform cross-section .... oO " upper fibres ^^ " lower " ... 21. — Central bending moment on beam, uniform load 22.- " 19. 20.- .... 50 52 centre load .... 5'-^ 23.- 24.- 25.- 26.- 27.- 28.- 29.- 30.- 31.- 32.- 33.- 34.- 35.- 30.- 37. - 38.- 39. ■ 40. ■ 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. hl^T OF KOU.MUL.K. ■ Bending moment, au}- point, several loads 53 » '< " " clieck to No. 23 . 53 " " cantilever, uniform load 53 '< " " end load 53 " " " any loading 54 - Safe deflection, beams, not to crack plastering .... 56 - " " cantilevers or uneven loads on beams, not to crack plastering . . . 5t - Comparative formula — beams, strength, different cross-sections, 60 « '< " deflection " " 61 " " " strength, different lengths and sections, 02 << " " deflection, " " 62 " " columns, different lengtlis ... 65 " " " different sections ... 65 - Approximate formula for plate girders 65 -Deflection, cantilever, uniform load 66 " " end load 66 " beam, uniform load 66 " " centre load 66 _ " " any load 67 " cantilever, " 67 -Point of greatest deflection of beam 67 -Strain on arch edge nearest to line of pressure . ... 78 " " farthest from " 79 - Safe load on long piles 93 - Amount of pressure again. • 1 , Factor-of- culate so closely. Besides we can never determine Safety. accurately the actual ultimate stress, for different jiieces of the same material vary in practice very greatly, as has been often proved by experiment. Therefore the actual ultimate stress might be very much less than that calculated. Again, it is impossible to calculate the exact strain that will alwavs take place at a certain point ; the applied forces or some other con- ditions might vary. Therefore, to provide for all possible emergen- cies, we must make our material strong enough to be surely safe ; that is, we must calculate (allow) for a considerably greater ultimate stress at every point than there is ever likely to be strain at that point. The amount of extra allowance of stress varies greatly, according to circumstances and material. The number of times that we calcu- late the ultimate stress to be greater than the strain is called the fac- tor-of-safety (that is, the ratio between stress and strain). If the elasticity of different pieces of a given material is practi- cally uniform, and if we can calculate the strain very closely in a given case, and further, if this strain is not apt to ever vary greatly, or the material to decay or deteriorate, we can of course take a low or small factor-of-safety ; that is, the ultimate stress need not exceed many times the probable greatest strain. On the other hand, if the elasticity of different pieces of a given material is very apt' to vary greatly, or if we cannot calculate the strain very closely, or if the strain is apt to vary greatly at times, or the material is apt to decay or to deteriorate, we must take a very high or large factor-of-safety, that is, the ultimate stress must exceed many times the probable greatest strain. Factors-of-safety are entirely a matter of practice, experience, and circumstances. In general, we might use for stationary loads : A factor of safety of 3 to 4 for wrought-iron and steel, " " " 6 for cast-iron, " " " 4 to 10 for wood, " " "10 for brick and stone. For moving-loads, such as people dancing, machinery vibratin"-, dumping of heavy loads, etc, the factor-of-safety should be one- half larger, or if the shocks are often repeated and severe, at least double of the above amounts. Where the constants to be used in formuliB are of doubtful authority (as is the case with most of them for woods and stones), the factor-of-safety chosen should be the hi"-h- est one. SAFE BUILPING. In building-materials we meet with four kind of strains, and, of course, with the four correspondiag sti-esses resisting them, viz.: — Compression, or crushing strains. Tension, or pulling strains, Shearing, or sliding strains, and Transverse, or cross-breaking strains. STRESSES. The resistance to Compression, or crushing-stress. The resistance to Tension, or pulling-stress. The resistance to Shearing, or sliding-stress, and The resistance to Transverse strains, or cross-breaking stress. Materials yield to Compression in three different ways : — 1. By direct crushing or crumbling of the material, or 2. By gradual bending of the piece sideways and ultimate rupture, or 3. By buckling or wrinkling (corrugating) of the material length- wise. ^laterials yield to Tension, 1. By gradually elongating (stretching), thereby reducing the size of the cross-section, and then, 2. By direct tearing apart. Materials yield to Shearing by the fibres sliding past each other in two different ways, either 1. Across the grain, or 2. Lengthwise of the grain. Materials yield to Transverse strains, 1. By deflecting or bending down under the load, and (when thia passes beyond the limit of elasticity), 2. Bv breaking across ti-ansversely. In calculating strains and stresses, there are certain rules, expres- sions, and formulae which it is necessary for the student to under- stand or know, and which will be here given without attempting elab- orate explanations or proofs. For the sake of clearness and simplic- ity, it is essential that in all formulae the same letters should alwaijs represent the same value or meaning ; this will enable the student to read every formula off-hand, without the necessity of an explanatory key to each one. The writer has further made it a habit to express, in all cases, his formuUe. in pounds and inches (rarely using tons or feet) ; this Avill frequently make the calculation a little more elabo- GLOSSAKY OV SYMBOLS. 5 rate, but it will be found to greatly simplify the formulae, and to make their understanding and retention more easy. In the following articles, then, a capital letter, if it were used, would invariably express a quantity (respectivel}'), either in tons or feet, while a small letter invuriabl>j expresses a quantity (respec- tively), either in pounds or inches. The following letters, in all cases, will be found to express the same meaning, unless distincthj otherwise staled, viz. : — a signifies area, in square inches. b " breadth, in inches. c " constant for ultimate resistance to compression, in pounds, per square inch. (See Tables IV and V.) d signifies depth, in inches. e " constant for modulus of elasticity, in pounds-inch, that is, pounds per square inch. (See Table IV.) f " factor of safety. g " constant for ultimate resistance to shearing, per square inch, across the grain. (See Tables IV and V.) g^ " constant for ultimate resistance to shearing, per scjuare inch, lengthwise of the grain. (See Table IV.) h " height, in inches. i " moment of inertia, in inches. (See Table I.) k " ultimate modulus of rupture, in pounds, per square inch. (See Tables IV and V.) I " length, in inches. 771 " moment or bending moment, in pounds-inch. (See Table IX.) n " constant in Rankine's formula for compression of long pillars. (See Table II.) " the centre. p " the amount of the left-hand re-action (or support) of beams, in pounds. q " the amount of the right-hand re-action (or support) of beams, in pounds. r " moment of resistance, in inches. (See Table I.) s " sti'ain, in pounds. t " constant for ultimate resistance to tension, in pounds, per square inch. (See Tables IV and V.) u " uniform load, in pounds. V " stress, in pounds. to " load at centre, in pounds. X, y, and z signify unknown quantities, either in pounds or inches. € SAFE BUILDING. (3 signifies total dejlectlon, in inches. p2 « square of the radius of gyration, in inches. jj " diameter, in inches. f " radius, in inches. ff = 3.14159, or, say, 3.1-7 signifies the ratio of the circun}ference and diameter of a circle. If there are more than one of each kind, the second, third, etc., are indicated with Roman numerals, as for instance, a, a^, a„, a„„ etc., or b, h„ 6,„ 6„„ etc. In taking moments, or bending mo.ments, strains, stresses, etc., to signify at what point they are taken, the letter signifying that point is added, as for instance : — m signifies moment or bending moment at centre. 7,)^ " " " " point A. m^ " " " " point B. m^ " " " " pomf X. s " strain at centre, s^ " " poiiit B. Sx " " " X. V " stress at centre. I'p " " point D. v^ « " X tv signifies load at centime. u\ " " " point A, CENTKE OF GRAVITY. (German, Scliwerpunkt; French, Centre de gravite.) The centre of gravity of a figure, or body, is that centre of point upon which the figure, or body, will balance Gravity. itself in whatever position the figure or body may be placed, provided no other force than gravity acts upon the figure or body. To find the centre of gravity of a plane figure, find two neutral axes, in different directions, and their point of intersection will be the centre of gravity required. NEUTRAL AXIS. (German, Neutrale Achse; French, Axe neutre.') The neutral axis of a body, or figure, is an imagin- Neutral Axis. ary line upon which the body, or figure, will always balance, provided the body, or figure, is acted on by no other force than gravity. The neutral axis always passes through the centre of gravity, and may run in anv direction. In calculating transverse strains, the neutral axis NEUTRAL AXIS. 7 designates an imaginary line of the body, or of the cross-section of the body, at which the forces of compression and tension meet. The strain on the fibres at the neutral axis is always naught. Extreme Fibres. On the upper side of the neutral axis the fibres are compressed, while those on the lower side are elongated. The amount of compression or elongation of the fibres increases directly as their distance from the neutral axis ; the greatest strain, therefore, being in the fibres along the upper and lower edges, these being farthest from the neutral axis, and tlierefore called the extreme fibres. It is necessary to calculate only the ultimate resistance of these extreme fibres, as, if they will stand the strain, certainly all the other fibres will, they all being nearer the neutral axis, and consequently less strained. Where the ultimate resistances to compression and tension of a material vary greatly, it is necessary to so design the cross-section of the body, that the " extreme fibres " (farthest edge) on the side offering the weakest resistance, shall be nearer to the neutral axis than the " extreme fibres " (farthest edge), on the side offering the greatest resistance, the distance of the " extreme fibres " from the neutral axis being on each side in direct proportion to their respective capacities for resistance. Thus, in cast-iron the resistance of the fibres to compression is about six times greater than their resistance to tension ; we must therefore so design the cross-section, that the distance of the neutral axis from the top-edge will be six- sevenths of the total depth, and its distance from the lower edge one-seventh of the total depth. To find the neutral axis of any plane-figure, some writers recom- /^7\ mt-^nd cutting, in ^^^ ^^ „„j, 7 ~'Y~ stiff card-board. Neutral Axis. T I aduplicateof the figure (of which -ts X — \ the neutral axis is sought), then — — Nto experiment until it balances )H^ -n J on the edge of a knife, the line c\ ^m ^ -- '-^ ■^ — ■ f Oil 7 Ltj^r — :i2L^ ij on which it balances beinsr, of L L dir I '^ y. I r ^ '^ \i [ \ -y course, the neutral axis. This Fig. I. is an awkward and unscientific method of procedure, though there may be some cases where it will recommend itself as saving time and trouble. The following general formula, however, covers every case : To find the neutral axis M-N in any desired direction, draw a line X-Y at random, but parallel to the desired direction. Divide the figure into any number of simple figures, of which the areas and cen^ 8 SAKE BUILDING. tres of gravity can be readil}- found, then the distance of the neutral axis M-N from the line X- Y will be equal to the sum of the productn ^ of each of the small areas, multiplied by the distance of the centre of gravity of each area from X-Y, the whole to be divided by the entire area of the whole figure. An Example will make this more clear. Find the horizontal neutral axis of the tross-section of a deck-beam, standing vertically on its bottom-Jlange. Draw a line {X-Y) horizontally (Fig. 1), then let (/„ J,„ c/,„, rep resent the respective distances from X-Y oi the centres of gravity of the small subdivided simple areas a„ a„, a,„, then let a stand for the whole area of section, that is : — «. + «,i + «m =a, then the required distance (d) of the neutral axis M-iVfrom X-Y, will be 7 a, d, 4- a,, d,, -\- a,,, d,,, a To find the centre of gravity of the figure, we might find another neutral axis, but in a different direction, the point of intersection of the two being the required centre of gravity. But as the figure is uniform, we readily see that the centre of gravity of the whole figure must be half-way between points A and B. Centres of '^^^ centre of gravity of a circle is always its cen- Cravity. tre. The centre of gravity of a parallelogram is al- ways the point of intersection of its two diagonals. The centre of orravity of a triangle is found by bi-secting two sides, and connecting these points each with its respective opposite apex of the triangle, the point of intersection of the two lines being the required centre of gravitv, and which is always at a distance from each base equal to one-third of the respective height of the triangle. Any line drawn through either centre of gravity is a neutral axis. MOMENT OF INERTIA. (German, Trdgheitsmoment ; French, Moment d'enertie, or Moment de giration.) Moment of iner- "^^^^ moment of inertia, sometimes called the mo- tia. cseeTabiei.) ment of gyration, is the formula representing the inactivity (or state of rest) of any body rotating around any axis. The reason of the connection of this formula with the calculation of strains and the manner of obtaining it cannot be gone into here, as it would be quite beyond the scope of these articles. The moment of ' If line X-Y is inside of (bisects) figure, take sum of products on one side only and deduct sum of products on other side. TIIK CKNTRK OF GYUATIOX AXI) KADIUS OK GYKATKJX. inertia of any body or figuro is llie sum of the products of each par- ticle of the body or figure multiplied by the S(]uare of its distance from the axis around which the body or figure is rotatin"-. A table of moments of inertia, of various sections, -will be "iven later on and will be all the student will need for practical purposes. THE CENTRE OF GYRATION AND RADIUS OF GYRATION. (German, Tragheitsmiltelpunkt ; French Centre de fjiration.) The centre of gyration "is that point at which, if the whole mass of a body rotating around an axis us^'of'^Cyratfon" or point of suspension were collected, a given force (^ce Table i.) applied would produce the same angular velocity as it would if ap- plied at the same point to the body itself." The distance of this cen- tre of gyration from the axis of rotation is called the radius of gyra- tion (German, TragheilsJialbmesser ; French, Rayon de giralion) ; this latter is used in the calculation of strains, and is found by dividin^r the moment of inertia of the body by the area or mass of the body, and extracting the square root of the quotient, or, 9=Sl^, or s — a- A table will be given, later on, of the " squares of the radius of o-yra- tion" (German, Quadrat des TragheUslialhmessers ; French, Carri du rayon de giralion). THE MOMENT OF Kp:SISTANCE. (German, Widerstandsmoment ; French, Moment de resistance). The moment of resistance of any fibre of a body, revolving around an axis, is equal to the moment of sistamfe'!*°ee'T'J inertia of the whole body, divided by the distance of ^^^ ^'^ said fibre from the (neutral) axis, around which the body is revolvin"-. A table of moments of resistance will be given later on. MODULUS OF ELASTICITY. (German, Elasticitats modulus, French, Module d'elasticite). The modulus of elasticity of a given material is the force required to elongate a piece of the mate- Elasticity."sceTi^ rial (whose area of cross-section is equal to one '"^ ^^'•* square inch) through space a distance equal to its primary length. Thus, if a bar of iron, twelve inches long, and of one square inch area of cross-section, could be made so elastic as to stretch to 10 SAFE BUILDING. twice its length, the force recjuired to stretch it until it were twenty- four inches long would be its modulus of elasticity in weight per scjuare inch. MODULUS OF RUPTURE. (German Bruchcoefficient ; French, Module de rupture.^ It has been found by actual tests that though the Modulus of ,.^ -, r . 1 T • Rupture. (SeeTa- diiierent ubres of materials under transverse strains "^^ "" ■' are either in compression or tension, the ultimate resistance of the "extreme fibres" neither entirely agrees with their ultimate resistance to compression nor tension. Attempts have been made to account for this in many different ways ; but the fact re- mains. It is usual, therefore, where the cross-section of the material is uniform above and below the neutral axis, to use a constant derived from actual tests of each material, and this constant (which should always be applied to the " extreme fibres," i. e., those along upper or lower edge) is called the modulus of rupture, and is usually expressed in pounds per square inch. TO FIND THE MOMENT OF INERTIA OF ANY CROSS-SECTION. Howtofindmo- Divide the cross-section into simple parts, and SflTnycross-Sc^fi^'i t'^^ moment of inertia of each simple part tion. around its own neutral axis (parallel to main neu- tral axis) ; then, if we call the moment of inertia of the whole cross-section i, and that of each part i„ t,„ t,„, z,„„ etc., and, fur- ther, it we call the area of each part a„ a,„ a,„, a,„„ etc., and the distance of the centre of gravity of each part from the main neutral axis of the whole cross-section, etc. Referring back to Figure 1, we should have for Part I : — t, = 11 1,*. (See Table I., column 8.) For Part II we should have : — '" — rr And for Part III: — '■"— 12 For the distances of individual centres of gravity from main centre of gravity we should have for Part I : d^-d. For Part II: d,,-d. And for Part III: d-d,„. TO FIND THE MOMENT OF INERTIA OF ANY CROSS-SECTION. 11 Therefore the moment of inertia, i, of the wliole deck-beam would be: — But a,=Bx,^ Further, a„ = &„ ft„» And a„. = 6,„ A„„ which, inserted above, gives for The following table (I) gives the values for the moment of inertia (i), moment of resistance (r), area (a), square of radius of gyration (0*), etc., for nearly every cross-section likely to be used in building. Those not given can be found from Table I by dividing the cross- section into several simpler parts, for which examples can be found in the table. Note, that it makes a great difference whether the neu- tral axis is located through the centre of gravity (of the part), or elsewhere. When making calculations we must, of course, insert in the different formulae in place of i, r, a, Q^ their values (for the re- spective cross-section), as given in Table I. 12 SAFE BUILDING. TABLE I. DISTANCE OF EXTREME FIBRES, MOMENTS OF INERTIA AND RESISTANCE, SQUARE OF RADIUS OF GYRATION, AND AREAS OF DIFFERENT SHAPES OF CROSS-SECTIONS. Kumber aud Form of Section. 5 • % •* W CO o C 1 d-4 TFWTcU ^^ '■K<^x- ■;-N 12 M3 12 12 &f/3-i/Z,3 12 11 r4^ ^3 6 6 6d hd^-h,d,^ Gd 11 rZ2 ?/(/ (/2-f/2 bd-b,d. 22 . 12 12 aS Number and Form of i^ X " a 5 f* s* a ■^, -r? o Section. ." s r' o »-*» M» S : •■'S' a § • !z! a ? fe! ? so' ■ S- i' iMTic^ "3-ct, IMiy 'M-;l--l-I\J l>i ? c; ^ " C .• C5 "co "r "1 -L 1" r "eo 14 SAFE BUILDI^XG. Number and Form of Section. b — -4j +1 + + ! " ' Si. ''r 4- + r + + TABLK I, CONTINUKD. Id Number and Form of Section. 2 S. w 17 MttN,-i l^-^J 18 r I^iticed I kthced + + + + + + Si »5 — t +1^ T^lA 12 16 SAFE BUILDING. Number and Form of Section. p S § 2. fe: ? ■^ w o ® + + ' a, + "K *^ + I ® + + , t? TAULE I, CONTINUED. 17 Number and Form of Distance tral axis from ex br o B » o S » a c a o n- CO o 3 Section. of Neu- M N treme fl- ea. M D a ^1 + I o + i (5- I + wo _i_ "O C^ hi I IB + + + + + t + -f + 18 SAFE BUILDING. Kuniber and Form of Section. ^ X a= » • 2 '^ B o C 3 -i Q A ?^ ,^ :^ r r 'I 1 M M t" X + r- C5 x\ + ;+ + + :+ + + : + +: + TABLE I, CONTINUED. 19 n if O S ■> CO O » X o o Number ami Form of ^ X " » S 2- f 3 S ? o > Section. B : a r^ ^* a o :::r » • !^ a : o Vi' — * » !z! c S' "T 5 2s M- M--^ M- -t-N M — Lower Fibres. iL 3 Upper Fibres. _2^ 3d + 3 12 MS i 36 4- 12 M^ 4 Lower Fibres. 12 TTpper Fibres. M^ 24 hd f/2 &(Z ?;./ 9 + + 3 6 18 20 SAFE BUILDING. Number and Frirni of Section. B !o « h_ b 1 ,/2" 1.4142 mIBIn T 37 M- 51 s frr --^ 12 + + ^,. c^' ^ ►^ : <— ' ^ •— * <-! t(^' l— i4— •— ' <-t --t tc\ ~£. rt . "^ 1^ ! ^ »£ r ^' i + ?K- 5-W 0.1179 h^\ Ifi + ?/2 12 4- I -J »^ TABLE I, CONTINUED. 21 Number and Form of Sectiou. »-»> r^ i-H g 2 d >= S- o .^ taiioe 1 axis m ex bi 2 , ;4 V a 3 ? > a 5 ? " = B '. :* w e ^» « 1 iz: « fl 3) i^* S P '^ CALCULATION OF STRAINS AND STRESSES. As we have already noticed, the stress should exceed the strain as many times as the adopted factor-of-safety, or : — _ — 11-^ factor-of-safety. Strain Or, stress = strain X factor-of-safety. This hokls good for all calculations, and can be expressed by the following simple fundamental formula : — Fundamental v^=.s.f (1) Formula- Where v =: the ultimate stress in pounds. " s = " strain in pounds. 22 SAFE BUILDING. And where /= the factor-of-safety. COMPRESSION. Compression, In pieces under compression the load is directly s or CO um . j^ppjj^j ^^ ^^xe material. In short pieces, therefore, -which cannot give sideways, the strain will just equal the load, or we have : — s = w. Where s = the strain in pounds. And where ic ^= the load in pounds. The stress will be equal to the area of cross-section of the piece being compressed, multiplied by the amount of resistance to com- pression its fibres are capable of.^ This amount of resistance to compression which its fibres are capable of is found by tests, and is given for each square inch cross-section of a material. A table of constants for the resistance to crushing of different materials will be given later on. In all the formulae these constants are represented by the letter c. We have, then, for the stress of short pieces under compression : — v^a.c Where v is the ultimate stress in pounds. Where a is the area of cross-section of the piece in inches. And where c is the ultimate resistance to compression in pounds per square inch. Inserting this value for v, and w for s in the fundamental formula (1), we have for short pieces under compression, which cannot yield sideways : — a.c = io ./, or : — t,= a.(^-^). (2) Short Columns. Where ?i> =:the safe total load in pounds. Where a = the area of cross-section in inches. And where ( -^ j ^ the safe resistance to crushing per square inch. Example. What is the safe load which the granite cap of a 12" x 12" pier tvill carry, the cap being twelve inches thick ? The cap being only twelve inches high, and as wide and broad as 1 This is not theoretically correct, as there is in every case a tendency for the material under compression to spread; but it is near enough for all practical purposes. COMPRESSION. 23 higli, is evidently a short piece under compression, therefore the above formiihi (2) applies. The area is, of course: a=: 12.12= 144 square inches. The ultimate resistance of granite to crushing per square inch is, say, fifteen thousand pounds, and using a factor-of-safety of ten, we have for the safe resistance : — c 15000 i-ArtU — = zrzloOOIbs. / 10 Therefore the safe load w on the block would be : — 7« = 144, 1500 = 216000 pounds. Where long pieces (pillars) are under compres- Compression, 1 ^ , • i. • IT 1 Long Columns. sion, and are not secured against yielding sideways, it is evident they would be liable to bend before breaking. To ascer- tain the exact strain in sucli pieces is probably one of the most dif- ficult calculations in strains, and in consequence many authors have advanced different theories and formulae. The writer has always preferred to use Rankine's formula, as in his opinion it is the most reliable. According to this, the greatest strain would be at the cen- tre of the length of the pillar, and would be equal to the load, plus an amount equal to the load multiplied by the square of the length in inches, and again multij^lied by a certain constant, ri, the whole divided by the " square of the radius of gyration " of the cross-sec- tion of the pillar. We have therefore for the total strain at the cen- tre of long pillars : — I w.l-n . = ..+ -^ Where s = the strain in pounds. " ii) = the total load in pounds. " Z = the length in inches. <' p- = the s(piare of the radius of gyration of the cross-section. " n = a constant, as follows : — TABLE II. VALUE OF n IN FORMULA FOR COMPRESSION. Material of pillar. Both ends of pillar smooth (turned or planed.) One end smooth (turned or planed) other end a pin end. Both ends pin ends. Cast-iron 0.C003 0.0001 0.00057 ■VVrought-iron 0.000025 0.000033 0.00005 Steel 0.00002 0.000025 0.0000.33 Wood 0.00033 0.00041 0.00067 Stone 0.002 Brick 0.0033 24 SAFE BUILDING. The stress of course ■will be as before : — V = a. c. Where r = the ultimate stress in pounds. " a = the area of cross-section in inches. " c = the ultimate resistance to crushing per square inch. Inserting the values for strain, s, and stress, v, in the fundamental formula (1) we have : — / , w.r-.7i\ ,. = a constant, d ■= the thickness of plate in inches, SAFE BUILDING. b = the unstiffened breadth of plate in inches. If a plate has stiffening ribs along both edges, use for b the actual breadth between the stiffening ribs ; if the plate is stiffened along one edge only, use 4&, in place of 6. Thus, in the case of the boxed girder, Figure 3, if we were considering the part of top plate between the webs, we should use for b in the formula, the actual breadth of b in inches ; while, if we were considering the overhang- ing part &, of top plate, -we should use 4&, in place of b in formula. For rectangular columns use 160,000 pounds for Wr', for tubular beams, top plates of girders, and single plates use 200,000 pounds for Wr- With a factor-of-safety of 3, we should have 160000 = 53000 pounds for rectangular columns, and 200000 66000 pounds for tubular beams, top plates of riveted girders and single plates. o p f\r\f\ For to we shall use, of course, — = 12000 pounds, which is the safe allowable compressive strain. This would give the following table for safe unstiffened breadth of wrought-iron jilates, to prevent wrinkling of plates. TABLE III. Safe breadth in inches of Plate Safe breadth n inches of Plato stiffened along both edges. stiffened along one edge only. (use 6.) (use 46,) Thickness of Rectan- Tubular Beams, Rectangular Riveted Girders Plate gular Col- riveted Girders, and Columns. and single Plates. In inches. umns. single Plates. 1 L -J K b TFT' V^ ^^ ^ Tbil t^I L h d b i r" Ih IHj nljt 1^4 lln^ "^-"H i 2-"- ^.3 '^4 8 if X 4' 7 \1 1'' 4 ^8 ' 1 6 '4 '■ 8 3 7 5 113 U 913 ' 16 8, ^ n 151 h\ h' 1 IVb 18| 3 4H f 14| 22H m 5| f 1^6 26i H 6t^6 1 191 301 4| "fe H 24f 37if H 9tV H 29i 45| h\ 11 A If 34^ 52^1 8t% 13A 2 39 60^ 9f 151 The above table will cover every case likely to arise in buildings. WRINKLING. 27 Two facts should be noticed in connection with wrinkling: 1. That the length of plate does not in any way affect the resist- ance to wrinkling, which is dependent only on the breadth and thick- ness of the part of {jlate unstiffened, and 2. That the resistance of plates to wrinkling being dependent on their breadth and thickness only, to obtain equal resistance to wrink- ling at all points (in rectangular columns with uneven sides), the thickness of each side should be in proportion to its breadth. Thus, if we have a rectangular column 30" X 15" in cross section and the 30" side is 1" thick, we should make the 15" side but ^" thick, for as 30" : 1" : : 15" : ^". Of course, we must also calculate the column for direct crushing and flexure, and in the case of beams for rupture and deflection, as well as for wrinkling. Example of Wrinkling. It is desired to make the top plate of a boxed girder as wide as pos- sible, the top flange is to be 1^" thick, and is to be subjected to the full amount of the safe compressive strain, viz : 1 2,000 pounds per square inch ; how wide apart should the ivebs be placed, and how much can the plate overhang the angles without danger of lonnkling ? Each xceb to be I" thick, and the angles 4" X 4" each ? For the distance between webs we use b in Formula (4). ''— ^4-\^ 12000/ —^4-02 — «i^6 » which is the safe width between webs to avoid wrinklino- For the overhanging part of top plate we must use 46, in place of h in Formula (4). ^^•= ^HtIU^J =37if, therefore, h. = ^ = 9,453, or say, 5. = 9^^". The total width of top plate will be, therefore, including 1" for two webs and 8" for the two angles, or 9", and remembering that there is an overhanging part, &„ each side, am— CTa \^-m= G511". By referring to Table III, we should have obtained the same result, with- F'£- 4 out the necessity of any calculation. Figure 4 will make the above still more clear. 28 SAFE BUILDING. LATERAL FLEXURE IN TOP FLANGES OF BEAMS, GIRDERS, OR TRUSSES, DUE TO COMPRESSION. The usual formulae for rupture and deflection assume tlie beam, girder or truss to be supported against possible lateral flexure (bending sideways). Now, if the top c-hord of a truss or beam is comparatively narrow and not supported sideways, the heavy, com- pressive strains caused in same may bend it sideways. To calculate this lateral flexure, use the formula given for long columns in com- pression, but in place of I use only two-thirds of the span of the beam, girder or truss, that is |/, and for w use one-third of the great- est compressive strain in top chord, which is usually at the centre. Inserting this in Formula (3) we have : "3" ^TTA^ transposing, we have, w =—Wn ^^^ where a the area of the cross-section of the top chord in inches, 0^ is the square of the radius of gyration of the top chord around its vertical axis ; we must therefore reverse the usual positions of h and rf, that is the breadth of top chord, becomes the depth or d, and the depth of top chord becomes the thickness, or h (both in formulae given ia last column of Table I.) — is the greatest allowable compressive strain in pounds at any point to resist lateral flexure safel}' at that point. ( — ) is the safe resistance of the material to compression per square inch in pounds. I is the total length of span in inches. n is given in Table II. Example. A trussed girder is 60' long between hearings, and not supported side- ways ; the top chord consists of two plates each 22" deep and 1" thick; the plates are 2" apart, as per Figure 5. The greatest compressive strain on top chord has previously been ascertained to be on the central panel, and to he 525000 pounds. Is there danger of the girder bending sideways? The girder is safe against lateral flexure so long as the strain at centre does not exceed -^ in Formula (5). Now, the area «^ 2.1.22 =44. LATERAL FLEXURE. 29 Usin"' 48000 pounds per square inch for ultimate resistance to com- pression of wrought-iron, and a factor-of-safety of 4, we have 48000 (7) = 12000 4 The length is GO', or 720", therefore Z2 = 518400. From Table II we have ?i = 0,000025. And from Table I, section Number 16, we have Fig- 5. for the above cross-section, § ~~ V2id-d,) As we are considering the section for bending sideways, we must, of course, take the neutral axis x---y vertically, therefore d becomes 4" and d becomes 2". This supposes the plates to be stiffly latticed or bolted together, with separators between. We have then 43-23 Then for w we have, 3.44.12000 «>= 4,518400 0.000025. 9- 21- _ 1584000 _ 1584000 _ 456 484 lbs. 1+2,47 3,47 Or, we find that there is danger of the girder bending sideways long before the actual compressive strain of 525000 pounds has been reached. It will, therefore, be necessary to re-design the top chord, so that it will be stiffer sideways. This subject will be more fully treated when considering trusses. TENSION. In tension the load is applied directly to the material, and it is, therefore, evident that no matter of what shape the material may be, the strain will always be the same. This strain, of course, will be just equal to the load, and we have, therefore: — s-=w. Where s = the amount of strain. Where w = the amount of load. The weakest point of the piece under tension will, of course, be where it has the smallest area of cross-section; and the stress at 30 SAFE BUILDING. such point will be equal to the area of cross-section, multiplied by the amount of resistance its fibres are capable of.^ The amount of resistance to tension the fibres of a material are capable of is found by experiments and tests, and is given for each material per square inch of cross-section. A table of constants for the ultimate and safe resistances to tension of different materials will be given later; in all the formulas these constants are represented by the letter /. We have, then, for tlie stress : — v = a. t Where i; = the amount of ultimate stress. Where a=:the area of cross-section. Where i= the ultimate resistance to tension, per square inch of the material. Therefore, the fundamental formula (1), viz. : v=rs.f, becomes for pieces under tension : — a. t = w. f, or : — t« = a.(-i) (6) Where 7i> = the safe load or amount of tension the piece will stand. Where a = the area of cross-section at the weakest point (in square inches). Where (A'j^the safe resistance to tension per square inch of the material. Example. A weight is hung al the lower end of a vertical wrought-iron rod, wMch is firmly secured at the other end. The rod is 3'' at one end and tapers to 2" at the other end. How much weight will the rod safely carry f The smallest cross-section of the rod, where it would be likely to break, would be somewhere very close to the 2" end, or, say, 2" in diameter. Its area of cross-section at this point will be : — 99 9*^ • t a= — — = 3i square inches. 7*4 ' ^ The ultimate resistance to tension of wrought-iron per square inch is, from forty-eight thousand pounds to sixty thousand pounds. We do not know the exact quality, and, therefor e, take the lower figure ; 1 This, again, is not theoretically correct, as a piece under tension is apt to stretcli and so reduce the area of its cross-section ; hut the above is sufiiciently correct for all practical purposes. SHEARING. 31 using a factor-of-safety of four, we have for the safe resistance to tension per square inch : — /J^X 48 000 _ 12 000 pounds. Therefore, the safe load will be : — w^3\. 12 000 z= 37 714 pounds. SHEARING. In compression the fibres are shortened by squeezing; in tension they are elongated by pulling. In shearing, however, the fibres are not disturbed in their individualities, but slide past each other. When this sliding takes place across the grain of the fibres, the action of shearing is more like cutting across. When this sUding takes place Fig. 6. '^'^'^ along the grain, tlie action of sliearing is more like splitting. Thus, if I very deep, but thin, beam is of short span and heavily loaded, it might not break transversely, nor deflect ex- cessively, but shear off at the supports, as shown in Figure 6, the ac- tion of the loads and supports being like a large cutting-machine, the weights cutting off the central part of beam and forcing it down- wards past the support. This would be shearing across tlie grain. If the foot of a main rafter is toed-in to the end of a tie-beam, and the foot forces its way outwardly, pushing away the block or part of tie-beam resisting it (splitting it out as it were), this would be shear- ing along the grain. In most cases (except in transverse strains) the load is directly ap- plied to the point being sheared off ; the strain will, therefore, just equal the load, and we have : — Where s^ the amount of the shearing strain. " if> = " " load. The stress will be equal to the area of cross-section (affected by the shearing strain) multiplied by the amount of resistance to sep- aration from each other that its fibres are capable of. This amount of resistance is found by tests and experiments, and is given for each material per square inch of cross-section. A table of constants for resistance to shearing of different materials will be given later ; in the formulae these constants are represented by the letter g for shearing across the grain, and g, for shearing along the grain. 32 SAFE BUILDING. We have, then, for the stress : — v = a. g. Where v = the amount of ultimate stress. Where a = the area of cross-section in square inches. Where (7 = the ultimate resistance to shearing across the grain per square inch. Therefore, the fundamental formula (1) v = s.f, becomes for pieces under shearing strains across the grain : — a. g = w.f, or: — (7) '=«•(!) /- And similarly, of course, we shall find : — ,=..(x.) // (') Where w = the safe-load. Where a = the area of cross-section in square inches, at the point where there is danger of shearing. Where {^^ = the safe-resistance to shearing across the fibres per square inch. Where ^ -^ j = the safe-resistance to shearing along the fibres per square inch. Example. At the lower end of a vertical wrouglit-iron fiat bar is suspended a load of eight thousand pounds. The bar is in two lengths, riveted to- gether with 07ie rivet. What diameter should the rivet be f The strain on the rivet will, of course, be a shearing strain across the grain, and will be equal to the amount of tension on the bar, which we know is equal to the load. We use Formula (7), and have: — w = 8000 pounds. The safe shearing for wrought-iron is about ten thousand pounds per square inch ; inserting this in formula, we have : — 8000 = a. 10000, or a. = ^^ =f The area of rivet must, therefore, be four-fifths of a square inch. To obtain diameter, we know that : — d=s/w^ =\/m =^n =^1^01818- This is, practically, equal to one ; therefore, the diameter of rivet should be 1". CROSS-SHEARING IX BEAMS. 83 In transverse strains the (vertical) cross-sliearing is generally not equal to the load, but varies at different points of the beam or canti- lever. The manner of calculating transverse strains, however, allows for straining only the edges (extreme fibres) up to the maximum ; so that the intermediate fibres, not being so severely tested, generally have a sufiicient margin of unstrained strength left to more than off- set the shearing strain. In solid beams it can, therefore, as a rule, be overlooked, except at the points of support. (In plate-girders it must be calculated at the different jjoints where weights are applied.) The amount of the shearing at each support is equal to the amount of load coming on or carried by the support. "We must, therefore, substitute for w in Formula (7) either /> or q, as the case may be, and have at the left-hand end of beam for the safe resistance to shearing : — .(5) And at the ridit-haud end of beam : — = a.U) (9) (10) Where /) = the amount of load, in pounds, carried on the left- hand support. Where q = the amount of load, in pounds, carried on the right- hand support. Where a = the area of cross-section, in inches, at the respective support. Where ^ -^ j = the safe resistance, per square inch, to cross-shear- ing. Example. A spruce beam of 5' dear span is 24" deep and 3" wide ; how much uniform load will it carry safely to avoid the danger of shearing off at either point of support? The beam being uniformly loaded, the supports will each carry one-half of the load ; if, therefore, we find the safe resistance to shear- ing at either support, we need only double it to get the safe load (in- stead of calculating for the other support, too, and adding the results). Let us take the left-hand support. From Formula (9) we have : — Now, we know that a == 24. 3 = 72 square inches. 84 SAFE BUILDING. The ultimate resistance of spruce to cross-shearing is about thirty- six hundred pounds per square inch ; using a factor-of-safety of ten, we have for the safe resistance per square inch: — We have, now : — p = 72.360 = 25920 pounds. Similarly, we should have found for the right-hand support : — ^ = 25920 pounds. And as: — u =p -^q = 51840 pounds, that will, of course, be the safe uniform load, so far as danger of shear- ing is concerned. The beam must also be calculated for transverse strength, deflec- tion and lateral flexure, before we can consider it entirely safe. These will be taken up later on. Should it be desired to find the amount of vertical shearing strain X at any point of a beam, other than at the points of support, use : — ovi—^w (11) Where a: = the amount of vertical shearing strain, in pounds, at any point of a beam. C „ ■) the reaction, in pounds, (that is, the share of the Where ^ or >- = total loads carried) at the nearer support to the ( '7 ) point. Where S ?« = the sum of all loads, in pounds, between said nearer support and the point. When X is found, insert it in place of w;, in Formula (7), in order to calculate the strength of beam necessary at that point to resist the shearing. Exa?nple. A spruce beam, 20' long, and 8" deep, carries a uniform load of one hundred pounds per running foot. WJiat should be the thickness of beam b' from either support, to resist safely vertical shearing? Each support will carry one-half the total load ; that is, one thou- sand pounds; so that we have for Formula (11) : — |or >-"=:1000 pounds. The sum of all loads between the nearer support and a point 5' from support will be : — S ivz=b. 100 = 500 pounds. HORIZONTAL SHEARING IN BEAMS. 35 Therefore, the amount of slieariiig at the point 5' from support will be: — a: =: 1000 — 500 = 500 pounds. Inserting this in Forrauhi (7) we have : — 500 500 = «. (y), or, (1= / £\ We have just found that for spruce, fJl-.^ = 360 pounds. Therefore, a = ' — i=: 1,39 square inches. 3G0 ' ^ And, as b. d = a, ov b = — , we have, b = -!— - = -— d 8 6 This is such a small amount that it can be entirely neglected in an 8" wooden beam. To find the amount of vertical shearing at any point of a canti- lever, other than at the point where it is built in, use : — x= S 10 (12) Where x the amount of vertical shearing strain, in pounds, at any point of canti-Iever. Where S lu the sum of all loads between the free end and said point. To find the strength of beam at said point necessary to resist the shearing, insert x for w in Formula (7). In transverse strains there is also a horizontal shearing along the entire neutral axis of the piece. Tliis stands to reason, as the fibres above the neutral axis are in compression, while those below are in tension, and, of course, the result along the neutral line is a tendency of the fibres just above and just below it, to slide past each other or to shear off along the grain. AVe can calculate the intensity (not amount) of this horizontal shear- ing at any point of the piece under transverse strain. If X represents the amount of vertical shearing at the point, then 3 X the intensity of horizontal shearing at the point is = . A a If this intensity of shearing does not exceed the safe-constant \f} for shearing along the fibres, the piece is safe, or : — 3 -M^) (»> 2' a —\f 36 SAFE BL'ILDIXa. Where X is found by formulic (11) or (12) for any point of beam or, ,„, \ ^' ,' the amount of supnortino; force, in pounds, W here x — -; uv ^ — ^ ^ ^^ ( q [ foi" either point of support. Where a = the urea of cross-section in square inches. Where ( 2; ) = the amount of safe resistance, per sqnare inch, to shearing along fibres. Example, Take the same beam as he/ore. The amount of vertical shearing 5' from support ive found to bejice laindred pounds, or: — X = 500. The area was 8" multiplied by thickness of beam, or : — a = 8b. The ultimate shearing along the fibres of spruce is about four hun- dred pounds per square inch, and with a factor-of-safety of ten, we should have : — \f) 10 Inserting this in Formula (13) ^ . — = 40 , 1500 0//Q1 1G.40 The beam should, therefore, be at least 2^" thick, to avoid danger of longitudinal shearing at this point. At either point of support the vertical shearing will be equal to the amount supported there ; that is, one-half the load, or one thousand pounds. Substituting this for X in Formula (13), we have: — l.i^^40,ori=:.^^-^ =4,"68. 2 8& 1G.40 The beam would, therefore, have to be 4|" thick at the points of support, to avoid danger of longitudinal shearing. The beam, as it is, is much too shallow for one of such span, a fact we would soon dis- cover, if calculating the transverse strength or deflection of beam, which will be taken up later on. It will also be found that the greater the depth of the beam, the smaller will be the danger from longitu- dinal shearing, and, consequently, to use thinner beams, it would be necessary to make them deeper. TABLK IV. 37 "3 ^IS o ooo o Sg 18 ooooo< O O O Q O < cc L-- 00 O o «; g |gg t« T]i ^ t* 00 too oo OO OQ OX) o o IS8SSI O loO-^-^OOTlO oo felS =^.ls Sg? oo oooo ;c l:: r t o -r O I- ^ :o •-I lO OOO ooo «5 O -t^ Oi-I L-J o I o o o o r1 i o ooo o o 1-^ O O O O o C^ CD O X O O O O o L-i Xl ESS OOOOOO |00 oooooo |oo o o o <^ w.-- w ^ J i --' _ o o o o loo — ■ en CO o -^ |;c»ooooo I 2 -^ «|S * (JJ JJC-ilO . 00 O X *-l O <£) C^ X c* o o o o o o ■X' o o t- o o :o X ^ X o I oooo loooooooooooo S ISSSio ll-OO^OOr-IOSOT-lCOCSliO lOOOOO lO lOOO 00!=0 lo lO-O cooot-o lio |=r--j) w M- a fcl « ~ p:« oap ss SAFE BUILDING. 1 b in! ■^t-'J' c» 0^ to to 01 •<3' 10 CO «0 "J" ^ •w 1010 iOtj< 10 to 'S p< s ^ 3 S ^i^ 0© 00 00 00 00 00 0000 "3 J 3 « 00 0000 00 cr, 1- -3 -1 1- 00 tot^ o 2 1-^ -t* £ rt t- e-1 00 ci 03 ^ 00 S 3 « 3 a 3 o © og lO 00 00 00 000 .> 1. 1-0000 0000 00 1,0 00 -y V cS I- at, — l-OOr-i '-' 1 1 ^" S '^' »-l "^ r-l S ,-H 1-1 D K tn Is 80 8 88 op 00 000 p g 00 CJ 3 •!« 1-0 (M !?; ■5 ■;3 05 W ■^ 00 -^ to »o 000 to I- to t- t-to«> t- 3 © fe|S ^ 010 0000 10 10 00000 IH u -:r L- 00 -r CO CO 1- S t3 ci to •* MOl-J- 00 ,e n m fc>IS 09 0) c5 a to 8 go gg og SSg88 oooop p^ooo C/5 v_> bl 10 (M to T-i CO 1-1 t- to ■>J< e-i i4 tk "^ 00 '■+3 to nil II nil nil IMIIMIII IIIIIIIMI CO 5 1— ( t-.5) to toto tot; tototototo to, to. to. to. ? 51 a> |gS§gl3S ISSSSSSSS 18 000000000 CO t- - 10 -r S » -^ v 1 t- t- ^ O) CO t- .I Ol .-n UO 00 1^ (M t- t- to C1(M rl 1 w rt 1 ' rtM rlli-lrtr-lr-l .-(Flli-1 "^ rl r1 1 ij H pa a ooooooo 100000000 10 8 |88888888 1 Q ■va C :o?=;i.oo 00000000 'n ir: L2 ?^ 10 1 1- 1--5 CO 10 -icoo-r 1 fH o X oc 00^ II-,— -T Cl IQ t- i-l Cl I .-1 00 at >- o5 L- - t- to Cl OJ rH 1 W ' P ^^ tH rl ® 55 0000100 1 1000000000000 c. c H 10 -M lOOOOCrOOl-fOOOOlTtl c: 10 coo 10 3 p-.^ Jo S ia< oc^ioto T}ii-lOOCOCOC^XIC^tOt-01 to W 00 CI I- rl< o w t-l rH eu a> a O. 3 000000 oooooooc 10000 000000 c « Ooooioo 000 = t-o OlOlOOClt:^ lOCOOOOW-^OOOOCM-T c: oc 100 C CO l- •w cseoioto '("•-lOoeoooNaJi-itDt-e^ to t- 00 »-i to ■<)< S "^ 11 p TO u s Cj ^ 4) - tfi to i 3 i' 1^* p a 8 r a c ii I 03s t> E of 1. " c c 1 X 2 c, 0: c * a § 1 op "5 '1 2 1 a u C a ^ s 1 II C 6 r 1^ " e t. 'l '-•S ^ c c- i- - z - - - = - - <• TABLE IV, CONTINUKD. S9 C3 < t3 a a C/J K -A H li|§i|i § |§ |§SS iss ISSS fr.s -K «ls )i-^lCOOOOOOO OO I oo ooo - CI -^ »o lO SI lisiss II II II II II II II nil II INI nil » ^tr: t^^t^O^^^' ^_ t: C)i g>. ^ O O O O O O O (-5 |00 80 I O O 10 Q 25 o o »oo ,00000000 JOOOOlOOO*"^ CO CJ O i-l t- o J c^ o 10 o o < to »b 10 1-* t- W 05 >oo 000000 > -^ 10 t- i-< o O i-H c> i CD O CXI - 1 t- W 5D 00 looooicooo 00 looooc^oioo t*iO I CD 10 lOr-tt-r-lQO I ^^-^oioooooc,,,^ ooomr-^oOT-o s ^ 2. -Si. e8 e8 MM - 4) *.«- OOPh PhP^ ,3 . • . ^ o o P^ P-lf4CM " o « «2 C g cS 40 SAFE BUILDING. [OOiflOO lOO ICO o lO O ^ I lO LO 00 CO 3 ^ .2 O O I oo o o o o ' o o loo OOOO I o o oooo o = OOOO o o o o o o ^K oo o o o oo o o oo o o o O is :o -r -; (M CNl 00 I— 'K CO .1 T-H .-I O) Tl < ooooooooooc OOOOOOOQOOC OOOOOOOOO I OQOQ OOOOOOOOOOOOOOOQOOOOOO I OOOO owilooooooooooooooooooo OOOlO coSeooooo^ocooo-7 I OO ooo loo 00-1 loo > O t~- I 1-0 O I *-t '^ 'X L' c sooooooooooooo oooooooooooooo O CO O 1-0 O O O O I- O O ro 1.0 o j O O CO CI l-O CO LO I OC5 COQO f OOCC "3 «1S 8 1 § l-O CI ■ o"^ ^ u ^-^ IM c^ ^5 c s 2 a ffl ►; 3 :e M t- « o is 2a -5 42 SAFE BUILDING. o CO fc. O 4^ (M OS 1 00 cot- 1 1 f- •*< to 00 CO 05 1 to o 1 » .2 ^ 1 t^ Swiai 1 iior-iTrociOi'j'i loicoi I Tj« 1 'T ^^111 „^^ „ 1^ j |„ 1 1 o s lA o OOO O lOOOOOOO o o OOO O JOOQOOOO O Q o "3 •53 sl « o 0&O-/3 O C^tOClOC^^O^^ "O 00 O S S s CfclO «3 eOO III ooooooo OOOlOOOOOOOOOO 1-h oioo-i" c: oo OOl- COOCJ-MOOOt-O a C^M lO CO C^ C-i c^ »U 0.-1 ■* lO » JU Ofc t- lO o 3> 1 +J a a - §S8 OOOCJOOO ooooooooooooo OOlO — OOO OOOIOOOOOOOOOQ 0*000-^ CO »0 OOh- C0ClC3Ci:-^OOt^Q H OOro CIOCO ta (M«s^s^iio oi-< • 1 llrllll © u &• c4 o ooooo o oooo o o a a « Oi-SOOO CO cooo o o oooioo c^ ol 5 57 3a c p II s > 43 » o, d >" g > w kJ < pa < rf) H O 3 a, oo coco o CT l§ OO o o g o oo I- o -Jj o o I oo o O I 1- O I- IT I CC O O o«o lO-H o cS ulS 1 o \a o g o ^ oo o o 1 S .§ C § o o ^ P "SyK MOOOC^>MO-t0100«MOlOlCC--l-^< I i-l-< o I- 00 CO L— O O O O ~ QOO O O O o =■ O SS o rH CO I O c: O I- O O Tl 03 I 00 -- CO CO Ol I C-1 O li«N oooooo o lOo loo roo i- OJCOrH JOt-COCJ QOOOOOOC SS50000000000000000000 _iOOCOOt-OCSOOOC-10C-lWL'-000»OCOl-CiOOC-lC-M-'^'0 ei3CO'-IOOt»COOJlOOC; o 't''J.* %3 au S^' ■£- '^ c- '^ p* fi^g£=ii^S eSSWS'^ = 11 TAUI.K V, CONTINUED. 45 O h- lO I CC O ^H C"1 O (M 0»-r- O o SiS P P p:?h;=;p,:=.= 'p:z:c;^ ft : ^ £"3 o ^■- Oi^Oh 46 SAFE BUILDING. The amounts given in Table IV for compression, tension and shear- ing are along fibres, except where marked across. It will also be noticed that the factors -of-sa£ety chosen are very dif- ferent; the reason being that where figures seemed reliable the fac- tor chosen was low, and became higher in proportion to the unre- liability of the figures. The tables, as they are, are extremely un. satisfactory and unreliable, though the writer has spent much time in their construction. Any one, who will devote to the subject even the slightest research, will find that there are hardly any two origi- nal experimenters who agree, and in most cases, the experiments are so carelessly made or recorded that they are of but little value. TABLE VI. WEIGHT PER CUBIC FOOT OF MA.TERIALS. (Xot included in Tables IV and V.) Material. Ashes Asphali Butter Camphor Charcoal Coal, solid " loose Coke Cork Cotton in bales Fat Gunpowder... . Hay in bales.. . Isinglass Lead, red Paper Weight. 59 150 60 63 23 93 64 60 15 20 58 66 17 70 660 65 Material. Peat Petritied wood Pitch Plumbago Pumice-ttone Resin Rock crystal Rubber Salt Saltpetre ' Snow, fresh fallen. " solid Sugar Sulphur Tiles Water Weight. So 145 75 131 66 68 172 62 134 130 6 20 82 125 115 63 REAC'IION OF SUPPORTS. 47 TRANS VKRSK STRKNGTII. — RL'PTURE. If a beam is supported at two en<]s, and loads are applied to the beam, it is evident : — 1st, that the beam will bend under the load, or deflect. 2d, that if the loading continues, the beam will eventually break, or be ruptured. Deflection when "^'"^ methods of calculating deflection and rup- non-important. ture diff'jr very greatly. In some cases, -where de- flection in a beam would do no damage — such as cracking plaster, lowering a column, making a floor too uneven for machinery, etc., — or where it would not look unsightly, we can leave deflection out of the question, and calculate for rupture only. Where, however, it is im- portant to guard against deflection, we must calculate for both. RKACTIOX OF SUPPORTS. If we imagine the loaded beam supported at both ends by two giants, it is evident that each giant would have to exert a certain amount of force upwards to keep his end of the beam from tipping. We can therefore imagine in all cases the supports to be res'isting Amount of Reac- *^'' reacting with force sufficient to uphold their re- *'0"' spective ends. The amount of this reaction for either support is equal to the load multiplied by its distance from the further support, the whole divided by the length, or w. n p: I (14) Where /J = the amount of the left hand reaction or supporting force. and q=z——. (15) Where q = the amount of the right hand reaction or supporting force. If there are several loads the same law holds good for each, the reaction being the sum of the products, or ?r, n I 20,, s (16) (17) andq = -Y + -^-f- As a check add the two reactions together and their sum must equal the whole load, that is, p-\-q = lo, -|- ii\, 48 SAFE BUILDING. Example. A beam 9' 2" long between bearings carries two loads, one of 200 lbs. 4' 2" from the left-hand support, and the other of 300 lbs. 3' 4" from the right-hand support. What are the right-hand^ arid left-hand reactions ? Referring to Figure 8 Fig. 8. we sliould have w, = 200 lbs., and ?r„ = 300 lbs., further Z =: 110" ; m = 50"; n = 60"; s = 40", and r=70", therefore the left-hand reaction would be : — 200. 60 , 300. 40 p: = 2 18^2_ pounds. 110 ' 110 and the right-hand reaction would be : — 200. 60 , 300. 70 „oi q i. o = + = 281 fir pounds. ^ 110 ' 110 11^ As a check add p and q together, and they should equal the whole load of 500 lbs., and we have in effect : — p-\- q = 218^2_ _|_ 281^9j- = 500 pounds. If the load on a beam is uniformly distributed, or is concentrated at the centre of the beam, or is concentrated at several points along the beam, each half of beam being loaded similarly, then each sup- port will react just one half of the total load. THE PRINCIPLE OF MOMENTS. Law of Lever. The law of the lever is well known. The distance of a force from its fulcrum or point where it takes effect is called its leverage. The effect of the force at such point is equal to the amount of the force multiplied by its leverage. Moment of a The effect of a force (or load) at any point of a force. beam is called the moment of the force (or load) at said point, and is equal to the amount of the force (or load) multi- plied by the distance of the force (or load) from said point, the distance measured at right angles to the line of the force. If therefore we find the moments — for all of the forces acting on a beam — at any single point of the beam we know the total moment at said point, and this is called the bending-momenl at said point. Of course, forces BENDING MOMENT. 49 actinr' in opposite directions will give opposite mo- Bend Ing oii »ri4i moment, nients, and will counteract each other; to hnd tue bending-moment, therefore, for any single i)oint of a beam take the difference between the sums of the opposing moments of all forces acting at that point of the beam. Now on any loaded beam we have two kinds of forces, the loads which are pressing downwards, and the supports which are resisting upwards (theoretically forcing upwards). Again, if we imagine that the beam will break at any certain point, and imagine one side of the beam to bo rigid, while the other side is ten ling to break away from the rigid side, it is evident that the effect at the point of rupture will be from one side only ; therefore we must take the forces on one side of the point only. It will be found in practice that no matter for what point of a beam the bending moment is sought, the bending moment will be found to be the same, whether we take the forces to the right side or left side of the point. This gives an ex- cellent check on all calculations, as we can calculate the bending moment from the forces on each side, and the results of course should be the same. Now to find the actual strain on the fibres of any cross-section of the beam, we must find the bending moment at the point where the cross-section is taken, and divide it by the moment of resistance of the fibre, or, ^= s r Where m = the bending moment in lbs. inch. Where r = the moment of resistance of the fibre in inches. Where s = the strain. The stress, of course, will be equal to the resistance to cross-break- ing the fibres are capable of. In the case of beams which are of uni- form cross-section above and below the neutral axis, this resistance is called the Modulus of Rupture (k). It is found by experiments and tests for each material, and will be found in Tables IV and V. AVe have, then, for uniform cross-sections : — V =k Where v = the ultimate stress per square inch. Where k =■ the modulus of rupture per square inch. Inserting this and the above in the fundamental formula (1), viz.: v^sf, we have : — , 771/. « = —./, or 50 SAFE BUILDING. Transverse ^ strength uni- — -. — = r (18) form cross sec- { -L\ tion. \fj Where ?/i = the bending moment in lbs. inch at a given point of beam. Where r = the moment of resistance in inches of the fibres at said 23oint. Where ( —.) =the safe modulus of rupture of the material, per square inch. If the cross-section is not uniform above and be- "'"sfre ngth ^ s e c- lo^^ the neutral axis, we must make two distinct tion not uni- calculations, one for the fibres above the neutral axis, the other for the fibres below ; in the former case the fibres would be under compression, in the latter under tension. Therefore, for the fibres above the neutral axis, the ultimate stress would be equal to the ultimate resistance of the fibres to com- pression, or v = c. Inserting this in the fundamental formula (1), we have: — ?« //6 y. Upper fibres. { — \ (19) Where ??i = the bending moment in lbs. inch, at a given point of beam. Where r = the moment of resistance in inches of the fibres at said point. AVhere ( — j = the safe resistance to crushing of the material, per square inch. For the Jihres below the neutral axis, the ultimate stress would be equal to the ultimate resistance of the fibres to tension, or, v=: t. Inserting this in the fundamental formula (1) we have : — t = — ./, or -^ = r (20) Lower fibres. / f \ '■ -' (7) Where m = the bending moment in lbs. inch at a given point of beam. GREATKST liK.NDlNG MOMKNT. 51 Where r = the moment of resistance in inches of the (il)res at said point. AYliere (— ) = the safe resistance to tension of the material, per square inch. The same formulaj apply to cantilevers as well as beams. Tlie moment of resistance r of any fibre is equal to the moment of inertia of the whole cross-section, divided by the distance of the fibre from the neutral axis of the cross-section. The greatest strains are along the upper and Greatest strains lower edges of the beam (the extreme fibres) ; we, on extreme fi- therefore, only need to calculate their resistances, as all the intermediate fibres are nearer to the neutral axis, and, consequently, less strained. The distance of fibres chosen iu calculating the moment of resistance is, therefore, the dis- tance from the neutral axis of either the upper or lower edges, as the case may be. The moments of resistance given in the fourth column, of Table I, are for the upper and lower edges (the extreme fibres), and should be inserted in place of r, in all the above formulae. To find at what point of a beam the greatest bend- Point of great- i"? moment takes place (and, consequently, the est bending greatest fibre strains, also), begin at either support moment. °^^ ^^^^ ^^^^^^ ^^^ ^^^^ towards the other sup- port, passing by load after load, until the amount of loads that have been passed is'cqual to the amount of the reaction of the support (point of start) ; the point of the beam wliere this amount is reached is the point of greatest bending moment. In cantilevers (beams built in solidly at one end and free at the other end), the point of greatest bending moment is always at the point of the support (where the beam is built in). In light beams and short spans the weight of the beam itself can be neglect'^d, but in heavy or long beams the weight of the beam should be considered as an independent uniform load. RULES FOR CALCULATING TRAXSVERSE STRAINS. 1. Find Reaction of each Support. Summary of If the loads on a girder are uniformly or sym- '*"'®®' metrically distributed, each support carries or re- acts with a force eciual to one-half of the total load. If the weights are unevenly distributed, each support carries, or the reaction of each support is equal to, the sum of the products of each load into its 62 SAFE BUILDING. distance from the other support, divided by the whole length of span. See Formula; (14), (15), (16), and (17). 2. Find Point of Greatest Benxling Moment. The greatest bending moment of a uniformly or symmetrically distributed load is always at the centre. To find the. point of great- est bending moment, when the loads are unevenly distributed, begin at either support and pass over load after load until an amount of loads has been passed equal to the amount of reaction at the support from which the start was made, and this is the desired point. In a cantilever the point of greatest bending moment is always at the wall. 3. Find the Amount of the Greatest Bending Moment. In a beam (supported at both ends) the greatest bending moment is at the centre of the beam, provided the load is uniform, and this moment is ecjual to the product of the whole load into one-eighth of the length of span, or rn=^^ , (21) Where m = the greatest bending moment (at centre), in lbs. inch, of a uniformly-loaded beam supported at both ends. Where u = the total amount of uniform load in pounds. Where I = the length of span in inches. If the above beam carried a central load, in place of a uniform load, the greatest bending moment would still be at the centre, but would be equal to the product of the load into one-quarter of the length of span, or m = — (22) Where m = the greatest bending moment (atxientre), in lbs. inch, of a beam with concentrated load at centre, and supported at both ends. Where w = the amount of load in pounds. Where I = the length of span in inches. To find the greatest bending moment of a beam, supported at both ends, with loads unevenly distributed, imagine the girder cut at the point (previously found) where the greatest bending moment is known to exist ; then the amount of the bending moment at that point will be equal to the product of the reaction (of either support) into its distance from said point, less the sum of the products of all the loads on the same side into their respective distances from said point, i. e., the point where the beam is supposed to be cut. To check the whole calculation, try the reaction and loads of the discarded side of the beam, and the same result should be obtained. RULKS TRANSVERSE STRAINS. 53 Amount of greatest bend- ing moment. To put tlie above in a formula, v;e should have : — 7)},,=p.x-'Z(u\ X, -f ?r„ x,, -)- «,•„, a:,,,, etc.) (23) And as a checL to above : ?«,v = q.(l-x)-^ ((i'„„ a-.„, + w, X, + 10^, x,,) etc. (24) Where A = is the point of greatest bending moment. Where m^^ is the amount of bending moment, in lbs. inches, at A. Where ;j = is the left-hand reaction, in pounds. Where 7 ^ is the right-liand reaction, in pounds. Where x and (/-.r) = tlie respective dis- tances in inches, of the left and right reac- tions from A. ® Where x,, a:,,, a;,,,, etc., = the respective distances, in inches, r-Xi- — [ — Xv- -^ Fig. 9. from A, of the loads to,, w„, w,„, etc. Where u\, w,,, ii\,„ etc., = the loads, in pounds. Where S z= the sign of summation. Tlie same formulfe, of course, would hold good for any point of beam. In a cantilever (supported and built in at one end only), the great- est bending moment is alwai/s at the point of support. For a uniform load, it is equal to the product of the whole load into one-half of the length of the free end of cantilever, or u.l , ^ '"==-2- (25) Where m z= the amount, in lbs. inch, of the greatest bending moment (at point of support). Where u =z the amount of the whole uniform load, in pounds. Where I =z the length, in inches, of the free end of cantilever. For a load concentrated at the free end of a cantilever, the great- est bending moment is at the point of support, and is equal to the product of the load into the length of the free end of canti- lever, or m = w. I (2G) Where m = the amount, in lbs. inch, of the greatest bending moment (at j)oint of support). 54 SAFE BL'ILDIXG. Where 26" :^ the load, in pounds, concentrated at free end. Where I = the length, in inches, of free end of cantilever. For a load concentrated at any point of a cantilever, the greatest bending moment is at the point of support, and is equal to the pro- duct of the load into its distance from the point of support, or m = w. X (27) Where m = the amount, in lbs. inch, of the greatest bending moment (at point of support). Where w z= the load, in pounds, at any point. Where x ■=:■ the distance, in inches, from load to point of support of cantilever. Note, that in all cases, when measuring the distance of a load, we must take the shortest distance (at right angles) of the vertical neutral axis of the load, (that is, of a vertical line through the centre of gravity of the load.) 4. Find the Required Cross-section. To do this it is necessary first to find what wiU be the required moment of resistance. If the cross-section of the beam is uniform above and below the neutral axis, we use Formula (18), viz. : — (I) k_ /' If the cross-section is unsymmetrical, tliat is, not uniform above and below the neutral axis, we use iov i\\Q fibres above the ntutral axis, formula (19), viz.: — and for the fibres below the neutral axis, Formula (20), viz. m (f) In the latter two cases, for economj', the cross-section should be so designed that the respective distances of the upper and lower edges (extreme fibres), from the neutral axis, should be proportioned to their respective stresses or capacities to resist compression and ten- sion. This will be more fully explained under cast-iron lintels. A simple example will more fully explain all of the above rules. EXAMPLE TRANSVERSE STRAINS 55 Example. Three weights of respectivehj 500 lbs., 1000 lbs., and 1500 lbs., are placed on a beam oflV 6" (^or 210") clear span, 2' 6" {or 30"), V 6" {or 90"), and 10' 0" (or 120") from the left-hand support. The mod- ulus of rupture of the material is 2800 lbs. per square inch. The fuc- tor-of safety to be used is 4. The beam to be of uniform cross-section. What size of beam should be used? 1. Find Reactions {see formulce 16 and 17). 500.180 , 1000.120 , 1500.90 Reaction p will be in pounds, = 1642f pounds. 210 Reaction q will be in pounds, = — irV" H~ 210 1000.90 + 210 1500.120 210 ' 210 ' 210 1357^ pounds. Check, p -\- q must equal whole load, and we have in effect: — p-\-q=lG42'f-}- 13571 = 3000, which being equal to the sum of the loads is correct, for : — 500-1- 1 000 -f- 1500 — 3000. 2. Find Point of Greatest Bending Moment. Begin at jo, pass over load 500, plus load 1000, and we still need to pass 142|^ pounds of ^500) 2q^j j.^ make up amount of reaction p (1642| lbs.); therefore, the greatest bending mo- ment must be at load 1500; check, he^xn at 9 and we arrive only at the first load (1500) be- fore passing amount of reaction q (1357J lbs.), (-3o^- ■l8o4 - ■- — -y^ - -=i 9^ •-\{2o- -4-- 9^- 210'- Fig. 10. therefore, at load 1500 is tiie point sought. 3. Find Amount of Greatest Bending Moment. Suppose the beam cut at load 1500, then take the left-hand side of beam, and we have for the bending moment at the point where the beam is cut. m = 1642f 120— (500.90 + 1000.30 -\- 1500.0)' = 197143— (45000-j-30000-f0) = 197143 — 75000 == 122143 lbs. inch. 56 SAFE BUILDING. As a check on the calculation, take the right-hand side of beam and we should have : — m = 13571-.90 — 1500.0 = 122143 — = 122143 lbs. inch, which, of course, proves the correctness of former calculation. 4. Find the Required Cross-section of Beam. We must first find the required moment of resistance, and as the cross-section is to be uniform, we use formula (18), viz. : — m r i-f) Now, m = 122143, and -^ = = 700, therefore, / 4 r = " ^ = 174,49 or say = 174,5 700 J J y Consuming Table I, fourth column, for section No. 2, we find r = —-, we have, therefore, 6 — = 174,5 or 5^2 _ 1047. 6 If the size of either 6 or rf is fixed by local conditions, we can, of course, find the other size (d or h) very simply ; for instance, if for certain reasons of design we did not want the beam to be more than 4" wide, we should have 6 = 4, therefore, 4.^/2=1047, and ^2_ 1047 _ .^^2^ therefore, tZ= (about) 16"', or, if we did not want the beam to be over 12" deep, we should have d = 12, and d'^ = 12.12 = 144, therefore, "&.144 = 104 7, and b = i^ = 7,2" or say 1\''. 144 The deepest One thing is verv important and, must be remem- beam the most , , , , , ' , , , economical. oered, that tiie deeper the beam is, the more eco- nomical, and the stiffer will it be. If the beam is too shallow, it might deflect so as to be utterly unserviceable, besides using very much more material. As a rule, it will therefore be necessary to calculate the beam for deflection as well as for its transverse strength. The deflection should not exceed 0,"03 that is. Safe deflection. ^^^reQ one-hundredths of an inch for ea(;h foot of span, or else the plastering would be apt to crack, we have then the formula : — 8 = 1. 0,03 (28) DEFLECTION. 67 Where 6 = the greatest allowable total deflection, in inches, at centre of beam, to prevent [)laster cracking. Where L =. the length of span, in feet. In case the beam is so unevenly loaded that the greatest deflection will not be at the centre, but at some other point, use : — 8 =- A'. 0,06 (29) Where 6 = the greatest allowable total deflection, in inches, at point of greatest deflection. Where X = the distance, in feet, to nearer support from point of greatest deflection. If the beam is not stiffened sideways, it should also be calculated for lateral flexure. These matters will be more fully explained when treating of beams and girders. COMPARATIVE STRENGTH AND STIFFNESS OF BEAMS AND COLUMNS. (1) If abeam supported at both ends and loaded uniformly will safely carry an amount of load =u; then will the same beam : (2) if both ends are built in solidly and load uniformly distributed, carry 1^. u, (3) if one end is supported and other built in solidly and load uni- formly distributed, cai-ry 1. u, (4) if both ends are built in solidly and load applied in centre, carry 1. w, (5) if one end is supported and other built in solidly and load ap- plied in centre, carry §. u, (G) if both ends are supported and load applied in centre, carry ^. u, (7) if one end is built in solidly and other end free (cantilever) and load uniformly distributed, carry ^. u, (8) if one end is built in solidly and other end free (cantilever) and load applied at free end, carry ^. it. That is, in cases (1), (3) and (4) the effect would be the same with the same amount of load ; in case (2) the beam could safely carry 1^ times as much load as in case (1) ; in case (5) the beam could safely carry only § as much as in case (1), etc., provided that the length of span is the same in each case.^ If the amount of deflection in case (1) were <5, then would the amount of deflection in the other cases be as follows : Case (2) (5„ = i. (5, Case (4) d,y = |. P Vi » II «r+ II CI- 11 w s o S o o 09 OO OO OO II » II II 93 II 85 WJH- S w| H- & w| i— Qc] 1— ^ a a o a S (I o a a o a 'B p. a u. TABLE VII. 59 VII. (See foot-uote p. CO.) OF BEAMS AND CANTILEVEKS FOR VARIOUS LOADS. -^ .-o* — HJ -0 — ^TrVfl? F^ Load at any point on beam sup- ported at both ends. Load at centre of beam support- ed at both ends. Uniform load on beam supported at both ends. Descrip- tion. If M smaller If w greater than than ±- 2 For Xi use: For x use: At load When X greater than l use (l-x) in place of x. "2" to) fi 'O] C; ? to|e 5§ in II HI r» '^ CD c II g"S|^ c^is ^ lis •-J II t*|S p II t*I 8 at support p or (f. Location and amount of greatest shearing- strain a. 00 00 cy? 1 II II i-h f «c SS § >l^l ^ ?= Oil » 1? g 5*^ «« ex c n, » a *■ a S- J^.i « >-« «.|S D . a a - =!■ a a '^'l 'es a> « c 01 t « 60 SAFE BUILDING. Strength of The comparative transverse strength of two or ferent cross-sec- ^^ove rectangular beams or cantilevers is directly *'°"®' as the product of their breadth into the square of their depth, provided the span, material and manner of supporting and loading are the same, or xz=bd^ (30) Where a; = a figure for comparing strength of beams of equal spans. AVhere h = the breadth of beam, in inches. Where d= the depth of beam, in inches. Example. What is the. comparative strength between a 3" x 12" beam, and a 6" X 12" beam? Also, between a 4" x 12" beam, and a 3" x 16" beam? All beams of same material and span, and similarly supported and loaded. The strength of the 3" x 12" beam would be ar,=:3. 12. 12 = 432. The strength of the 6" x 1 2" beam would be a:„ = 6. 12. 12 = 864, therefore, the latter beam would be just twice as strong as the former. Again, the strength of the 4" x 12" beam would be x„. = 4. 12. 12r=576 and the strength of the 3" x 16" beam would be a;.,„ = 3. 16. 16=768. The latter beam would therefore be or iust li times as strono' as 576 •■ ^ ° the former, while the amount of material in each beam is the same, as 4. 12 = 3. 16 = 48 square inches in each. The reason the last beam is so much stronger is on account of its greater depth. Strength of The comparative transverse strength of two ur beams of dif- , .-i c • , ferent lengths, niore heams or cantilevers of same cross-section and material, but of unequal spans, is inversely as their lengths, provided manner of supporting and loading are the same. That is, a beam of twenty-foot span is only half as strong as a beam of ten-foot span, a quarter as strong as one of five-foot span, etc. AH measurements in Table VII are in inches; all weights in pounds; c = mod- ulus of elasticity in pounds inch ; i=: moment of inertia of cross-section of beam or cantilever around its neutral axis in inches ; ??i = banding-moment in pounds Inch; s= amount of shearing strain in pounds; 6 = total amount of deflection in Inches. COMPAKATIVK STIFFNESS. 61 Stiffness Of The stiffness of beams or cantilevers of same ferent lengths.' fross-section and material (and similarly loaded and supported), however, diminishes very rapidly, as the length of span increases, or what is the same thing, the dellection increases much more rapidly in proportion than the length; the comparative stiff- ness or dellection being directly as the cube of their respective lengths or Z*. That is if a beam 10 feet long deflects under a certain load one- third of an inch, the same beam with same load, but 20 feet long will deflect an amount x as follows : a; : i = 208: 103, or a: = ?^ = ^^= 2r ^ 1U3 3000 ^ Stiffness of beams of dif- J he comparative stiffness, that is amount of de- ferent cross-sec- n t- c . i, -i • • tions. uection ot two or more beams or cantilevers, simi- larly supported and loaded, and of same material and span, but of different cross-sections, is inversely as the product of their respective breadths into the cubes of their respective depths or Where x = a figure for comparing the deflection of beams of same material, span and load. Where 6=^ the breadth of beam, in inches. Where c?= the depth of beam, in inches. Example. If a beam 3" x 8" deflects ^" under a certain load, ivhat will a beam 4" X 12" dcjlect, if of same material and span, similarly supported and with same load f For the first beam we should have For the second beam we should have X.. = -i— = ^ = 0,00014 " 4.123 6912 ' The deflection of the latter beam will be as S : 0",5 = 0,00014 : 0,00065, or S = 0",108 Strength of The comparative strength of rectangular beams fere^n^ngths & °^ cantilevers of different cross-sections and spans, cross-sections. but of same materials and similarly loaded and snp- ported, is, of course, directly as the product of their breadth into the 62 SAFE BUILDING. squares of their depths, divided by their lengtli of span, or x = ^ (32) Where x = a, figure for comparing the strength of different beams of same material, but of different cross-sections and spans. Wliere i=rtlie breadth, in inclies. AVhere d = thc depth, in inches. Where i = the length of span, in feet. Stiffness of "^^^^ comparative stiffness or amount of deflection beams of dif- of different rectangular beams or cantilevers of ferent lensths& , . , , , n , i i cross-sections, same material, and similarly loaded and supjiorted, but of different cross-sections and spans, would be directly as the cubes of their respective lengths, divided by the product of their respective breadths into the cubes of their depths or Where a; = a figure for comparing the amount of deflections of beams of same material and load, but of different spans and cross- sections. Where L = the length of span, in feet. Where b = the breadth, in inches. Where (I = thc depth in inches. Strength and ^^ ^^ '^ desired to calculate a wooden girder sup- deflection of iiorted at both ends and to carr\- its full safe uniform wooden beams, , , , , ^ , one inch thick, load, and yet not to deflect enough to crack plaster, the following will simphfy the calculation : TABLE VIII. Spruce. Georgia piue. "White pine. White oak. Hemlock. Calculattt (x) for transverse strength only if d is greater than HL L l-^L IIL \-t^ Calculate (j-,) for deflection .mly if d is less than H^ L l^\L HL HL Where L = the length of span, in feet. Where d = the depth of beam, in inches. Where a: or x, = is found according to Table IX. To find the safe load (.r) or (r,) per running foot of span, which a beam supported at both ends, and 1" ihick will carry, use the follow- ing table. (Beams two inches thick will safely carry twice as much WOODEN BEAMS. 63 per runnin Fig. 20. 72 SAFE BUILDING. measured / ■e' by the same scale as e d, is the amount of force re- quired for the stick D A to exert, while a e, 1 ^^ measured by the same scale, is the amount of / ^\ force required for the stick A E to exert. If ' \ in place of the force E D we had had a load, the same truths would hold good, but /'■' / we should represent the load • ' / by a force acting downward Q<'' / in a vertical and plumb line. \ / Thus, if two sticks, B A and \ ' A C, Fig. 23, are supporting A a load of ten pounds at their Figs. 21 and 22. summit, and the inclination of each stick from a horizontal line is 45°, we proceed in the same manner. Draw c 6, Fig. 24, at any scale equal to ten units, through h and c draw h a and a c at angles of 45° each, with c b, then meas- ure the number of (scale measure) units in 6 a and a c, which, of course, we find to be a little over seven. Therefore, each stick must resist with a force equal to a little over seven pounds. Now, to find the di- rection of the forces. In Fig. 23 we read CB,BAand AC,the corresponding parts in the strain diagram, Fig. 24, are c b, h a and a c. Now the direction of c 6 is downwards, therefore C B acts downwards, which is, of course, the effect of a weight. •"'£«■ 23 and 24. The direction, however, of 6 a and a c is upwards, therefore B A and A C must be pushing upwards, or towards the weight, and therefore they are in compression. The same truths hold good no matter how many forces we have acting at any point; that is, if the point remains in equilibrium (all the forces neutralizing each other), we can con- struct a strain diagram which will always be a chimed polygon with as many sides as there are forces, and each side equal and parallel to one of the forces, and the sides being in the same succession ANALYSIS OF TRUSS. 73 Roof Trusses, simple truss. Fig. 25. to each other as the forces are. We can now proceed to dissect a Take a roof truss with two rafters and a single tie-beam. The rafters are sup- posed to be loaded uni- formly, and to be strong enough not to give way transversely, but to *v transfer safely one-half \Cp^3,' l^of the load on each rafter to be supported on each joint at the ends of the rafter. Wo consider each joint separately. Take joint No. 1, Fig. 25. We have four forces, one O A (the left-hand reaction), being equal to half the load on the whole truss ; next, A B, equal to half the load on the rafter B E. Then we have the force acting along B E, of which we do not as yet know amount or direction (up or down), but only know that it is parallel to B E; the same is all we know, as yet, of the force E O. Now draw, at any scale, Fig. 26, No. 1, a =:and parallel to O A, then from a draw a b = and parallel to A B (a 6 will, of course, lap over part of a, but this does not affect anything). Then from b draw h e parallel to B E, and through o draw e o par- ^ allel to E O. Now, in read-|^ ing off strains, begin at O A, then pass in succession to A B, B E and E O. Follow on the strain diagram Fig. 26, No. 1, the direction as read off, with the finger (that j is, a, a h, b e and e o), and -i^ we have the actual direc- tions of the strains. Thus o a is up, therefore pushing up; a 6 is down, therefore pushing down; & e is downwards, therefore pushing against joint No. 1 (and we know it is compression) ; lastly, e o is 74 BAFE BUILDING. pushing to the right, therefore puhing away from the joint No. 1, and -we know it is' a tie-rod. In a similar manner we examine tlie- joints 2 and 3, getting the strain diagrams No. 2 and No. 3 of Fig. 26. In Fig. 27, we get tlie same results exactly as in the above three diagrams of Fig. 26, only for simplicity they are combined into one diagram. If the single (combination) diagram. Fig. 27, should prove confusing to the student, let him make a separate diagram for each joint, if he will, as in Fig. 26. Tlie above gives the principle jQ of calculating the strength of trusses, graphically, and will be more fully used later on in practical examples. Should the student desire a fuller knowledge of the subject, let him refer to " Greene's Analysis of Roof Trusses," which is simple, short, and by far the best manual on the subject. In roof and other trusses the line Line of Pressure ^ , • n i Central. of pressure or tension will always be co-incident with the central line or longitudinal axis of each piece. Each joint should, therefore, be so designed that the central lines or axes of all the pieces will go through one iwint. Thus, for in- Btance,^he foot of a king post should be designed as per Fig. 28. In roof-trusses where the rafters support purlins, the rafters must not only be made strong enough to resist the compressive strain on Fig. 27. N^ ^ M i y ^ 7 them, but in addition to this enough material must be added to stand the transverse strain. Each part of the rafter is treated as a separate beam, supported at each joint, and the amount of reaction at each joint must be taken as the load at the joint. The same holds good of the tie-beam, when it '^* has a ceiling or other weights suspended from it ; of course these weights must all be shown by arrows on the drawing of the truss, so a°s to get their full allowance in the strain diagram. Strains in opposite directions, of course, counteract each otlier ; the stress, therefore, to be exerted by the material need only be equal to the difference between the amounts of the opposing strains, and, of course, this stress will be directed against the larger strain. ANALYSIS OF ARCH. 76 Fig. 29. We consider an arch as a truss with a 8 ucces sion of straight pieces; w e can calculate it graphically the same as any other ■"""''1°'^ truss, only we will find that the ab- sence of central or inner members (struts and ties) "will force the line of pressure, as a rule, far away from the central axis. Thus, if in Fig. 29, we con- sider A B C D as a loaded half-arch, we know that it is held in place by three forces, viz.: 1. The load B C L M which acts through its centre of gravity as indicated by arrow No. 1. 2. A horizontal force No. 2 at the crown C D, which keeps the arch from spreading to the right. 3. A force at the base B A (indicated by tlie arrow No. 3), which keeps the arch from spreading at the base. Now we know the direc- tion and amount of No. 1, and can easily find Nos. 2 and 3. In an arch lightly loaded, No. 2 is always assumed to act at two-thirds way down C D, that is at F (where CE=:EF = FD = |C D). In an arch heavily loaded, No. 2 is always assumed to act one-third way down C D, that is at E; further the force No. 3 is always assumed to act through a point two-thirds way down B A, that is at H (where BG = GII=:HA=:^B A). The reason for these assumptions need not be gone into here. Therefore to find forces Nos, 2 and 3 proceed as follows : If the arch is heavily loaded, draw No. 2 hori- zontally through E (C E being equal to ^ C D), prolong No 2 till it intersects No. 1 at O, then draw O H (H A being equal to ^ B A), •which gives the direction of the resistance No. 3. We now have the three forces acting on the arch concentrated at the point O, and can easily find the amounts of each by using the parallelogram of forces. Make O I vertical and (at any scale) equal to whole load (or No. 1), draw I K horizontally, till it intersects O II at K; then scale I K, 76 SAFE BUILDING. and K O (at same scale as O I), which will give the amount of the forces Nos. 2 and 3. The line of pressure of this arch A B C D, is Line of pressure therefore not through the central axis, but along not central. E O H (a curve drawn through E and H with the lines Nos. 2 and 3 as tangents is the real line of pressure). Now let in Fig. 30, A B C E F D A represent a half-arch. We can examine A B C D same as before, and obtain I K = to force Fig. 30. , , , . No. 2; KO = to resistance (and direction of rfe^'^s..^ I / same) at U, where U C = J C D;OI being equal to the load on D A. Now if we consider the whole arch from A to F, we proceed similar- ly. L G is the neutral axis of the whole load from A to F, and is equal to the whole load, at same scale as O I. That L G passes through D is accidental. Make E M = ^ E F and draw L M ; also G H horizontally till it intersects L M at H, then is G II the horizontal force or No. 2. AVe now have two different quantities for force No. 2, viz. : I K and G H, I K in this case being the larger. It is evident that if the whole half- arch is one homogeneous mass, that the greatest horizontal thrust of any one part, will be the horizontal thrust of the whole, we select therefore the larger force or I K as the amount of the horizontal thrust. Now make S P = to I K and P Q = to No. 1 , or load on A D and P R = to L G or whole load on F A, at any scale, then draw Q S and R S. Now at O we have the three forces concentrated, which act on the part of arch A B C D, viz. : Load No. 1 (= P Q), horizontal force No 2 (= S P) and resistance K O (= Q S). Now let No. 3 repre- sent the vertical neutral axis of the part of whole load on F D, then LIXE OF TRESSURE. 77 prolong K until it intersects No. 3 at T; then at T we have the three forces acting on the part of arch E CDF, viz. : The load No. 8 (= Q 11), the thrust from A B C D, viz. : O T (= S Q), and the resistance N T (= R S). To obtain N T draw through T a line par- allel to R S, of course R S giving not only the direction, but also the amount of the resistance N T. The line of pressure of this arch therefore passes along P O, O T, T N. A curve drawn through points P, U and N — (that is, Avhere the former lines intersect the joints A B, D C, F E) — and with lines P O, O T and T N as tan- gents is the real line of pressure. Of course the more parts we divide the arch into, the more points and tangents will we have, and the nearer will our line of pressure approach the real curve. Now if this line of pressure would always pass through the exact centre or axis of the arch, the compression on each joint would of course be uniformly spread over the whole joint, and the amount of this compression on each square-inch of the joint would be equal to the amount of (line of) pressure at said joint, divided by the area of the cross-section of the arch in square inches, at the joint, but this rarely occurs, and as the position of the line of pressure varies from the central axis so will the strains on the cross section vary also. Stress at intra- Let the line A B in all the followinf fio-ures renre- dos and extra- . ., ,. r . ■ e " , ^, , , dos. sent tlie section of any joint of an arch (the thick- ness of arch being overlooked) C D the amount and actual position of line of pressure at said joint and the small arrows the sti-ess or resistance of arch at the joint. We see then that when C D is in the centre of A B, Fig. 31, the stress is uniform, that is the joint is, uniformly compressed, the > — V amount of compression (C ) being equal to the aver- 1^^ age as above. As the I line of pressure C D ap- A Id . fe Proachos one side, Fig. illttttttttt t M tft tit tttttti ^^' ^^^^ amount of com- Pjg^ 3 1^ pression on that side in- creases, while on the fur- © ther side it decreases, until the line of pressure CD, reaches one-third Fig. 32. see there is no compres- 78 SAFE BUILDING. A • *♦■»♦! tttt ft t I h sion at A, but at B the compression is equal to just double the average as it was in Fig. 31. Now, as C D passes beyond the central third of '^ilN A B, Fig. 34, the com- ' pression at the nearer side increases still fur- ther, while the further side begins to be sub- jected to stress in the opposite direction or ten- sion, this action increas- ing of course the further C D is moved from the t r t t L) central third. This means that the edge of arch section at B would be^subject to very severe crushing, while the other When the line C D ng. 33. AiUi I n Fig. 34. edge (at A) would tend to separate or open, passes on to the edge B, the nearer two-thirds of arch joint will be in compression, and the further third in tension. As the line passes out of joint, and further and further away from B, less and less of the joint is in compression, while more and more is in tension, until the line of pressure C D gets so far away from the joint finally, that one-half of the joint would be in tension, and the other half in compression. Tension means that the joint is tending to open upwards, and as arches are manifestly more fit to resist crushing of the joints than opening, it becomes apparent why it is dangerous to have the line of pressure far from the central axis. Still, too severe crushing strains must be avoided also, and hence the desirability of trying to get the line of pressure into the inner third of arch ring, if possible. But the fact of the line of pressure coming outside of the inner third of ai'ch ring, or even entirely outside of the arch, does not necessarily mean that the arch is unstable ; in these cases, however, we must cal- culate the exact strains on the extreme fibres of the joint at both the inner and outer edges of the arch (intrados and extrados), and see to it that these strains do not exceed the safe stress for the material. The formulae to be used, are : For the fibres at the edge nearest to the line of pressure And for the fibres at the edge furthest from the line of pressure PRESSUKE ON JOINT. 79 a a.d ^ ' Where v = the stress in lbs., required to be exerted by the extreme edge fibres (at intrados and extrados). Where x = the distance of line of pressure from centre of joint in inches. Where a=:the area of cross section of arch at the joint, in square inches. Where p = the total amount of pressure at the joint in lbs. Where cf = the depth of arch ring at the joint in inches, measured from intrados to extrados. When the result of the formulae (44) and (45) is a positive quan- tity the stress v should not exceed (y\ that is the safe compressive stress of the material. When, however, the result of the formula (45) yields a negative quantity, the stress v should not exceed ('-> Y.that is the safe tensile stress nf the material, or mortar. The whole subject of arches will be treated much more fully later on in the chapter on arches. A Fig. 35. 80 SAFE BUILDING. TO ASCERTAIN AMOUNT OF LOADS. Let A B C D be a floor plan of a building, A B and C D are the walls, E and F the columns, with a girder between, the other lines being floor beams, all 12" between centres; on the left side a well- hole is framed 2' x 2'. Let the load assumed be 100 pounds per square foot of floor, which includes the weight of construction. Each Load on ^^ *'^® right-hand beams, also the three left-hand Beams, beams E L, K E and F Q will each carry, of course, ten square feet of floor, or 10.100= 1000 pounds each uniform load. Each will transfer one- half of this load to the girder and the other half to the wall. Tlie tail beam S N will carry 8 square feet of floor, or 8.100 = 800 pounds uniform load. One-half of this load will be transferred to the wall, the other half to the header R T, which will therefore carry a load of 400 pounds at its centre, one-half of which will be transferred to each trimmer. The trimmer beam G M carries a uniform load, one-half foot wide, its entire length, or fifty pounds a foot (on the off-side from well- hole), or 50.10=: 500 pounds uniform load, one-half of which is transferred to the girder and the other half to the wall. The trimmer also carries a similar load of fifty pounds a foot ou the well-hole side, but only between M and R, which is eight feet long, or 50.8 = 400 pounds, the centre of this load is located, of course, half way between M and R, or four feet from support M, and six feet from support G, therefore M will carry (react) = 240 pounds and G wiU carry 10 4.400 160 pounds. 10 See FormuljE (14) and (15). We also have a load of 200 pounds at R, transferred from the header on to the trimmer ; as R is two feet from G, and eight feet from M, we will find by the same formulae, that G carries 8.200 10 2.200 = IGO pounds and M carries = 40 pounds. 10 So that we find the loads which the trimmer transfers to G and M, as follows : At M=250 + 240-j- 40 = 530 pounds. « G = 250 + 160 + 160 = 5 70 pounds. DISTIilBUTION OF LOADS. 81 The loads which trimmer O I transfers to wall and girder will, of Load on course, be similar. We therefore find the total load- Walls, ing^ as follows: On the wall A B : At L = 500 pounds. " M= 530 pounds. " N= 400 pounds. « O = 530 pounds. " P = 500 pounds. " Q = _500 pounds. Total on wall A B == 2'JCO pounds. On the wall C D we have six equal loads of 600 pounds each, a Load on Girder, total of 3000 pounds. II O o I o II o o o o I I o o o It i ' ' I I'o— i I — 4'd"- • W H If- — ji^ — 2^o-- o --- _^.o-— f- I'O F -:>, f Fig. 36. On the girder E F, we have: At E from the left side 500 pounds, from the right 500 pounds. At G from the left side 570 pounds, from the right 500 pounds. At II from the left side nothing, from the right 500 pounds. At I from the left side 570 pounds, from the right 500 pounds. At K from the left side 500 pounds, from the right 500 pounds. Total 1000 pounds. Total 1070 pounds. Total 600 pounds. Total 1070 pounds. Total 1000 pounds. 82 SAFE DUILDING. At F from the left side 500 right 500 pounds. pounds, from the Total 1000 pounds. Total on girder 5640 pounds. As the girder is neither uniformly nor symmetrically loaded, we must calculate by Formulse (16) and (17), the amount of each reac- tion, which will, of course, give the load coming on the columns E and F. (These columns will, of course carry additional loads, from the girders on opposite side, further, the weight of the column should be added, also whatever load comes on the column at floor above.) Girder E F then transfers to columns, At E = 1000+ (f 1070) + (|. 500) 4- (|.1070) + (4. 1000) -}- (0. 1000) = 2784 pounds. At F = 1000 + (|. 1000) + (f . 1070) + (|. 500) + (i. 1070) + (0. 1000) = 2856 pounds. As a ciieck the loads at E and F must equal the whole load on the girder, and we have, in effect, 2 784-}- 2856 = 5640. Now as a check on the whole calculation the load on the two col- umns and two walls should equal the whole load. The whole load being 20'x6'x 100 pounds minus the well-hole 2'x2'x 100 pounds, or 12000 — 400 = 11600 pounds. And we have in effect. Load on A B = 2960 pounds. "CD = 3000 pounds. " two columns = 5640 pounds. Total loads =11600 pounds. We therefore can calculate the strength of all the beams, headers and trimmers and girders, with loads on, as above eiven. For the columns and walls, we must however add, the weight of walls and columns above, including all the loads coming on walls and columns above tlie point we are calculating for, also what- ever load comes on the columns from the other sides. If there are openings ... .. in a wall, one-half the Load over Wall openings. load over each opening goes to the pier each side of the opening, including, of course, all loads on the wall above the opening. Thus in Fi"^ure 37, the weight of walls would be distributed, as Fi2. 37, WIND AND SNOW. 83 indicated by etched linos; where, however, the opening in the wall is very small compared to the mass of wall-space over, it would, of course, be absurd to consider all this load as on the arch, and pract- ically, after the mortar has set, it would not be, but only an amount about equal to the part enclosed by dotted lines in Figure 38, the inclined lines being at an angle of 60° with the horizon. Where only part of the wall is calculated to be carried on the opening, the wooden centre should be left in until the mortar of the entire wall has set. In case of beams or lintels the wall should be built up until the intended amount of load is on them, leav- ing them free underneath; after the intended load is on them, they should be shored up, until the rest of wall is built and thoroughly set. Wind-pressure on a roof is Wind Pressure ,, , and Snow, gcnerallyassumcdat a certain load per square foot superficial measurement of roof, and added to the actual (dead) weight of roof ; except in large roofs, or where one foot of truss rests on rollers, when it is im- portant to assume the wind as a separate force, acting at right angles to incline of rafter. ^'^' ^^' The load of snow on roofs is generally omitted, when wind is allowed for, as, if the roof is very steep snow will not remain on it, while the wind pressure will be very severe ; while, if the roof is flat there will be no wind pressure, the allowance for which will, of course, offset the load of snow. If the roof should not be steep enough for snow to slide off, a heavy wind would probably blow the snow off. In case of " continuous girders," that is, beams or girders sup- ported at three or more points and passing over the intermediate supports without being broken, it is usual to allow moi'e load on the central suj)ports, than the formulae (14) to (17) would give. This subject will be more fully dealt with in the chapter on beams and girders. FATIGUE. If a load or strain is applied to a material and then removed, the material is supposed to recover its first condition (provided it has not been strained beyond the limit of elasticity). This practically, however, is not the case, and it is found that a small load or strain often applied and removed will do more damage (fatigue the mate- rial more) than a larger one left on steadily. Most loads in buildings 84 SAFE BUILDING. are stationary or " dead " loads. But where there are " moving " loads, such as people moving, dancing, marching, °* "Loads. etc., or machinery vibrating, goods being carted and dumped, etc., it is usual to assume larger loads than will ever be imposed ; sometimes going so far as to double the actual intended load, or what amounts to the same thing, doubling (or increasing) the factor-of-safety, in that case retaining, of course, the actual intended load in the calcu- lations. This is a matter in which the architect must exercise his judgment in each individual case. you^•DATION8. 85 CHAPTER II. FOUNDATIONS. Nature of Soils. The nature of the soils usually met with on build- in"' sites are: rock, gravel, sand, clay, loamy earth, "made" ground and marsh (soft wet soil). If the soil is hard and practically non-compressible, it is a good foundation and needs no treatment ; otherwise it must be carefully prepared to resist the weight to be superimposed. The base-courses of all foundation walls must be Stepping Courses, spread (or stepped out) sufficiently to so distribute the weio-ht that there may be no appreciable settlement (compression) in the soil. Two important laws must be observed : — 1. All base-courses must be so proportioned as to produce exactly the same pressure per square inch on the soil under all parts of buUdin"' where the soil is the same. Where in the same building we meet with different kinds of soils, the base-courses must be so pro- portioned as to produce the same relative pressure per square inch on the different soils, as will produce an equal settlement (compres- sion) in each. 2. "Whenever possible, the base-course should be so spread that its neutral axis will correspond with the neutral axis of the superim- posed weight ; otherwise there will be danger of the foundation walls setthng unevenly and tipping the walls above, producing unsightly or even dangerous cracks. Example. In a church the gable wall is 1' G" thick, and is loaded (including weight of all walls, floors and roofs coming on same) at the rate of b2 lbs, per square inch. The small piers are 12" x 12" and 5' high, and carry a floor space equal to 14' x 10'. What should be the size of base- courses, it being assumed that the soil will safely stand a pressure of 30 lbs. per square inch f If "we were to consider the wall only, we should have the total pressure on the soil per running inch of wall, 18.52 = 93G lbs. 86 SAFE BUILDIXG. Dividing this by 30 lbs., tlie safe pressure, we should need -^ =: 31,2" or say 32" width of foundation, or we should step out each side of foundation wall an amount — — - =: 7" each side. 2 Xow the load on pier, assuming the floor at 100 lbs. per square foot, would be 14 X 10 X 100 = 14000 lbs. To this must be added the weight of the pier itself. There are 5 cubic feet of brickwork (weighing 112 pounds per foot) = 5.11 2^500 lbs., or, including base- course, a total load of say 15000 lbs. This is distributed over an av- erage of 144 square inches ; therefore pressure per square inch under pier. 15000 , „ . ,rto „ = 104 or, say, 100 lbs. 144 We must therefore make the foundation under pier very much wider, in order to avoid unequal settlements. The safe pressure per square inch Ave assumed to be 30 lbs.; therefore the area requiisd would be = — orp ^= 500 square inches, or a square about 22" x 22". We therefore shall have to step out each side of the pier an amount "^ = 5 . 2 The safe compressions for different soils are given in Table V, but in most cases it is a matter for experienced judgment or else experi- ment. Testing ^^ ^^ usual to bore holes at intervals, considerablj' Soils, deeper than the walls are intended to go, at some s])Ot where no pressure is to take place, thus enabling the architect to judge somewhat of the nature of the soil. If this is not sufficient, he takes a crowbar, and, running it down, his experienced touch should be able to tell whether the soil is solid or not. If this is not suffi- cient, a small boring-machine should be obtained, and samples of the soil, at different points of the lot, bottled for every one or two feet in depth. These can be taken to the office and examined at leisure. The boring should be continued if possible, until hard bottom is struck. If the ground is soft, new made, or easily compressible, experi- ment as follows: Level the ground off, and lay down four blocks each, say 3" x 3" ; on these lay a stout platform. Alongside of jilai- form plant a stick, with top level of platform marked on same. Xow pile weight onto platform gradually, and let same stand. As soon as platform begins to sink appreciabl}' below the mark on stick, )0U WET FOUXDATIOXS. 87 have the practical ultimate resistance of the foundation ; this divided by 36 gives the ultimate resistance of the foundation per square inch. One-tenth of this only should be considered as a safe load for a per- manent building. Drainage of I^rainage is essential to make a building healthy, but -oi'- can hardly be gone into in these articles. Sometimes it is also necessary to keep the foundations from being undermined. It is usual to lead off all surface or spring water by means of blind drains, built underground with stone, gravel, loose tile, agricultural- tile, half-tile, etc. To keep dampness out, walls are cemented and then asphalted, both on the outsides. If the wall is of brick, the Damp-proof- cementing can be omitted. Damp-courses of slate or '"£• asphalt are built into walls horizontally, to keep damp- ness from rising by capillary attraction. Cellar bottoms are con- creted and then asphalted ; where there is pressure of water from c & I l^'^- 39. Fig. 40. underneath, such as springs, tide-water, etc., the asphalt has to be sufficiently weighted down to resist same, either with brick paving or concrete. Where there are water-courses they should be diverted from the foundations, but never dammed up. They can often be led into iron or other wells sunk for this purpose, and from there pumped into the building to be used to flush water-closets, or for manufacturing or other purposes. Clay, particularly in vertical or inclined layers, and sand are the foundations most dangerously a£fected by water as they are apt to be washed out. Where a very wide base-course is required by the nature of the soil, it is usual to step out the wall above gradually ; the angle of stepping should never be more acute than 60°, or, as shown in Figure 39. Care must also be taken that the stepped-out courses are sufliciently wide to project well in under each other and wall, to prevent same break- ing through foundation, as indicated in Figure 40. Where, on account of party lines or other buildings, the stepping 88 SAFE BUILDING. out of a foundation wall has to be done entirely at one side, the step- ping should be even steeper than 60°, if possible ; and particular at- tention must be paid to anchoring the walls together as soon and as thoroughly as possible, in order to avoid all danger of the foundation wall tipping outwardly. Where a front or other wall is composed of isolated piers, it i> well to combine all their founda- tions into one,and to step the pier? down for this purpose, as shown ^ in Figure 41. Where there is not F\i,. 4i. sufficient depth for this purpose, inverted arches must be resorted to. The manner of calculating the strength of inverted arches will be given under the article on arches. Inverted arches are not recom- _ . mended, however (except where the foundation wall is Inverted '. , ,, ^ . arches, by necessity very shallow), as it requires great care and good mechanics to build them well. Two things must particularly be looked out for : 1, That the end arch has sufficient pier or other abut- ment; otherwise it will throw the pier out, as indicated in Figui-e 42. (This will form part of calculation of Fig. 42. sti-ength of arch.) "Where there is danger of this, ironwork should be resorted to, to tie , ^, back the last pier. 8lZ6 of Sk6W* back, 2. The skew-back of the arch should be sufficiently wide to take its proportionate share of load from the pier (that is, amount of the two skew-backs should be proportioned to balance of pier or centre jiart of pier, as the width of opening is to width of pier) ; otherwise the pier would be apt to crack and settle past arch, as shown in Figure 43. An easy way of getting the graphically is given below. In Figare 44, width of Fig. 43. skew-back INVERTED ARCHES. 89 draw A B horizontally at springlng-Iine of inverted arch; bisect A C at F, and C B at E. Draw E O at random to vertical through F ; then draw C, and parallel to C draw G D ; then is C D the required skew-back.* A good way to do is to give the arches wide skew-backs, and then to introduce a thick granite or blue, stone pier stone over them, as shown in Figure 45. This will force all down evenly and avoid cracks. The stone must be Fig. 44. thick enough not to break at dotted lines, and should be carefully bedded. Example. A foundation pier carrying 150000 lbs. is 5' wide and 3' hroad The inverted arches are each 24" deep. What thickness should the granite block have ? W e have here vh'tual- ly a granite beam, 60" long and 36" broad, s u p - ported at two points (the centre lines of s k e w- backs) 30" apart. The \ Ffg. 45. load is a uniform load of 150000 lbs. The safe modulus of rupture, according to Table V, for average granite is ( -— j = 180 lbs. iln reality C D should be somewhat larger than the amount thus obtained; but this can be overlooked, except in cases where the pier approaches in widtli the width of opening. In such cases, however, stepping can generally be resorted to in place of inverted arches. Then, too, if the opening were very wide and the line of pressure came very much outside of central third of C D, it might be necessai-y to still f ui-thcr increase the width of skew-back, C D. 90 SAFE BUILDING The bending moment on this beam, according to Formula (21) is ; 675 000 u.l 150 000.36 m = -s- = 5 The moment of resistance, r, is, from Formula (18) m G 75 000 (r) 180 -=3 750 From Table I, No. 3, we find r= -Q-, therefore bd^ = 3750 ; now, as 6 = 36, transpose and we have (Z2 — 3750.6 36 625. Therefore d = 25" or say 24". have to be 5' x 3' x 2'. Fig. 46. The size of granite block would As this would be a very un- wieldy block, it might be split in two lengthwise of pier ; that is, two stones, each 5' x 18" x 2' should be used, and clamped together. Before building piers, he arch should be allowed to get thoroughly set and hardened, to avoid any after shrinkage of the joints. A parabolic arch is best. Next in order is a pointed arch, then a semi-circular, next elliptic, and poorest of all, a segmental arch, if it is very flat. But, as before mentioned, avoid inverted arches, if pos- sible, on account of the difficulty of their proper execution. „ . A rock foundation makes an excellent one, and needs Rock , foundations. UtQe treatment, but is apt to be troublesome because of water. Remove all rotten rock and step off all slanting surfaces, to make • Co/sciJETE: Fig. 47. level beds, filling all crevices with concrete, as shown in Figure 47 : In no case build a wall on a slanting foundation. Look out for springs and water in rock foundations. "\^Tiere soft ROCK FOUXDATIOXS. 91 soils are met in connection with rock, try and dig down to solid rock, or, if this is impossible, on account of the nature of the case or ex- pense, dig as deep as jwssihle and put in as wide a concrete base- course as possible. If the l)ad spot is but a small one, arch over from rock to rock, as shown in Fl"-ure 48 : Fig. 48. _ . , Good hard sand or even quicksand makes an excel- Sand, gravel . , . ,. '■ , and clay, lent foundation, if it can be kept from shifting and clear of water. To accomplish this purpose it is frequently " sheath- piled " each side of the base course. Gravel and sand mixed make an excellent, if not the best founda- tion ; it is practically incompressible, and the driest, most easily drained and healthiest soil to build on. Clay is a good foundation, if in horizontal layers and of sufficient thickness to bear the superimposed weight. It is, however, a very treacherous material, and apt to swell and break up with water and frost. Clay in inclined or vertical layers cannot be trusted for im- portant buildings, neither can loamy earth, made ground or marsh. If the base-course cannot bo sufficiently spread to reduce the load to a minimum, pile-driving has to be resorted to. This is done in many Short different ways. If there is a layer of hard soil not far piles, down, short piles are driven to reach down to same. These should be of sufficient diameter not to bend under their load ; they should be calculated the same as columns. The tops should be well tied together and braced, to keep them from wobbling or spreading. Example. Georrjia-pine piles of IG" diameter are driven through a layer of soft soil 15' deep, until they rest on hard bottom. What will each pile safely carry f The pile evidently is a circular column 15' long, of IG" diameter, solid, and we should say with rounded ends, as, of course, its bear- 92 SAFK BUILDING. ings cannot be perfect. From Formula (3) we find, then, that the pile will safely carry a load. a. w = — / c \ -^j2j^f now from Table I, Section No. 7, and fifth 1+ ^ column, we have 22 „ 22.82 From the same table we find, for Section No. 7, last column, t^ 82 From Table IV we find for Georgia pine, along fibres, ^■4^ = 750. And from Table 11, for wood with rounded ends, n = 0,00067, therefore : 201.750 150750 ^~. , 1802.0,00067 ~ "2;357~ ~ ^^958, •■ 16 or say SO tons to each pile. gj,^^ Sometimes large holes are bored to the hard soil and piles, filled with sand, making " sand piles." This, of course, can only be done where the intermediate gi-ound is sufliciently firm to keep the sand from escaping laterally'. Sometimes holes are dug down and filled in with concrete, or brick piers are built down ; or large iron cylinders are sunk down and the space inside of them driven full of piles, or else excavated and filled in with concrete or other masonry, or even sand, well soaked and packed. If filled with sand, there should first be a layer of concrete, to keep the sand from possibly escaping at the bottom. Where no hard soil can be struck, piles are driven over a large area, and numerous enough to consohdate the ground ; they should not be closer than two feet in the clear each way, or they will cut up the ground too much. The danger here is that they may press the ground out laterally, or cause it to rise where not weighted. Some- times, by sheath-piling each side, the ground can be sufficiently com- pressed between the piles, thereby being kept from escaping laterally. But by far the most usual way of driving piles is where Long •. J r, I pi les. they resist the load by means of the friction of their sides against the ground. In such cases it is usual to drive experimental piles, to ascertain just how much the pile descends at the last blow of the hammer or ram ; also the amount of fall and weight of ram, and riLKS. 93 then to compute the load the pile is capable of resisting : one-tenth of this might be considered safe. The formula tlien is : — «' = nrf W "Where w = the safe load on each pile, in lbs. " r = the weight of ram used, in lbs. " f= the distance the ram falls, in inches. « s = the set, in inches, or distance the pile is driven at the last blow. "VVliorc there is the least doubt about the stability of the pile, use three-fourths w, and if the piles drive very unevenly, use only one- half w. Some engineers prefer to assume a fixed rule for all piles. Profes- sor Rankine allows 200 lbs. per square inch of area of head of pile. French engineers allow a pile to carry 50000 lbs., provided it does not sink perceptibly under a ram falling 4' and weighing 1350 lbs., or does not sink half an inch under tliirty blows. There are many other such rules, but the writer would recommend the use of the above formula, as it is based on each individual experiment, and is therefore manifestly safer. Example. An experimental pile is found to sink one-half inch under the last blow of a ram weighing 1500 lbs., and falling 12'. What will each pile safely carry ? According to formula (46), the safe load w would be w = r^r-we have, 10. s r== 1500 lbs. /= 12.12 = 144", and s = ^", therefore 1500.144 .„„^^,, It; = -j^ — =43200 lbs. ^ If several other piles should give about the same result, we would take the average of all, or else allow say 20 tons on each pile. If, however, some piles were found to sink considerably more than others, it would be better to allow but 10 tons or 15 tons, according to the amount of irregularity of the soil. All cases of pile-driving require experience, judgment, and more or less experiment ; in fact all foundations do. All piles should be straight, solid timbers, free from projecting 94 SAFE BUILDING. branches or large knots. They can be of hemlock, spruce or wlute pine, but prcferal)ly, of course, of yellow pine or oak. There is danger, where they are near the seashore, of their being destroyed by worms. To guard against this, the bark is sometimes shrunken on ; that is, the tree is girdled (the bark cut all around near the root) before the tree is felled, and the sap ceasing to flow, the bark shrinks on very tightly. Others prefer piles without bark, and char the piles, coat them with asphalt, or fill the pores with creosote. Cojiper sheets arc the best (and the most expensive) covering. Piles should be of sufficient size not to break in driving, and should, as a rule, be about 30' long, and say 15" to 18" diameter at the top. They should not be driven closer than about 2' 6" in the clear, or they will be apt to break the ground all up. The feet should be shod with wrought-iron shoes, pointed, and the heads pro- tected with wrought-iron bands, to keep them from sjilitting under the blows of the ram. Sheath- ^^ sheath-piling it is usual to take boards (hemlock, piling, spruce, white pine, yellow pine or oak) from 2" to 6" thick. Guide-piles are driven and cross pieces bolted to the insides of them. The intermediate piles are then driven between the guide piles, mak- ing a solid wooden wall each side, from 2" to G" thick. Sometimes the sheath-piles arc tongued and grooved. The feet of the piles are cut to a point, so as to drive more easily. The tops are covered with wrought-iron caps, which slip over them and are removed after the piles are driven. Piles are sometimes made of iron : cast-iron beincr T)ref- Iron ^ ' ^ i piles, erable, as it will stand longer under water. Screw-piles are made of iron, with lai'ge, screw-shaped flanges attached to the foot, and they are screwed down into the ground like a gimlet. Sheath-piling is sometimes made of cast-iron plates with vertical strengthening ribs. Where piles are driven iinder water, great care must be taken that they are entirehi immersed, and at all times so. They should be cut off to a imiform level, below the lowest low-water mark. If they are alternately wet and dry, they will soon be destroyed by decay. _ After the piles are cut to a level, tenons are often cut Base-courses ^ * ' over piles, on their tops, and these are made to fit mortises in heavy wooden girders Avhich go over them, and on which the superstructure The sizes given on this page are for heavy huildiiigs ; for very light work use piles of 8" to 9" diameter at tops, about 20 feet long aud 16" apart. SLII'-.IOINTS. 95 rests. This is usiuil for docks, ferry -houses, etc. For other buildings we fretjuently see concrete packed hetwecn and over their tops; this, however, is a very bad practice, as the concrete surrounding the tops is apt to decay tliem. It is better to cover the ])iles with 3" x 12" or similar jjlanks (well lag-screwed to piles, where it is necessary to steady the latter) and then to build the concrete base course on these planks.^ Better yet, and the best method, is to get large-sized building- stones, with levelled beds, and to rest these directly on the piles. In this case care must be taken that piles come at least under each cor- ner of the stone, or oftener, to keep it from tipping, and that the stone has a full bearing on each pile-head. On top of stone build the usual base-courses. Piles should be as nearly uniform as possible (particularly in the case of short piles resting on hard ground), for otherwise their re- spective powers of resistance will vary very much. It is well to connect all very heavy parts of buildings joints, (such as towers, chimneys, etc.) by vertical slip-joints with rest of building. The slip-joint should be carried through the foundation-walls and base-courses, as well as above. Where there are very high chimneys or towers, or unbraced walls, the foundation must be spread sufficiently to overcome the leverage produced by wind. These points will be more fully explained in the chapter on " Walls and Piers." All base-courses should be carried low enough to be frost, below frost, which will penetrate from three to five feet deep in our latitudes. The reason of this is that the frost tends to swell or expand the ground (on account of its dampness) in all direc- tions, and does it with so much force that it would be apt to lift the base-course bodily, causing cracks and possible failure above. 1 The strength of planks is calculated the same as for beams laid on their fiat sides. 96 SAFE IIUILUIXG. CHAPTER III. CELLAR AND RETAINING WALLS. PCB f Fig. 49. ANALYTICAL SOLUTION. rHE architect is sometimes called upon to build retaining-waUs in connection with ter- races, ornamental bridges, city reservoirs, or similar problems. Then, too, all cellar walls, where not adjoining other buildings, become re- taining walls ; hence the necessity to know how to ascertain their strength. Some writers dis- tinguish between " face-walls " and " retaining- waUs " ; a face-wall being built in front of and against ground which has not been disturbed and is not likely to slide ; a retaining-wall being a waU that has a filled-in backing. On this the- ory a face-wall would have a purely ornamental duty, and would receive no thrust, care being taken during excavation and building-operations not to allow damp or frost to get into the ground so as to prevent its rotting or losing its natural tenacity, and to drain off all surface or underground water. It seems to the writer, however, that the only walls that can safely be considered as " face-walls " are those built against rock, and that all walls built against other banks should be calculated as retaining- walls. Most Economi- The cross-section of retaining-walls vary, accord- cal Section, ing to circumstances, but the outside surface of wall is generally built with a " batter " (slope) toicards the earth. The most economical wall is one where both the outside and back sur- faces batter towards the earth. As one or both surfaces become near- ly vertical the wall requu-es more material to do the same work, and KETAINING WALLS. 07 the most extravagant design of all is where the back face batters away from the earth ; of course, the outside exposed surface of wall must eitlicr batter towards the earth (A B in Figure 4D) or be vertical, (A C) ; it cannot batter away from the ground, otherwise the wall would overhang (as shown at A D). "Wliere the courses of masonry are built at right angles to the outside surface the wall will be stronger than where they are all horizontal. Thus, for the same amount of material in a wall, and same height, Figure 50 will do the most work, or be the strongest retaining-wall, Figure 51 the next strongest, Figure 62 the next, Figure 53 next, w Fig. 51. Fig. 52. Figure 54 next, and Figure 55 the weakest. In Figure 50 and Fig- ure 52 the ioints are at right angles to the outside surface; in the Fig. 55. other figures they are all horizontal. For reservoirs, however, the shapes of Figures 53 or 55 are often employed. 08 SAFE BUILDING. To calculate the resistance of a retaining-wall proceed as follows ' Height of Line The of Pressure, c e n - tral line ov axis of the pressure O P or p of backing will be at one- third of the height of back surface, meas- ured from the ground lines,! tliat is at O in Figure 5G, where A O = i AB. The direction of the pressure-line (except for reservoirs) is usu- ally assumed to form an angle of 57° with the back surface of wall, or L POB = 570. For water it is as- sumed normal, that is, at right angles to the back surface of wall. If it is desired, how- F'g- 56. ever, to be very exact, erect O E perpendicular to back surface, aud make angle E O P, or (Z. x) = the angle of friction of the filling-in or backing. This angle can be found from Table X. Amount of pres- The amount (p) of the pressure P O is found -General £j,qjjj ^j^g followino; formulae : sure- case. If the backing is filled in higher that the wall,'' Bacl 100 40° 100 38° 110 370 112 32° 110 32° 125 27° 125 24° 125 24° 110 23° 130 17° 125 17° G4 0° C2J 0=" Even those who do not understand trigonometry can use the above formulae. It will simply be necessary to add or subtract, etc., the numbers of degrees of the angles y and x, and then find from any table of natural sines, cosines, etc., the corresponding value for the amount of the new angle. The value, so found, can then be squared, multi- plied, square root extracted, etc., same as any other arithmetical problem. Should the number of degrees of the new angle be more than 90°, subtract 90° from the angle and use the positive cosine of the difference in place of the sine of whole, or the tangent of the difference in place of the co-tangent of the whole ; in the latter case the value of the tangent will be a negative one, and should have the negative sign prefixed. Thus, if a: = 33° and 7/ = 50°, formula (47) would become: _ tc. L2 sin.2 (1 7)° p: sin.2 50.0 giu, 88° 'Ab )ve table of friction angles is taken from Klasen's " Jlockbau und Briick- eiibau - Conslructlonen." As a rule it will do to assume the angle of friction at33** and tlie weight of backing at 120 lbs. per cubic foot, except in the case of water. 100 SAFE BUILDING. The values of wliicli, found in a table of natural sines, etc., is • _ u: U- 0,2924-^ P~ 2 ' 0,7G(i-^. 0,999-4 ■1l. 0,1458 = 0,729. ?o.L2 2 Similarly, in formula (48), we shoul d have for tlie quan tity : y/cot.a;-cot. (^+2x-; = y/cot. 33° - cot. 110° = y/cot. 33° - [-tg. (110" - yo-;j = y/cot. 33° + tg. 26° __y/l,5399 4-0,4877 = y/2,0276 =1,424 Average Case. As already mentioned, however, the angle of fric- tion — (except for water when it is=r 0°, that is, normal to the back surface of wall) — is usually assumed at 33° ; this would reduce above formulae to a very much more convenient form, viz. : For the average angle of friction (33°) If the backing is higher than the wall : Backing higher ^^_r.L2. (10 -77. 0,55)^. Vl44 -}- n^ (49) than Wall. ^ 2 (10 + ?i. 0,55). 144 Or, if the backing is level with top of wall: Backing level „ _ !^^ Vl44-fn2 / V A 5i_ " • M- H ..._ with Wall. '' 2 ' 9-}-n.l,7 V^ ' 5-|-n.0,9 i/n _ n.0.4-ri y (50) V 12 5 4- n. 0,9/ Where p, to and L same as for formulae (47) and (48). "\Miere ?i = amount of slope or batter in inches (per foot height of wall) of rear surface of wall. Thus, if the rear surface sloped towards the backing three inches (for each foot in height) we should have a positive quantity, or 71 = -f 3. If the rear surface sloped aiva^ from the backing three inches (per foot of height), n would become negative, or n = — 3. When the rear surface of wall is vertical, there would be no slope, and we Avould have Cellar Walls. n=0. The latter is the case generally for all cellar walls, which would still further simplify the formula, or, for cellar loalls where weight of soil or backing varies materially from 120 pounds per cubic foot. CELLAU WALLS. 101 Cellar Walls- p = w. JA O.l^iS. (51) general case. ^ For ceUnr walls, where the wjight of soil or backing can be safely assumed to weigh 120 pounds per cubic foot Cellar Walls- p=lG?i.L^. (u2) usual case. ^ •» Where p = the total amount of pressure, in pounds, per each running foot in length of -wall. Where ?r=:the weight, in pounds, per cubic foot of backing. AVhere L = the height, in feet, of ground line above cellar bottom. For different slopes of the back surface of retaining walls (assum- ing friction angle at 33°) we should have the following table ; -(- denoting slope towards backing, - denoting slope away from backing. TABLE XI. Slope of back sur- Value of p for backings of Value of p for the average face of wall in inches different weights per cubic backing, assumed to weigh per foot of height. foot. 120 lbs. per cubic foot. +4" j5=0,072. w. Ifl V- H' L= +3" p— 0,088. IV. \? p—W , U +2" »=0,098. xo. L2 P=i2 . y +1" »=0,112. w. L2 ;;=13i. L2 0" j9=0,133. «•. L2 w=163. L2 —1" ;7=0,157. w. \? p^lQ. L« orr p=0,185. XL\ \? n=:22i. 1.2 3^/ ^)=0,205. XL\ \? p=2ii. y —i" p=0,258. M?. L3 p=31 . Ifl Now having found the amount of pressure p from the most conven ient formula, or from Table XI, and referring back to Figure 56, proceed as follows : To f i nd C u rve of ^'^^^ *^^^ centre of gravity G of the mass ABC D,i Pressure. from G draw the vertical axis G H, continue P O till it intersects G H at F. ]\Iake F H equal to the Aveight in pounds of the mass A B C D (one foot thick), at any convenient scale, and at same scale malce II I = p and paral- lel to P O, then draw I F and it is the I '^c ■ resultant of the pressure of the earth, and the resistance of the retaining walk Its point of intersection K with the Amount of base D A is a point of Stressat Joint, the cui'vc of pressure, j^cl— To find the exact amount of pressure on the joint D A use formula (44) for the edge of joint nearest to the point K or edge D, and formula (45) for the edge of joint farthest from the point K, or edge A. 1 To lind the centre of gravity of a trapezoid A B C D, Fig. 57, prolong CB until B F = CI = D A and prolong D A until A E = D H = C B.draw F I and U Fa:id their point of intersection G is the centre of gravity of the whole. 102 8AFK BUILDING. •^ a.d Formula (44) was v=JL a If ]\I be the centre of D A, that is D M= M A = |. D A, and re- membering that the piece of wall we are calculating, is only one run- ning foot (or one foot thick), we should have For a; = K INI ; exjDressed in inches. For a = A D. 12 ; (AD exjjressed in inches). For (/ = A D ; in inches, and For jo = F I, in lbs., measured at same scale as F H and H I ; or, the stress at D, (the nearer edge of joint) would be v, in pounds, per square inch, AI>. 12~ 12.ADa Remembering to measure all parts in inches except F I, which must be measured at same scale G B^/'"^'^ ^^ '^^'^ "sd to lay out F II and II I. Similarly we should obtain the stress at A in pounds per square inch. FI 6. KM.Fl AD. 12 "12.AD^ V should not exceed the safe crushing strength of the ma- terial if positive ; or if y is negative, the safe tensile strength of the mortar. If we find the wall too weak, we must enlarge A D, or if too strong, we can diminish it ; in either case, finding the new centre of gravity G of the new mass A B C D and repeating the operation from that jioint ; the pres- sure, of course, remaining Fig. 58. the same so long as the sloiie of back surface remains unaltered. If the wall is a very high one, it should be divided into several sections in height, and each section ex- amined separately, the base of each section being treated the same as if it were the joint at the ground line, and the ichole mass of wall bi the section and above the section bcinir taken in each time. RESEKVOIR WALLS. 103 Thus in Figure 58, when examining the part A, BCD, we t^houM find O, P, for the part only, using L, as its height, the point O, being at one third tlie heiglit of A, B or A, O, = ^. A, B ; G, would be the centre of gravity of A, B C D„ while F, II, would be equal to the weight of its mass, one foot thick ; this gives one point of the curve of pressure at K„ with the amount of pressure = F, I„ so that we can examine the pressures on the fibres at D, and A,. Similarly when com- paring the section A„ B C D„ we have the height L,„ and so find the amount of pressure 0„ P,„ applied at 0,„ where A„ 0„ := ^ A„ B ; G„ is centre of gravity of A„ B C D„ while F„ H„ is equal to the weight of A„ B C D,„ one foot thick, and F„ I„ gives us the amount of pressure on the joint, and another point K„ of curve of pressure, so that we can examine the stress on the fibres at D„ and A„. For the whole mass A B C D we, of course, proceed as before. Reservoir For reservoirs the line of pressure O P is always Walls, at right angles to the back surface of the wall, so that we can simpUfy formula (50) and use for rain water : ;j = 31f L2 (53) For salt water : ;; = 32. LS (54) "Where p = the amount of pressure, in pounds, on one running foot in length of wall, and at one-third the height of water, measured from the bottom, and p taken normal to back surface of wall, "Where L = the depth of water in feet. If Backing Is "Where there is a superimposed weight on the back- Loaded * ing of a retaining-wall proceed as follows : In Figure 59 draw Si^liil^^»giSf ^l^:^ &ii^ the angle C A D = x, the angle of friction of the material. Then take the amount of load, in pounds, com- ing on B C and one running foot of it in thickness (at right angles to B C), divide this by the area, in feet, of the triangle ABC and add the quotient to w, the weight of the backing per cubic foot, then proceed as before inserting the sum w, in place of w in formulaa (47) to (51) and in 2 ~ Table XT, when calculating p ; or u\ = w -}- =^^ (55) 15. Li A D" Fig. 59. 104 SAFE BUILDING. "Where w, = the amount, in pounds, to be used in all the formulae (47) to (51) and in Table XI, in place of w. "Where i« = -weight of soil or backing, in pounds, per cubic foot. "\Miere z = the total superimposed load on backing, in pounds, per running foot in length of wall. "Where B = length, in feet, of B C, as found in Figure 59. "Where L = the height of wall, in feet. "Where there is a superimposed load on the backing, the central line of pressure j5 should be assumed as striking the back surface of wall higher than one-third its height, the point selected, being at a height X from base ; where X is found as per formula (56) ^^ L.(e..-§z.) 2w^-W ^ ^ "Wliere X = the height, in feet, from base, at which pressure is ap- pUed, when there is a superimposed load on the backing. "Where L =: the height, in feet, of wall. "Where w = the weight, in pounds, per cubic foot of backing. "Where w,:='is found from formula (55) "When calculating the pressure against cellar walls, only the actual weight of the material of walls, floors and roof should be assumed as coming on the wall, and no addition should be made for wind nor for load on floors, as these cannot always be reUed on to be on hand. The additional compression due to them should, however, be added afterwards.^ OBAFHICAL METHOD. X ^ ^ ^ The graph- ^\ ical method of \ calculating re- \ taining-walls is __A-niuch easier — — """ x'' Lthan the ana- ^^ lytical, b e i n g less liable to cause errors, and is recom- mended for office use, though the an- alytical method 1 Where a wall 13 not to be kept braced until the superimposed wall, etc. is on it, these should of course be entirely omitted from the calculation, and the wall must be made heavy enough to stand alone. GRAPHICAL METHOD. 105 might often serve as a check for detecting errors, when undertaking important work. If A B C D is the section of a retaining wall and 13 I the top line of backing, draw angle F A M = x = the angle of friction, usually assumed at 33° (except for water) ; continue B I to its intersection at E with A M ; over B E draw a semi-circle, with B E as diameter ; make angle B A G = 2 x (usually 6G°), continuing line A G till it intersects the continuation of B I at G ; draw G II tangent to semi- circle over BE; make G I = G II ; draw I A, also I J parallel to B A ; draw J K at right angles to I A ; also B ]\I at right angles to A E. Now for the sake of clearness we will make a new drawing of the wall A B C D in Figure 61, Calling B M = Z and K J = Y (both in Figure 60) make A E = Q (Figure 61) where Q is found from formula (57) followino-: — ■\r r/ \ y a 1. Z.S Q = Where Q = the length of A E in Figure 61, in feet, "Where Y = the length of K J in Figure 60, in feet,i Where Z = the length of B M in Figure 60, in feet, C (57) Figs. 61 and 62. Where s = the weight of one cubic foot of backing, in lbs. ^Vhere m = the weight of one cubic foot of wall, in lbs. Where L = the height of backing, in feet, at walk JfR fw ^^d « f I ?-® ^ } ^? ^^"^ lionzon 13 equal to the augle of friction, as is often the case, faml A G as before ami use tliis length in place of K J or Y which of course, it will bo impossible to flnd, as A M and B I would bo oaral'lel and would have no point of intersection ; of course, B 1 should never be steeper than A INI or else all of the soil steeper than the lino of angle of friction would be apt 106 SAFE BUILDING. Sections must Draw E B, divide the wall into any number of sec- helg^hts.^ tions of equal height, in this case we will say three sections, A A. D, D ; A, A„ D„ D, and A„ B C D„. Find the cen- tres of gi-avity of the different parts, viz. : G, G, and G,„ also F, F, and F„. Bisect D D, at S, also D. D„ at S, and D„ C at S„- Draw S N, S, N, and S„ N„ horizontally. Through G, G. and G„ draw vertical axes, and through F, F, and F„ horizontal axes, till they in- tersect A B at O, O, and 0„. Draw O P, O. P, and 0„ P„ parallel to M A, where angle M A E = x = angle of friction of soil, or back- ing. In strain diagram Figure 62 make a 6, := R„ N„ ; also 6, rf, =-: R, N, and d, / = R N. From h„ d, and / draw the vertical lines. Now begin at a ; draw a b parallel to M A ; make 6 c = S„ R„ ; draw c d parallel to INI A ; make d e = S, R, ; draw e f parallel to M A and make f g =z S R. Draw a c, a d, a e, a /and a g. Now returning to Figure Gl, prolong P„ 0„ till it intersects the vertical axis through G„ at H„ ; draw II„ H, parallel to « c till it intersects P, O, at 11, ; draw II, I. parallel to a d till it intersects the vertical axis through G, at I, ; draw I, II parallel to a e till it intersects P O at H; draw H I parallel to a /till it intersects the vertical through G at I ; draw I K parallel to a g. Then will points K, K, and K„ be points of the curve of pressure. The amount of pressure at K„ will be a c, at K, it will be a e, and at K it will be a g, from which, of course, the strains on the edges D, D, and D„, also A, A, and A„ can be calculated by formulaj (44) and (45). To obtain scale, by which Scale of strain ^o measure a c, a e and a g, make g h, Figure 62 at diagram. anj scale equal to the weight, in pounds, of the part of wall A A, D, D one foot thick, draw h i parallel /or, then g i meas- ured at same scale as g h, is the amount of pressure, in pounds, at K. Similarly make e k =: weight of centre part, and c m =l weight of upper part, draw k I parallel d a, and m n parallel b a, then is e Z the pressure at K, and c n the pressure at K„, both measured at same scale ; or, a still more simple method would be to take the weight of A A, D, D, in pounds, and one foot thick, and divide this weight by the length of ^ /in inches; the result being the number of pounds per inch to be used, when measuring lengths, etc., in Figure 62. The above graphical method is very convenient for high walls, where it is desirable to examine many joints, but care must be taken to he sure to get the parts all of equal height, otherwise, the result would be incorrect. If backing Ji^ <2^se of a superimposed weight find w„ as di- loaded. rected in formula (55), make A T at any scale equal to w and A U = «;„ draw T E and parallel thereto U V, draw V B, BUTTRESSED WALLS. 107 tf- parallel to E B and use V B, in place of E B, proceeding otherwise as before. The points O of application of pressure P O, will be sliglitly changed, particularly in the upper part, as they will be hori- zontally opposite the centres of gravity of the enlarged trapezoids, and in the upper case this point would be much higher, the figure now being a trapezoid, instead of a triangle as before. Buttressed ^Vliere a wall is made very thin and then buttressed walls, at intervals, all calculations can be made the same as for walls of same thickness throughout, but the vertical axis through centre of gravity of wall should be shifted so as to pass through the centre of gravity of the whole mass, in- .-2ft-.ij o.^ eluding buttresses ; and the weight of thia part of wall should be increased proportion- ately to the amount of buttress, thus : If a 12" wall is buttressed every 5 feet (apart) with 2' X 2' buttresses, proceed as follows : Find the centre of gravity G of the part of wall A B C D (in plan) Figure 63, also centre of gravity F of part E I II C, draw lines through F and G parallel to wall. Now make a b parallel to wall and at any scale equal to weight or area of A B C D jcai« of strains ovd^"^ ^ '; ^^"^^ *« that of E I H C. From Fig. 63. any point o draw the lines oa,ob and o c ; now drawK L (anywhere between parallel hues F and G), but paral- lel to b 0, and from L draw L M parallel to o c, and from K draw K M parallel to a o, a line through their point of intersection M drawn parallel to wall is the neutral axis of the whole mass. When \) C drawing the vertical section of wall-part A B C D, Figure 64, therefore, instead of locating the neutral axis through the centre of wall it will be as far outside as M is from B C, in Figure 63 ; that is, at G H, Figure 64. When considering the weight per cubic foot of wall, we [^ add the proportionate share of buttress; now in Figure 63 "^ there are 4 cubic feet of buttress to every 7 feet of wall, "^'s- ^•*' so that we must add to the usual weight w per cubic foot of wall ^ w, or w. (1 -|-f) To put this in a formula. 5oo loop geoo W,, = W (1 + J-) (58) 108 SAFE BUILDING. "Wliere w„ = the weight per cubic foot, in pounds, to be used for buttressed walls, after finding the neutral axis of the whole mass. Whei*e ?^ = the actual weight, in pounds, per cubic foot of the material. ■WTiere A = the area in square feet of one buttress. Where A,= the area in square feet of wall from side of one but- tress to corresponding side of next buttress. Walls with Buttresses, however, will not be of very much value, counterforts, unless they are placed quite close together. But- tresses on the back surface of a wall are of very little value, unless SJ5W^'^' thoroughly bonded and anchored to walls ; these latter are called counterforts. It is wiser and cheaper in most cases to use the additional masonry in thickening out the lower part of wall its entii-e length. Resistance Where frost is to be resisted the back to frost, part of wall should be sloped, for theTS^ depth frost is hkely to penetrate (from 3 to 4 feet in 'p.g. 65. our chmate), and finished smoothly with cement, and then asphalted, to allow the frozen earth to slide upwards, see Figure G5. Example I. Cellar wall to -^ ^'■'-'o story and attic frame house has a 12" brick frame It should really be the vertical neutral axis of the whole weight, which would be a trifle nearer to D C thau centre of wall. 110 SAFE UUILDING. the curve of pressure, and F M, at same scale the amount of pressure on D A for the bearing walls of house; E N we find measures 2h", and F M measures 4600 units or pounds. The stress at A, then, would be : 4600 , p 2i.460O 144 ~ 144. 12 = -j- 72 pounds. and the stress at D would be : _ 4600_g 2h 4000 = — 8 pounds. 144 144. 12 There will, therefore, be absolutely no doubt about the safety of bearing walls. Example IT. Cellar wall A cellar wall A B C D is to be carried 15 feet be- lofn\^%t>uf\a^ns'low the level of adjoining cellar; for particular rea- sons the neighboring wall cannot be underpinned. It is desirable not to mate the loall A B C D over 2' 4" thick. Would this be safe f The soil is wet loayn. In the first place, before excavating we must sheath- pile along line C D, then as we excavate we must secure horizontal timbers along the sheath-piUng and brace these from opposite side of excavation. The sheath-piling and horizon- tal timbers must be built in and left in wall. The braces will have to be built around and must not be removed until the whole weight is on the wall. The weight of the wall C G, per running foot of length, including floors and roofs, we find to be 13000 pounds, but to this we must add the possible loads com- ing on floors, wliich we find to be 6000 pounds addi- tional, or 19000 pounds total, possible maximum load. This load will *», 4a 60 72 84 96 Jcale of= Lenjrhi Unchu) »0 loocq ^?°^ ^ ^Q P*^^ ■-— , JcaTe of JtroinJ(lk>J) Fig. 67. DEEP CELLAU WALL. HI be distributed over the area of C D E. Tii calculating the weight of A B C D resisting the pressure, we must take, of course, only the min- imum weight ; that is, the actual weight of construction and omit all loads on floors, as these may not always be present. The weight of walls and unloaded floors coming on A B C D, and including the weight of A B C D itself, we find to be 21500 j)ounds per running foot. Now to find the pressure p, proceed as follows : Make angle E, D M= 17°, the angle of friction of wet loam (Sec Table X), and prolong D E, till it intersects C E„ at PI Now C E, we find, measures 52 feet; CD or L is 15 feet ; then, instead of using w iu formula (51), we must use «?,, as found from formula (55), viz. : , 2. 19000 "■ = "+(7^7^ w for wet loam (Table X) is 130 pounds; therefore, ry,= 130 + ^1^ = 130 + 48, 7 = 179. 52. 15 ' Inserting this value for lo in formula (51) we have : ;> = 179. 152. 0,138 = 5558. The height X from D at which P O is applied is found from formula 5G, and is : V _ 15. (179 — |. 150) __ 15. (179 — 100) _ 2.179 — 150 "~ 358 — 150 ~ 5',697==:5'8i" Make D = X = 5'8r ; draw P O parallel to E D till it inter- sects the vertical neutral axis of wall (centre line) at F ; make F II (vertically) at any scale equal to 21500, draw H I parallel to P O and make II 1=7^ = 5558 pounds at same scale, draw I F, then is its point of intersection with the prolongation of A D at K a point of the curve of pressure, and F I measui-ed at same scale as F H is the amount of pressure on joint. The distance K from centre of joint N we find is 14^", F I measures 23800 units or pounds; the stress (y) at A, therefore, will be, formula (44) : ^ _ 23800 14^^^3800 _ 28.12 ~ 28.12.28 ~ "While the stress at D would be formula (45) 23800 „ 14^.23800 — ii. —^ = — 150 28.12 28.12.28 Or the edge at A would be subjected to a compression of 292 pounds, while the edge at D would be submitted to a tension of 150 pounds per square inch, both strains much beyond the safe limit of even the best masonry. The wall will, therefore, have to be thickened and a new calculation made. 112 SAFK BUILDING. Examjde III. Wall to -^ stage-pit ZO feet deep is to le enclosed by a stone- stage pit. icall, 3 feet thick at the top and increasing 4 inches in thickness for every 5 feet of depth. The wall, etc., coming over this wall weighs 25000 pounds per running foot, hut cannot he included in the calculation, as peculiar circumstances will not allow hraces to he kept against the cellar wall, until the superimposed loeight is on it. The surrounding ground to be taken as the average, that is, 120 pounds weight per cubic foot, and with an angle of friction of 33°. c > S" //A rM- \;^-P' Ff:4. ^^■^^ih a<:. /Pu \ t^. :^ y^\ ^r-'-'f o i2 Zi-^6 48 (fa 77 S » Jcale of Lengfhs (inches) Fig. 68. Find B M (= Z) and Q T (= Y) by making angles T A E = 330 and B A U = 66° and then proceeding as explained for Figure 60. We scale B M and Q T at same scale as height of wall A B is drawn, and find : BM = Z = 25ft. 6" = 25J QT = Y = 9 ft. 8" = 9|; assuming each cubic foot of wall to weigh 150 pounds we find Q from Formula (57) 8TAGK-PIT WALL. 113 Q = 93- 25^.120 _ Qr^ yg _ g, y,,^ ^ 30. 150 ' Make A E = Q = G' 7" and draw B E. At equal heighls, that is, every 5 feet, in this case, draw the joint nes D E, D, E„ D„ E,„ etc. Find the centres of gravity F, F„ F,„ etc., of the six parts of A E B, "^'^(see foot-note, p. 101) and also the centres of gravity G, G„ G,„ etc., of the six parts of the wall itself, which, in the latter case, will be at the centre of each part. Horizontally, opposite the centres F, F„ F,„ etc., apply the pressures P O, P, 0„ etc., against wall, and parallel to M A. Through centres G, G„ G,„ etc., draw vertical axes. Draw the lines S N, S, N,, S„ N,,, etc., at half the vertical height of each section. Now in Figure G 9 make a ft, = Rv Nv ; fe. d, = B.v N,v ; (1, /, = R,„ N„, ; /, /j, = R„N„; /^y, = R.N.; and j\ I, = R N. Draw the vertical lines through these points. Now begin at a, make a b parallel to P O, make b c=i Rv Sv ; draw c d parallel P ; make de = R,v S,v ; draw e f Jcale op Lengths (Inches) Fig. 69. parallel P 0, make fgz= R„, S„„ and similarly gh,i j and k I par- allel P O, and h i = R„ S,. ; y A: = R, S, and Z m = R S. Draw from a lines to all the points c, d, e, f, g, etc. Now in Figure G8 begin at Pt Ov, prolong it till it intersects vertical axis Gy at Iv, draw I^ H^ parallel a c till it intersects P,y 0,v at 11,.; draw Ily T,v parallel to a d till it intersects vertical axis G,v at T,v ; draw 1,^ Il.y parallel to a e tm it intersects P,„ 0„, at H,^ and similarly II.v I,„ parallel af; I,„ 11,,, parallel ag: II,„ T„ parallel a h; I„ II„ parallel to a i; II,. I, parallel to a j ; 1, 11, parallel to a k ; II, I parallel to a I, and I K parallel to a m. The points of intersection K, K„ K,„ etc., are points of the curve of pressure. To find the amount of the pressure at each point, find 114 SAFE BUILDING. weif^ht per running foot of length of any part of wall, say, the bottom part (A A, D, D) the contents are 5' high, 4' 8" wide, 1' thick =: 5.4§. 1 = 23^ cubic feet a 150 pounds = 3500 pounds. Divide the weight by the length of m I in inches, and we have the number of pounds per inch, by Avhich to measure the pressures. As m I measures 50 inches, each inch will represent ^|§^ = C2|- lbs. Xow let us examine any joint, say, A,„ D„, ; I„, II„, which inter- sects A„, D,„ at K,„ is parallel to a g. Now a g scales 166 inches, therefore, pressure at K,„=: 1G6. 02^ = 10375 pounds. In measur- ing the distance of K,„ from centre of joint in the following, remem- ber that the width of A,„ D„, is 44 inches, the width of masonry above joint, and not 48" (the width of masonry below). A,„ K,„ scales 38", therefore, distance x of K,„ from centre of joint is x = 38 — 22 :^ 16". We have, then, from Formula (44) ,-p, 10375,. 16.10375 , .„ , stress at D,„ v ^= ^ , ^ - + 6. . , ^ . ,, = -4- 63 pounds. 44.12 44.12.44 ' ^ and from formula (45) , . 10375 „ 16.10375 „„ , stress at A... ; v = ^^^^^ — 0. -^^j^j^ = — 23 pounds. The joint A„, D„„ therefore, would be more than safe. Let us try the bottom joint A D similarly. I K is parallel to a m now a m scales 480", therefore, the pressure at K is ^ ^=480. 62^^ =z 30000 pounds. Now K is distant 53 inches from centre of joint, therefore, stress • T^ • 30000 , . 53.30000 , ^„„ , ^' ^ ^^ " = 3gT2- + '^^ 5(U23(r = + -^^ P°"^^^- , , ... 30000 „ 53.30000 „^„ , and stress at A is u = — 6. = — 209 pounds. oG.12 5G. 12.56 ^ The wall would evidently have to be thickened at the base. If we could only brace the wall until the superimposed weight were on it, this might not be necessary. If we could do this we should lengthen 6 c an amount of inches equal to the amount of this load divided by 62J (the number of pounds per inch), or 6 c instead of being 36 inches long would be : 36 + -r^^ = 436 inches long. While this lengthening of /; c would make the lines of pressure a c,ae, af, etc., very much longer, and consequently the actual pressure verv much greater, it will also make them very much steeper and conse- quently bring this pressure so much nearer the centre of each joint. EXAMTLE-KESERVOIR. 115 C 5 that the pressure will distribute itself over the joint mui'h more evenly, and the worst danger (from tension) -will probably be entirely removed. Example IV. Reservoir ^ stone reservoir loall is plumb on the outside, 2 feet Walli wide at the top and 5 feet loide at the bottom; the wall is 21 feet high, and the possible depth of loater 20 feet. Is the wall safe ? Divide the wall into three parts in height ; that is, D D, = D, D„ r=: D„ C. Find the weight of the parts from eaeh joint to top, per run- ning foot of length of wall, figuring the stone- work at 150 pounds per cubic foot, and we have: Weight of A„ BCD,, = 2975 pounds. Weight of A. BCD. = 7140 pounds. Weight of A B C D = 12495 pounds. Find centres of grav- ity of the parts A B C D (at G), of A, B C D, (at G.) and of A„ B C D„ (at G„). Apply the pressures P O at ^ height of A E ; P, O. at ^ height of A, E and P„ 0„ at A height of A„ E, where E top level of water. The amount of pres- sures will be from form- / ula (53). r"' For part A„ E. P.. 0„ = 31|. L2, ~ 31|. C2=1125 poimds. — \^- Dii binii°Gi r , 1 Xl- ' o / 1 I if — .__;l--\ai-— ^-ip ■7% \ 1' \ \ 1 A D in A H JcQle of itTfiinj ()lps) J<.ale of- Lengths (.m'hfs) Fig. 70. For part A, E ; P. O, = 31f L^, = 31f 132 _ 5251 pounds. For part A E ; P O = 31^. 1.2= 31 1. 202 _ JOSOO pounds. • The pressures P O, P, 0„ et-., will be applied at right angles to A E, prolong these lines, till they intersect the vertical axes through 116 SAFE BUILDING. (the centres of gravity) G, G, and G„ at F, F, and F„. Then make F„ II„ = 2D 75, weight of upper part. F, II, = 7140, weight of A, B C D„ and F II = 12405, weiglit of A B C D. Draw through II, II, and II„ the hncs parallel to pressure lines making H„I„==P„0„= 1125 II, I, = P, O, = 5281 II I ^P O ==12500 Draw I„ F,„ I, F, and I F, then will their lengths represent the amounts of pi-essure at points K,„ K, and K on the joints A„ D„, A, D, and A D. F„ I„ measures 3300 units or pounds. F, I. " 9500 " " F I " 18800 " «« By scaling we find that K„ is 10^ inches from centre of D„ A„ K, is 87" » " D, A, K is 74 " "DA The stresses to be exerted by the wall will, therefore, be .tD.;.= -i + S^S^ = +03 pounds. atA ;. = i^8^-6.iH8800_ = _lG7 pounds. ' GU.12 60.12.G0 '■ From the above it would appear that none of the joints are subject to excessive compression : further that joint D„A„ is more than safe, but that the joints D, A, and D A are subject to such severe tension that they cannot be passed as safe. The wall should, therefore, be redesigned, making the upper joint lighter and the lower two joints much wider. WALLS A^D riEBS. 117 CHAPTER IV. WALLS AND PIERS. WALLS are usually built of brick or stone, wliicli are sometimes, tliougli rarely, laid up dry, but usually with mortar filling all the joints. The object of mortar is threefold : Object of 1. To keep out wet and changes of temperature by mortar, filling all the crevices and joints. 2. To cement the whole into one mass, keeping the several parts from separating, and, 3. To form a sort of cushion, to distribute the crushing evenly, tak- ing up any inequalities of the brick or stone, in their beds, which might fi-acture each other by bearing on one or two spots only. To attain the first object, " grouting " is often resorted to. That is, the material is laid up with the joints only partly filled, and Hquid cement-mortar is poured on till it runs into and fills all the joints. Theoretically this is often condemned, as it is apt to lead to careless and dirty work and the overlooking of the filling of some parts ; but practicalUj it luakcs the best work and is to be recommended, except, of course, in freezing Aveather, when as little water as possible should be used. To attain the second object, of cementing the Avhole into one mass, it is necessary that the mortar should adhere firmly to all parts, and this necessitates soaking thoroughly the bricks or stones, as other- wise they will absorb the dampness from the mortar, which will crumble to dust and fail to set for want of water. Then, too, the brick and stone need washing, as any dust on them is apt to keep the mortar from clinching to them. In winter, of course, all wetting must be avoided, and as the mortar will not set so quickly, a little lime is added, to keep it warm and prevent freezing. Thickness of To attain the third object, the mortar joint must be Joints, matle thick enough to take up any inequalities of the brick or stone. It is, therefore, impossible to set any stanJard for 118 SAFE BUILDING. joints, as the more irregular the beds of the brick or stone, the hirger should be the joint. For general brickwork it will do to assume that the joints shall not average over one-quarter of an inch abwe irregu- larities. Specify, therefore, that, say, eight courses of brick laid up *' in the wall " shall not exceed by more than two inches in height eight courses of brick laid up " dry." For front work it is usual to gauge the brick, to get them of exactly even width, and to lay them up with one-eighth inch joints, using, as a rule, " putty " mortar. While this makes the prettiest wall, it is the weakest, as the mortar has little strength, and the joint being so small it is impossible to bond the facing back, except every five or six courses in height. " Putty " mortar is made of lime, water and white Isad, care being taken to avoid all sand or grit in the mortar or on the beds. Quality of "^^^^ ^'^^^ mortar consists of English Portland ce- mortars. ment and sharp, clean, coarse sand. The less sand the stronger the mortar. Sand for all mortars should be free fi'om earth, salt, or other impu- rities. It should be carefully screened, and for very important work should be washed. The coarser and sharper the sand the better the cement will stick to it. English Portland cement will stand as much as three or four parts of sand. Next to English come the German Portland cements, which are nearly as good. Then the American Portland, and lastly the Rosendale and Virginia cements. Good qualities of Piosendale cements wiU stand as much as two-and-a-half of sand. Of lime?, the French lime of Tcil is the strongest and most expensive. Good, hard-burned lime makes a fairly good mortar. It should be thoroughly slacked, as otherwise, if it should absorb any dampness afterwards, it will begin to burn and swell again. At least forty-eight hours should be allowed the lime for slacking, and it is very desirable to strain it to avoid unslacked lumps. Lime will take more sand than oement, and can be mixed with from two to four of sand, much depending on the quality of the sand, and particularly on the "fatness " of the lime. It is better to use plenty of sand (with lime) rather than too little ; it is a matter, however, for practical judgment and experiment, and while the specification should call for but two parts of sand to one of lime, the architect should feel at lib- erty to allow more sand if thought desirable. Lime and Rosendale cement are often mixed in equal proportions, and from three to five parts of sand added ; that is, one of lime, one of cement, and three to five of sand. It is advisable to specify that all parts shall bs actually measured in barrel.-:, to avoid such tricks, for instance, as hiring a M0RTAK8. 119 decrepit laborer to sliovcl cement or lime, while two or tlirec of the strongest laborers are shovelling sand, it being called one of cement to two or three of sand. A little lime should be added, even to the very best mortars, in winter, to prevent their freezing. Frozen When a wall has been frozen, it should be taken walls. down and re-built. Never build on ice, but use salt, if necessary, to thaw it; sweep off the salt-water, which is apt to rot the mortar, and then take off a few courses of brick before continu- ing the work. Protect wallf from rain and frost in winter by using boards and tarpaulins. Some writers claim that it does no harm for a wall to freeze ; this may be so, provided all parts freeze together and are kept frozen until set, and that they do not alternately freeze and thaw, which latter will undoubtedly rot the inortar. Plaster-of- Paris makes a good mortar, but is expensive and cannot stand dampness. Cements or limes that; will set under water are called hydraulic. Quickness of setting is a very desirable point in cements. All ce- ment-mortars, therefore, must be used perfectly' fresh; any that has begun to set, or has frozen, should be condemned, though many con- tractors have a trick of cutting it up and using it over with fresh mortar. To keep dampness out of cellar-walls the outside should be plastered with a mortar of some good hydraulic cement, with not more than one part of sand to one part of cement ; this cement should be scratched, roughened, and then the cement covered outside with a heavy coat of asphalt, put on hot and with the trowel. In brick walls, the coat of cement can be omitted and the joints raked out, the asphalt being applied directly against the brick. This asphalt should be made to form a tight joint, with the slate or asphalt damp- course, which is built through bottom of wall, to stop the rise of dampness from capillary attraction. In ordinary rubble stonework the mortar should be as strong as possible, as this class of work depends entirely on the mortar for its strength. For the strengths of different mortars, see Table V. Some cements are apt to swell in setting, and should be avoided. Smoke Where flues or unplastered walls are built, the joints flues, should be "■ strucl:" that is, scraped smooth with the trowel. No flues should be " pargetted " ; that is, plastered over, as the smoke rots the mortar, particles fall, and the soot accumulating in the crevices is apt to set fire to the chimney. Joints of chimneys are liable to be eaten out from the same reason, and the loose por- 120 SAFE BUILDING. tions fall or are scraped out when the flues arc cleaned, leaving dan- gerous cracks for fire to escape through. It is best, therefore, to line up chimneys inside with burned earthenware or fire-clay pipes. If iron pipes are used, cast-iron is preferable; wrought-iron, unless very thick, will soon be eaten away. Where walls are to be plastered, the joints are left as rough as possible, to form a good clinch. ^g,, Outside walls are not plastered directly on the in- furrings. side, unless hollow; otherwise, the dampness would strike through and the plaster not only be constantly damp, but it would ultimately fall o£E. Outside walls, unless hollow, are always " furred." In fireproof work, from one to four-inch thick blocks are used for this purpose. These blocks are sometimes cast of ashes, lime, etc., but are a very poor lot and not very lasting. Generally they are made of burnt clay, fire-clay or porous terra-cotta. The latter is the best, as, besides the advantages of being lighter, warmer and more damp-proof, it can be cut, sawed, nailed into, etc., and holds a nail or screw as firmly as wood. These blocks are laid up inde- pendently of the wall, but occasionally anchored to the same by iron anchors. The plastering is applied directly to the blocks. In cheaper and non-fireproof work, f urrings are made of vertical strips of wood about two inches wide, and from one to two inches tliick, according to the regularity of the backing. For very fine work, sometimes, an independent four-inch frame is built inside of the stone-wall, and only anchored to same occasionally by iron an- chors. Where there are inside blinds, a three or four inch furring is used (or a fireproof furring), and this is built on the floor beams, as far inside of the wall as the shutter-boxes demand. To the wooden furrings the laths are nailed. Furrings arc set, as a rule, sixteen inches°apart, the lath being four feet long; this affords four nailings to each lath. Sometimes the furrings are set twelve inches apart, affording five nailings. All ceilings arc cross-furred every twelve inches, on account of stiffness, and the strips should not be less than one-and-thrce-eighths inches thick, to afford strength for nailing. Furring-strips take up considerable of the strain of settlements and shrinkage, and prevent cracks in plastering by distributing the strain to several strips. To still further help this object, the -'heading- joints " of lath should not all be on the same strip, but should bo fre- quently broken (say, every foot or two), and should then be on some other strip. Laths should be separated sufficiently (about three-eighths inch) to allow the plaster to be well worked through the joint and get a strong grip or " clinch " on the back of the laths. If a building is SIIUIXKAGE OP' WALLS. 121 properly built, llieoretically correct in every respect, it should not show a single crack in plastering. Practically, however, this is impossible. "^ But there never need be any fear of shrinkage or Shrinkage ,, ', i -i i- i ^.^ ^ of joints, settlement, in a well-constructed building, where tLe foundations, joints and timbers arc properly proportioned. The danf^er is never from the amount of settlements or shrinkage, but from the °inequnlluj of same in different parts of the building. Inequalities in settlements are avoided by properly proportioning the foundations. Inequalities in shrinkage of the joints, though quite as important, are frequently overlooked by the careless architect. lie will build in the same budding one wall of brick with many joints, another of stones of aU heights and with few joints, and then put iron columns in the centre, miking no allowance whatever for the difference in shrink- age. If ho makes any, it is probably to call for the most exact set- tincr of the columns, for the hardest and quickest-setting Portland ce- ment for the stonework, and probably be content with lime for the brickwork. To avoid uneven shrinkages, allowances should be made for same. Brickwork will shrink, according to its quality, from one- sixteenth to one-eighth inch per story, ten to twelve feet high, and ac- cording to the total height of wall. The higher the wall, the greater the weight on the joints and the greater the slirinkagc. Iron col- mnns should, therefore, be made a trille shorter than the story re- quires, the beams being set out of level, lower at the column. The plan should provide for the top of lowest column to be one-sixteenth or one-eighth inch low, while the top of highest column would be as many times one-sixteenth or one-eighth inch low as there were sto- ries ; or if there were eight stories, the top of bottom column for the very best brickwork would be, say, one-sixteenth inch low, and the top of highest column would be one-half inch low. Stone walls should have stone backings in courses as high as front stones, if possible ; if not, the backing should be set in the hardest and quickest-setting ce- c,i . ment. Stone walls should be connected to brick " joints, walls by means of slip-joints. By this method the writer has built a city stone-front, some 150 feet high and over 50 feet wide, connected to brick walls at each side, without a single stone sill, or transom, or Untel cracking in the front. The slip-joint should carry through foundations and base courses where the pressure is not equal on all parts of the foundation. If for the sake of design, it is necessary to use long columns or pilasters, in connection with coursed stone backings, the columns or pilasters must either be strong enough to do the whole work of the wall, or else must be bedded in putty- 122 SAFE BUILDING. mortar with generous top and bottom joints, to allow for shrinkage of the more frequent joints behind them ; otherwise, they are apt to he shattered. Such unconstructional designs had, however, better be avoided. In no case should a wall be built of part iron uprights and part masonry; one or the other must be strong enough to do the work alone ; no rehance could be placed on their acting together. In Shrinkage frame walls, care should be taken to get the amount of timber, of " cross " timbering in inner and outer walls about equal, and to have as little of it as possible. Timber will shrink " across " the grain from one-fourth to one-half inch per foot. Where the outer walls are of masonry, and inner partitions or girders are of wood, great care must be taken that the shrinkage of each floor is taken up by itself. If the shrinkage of all beams and girders is trans- ferred to the bottom, it makes a tremendous strain on the building and will ruin the plastering. To effect this, posts and columns should bear cUrecthj on each other, and the girders be attached to their sides or to brackets, but by no means should the girder run between the upper and lower posts or columns. If there are stud-partitions, the head pieces should be as thin as possible, and the studs to upper par- titions should rest direclly on the head of lower partitions. All beds ^"^ masonry, all beds should be as nearly level as level, possible, to avoid unequal crushing. Particularly is this the case with cut stonework. If the front of the stone comes closer than the backing (which is foolishly done sometimes to make a smaU-looking joint), the face of the stone will surely split off. If the back of a joint is broken off carelessly, and small stones inserted in the back of a joint to form a support to larger stones, they will act as wedges, and the stone will crack up the centre of joint and wall. Stones should be bedded, therefore, perfectly level and solid, except the front of joint for about three-fourth inches back from the face, which should not be bedded solid, but with " putty "-mortar. Light- Cement colored stones, particularly lime-stones, are apt to stains, stain if brought in connection with cement-mortar. A good treatment for such stones is to coat the back, sides and beds with lime-mortar, or, if this is not efficacious, with plaster-of-Paris. Natural ^^^ stones should be laid on their " natural beds " ; bed. that is, in the same position as taken from the quarry. This will bring the layers of each stone into horizontal positions, on top of each other, and avoid the " peeling " so frequently seen. Ash- lar should be well anchored to the backing. The joints should be filled with putty-mortar, and should be sufficiently large to take up STONE-WORK. 123 the slirinkase of the backing. Stones should not be *'" °stones. so hirsc as to risk the danger of their being improp- erly bedded and so breaking. Professor llankine recommends for soft stones, such as sand and lime stones, which will crush with less than 5000 pounds pressure per square inch, that the length shall not exceed three times the depth, nor the breadth one-and-a-half times the depth. For hard stones, which will resist 5000 pounds' compres- sion per square inch, he allows the length to be from four to five times the depth, and the breadth three times the depth. Stones are sometimes joined with "rebated" joints, or " dove-tail" joints, the latter particularly in circular work, such as domes or light-houses. j,j,gt, ' All sills in either stone or brick walls should be hollow, bedded at the ends only, and the centre part left hol- low until the walls are thoroughly set and settled; otherwise, as the piers ^^ ^ ''"^^' ^^^ ^'^ count on the front work the best. fo,. strength. We fretiuently sec masons laying up bi'ick walls by first hiying a single course of Iieaders or stretchers on the outside of the wall, and then one on the inside, and then filling the balance of wall with bats and all kind of rubbish. This makes a very poor wall. The specification should provide that no bats or broken brick will be allowed, leaving it to the architect's discretion to stop their use, if it is being overdone ; of course, some few will have to be used. But, after all, the best wall is that one which is built the most regularly and with the most frequent bonds, and no architect should be talked out of good, regular work, as being too theorelical, by so-called " practical " men. The necessity for regular- ity and bond is easily illustrated by taking a lot of bricks of different sizes, or even toy blocks, and attempting to pile them up without regularity ; or, even if piled regularly, without bond. It will quickly be seen that the most regular and most frequently-bonded pile will go the highest. By " bond " is meant alternating headers and stretch- ers with regularity, and so as to cover and break joints. Use cf bond- "^^^ ^^^^ '^^ " bond-stones " at intervals onbj is bad ; stones, they should be carried through the whole surface (width and length) of wall and be of even thickness, or else be omitted. Using bond-stones in one place only tends to concentrate the compres- sion on one part of the walL Thus, bond-stones built under each r other at regular intervals, as ,JL , shown in Figure 80, are bad, V" ' '\ as they give the pressure no — i ' ' ■ chance to spread, but keep con- ' I ^■• ;■ ' I ~~- centrating it back onto the part | i ' i of wall immediately under . 1 'i , bond-stones, whereas, in Fi feet wide by 18 feet high each. There are S +-**^ feet of solid loall over openings. \ What should be the thickness I I S^ of belfry piers ? The mason- I ^ ry is ordinary rubhle-work. 'x In the first co place we will 1 Tower Walls. try Formula (60) giving ' strength of whole tower at ■ f% base of belfry piers. The _i _ load will be: 4.(2G.16 — 18.8 — 2(i. If) =892 superfi- ^^^ cial feet of masonry 20" thick Fig. 35. and weighing 250 lbs. i)cr superficial foot = 223000 lbs. (see Figures Fig. 84 136 SAFE BUILDINa. 84 and 85) : add to tliis spire and we have at foot of belfry piers : Actual load = ^oSOOO lbs. or = 126 tons. XoTv P2 (the square of the radius of gyration) would be P^ = --j- ; the area A:=162 — (12§2-j-4.8.1f) =43 ; the moment of inertia 1 _ j^. (1G4_12|*— 3^.83 — 8.1G3+8.12§3) =1799. Therefore P2=?^ = 41,8. Xo.v for rubble -work, Table Y, / ^ '\ =r 100 ; and, from Formula (GO), the safe load would be : ^ 43.100 4300 18,"8; or, say, ?/ = 19". , -r ^'^=14 + 0,046.2^ 1^''' *^ -^ ^ 41,8 -^ l^ ^ 291 tons; or more than strong enough. M :^ rz'^ '■' Piers at Now let us examine the ^ >^ opening, strength of each pier by ~\)'APs- -^ itself. Figure 86. In the first place we Pj gg^ must find the distance y of the neutral axis M-N from say the line A B. This from Table I, Section Xo. 20 is : ig|l'+.20..S.(20 + ¥) _ y ~ 48.20 + 20.28 Now i = ?M^i+^5^i^±^^^'= 272347 (in inches) and a = 20.48 -|- 20.28 =: 1520 square inches, therefore q^ = — = 179 (in inches). The len^i-th of each pier is IS feet, or ' L=18. Therefore, from Formula (59) we have the safe load : ^^_ 1520.100 _ gjg^Q pounjs^ , , r^ , ~- 18.18 1 + 0,4-0.^^ or say the safe load on each pier would be 41 tons. we know is — ^=: 31 1 tons, or 4 Xow let us see how far down it would be safe to of walls, carry the 20" walls. We use formula (60) and have from Section Number 4, of Table 1 : P2 ^ 16^+l .!r =, 34| (in feet). 126 :ual load we Jinow is than safe The actual load we know isi+-= 31 ^ tons, or the pier is more 4 TOWER WALLS. 137 The area would be A = 1G2— 12|2 = 9G square feet. The load for each additional foot under belfry would be then : 90.150 = 14400 lbs., or 7,2 tons. The whole load from top down for each additional foot would be, in tons : W. = 12G + (L — 26).7,2 = 7,2.L — 61 While the safe load from Formula (60) would be : IV = ^li^O 14 + 0,046.3^ Now trying this for a point 50 feet below spire, we should have the actual load : W. = 7,2.50 — 61 = 299 tons. and the safe load : W= ^G-'i^^O 554 toTjg or, we can go still 14+0,046.^ lower with the 20" work. For 70 feet below spire, we should have actual load : W,= 7,2.70 — 61 =443 to while the safe load : W— ^"^--^^^ =468 tons, or, 70 feet would be 70.70 14X0,046.1|^ about the limit of the 20" work. If we now thicken the walls to 24", we should have A= 112 square feet. P2 from Section 4, Table I, = 33^ (in feet). The weight per foot would be 112.150 = 16800 lbs. (or 8,4 tons) additional for every foot in height of 24" woi-k. Therefore the actual load would be, 443 -f (L— 70).8,4 or "W. = Z.8,4— 145. Now, for X=: 80 feet, we should have the actual load : W, = 527 tons, while the safe load would be : W=—^^^^ = 491 tons. U + 0,046.«ig? This, though a little less than the actual load, might be passed. Rubble stone work, however, should not be built to such height, good 138 SAFE BUILDING. brickwork in cement would be better, as it can be built ligbter ; for /'£_\_200, would give larger results, and brickwork weigbs less, besides ; then, too, we have the additional advantage of saving con- siderable weight on the foundations. Thickening the walls of a tower or chimney on the inside does not strengthen them nearly so much as the same material applied to the outside would, either by offsetting the wall outside, or by build- ing piers and buttresses. It is mainly for this reason, and also to keep the flue uniform, that chimneys have their outside dimensions increased towards the bottom. Example. Calculation of A circular brick chimney is to he built 150 feet high, chimneys, ^/^g y^„g entering about 6 feet from the base; the horse- poioer of boilers is 1980 HP. What size should the chimney be? The formula for size of flue is : _^^ 0,3.//P+10 (61) VL Where A = the area of flue, in square feet. Where L = the length of vertical flue in feet. Where HP =the total horse-power of boilers. Size of flue. A circular flue will always give a better draught than any other form, and the nearer the flue is to the circle the bet- ter will its shape be. In our case the flue is circular, so that we will have A=^.E^ (see Table I, Sec. No. 7) or V 22 Inserting the value of A from formula (61) we have: R= /7 0,3.1080 -f 10 ^ M. 50,3 = 4 V22- VUT V^^ or the radius of flue will be 4 feet (diameter 8 feet). Now making the walls at top of chimney 8" thick and adopting the rule of an outside batter of about f to the foot, or say 4" every 15 feet, we get a section as shown in Figure 87. Let us examine the strength of the chimney at the five levels A, B, C, D and E. The thickness of the base of each part is marked on the right-hand section, and the average thickness of the section of the part on the left-hand side. CHIMNEY WALLS. 189 Take the part above A ; the average area is (^^.5^ — flue area) or, 78 — 50 = 28 square feet. This multiplied by the height of the part and the ■weight of one cubic foot of brickwork (112 lbs.) "■ives the weight of the ■whole, or actual load. W. = 28.30.112 = 94080 lbs., or 47 tons. The area of the base at A would be : A =(=^.51^ — flue area); or A = 89 — 50 = o9 square feet. The height of the part is 2: = 30. The square of the ra- dius of gyration, in feet, ''p2=5iMll^_ 11,11 4 Inserting these values in Formula (GO) the safe load at A would be : 39.200 = 440 tons, or about nine times the actual load. Now, in examining the joint B we must remem- ber to take the whole load of brickwork to the top as well as whole length L to top (or CO feel). The load on B we find is : W, = 131 tons, while the safe load is : vO/-zMf itfTK/iSCZ Fig. B7. 140 SAFE BUILDING. n.= _5t?«» =472t„n.. 14 + 0,046.5H2 Similarly, we should find on C the load : W, = 259 tons, while the safe load ia : jy. 89.200 ... . „„ ^ ' 15,11 On D we should find the load : "W, = 432 tons, while the safe load is : W= '-1^:^ = 448 ton«. 14 + 0,046.^^:1^ ~ ' 17,44 Below D the wall is considerably over three feet thick, and is soUd, therefore we can use ( — r ) = 300, provided good Portland cement is used and best brick, which should, of course, be tlie case at the base of such a high chimney. We should have then the load on E : W,= 657 tons, while the safe load is : W= 151:300 ^ ggQ ^ 14 + 0,046.1^^ The chimney is, therefore, more than amply safe at all points, the bottom being left too strong to provide for the entrance of ilue, which will, of course, weaken it considerably. "We might thin the upper parts, but the bricks saved would not amount to very much and the offsets would make very ugly spots, and be bad places for water to lodge. If the chimney had been square it would have been much stronger, though it would have taken considerably more mate- rial to build it. It is generally best to build the flue of a chimney plumb from top to bottom, and, of course, of same area throughout. Sometimes the flue is gradually enlarged towards the top for some five to ten feet in height, which is not objectionable, and the writer has obtained good results thereby ; some writers, though, claim the flue should •w r #»i.i be diminished at the top, which, however, the writer Tops Of Chlm- '■ \ ney Flues. has never cared to try. Galvanized iron bands should be placed around the chimney at intervals, particularly around the top part, which is exposed very much to the disintegrating effects of the weather and the acids contained in the smoke. Xo smoko flue should ever be pargetted (plastered) inside, as the acids in the smoke will eat up the lime, crack the plaster, and cause it to faU. The BULGING OF WALLS. 141 crevices will fill with soot and Ije liable to catch fire. The mortar- joints of flues should bo of cement, or, better yet, of fire-clay, and should be carefully struck, to avoid being eaten out by the acids. . Where walls arc long, without buttresses or cross- Calculation of , ,, . , n r i m r « Walls.- Bulging, walls, such as gable-walls, side-walls ot building^ etc., we can take a slice of the wall, one running foot in length, and consider it as forced to yield (bulge) inwardly or outwardly, so that for p^ we should use : Q- _ i^ ; where d the thickness of wall in inches. The ^ 12 area or a would then be, in square inches, a = \1.d. Inserting these values in formula (59) we have for BRICK OR STONE WALLS. fl (62) (t) M7 = - L3 0,0833 + 0,475.-^ Where w = the safe load, in lbs., on each running foot of wall (d" thick). Where J = the thickness, in inches, of the wall at any point of its lieight. Where L=the height, in feet, from said point to top of wall. Where (-^ ) = the safe resistance to crushing, in lbs., per square inch, as given in Table V. (See page 135.) If it is preferred to use tons and feet, we insert in formula (GO) : for A = D, where D the thickness of wall, in feet, and we have : P2=±--: therefore 12 ' o U + 0,552.jp Where AV = the safe load, in tons, of 2000 lbs., on each running foot of wall (D feet thick). Where D = the thickness of wall, in feet, at any point of its height. Where L = the height, in feet, from said point to the top of wall. "Where (-^) = the safe resistance to crushing of the material, in lbs., per square inch, as found in Table V. (See page 135.) Anchored Walls. AVliere a wall is thoroughly anchored to each tier 142 SAFE BUILDING. of floor beams, so that it cannot possibly bulge, except between floor- beams, use the height of story (that is, height between anchored +-- ' QrfY rLOon. ID +-■ ; 7 ^ 2& 1 i 5 ■!o 32 36 -In 40 r F/krmooJi Fig. 88. Fig. 89. Fig. 90. beams in feet) in place of L and calculate d or D for the bottom of •wall at each story. The load on a waU consists of the wall itself, from the point at which the thickness is being calculated to the top, plus the weight of EXAMPLES OK WALLS. 143 one foot in widtli by half the span of all the floors, roofs, partitions, etc. Where there are openings in a wall, add to pier the proportion- ate weight Avhich would come over opening ; that is, if we find the load per running foot on a wall to be 20000 lbs., and the wall con- sists of four-foot piers and three-foot openings alternating, the piers will, of course, carry not only 20000 lbs. per running foot, but the 60000 lbs. coming over each ojoening additional, and as there are four feet of pier we must add to each foot ^^^i^z= 15000 lbs. ; ^vc therefore calculate the pier part of Avail to carry 35000 lbs. per run- ning foot. The actual load on the wall must not exceed the safe load as found by the formula (G2) or (03). Example. Wall of Country A two-Hlorij-and-atlic dioelling has hriclc icalls 12 ^"®®" inches thick; the walls carry two tiers of heanis of 20 feet span; is the wall strong enough f The brickwork is good and laid in cement mortar. We will calculate the thickness required at first story beam level, Figure 88. The load is, pei running foot of wall : Wall =22.112 = 2464 lbs. Wind =22. 15= 330 lbs. Second floor =10.90= 900 lbs. Attic floor = 10. 70 = 700 lbs. Slate roof (inch wind and snow) = 10. 50 = 500 lbs. Total load =4«94 lbs. For the quality of brick described we should take from Table V : ('-^) = 200 lbs. J The height between floors is 10 feet, or L=10, therefore, using formula (02) we have : d.\ Kj) 12-200 =35807 lbs. 0,0833 + 0,475. ^^ 0,0833 + 0,475. 1M2 So that the wall is amply strong. If the wall were pierced to the extent of one-quarter with openings, the weight per running foot would be increased to 6525 lbs. Over 700 lbs. more than the safe load, still the wall, even then, would be safe enough, as we have allowed some 330 lbs. for wind, which would rarely, if ever, be so strong; and further, some 1200 lbs. for loads on floors, also a very 144 SAFE UUILDING. ample allowance; and even if the two ever did exist together it would only run the compression T— j up to 225 lbs. per inch, and for a temporanj stress this can be safely allowed. The writer would state here, that the only fault he finds with for- mulae (59), (60), (62) and (63), is that their results are apt to give an excess of strength ; still it is better to be in fault on the safe side and be sure. Example. Walls of City ■^'''^ hrick walls of a warehouse are 115 feel high, Warehouse, the 8 stories are each 14 feet high from floor to floor, or 12 feet in the clear. The load on floors per square foot, including the fire-proof construction, loill average 300 lbs. What size should the walls be f The span of beams is 26 feet on an average. (See Fig. 89, page 142.) According to the New York Building Law, the required thicknesses would be: first story, 32" ; second, third, and fourth stories, 28"; fifth and sixth stories, 24" ; seventh and eighth stories, 20". At the seventh story level we have a load, as follows, for each run- ning foot of wall : Wall =30.1|.112 = 5600 Wind = 3.0.30= 900 Eoof = 13.120=7 1560 Eighth floor = 13.300 = 3900 Total = 11960 lbs., or 6 tons. The safe load on a 20" wall 12 feet high, from formula (63) is: W— l|-.200 333 - .^, ,33, 12.12 = 14 + 0,552.51,84 = ^'^^^ *°"^' °^ ' 'If-lf 15622 lbs. If one-quarter of the wall were used up for openings, slots, flues, etc., the load on the balance would be 8 tons per running foot, which IS still safe, according to our formula. At the fifth-story level the load would be : Load above seventh floor = 11960 Wall =28.2.112=1: 6272 Wind =28.30 = 840 Sixth and seventh floors = 2.13.300 = 7800 Total = 26872 lbs. or 13i tons. WAREHOUSE WALL. 145 The safe load on a 24" wall, 12 feet higli, from formula (G3) is : w 2.200 400 14 + 0,5o2 ' ' ^ ' 2.2 or 23C18 lbs. This is about 10 per cent less than the load, and can be passed as safe, hut if there -were many ilucs, openings, etc., in wall, it should be thickened. At the second-story level the load would be : Load above fifth floor = 26872 Wall =42.2^.112 = 10976 Wind = 42. 30 = 1260 Third, fourth and fifth floors = 3.13.300 = 11700 Total = 50808, or 25 tons. The safe-load on a 28" wall, 12 feet high, from formula (63) is : ,Tr 2i.200 467 ,c oo * VV = ^ ; = 16,33 tons, 14 -I- 559 IM? "" 14 + 0'552.26,45 or 32660 lbs. Or, the wall would be dangerously weak at the second-floor level. At the first-floor level the load would be : Load above second floor = 50808 Wall = 14. 2|.112 = 4181 Wind = 14.30 = 420 Second floor = 13.300 = 3900 Total = 59309, or 29^ tons. The safe-load on a 32" wall, 12 feet high, from formula (63) is: W = '-^^ r-^^^ 21,169 tons, 14-f0,552.1^2~l^ + «'-5^-20,25- ' 2f.2|- or 42338 lbs. The wall would, therefore, be weak at this point, too. Now while the conditions we have assumed, an eight-story ware- house with all floors heavily loaded, would be very unusual, it answers to show how impossible it is to cover every case by a law, not based on the conditions of load, etc. In reality the arrangements of walls, as required by the law, are foolish. Unnecessary weight is piled on top of the wall by making the top 20" thick, which wall has nothing to do but to carry the roof. (If the span of beams were increased to 31 feet or more the law compels this top wall to be 24" thick, if 41 feet, it would have to be 28" thick, an evident waste of material.) It 146 SAFK BUILDING. ■would be inucli better to make the top walls lighter, and add to the bottom ; in tliis case, the writer would suggest that the eighth story be 12" ; the seventh story IG" ; the sixth story, 20"; the fifth story, 24"; the fourth story, 28"; the third story, 32"; the second story, 3G", and the first story 40", see Figure 90. (Page 142.) This would represent Jmt 4-| cubic feet of additional brickwork for every running foot of wall ; or, if we make the first-story wall 3(i" too, as hereafter suggested, the amount of material would be exactly the same as required by the law, and yet the wall would be much better proportioned and stronger as a whole. For we should find (for L = 12 feet), Actual load at eighth-floor level, 3832 ) Safe load on a 12" wall from Formula (G2) 42D8 \ Actual load at seventh-floor level, 10243 ) Safe load on a 16" wall from Formula (G2) 9135 | Actual load at sixth-floor level, 1717G ) Safe load on a 20" wall from Formula (G2) 15729 ^ Actual load at fifth-floor level, 24G32 [ Safe load on a 24" wall from Formula (G2) 23750 ) Actual load at fourth-floor level, 32G11 } Safe load on a 28" wall from Formula (G2) 32825 j Actual load at third-floor level, 41112 Safe load on a 32" wall from Formula (62) 42638 Actual load at second-floor level, 5013G Safe load on a 3G" wall from Formula (62) 5290 Actual load at first-floor level, 69683 Safe load on a 40" wall from Formula (62) 6349 The first-story wall could safely be made 36" if the brickwork is good, and there are not many flues, etc., in walls, for then we could use (7) . = 250, which would give a safe load on a 36" wall = 65127 lbs., or more than enough. The above table shows how very closely the Formula (62) would agree with a joractical and common - sense arrange- ment of exactly the same amount of material, as required by the law. 56) )2| )2; Fig. 91. TIIUUST OK DAKUKLS. 147 _. ^ , Now, if the upper (loor were liulen with l);uTel«, Thrust of » ' barrels, there might be some danger of these thrusting out the wall. We will suppose an extreme ease, four layers of flour barrels l)aekeil against the wall, leaving a 5-foot aisle in the centre. We should have 20 barrels in each row (Fig. 91), weighing in all 20.190 = 3920 lbs. These could not well be placed closer than 3 feet from end to end, or, say, 1307 lbs., per running foot of wall; of this amount only one-half will thrust against wall, or, say, G50 pounds. The ra- dius of the barrel is about 20". If Figure 92 represents three of the barrels, and we make A B = ti\ = ^ the load of the flour barrels, per running foot of walls, it is evi- dent that D B will represent the horizontal thrust on wall, per run- ning foot. As D B is the radius, and as we know that A D = 2 D B or = '2 radii, we can easily find A B, for : A D2 Fie. 92. DB2 — AB2 DB2r=-^or 3 DBr=^^z= or 4. D B'^ — D B2 : :0,578.tt'. •- /I /I \\ IHH I I ^ O ^3" i,n Or, A = 0,578. ?o, (64) Where h = the horizontal thrust, in lbs., against each running foot of wall, u!,=r one- half the total load, in lbs., of barrels coming on one foot of floor in width, and half the span. In our case we should have : 74 = 0,578.650 = 375 lbs. Now, to find the height at which this thrust would be applied, we see, from Fig- ure 91, that at point 1 the thrust would be from one line of barrels ; at point 2, from two lines ; at point 3, fi-om three lines, etc. ; therefore, the average thrust will be at the centre of gravity of the triangle A B C, this we know would be at one-third = the height A B from its base A C Now B C is eciual to 6?- or six radii of >- » > ^ SCALE OF X'-'erGHrs.CLB5j i2 24 36 iCALE OFL£AfOr/fJ (/NCffEiJ O ISOO 3O0O 45QO the barrels ; further, A C = 3r, therefore : ng.93. A B2 = 3C.r2 — 9r2 AB 25.?-2, and A B ^ 5?-; therefore, -— =: l|r. 148 SAFE KLILDING. To this must be ailJetl the radius A D (below A) so tliat the central point of thrust, O in this ease, would be above the beam a distance 2/ = 2|.r. Where y = the height, in inches, above floor at which the average thrust takes place. AVhere r == radius of barrels in inches. Our radius is 10", therefore : 7/ = 202-" Xow, in Figure 93 let A B C D be the 12" wall, A the floor level, G M the central axis of wall, and A O = 26|; draw O G horizon- tally ; make G II at any scale equal to the permanent load on A D, which, in this case, would be the former load less the wind and snow allowances on wall and roof, or 3832 — (16. 30 4" 13.30) = 2962, or, say 3000 lbs. Therefore, make G II = 3000 lbs., at any scale ; draw B.l = h = 375 lbs., at same scale, and draw and prolong G I till it intersects D A at K. The pressure at K will be /; = G I = 3023. *We find the distance M K measures MK = x = 3i". Thei'efore, from formula (44) the stress at D will be : 3023 I (3313023 _ I 5(3 ii3s_ ("oi, compression) 144 ~ 12.144 n- V i- y While at A the stress would be, from formula (45) : 3023_g3|^3023__j^ ^^g_ , tension), so that 144 12.144 ^ the wall would be safe. The writer has given this example so fully because, in a recent case, where an old building fell in New York, it was claimed that the walls had been thrust outwardly by flour barrels piled against them. Narrow Piers. Where piers between openings are narrower than they are thick, calculate them, as for isolated piers, using for d (in place of thickness of wall) the width of pier between openings ; and inP- place of L the height of opening. The load on the pier will consist, besides its own weight, of all walls, girders, floors, etc., coming on the wall above, from centre to centre of openings. Wind-pressure. To calculate wind-pressure, assume it to be normal to the wall, then if A B C D, Figure - 94, be the section of the whole wall above ground (there being no beams or braces against wall). DMC K Fig. 94. WIND-PKESSUKE. 14l> Make O D = ^. A I> = — ; draw G II, the vertical neutral axis of the -whole mass of wall, make G II, at any convenient scale, equal to the whole weight of wall; draw II I horizontally equal to the total amount of wind-pressure. This wind-pressure on vertical surfaces is usually assumed as being equal to 00 lbs. per square foot of the surface, provided the surface is flat and normal, that is, at right angles to the wind. If the wind strikes the surface at an angle of 45° the pressure can be assumed at 15 lbs. per foot. It will readily be seen, therefore, that the gi-eatest danger from wind, to rectangular, or square towers, or chimneys, is when the wind strikes at right angles to the widest side, and not at right angles to the diagonal. In the latter case the exposed surface is larger, but the pressure is much smaller, and then, too, the resistance of such a structure diagonally is much greater than directly across its smaller side. In circular structures multiply the average outside diameter by the height, to obtiin the area, and assume the pressure at 15 lbs. per square foot. In the examples already given we have used 15 lbs., where the building was low, or where the allowance was made on all sides at once. AVhere the wall was high and supposed to be normal to wind we used 30 lbs. Referring again to Figure 94, continue by drawing G I, and prolonging it till it intersects D C, or its prolonga- tion at K. Use formulae (44) and (45) to calculate the actual pres- sure on the wall at D C, remembering that, x = M K ; where M the centre of D C, also that, p = G I; measured at same scale as G H. Remember to use and measure everything uniformly, that is, all feet and tons, or else all inches and pounds. The wind-pressure on an isolated chimney or tower is calculated similarly, except that the neutral axis is central between the walls, instead of being on the wall itself; the following example will fully illustrate this. Example. Is the chimney, Figure 87, safe against wind- Wind- pressure on Chimney, jjressure? We need examine the joints A and E only, for if these are safe the intermediate ones certainly will be safe too, where the thickening of walls is so symmetrical as it is here. The load on A we know is 47 tons, while that on E is 057 tons. 150 SAFE BUILDING. Now the wind-pressure down to A is : Pa = 10.30.15 = 4500 lbs., or = 2^ tons. On base joint E, the wind-pressure is : P = r2f.l50.15 — 28500 lbs., or = U\ tons. We can readily see that the wind can have no appreciable effect, but continue for the sake of illustration. Draw Pa horizontally at half the height of top part A till it intersects the central axis G, ; make G, 11, at any convenient scale = 47 tons, the load of the top part ; draw H, I, horizontally, and (at same scale) = 2\ tons = the wind-pressure on top part ; draw G I, ; then will this represent the total pressure (from load and wind) at K, on joint A A,. TTmi Formula (44) to get the stress at A, where x =. K, M, = 9", or 4 feet; and p = G, I„ which we find scales but little over 47 tons ; and Formula (45) for stress at A. For d the width of joint we have of course the diameter of base, or lOf feet. Therefore Stress at A, = |^ + 6. ^^ = + 1,71 tons, per square foot, or^-iIi^^^ = -f-24 lbs. (compression), per square inch, and 14: -i Stress at A =1^ — 6. 4^2 = + 0,7 tons, per square foot, or 0,7.2000 — _j_^^ j|^g_ (compression), per square inch. To find the pressure on base E E. ; draw P G horizontally at half the whole height; make G M = 657 tons (or the whole load)i and draw U K horizontally, and == 141 tons (or the whole wind-pressure). Draw G K and (if necessary) prolong till it intersects E E, at K. From formula (44) and (45) we get the stresses at E, and E : p being = G K = 658 tons ; and x = M K = 20", or If feet. For d the width of base we have the total diameter, or 16 feet. Therefore stress at E - ^^ 4- 6. ^iHl = -4- 7 tons per sq. ft., or ^4^ = +07 lbs. ■^'"~151~ 151.16 ' ^ I'l-i , T G58 p 658.1|_ (compression), per square inch, and stress at t. _ — — o.~— — 4- 1| tons (compression), per square foot, or ^j-^ = -f- 23 lbs. (compression) per square inch. There is, therefore, absolutely no danger from wind. Strength of Corbels carrying overhanging parts of the walls, Corbels, etc., should be calculated in two ways, first, to see > Scale of weights. Fig. 87, applies to G, H,, etc., but not to G IM. STRENGTH OF CORBELS. 151 wlicthcr the corbel itself is strong enough. We consider the corbel as a lever, and use either Formula (25), (2G) or (27) ; according to how the overhang is distributed on the corbel, usually it will be (25). Secondly, to avoid crushing the wall immediately under the corbel, or possible tipping of the wall. "Where there is danger of the latter, long iron beams or stone-blocks must be used on (op of the back or wall side of corbel, so as to bring the weight of more of the wall to bear on the back of corbel. To avoid the former (crushing under corbel) find the neutral axis G H of the whole mass, above corbel. Figure 95 ; continue G II till it intersects A B at K, and use Formulas (44) and (45). If M be the centre of A B, then use a: = K M, and p = weight of corbel and mass above ; remembering to use and measure all parts alike ; that is, either, all tons and feet, or all pounds and inches. Example. Calculation of '^ brick tower has pilasters 30" wide, projecting corbel. jg". Qf^ Q^e side the toioer is engaged and fur rea- sons of planning the pilasters cannot be carried down, but must he sup- ported on granite corbels at the main roof level, which is 36 feet below top of tower. The loall is 24" under corbel, and averages from there to top IG", offsets being on both sides, so that it is central over 2i" wall. What thickness should the granite corbel be ? If we make the pilaster hollow, of 8" walls, we will save weight on the corbel, though w^e lose the advant- [* 5"Cr ■*! age of bonding into wall all way up. Still, we will make it hol- low. First, find the dis- K Vl S I I ! tancc of neutral axis M-N (Figure 9G), of pilaster from 24" wall, which will give the central point of H M^- -4:_ ~'T' _N % Ro^, 1 .1. F ig. 96. We use rule given in the first article, Fi2. 95 application of load on corbel and have ^ _ 8.30.12 + 8.8.4 + 8.8.4 ^ , „ 8.30+8.8 + 8.8 ^ The weight of pilaster (overlooking corbel) will be GG24 lbs., or 3^ tons 152 SAFE BUILDIXG. Assuming that only 30" in length of the wall will come on back part of corbel, the weight on this part would be 8640 lbs., or 4^ tons. The bending moment of 6624 lbs. at 9^" from support on a lever is (see Formula 27) : 7)1 = 6624.9A- = 61824 (lbs. inch). From Formula (18) we know that: 771 (7) From Table I, Section Number 3, we have: And from Table V, we have for average granite : ^ (7)="" Inserting these values in Formula (18) we have: 61824 , „ „ 61824 5.d^, or d^ : = 68,7, thei-efore, d 180 ' 900 We should make the block 10" deep, however, to work better with brickwork. This would give at the wall a shearing area = 10.30 = 300 square inches, or — — = 22 lbs., per square inch, which granite will certainly stand ; still, it would be better to corbel out brickwork under granite, which will materially stiffen ,5^ 4/^ Pressure ^^^ strengthen the block. Q Q) under corbel. To find the crushing strain !<■ S^AZ~~^ on bi'ick wall under corbel, find the central axis of both loads by same rule as we find centre of gravity ; that is, its distance A Kf from rear of wall will be (Figure 97) 41.12 + 31.331 ^* + ^3 or, say, 22", the pressure at this point will, 224" "SI \ IM •*" 22 'A •24—^ Fig. 97. of course, be = 4J -j- 3J = 7J tons, or p :^ 15340 lbs. The area will be = 24.30 = 720 square inches, while K M meas- ures 10" ; we have, then, from Formulte (44) and (45) stress at B = 15340 15340.10 stress at A = 720 15340 720.24 15340.10 74 lbs. (compression), and 31 lbs. (tension). 720 720.24 There would seem to be, therefore, some tendency to tipping, still PRESSURE UNDER CORBELS. 153 we can pass it as safe, particularly as much more than 30" of the wall will bear on the rear of corbel. If the wind could play against inside wall of tower, it might help to upset the corbel, but as this is impossible, its only effect could be against the pilaster, which would materially help the corbel against tipping. CHAPTER V. 3p / Fig. 98. HE manner of laying arches has been de- scribed in the previous chapter, while in the first chapter was given the theory for calcu- lating their strength ; all that will be necessary, therefore, in this chapter will be a few practical examples. Before giving these, however, it will be of great assistance if we first explain the method of obtaining grapldcally the neu- tral axis of several surfaces, for which the arithmetical method has already been given Neutral axis (P- ^O- To find the vertical neu- found graphi- tral axis of two plane surfaces *^^"^' A B E F and B C D E, proceed as follows : Find the centres of gravity G„ of the former surface and G, of the latter surface. Through these centres draw G„ II„ and G, H, vertically. Draw a vertical line a c anywhere, and make ab at any scale equal to surface A B E F, and at same scale make 6 c = B C D E. From any point o, draw o a, ob and oc. From the point of intersection g,oi oc with H, G„ draw ^r, 0 units and e/= 30 units. Draw oa, oh, oc, etc. Drawing the lines parallel thereto, beginning at a we get the line a 1 t, h h u lU, same as before Imagining a joint -r 2o1-o i\oo 200 ^oo ^oo yoo t OF POUMDJ. SC/>^UE op FEBT. Fig. 105. at C D this would evidently be the joint with greatest stress, for the same reasons mentioned before. We find C K^ scales 2§", and as C D scales 7^" the point K^ is distant from the centre of joint. (X = )3|-2|r=l" as o/ scales 2100 units or pounds, and the joint is 7;i" deep with area = 7^. 12 = 87 square inches, we have : „^ ^ r, 2100 , „ 2100. 1 , A, ,. 1 A Stress at C = -—- + 6. ^ =-}- 44,14 pounds and 0/ o7. i-^ Stress at D 2100 6. 2100.1 4,14 pounds. 87. 7i The arch, therefore, would seem perfectly safe. But the blocks are not solid ; let us assume a section tlirough the skew-back joint C D to be as per Fig. 106. We should have in Formulte (44) and (45) X, p, and the depth of joint same as before, but for the area we should use a == 3.1^.12 = 54 square inches, or only the area of soUd parts of block. Therefore we should have : FimU'nOOF FLOOK-AIICII. IGl Stress at C = 2100 Stress at D = 54 2100 G. -j- 71 pounds, and 2100. 1 . 54. 7f 21*^0.^= _^ 6,71 pounds. 54 54. 7i There need, therefore, be no doubt about the safety of the arch. Example IV. Over a 20-inch brick arch of 8 feci clear span is a centre pier 16' wide, carrying some ttco tjns tveif/ht. On each side of pier is a window opening 2^ feet wide, and beyond, piers similarly loaded. Is the arch safe? . ^ , , We divide the half arch into five equal voussoirs. Arch In front t wal I concen- The amounts and neutral axes of the dillcrent vous- trated loads. gQjj,g^ j^jjj lyads coming over each, are indicated in circles and by arrows; thus, on the top voussoir E B (Fig. 107) we have a load of 2100 pounds, another of 02 pounds, and the weight of voussoir or 228 pounds. The neutral axis of the three is the verti- cal through G, (Fig. 108). Again on voussoir E F (Fig. 107) we UCAI-E op FEBT, Fig. 107. have the load 174 pounds, and weight of voussoir 228 pounds; the vertical neutral axis of the two being throujih G„ (Fig. 108). Simi- larly we get the neutral axes G„„ G,v and Gy (Fig. 108) for each of the other voussoirs. Now remembering that 1 r/, (Fig. 107) is tho 162 SAFE UUILUING. neutral axis of and equal to the voussoir B E and its load ; 2 g^ the neutral axis of and cipal to the sum of the voussoirs B E and E F and their loads, etc., we find the horizontal thrusts ^r, /i„ g^ lu, gj hi, etc. The last g^ Ih is again the largest, and we find it scales 7850 units or pounds. The arch being heavily loaded we selected a at one-third from the top of A B. We now make (Fig. 108) a o= 7850 pounds or units at any scale ; and at same scale make ab= 2390 pounds ; b c = 402 pounds; c d = 432 pounds ; d e = 2956 pounds, and e f= 18G8 pounds. Draw o b, o c, o d, etc. Now draw as before a 1 parallel with a to axis G, ; also 1 i, parallel with o & to axis G„ ; i, h parallel with c to G,u, etc. We then again have the points a, /f„ K«, K^, Kt O t oo 2oo 5oo 4-00 ?oo fto o 5CA1.E OF tOtJtiOf /C:is Gg. Returning now to the arch, we go through the same iirocess as before. AVe find the horizontal pressu.es (Fig. 109) ff, h„ g.^ h, etc. In this case we find that the last pressure f/; 1^ is not as large as (j^ h ; therefore we adopt the latter; it scales 1425 units or pounds. We now make (Fig. 110) a o =^ 1425 pounds ; and a 6 = 251 pounds ; 6 c r= 280 pounds ; c (/ = 373 pounds, etc. ; g h is equal to the last section of arch or 1 782 pounds. We continue, however, and make 7t i = 4G00 pounds = the weight of abutment. Draw o a, o b, o c, etc., to i. Then get the tangents to the curve of pressure, as be- fore, viz. : a 1 /, h h u h u K^ ; we now continue i^ K^, which is par- allel with h till it intersects the vertical axis G^ of the abutment at t, and from thence we draw ?j Kg parallel with o i. We will now examine the base joint I H of pier. "'"^''"I\)utment. I JQ scales 10;^", and as the pier is 3G" wide, Ka is 7|" from the centre of joint. The area is a = 12.36 = 432 and the pressure isp = o i = 0100 pounds. Therefore, Stress at I = ^^ + 6. ^'^ = + 48 pounds, and StressatH = ^^-6.|l|?fII = -G pounds. 432 432. oO There is, therefore, a slight tendency for the pier to revolve around the point I, raising itself at H ; still the tendency is so small, only 6 pounds per square inch, that we can safely pass the pier, so far as dano-er from thrust is concerned. Joint C D at the spring of the arch looks rather dangerous, how- ever, as lo h cuts it so near its edge D. Let us examine it. D A'; measures If, therefore K^ is U" from the centre of joint, which is 12" wide. The area is, of course, a = 12. 12 = 144 and the pressure J) = h = 4G00 pounds. Therefore, Stress at D = ^^ + G.^^* = + 104 pounds, and Stress at C = ^ - G. ^ = -40 pounds. 144 It is evident, therefore, that the arch itself is not safe, and it should be designed deeper ; that is, the joints should be made deeper (say, 16"), and a new calculation made. TIE-RODS TO AUCIIES. 1G5 Tie-rods to ^^' instead of an abutment-pier, wo had used an arches, ji-on tic-rod, its sectional area would have to be suf- ficient to resist a tension equal to the greatest horizontal thrust o a; /o /o i / •m !/■ / ■h ; i / ! / ^ . Fig. I 10. ■JCAI-EOFFEET- •• and care should be taken to proportion the washers at each end, large 166 SAFE BUILDING. enough that they may have sufficient bearing-surface so as not to crusli the material of the skew-backs. Thus, in Example III, Fig. 105, if we should place the iron tie-rods to the beams 5 feet apart, they would resist a tension equal to five times the horizontal thrust o a, which, of course, was calculated for 1 foot only, or t = 5. 2040 = 10200 pounds. The safe resistance of wrought-iron to tension is from Table IV, 12000 pounds per square inch ; we need, therefore : 10200 „„. . , — — — = 0,8o square inches of area in the rod ; or the i-od should be 1 ^V" diameter. A 1" or even |" rod would probably be strong enough, however, as such small iron is apt to be better welded, and, consequently, stronger, and the load on the arch would probably be a " dead " one. As the end of rod will bear directly against the iron beam, the washers need have but about ^" bearing all around the end of the rod, so that the nut would probably be large enough, and no washer be needed. Example VI. A pier 28" ivide and 10' Jiir/h supports iv:o abull'mg semi-circular arches ; the rifjht one a 20" Irick arch of 8' span ; the left one an 8" brick arch of 3' span. The loads on the arches are indicated in the Figure 111. Is the pier safe f Uneven arches The loads are so heavy compared to the weight of with central pier, the voussoirs, that we will neglect the latter, in this case, and consider the vertical neutral axis of and the amount of each load as covering the voussoirs also ; except in the case of the lower two voussoirs, where the axes are considerably affected. We find the curve of pressure of each arch as before. For the large arch we would have the curve through a and i, for the small one through a, and i, ; the points i and z, being the inter- sections of the curves with the last vertical neutral axis of each arch. Now from i draw i x parallel with of, and from z, draw i, x (back- wards), but parallel with o^f till the two lines intersect at x. Now make f g parallel with and = o, f, and draw g h vertically = 2600 pounds z= the weight of the pier from the springing line to the base (1' thick). Draw o g and o h. Now returning to x, draw x y paral- lel with g till it intersects the neutral axis of the pier at y, and from y draw y z parallel with o h till it intersects the base joint C D of THRUST ON CENTRAL PIER. 1G7 pier (or its prolongation) at K,. Continue also x y till it intersects the springing joint A B and K. Now, then, to get the stresses at Fig. 111. joint A B we know that the width of joint is 28", therefore a = 28.12 = 336 • further p = o g-= 19250 pounds, and as B K scales one inch, K is distant 13" from the centre of joint, therefore Stress at B = 1?S^ + 6. ^'^:^ = -\-2ll pounds. and 336 , . 19250 p Stress at A = — 6 336. 28 19250. 13 ■102 pounds. 336 ' 336. 28 The arch, therefore, cannot safely carry such heavy Thrust on ,, ^ ^ c i central pier, loads. The pier we shall naturally expect to lind still more unsafe, and in effect have, remembering that joint D C is 28" wide, therefore, area 336 square inches, and as K, distant 54" from centre of joint, and p = o A = 21 750 pounds. Stress at D = ^V 6. ^S^ = + ^13 pounds. 336 336. 28 168 SAFE BUILDING. and Stress at C = 21750 21750.54 = — 684 pounds. 336 336. 28 Relief through The construction, therefore, must be radically iron-work. changed, if the loads cannot be altered. If the arches are needed as ornamental features, they should be constructed to carry their own weight only, and iron-work overhead should carry the loads, and bear either nearer to, or directly over, the piers, as farther trials and calculations might call for. If this is done the wall To avoid should be left hollow under all but the ends of iron- cracks, work until it gets its " set " ; that is, until it has taken its full load and deflection ; and then the wall should be pointed with soft " putty " mortar. Example VII. The foundations of a building rest on brick piers 6' wide and 18' apart. The piers are joined by 32" brick inverted arches and tied to- gether 8' above the spring of the arch. Piers and arches are 3' thick. Load on central piers is 72 tons ; on end pier 60 tons. Is this con- struction safe ? Inverted We will first examine the inner or left pier. The arches, pjer being 3' thick and the load 72 tons, each 1' of thickness will, of course, carry -''^ = 24 tons. The width from centre to centre of piers is just 24', so that each running foot of wall under ^CAUB «F To J arch. Wc make a b = 2 tons ; b c ^= 2 tons, etc., and nnd the horizontal pressures 46 or, say, lU square feet. As A B (the radius) measures 2' 9", the circumference of its semi-circle would be 8',G. Taking the weight of the stone at IGO pounds per cubic foot, we should then have the weight of No. I = 11, .5. 8, G. IGO = 15824 pounds, or, say, 8 tons. Similarly we should have : No. II = 12,4. 25,1. IGO = 49798 pounds, or, say, 25 tons. No. Ill = 14,2. 40. 1G0= 90880 " " 45 " No. IV = 17,4. 52,7. 1G0 = 14G71G '= " 73 " No. V = 21,2. 62,8. IGO = 213017 " " lOG " No. VI = 28,7. 69,1. IGO = 317307 " " 153 " 172 SAFE BUILDING. We now make a b = 8 tons, & c = 25 tons, c d ^ io tons, d e = 73 tons, e/= 106 tons andfg = 158 tons. We find the horizontal pressures g, h„ g^ ho, etc. (in Fig. 113), same as before. In this case N^VIN'VNMV mi" "c" B A 70 o I ' Fig. 114. •u-e find that the largest pressure is not the last one, but g^ Jh which measures 78 units; we therefore select the latter and (in Fig. 114) make a o = g-^ h^ z= 78 tons. Draw o b, o c, o d, etc., and construct the line a 1 i, i^ is U h IQi same as before. In this case we cannot tell at a glance which is the most strained voussoir joint, for at joint U I the pressure is not very great, but tlie line is farthest from the centre of joint. Again, while joint J L has not so much pressure as STRESS AT .JOINTS. 173 the bottom joint M N, the line is farther from the centre. We must, therefore, e>:amine all three joints. Wo will take II I first. The width of joint is 2' 5" or 29". The pressure is o c, which scales 88 tons or 17G000 pounds. The distance of K« from the centre of joint P is 10". The area of the joint is, of course, the full area of the joint around one-half of the dome, or equal to II I multiplied by the circumference of a semi-circle with the dis- tance of P from a ^, stopped by grav- ity, resistance of the atmosphere, N friction or some other force ; similarly, matter, if once at rest, will so remain unless started into motion by some external force. Formerly it was believed, however, that all matter had a certain re- pugnance to being moved, which had to be first overcome, before a body could be moved. Probably in connection with some such theory the term arose. In reality matter is perfectly indifferent whether it be in motion or in a state of rest, and this indifference is termed " Inertia." As used to-day, however, the term Moment of Inertia is simply a symbol or name for a certain part of the formula by which is calculated the force necessary to move a body around a certain axis with a given MOMKXT OF INKUTIA. 181 velocity in a certain space of time ; or, what amounts to the same thing, the resistance necessary to stop a body so moving. In making the above calculation the " sum of the product of the weight of each particle of the body into the s(iuare of its distance from the axis " has to be taken into consideration, and is part of the formula ; and, as this sum will, of course, vary as the size of the body varies, or as the location or direction of the axis varies, it would be difficult to express it so as to cover every case, and there- fore it is called the " Moment of Inertia." Hence the general law or formula given covers every case, as it contains the ^Moment of Inertia, which varies, and has to be calculated for each case from the known size and weight of the body and the location and direction of the axis. In plane figures, which, of course, have no thickness or weight, the area of each ])article is taken in place of its weight ; hence in all plane figures the Moment of Inertia is equal to the " sum of the prod- ucts of the area of each particle of the figure multiplied by the square of its distance from the axis." Calculation of Thus if we had arectangular figure (HO) finches "^"tVa."* °^ '"' "wide and d inches deep revolving around an axis M-N, we would divide it into many thin slices of equal height, say n slices each of a height = 2. X. The distance of the centre of gravity of the first slice from the axis M-N will, of course be = ^. 2. X. = 1. X The distance of the centi'e of gi-avity of the second slice will be = 3.x, that of the third slice will be= 5. X, that of the fourth slice will be= 7. X, that of the last slice but one will be = (2 n — 3). X. and that of the last slice will be i= (2 n — 1). X The area of each slice will, of course, be = 2. X. 6 ; therefore the Moment of Inertia of the whole section around the axis M-N will be (see p. 8), i = 2. X. h. (1. X)2 + 2. X. h. (;3. X)2-|- 2. X. h. (5. X)2 + 2. X. h. (7. X)--f etc + 2. X. 6. [(2;i — 3).Xp -f 2. X. ?;.[(2n — l).X]-or, i=2.X.3Z<.[l^+3-2 + 52+72 + etc + (2;i — 3)2 + (2u-l)2] now the larger n is, that is the thinner we make our slices, the nearer will the above approximate : 182 SAFE BUILDIKG. t = 2. X8. 6, [t--] = 8. X3. nS Therefore, as : 2. X. n = rf we have, by cubing, 8. X^. n^ = d^; inserting this in above, we have ; J3. h. h. — or - The same value as given for i in Table I, section No. 29. Of course it would be very tedious to calculate the Moment of Inertia in every case ; besides, unless the slices were assumed to be very thin, the result would be inaccurate; the writer has therefore given in Table I, the exact Moments of Inertia of every section likely to arise in practice. The Moment of Inertia applies to the whole .()X I-BEAM GIUDKKS, BUACEI) SIDEWAYS. "si's 03 SPAN OF GlIiDEU IN FKET. ; . . . . ;i| rt rH rH tH T-( tH rH rl r-( S-1 (M (>» (M -M "M 'M ri 'M ■>! r: r: JQ c: 1. . 2. . 3 . . 4. . 5.. 6.. 7. . 8. . 9.. 10.. 11. . 12. . 1 ;) . . 14. . 15.. 10. . 17. . 15. . 19. . 20. . 21. . 22. . 23. . 24. . 25 . . 2(i.. 27.. 28.. 29.. 30.. 31.. 32.. 33.. 34.. 35.. 36.. 37.. 38. . 39.. 40.. 41.. 42.. 43.. 44. . 45. . 46. . 47. . 1 1 1 mm\\\ I 1 1 1 II 1 1 1 ! 1 1 1 1 1 1 1 1 1 ) 1 - >■ << -f ^^ -^ ..^fo^ top J. -4 01 iti -h 4 4 4 + 4 4 4- 4- + 4 4 1- t 4 — - ^ ^ ^ -f -t ■r ^ [i^ + .- < +, ^ 1 ^ t g=i ov^ °>- K .•50- CSf tn 4 ^-4 Vb^I- i- •+ — — -H H? 4 +/ 4 4 ^ -1- "^L^- -w^ % 1% _i- -i-^ *r 1 m^ 4 ^! — — -I- ^- ■y \ -V, ^ 4> ^ 7^ ^ 4 :i- -^ <: -4- \'^ -4 4 > ^^ m^ -^ — — 4^ ^ -1- 4 ^ > 4 -t- -> ^ ic ^ ''4 jfe fr ^s^^\ - i- + ^ 4 ^ + ^> -^ ^ -^ ^ 4- >- ^ 4^-1 >. 1^ ^ 4 ir 44 +n + -1- w q^ -r / + ^ >^ ;^ 4- -^ s^ 4- -+ ^ri \^ 4- ;^ 4 4 4 +1+ -t-i— + =r 2! 2 ±. ^ ^ ^ ^ -4- ^ 4 ^-^ ~.^ -4 4- 4 4 "'I'ote^H'^ _ -1- 4- 7 -+ -y- 4- 4- -t-. ^ ^ 4 ^- ^ > 4 ^"4 4 4 -t 4- r-^ >i^^|-r- — -t- +. di ^ -t- -t^ -H 4 1? 4 ^ 4=- 4 -4- -t -4- 4- 4- ;^ 4 4~ +-j4:]4-'.— + 4 -t- -h 4- 4 -^ '+ -h -t-^ ^ ■4- -1- -y ^-^. 4 /^ -h H- i- -t -4 ;^ 4 4 -4- 4_ J. j-+-j— !— — -1- + + / -1- -t> -^ -t- -1- T*' -V ^ ^^ -1- 4- 4-4 v-^ -1- H -f 4 + ,Tm-~i -H =r T T ■q^ ^ H + y -y 't A 4 -H -1- -(- l'^ - 4- -1- 4- 4 V^ ->' 4-|-*-! — — i- '-t- -+ + / -h r 4, /+ ^ 4- 4 -1- 4- ;^ 4--I - -t- 4- -1- >i ^ ir 4- 4H-I — — -p 4- + -1- + Y -h -(y ^+ A- / 4 4 4 4- / 4 -^ i - 4- 4- -;y ^-4 4 i 4- i-|4-'— — -^ =t a^ =t -U A- -1- ^ -^ y 4- 4 -^ ■^ .^ 4- H- 4-4 - -1- ^ 4- 4- -+ 4- 4- ,A)^^ T -1- 4 t 4 -t- ^/ /-f + 4- 4 / -+ -1- 4 + 4 -/ 4 4- -( -^ + J^4 ^- -t- -h -(- ±. i j^ i^ / -t- 4 4 ¥ ^ -t- -1- 4 ^.-^ -4 4- 4 -+ > p _!_ E -t- + -,- -+ 4 4: y -^ -t- 4 4 V 4 4- 4- -\ 4-H \ 4 4 4 4 ^ '+ + -t- — 4- -t- 4 -H + ^ 4 -r + A\ + + 4 -i- A 4 H \ 4 4 4 ;^ -t- + 4-1 — — H- -V- -^■ + -1- ^ / 4 -f 4 / -It + 4 4 f 4- + - ^ 4- 4 7f -t 4 -+ -^ 4!-r — + 4 + + i S /-t -f -1- ^i < 4- 4 + / 4 4 ■1- - 1- + / -t 4 4- i- -t- — — 4- -+■ -1- -r ^ ^ -t- 4 -t / -r + 4 +/ '+ 4- 4- 4 ^r 4- -t 4- +- 4 4- T'-r — + ^ -h + j- -f- -1- i- ^y '-+ + - + ^ + + 4- 4- - »-/+ 4 -\ 4 4 +lt 4 -j- — m + -\ ,-t r + + -+ + J -1- i- +- f 4 4 t 4- +-/ f ^ '4- -^ -» + + 4- 4 4 ~ -f- -1- 4 -t- + + -1- -Hj, '+ 4 4 4- -t- 4 4 + 4- /- f i -t- 4 -t- 4 + + tl z: ; + -t- 4 -l- 4 -h -\- jf -(- 1- -^ i- -f 4 - — HT T T -1- i- ^ 4 i 4^ 4 4- T 4 T " s"o Span of Girder ix Feet. .6 1 * — + + -t- -t- + 4 4 + ^ -^ 4- i- 4 -+ — +- 4 -^ 4 -+ -1- + 4 -t- -t + 4- + 4 'i.£g 'M CO -4< '^ ^ "^ X' >— 1 "M ?tcceCJ0cccccC'CC-*-l-'r ■ — + -r -I- + -r + + 4 -r -1- 4- 4- + 4 — + -1- r + + + + + + 4 4 4 4 / . + + Jl 4 4 -h f + 4 -r \ ¥ -1- /+- -4- i- -t- -t- -1- + -1- + 4 4- 4 -1- -r 4- 10.. 11.. 12.. 13.. 14.. 15.. 16.. 17.. 18.. 19.. 20. . - 1^+ + + 4 4 1 + 4 i^s — -t- + -t- t -f i- + 4 4 4 4 -4- - •f 4 4 4 -h 4 ir ■il p^ 1-, ' — -h -t- 4- -1- ■f -t- + + -^ 4- 4- -y t 4 - 4- + t 4 4 M^ 4- 4 -t- — + 4 + + 4 -t- -+ + 4 t + yf 4 + - + 4 t *- < - -t- 4- T 1- — i- -v -t- 4 -t- 1- t -1- + -1- 4 4- 4 -1- — 1- >- < -1- -rh -t- -1- ;tv ^ — -h ■1- 4 -1- -h + -t -+ 4 -1 -j 4- ± -1- -< 4 4 -t- 4- t y ir — -+- +- i- -f 4 -t 4 -1- 4- -4 [\_ 4 -f- ± -+ 4- 1- t- 4- -t- > -1- -f 4 — -t- -t- i- 4 + 4 -t- -f- + ^/ -^ 4 -+ -t - -1- -K 4 -1- > '■+ 4 4- 4 4 — -t- -t- 4 ■+ -+ -h 4 + -+ / 4 4- 4- 4 - --4- 4- > ^ 4 4- 4 4 4 -1- ^ 4 + -+ J_ 4- -f + + 4- /; 4 4 + 4- :;^ ^ 4- -h 4- 4- -t- -t- -t- 4 — — 1 -1- 4- + 4 - 4- 4- -1- For Ste^l Bpams, add one-quarter to safe uniform load on Iron Beams ; but length of span in feet inu 4 'fe\ ti 1_ i- ,1 yU^ Fig. 137. 6 i'6 8 Fig. 138. longitudinal shearing along this line. For this reason the bolts are placed, alternating, above and below the line, forming two lines of bolts, as shown in Fig. (137), The end bolts are doubled as shown; the horizontal distance, a-l, between two bolts should be about equal to the depth of the beam. If we place the bolts in /our exam- 210 SAKK lU'ILDING. pie, say 3" above and o" below the neutral axis, we can readily cal- culate the size required. Take a cross-section of the beam (Fig. 138) showing one of the upper bolts. Now the fibre strains along the upper edge of the girder, we know are ( -; ) or 1200 pounds per sqiuire inch, for the wood, and we just found the balance of the load coming on the iron would strain this on the extreme upper edge = 1125G pounds 2"»er square inch. As the centre line of the bolt is only 3" from the neutral axis or f of the distance from neutral axis to the extreme upper fibres, the strains on the fdjres along this line will be, of course, on the wood | of 1200, or 450 pounds per square inch : and on the iron | of 112.jG=4221 ])ounds per square inch. Now, supposing the bolt to be 1" in diameter. It then presses on each side against a surface of wood = 1" X 6" or = six stpiare inches. The fibre strain being 450 pounds jier s(piare inch, the total pressure on the bolt from the wood, each side, is : G.450 = 2700 jjounds. On the iron we have a surface of 1" X |" ^ | square inches. And as the fibre strain at the bolt is 4221, the total strain on the bolt from the iron is=|. 4221 =: 2G38 pounds. Or, our bolt virtually becomes a beam of wrouglit-iron, circular and of 1" diameter in cross-section, supported at the points A and B, which are C|" apart, and loaded on its centre C with a weight of 2G38 pounds. Therefore we have, at centre, bending-moment (Formula 22) 2G38. G| in = T— 2 = 43G9. 4 From Table I, Section Xo. 7, we know that for a circular section, the moment of resistance is. ■=n-'=;fa)=«. 098 Xow for solid circular bolts, and which are acted on really along their whole length it is customary to take { -i= ) the safe modulus of rupture rather higher than for beams. Where the bolts or pins have heads and nuts at their ends firmly holding together the parts acting across them they are taken at 18000 pounds for steel and at 15000 pounds for iron. We have therefore transposing (Formula IS) for the required moment of resistance 43G9 T . , . , . . , 0,291. Inserting this value for r in the 15000' felZK OK ISOLTS, 2Ji abovi; wu li;ive for tlie radius of bolt, - .-. r''=0,2;U and ^= Vir ^''2''i = \'/o,3 7o-t = 0,718" Or the diameter of liolt should be 1,43G" or say 1 7-16". But 1" will be quite ample, as we must remember that the strains calculated will come only on the one bolt at the centre of span of beam ; and that, as the beam remains of same cross section its whole length the extreme fibre strains decrease rapidly towards the supj)orts, and therefore also the strains on the bolts. The end bolts are doubled however, to resist the starting there of a tendency to longitudinal shearing. We might further calculate the danger of the bolt crush- ing the iron plate at its bearing against it; or crushing the wood each side ; or the danger of the iron bolt being sheared off by the iron plate between the wooden beams ; or the danger of the iron bolt shearing off the wood in front of it, that is tearing its way out through the wood ; but the strains are so small, that we can readilv see that none of these dangers exist. Keyed Girders. Another method of adding to the sum of the separate strengths of the girders is to j)lace one under the other mak- ing a straight joint between the two parts and to drive in hard wood keys, as shown in Fig. l-iO. The keys can either be made a trifle thicker than the holes and the beams then firmly bolted together so as to take hold of keys securely; or, keys can be shaped in two wedged-shaped pieces to ^ each key, and driven into the hole from op- posite sides, after the beams are fi r m 1 y bolted. In the latter case, care must be taken that the joint Pig, 139, between the opposite wedges is slanting or diagonal, and 7iot liorizontal or else, of course, the keys would be useless. Either method allows, for tightening up, after shrinkage has taken place. Iron bands are frequently used in place of bolts but they are more clumsy, less liable to all fit exactly, and besides do not allow for tightening up so easily as with bolts. Where beams are very wide, however, the I if: r 212 SAKK ISril.DINC.. bands are -very advantageous. Tredgold says the keys should be twice as wiik', as high; and tliat tlie sum of all their heights should t-ijual one and a third times the dejith of girder. They can be easily calculated, however. As the main strain on them is a horizontal shearin 850000 From Table T, Section No. 1 7, we have for' the weakest section of the girder, which would be tliroiigh a key, and as shown in Fig. 141, f = ^-0/'-'A^) 10 _ {^ . (--243 — 63) = 11340, therefore — 12 ^ ^ I4i_ inserting these values in the transposed Formula(40) 0,9.48.8.50000.11340 '"= 3603 = 8925 Or the safe centre load, not to crack plastering, would 1)0, say 9000 pounds. End Keys. Now let us try the keys. We first take the great- est horizontal shearing, which will be at the end keys. The vertical shearing at these keys will be equal to the reaction (see Table VII, or Formula 11.) As the load is central, each reaction will, of course, be one-half the load, or 4500 pounds, therefore the vertical shearing strain at end key, will be (a little less than) x = 4500 Now from Formula (13) we know that the horizontal shearing strain at the same point is : 3 X T ■ Ti For the area we take the full area of cross-section or = 10.24 = 240, therefore horizontal shearing strain : ■ ' = 28,125 pounds per square inch The 2. 240 amount of this strain that will act on each key is, of course, equal to the area at the neutral axis from centre to centre of key, or 40. 10 = 400 stjuare inches multiplied by the strain per square inch, or 400. 28,125 = 11250 pounds. To resist this we have a key 12" X 10" = 120 scpiare inches, being sheared across the grain. From Table IV we know that the safe shearing stress of Georgia pine across the grain or fibres, is : (X\^blO pounds so that the key could safely stand an amount of horizontal shearing ^570. 120 r= GS400 pounds SriEAlUNt; OS KKYS. 215 or more than six times the actual strain. Had we, liowcvcr, placed the grain of the key horizontal, the shearing would he with the grain or along the fibres; the safe shearing stress this way for Georgia pine (Table IV) is only BO pounds per scpiare inch, so that the key would only have resisted = 50.120 == (1000 pounds, or it would have been in serious danger of splitting in two. Central Keys. Now take the Key A B immediately to the right of the weight. The bending-moment at A will be (Taljle VII) m,^= 4500.166 = 747000 and at B m„= 4500.154 = 69.3000 Now at A and !>, the girder being uncut, the moment of resistance will be (Table I, Section No. 2) 10. 242 r= —. = 960 Dividing the bending moment by the moment of resistance (Formula 18 transposed) gives the extreme fibre strains, . 74 7000 at A =:= - Q .„ =r 778 pounds. 69.3000 and at B = ^g^ — = 722 pounds. Now the centres (x and y, see Fig. 139) of each side of key will be 1|" from neutral axis, the extreme fibres being, of course, 12" distant from neutral axis, therefore average strain on side of key at A. (Or 1^ X, Fig. 139) := -5 . 778 = 97 pounds, and at B (or >j, Fig. 139) = lf . 722 — 90 jiounds. The extreme compression, will, of course, be on the lower edge of key, at A and will be = 2.97 = 194 pounds per square inch. From Table IV we find that Georgia pine will safely stand a pressure of 200 pounds per scpiare inch, across the fibres, so that we are just a little inside of the safety mark. We now have to consider our key as a cantilever with cross-section 10" wide and 12" deep, projecting 3" beyond the support and loaded uniformly with a weight e(|ual to 90 pounds per square inch, or u = 90.3.10 = 2700 pounds. Now the bending moment at support is, (Formula 25.) 2700.3 m = — ^ — =r 4050 pounds. 216 SAFE BUn.DING. Till." inoiiR'nt of resistant'o (Table T, Section 2) is 10.122 6 = 240 Tliercfore (Formula 18) the extreme fibre strains on key 111 4050 = y = 240 = ^ ' po""'^^- Or not enough to be even considered seriously. Another method of conibining and strenijthening Notched , . ,1 1 1 girders, wooden girders, is to cut them with saw-siiaped notches, as shown in (Fig. 142) and fit the teeth closely together, firmly bolting the two parts together, so as to force them to act [J Fig. 142. together as one girder. Sometimes the top surface slants towards each end, and iron bands are driven on towards the centre, till they are tight. But bolts are more reliable, and not likely to slip ; where the gh-der is broad, they should be doubled, that is, placed in pairs across the width of girder. The distance between bolts should not exceed twice the depth of girder. Great care must be taken to get the right side up.i Many text-books even being careless in this matter". It must be remembered that the upper fibres are in com- pression, crowding towards the centre, while the lower ones, in ten- sion, are pressing away. The girder must therefore be placed, as shown, so that the two sets of fibres will meet at the short joints and oppose each other. The girder is easily calculated similarly to the former example. The crushing on C D or A B can be found, and also the stress on their extreme edges ; this must not exceed the safe stress of the material for compression along the fibres. Then D B or C A must have area sufficient to resist the horizontal shearing strain. ^^^^^ In all these girders the most careful fitting of Bearings, joints is necessary; then, too, the ends must have sufficient bearing not to crush under the load. Thus, take the former example, the reaction was 4500 pounds ; the safe resistance of spruce to crushing across the fibres is (Table IV) = 75 pounds. We need therefore an area — ^^ = 60 sejuare inches, and as the girder is 10" broad it should bear on each support f f = 6 1 The reasons for placing the notched girder in the position shown in Figure 142 have been fully stated by me in a letter published May 10, 1890, in the American Architect, Vol. XXYIII, No. 750. CONTIXL'OUS OIIJDERS. m inches. The end of girder should Ijc deep enough to resist vertieal shearing. In our case it is trilling, and we need not consider it. In all of these examples wo have omitted the weight of the girder, to avoid complication. This should really he taken into account, iu such a long girder, and treated as an additional but uniform load. Continuous Wlien girders run over three or more supports in Girders, one piece, that is, are not cut apart or jointed over the supports, the existing strains and reactions of ordinary girders, are very much altered. These are known as "continuous girders." If Tve have (Fig. 143) three supports, and run a continuous girder over them in one piece and load the girder on each side it will act as shown iu Fig. 143 ; if the girder is cut it will act as shown in Fig. 144. Very little thought will show that the fibres at A not being able to separate in the first case, though they Avant to, nnist cause considerable tension in the upper fibres at A. This tension, of course, takes up or counterbalances part of the compression existing there, and the result is that the first or continuous girder (Fig. 143) is considerably stronger, that is, it is less strained and considerably stiff er, than the sectional or jointed girder (Fig. 144). Again we Fig. 144. can readily see that the great tension and conflict of the opposite stresses at A would tend to cause more pressure on the central post in Fi"". 143, than on the central post in Fig. 144, and this, in fact, is the case. 218 SAFE BUILDING. --» CO ~" 1 ^ r-N S i. S "" K — 1 c s" c o ^ ./ ^ + 5-2 S^T 5J C 1 'S '-^ ^ §q o .2 <« Ql 1 y 1 .2 g <5J o £L o "M -M •1 s 'n ^ 1 . ^ '- ^ li S^ II q CO (^- CO »S CO C^ CO .,-^ ^^ s" /^~\ S o i2 s^ + i. a § c O c g c S --J v-x to a -9 -^ K a "^ o " Ico o o S o o a ^ II II II ^ II 8 CO s & O t3 g H O o 1 1-5 03 1 o 3 reaction. reaction, to.. — 3 . n\ 2 •§ i .2 + 2 + 2 1— ( > X a 3 o s -^ 1 l::i";5i^ ^ II ^ II ^ II II 1 II a. II II i II n 11 pq C3 ~ F^ a" + ^ + -:: + gi + 4 + 1 f ^ s" "P ^ ^- 3 '^ + V e •- L. '^ ■- w ~ ,^ s n o Ss. w • -^ a •^ •- fi ~ y ■~:-s«. w f 0? 1 1 "i 1 so o ■ ■a ^ 1 O •M CO H 1 ts 1^ 1 ^ 60 II t^ CO J ^ CO 1 q 6o 1 K ^-> *^ ^- !^ ^-N CO g o g- s" o l_ S + + s- s" ^ s" e a -^ + « + .^fw ^ S o S r* o-\ 11 O CO -^IS 3 .=>;-> H 11 ^ 1! II S ^ /""^ , /"^^ !- /"^^ '~^ ^^-.^ ^ g= ^ £ o o s S . L ''—• + ^ + + .: + reaction. . 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The architect can, if he wishes, neglect to allow for the additional strength and stiffness of continu- ous girders, as both are on the safe side. But he must never ovei-- Variation of look the fact tJiat the central reactions are much Reactions! {greater, or in other words, that the end supports carry less, and the central supports carry more, than Avhen the girders are cut. Bending moments can l)e figured, at any desired point along a continuous girder, as usual, subtracting from the sum of the reactions on one side multiplied by their respective distances from the point, the sum of all the weights on the same side, multiplied by their re- spective distances from the point. Sometimes the result will he negative, which means a reversal of the usual stresses and strains. Otherwise the rules and formula? hold good, the same as for other girders or beams. Table XYII gives all necessary information at a glance. Strength is frequently added to a girder or beam by trussing it, as shown in Table XVIII, pages 220 and 221. One or two struts are placed against the lower edge of a beam and a rod passed over them and secured to each end of the beam ; by stretching this rod the beam becomes the compression chord of a truss and also a Trussed continuous girder running over one or tAvo supports. Beams. There must therefore be enough material in the beam to stand the compression, and in addition to this enough to stand the transverse strains on the continuous girder. If the loads are concen- trated immediately over the braces, there will be no transverse strain wdiatever, but the braces will be compressed the full amount of the resj)ective loads on each. In the case of uniform loads, transverse strains cannot be avoided, of course, but where loads are concen- Struts placed trated the struts should always be placed immed- under load. iately under them. Even where loads are placed very unevenly, it is better to have the panels of the truss irregular, thus avoiding cross or transverse strains. This same rule holds good in designing trusses of any kind. Necessary Table XVIII shows very clearly the amount and Conditions, kind of strains in each part of trussed beams. Where there are two struts and they are of any length care must be taken by diagonal braces or otherwise, to keep the lower ends of braces from tipping towards each other. Theoretically they cannot tip, but TKUSSKl) ItKAMS. 223 practically, sometimes, llicy do. Can- must be taken that the heam is braced sideways, or else it must lie figured fur its safety against lateral flexure (Formula 5.) 'J'lien it must have material enough not to shear off at supports, nor to crush its under side where lying on support. The ends of rods must have suflicient bearing not to crush the wood. Iron shoes are sometimes used, l)ut if very large are apt to rot the wood. In that case it is well to have a few small holes in the shoes, to allow ventilation to end of timber. If iron sti-aps and bolts are used at the end, care must be taken that the strap does not tear apart at bolt holes ; that it does not crush itself against bolts ; that it does not shear off the bolts, and that it does not crush in the end of timber. Care must also be taken to have enough bolts, so that they do not crush the wood before them, and to keep the bolts frora shearino- out, that is tearing out the wood before them. In all truss- es and trussed works the ioints must be carefully liTiporx^ncQ •' of Joints, designed to cover all these points. Many architects give tremendous sizes for timbers and rods in trusses, thus adding unnecessary weight, but when it comes to the joint, they overlook it, and then are surprised when the truss gives out. The next time they add more timi)er and more iron, till they learn the lesson. It must be remembered that the strength of a truss is only e(|ual to the streno-th of its weakest part, be that part a member or only a part of a joint. This subject will be fully dealt w^ith in the cha[)ter on Trusses. The deeper the truss is made, that is, the further Depth '■ Desirable, we separate the top and bottom chords, the stronger will it be ; besides additional depfh adds very much to the stiffness of a truss. _ „ ^. All trussed beams, and all trusses should be "cam- Deflection of Girders bered up," that is, built up above their natural lines and Beams, sufliciently to allow for settling back into their cor- rect lines, when loaded. The amount of the camber should equal the calculated deflection. For all beams, girders, etc., of uniform cross- section throughout, the deflection can be calculated from FormuliB (37) to (42) according to the manner of loading. For wrought-iron beams and plate-girders of uniform cross-section throughout, the de- flection can be calculated from the same formulae ; where, however, the load is uniform and it is desired to simplify the calculation, the deflection can be cpute closely ealculated from the following Fornuda: Uniform Cross- ^ . -^ ' Ci^^ section and Load. ,(/ 224 SAFK BUILDING. Where 3=^ the greatest tlelleeticin ;it centre, in inches, of a ■wrou"-ht-iron beam or ])Uite girder of uniform cross-section through- out, and carrying its total safe uniform load, calculated for rupture only. Where L = the length of span, in feet. "\Vhei-e r/ = the total depth of beam or girder in inches. If beam or plate girder is of steel, use 64^ instead of 75. If the load is not uniform, change the result, as jarovided in cases (1) to (8), Table YII. For a centi-e load we should use 93 1 in jalace of 75 or Uniform Cross- 5; L'^ ^oa\ section, Centre " — ; ,...; ,/ ^^^^ Load. ^'"'"f " Where values are the same, as for Formula (79) except that beam or girder carries its total safe centre load, calculated for rupture only. If beam or girder is of steel use 80§ instead of 93|. Therefore not to crack plastering and yet to carry their full safe loads, wrought-iron beams or plate girders should never exceed in length (measured in feet) twice and a quarter times the depth (measured in inches), if the load is uniform, or Safe length, uniform Cross- 2^.d = L (81) section and " i' Load. Where L = the ultimate length of span (not to crack plastering), in feet, of a wrought-iron beam or plate girder, of uniform cross- section throughout and uniformly loaded with its total safe load. Where rf^the total depth of beam or girder in inches. If beam or girder is of steel, use 2 instead of 2|. If the load is central the length in feet should not exceed 2 1 times the depth in inches, or Safe length, uniform Cross- 9 4 ^ -— /^ (82) section, Centre - 5" • ^ -' Load. Where X = the ultimate length of span in feet (not to crack plastering), of a wrought-iron beam or plate girder, of uniform cross- section throughout, and loaded at its centre with its total safe load. Where d = the total depth of beam or girder in inches. If beam or girder is of steel use 2-| instead of 2|. Deepest beam One thing should always be remembered, when eccfnomical. "^ing iron beams, and that is, that the deepest beam is always not only the stiffest, but the most economical. For instance, if we find it necessarv to use a loi" beam — 105 pounds per yard, DEKLFXTIOX OF TKUSSES. 'i'lo it will be cheaper to use instead the 12" beam — 96 pounds per yard. The latter beam not only weighs 9 pounds per yard less, but it will carry more, and deflect less, owing to its extra two inches of depth. This same rule holds good for nearly all sections. ^ „ ^. To obtain the deflections of trussed beams or Deflection of Trusses, girders by the rules already given would be very complicated. For these cases, however, Box gives an approximate rule, which answers every purpose. lie calculates the amount of extension in the tension (usually the lower) chord, and the amount of contraction in the compression (usually the upper) chord, due to the strains in each, and from these, obtains the de- flections. Of course the average strain in each chord must be taken and not the greatest strain at any one point in either. In a truss, where each part is proportioned in size to resist exactly the com- pressive or tensional strain on the part, every part will, of course, bo strained alike ; the strain in the comi)rcssive member being = ( ,, j per square inch, throughout the whole length, and in the tension member ^= I— A per square inch, throughout the whole length. The same holds good practically for plate girders, where the top and bottom flanges are diminished towards the ends, in proportion to the bending moment. But where, as in wrought-iron beams (and in many trusses), the flanges are made, for the sake of convenience, of uniform cross-section throughout their entire length, the " average " strain will, of course, be much less, and consequently the beam or girder stiffer. -.. • If we construct the graphical representation of the Average strain o i •■ in Chords, bending moments at each point of beam (as will be explained in the next Chapter) and divide the area of this figure in inch-pounds by the length of span in inches, we will obtain the average strain in either flange, provided the flange is of uniform cross-section throughout, or Uniform Cross- f = iL (83) section. I Where v = the average strain, in pounds, on top or bottom flange or chord, where beam or girder is of uniform cross-section through- out. Where I = the length of span, in inches. A\ here a = the area in pounds-inch of the graphical figure giving the bending mcment at all points of beam. 226 SAFE BUILDING. To obtain the dimensions of this figure measure its base line (or horizontal measurement) in inches, and its lieight (or vertical meas urement) in pounds, assuming the greatest vertical measurement as r= ( -L ) or = ( -^ ), in i oiinds, according to which flange we are ex- amining. Thus, in the case of a uniform load, this figure -would be a parabola, with a base of length equal to the span measured in inches, and a height equal to the greatest fibre strains in pounds ; the average strain therefore in the compression member of a beam, girder or truss, of uniform cross-section throughout would be, — (remembering that the area of a parabola is ec[ual to two-thirds of the product of its height into its base). Uniform load '' / ( "T" ) and Cross- i!^l_AZ__or section! I (84) |-(f) J "Where v = the average strain, in pounds, in compresf^ion flange or chord of a beam, gh-der or truss of uniform cross-section through- out and carrying its total safe uniform load. "Where ( -^ ) ^ the safe resistance to compression per square inch of the material. It is supposed, of course, that at the point of greatest bending moment — or where the greatest compression strain exists — that the part is designed to resist or exert a stress = ( — j per square inch. If the greatest compression stress is less, insert its value in place oi (^\ Of course, it must never be greater than (-^V Similarly we should have Uniform Load and Cross- section. -f (7) "Where v = the average strain, in pounds, in tension flange or chord of a beam, girder or truss of uniform cross-section throughout, and carrying its total safe uniform load. Where ( —^ ) = the safe resistance to tension, per square inch, of the material. It being understood that at the point of greatest bending moment — or where the greatest tension strain exists — that the part is de- Centre Load Uniform Cross-Section. (86) (8 7) DKKI.KCTION, KOUML'LJ:. 227 signed to resist or exert a stress = ( y) per s.iiiare indi. If tliis greatest tensional stress is less than {-L^ insert its value in its j.jaee in Formula (85). Of eonrsL-, it must never be greater than ( i Y For a beam, girder or truss with a load eoncentrated at the cen- tre, but with flanges or chords of uniform cross-section throughout, the average strain would be just one-half that at the centre ; for, the bending-moment graphical-figure will be a triangle, and inserting the values in Formula (83) would give for the compression member ■" -■(7) and for the tension member : ■■='■(7) The meaning of letters being the same as in Formula? (84) and (85), but the total safe load being concentrated at the centre instead of uniformly distributed. To obtain the amount of contraction or expansion due to this average strain, use the following Formula : Expansion or v. I Contraction X^ — (^gg) from Strain. ^ Where ?; r= the avercuje strain, in jiounds ^ler square inch, in eith- er chord or flange. Where I = the length of span, in inches. Where e = the modulus of elasticity of the material, in jjounds- inch. A\ here x^the total amount of extension or contraction, in inches, of the chord or flange. Now let us apply the above rules to beams, plate girders, and trussed beams. Taking the case of a beam or plate girder or truss with parallel flanges or chords. Figure 145 shows the same, after the deflection has taken place. We can now assume approximate- H ly, that C A is equal to one- half the difference between the contraction of G C and '^'^" ^'^^' the elongation of H D, or, what amounts to the same thing, that CA 228 SAKK HIILDIXG. is c:(jnal to oiu'-lialf the sum of tho contraction of the one and the elongation of tlie other. Further, we can assume tliat apjiroximately, A B.=^ d or the depth of beam, and C D = -^ or one half ihe span. The curve C E C will approximate a parabola, so that if we draw D F a tanijcnt C F to the same at C, we know that D E = E F^ — ^ — or J) i''=2. D E. But as DE represents the deflection ( 8 ) of the beam, we have D F= 2. S Xow as C F is normal to C B, and C D normal to .4 B, we know- that angles D C F = A B C; further, as both triangles are right angle triangles, we know that they are similar, therefore : D F: C A.:D C: A B, or 2. 8 : C A .-.J. : d or S ^'2 CA.l " — 2.d — 4.d. If now we assume the sum of the extension and contraction of the two flanges or chords to be = x. We have C A ^ — or ■2 Deflection of Par- o x I allel Flanges 0=^ (89) orChords, any 8. a Cross-section. "Where 8 = the deflection, in inches, of a beam, plate girder or truss, with parallel flanges or chords. Where x = the sum of the amount of extension in tension chord, plus the amount of contraction in compression chord. Where I = the length of span, in inches. "\Miere d = the total depth of beam, girder or truss in inches. Take the case of a wrought-iron plate girder or beam of uniform cross-section throughout carrying its full uniform load, we should have the strain at the centre on the extreme fibres = 12000 pounds per square inch. Now the average strain on both upper and lower flanges would be. Formula (84) and (85). v = §. 1 2000 = 8000 pounds per s(juarc inch. Therefore amount of contraction in upper flange DKKLKCTION CONTINIED. 229 Formula (SS), (and reincmberin;j; that, from Table IV, c — 27000000) __8000. ^ _ I ^UOUOOO 3375 The elongation of the bottom (lange Avould be an ecjual amount, therefore the sum of the two '' I .r, = 2.x= " . 3375 _ I 1G87,5 Inserting these values in Fornuila (S'J) we have the deflection 8- I' - I' 8.1G87, 5.^ 135U0.(i and inserting for /-=:144. Z^, we have ^_ 144. Z'-' 1350U. (/. ~ 93f . d. Had we assumed that the area of flanges or chords diminished to- wards the suiaports in proportion to the bending moment or actual stresses required, the average strain would, of course, be 12000 pounds per square inch throughout the entire length, no matter how the load might be applied. Inserting this value in Formula (88) we should have had, for the amount of contraction of top flange ,_ 12')00. Z _ / 27 )U0O00 2250 The same for the extension of bottom chord, or ^' ■ ""2250 1125 Inserting this in Formula (89) we have for the deflection : 8- I' - I' 8.1125. (/ yooo. (/ Inscrtii.g 144 L'^r^l'^ wq have S 144.Z2 OOUO. (/ Parallel flanges orChords, Dl- 5 j^2 (90) minished Cross- '' — section, any loads. "Where 8 = the greatest deflection, in iiuhes, of a wrought-iron plate girder, or wrought-iron truss, with i)arallel flanges or chords, and where the areas of flanges or chords are gradually diminished 230 SAFE HUILDIXG. towards supports, and no matter how the load is applit'd ; in no part however must the stresses, per scpiare inch exceed respectively either Where L = the length of span, in feet. Where f/=r the total depth (height) in inches, from top of top ■flange or chord to bottom of bottom flange or chord. If girder or truss is of steel, use 53§ instead of 62^. From Formula (90) and Formula (28) we get the rule that (no mat- ter how the load is applied) if we want to carry the full safe load and not have deflection enough to crack plastering the length in feet must not exceed 1| times the total depth in inches. For: ^- ''''= G^°^ Z = 621. 0,03. (Z =r 1,875. d or say Safe Length, Di- minished Cross- section, any L = lLd (91) Load, Parallel ^ ^ ^ Flanges or Chords. Where Z = the length, in feet, of a wrought-iron plate girder or wrought-iron truss, with parallel flanges or chords and with area of flanges or chords diminishing gradually towards supports and no matter how the load is applied; in no part however must the stresses, per square inch, exceed respectively either (^—rj or ^ —^j. Where f/ = the total depth (height), in inches, from top of top flange or chord to bottom of bottom flange or chord. If girder or truss is of steel, use 1 f instead of 1|. We see therefore that a beam of diminishing cross-section through- out is only about | as stiff, as one with uniform cross-section, as its amount of deflection will be one-half more than that of the latter. Both deflections are approximate only, however, as we see by comparing the amount for the uniform cross- section to that obtained from Formula (79). The deflection for varvina; cross-sections how- SAFK LKNGTII. 231 ever can be assumed as nearly enough correct, as these are never diniinislied so nuich practically as we have assumed in theory. Xow taking the case of a trussed beam. Deflection ^^ Figure 140, let A B be one half of a trussed Trussed Beam, b^am, let B C be the strut and A C the tie. We •will consider the load concentrated at B. Xow tlie first effect is to shorten A B by compression, let us say to D B. Then, of course, A D will represent one half of the contraction in the whole beam A G. Now the end of rod .1 moving to J) will, of course, let the point C down to i?, if we make D E = A C. But there will be an elongation m D E besides, due to the tension in it, Avhich will let it down still further, say to F, if D F= A C -\- elongation in A C, of course the point B will move down too, but we can overlook this to avoid complication. We now have C F repre- senting the amount of the deflection. To this should be added the amount of contraction of B C due to the compression in it. We can readily find C F. We know that BF=y^)F^"-^^' Now D F we know is = .4 C plus the elongation oi A C due to the tension in it, which we can find from Formula (88). From same for- mula we find the amount of contraction m A G oi which ^ D is one- half, subtracting this from A B or —leaves, of course, D B. Now having found B F we substract from it B C, the length of which is known, and the balance is of course the deflection C F ; to this we add the contraction of B C and obtain the total deflection of the whole trussed beam. If the load had been a uniform load, instead of a concentrated one over the strut, there would be a deflection in that part oi A G which would be acting as a continuous girder. But this deflection would take place between B and G and between B and A and would not af- fect the deflection of the -whole trussed beam. An example will make much of the foregoing more clear. Example. Trussed Beam. A trussed Georgia Pine beam is 16" deep and of 2i feet dear span; it bears IC" on each support and is trussed as 232 SAFE BUILDING. shown in Figure 147. The ^ beam carries a uniformly distributed load of 40800 'is pounds on the whole span r including weight of beam and trussing. Of what size should the parts be ? Fig. 147. We draw the longitudinal neutral axes of each part, namely A B, B C and A C. The latter is so drawn that the neutral axis of the reaction, which is of course half way between end of girder and E (or 8" from E) will also jiass through A. In designing trusses this should always be borne in mind, that so far as possible all the neutral axes at each joint should go through the same point. ^ . The beam A F virtually becomes a continuous Cross-strains e r in Beams. (birder, of two equal spans of 12 feet or 144 ' each, uniformly loaded with 20400 pounds each, and supported at three points A, B and F. From table XVII we know that the greatest bending moment is at B and _tU_20400^_3,_200 pounds-inch. 8 8 ^ The modulus of rupture for Georgia pine (Table IV) is ^ ^ -200, therefore moment of resistance (r) from Formula (18) (|) = .K or and Table I, section No. 2, __?;.(/ 2 _ 367200 ' ~~ 13 1 200 b.d -=1S3G Xow we know that fZ= IG, or d'^: 1836 256, therefore b:=—^ = 7,2 or sav we need abeam 71" x 16" for the 256 ' * KXAMPI.K, TUUSSKD liEAM. 233 transverse strain. We must add to this however for the additional conipression due to the trussing. Compression "^''^ amount of the load carried by strut C B, see in Strut. Ta])le XVII, is = |. u from each side, or = 25500 on the strut B C, of Avhich = 12750 from each side. If now we make at any scale a vertical line b c=zhalf the load Compression carried at j^oint B or = 12750 in our case, and in Beam. draw b a horizontally and a c parallel to A C, we find the strain in B A by measuring b « = (32300 pounds) or in A C by measuring a c=:(34G38 pounds) both measured at same scale as h c. We find, further, in passing around the triangle c b a c — (c b being the direction of the reaction at A'), that b a is pushing to- wards A, therefore compression ; and that a c is pulling away from .1, therefore tension. Using the usual signs of -(-for compression, and — for tension, we have then : J. -B = -[- 32300 pounds, A C= — 34G38 pounds. BC = -{- 25500 pounds. Had we used Table XVIII we should have had the same result for : Compression in A ^= ?i^|^, ^^ = -4- 32300 pounds and 2 Z> C rn ■ • i n 25500 A C „.„.,„ lension in A C^ — - — -jT-r,^= — ^34038 pounds. 2 x> C Now the safe resistance of Georgia jiine to compression along fibres (Table IV) is ( -7; j= 750 2)uunds. If A B were very long, or the beam very shallow or very thin, we should still further reduce {-^, j by using Formulae (3), or (5). But we can readily see that the beam will not bend much by vertical flexure due to compression, nor will it deflect laterally very much, so we can safely allow the maximum safe stress per square inch, or 750 pounds, that is, consider A B a, short column. The necessary area to resist the compression, Formula (2) is : 32300 = 0. 750 or 32300 a = — .— — ^43 sciuare inches. 7y0 ^ 234 SAFE BUILDING. As the beam is IG" dofp, this would mean an additional thickness 4 3 — o 1 i_ IB' 16 Adding tliis to the 7^" already found to be necessary, we have 7J._L 911 — 915 '4 T^ -16 — ^16 or the beam would need to be, say 10" x 16". Size of Strut. Xow the size of B C must be made sufficient not to crush in the soft underside of the beam at B. The bearing here would be across the fibres of the beam, and we find (Table lY) that the safe compressive stress of Georgia pine across the fibres ij / — ^ ) ^= 200 pounds. We need therefore an area 25500 ,„Q . , a = =128 inches. 200 As the beam is only 10" wide the strut B C will have to measure, 128 =z 12+ inches the other way, or we will say it could be 10" x 12". 10 ^ •" •' This strut itself might be made of softer wood than Georgia pine, say of spruce ; the average compression on it is 25500 „,„ , . , =212 pounds per s(iuare men. 10.12 ^ ^ ^ Xow spruce will stand a compression on end (Table IV) of ( —^ )r=r650 or, even if spruce is used, the actual strain would be less \J ^ than one-third of the safe stress. At the foot of the strut B C yve put an iron plate, to prevent the rod from crushing in the wood. The rod itself must bear on the plate at least Iron Shoe to Z — i— =rr:2,l square inches, or it would crush the Strut. 12OU0 '■ iron — (120U0 pounds being the safe resistance of wrought-iron to crushing). Size of Tie-rod. The safe tensional stress of wro :ght-iron being 1 2000 pounds per s) 34638 = a. 12000 or a= — := 2,886 square inches. 12000 ' ^ From a table of areas we find that we should re([uire a rod of 1 \l" diameter, or say a 2" rod. The area of a 2" rod being =3,14 square inches the actual ten- sional stress, per sipiare inch on the rod, will be only ?i^ = 11312 pounds per scniare inch. SIZE OF ROD. 235 Size Of Washer. We must now proportion the bearing of the wasli- er at "A" end of tic-rod. Tlie amount of tlic crusliing coming on washer will he whichever of the two strains at A, (viz. B A and A C) is the lesser, or i? ^ in our case, which is 32300 pounds. Wc must therefore have area enough to the washer not to crush the end of beam (or along its fibres), the safe resistance of which we already found to be: T —^ j = 750 pounds per squave inch; we need there- fore 32300 .„ . , ■ = 43 square inches. The washer therefore should be about Upset Screw- "^'^^ ^"^^ *-*^ *^^^ ^'°^ must have an "upset" screw- end, end ; that is, the threads are raised above the end of rod all around, so that the area at the bottom of sinkage, between two adjoining threads, is still equal to the full area of rod. If the end is not " upset " the whole rod will have to be made enough larger to allow for the cutting of the screw at the end, which would be a wilful extravagance. Tt is unnecessary to calculate the size of nuts, heads, threads, etc., as, if these are made the regulation sizes, they are more than amply Central Swivel. strong. It should be remarked here that in all trussed beams, if there is not a central swivel, for tightening the rod, that there should be a nut at each end of the rod ; and not a head at one end and a nut at the other. Otherwise in tightening the rod from one side only it is apt to tip the strut or crush it into the beam on side being tightened. We nmst still however calculate the verti- cal shearing across the beam at the supports, which we know equals the reaction, or 20400 pounds at each end. To resist this we have 10" X Id" = 160 square inches, less 3" x 16", cut out to allow rod end to pass, or say 112 square inches net, of Georgia pine, across the grain ; and as r -^ j = 570 pounds per square inch (see Table IV); the safe vertical shearing stress at each support would be (Formula 7) 112.570:= 63840 pounds or more than three times the Bearinsof actual strain. Then, too, Ave should see that the Beam, bearing of beam is not crushed. It bears on each re- action 16 inches, or has a bearing area= 10.10 = 160 scjuare inches. 236 SAFE BUILDING. ( -£j ) for Georgia pine, across the fibres, Table IV, is ('-£;. j = 200, therefore the beam will bear safely at each end 160.200= 32000 pounds or about one-half more than the reaction. There will be no horizontal shearing, of course, except in that part of beam under transverse strain, and this certainly cannot amount to much. The beam is therefore amply safe. _ ,, ^._ Now let us calculate the deflection. The modulus Deflection , of Beam, of elasticity for Georgia pine, Table IV is: ez= 1200000 pounds-inch. The average compression strain in A F was 750 pounds per square inch, therefore the amount of contraction (Formula 88) i x = -I^MOi = 0,19 inches. 1200000 ' Now A D (in Fig. 14G) will be one-half of this, or 0,095 inches. The amount of elongation in A C will be, remembering that we found the average stress to be only 11312 pounds per square inch, and that for wrought-iron e = 27000000 (Formula 88) 11312. 1G3 ^ Ai-oo x = = 0,0o82 27000000 ' The exact length of yl C (Fig. 147 should be 1G3,41 not 163"). Therefore D F (Fig. 140) will be DF= 163,41 + 0,0682 = 163,4782" , X)£ = 152 — 0,19 = 151", 81 Therefore (Fig. 146) 2 ^i^=\/ 163,4782 2 — 151,812 = 60", 655 Now B C (Fig. 147) would be = 60", deducting this from the above we should have a deflection = 0", 655. To this we must add the contraction of B C. The strut will be less than 60" long, say about 50''. The average compressive stress per square inch we found =: 212 pounds. The modulus of elasticity 1 111 reality the contractiou of A i** would be much less, as the part figured for transverse strain only would very materially help to resist the compression, one- half of it beiDg in tension. TABLKS XIX TO XXV. 237 for spruce, Table IV, is e = 850000, therefore contraction in strut (Formula 88) d^::^^ =0,0125 850000 Adding tliis to the above we should have the total deflection 3 = 0,G55-f 0,0125 = 0,GG75 This would be the amount we should have to " camber " up the learn, or say J". The safe deflection not to crack plastering, would be (Formula 28) S = L. 0,03 = 24.0,03 = 0,72 So that our trussed beam is amply stiff. Explanation of Table XIX gives all the necessary data in regard ?o^xxv.''"' to the use of Tables XX, XXI, XXII, XXIII, XXIV and XXV. These tables give all the necessary information in regard to all architectural sections which are rolled. Where, after the name of the company rolling the section there are several letters, it means that practically the same section is rolled by several com- I-sections not panies. It should be remarked that except in the economical, ^.j^gg ^f ^[^q simplest kind of beam work, it is cheaper to frame up plate girders, or trusses, of angles, tees, etc., as there is a strong pool in the rolling of I-beams and channel sections, which keeps the price of these two sections unreasonably high, in pro- portion to other rolled sections. Steel beams and sections are sold as cheap as iron, (they are really cheaper to manufacture), and where their uniformity can be relied on, should be used in prefer- ence, as they are much stronger and also a trifle stifEer. As a rule, however, the uniformity of steel in beams and other rolled sections cannot be relied on. One example of an iron beam will make the application of the Tables to transverse strains clear, and help to review the subject, be- fore taking up the graphical method of calculating transverse strains. Example. Use of Tables ^ wrought-iron I-beam of 25-foot clear span, car- XIX to XXV. j.igg a uniform load of 500 ^^ounds per foot including weight of beam; also a concentrated load o/lOOO pounds \Q feet from 238 SAKE KL'ILDING. the right hand support. The beam is not supported sideways. What size heam should he used? The total uniform load m = 500.25 = 12500 pounds of which one- half or G250 pounds will go to each reaction ; of the 1000 pounds load ISO 300 or I will go to the nearer support q (Formula 15), therefore 7 = 6250 + f. 1000 = 0850 Similarly we should have (Formula 14) p = 6250 + f . 1000 = G650 As a check the sum of the two loads should = 13500, and we have, in effect : G850-f 0650=13500 To find the point of greatest bending moment begin at q pass to load 1000, and we will have passed over ten feet of uniform load or 5000 pounds, add to this the 1000 pounds making 6000 pounds, and we still are 850 pounds short of the reaction, we pass on therefore towards j3 one foot, which leaves 350 pounds more, and pass on another ^'^ of l&o -l2o- -"■ [~^ 5oo]i>s.}>er ft.' la^ oolhs. l6o A l4o (f't>656) 3oo ('\'6€>^6) Fig. 148. afoot (to A) which very closely makes the amount. The point of o-reatest bending moment therefore is at A, say 1' 8" to the left of the weight, or 140" from q: As a check begin at p and we must pass along 160'' or 13' 4" of uniform load before reaching the point i4, at 500 pounds a foot this would make 13^.500. = 6666 or close enough to amount of reaction p for all practical purposes. The uniform load per inch will be^ =41§ pounds. Now the bending moment at A will be, taking the right-hand side (Formula 24) ??i^=6850.140— 41 §.140.70 — 1000.20 = 530 667 pounds-inch. As a check take the left-hand side (Formula 23) nij, = 6650.160 — 41f.lG0.80 USE OF TABLKS. 230 = 530014 j)oun(Is-inch, or near enough alike for all practical purposes. Now the safe modulus of rupture for wrought-iron (Table IV) is (-rr y^= 12000 pounds, therefore the recpiired moment of resistance r from Formula (18) ,._530GG_7_ 12000 Looking at the Table XX we find the nearest moment of resistance to be 46,8 or we should use the 12" — 120 pounds per yard I-beam. But the beam is unsupported sideways. The width of top flange is h = 5^". We now use Formula (78) to find out how much extra strength we require. Reduction for In inserting value for y, we use the second column uref"^^ ^^ °^ Table XVI, as the beam is, of course, of uniform cross-section throughout, and have y = 0,0102. In place of to we can insert the actual value r of the beam, and see what proportion of it is left to resist the transverse sti-ength, after the lateral flexure is attended to, or r,= , 0,0192.252 1 _|_ 0,3966 ^ 46,8 _, ' l,3yGG enough. The next size would be the 12-^" — 125 pounds per yard beam, but as the 15" — 125 pounds per yard beam would cost no more and be much stronger we will try that. Its width of flange is b = 5" and moment of resistance ?- = 57,93. Inserting these values in (Formula 78) and using r in place of lo we have _ 57,93 5 7.93 ^' "~ J I 0,0192 "."252 — 1,48 ' P = 39,14 The required moment of resistance was r^=:44,2 so that this is still short of the mark, and we should have to use the next section or the 15" — 150 pounds per yard beam. The moment of resistance of this beam is r^69,8 its width of flange the same as before, therefore : G9,8 ,„, ' 1,48 ' 240 SAFE CUILDIXG. Or this beam would be a trifle too strong even if unsupported fide- ways. We need not bother with deflection, for the length of beam is only 1§ times the span, and besides not even f of the actual transverse strength of the beam is required to resist the vertical strains, and, of course, the deflection would be diminished accordingly. » . .. -r ,^ The column in Table XX headed " Transverse Safe Uniform , , • i -c Load. Value," gives the safe uniform load, in pounds, it divided by the span in feet, for beams supported sideways. Of course the result should correspond with Table XV, except that the uniform load will be expressed in pounds here, while it is expressed in tons of 2000 pounds each in that table. For Tables XXI, XXII, XXIII, XXTV and XXV the use of the " Transverse Value " is similar, and as more fully explained in Table XIX. CHAPTER VII. Gbaphical Analysis of Tkaxsverse Strains. W Fig. 149. LL the (lif- erent cal- culations to ascertain the amounts of ;^J* bending-mo- ments, the re- quired moments of resistance and inertia, the amounts of re- actions, vertical shearing on beam, d e fl e c- tions, etc., can be done graph- ically, as -well as arithmetical- ly. In cases of complicated loads, or where it is desired to economize by reducins: size of flanges, the graphical method is to be preferred, but in cases of uniform loads, or where there are but one or two concentrated loads, 242 SAl-E BUILDING. the arithmetical method will probably save thne. As a check, how- ever, in important calculations, both methods might be used to advan tage. Basis of Cra- If ^^ have three concentrated loads w, w„ and to,, phical Method. ^^ g^ beam A D (Fig. 149), as represented by the arrows, we can also represent the reactions p and q by arrows in op- posite directions, and we know that the loads and reactions all counterbalance each other. The equilibrium of these forces will not be disturbed if we add at £■ a forcc = -|-Z/' pi"oviding that at Fwe add an equal force, in the same line, but in opposite direction or = —y- We have now at E two forces, -f y and p. If we draw at any scale a triangle a o x (or I) where a o parallel and =p, and where o x paral- lel and = -[-?/, we get a force x a, which would just counterbalance them, or a x, which would be their resultant. That is, a force G E thrusting against E with an amount a x (or x,) and parallel a x would have the same effect on E as the two f orccs + ?/ and/). Continuing X, till it intersects the vertical neutral axis through load w at G, wa obtain the resultant x. of the two forces acting at G, namely x, and w (see triangle 6 a x or II). Similarly we get resultant x, at H, of load t(j,andx2, (see triangle c & x or III) ; also resultant x, at I of loadti-,, and Xj (see triangle dcxov IV) ; and finally resultant -{-y,^tF of re- action of q and Xi (see triangle odxovY). As this resultant is + ?/ it must, of course, be resisted by a force — ?/ that the whole may remain in equilibrium. By comparing the triangles I, II, III, IV and V, we see that they might all have been drawn in one figure (Fig. 150) for q -\-p = ii\, -\- IV, -{- w, therefore : do-{-oa=idc-\-cb-\-ba, further both V and IV contain J x = x^ u " V " I " ox = y « « n " I " ax = x, " " II " III " bx = X2 " " III " IV " CX = X3 We know further that the respective Unes are parallel with each other. In Fig. 150 then, we have dc = ii\, cb = u\ ba=.w ao=-p and od^=q roi.E AXl) LOAD LIXK. 243 The distance xy of pole x from load line da being arliitrary, and the position of jiole x tlic same. Tlie figure E G H I F E (Fig 149) has many valuable qualities. If at any point K of beam we draw a vertical line K L M, then L M will represent (as compared with the other vertical lines) the pro- portionate amount of bending moment at A'. If we measure L M in parts of the length of A D and measure xy (the distance of pole, Fig. 150) in units of the load line da, then will the product of L M and xy represent the actual bending moment at A'. That is, if we measure L M in inches and — (having laid out d c, c h, Fig. 150. etc., in pounds) — measure x y in pounds, the bending moment at A' will be^a; y. L M (in pounds-inch.) Similarly at w the bending moment would be ■=x y. N G (in pounds-inch.) and at w;, it would be ^=x y. R II " " " and at ty„ it would h(i^=x y. S I " " " measuring, in all cases, x y m pounds and N G, R H and S I in inches. Average Strain The area of E G II IF E, divided by the length Fibres. ^^ span in inches will give the average strain for the entire length on extreme top or bottom fibres of beam, providing the beam is of uniform cross-section throughout. The area should be figured by measuring all horizontal dimensions in inches, and all vertical dimensions in parts of the longest vertical {R Hin our case), this longest vertical being considered = ( — ^ ) ^or top, or (— ;- j for bottom fibres, or where these are practically equal = (-^j. The greatest bending moment on the beam will occur at the point where the longest vertical can be drawn through the figure. From this figure can also be found the shearing strains and deflection of beam, as we shall see later. Distance of If now instead of selecting arbitrarily the distance '*° X y oi the pole from load line d a (Fig. 150) we had made this distance equal the safe modulus of rupture of the material, or X y=(—^ \ — measuring x y in pounds at same scale as the load line (Z a — it stands to reason that any vertical through the Figure SAKE BUILDING. E G H I F E (Fig. 140) measured in inches, will represent the re- quired moment of resistance, for if L M. x 7/ = m, we know from Fig. 151. Formula (18), that ?n = ?•.(__ j and as we made x y= ( — J, we have, inserting values in above : L3/.(|)=,.(|)or LM=. SEVKRAL LOADS. 245 Having thus sliown the basis of the graphical method of analyzing transverse strains, wo will now give the actual nie'Jiod without wasting further space on proofs. Several Concen- ^^ there are three loads tc,u\ and ?o„ on a beam trated Loads. ^ ^ (Fig. 151) we proceed as follows : at any conven- ient scale — to be known as the pounds-scale — lay off in pounds, dc=u\^; ahocb = w, and ba=zw. Let A i3r=Z measured in inches - this scale being called the inch-scale. Kow select pole x at random, Strain Diagram, but at a distance (measured with pounds-scale) xy=(-^J=:the safe modulus of rupture of the material. Draw x d, xc, xb and x a. Now begin at any point G of reaction q, draw G F parallel d x, till it intersects vertical w,, at-F: then from F draw F^ paralk-l c x to vertical w, : then draw E D parallel 6 x to vertical w ; and then D C parallel a a; to reaction p. From C draw C G, and through X draw x o parallel C G. Reactions. We now have the following results : J = reaction q (measured with pounds-scale.) ao= " p " " " " any vertical through figure C DE F G C, (measured with inch-scale) gives the amount of rr= required moment of resistance in inches, at point of beam where vertical is measured. The longest vertical passes through the point of greatest bending-moment in beam, ^lul- tiply any vertical (in inches) with x y (in pounds) to obtain amount of bending-moment at point of beam through which vertical passes. Moment of or we should have : r=: y (92) Resistance. ^ ■' Where r = the required moment of resistance, in inches, at any point of beam, provided pole distance xy=z(JL\. Where v =: the length (measured with inch-scale) of the vertical through upper figure C D EFGC at pohit of beam for which r is sought. And further : Bending- „j__^,3.„ --ggx moment. '' ^ ■' Where ni =z the bending moment at any point of beam in pounds- inch. Where u = the same value as in Formula (02) Whei-e x ?/ = the length, (measured with jiound-scale) of distance of jiole X from load line, in upper strain diagram x a d. ■246 SAFE BUILDING. If now we draw horizontal lines through d, c, b and a ; and through ■0 the horizontal line for horizontal axis ; and continue these lines until thev intersect their respective load verticals ic\„ il\ and tv, the shaded figure 0, HIJ KLMN 0, will give the vertical shearing strain along beam. Any vertical (as ii, S) drawn through this figure to horizontal axis and measured with jjounds-scale, gives the amount of vertical shearing at the point of beam {R) through which vertical is drawn. Or, Vertical Cross- shearing. ^ — ^'n (94) Where .s = the amount of vertical shearing strain in pounds, at zx\y point of beam. Where u,, == the length (measured with pounds-scale) of vertical through figure O.HIJ KLM N 0, dropped from point of beam for which strain s is sought. We now divide G C into any number of equal parts — say twelve in our case — and begin with a half part, or G to 1 r=r 12 to C=: 2V ^ ^^ ^IsO lto2 = 2to3 = 3to4 = 4to5,etc. = J2- ^ ^ and make the new lower load line gc with inch-scale so that Deflection^^^ ^ to I = length of vertical 1 e further I to II = length of vertical 2/ " II " 111= » " " 3 A " III " IV = " « " 4 i, etc. unti. " XI " c = " " " 1 2 k Now select arbitrarily a pole z at any distance zj from load line g c. Now draw below the beam where convenient (say I. Fig. 151) beginning at .7,, the Hne g, e, parallel g z till it intersects the pro- longation of 1 e (from above) at e, ; then draw e,/, parallel I 2 tiU it intersects vertical 2/at/; and similarly draw /, /i, parallel II 2; also //, 2, parallel III z, etc., to m^ l\ parallel XI z and finally Jr, c, parallel c z. The more parts (/,) we divide the beam into, the nearer will this line g, e,/, m^ k, c, approach a curve. The real line to measure deflections would be a curve with the above fines as tang- ents to it ; we need not, however, bother to draw this curve for prac- tical work. Now draw c, + 1200.96.48.(144 + 24).V/^gg^^gi> *'"■ 9.144.850000.182 If we figure out the above tedious formula we should have 8 = 0,422" or practically the same result as we obtained graphically. The safe deflection, were the beam to carry plastering, should not exceed Formula (28) 8=12.0,03=0,36" Our beam is therefore not nearly stiff enough, and we must make it thicker ; or else if we wish to save material, we will make it thinner, but deeper ; and then brace it sideways, see Formula (31). Example IV. Five Concentra- A spruce girder A B of 18/oo< span carries five ted Loads, i^adg^ ^g sliown in Figure 155. What size should the girder be f We draw AB = 2W (inch scale) ; further 6a=:2700 pounds (sum of loads at pounds scale) ; make bh = Wy = 540 pounds, ii g = w,v = 1 80 pound?, ed = w,,, = 360 pounds, dc = to„ =720 pounds, and c a = iOi = 900 pounds. 258 SAFE BUIT-DING. At C :4& ?»^-°^ ooooo«o«<> Fig. 155. FIVE CONXENTUATKI) LOADS. 259 Select X distant x y^ 1000 pounds from h a, (as 1000 =^( -^) for spruce, see Table IV). Draw x/), x//, are, etc., and figure C D G. Draw xo parallel C G; it divides load line as follows: a 0=- 1580 pounds or reaction at A. o& = 1120 pounds or reaction at B. We find longest vertical through C D G, is at load w,„ therefore greatest bending-moment on beam at w,^ ; now D E scales 70^", there- fore Formula (93) : m,y„ = 70i. 1000^:70500 and Formula (92) ?-=70, 5 From Table I, Section No. 2 b il^ r = -^- = 70, 5 and ii b = 5, we have 6 5. c?2 = 6. 70, 5 or rf2 = 84,Gand d =i/84, 6 =9,2" or say 10" which is the nearest size larger than 9,2", and of course Avooden beams are never ordered to fractions of inches. Had we worked arithmetically we should have had practically the same rer-ults. From Formulae (IG) and (17) we should have had: reaction at A = 1580 pounds, reaction at jB = 1 1 20 pounds. From rule for finding greatest bending-moment we should have located it at w„ and then had Formula (23) TOv^,, = 1580, 72 — 48.900 = 70560 and from Formula (18) 705G0 r,n nr r= = 70, 5b 1000 We now draw the shearing diagram 0, II I J K L M N P and find as follows : Cross-shearing A to w, =11 = 1580 pounds. Cross-shearing ^o^ to 20„ = J K = 680 pounds. Cross-shearing w„ toiv^,^=KL= 40 pounds. Cross-shearing 2t',„ to tL\y = M E= 400 pounds. Cross-shearing w,v to Wy =N S = 580 pounds. Cross-shearing tOy toB = P =1120 pounds. We need not bother with it, therefore. For deflection we now^ 2G0 SAFE BUILDIXG. divide C (? again into eight equal parts, (or l^ = ^^--=27") beginning 8 ■with half parts at C and G. We now make lower load line g c = the sum of the eight verticals, putting the right vertical at the top from g down. We select pole z at a distance zj = 120" from g c and draw zg, zc, etc. We construct figure g,f, c, and draw zo parallel to '^i 9i- ^^6 now divi le c, (/, at/, so that g^f: fc^r=iCO'. g, carrying // up to beam, we have the point F, distant 102" from B, and 114" from A, which is the point of greatest deflection. We find that//, scales 102", remembering that e = 850000 for spruce (Table IV), and that iz=±^ =417 (See Table I, Section No. 2) we have, Formula (95). g^ 102. 27. 120. 1000 ^ 850000.417 This would be too much for plastering, for if the girder supported plastering, the deflection should not exceed Formula (28) S = 18. 0, 03 = 0,54" We must therefore deepen the beam very materially. We use Formula (31), _ 1 In our case it would be x= ^ „ = ^— = 0,0002 5.103 5000 Supposing we were to make the beam 4" x 1 2", then we should have a; = -1^ = 0, 000 144 4.123 ' The deflection of the latter, then, would be §: 0,93 = 0,000144: 0, 0002 or S = 0>93. 0.000144^Qg^,, g^jjj ^^^ ^^^^^ deflection. " 0,0002 Were we to make the beam 3" x 14", we should have : x=— 1^=0,0001215 3.143 The corresponding deflection for this beam would be : S: 0,93 = 0,0001215: 0,0002 or g _ 0,93. 0,0001215 _,, ° ~ 0,0002 ' FIVE CONCENTHATKD LOADS. 261 or just about what would be required iu the way of stiffness. ©»r ^•♦V* Fig. 156. Had we used Formula (95) we should have had, remembering that now 12 8 = 102.27.120.1000 = 0,5G8" 850000.686 showing that we have made no mistake in applying Formula (31). •262 SAFE BUILDING. If we have any doubts as to whether a 3" x 14" stick is as strong as a 5" X 10" we use Formula (30) and have for tne former a; = 3. 142 — 588 while for the latter a; = 5. 102 = 500, so that the 3" x 14" stick is actually much stronger, as well as much stiffer than the 5" x 10". It is, how- ever, a very thin beam, and would be apt to warp or twist, unless braced sideways about every five feet of its length. To attempt to get the deflection of the girder arithmetically would be a very tedious operation. It could be done, however, by inserting in Formula (41) the different values for n and ?«, remembering every time to make n the distance from each weight to the nearer support to respective weight, and m the distance from same weight to the further support. Example V. Uniform Load. A. wr^aghi-iron beam of 25-foot span (Figure 156) carries a uniform load of 800 pounds per running foot of beam, in- cluding weight of beam. The beam is tliorougldy braced sideways. What beam should be used f We draw A B = 300" at inch scale, and then divide our uniform load into a number of equal sections, say eight, each l, = ^ = d7i" long. The total load on beam is u = 25. 800 = 20000 pounds. Each section therefore carries : U 20000 T-nn 1 _ = = 2o00 pounds. 8 8 ^ We place our arrows iv„ w,„ etc., at the centre of each section, which will bring the end ones at ^ distant from each support, so that these same verticals will answer when obtaining deflection figure. We now make i a=: 20000 pounds at pound scale, and divide it into eight equal parts, each equal w^ = il\,^=io,„ etc.,= 2500 pounds. We make XT/ = 12000 pounds, which is the r -^ j for wrought-iron, see Table IV. We draw xb,xa, etc., and construct figure C E G, which will approach a parabola in outline. The more parts we take the nearer will it be to a parabola. We draw x parallel C G and find it bisects b a, or each reaction UNIFORM LOAD. 263 is one-half the load or= 10000 pounds. This we know is the case. The longest vertical will, of course, be at the centre Z) of C G, or greatest bending-moment will be at the centre, this we know is the case. D E scales (inch scale) 62^" which will be the required r or moment of resistance (Formula 92). The bending-moment at the centre will be. Formula (93). m=G2^. 12000 = 750000 Had we used Formula (21) we should have had m = -^ == 750000 or same result, and from For- 8 mula (18) for r=i ■ — - — =62i also the same as before. From 12000 ^ Table XX we find the nearest r to our required r (62,5) is 69,8 which calls for a 15" — 150 pounds beam; as the beam is braced sideways this will do, if sufficiently stiff. In regard to shearing, we draw the figure O^H I J K L M N P R S and find shearing on bolh sides of beam similar, increasing gradually from the centre to ends.^ It would be : Cross shearing from A to ?7 square inches next triangular part AX 3 The web parts 5 ., ^11 = f'lii = «iv= «v = — -: — = 0,4 square inches each. Leaving for the roll at top a, = 1,3 We now make the horizontal line a /t = 5,5" and divide it, so that a&= 1,3 inches bc = cd = de = ef= 0,4 inches fg = 0,9 inches and gh = 1,7 inches. Select X at random and draw xa, xb, xc, etc. Draw the horizontal neutral axes T, II, III, etc., through their re- spective parts. Begin anywhere on I and draw j k parallel & a: to line II; then kl parallel ca: to III; then Im parallel dx to IV; then JTjn parallel ex toY; then np parallel /x to VI ; then p^ parallel HOI.LKD DECK-BKAM. 2G9 gx to VII; Now draw from 7 the line qo parallel xh, and f rom y „ . ^ , the line jo parallel ax till thev intersect at 0. A Horizontal j i ' 1 • r 1 1 Neutral Axis, horizontal through is the neutral axis of whole beam. We will now make a new drawing of figure j oq for the sake of clearness. Draw horizontals through it every inch in height be- ginning at both top and bottom with one-half inch. The top one scales nothing, the next i", then f , then U", then 2|", then 1^", Area of Dia- ^"'^^ t^^^ bottom one ^", the sum of all being 6^V' gram, or 6,41G". This multiplied by the height of the parts, which is one inch, would give us, of course, 6,416 square inches area. Multiplying this area by the area of the cross-section of deck beam 5,5 square inches, we should have 1 = 5,5.6,416 = 35,288. Moment of In- I" '^ablc XX it is given as 35, 1 so that we are ertiaof Beam, not very far out. If we had taken more parts, of course the result would have been more exact. Reducing When constructing plate girders of large size, ^"^^^'Cirdefs. much material can be saved by making the flanges heaviest at the point of greatest bending-moment, and gradually re- ducing the flanges towards the supports. This is accomplished by making each flange at the point of great- est bending-moment of several thicknesses or layers of iron, the outer layer being the shortest, the next a little longer, etc. Of course the angles, which form part of the flange are kept of uniform size the whole length, as it would be awkward to attempt to use different sized angles. Generally (though not necessarily) the inner or first layer of the flange plates, is also run the entire length. Of course; where the flanges are gradually reduced in this way, it becomes ne- cessary to figure the bending-moment and moment of resistance at manv points along the plate girder to find where the plates can be re- duced. This would be a wearisome job. By using the graphical method, however, it can be easily accomplished. Referring back to Fi 1 1 \ 1 I I _\ 1 1 16 12 14 16 16 12 16 14 14 14 12 16 12 16 16 12 16 14 14 14 14 12 12 12 16 16 12 16 16 14 14 14 14 12 14 12 12 16 16 * ! 4 ,- , 1 \ , 1 \ _, , »? 1 1 10 10 \ ' [ 10 10 { ' 1 10 11 J \ |U 2 U 4 u c '11 c U 9 '11 9 >?'._ [ ^, , >> 1 4 , ; ? 1 * 1 11 11 ^ 1 1 1 ^ ' 1 'l2 ^J ' i ' 1 \ 1 12 b k , ; * , ) 1^ 13 1 1 1 13 1 ' j_ — 9 ' ' V r Q ■ b 1 . ' 1 1 ' 1 bl t 1 ' ' , 1 ' T ' , . , 1 1 ' , p. 1 1 1 1 ' r 1 i ■ 6. ' 1 1 Q 1 ' , 1 Q 1 1 , , 1 .1 . ' [ ■ 1 ' b| 1 ?' ' t' ' i 9 1 i^ ' , ^ ? ' , ' <^ 1 , ■ ' Y ' ' . ' ■ ' ' ' . 1 ' ^ ' ' ' ; ' ' , • , 9 , , ' ' 1 ■ ' 1 1 1 , ' ' ■ . ' 1 ' 1 ' ' 1 1 , ■ 1 ' ' 1 ' ' 1 ■ ' 1 ; . 1 ' ' 9 1 , , 1 1 \ 1 1 , ■ * , V 1 1 57 > 1 b t ' ? ' 1 K: ^ "^ !> . ) r ; 4: ' X >? ' 1 L *! ; 1 1 V , 1 -c^, XIII. 5RACED SIDEWAYS. SPAN IN FEET. Ol O ^ ! C^4 CO IC CO CO CO CO Safe Uniform Load in Lbs. Foii Full Lines. Goorsia Pine. White Spruce or Onk. W. Pine. nemlock. 500 475 445 1000 f)50 890 1500 1425 1335 2000 1900 1780 2500 23 75 2225 3000 2850 2670 3500 3325 3115: 4000 3800 3560 4500 4275 4005 5000 4 750 4450 5500 5225 4895 6000 5700 5340 / Table XIII. WOODEN GIRDERS, BRACED SIDEWAYS. KlY, FOR FLOORS. LENGTH OF SPAN OF BEAM IN FEET. »r Iron Beams ; but length of span in feet must not exceed twice the depth ^ Table XIV. IRON I BEAMS FOR FLOORS. r&T ,.»„!, .~ J ITS ;;:,E ''■'"■ 1 •«•'"' : i : i : :""";"""1""~T:T'T:':":1": "":""" """""111''' "■ 'i"ji 1 113 1 a 1 9 2 2 3 2 7 2 10 3 2 3 n 3 9 4 4 3 4 10 5 2 5 r, 5 9 G G 3 G 7 e 10 7 2 7 .0 7 9 8 8 3 8 7 1 '"s 2 2 4 2 8 3 3 4 3 8 4 4 4 4 8 5 6 4 .■j 8 G 6 4 G 8 7 4 7 8 8 8 4 8 8 9 9 4 9 8 10 2 "' 2 5 2 10 3 2 3 7 4 4 5 4 10 r, 2 5 7 6 5 G 10 7 2 7 7 8 8 S 8 10 9 2 9 7 10 1,'°'c * ^^-^- + ^J^-^-fk--t-^f+■'-^• + •l-^■+ + ^-^•■l-^- + + +4•^t44■4^■4-r--t^■ + +4 .4' I „ + 4 4 4 +, 1 .| 1- !p J- 1.^ 1 ■( fi 1 -1 1- 1- 1 (■ 1- H- i-.t. K f > 1 1 4 1 -1 1 -I- 1 4 -I- i- -1 1 t I n 1 5 1 10 2 2 2 2 3 2 7 2 10 .1 S 2 3 !> 3 7 8 10 4 2 4 10 a 2 5 5 r> 7 a 10 1 9 2 2 3 2 S 2 9 3 3 3 3 G 3 9 4 3 4 e 4 9 5 3 5 G 5 9 G 6 3 G G G 9 7 7 3 7 G ^ 4444;?4 4,V^t^S4gt^5 4:^44K4S^**£^•^4 4,stS^^4^.4 4 444, } ; 4 * * '^i-ffi25^;^Z4^ti-?^+-.-4^4 e-J^r 1 8 1 10 2 2 2 2 4 2 (1 2 a 2 10 3 3 2 3 4 3 3 8 3 10 4 4 2 4 4 4 6 4 8 4 10 5 ^--"|^t|'/4if4?'4gj?4r5P|^|:e.K'fir4 4 4j^.-t4,*-r4{r 5 6^ + :i('?W->^-|z:^ZtZ^^S*i^5?*?-i?P^^- , j?^^^^^?.,4^5 6 ' ^7^aiEdEir^Sf^2i^;iPli-£5£5'' -i* ^ - ^ ^^ ♦ ' i?;*^ 6 6**^i.li:4'i + t,i*i^*'-?l^iZ'-*IS*^^»i:'^*;^**+4,^^** 7 + '-iM£Oit-§<-0'-^5Z;2^i47?^. + 4^f.. .^. ,, ^..^^^ 7 „.44ts-!r^tv^!?'Jj^t^g^t|r4T7Z^-47,»»4'r.^-t5',•^4Jij 8 M*iM5*'^4<'-|Z*-ZZ^;K^*Z2^ii^*-'- — PHCENIX IRON COMPANY. Philadelphia, Pa. D — POTTSVILLEIRON AND STEEL COMPANY, Pottsville, Pa. Pa. F— PASSAIC ROLLING MILL COMPANY, Passaic, N. J. I ;es the Mill which rolls the e.N^act shape given in the Table; the other letters give the TEE. Flanee Ranoe. DECK BEAM. Taule XIX. EXPLAIXINCt USE OF TABLES XX TO XXV. ^ EEi!^ 8=^. 8=^ _^^ F^ ^3 !2J.rf '- = '^■"1^-0^1 ; X = leiigt!iinf. J given la Tables XX to XXV ; Note. — If the transverse values (y), given for steel, are used, test each piece carefully, as steel varies greatly in strength. For equal deflections of steel and iron, add only 7A% to iroa transverse values, instead of 35% as given in Tables. In calculating the transverse values, the moduli of rupture used were: for iron, 12000 pounds per sfjuarQ inch, and for steel, 15000 pounds per square inch. From the "Safe Load" as obtained above, tMucl the weight of beam in pounds, as follows: — In case of uniform load deduct enfiVc weight of beam; in case of centre load, deduct one-half the weight of beam; in case of loads at thirds of span, deduct from each load | the weight of beam; in ease of loads at quarters of span, deduct from each load J tlio weight of beam. ■ (about 1%) than iro cxactlv the In onlei steel . Steel sections will be slightly ii pijuired weiglit of section, hcct hoih. Tlie capital letters in the lirst column Iicadod "Mills Rolling Shape" in each Table represent, tlip following Rolling Mill: on. N. J n, give either the requiretl dir ■ the R — PHCENIX IRON COMPANY, Philadelphia, Pa. I)_POTTSVILLE RON AND STEEL COMPANY. Pottsville. Pa. F— PASSAIC ROLLING MILL COMPANY, Passaic, N. J. A — NEW JERSEY STEEL AND IRON COMPANY, T C— PENCOYD IRON WORKS, Philadelphia. Pa. K— UNION IRON MILLS and HOMESTEAD STEEL WORKS, Pittsburgh, Pa. In the columns hcadtsd "Mills Rolling Shape" the drst letter on each line indicates the Mill which rolls tlie e.\;iLt shape given in the Table; the other letters give the Mills which roll an approximately similar shape. In gel of parts of i , thev ! taken as shown below r XX. D STEEL I BEAMS. OF THIS Table, see Table XIX.) al to Web. n J N :.6 .5 .5 ! ► !.5 5 ^4 5 '.3 t.5 I 7 1.7 r.i 1.9 .3 ^.2 5 3 4 2 r.5 [.7 ),2 1.7 .3 ,5 ,4 t.9 9 1.6 L 5«0 60.67 60.42 61.99 59.68 34.80 35.32 35.15 32.52 31.24 34.70 31.47 35.20 35.76 34.40 23.32 22.60 23.35 18.92 22.46 22.60 23.98 21,80 24.22 17.49 17.77 17.39 18.42 13.91 15.52 15.35 16.40 12.63 11.70 12.23 11.83 13.10 13.16 13.41 8.64 10.44 Transverse Value (f) in lbs. For Iron. For Steel. 1320000 1160000 990000 917000 552000 748000 460000 864000 800000 740000 655000 563000 556000 458000 508000 504000 376000 454000 510000 377000 375000 290000 306000 356000 282000 251000 250000 300000 278000 258000 237000 336000 282000 268000 208000 211000 199000 167000 181000 168000 1650000 1450000 1238000 1146000 690000 935000 575000 1080000 1000000 925000 819000 704000 695000 573000 635000 630000 470000 567000 637000 471000 409000 363000 382000 445000 353000 314000 312000 375000 347000 322000 297000 420000 353000 335000 260000 264000 249000 209000 226000 210000 Axis Parallel to Web. «—■ H=^-H 46.50 51.78 26.62 31.50 15.29 27.46 11.64 40.84 29.90 33.79 20 18.34 16.91 13.13 25.41 20.90 11.54 15.50 24.08 12.98 16.76 8.74 11.66 15.80 9.43 8.01 8.09 11.30 10.64 11.08 8.09 23.16 14 11.23 7.14 8.44 7.35 4.92 6.96 7.55 13.78 16.57 8.87 10.08 6.12 9.55 4.66 13.90 10.29 12.12 7.50 7.34 6.00 5.34 9.24 7.96 4.82 6.09 8.76 5.46 6.10 4 4.44 6.32 4.19 3.53 3.59 4.74 4.60 4.79 3.69 8.62 5.67 4.99 3.30 3.80 3.27 2.46 3.25 3.35 1.71 2.16 1.33 1.04 1.02 1.37 .94 1.62 1.25 1.69 1.02 1.22 1.17 1.05 1.52 1.23 .93 .86 1.42 1.04 1.43 .87 1.23 1.18 .90 .85 .91 .83 .96 1.05 .90 1.54 1.02 .91 .72 .92 .86 .70 .67 .93 Transverse Value (r) in lbs. 110000 132600 71000 80600 49000 76000 37000 111200 82300 97000 60000 58700 52800 42700 74000 63700 38500 48700 70100 43700 49000 32000 I 35500 50500 33500 28200 28700 37900 36800 38300 29500 69000 45400 40000 26400 30900 26000 19700 26000 96800 137800 165700 88700 100800 61200 95500 40600 139000 103000 121000 75000 73400 66000 53400 92400 80000 48200 61000 87600 54600 61000 40000 44400 63200 41900 35300 35900 47400 46000 48000 36900 86000 56700 49900 33000 38600 32700 24600 32500 SS.'inO Table XX. LIST OF IRON AND STEEL I BEAMS. (For IsffoaMATiox as to thk Use of this Table, sbc Table XIX.) L P 1 J .Is- f is •3 I i i r A.,.^ — "-I-- 1 A.,. P„.„e, ..web "--■(MF- ! jr Ir P^ '(" tli'lb..'"" 1 tL 11^ P P 0) in 1bl'.""° Fo,In.«. F„»,„l. r„,w ro,s,«. 20 20 20 20 lit \'} 15 15 16 15 15 16 12 12 12 12 12 12 12 12 lOJ lOi lOJ '"i 10 10 10 10 9 9 9 9 9 9 9 8 e 8 7 7 7 7 7 6 G 6 C e 5 5 3 3 3 3 U 272 240 200 192 160 200 125 250 240 200 195 1,50 145 125 170 170 125 180 170 125 120 100 96 135 105 95 90 135 112 105 90 150 135 125 99 90 85 70 105 SO 05 75 69 C5 CO 55 62 120 90 64 60 40 40 36 34 30 37 30 28 24 18 27 23 21 17 M 6.75 7 6 6.25 5 5.75 5 5.875 5.81 5.56 5.33 5 5.125 5.50 5.26 4.79 5.09 5.50 4.76 5.50 4.44 5.25 5 4.50 4.54 4.60 4.77 4.625 4. 025 4.376 5.375 4.94 4.50 4.33 4.375 4.50 4 4.29 4.60 4 3.91 4 3.81 3.50 3.75 3.61 5.25 5 3. 46 3.60 3 3 3 2.844 2.75 3 2.76 2.76 2.25 2 2.52 2.60 2.32 2.25 1.50 0.69 0.60 0.50 0.50 0.50 0.60 0.42 0.876 0.93 0.625 0.77 0.47 0.44 0.44 0.6U 0.66 0.47 0.96 0.69 0.49 0.39 0.44 0.31 0.47 0.S75 0.41 0.31 0.77 0.50 0.50 0.S4 0.60 0.75 0.67 0.53 0.41 0.375 0.30 0.79 0.375 0.30 0.53 0.375 0.44 0.40 0.30 0.234 0.625 0.50 0.46 0.30 0.25 0.31 0.30 0.31 0.25 0.81 0.25 0.25 0.81 0.19 0.39 0.26 0.19 0.166 0.126 7.87 G.63 5.65 5.09 4.34 6.44 3.37 7.22 6.29 6.06 4.84 4.47 4.48 3.35 5.47 5.50 3.78 4.26 6.77 3.80 3.87 2.71 3.09 4.84 3.07 2.96 3.11 3.67 3.01 3.24 3.14 5.59 4.22 4.41 2.85 3.07 2.93 2.40 2.74 2.82 2.21 2.29 2.47 2.08 1.88 1.90 1.92 4.78 3.31 1.61 1.77 1.42 1.86 1.20 1.06 .99 1.40 1.08 1.07 .71 .68 .93 .86 .85 .68 .19 11.46 10.64 8.67 9.02 6.36 7.14 5.62 10.56 11.42 7.88 9.82 6.06 6.59 6.80 6.83 6.88 4.77 9.48 6.46 4.90 3.99 4.53 3.28 3.68 3.10 3.58 2.68 6.16 3.95 4.02 2.70 3.82 6.00 3.61 4.20 2.93 2.64 2.20 6.02 2.30 1.96 2. 92 1.90 2.42 2.24 1.70 1.30 2.28 2.08 2.18 1.37 1.17 1.18 1.20 1.26 1. 01 .86 .76 .76 .99 .61 ,84 .53 .40 .35 -14 27.20 24 19.97 19.20 16.04 20.02 12.36 25 24 20 19.60 15 14.56 12.50 16.77 17 12.33 18 17 12.60 11.73 10 9.46 13.36 10.14 9.50 8.90 13.50 11.17 10.50 9.04 15 13.60 12.33 9.90 9.07 8,50 7 10.50 8.03 6.37 7.50 6.90 6.58 6 5.50 6.14 11.84 8.70 5,40 4.91 4.0! 3.90 3.60 3.38 2.99 3.66 2.91 2.90 2.40 1.77 2.70 2.26 2.10 1.71 .52 1650.3 1460 1238 1146 523.6 707.1 434.6 813 750 694 614 523 621.2 430 891.2 385 288 340 381.9 282.6 281.3 218 229.2 233.7 185.6 165 164 187 173.6 161 148.3 189.1 159 150.8 117 118.8 111.9 93.9 90.4 83.9 67.4 54.3 55.7 49.8 46 44.8 43.1 64.9 49.8 28.4 29 23.5 15.4 14.9 13.4 12.1 9.2 7.6 7.7 5.6 4.6 3,64 3.29 3.09 2.66 .135 105 146 123.8 114.6 69 93.5 67.6 108 100 92.6 81.5 70,4 69.6 67.3 63.6 63 47 56.7 63.7 47.1 46.9 36.3 38.2 44.5 35.3 31.4 31.2 37.5 34.7 32.2 29.7 42 35.3 33.5 26 26.4 24.9 20.9 22.6 21 16.9 15.6 15.9 14.2 12.9 12.7 12.3 21.6 16.6 9.6 9.6 7.8 6.1 5.96 5.4 4.8 4.6 3.75 3.84 2.80 2.25 2.36 2.16 2.06 1.77 .21 60.67 60.42 61.99 59.68 34.80 36.32 35.15 32.52 31.24 34.70 31.47 35.20 35.76 34.40 23.32 22.60 23.35 18.92 22.46 22.60 23.98 21,80 24,22 17,49 17,77 17.39 18.42 13.91 16.52 15.35 16.40 12.03 11.70 12.23 11.83 13.10 13.16 13.41 8.64 10.44 10.58 7.24 8.08 7.56 7.60 8.0.5 8.35 5.48 5.72 5.29 5.90 5.86 3.95 4.14 3.96 4.04 2.61 2.57 2.65 2.33 2.54 1.32 1.46 1.46 1.56 .26 1320000 1160000 990000 917000 562000 748000 460000 864000 800000 740000 656000 663000 556000 458000 608000 604000 376000 464000 610000 377000 375000 290000 306000 356000 282000 251000 260000 300000 278000 258000 237000 336000 282000 268000 208000 211000 199000 167000 181000 168000 135000 124000 127000 U3800 103200 101000 98500 172000 132000 76000 76800 62400 4S800 47700 42800 38400 36800 30000 30700 22400 18000 18900 17300 16500 14200 1650000 1450000 1238000 1146000 690000 935000 676000 1080000 1000000 925000 819000 704000 696000 573000 635000 630000 470000 567000 637000 471000 469000 363000 382000 446000 363000 314000 312000 375000 347000 322000 297000 420000 353000 836000 260000 264000 249000 209000 226000 210000 169000 155000 169000 142000 129000 127000 123000 216000 160000 95000 96000 78000 61000 59600 63500 4S000 46000 37600 ,38400 28000 22500 23000 21600 20600 17700 2160 46.50 51.78 26.62 31.60 16.29 27.46 11.64 40.84 29.90 33.79 20 18.34 16.91 13.13 25.41 20.90 11.54 15.50 24.08 12.98 16.76 8.74 11.66 15.80 9.43 8.01 8.09 11.30 10.64 11.08 8.09 23.16 14 11.23 7.14 8.44 7.35 4.92 6.96 7.55 4.56 4.87 5.42 4.16 3.15 3.90 3.43 18.59 10.78 2.61 2.74 1.61 1.G8 1.74 1.21 1.04 1.74 1.11 1.17 .58 .31 .84 .77 .66 .48 .069 13.78 16.57 8.87 10.08 6.12 9.66 4.66 13.90 10.29 12.12 7.50 7.34 6.60 6.34 9.24 7.96 4.82 6.09 8.76 5.46 6.10 4 4.44 6.32 4.19 3.53 3.69 4.74 4.00 4.79 3.69 8.62 5.67 4.99 3.30 3.86 3.27 2.46 3.25 3.85 2.27 2,50 2.71 2.18 1.80 2.08 1.90 7.08 4.32 1.45 1.57 1.07 1.12 1.16 .85 .76 1.16 .81 .85 .62 .31 .67 .62 .47 .43 .092 1.71 2.16 1.33 1.64 1.02 1.37 .94 1.62 1.25 1.69 1.02 1.22 1.17 1.05 1.52 1.23 .93 .86 1.42 1.04 1.43 .87 1.23 1.18 .90 .85 .91 .83 .96 1.05 .90 1.54 1.02 .91 .72 .92 .86 .70 .67 .93 .71 .66 .786 .62 .53 .71 .67 1.57 1.24 .46 .66 .40 .43 .483 .36 .35 .48 .38 .40 .22 .175 .31 .35 .30 .28 .013 110000 132600 71000 80600 49000 76000 37000 111200 82300 97000 60000 68700 62800 42700 74000 63700 38600 48700 70100 43700 49000 32000 35500 50500 33500 28200 28700 37900 36800 38300 29500 69000 46400 40000 26400 30900 26000 19700 26000 20800 18200 20000 21680 17400 14400 10600 15200 66600 34500 11600 12600 8500 9000 9260 6800 6000 9500 6500 6800 4160 2500 6360 4960 3760 3440 137800 165700 88700 100800 61200 95500 46600 139000 103000 121000 75000 73400 66000 53400 92400 80000 48200 61000 87600 54000 61000 40000 44400 63200 41900 35300 35900 47400 46000 48000 36900 86000 66700 49900 33000 38600 32700 24600 32500 33500 22700 25000 27100 21800 18000 20800 19000 70800 43200 14500 15700 10700 11200 11580 8600 7600 11600 8100 8600 6200 3100 6700 6200 4700 4300 920 A,B E D,B.C,E.... D.B.E D.B.E A B.C.D R,I),E A,C A,U,E A.B.C.D.E.F. A,B,C,D,F... A.B.C.D.F... C,D,E A,B,D,F A,1J,C,D.E,F, A.B.CD.F... A,B,C,D,E,F. A,B,D,E .... A.B.CD.F... A B,C,D,E,F. A,D,E,F A B.C.D.E.F, A,F A,B,D,E,F .. A.B.C.D.F .. r Tablk XXI. r AND STEEL CHANNELS. Use of this Tabi.k, skk T.abi-k XIX.) Ixis Normal "T^ to Web. " J 1 Axis Parallel . ^ ! to Web. " g***^ " > . * V Transverse '~^_. Z '-^ Trjiiisverso i'ls Hi s c==--r Value (v) in lbs lie Sjs'ta vis Valine (/•) in lbs. »-< 3 5 CO For Iron . For Steel 1- £ '5 ^ :»«5' «ll= For Iron. For Steel. J4.0 70.0( ) 29.20 560000 ) 700000 21.60 6.93 1.21 1.06 55600 69500 )7..5 62.3f ) 31.30 49840( ) 623000 19.70 6.84 1.32 1.09 54720 68400 ;o.7 88.0f > 28.73 70500C 881000 3 7.56 10.05 1.63 1.28 8050< 100).JOO i4.G 7;{.9^ 27.73 59150C 739400 23.61 7.15 1.18 1.08 5720f 71500 SCO 78.13 30.84 62500C 781300 32.25 9.24 1.70 1.26 74O0f 92JOO (7.0 70.2(j 30.03 562080 702600 31.41 8.92 1.80 1.23 7136( 89200 9.1 59.88 29.94 479000 598800 18.27 6.09 1.22 1.00 48720 60900 6.0 50.13 31.33 401000 501300 14.47 4.74 1.21 0.95 38000 4 7400 9A 55.41 19.07 443300 554100 20.65 6.55 1.16 1.16 5250)0 65500 1.6 47.64 20.83 381000 476400 17.87 6.20 1.28 1.12 496O0 62000 4.3 35.00 21.40 280000 350000 8.93 3.68 0.88 0.86 29440 36800 8.9 27.70 21.10 221600 277000 5.44 2.39 0.67 0.73 19120 2390)0 3.2 25.03 21.89 200000 250300 5.04 2.24 0.72 0.755 18000 22400 5.7 39.29 15.72 314200 392900 8.44 3.13 0.56 0.80 250)40 31300 1.5 30.25 20.16 242000 302500 7.11 3.29 0.79 0.84 26320 32900 3.7 20.62 20.79 164960 206200 3.22 1.62 0.55 0.62 12960 1620)0 9.4 24.64 12.32 197000 246400 4.96 1.98 0.47 0.69 16000 1980)0 6.0 22.10 15.50 176800 221000 5.20 2.55 0.71 0.80 20400 25500 8.4 16.84 14.73 134700 168400 3.84 1.81 0.64 63 14500 18100 7.0 16.60 15.90 132800 166000 3.20 1.70 0.58 0.70 13600 1 7000 0.0 28.00 10.82 224000 280000 7.79 2.93 0.61 0.84 23440 29300 8.6 25.72 11.59 205760 257200 5.26 2.35 0.47 0.76 18800 23500 0.5 20.10 10.92 161000 201000 3.02 1.31 0.33 0.626 10500 1310)0 7.4 19.4 7 12.98 155760 194700 3.51 1.78 0.47 66 14240 1 7800 3.2 14.63 12.84 117040 146300 1.97 1.15 0.35 0.53 9200 11500 4.0 12.80 13.33 102400 128000 2.20 1.14 0.46 0.565 9100 11400 8.7 24.16 9.97 193300 241600 7.30 2.75 0.67 0.91 22000 2750)0 9.1 19.80 9.92 158400 198000 4.25 2.02 0.47 0.73 16160 20200 2.1 18.24 11.73 146000 182400 5.35 2.35 0.76 0.85 18800 23500 8.8 13.07 11.76 104600 130700 2.53 1.35 0.51 0.63 10800 13500 2.0 9.33 11.42 74640 93300 1.52 0.99 0.41 0.55 7920 9900 4.5 16.14 7.67 129100 161400 4.00 1.98 0.48 0.73 15840 19800 0.1 15.03 8.12 120200 150300 3.53 1.69 0.48 0.755 13500 16900 8.4 12.10 7.81 97000 121000 1.94 0.98 0.31 0.584 7840 9800 4.5 11.12 9.89 89000 111200 2.54 1.46 0.56 0.76 11700 14600 8.2 7.06 9.55 56480 70600 1.06 0.71 0.36 0.50 5680 7100 7.9 10.83 6.30 86640 108300 2.85 1.53 0.48 0.68 12240 15300 5.6 7.31 5.5 7 58500 73100 1.11 0.62 0.24 0.51 4960 6200 7.1 7.74 7.53 62000 77400 1.96 1.10 0..54 0.715 8800 11000 7.6 5.03 7.05 40240 50300 0.75 0.49 0.30 0.470 3920 4900 8.0 9.33 4.24 74600 93300 2.95 1.43 0.45 3.788 11440 14300 3.4 7.80 4.33 62400 78000 1.84 0.95 0.34 3.662 7600 9500 1.7 7.23 4.82 58000 72300 2.12 1.20 0.47 3.725 9600 12000 7.2 5.73 5.21 46000 57300 1.30 0.80 0.39 3.630 6400 8000 0.4 3.47 4.74 27760 34700 0.62 0.46 0.28 ( 3.400 3680 4600 3.4 5.34 3.17 42720 53400 1.50 0.94 0.36 X640 7520 9400 0.3 4.12 3.03 33000 41200 m 0.44 0.19 ( 3.493 3520 4400 A,F. F,A. F,A. A,F. F,A. A,P. A,B,C,D,E . B,A,C,D,E . B,A,C,U,E . A,D,E A E,A,B,C.... A,B,C,D,E,F. 1 A,B,C,D,F.. A,C,E A,C,D,E . . . A,B,C,D,E . C,B,D E,C A,B,C,D.E . A,B,C,D,E . B,A,C,1) . . . A A,B,C,F.... A,C,D,E,F . A,B,C,D,E,F. B,A,C,D,E,F. I E,C A,C,D,E . . . D,A,C,E . . . B,D,F U.C.E B.A'.Cn.E. A.B.I) I' o.so ".!■<•> 1. 00 HM ItMl a.fti 11.46 S.liO 0..19 ■?.m 0.3.1 1.(19 1 .00 a.M 0.44 ■?..n 0.2« 1.44 O.HO l..iO 0.4 a l.SO 0..')75 l.liO 0.30 1.30 l.Oli 1.7S 0.875 1.83 1.17 HM s.oo u;h,;i .1.62 7.00 l.iS.2 (i.ai. lii.OO 233.7 4.l!( 9.00 181.5 3.0B 5.94 12S.7 7..'iO 10.50 129.4 .3.90 7.50 U6.0 3.60 c.ou HK.4 :■»" 0.545 1.03 0.60 0.65 0.25 0.80 0.19 0.44 0.44 0.91 0.31 0.71 0.20 0.60 0.17 0.4? 0.53 0.67 0.376 0.48 0.20 O.flS 0.26 (1.34 0.27 0.27 0.13 O.U 117040 102400 193300 158400 146000 104600 74640 129100 120200 97000 89000 56480 86640 .58500 62000 40240 74600 62400 68000 46000 27760 42720 33000 30400 20320 28000 22160 15600 12960 14400 12080 IU640 5680 3840 1520 -liUlO 14.47 4.74 1.21 0.95 380(111 .17100 . tlud 20.65 6.55 1.16 1.16 526(1(1 ll.i.'iOO 17.87 6.20 1.28 1.12 49600 *;■_'( 1(10 (t 8.93 3.68 0.88 >.»6 294411 IldSllO ) 5.44 2.39 0.67 1.73 19120 23900 5.04 2.24 0.72 ).7.66 18000 22400 8.44 3.13 0.66 1.80 25040 31300 _■ ,H(I 7.11 3.29 0.79 1.84 26320 32900 .Jnll 3.22 1.62 0.55 ).62 12960 16200 ', luo 4.96 1.98 0.47 1.69 16000 19800 jinul) 5.20 2.65 0.71 1.80 204011 25500 ' . KM) 3.84 1.81 0.64 163 14600 18100 .i.UUd 3.20 I.!0 0.68 1.70 I36O0 17000 -, 1,1(1(1 7.79 2.93 0.01 1.84 23440 29300 7'iiil 5.26 2.35 0.47 J.76 188O0 23500 ^JM 1(1(1(1 3.02 1.31 0.33 0.626 1O5O0 13100 ;»4 7<)0 3.51 1.78 0.47 (166 1424C 17800 46300 1.97 1.15 0.35 I1..63 920C 11600 28000 2.20 1.14 0.46 11.565 91 IK 11400 241600 7.30 2.76 0.67 0.91 2200f 2/500 98000 4.25 2.02 0.47 (1.73 1616( 20200 ,82400 5.36 2.36 0.76 (1.86 1880f 23500 .30700 2.53 1.35 0.51 0.63 10801 13500 93300 1.32 0.99 0.41 0.55 792( 9900 61400 4.00 1.98 0.48 0.73 16S4( 19800 50300 3.53 1.69 0.48 0.756 13.50( 16900 ,21000 1.94 0.98 0.31 0.58J 784( 9800 ,11200 2.64 1.46 0.56 0.76 11701 14600 70600 1.06 0.71 0.36 0.50 668< 7100 108300 2.85 1.53 0.48 0.68 1224( 1.6300 73100 l.Il 0.62 0.24 (1.61 490( 6200 77400 1.96 1.10 0.64 0.716 880( 11000 60300 0.75 0.49 0.30 0.4 7( 392( 4900 93300 2.95 1.43 0.45 0.781 U44( 14300 78000 1.84 0.96 0.34 0.662 760( 9500 72300 2.12 1.20 0.47 0.72.'- 960( 12000 57300 1.30 0.80 0.39 0.63( 640( 8000 34700 0.62 0.46 0.28 0.40( S68( 4600 53400 1.50 0.94 0.36 0.64( -621 9400 41200 0.63 0.44 0.19 0.49.' 352( 4400 38000 0.87 0.68 0.34 0.61 ( 544( 6800 25400 0.43 0.37 0.25 0.4 7( 292f 3660 35000 1.14 0.83 0.36 0.68f 664( 8300 27700 0.79 0.66 0.33 0.60t 4512 5640 19500 0.32 0.31 0.19 0.46i: 2481 3100 16200 0.28 0.22 0.17 0.38( 176( 2200 18000 0.47 0.37 0.19 0.56S 296C 3700 15100 0.36 0.33 0.20 ().63(: 2616 8270 18300 0.29 0.29 0.19 0.5 IC 232C 2900 7100 0.21 0.24 0.185 (1.46(: 192C 2400 4800 0.08 0.11 0.096 :).S7(1 88C 1100 1900 0.014 0.032 0.036 0.200 256 360 c Tl XXII. EVEN- -LEGGED ANGLES. THIS TaBLB E, SBB TaBLB XIX.) Ixis Parallel to One Side. M— tr-N Axis ^ at 46°. ^' 1 i 05 TrsujsverHe Value (y) iu lbs. 5i II'" For Iron. For Steel. i6 7.54 3.24 1.860 60350 75400 15 1.37 31 7.01 3.28 1.820 56100 70100 13.10 1.34 91 4.61 3.46 1.685 36900 46100 7.75 1.35 22 3.90 3.42 1.580 31200 39000 6.77 1.35 34 4.97 2.19 1.010 39760 49700 8.67 .96 70 3.94 2.37 1.550 31550 39400 6.07 .98 Jo 2.64 2.52 1.460 21100 20400 3.77 1.02 20 3.28 1.80 1.396 2G250 32800 4.88 .772 20 2.24 1.92 1.286 18000 22400 2.05 .707 5G 2.47 1.40 1.271 19800 24700 3.45 .634 18 2.32 1.39 1 . 220 18550 23200 3.01 .624 56 1.52 1.52 1.138 12200 15200 1.86 ,650 J8 2.40 1.21 1.220 19200 24000 2.40 .470 58 1.74 1.08 1.122 13920 17400 2.04 .470 56 1.15 1.15 1.013 9200 11500 1.20 . 184 50 .90 1.12 .930 7200 9000 .95 162 n 1.13 .76 .996 9040 11300 1.05 338 13 .96 .79 .930 7080 9000 .95 .336 14: .58 .86 .842 4640 5800 .52 .361 13 .79 .G6 .887 6320 7900 .61 .300 .5 .59 .71 .802 4720 5900 .50 .309 »5 .48 .72 .780 3840 4800 .39 .303 !2 .61 .52 .770 4880 6100 .52 .221 !8 .66 .57 .806 5280 6600 .51 .227 '0 .39 .59 .717 3120 3900 .30 .252 12 .34 .59 .700 2720 3400 .25 .240 !2 .46 .45 .740 3680 4600 .35 .194 '8 .45 .44 .720 3600 4500 .35 .197 iO .31 .47 .654 2480 3100 .22 .208 :0 .26 .50 .690 2080 2600 .17 .212 ■8 .51 .38 .730 4080 5100 .28 .150 :5 .30 .33 .634 2400 3000 .21 .154 11 .22 .33 .580 1765 2200 .13 .158 17 .19 .38 .570 1520 1900 .11 .160 8 .35 .31 .640 2800 3500 .18 .120 17 .20 .27 .550 1600 2000 .12 .120 8 .15 .29 .507 1200 1500 .08 .129 9 .17 .19 .510 1360 1700 .09 .096 6 .14 .19 .487 1120 1400 .07 .083 4 .12 .19 .450 960 1200 .06 .081 1 .10 .21 .444 800 1000 .05 .094 9 .085 .21 .440 680 850 .04 .081 123 .126 .152 .400 1010 1260 .05 .070 ■77 .079 .138 .404 632 790 .04 .071 44 .050 .147 .358 400 500 .02 .067 37 .047 .084 .340 376 470 .02 .045 130 .040 .084 .310 320 400 .01 .040 f22 .031 .096 .296 250 310 .01 .043 19 .029 .066 .286 232 290 14 .023 .070 . 264 184 230 12 .021 .048 .254 168 210 1 09 .017 .053 .233 136 170 " 1 Table XXIl. LIST OF IRON AND STEEL EVEN-LEGGED ANGLES. , L,B,C,D,E,F. I a,b,6,'d,e,f. A,B,U,D,E,F. C,B E,F A,B,C,D,E,F. A,B,C,D,E,F. C,E,F A,t:,E,F 1,A,C,E.F... A,B,C,D,E,F. B,D P A.CE.F A,B,C,D,E,F. A,C,F B.A.CD.E.F. A,B,C,D,E,F. A,F A,F A,F A,F 3.500 3.438 3.250 2.781 2.750 2.500 2.500 1.C25 1.504 1.500 1.500 1.438 1.375 1.250 1.125 1.0C3 35.46 31.01 19.91 17.22 19.04 U.70 9.33 1.43 2.17 1.24 1.025 1.825 1.405 .72 1.385 5.44 5.1C 2.8C 5.10 2.81 1.44 2.77 .145 .100 .125 .085 3.24 3.28 3. 40 1.92 1.40 1.39 1.52 |4lll 1.8G0 1.820 .685 1.580 I.GIO 1.550 1.4G0 1.396 1.286 1.271 1.230 1.138 1.220 1.122 1.013 .930 .996 .56100 36900 31200 39760 31550 21100 26250 18000 19800 18550 12200 19200 13920 9200 7200 9040 7680 4640 6320 4720 3840 4880 6280 3120 2720 2480 2080 4080 2400 1765 1520 2800 1000 1200 1360 1120 75400 70100 40100 39000 49700 39400 26400 32800 22400 24700 23200 15200 24000 17400 11500 SOOO 11300 9000 6800 7900 5900 4800 6100 6600 3900 3400 4600 4500 3100 2600 5100 3000 2200 1900 3500 2000 1500 1700 1400 1200 1000 1.37 1,34 1.35 2.40 2.04 1.20 r Table LND STEEL GLES. nON AS TO USK Oi ■ Axis Parai'l to a-b. Parallel to Short I.ogto Loii^ 1-g. , A- . . . - rt C<<'. <\-^. 1 . .\>. ,.^ o » ° 1^ |I5'^ "So. 2 S J:.5'e Transver.ie Value ((■> in lbs. I'fc C « 2 For Iron. For Steel. 9.72 4.80 2. .77 .96 20880 26100 6.70 0.71 G.82 4.88 2. .85 .82 15680 19G00 4.45 0.72 8.25 4.08 2. 1.17 1.13 25680 32100 8.35 0.86 6.87 4.24 2. 1.19 1 . 03 21600 27000 6.75 0.90 4.07 4.33 2. 1.32 .91 14000 1 7500 3.60 0.88 7 3.42 2. 1.19 1.17 25360 31700 7.46 0.83 6.75 3.53 2. '1.25 1.08 21450 26800 5.72 0.83 3.83 3.70 1.1. 34 .96 14720 18400 3.55 0.85 3.40 3.72 2 1.37 .97 13200 16500 2.89 . 79 ■ 6. Id 3.46 2. .85 1.01 18880 23600 5. 75 0.66 4.!)2 3.57 2. .92 .90 14480 18100 3.78 0.67 3.24 3.72 2. 1 .82 10100 12600 2.11 0.62 4.09 2.99 1.: .94 .91 14000 1 7500 3.35 0.64 2.75 3.13 i.a.io .82 9760 12200 2.14 0.66 4.G8 2.28 1. 1.28 1.25 24000 30000 6.10 0.74 3.49 2.40 l.<1.37 1.11 18480 23100 3.93 0.74 2.3;! 2.53 l.il.44 1.04 12400 15500 2.20 0.69 3.82 2.39 1. .93 .99 14880 18600 3.72 0.64 2.30 2.55 1 . 1 1 . 05 .86 9680 .12100 1.96 0.64 2.03 2.59 1 . . 1 . 06 .79 8560 10700 1.94 0.71 3.72 2.40 l.f .64 .84 11040 13800 2.58 0.48 3.32 2.43 l.f .66 .81 10000 12500 2.11 0.45 1.88 2.56 l.j .72 .74 6160 7700 .97 0.41 3.81 2.10 l.< .77 .94 14800 18500 2.54 0.48 2.6G 1.96 1.; .67 .83 9600 12000 2.45 0.48 1.83 2.05 1.- .74 .74 7040 8800 1.25 0.47 3.08 1.66 1-1.21 1.16 20000 25000 2.80 0.57 1.99 1.49 1 . : 1 . 06 .99 12480 15600 2.29 0.58 1.43 1.56 i.n.io .89 9120 11400 1.61 0.61 3 1.69 l.'^ .86 .99 14400 18000 2 . 05 0.40 1.97 1 54 1.; .74 .86 9120 11400 1.56 0.42 1 .23 1.61 1.'- .78 .76 5856 7320 . 77 0.37 2.40 1.28 1.1 .88 1 . 04 14400 18000 1.86 0.39 1.48 1.14 1.1 .76 .88 8640 10800 1.35 0.40 .74 1.23 l.< .83 .79 4720 5900 .61 0.39 1.35 1.21 1.1 .52 .72 5760 7200 .80 0.28 .79 1.25 1.] .53 .62 3440 4300 .43 0.29 .G9 1.26 1..^ .14 .32 1150 1440 .16 0.134 1.30 1.14 1.1 .36 .61 4480 5600 .80 0.31 .G3 1.08 1.1 .32 .48 2080 2600 .41 0.32 1.02 .84 l.< .52 .77 5680 7100 .91 0.33 .56 .89 .S .57 .66 3220 4020 .36 0.28 .87 .86 1 . ( .28 .58 2920 3650 .47 0.21 .73 .86 l.( .31 .53 2800 3500 .39 0.22 .48 .93 A .U .48 1840 2300 .24 0.23 .58 .55 .>■ .31 .62 3100 3860 .37 0.18 .4] .66 .7 .35 .54 2030 2530 . 20 0.18 .31 .40 .!■ .16 .44 1350 1680 .15 0.12 .23 .50 .7 .18 .38 920 1150 . 08 0.12 .3G .42 .7 .17 .44 1520 1900 .11 0.095 • . 2;! .45 .C .15 . 3 7 960 1200 .08 0.106 .21 .35 .7 .17 . 32 712 890 .08 . nX4 .IG .40 .6 .15 .31 GOO 750 .05 0.096 Table XXIII. LIST OF IRON AND STEEL UNEVEN-LEGGED ANGLES. (For iNFORMATrox as to Usr of this Tadls, asB Taulk XIX.) XXIV. VND STEEL TEES. E OF THIS Table, ske Table XIX.) to Web. /v\ ,^JL^j n gpo 1.32 1.42 1.10 .(i4 .(;2 .43 .44 .48 1.14 .77 2.46 1.93 1.48 1.10 .76 .49 .28 1.51 1.12 1.14 .77 1 . 54 1.12 1.10 .80 .83 .82 .58 .52 .53 .31 .32 .12 .17 .40 .30 .86 .71 .61 .57 .58 .10 .47 .99 1.08 1.05 73 . 77 .61 .58 .66 1.13 .76 1.57 1.37 1.18 1.09 .80 .62 .46 1.24 1.06 1.03 .85 1 . 35 1.15 1.10 .93 .89 .84 .82 .70 .69 .54 .52 .40 .37 .75 .66 .96 .84 .75 .76 .74 .30 .66 Transverse Value (i') in lbs. For Iron. For Steel. 16560 16640 17360 8800 8880 6400 5680 6480 17440 6880 24400 19840 15760 15520 7680 4800 2800 14060 11920 10560 8000 16800 13360 10000 8560 6670 6080 6800 4800 4160 2960 2590 1600 1360 5280 4080 6480 5280 6000 4500 3860 800 2820 20700 20800 21700 110(10 1 1 1 00 8000 7100 8100 21.S00 8600 30500 24800 19700 19400 9600 6000 3500 18700 14900 13200 10000 21000 16700 12500 10700 8340 7600 8500 6000 5200 3700 3240 2000 1700 6600 5100 8100 6600 7500 5630 4830 1000 3520 Axis Parallel to Web. ^\- -¥- A 9.04 5.25 5.31 5.70 5 . 24 5 . 23 4.60 3.94 3.90 2.39 2.70 2.70 2.62 3.23 2 . 30 2.00 1.80 1.80 1.82 1.53 1.60 1 .30 1.40 1 .06 1.15 .97 .75 .94 .89 .74 .85 .68 .57 .56 .63 .62 .50 .50 .58 .49 .40 .33 .26 3.01 2.10 2.12 2.30 2.10 2.09 1.80 1.58 1.70 1.06 1.40 1.40 1 .31 1.61 1.10 1 .91 1 1.04 .87 . 93 .87 .92 .71 .77 .65 .50 .63 .60 .49 .567 .453 .455 .372 .460 .450 .400 .400 .460 .390 .320 . 260 .231 P5" 1.46 1.49 1.49 1.46 1.37 .86 .96 .64 .67 .70 .77 .81 .86 .92 .52 .56 .53 .59 .36 .38 .37 .42 .46 .38 .40 .44 .42 .49 .47 .38 .50 .30 .34 .26 .28 .30 .28 .27 .37 Transverse Value (r) in lbs. For Iron. For Steel 24080 16800 16960 18400 16800 16720 14400 12640 13600 8480 11200 11200 10480 12880 8800 8000 7280 8000 8320 6960 7440 6960 7360 5680 6160 5200 4000 5040 4800 3920 4540 3630 3640 2976 3680 3600 3200 3200 3680 3120 2560 2080 1850 30100 21000 21200 23000 21000 20900 18000 15800 1 7000 10600 14000 14000 13100 16100 lloOO 10000 9100 lOOdO 10400 8700 9300 8700 9200 7100 7700 6500 50O0 6300 6000 4900 5670 4530 4550 3 720 4600 4500 4000 4000 4600 3900 3200 2600 2310 A,C,F 1,B,D,E,K. A,C,E Talm.k .WIV. LIST OF IRON AND STEEL TEES. (Foe Ikfobmation as to thb Usb of this Table, sbk Table XIX.) 2.05 1.75 1.93 1.03 n.53 )1.G4 • 1.63 3 1.50 51.13 J 1.03 9 1.20 U.21 jl.02 J 1.12 50.944 0.440 0.376 » Normal to Web. 2.08 2.17 l.IO 1.11 3.64 3.20 2.14 5.55 2. 38 2. OS 1.76 1.73 1.4G 1.51 1.12 2.10 1.88 1.S5 1.80 1.9S 1.73 1.47 l.(ie 1.2G 1.12 1.93 1.48 1. 10 1.51 1.12 1.14 --.& 1.57 1.37 1.18 10560 16G40 1 7360 6800 24400 19840 157GO 15520 7680 4800 2800 14900 1 1 920 10560 8O00 16800 13360 10000 8560 6670 6080 6800 4800 4100 2900 2590 1600 1360 6280 4080 6480 6280 6000 4600 3860 20800 21700 UOOO 11100 HOOO 7100 8100 21800 8600 305O0 24800 19700 19400 9600 6000 3500 18700 14900 13200 10000 21000 16700 12500 10700 8340 8480 10600 2420 3020 1980 2470 1360 1700 430 540 2080 2600 1536 1920 1344 1680 1000 1250 --IH- ih 2.30 2.10 2.09 1.80 1.58 1.70 1.06 1.40 1.40 1.31 1.61 1. 10 24080 16800 1 0900 18400 16800 16720 14400 12640 13600 8430 11200 11200 10480 12880 6800 8000 3010 21000 21200 23000 21000 2O900 18000 16800 17000 10600 14000 14000 13100 16100 llOOO 10000 9100 lOOOO 10400 1700 5680 6160 5200 4000 5040 4800 3920 1540 36S0 3640 3120 2560 2080 1850 1712 1280 1704 1120 1680 1280 1440 7100 7700 6500 5000 6300 6000 4900 6670 4530 4550 3720 4600 4600 4000 4000 4600 3900 3200 2600 2310 2140 1000 2130 1400 2100 1600 1800 1400 1700 12.50 1140 fA r^ University of California SOUTHERN REGIONAL LIBRARY FACILITY 305 De Neve Drive - Parking Lot 17 • Box 951388 LOS ANGELES, CALIFORNIA 90095-1388 Return this material to the library from which it was borrowed. ; SOUTHERN REL,t;S?,t;?ii,!itii{i-|{nii A 000 020 824 9 ir U ^.f * \' ;4 ^ ^^ 5^ '^ ?V