. 4 Franklin Institute Library FHIL/IDELPHIfl Class Book Accession iiisiuuLe, ur luemuciB ui mjiuuio uv.v,v/..v. ^ , obtained the sanction of the Committee. The second class shall include those books intended for circulation. Article VI. — The Secretary shall have authority to loan to Members and to holders of second class stock, any work belonging to the second CLASS, subject to the following regulations : Section 1. — No individual shall be permitted to have more than two ])Ooks out at one time, without a written permission, signed by at least two members of the Library Committc ; nor shall a book be kept out more than two wkeks ; but if no one has applied for it, the former bor- rower may renew the loan. Should any person have applied for it, the latter shall have the preference. Section 2. — A fine of ten cents per week shall bo exacted for the detention of a book beyond the limited time ; and if a book be not re- turned within three months it shall be deemed lost, and the borrower shall, in addition to his fines, forfeit its value. Section 5.— Should any book be returned injured, the borrower shall pay for the injury, or replace the book, as the Library Committee may direct ; and if one or more books, belonging to a set or sets, be lost, the borrower shall replace them or make full restitution. Article VII. — Any person removing from the Hall, without permis- sion from the proper authorities, any book, newspaper or other property in charge of tlie Library Coramittefe, shall be reported to the Committee, who may inflict any fine not exceeding twenty-five dollars. Article VIII. — No member or holder of second class stock, whose annual contribution for the current year shall be uii])aid or who is in arrears for fines, shall be entitled to the privileges of the Library or Reading lioom. Article IX. — If any member or holder of second class stock, shall refuse or neglect to comply with the foregoing rules, it shall be the duty of the Secretary to report him to the Committee on the Library. Ar.TicLE X. — Any Member or holder of second class stock, detected in mutilating the newspapers, pamphlets or books belonging to the Insti- tute shall be deprived of his right of membership, and the name of the offender shall be made public. NOTES ON BUILDING CONSTRUCTION PART IV. CALCULATIONS FOR BUILDING STRUCTURES— COURSE FOR HONOURS NOTES ON BUILDING CONSTRUCTION Arranged to meet the requirements of the Syllabus of the Science and Art Department of the Committee of Council on Education, South Kensington. In Four Parts. Medium %vo. Sold separately. Part I. — First Stage, or Elementary Course. With Illustratiom, los. 6d. Part II. — Second Stage, or Advanced Course. With 479 Illustra- tions. \os. 6d. Part III. — Materials. Course for Honours. With 188 Illustrations. 21s. Part IV. — Calculations for Building Structures. Course for Honours. With 597 Illustrations. Official Report on the Examination in Building Construction, held by the Science AND Art Department, South Kensington, in May 1875.— "The want of a text-book in this subject, arranged in accordance with the pubHshed Syllabus, and therefore limiting the students and teachers to the prescribed course, has lately been well met by a work published by Messrs. Rivingtons, entitled ' Notes on Building Construction.' " "Those things which writers of elementary books generally pass over are here explained with minuteness. . . . Altogether the book is one which it is a pleasure to recommend. Its primary object may be to support the Science and Art Department, but it will be found to be of wider use ; and if the parts which are to follow are prepared as carefully as this is, the ' Notes on Building Construc- tion' will far surpass any work of the kind hitherto published." — Architect. " Something of the sort was verj' much needed. . . . The whole series when published will be a great boon to young students." — Builder. " One of the most sensible and really reliable aids to students of construction we have seen for a long time. If the remaining Parts are up to the standard successfully aimed at in Part I., the work cannot fail to become the standard text-book for students." — Building News. " It very rarely happens that explanations are given with such clearness as those in ' Notes on Building Construction,' and the dullest student cannot fail to grasp the idea intended to be conveyed. ... As a work of reference it will at once take a leading place."— Builder's Weekly Reporter. " Certainly the four parts will, judging from the first, form the besl text-book on the subject ex- tant." — English Mechanic. "The work throughout is got up in the most admirable style, and is profusely illustrated with well-drawn engravings." — Timier Trades' lournal. " The whole will form a compendious series of volumes of very great value to 'practical' men. The text is prepared in an extremely simple and consecutive manner, advancing from rudimental and general statements to those which are comparatively advanced ; it is a thoroughly coherent self- sustained account. . . . We can testify that its contents justify the promises of the title, that we have missed nothing which we looked for" and had a right to expect would be included in the volume." — Athenceimi. LONDON : LONGMANS, GREEN & CO. RIVINGTON'S SERIES OF NOTES ON BUILDING CONSTRUCTION BUILDING CONSTRUCTION ARRANGED TO MEET THE REQUIREMENTS OF THE SYLLABUS OF THE SCIENCE ART DEPARTMENT OF THE COMMITTEE OF COUNCIL ON EDUCATION, SOUTH KENSINGTON PART IV. CALCULATIONS FOR BUILDING STRUCTURES- COURSE FOR HONOURS LONDON LONGMANS, GREEN, AND CO. AND NEW YORK : 15 EAST i6th STREET 1891 N ON A II rights reserved PREFACE TO PART IV. rpHE writer desires to express to the Publishers, and to those who have done him the honour of readinaf the p)revious Parts of this work, his regret for the great delay tJhat has occurred in the preparation of this volume for the Press. More than one serious illness and the pressure of his dluties have obliged him again and again to lay this work emtirely aside. Finding that he was not likely to have Leisure or strength to complete the volume, he has, with t;he consent of the Publishers, entrusted most of it to ai friend, by which arrangement he feels that he has con- fi'erred a great benefit upon his readers. The object in view has been to explain fully and c^learly the simplest and best methods for calculating the p)roper forms and dimensions to be given to Timber and Eron Beams, Cantilevers, Plate, Box, and Trussed Girders, Eloofs, Walls, Arches, Water pipes, etc. etc., showing by eixamples the calculations required even for the details. In carrying out this object the use of Advanced Mathe- rmatics has been as far as possible avoided. The methods e3xplained are chiefly those that can be effected almost e3ntirely by drawing and measuring lines, and the calcu- kations given require but a very slight knowledge of IMathematics. Short rules are added for the use of prac- vi PREFACE tical men, and also tables by which many calculations are altogether avoided. The great advantage of using simple graphic methods instead of elaborate calculations is that the former, if in- correctly carried out, at once proclaim the fact by the polygons of forces refusing to " close," so that an error has to be faced at once and rectified ; whereas an error in a figure, or even of a decimal point, in calculations quite correct in other respects, may escape notice, and lead to an utterly erroneous result, which, if acted upon, may en- danger the stability of the structure to which it is applied. It is hoped that this volume will be found to explain not only the calculations that may be called for in the Honours Examination at South Kensington, but also all that can possibly be required in connection with ordinary buildings. Upon the more complicated engineering struc- tures it does not profess to enter. PREFACE vii NOTE. The following is an extract from the Syllabus of the Science and Art Department of the Committee of Council on Education, South Kensington. It shows the heads of the examination in connection with the calculation of structures for Honours, and opposite to each subject the portion of this volume in which the information required is to be found. FIRST STAGE, OR ELEMENTARY COURSE. No examination as to the calculation of structures. SECOND STAGE, OR ADVANCED COURSE. All that is required Avill be found in Part II. IRequirements of Syllabus. He must he able to solve siimple problems in the theory ojf construction, and to deter- miine the safe dimensions of irron or luooden beams subjected tio dead loads. EXAMINATION FOR HONOURS. Where dealt with in this Volume. Simple problems in the theory of con- struction, equilibrium, chap. ii. ; beams, chap. iii. ; dimensions of timber beams, pp. 54, 75-80 ; wrought iron beams, chap, iv. ; cast iron beams, chap. v. In ordinary roof trusses and firamed structures of a similar dlescription, he must be able to tirace the stresses, brought into aiction by the loads, from the 20i>oints of application to the ptoints of support, as well as too determine the nature and aimount of the stresses on the different members of the truss, amd, consequently, the quantify ojf material required in each poart. Stresses on frames in general, chap, ix.; open-webbed girders, chap. x. ; trussed beams, chap. xi. ; roofs, chap. xii. ; deter- mination of the quantity of material re- quired in each part, i.e. of the dimensions of the parts, tension and compression bars, chap, vi.; joints and connections, chap. vii. viii PREFACE In ordinary ivalls and re- Stability of walls, piers, chimneys taining walls, he must he able (house), and enclosure walls, cha^). xiii. ; to ascertain the conditions retaining walls for water and eartli, chap. necessary to stability, neglect- xiv. ing the strength of the mortar. In addition to the above-mentioned chapters, which give all the information required for the examinations, other chapters have been added upon Plate Girders, Arches, and Hydraulics, in order that the volume may be practically useful as an aid in solving all the calcula- tions that can be required in connection with ordinary Building Construction. NOTATION AND WORKING STRESSES. N.B. — In order to prevent constant repetitions in the body of the work, and to assist the student, the meaning of tlie letters used in the notation, and the working stresses used in the calculations in this book, are given in a collected and complete form in Tables XXV. and XXVI. at the end of the volume (pp. 348-351), so that they can easily be referred to at once without consulting the index. LIST OF BOOKS CONSULTED. Adam's Designing Cast and Wrought Iron Structures. Aide Memoire for the use of Officers of the Eoyal Engineers. Bow's Economics of Construction. Clark's Manual of Rules, Tables, and Data. Cunningham's Applied Mechanics. Fidler's Practical Treatise on Bridge Construction. Hurst's Architectural Surveyor's Hand-Book. Molesworth's Pocket-book of Engineering Eormulse. Proceedings of the Institute of Civil Engineers. ^ Eankine's Civil Engineering. Rankine's Useful Rules and Tables. Ritter's Elementary Theory and Calculation of Iron Bridges and Roofs. Stoney on Strains. Transactions of the Society of Engineers. TJnwin's Elements of Machine Design — Part I. Unwin's Wrought-Iron Bridges and Roofs. AVray's Application of Theory to the Practice of Construction (revised by Seddon). CONTENTS OF PART IV. Chapter I. introductory- Terms IN Use. — Load — Dead Load — Live Loads — Comparative effect of Dead and Live Load — Breaking Load — Working Load or Safe Load — Strain — Stress — Tension — Compression — Transverse Stress — Shearing Stress — Bearing Stress — Torsion — Stresses classified as regards Descrip- tion — Stresses classified as regards Intensity — Limiting Stress — Safe Working Stress — Intensity of Stress — Factor of Safety — Fracture — Table of Stresses and Mode of Fracture — Elasticity — Elastic Limit — Re- peated Stresses — Elongation and Shortening — Modulus of Elasticity — Deflection — Stiffness .... Pages 1-13 Chapter II. EQUILIBRIUM. External Forces acting on a Structure. — Reactions — Rules for find- ing Reactions at Supports- — Examples. Stresses produced by the External Forces . Pages 14-23 Chapter III. BEAMS. Bending Moments. — Rule for finding Bending Moment — Bending Mo- ment under various conditions. Moment of Resistance. — Method of ascertaining by reasoning — by Graphic Method — by Experiment — by Calculation — General Formulae — Practical Formulte — Neutral Layer — Neutral Axis — Equivalent Area of Resistance — Modulus of Rupture. Shearing Stress. — Vertical — Horizontal — Rules for finding amount of Vertical Shearing Stress — Graphic Representations, various cases — Dis- tribution of Shearing Stress. Beams of Uniform Strength. xii CONTENTS Deflection. — Formulae — Simpler Formulee. Fixed Beams. — Various Cases — Continuous Girders. Examples. — Timber Cantilevers — Joist — Useful Notes . Pages 24-80 Chapter IV. WROUGHT IRON ROLLED BEAMS. Finding Moment of Resistance — Graphic Method — Mathematical Method — Moment of Inertia — Practical Formulae — Examples — Remarks on Rolled Beams — Girders built up with the aid of X Beams. Pages 81-93 Chapter V. CAST IRON GIRDERS. Finding Moment of Resistance — Approximate Formulae — Graphic Method — Mathematical Method — Practical Points connected with the form of Cast Iron Girders — Examples — Cantilever — Girder . Pages 94-104 Chapter VI. TENSION AND COMPRESSION BARS. Tension Bars. — Symmetrical Stress — Effective Area — Unsymmetrical Stress — Examples. Compression Bars or Struts. — Fidler's Rules for Struts — Long and Short Compression Bars— Gordon's Formula — Formulae and Table for Wooden Struts — Examples — Cast Iron Columns — Timber Struts — Wrought Iron Strut — X Iron Principal . . Pages 105-121 Chapter VII. JOINTS AND CONNECTIONS. Points to be observed in making Joints. Riveted Joints. — Riveted Joints in Tension — Lap Joint — Butt Joint with Single Cover Plate— Butt Joint with Double Cover Plate — General Formulae for Riveted Joints other than Grouped Joints — Formulae for Grouped Joints — Examples — Oblique Riveted Joints — Riveted Joints in Compression — Dimensions for Riveted Joints — Note upon Figured and actual Sizes of Rivets and Rivet Holes. Pin Joints. — Example — Joint in Tie-rod Iron Roof — Joint at Apex and Feet of Truss. Screws. — Practical Formulae — Example — Tie Rod. CONTENTS xm CoTTKR Joints — Practical Formulae — Example. Joints in Wooden Structures.— i;xamj9Zes— Fished Joint— Scarfed Joint— Joint at foot of Principal Kafter of a Roof . Pages 122-153 Chapter VIII. PLATE GIRDERS. General Form — Practical points to be attended to. Flanges.— Graphic Method of finding length of Plates— Example. Web.— Shearing Stress— Example— Joints in the Web— Example— Con- nection of the Web with the Flanges— Example. Example op Plate Girder with Single Web. Example of Box Girder. Hog-backed Girders . - ■ • Pages 154-176 Chapter IX. BRACED OR FRAMED STRUCTURES. General Remarks.— Simple Frames— Triangle of Forces— Polygon of Forces— Different Forms of Framed Structures— Perfectly Braced Frame — Overbraced Frame. Maxwell's Diagrams— Method of Drawing— Bow's System of Notation. Method of Sections.— Rule— Definitions . • Pages 177-186 Chapter X. BRACED OR OPEN WEBBED GIRDERS. Warren Girders. Lattice Girders. Whipple-Murphy Girder.— Bow-string Girder. Examples.— Practical Formula— Graphic Method . Pages 187-194 i Chapter XI. TRUSSED BEAMS. ExAMVh-E— Graphic Method— Method of Sections . . Pages 195-198 Chapter XII. BOOFS. Loads to be borne by Roofs.— Permanent Load — Occasional Loads — Snow — Wind Pressure — Stresses in Roofs of Special Shape — in Ordinary Roofs. xiv CONTENTS Distribution of the Loads.— Example— J^eaciions at AUtMs—Wmdi on one side of Eoof. Example— Iron Eoof Truss.— Finding Stresses by Maxwell Diagrams — Calculation of PrinciiDal when carrying a number of Smal Purlins — as Jointed — as Continued. Diagrams for various Eoof Trusses . , Pages 199-217 Chapter XIII. STABILITY OF BEICKWOEK AND MASONEY STEUCTUEES. Single Block — Centre of Pressure — Distribution of Pressure. Uncemented Block Structures.— Conditions of Stability— ^mm_?jfe— Brick Pier — House Cbimney — Enclosure Walls . Pages 218-229 Chapter XIV. EETAINING WALLS. Usual Cross Sections for Eetaining Walls. Eetaining Walls for Water.— Example by Graphic Method— Practical Eules. Eetaining Walls for Earth. — Mathematical Formuh^ — Graphic Method— Practical Formulae— Examples— Baker's Practical Eules— Foundations ..... Pages 230-245 Chapter XV. AECHES. Graphic Method of Determining the Stability of an Arch Line of Eesistance— Moseley's Principle of Least Eesistance— Thickness of Arch Eing — Abutment — Table of Thickness of Arches. Example ...... Pages 246-259 » Chapter XVI. HYDEAULICS. Definitions.— Pressure— Head of Pressure and Elevation— Loss of Head — Wetted Perimeter — Hydraulic Mean Depth. Motion op Liquids in Pipes. — Discharge from a Pipe running full Formula— Examples — Practical Formulaj — Losses of Head — Due to Orifice of Entry — Due to Velocity— Due to Bends and Elbows — Ex- amples — Water Supply to a House — Intermittent — Constant — Examples CONTENTS XV — Ditto to a Street — Examples — Discharge from a Pipe running partially full — Formulse — Example — Drain Pipes — Example of Simple System of Drains — Egg-shaped Sewer — Example. Jets. — Height and Eange — Nozzles — Examples . . Pages 260-294 APPENDICES. NUMBER PAGE I. Factors of Safety ......... 295 II. Note on Equilibrium . . . . . . . 296 III. To Draw a Parabola . . . . . , . 297 IV. To show that the Neutral Axis in a Beam passes through the Centre of Gravity of the Cross section . . . 298 V. Shearing Stress Diagrams . . . . . . 298 VI. Graphic Method of finding Stresses in Beam . . . 299 VII. Stresses upon Beams with various Distributions of Load . 302 VIII. Comparison of Strength and Stiffness of Supported and Fixed Beams . . . . . . • • 303 IX, Deflection of Rectangular Beams ..... 304 X. Strength of Beams of Different Sections . . . . 304 XI. Continuous Beams, Eeactions and Bending Moments . . 305 XII. Centre of Gravity 308 XIII. Graphic Method of finding Centre of Gravity ... 309 XIV. Moment of Inertia of various Sections .... 310 XV. Formulse for finding approximately the Weight of Girders 311 XVI. Rules for Drawing Maxwell's Diagrams . . . . 311 XVII. Bow's System of Lettering Maxwell's Diagrams . , 313 XVIII. Calculation of Stresses on Iron Roof by the Method of Sections ......... 314 XIX. Centre of Pressure in Masonry Joints . . . . 321 XX. Arch with Unsymmetrical Load ..... 322 XXI. Formulse to be remembered . . . . . . 324 XVI CONTENTS TABLES. NUMBER PAGE A. Factors of Safety ......... 9 B. Stresses and Modes of Fracture ...... 9 C. Factors for Deflection of Beams — given Load . . . . 67 D. „ „ „ — given Limiting Stress . . 68 E. Distance of Points of Contra-flexure . . . . . 70 F. Fixed Beams — Bending Moments for various Cases . . . 74 G. Gordon's Formula for Long Columns — Values of a . . . 114 H. Iron Roof — Tabulation of Stresses . . . . . . 213 J. Flovp- of Liquids in Pipes — Darcy's Formula — Value of C . . 265 L Ultimate Resistance of Materials . . . . . 327 la. Safe Resistance of Materials . . . . . . 328 II. Safe Loads for Rectangular Beams of Northern Pine or Baltic Fir 329 III. Deflection of Rectangular Beams of Northern Pine or Baltic Fir 330 IV. Deflection of Wrought Iron Rolled Beams . . . . 331 V. Strength of Struts (Wrought Iron, Cast Iron, and Steel) . 332 VL „ Wooden Struts 334 VIL „ Timber Columns 334 VIIL „ Rivets 335 IX. Dimensions of Eyes of Wrought Iron Tension Bars . . 335 X. Weight of Angle Iron and Tee Iron . . . . 336 XL Web Plates— Thickness of 337 XII. Weight of Roof Framing, ceilings, etc. . . . . 337 XIII. „ Coverings 339 XIV. Wind Pressure 339 XV. Scantlings for Wooden Roofs 340 XVa. Coefiicients of Friction . . . . . . . 341 XVI. Retaining Walls (Angle of Repose of Earths and Value of K) 342 XVII. Weights of Earths, Stone, etc 342 XVIIa. Thickness required for Arches . . . . . . 343 XVIII. Flow of Water in Pipes running full (gallons per min.) . 343 XIX. „ „ „ (cubic feet per min.) . 344 XX. Cast Iron Pipes — Thickness, Weight, and Strength . . 345 CONTENTS xvii NUMBER PAGE XXI. Loss of Head due to Bends ...... 346 XXII. „ „ Elbows 346 XXIII. Pipes flowing partially full 347 XXIV. Jets — Factors to find Initial Velocity . . . . 347 Notation and Working Stresses. XXV. Notation employed in this Volume ..... 348 XXVI. Working Stresses „ ...... 350 EXAMPLES. 1. External Forces acting on a Girder . . . . . 21 2. External Forces acting on a Cantilever . . . . . 22 3. „ a Roof 23 4. Bending Moment for a Beam . . . . . . 27 5. Timber Cantilever for Balcony, Uniform Cross Section . . 76 5a. ,, ,, „ Strength and Width . 77 6. Timber Beam Loaded in Centre . : . . . . 78 7. Fir Joist for Given Load 79 8. Safe Distributed Load for an I Wrought Iron Joist of given Dimensions 8 8 9. „ „ „ „ weight, etc. 89 Qa. „ „ „ section 89 10. IE Joist required for a given Distributed Load ... 89 11. Cast Iron Cantilever . . . . . . . . 101 11a. „ Girder ... 103 12. Eectangular Tension Bar . . •. . . . . 108 13. Tension Rod - . . . 108 14. Rectangular Timber Strut . . . . . . . 115 15. Cast Iron Column . . . . . . . . 116 16. „ „ 117 17. Timber Strut . 118 18. „ „ .118 19. Wrought Iron Strut 119 20. T Iron Principal of a Roof 120 21. Lap Joint (Riveted) 124 22. Double-Cover Riveted Joint 128 23. „ Grouped Joint : . 131 24. „ „ for Plate Girder . . . . 132 h xviii CONTENTS NUMBER PAGE 25. Oblique Riveted Joint 134 26. Pin Joint 140 26a. Joints at Apex and Feet of Roof Truss . . . • 144 27. Screw of Tie-Rod 147 28. Cotter Joint 1^8 29. " Fished " Joint for Wooden Beam 149 30. "Scarfed" „ „ 151 31. Joint at foot of Principal Rafter (Wood) 152 32. Flanges for Plate Girder of 20 feet Span .... 156 33. Web for Plate Girder 159 34. Joint in the Web of a Plate Girder 161 35. Pitch of Rivets to connect the Web to the Flanges . . 161 36. Plate Girder Avith Single Web 162 37. Box Girder I'J'O 38. Stresses in Warren Girder . . . . • • • 187 38a. „ „ hy Graphic Method . . . 194 39. „ Whipple- Murphy Girder 190 39a. „ „ „ M Graphic Method . . 194 40. Trussed AVooden Beam . ...... 196 41. Loads on an Iron Roof .... ... 204 42. Stresses in an Iron Roof Truss 209 43. Pressure on a Masonry Joint . . . • • • 219 44. Brick Pier 222 45. Chimney 225 46. Enclosure Wall 227 47. Reservoir Wall 234 48. Retaining Wall for Light Vegetable Earth .... 241 49. „ ,, London Clay 243 50. Surcharged Retaining Wall, Loamy Earth .... 243 50a. Foundations 244 51. Semicircular Brick Arch 254 51a. Abutment for Arch 259 52. Velocity of Water flowing through a Pipe . . . . 265 53. „ „ „ „ .... 266 54. „ „ „ „ .... 266 55. „ „ „ „ .... 266 56. Loss of Head due to Orifice of Entry 269 57. „ „ Velocity 270 CONTENTS xix NUMBEIl p^^Qj, 58. Loss of Head due to a Bend 270 59. „ an Elbow ...... 270 60. „ the Resistance of the Pipe . . . 271 61. Water Supply for a House, Intermittent Service . . . 271 62. „ „ Constant Service . . . . 274 63. "Water Supply to eight Houses in a Street . . . . 277 64. Velocity of Sewage in a Drain Pipe 282 65. Simple System of Drains . . . . . . , 283 66. Egg-shaped Sewer . . 9 3g 67. Jet of Water 289 68. Vertical Jet of Water, effect of Nozzle . . . ■. . 291 69. Water Supply for House and Jet 292 IN APPENDICES. 70. Graphic Method of Finding Stresses in Loaded Beam . . 300 71. Drawing Maxwell's Diagrams 312 72. Bow's System of Lettering Maxwell's Diagrams . . . 313 73. Stresses on Roof found by Method of Sections . . . 314 Chapter I. INTRODUCTORY. THE object of this Part of the Course is to teach a student how to design those parts of buildings which require to be carefully proportioned, in order that the cost of material and labour may be a minimum, consistently with the structure being of sufficient strength and permanence ; but it does not deal in any way with the question of artistic design or appearance. It is assumed that the student is familiar with Parts I., II., and III. of this Course, in fact, that he understands the usual forms of parts of ordinary buildings, and the nature of the various materials in use for the construction of buildings. In designing any structure care should be taken that each jpart is strong enough to resist the forces that may act upon it. To insure this result it is necessary to know — 1. The nature and direction of the stresses ^ that each part may be called upon to resist. In ordinary buildings these stresses are produced by : (a) The force of gravity {i.e. the weight of the parts of the structure and of any load they may have to carry). (6) The force of the wind. 2. The most suitable description of material. 3. The lest form and dimensions to he given to each part in order that it may be able to resist the stress that may act upon it. When the student has ascertained these particulars, and de- signed the structure accordingly, he may be sure not only that each part will be strong enough, but that no more material will be used than is necessary to give the strength required ; in fact, that the materials will be used without waste. It should, however, be remembered that in practice cases frequently occur, especially where the parts are small, in which the minimum dimensions that can practically be allowed are greater than the maximum required by theory. ^ See p. 6 for definition of this term. B.C. IV. B 2 NOTES ON BUILDING CONSTRUCTION It is of course easy in many cases to make a structure strong enough by using plenty of material; but an engineer or archi- tect who understands his work will use only sufticient material to make sure of a safe and permanent structure, taking care to dispose it so as to obtain a maximum of strength at a mini- mum cost. The calculations required for designing the different parts of buildings are not difficult, and they require very little knowledge of mathematics. In many books on the subject formulae are given without any explanation as to how they are obtained, and are therefore rather bewildering, and tend to wrap the subject in a veil of mystery which does not properly belong to it. An attempt will be made in this chapter to explain the principles upon which the calculations depend, and then, in subsequent chapters, to show how these principles are applied to ordinary cases which occur in practice. If the student under- stands the principles, he will always be able to apply them to any unusual case v/hich may arise in his own experience. TEEMS IN USE. In order to clear the ground, it is desirable 'first to explain the meanings of the various terms that have to be used in con- sidering the subject. Many of these definitions have been given in Part III., but they are now repeated in a somewhat different form. Load The external forces that act upon any structure, tofyether with the weight of the structure itself, are called the "load." Thus in the case of a beam AB (Fig. i) supported at the ends, and carrying a load W at its middle point, the external forces acting are (1) the weight of the beam itself, (2) the weight of the load. The direction of these forces is shown by the arrows. In Fig. 2 the beam is uniformly TERMS IN USE 3 loaded througliout part of its length, the forces acting being (1) its own weight, w w tv, etc. (2) the weight of the tank and its contents, ttt, etc. On an inclined rafter (Fig. 3) there may be the weight of the rafter and roof-covering acting vertically, and the force of the wind acting normally, i.e. at right angles to the slope (air, being a fluid, exerts a normal pressure). Fig. 4 shows two men carrying a ladder with a boy sitting upon it, the load they have to bear is (1) the weight of the ladder, (2) the weight of the boy. Again, in the case of a column carrying a girder (Fig. 5) which supports a building, the weight (10 tons) on the head of the column is the load to which, it is subjected. The load on the foundation of the column, will be 10 tons + the weight of the column. Or in the case of a wire from which a lamp is suspended (Fig. 6), th.e weight of the lamp, say 50 lbs., is the load upon the wire. Tlie load in each case is the total of the external forces, i.e. Fig. 4. of the weight of beam + the weight of tank and its contents ; the weight of rafter + force of wind, etc. The weight of the structure itself is in some cases small com- pared with the load upon it and can in practice be ignored, as for instance that of the beam supporting a tank of water, or the wire supporting the lamp. In others the weight of the structure itself is important, as in the case of the ladder, which forms a consider- able part of the total load. Distribution of Load. — The load may be concentrated at the centre of the beam (Fig. i), or concentrated at any point (Fig. 18), 4 NOTES ON BUILDING CONSTRUCTION or uniformly distributed over the whole of the beam (Fig. i6), or over a ^portion (Figs. 2 and 20). Dead Load is that which is very gradually applied, and which remains steady. Thus the water might be poured into the tank very gradually, and when once poured in would remain quiet, forming a dead load. In the same way the weight quietly placed and remaining steadily on the column, and the lamp on the rod, would be dead loads. Of course the weight of a structure itself is a dead load. yri \ ■•" ;___,,_,,.C__J Fig. 6. Fig. 5. *LiVE Loads are those which are suddenly applied or are ac- companied by shocks or vibration. Thus the boy might jump suddenly upon the ladder (Fig. 4), causing a shock. An excited crowd upon a balcony or floor might make sudden and fitful movements causing jars and vibration — in either case the load would be called a " live load," not a dead or steady load. There are but few cases in which the parts of ordinary buildings are subject to live loads — the chief instance being in floors subject, as before mentioned, to suddenly moving loads. In engineering structures, however, railway bridges are subject to the live loads caused by swiftly passing trains, breakwaters and sea walls are liable to the sudden impact of the waves ; and so in parts of TERMS IN USE 5 machines, and in other cases, the loads acting upon a structure are often suddenly applied. Comparative Effect of Dead and Live Loads. — It has been found by experiment that a live load produces nearly twice the effect that a dead load of the same weight would produce. Therefore to find the dead load which would produce the same effect as a given live load, the latter must be multiplied by 2. This operation is called converting the live load into an equivalent dead load. Thus a floor girder may weigh 30 lbs. per foot run ; if the load upon it of 250 lbs. per foot run be a live load, the total equivalent dead load will be (30 lbs. + 2 X 250 lbs.) = 530 lbs. per foot run. The Breaking Load for any structure or piece of material is that dead load which will just produce fracture in the structure or material. The Working Load or Safe Load is the greatest dead load which the structure or material can safely be permitted to bear in practice. It may be useful to know the load that would cause rupture in the structure, i.e. the breaking load, but the load that should actually be applied to it, i.e. the working load, must be so much smaller as to put all danger of rupture out of the question. The breaking load or the working load may be either live load or dead load, or a combination of both ; but for convenience it is usual to reduce it all to an equivalent dead load, by doubling the live load and adding it to the dead load, as in the example given above. Strain is the alteration in the shape of a body produced by a stress. Thus a stress of tension, or tensile stress, produces a stretching or strain of elongation, a compressive stress leads to a shortening or squeezing strain, a transverse stress to a bending strain, and so on. At one time the word " strain " was generally used instead of the word "stress" to denote the forces of tension, compression, etc., and it is still so used in many treatises on applied mechanics ; but under the high authority and guidance of the late Professor Rankine, the best writers on the subject use the word " stress " to signify the forces acting upon a body,^ and the word "strain" to mean the alteration of figure that takes place under the action of those stresses. ^ "The word 'stress' has been adopted as a general term to comprehend various forces which are exerted between contiguous bodies or parts of bodies, and which are distributed over the surface of contact of the masses between which they act." — Civil Engineering, by Professor Rankine, p. 161. 6 NOTES ON BUILDING CONSTRUCTION It should be noticed that this use of the word " strain " is different from the more usual one, in which it is implied that some injury has been done to the material of the body which has been strained. Stress. — When a force acts upon a structure or piece of material it produces an alteration of form or strain which may be merely temporary, lasting only while the force is applied, or it may be permanent, or may eventually end in rupture. This alteration in form calls out a resistance in the internal structure of the material which is called the stress upon the body.-^ Thus the weight of the lamp (Fig. 6) tends to alter the form of the rod which supports it by making it elongate, tending to tear it in two ; but the strength of the fibres composing the rod resists this tendency to alter its form ; and it is subject to a stress in the direction of its length. ^ Again, the weight on the girder in Fig. 5 tends to shorten the column, and would, if sufficiently great, crush it. The weight (Fig. i) tends to bend the beam, and would, if sufficiently great, break it (Fig. 31). The tendency, however, in each case is resisted by the internal strength of the body. The various kinds of stress are as follows : — Tension. — This stress elongates the body upon which it acts, and tends to cause rupture by tearing asunder. There are many parts of ordinary structures subject to a tensile stress ; e.g. the tie beam TT of a roof (Fig. 7) is in tension, because the rafters tend to spread out at the feet and thus pull uj)on the ends of the tie beam. CoMPEESSiON. — This stress shortens the body upon which it acts and tends to cause rupture by crushing. Many parts of ordinary buildings constantly undergo compression, for example, columns and story posts, also the struts of roofs, which are com- pressed in the direction of their length by the weight of the roof that they support. ^ Theoretically the smallest force acting upon a body produces a permanent altera- tion in its form (see p. 10). Practically, however, the permanent change of form produced by forces which are very small in proportion to the strength of the struc- ture may be ignored. 2 Strictly, the stress in the rod increases towards the supporting beam, owing to the weight of the rod itself, and in the same way the stress in the column is greater near the ground than at the top, on account of the weight of the column itself, but the increase is so small that it may practically be neglected. Fig. 7. TERMS IN USE 7 TiRANSVEESE STRESS. — This stress bends the body on which it acts and tends to break it across. Instances of this stress in connection witli buildings will at once occur to the student, e.g. in lintels or bressumers carrying walls, joists and girders of floors, etc., tlie rafters of a roof under the influence of wind and load. Shearing Stress is that produced when one part of a body is forcibly pressed or pulled so as to tend to make it slide over another part. As, for example, when two plates riveted together (Fig. 8), are severed A Fig. 9. by pulling or pushing in opposite directions, the rivet r is sheared — one plate sliding upon the other (Fig. 9). Bearing Stress is that which occurs when one body presses against another, so as to tend to produce indentation or cutting.^ As, for example, when a rivet holding a plate cuts into the plate, making the hole larger: thus in Fig. 10 the plates A, B, being pulled in opposite Tiff. 10. Fig. 11. directions, the rivet c, being of harder iron than the plate B, has borne upon it and cut into it, making the hole larger as shown at d, Fig. 11. Torsion is the stress produced by twisting the parts of a body in opposite directions, or, what comes to the same thing, by fixing one part and twisting another around the axis of the fixed part.^ This stress is common in machinery, but not in ordinary building struc- tures, and it will therefore not be further considered. Stresses classified as regards Description. — Summing up therefore, it appears that there are six different kinds of stress, namely — Tensile Shearing Transverse Compressive Bearing Torsional ^ Bearing stress resolves itself into stresses of compression and shear, i.e., in Fig. 1 1, compression of the metal in front of the bolt, and shearing of the metal at the sides. 2 It can be shown that this stress resolves itself into tension, compression, and shear. 8 NOTES ON BUILDING CONSTRUCTION The three latter can, however, be resolved into combinations of the three former. The various stresses are also classified as follows : — Stresses classified as regards Intensity. — Beeaking Stkess is that stress at which a material will just give way. Limiting Steess is the term applied to a stress which is pur- posely limited to less than the breaking stress of the material. Thus it may be known that a bar of iron will bear a stress of 20 tons per square inch before breaking, but yet it may be determined, in order to be quite sure that the bar will be safe, not to load it with more than 5 tons per square inch. In this case, 5 tons would be the limiting stress ; any less stress might be applied, but not a greater stress than the limiting stress, i.e. 5 tons per square inch. The limiting stress should in all cases be less than the elastic limit of the material (see p. 10). Safe Woeking Steess is the stress that may in practice be safely allowed upon the parts of a structure. The amount of this stress depends upon the material, the nature of the stress, or the nature of load, whether live or dead, to which the structure is subjected. It is ascertained by dividing the breaking stress by a factor of safety, which varies under the several conditions mentioned above, and is determined by experience (see p. 9 and Appendix I.) Intensity of Stress is the amount of the stress, expressed in units of weight (such as tons), divided by the area of the surface over which it acts, expressed in units of area (such as inches); provided of course that the stress is uniformly distributed over the surface. If it is not uniformly distributed, the intensity of stress will of necessity be greater at some points than at others. It is generally expressed in tons per square inch, sometimes in lbs. per square inch. Thus in the case of the lamp hanging by a rod, or rather wire (p. 4), the stress is uniformly distributed over the cross section of the rod. Assuming that this cross section has an area of -wi-jth. square inch, T . c . weight of lamp 50 lbs. Intensity of stress = ^ — = — area of cross section -^^-^ square inch = 50 X 224 lbs. per square inch, or 5 tons per square inch. Again, if a bar 2 inches square, that is, with a sectional area of 4 square inches, has 40 tons hanging from it, the total tensile stress on the bar is 40 tons, but the intensity of stress is 10 tons per square inch. The cast iron column shown in Fig. 5 may be taken as another example. The load is lO'O tons, and assuming that the area of the cross section is 8 square inches, the average intensity of stress is ^= 1-25 ton per square inch.^ ^ As will be explained in Chapter VI. , unless the column is very short, it cannot be assumed that the stress is uniformly distributed over the cross section, and there- fore the intensity of stress will vary from point to point, and the maximum intensity may be far greater than the average. TERMS IN USE 9 The Factor of Safety is the ratio in which the breaking load exceeds the working load. This ratio depends upon the nature of the load and that of the material, and it is found by experience. The following Table shows the Factors of Safety given by Professor Kankine in bis Useful Mules and Tables. TABLE A. Factors of Safety. Dead Load. Live Load. For perfect materials and work- manship ..... For good ordinary ( Metals . materials and I Timber workmanship ( Masonry 2 3 4 to 5 4 4 6 8 to 10 8 It will be seen that, for the reasons given above, the factor of safety for a live load is taken at double that for a dead load. When a load is mixed, i.e. partly live and partly dead, the live portion may be converted into an equivalent amount of dead load, and the factor of safety for dead load then applied to the whole. The factors of safety shown in Table A are lower than can be safely used for the work ordinarily met with. Factors of safety which have been used in practice for different special structures are given in Appendix I. Fracture. — When a body is subjected to a stress, and the stress at any point is greater than the material can withstand, fracture ensues. The nature of the fracture depends of course upon the kind of stress to which the body has been subjected. Thus a tensile stress carried far enough produces fracture by tearing, a compressive stress by crushing, a transverse stress by cross breaking, etc. Table of Stresses and Strains and Modes of Fracture. The following Table shows the various stresses and the strains, and description of fracture to which they lead when carried suffi-ciently far. TABLE B. stresses. Strains. Modes of Fracture. Tensile or pulling . Compressive or thrusting Transverse .... Shearing .... Bearing ..... Torsional .... Extension . Compression Bending Distortion . Indentation . Twisting Tearing. Crushing. Breaking across. Cutting asunder. Cutting and crushing. "Wrenching asunder. lo NOTES ON BUILDING CONSTRUCTION Elasticity is the property which all bodies have (in a greater or less degree of perfection) of returning to their original figure after being strained by the application of any kind of stress.-^ When the original figure is completely and quickly recove:ed, the elasticity is said to be perfect. Thus if an ivory billiard ball be dropped upon a hard smooth sbne, smeared with black, a black spot of considerable size will be found on the ball, showing that the ball was to some extent flattened at the momer.t of impact. Upon examination of the ball, however, it will be found that all flataess has disappeared, showing that it has returned to its spherical shape ; oj, at any rate, that no deviation from that shape can be detected by the eye o: by ordinary modes of measurement. When the original figure is not completely recovered, but remains per- manently altered, the elasticity is said to be imperfect, and the disfigvirenent produced is called a permanent set or set. Although there is reason to believe that the elasticity of every solid is imperfect, so that the slightest strain produces a set, the elasticity of most building materials is practically perfect, up to a certain point; that is to say, that stresses up to a certain limit can be applied and removed, and the resulting disfigurement, or change of figure, is only temporary; there is no appreciable permanent set — that is, no set that can be measured with any ordinary instruments. Stresses above this limit, however, cause permanent sets. The Elastic Limit of any material is the maximum stress per square inch that can be applied to it without causing any appre- ciable permanent set. An illustration will perhaps make the meaning of the preceding terms more clear than the definitions alone would do. If a weight be suspended from a rod so as to cause a tensile stress in the direction of its length, the rod will at once be elongated. It will stretch a certain proportion of its own length. This proportion will vary according to the description and quality of the material and to the amount of weight applied. Taking the case of wrought iron as an illustration : If a weight of 1 ton be hung from the end of a wrought-iron rod of aver- age quality, having a cross-sectional area of 1 square inch, the bar will stretch about x^^Fo^^ P^^^ original length. If the weight be removed, the bar will immediately recover itself, that is, it will return to its original length. If measured by any ordinary means of measurement, it will be found to be of the same length that it was before the weight was applied. ^ This must not be confounded with the popular meaning of the word "elasticity," i. e. the property of being easily stretched. TERMS IN USE II This recovery of the bar occurs, however, only up to a certain point. If the load amounts to a considerable proportion of the breaking weight, the result produced is very different. For example, if instead of 1 ton, a weight of 1 2 tons be applied to the bar just mentioned, the iron will stretch about xoV (1*1^ length. Upon removal of the weight, however, it will not entirely recover itself, and upon measurement it will be found to be a minute portion longer than it was before. This slight increase in the length of the bar is called the permanent set. It is evident, then, that there is a very important line to be drawn — on one side of it are the stresses the application of which will produce no appre- ciable permanent set, on the other side are the stresses which do produce an appreciable permanent set. This line of demarcation is the Elastic Limit or limit of elasticity, and its value is generally stated in lbs. or tons per square inch, and it is important that materials should not be subjected to stresses exceeding the elastic limit. The above remarks have been made with regard to a tensile stress, but the same thing occurs with a body under compression — weights placed upon the end of a strut produce no permanent shortening, or set, up to a certain limit ; weights greater than this permanently shorten the strxit to an appreciable extent. This point is called, as before, the elastic limit or limit of elasticity. Repeated Stresses — Variations of Stress in amount and kind. — It was shown long ago by experiment that a stress not much exceeding half the breaking stress for iron or steel would cause rupture after being applied and removed a great number of times. This used to be explained by pointing out that if the stress exceeded the elastic limit each application of the load would cause a slight increase in the permanent set, i.e., the length of the bar would be slightly but permanently increased each time the load was applied until at last rupture would take place. More recent experiments on iron and steel have shown, however, that the action of repeated stresses is not so simple. Wohler has shown that if a bar be subjected to repeated stresses subject to variation in amount or in kind, the load nuist be much less than that which the bar will bear when it is steadily applied. Thus if t will just break a bar when steadily applied in tension, a tensile stress varying from 0 to | will break it if applied thousands of times, and | will break it if repeatedly applied as compression and tension alternately. The cases in which repeated stresses vary in kind are rare and unimportant in building construction. They do, however, frequently vary in amount, and the abov-e shows that the working stress for repeated loads should on no accoimt exceed half the breaking stress — in practice it is much less. Elongation and Shoetening. — Another point to be noticed is that in wrought iron, and in most building materials, the temporary elongations or shortenings produced before the limit of elasticity is reached are proportional to the loads which produce thoise elongations or shortenings. Thus in the bar above referred to — if a load of 1 ton produce an elonga- tion, of 12 ^0 0^^ length, 2 tons will produce j-gWo"^^^' ^ t^Io o^^^^' and so on until 12 tons produce an elongation of ^--^^(j^ths = ^ ^ ths the length. At this point, however, the permanent set becomes appreciable, and beyond it the elongations are not in proportion to the load, but increase more 12 NOTES ON BUILDING CONSTRUCTION rapidly than the load. Thus 13 tons will produce an elongation slightly- greater than j^'^^Q-^hs of the length, and so on in an increasing ratio. The Modulus of Elasticity is a number representing the ratio of the intensity of stress (of any kind) to the intensity of strain ^ (of the same kind) produced by that stress, so long as the elastic limit is not passed. Thus the modulus of tensile elasticity of any material is found by dividing the tensile stress in lbs. per square inch of sectional area by the elongation (produced by that stress) expressed as a portion of the length of the body. For instance, if a weight of 1 ton hung from an iron bar produce an elongation of xto-tto^^ length of the bar, the modulus of elasticity of the bar will be 2240 Ibs.-h 3-2^00 = 26,880,000 lbs. This is rather lower than the modulus of average wrought iron. Similarly, the modulus of compressive elasticity is found by dividing the compressive stress in lbs. per square inch of section by the shortening (produced by that stress) expressed as a fraction of the length of the body. In most building materials the modulus of tensile elasticity and that of compressive elasticity are practically equal to one another so long as the stresses do not exceed the elastic limit. In advanced works on applied mechanics other moduli are used which, however, are not required in ordinary calculations, except the modulus of rupture, see p. 52, and need not be further referred to in these notes. Moduli of elasticity for different materials are given in Table I. Deflection is the bending of any body caused by a transverse stress. Thus the ladder shown in Fig. 14 is bent by the weight of the boy sitting upon it. If the intensity of the stress be below the elastic limit, the deflection will disappear when the stress is removed ; but if the intensity of stress be in excess of the elastic limit, a permanent set will remain. Stiffness in a material is the power that it may possess to resist being strained out of its proper shape ; this is a very different thing from the strength, or power to resist rupture. Compare, for instance, glass and wrought iron. The stiffness of the material depends upon its elasticity; the greater the value of the modulus of elasticity, the greater the stiffness of the body. ^ By "intensity of strain" is meant the amount of alteration of figure per unit of length ; thus if a bar be elongated under tensile stress, the intensity of strain would be the amount of elongation (expressed as a fraction of a foot) per foot of the bar. Expressing the total elongation as a fraction of the length of the bar amounts to the same thing. TERMS IN USE 13 T]u stiffness of a structure depends not only upon the elasticity of the material of which it is composed, bnt upon its arrangement. Thus a floor with deep narrow joists is much stiffer than one of the same strength with wide and shallow joists. This quality has an importance only second to that of strength in fact, strength and stiffness must be considered together. In a floor, for example, although the joists may be strong enough to resist breaking, if they are not stiff enough, the floor will be springy and uncomfortable, and if they have a ceiling attached to them it will be cracked, and may be rendered dangerous by the deflection or bending of the joists. In roofs, the rafters, if not stiff enough, will bend or sag, causing ugly hollows, and in some cases lodgment of wet upon the roof-covering. Other Physical Properties of bodies do not affect the calcula- tions about to be entered upon, a discussion of them would tend only to confuse the student, and they will therefore not be alluded to in this volume. Chapter II, EQUILIBRIUM. HEIST external forces act upon a body, its powers of resistance T t are called out until they are just sufficient to balance the external forces ; when this balance is maintained, the external forces and those of resistance are said to be in equilibrium. When the external or active forces require, for equilibrium, greater resistances than the body can offer, movement ensues in the form of rupture of the material if the internal resistances are ineffectual, or of movement of the body itself if the external resistances are insufficient. Thus if a rope be fastened to a man, and a boy pull at it with a force of 30 lbs., the man passively resists until there is a stress upon him of 30 lbs., and equilibrium is established. If, however, the man begins to pull the rope with a force of 80 lbs., equilibrium is disturbed and movement commences — the boy is dragged forward ; or if he ties the rope to a wall, and the man pulls with sufficient force, the rope is broken. Every structure is under the action of two distinct sets of forces, namely — 1. The external forces. 2. The internal forces, or the stresses. And in order that the structure may not move, or, as it is termed, may be " in equilibrium," it is necessary in the first place that the external forces be in equilibrium amongst themselves; and secondly, that the internal forces, or stresses, be not greater than the resistance the material of the body is able to offer. The external forces on a structure consist of the various loads it has to bear, and of the reactions at the points of support. It will thus be seen that, to design a structure as regards strength, three distinct operations are required, namely — 1. To find the external forces acting on the structure. 2. To find the stresses produced by those forces. 3. To calculate the dimensions of the various parts of the structure so that the stresses (or, more strictly speaking, the inten- EXTERNAL FORCES ACTING ON A STRUCTURE 15 sity of stress) shall nowhere be greater than the material is safely- able to resist. The resistance is always exactly equal and opposite to the stress up to the point of rupture ; and the resistance offered at that moment is called the ultimate resistance. This introduces the important subject of resistance of materials on which ultimately depend all calculations made in connection with the strength of structures, and unfortunately it is a subject of which our knowledge is by no means as full as might be wished. At the end of this book will be found a Table giving the ultimate resistance to the various stresses of tension, compression, shearing, etc., of some of the materials usually employed in building construction, ^ but more detailed information is given in Part III. These operations involve the application of the conditions of equilibrium of a body, the general consideration of which requires a greater knowledge of mathematics than is assumed for this Course ; the large majority of cases that occur in building practice can, however, be treated in an elementary manner, which we now proceed to do in the order mentioned above. EXTEENAL FOECES ACTING ON A STEUCTUEE. The loads a structure has to bear may be given, but usually they have to be ascertained in each case, and it will be shown in the sequel how to do this. For instance, it might be given that a certain beam has to bear a weight of 10 tons ; but in designing a roof, the weight of the roof-covering, the loads caused by the wind and snow, etc., would have to be ascertained or estimated. The reactions at the points of support are determined, as soon as the loads are known, by the application of the following principle : Whenever a force acts upon a body tending to move it in a particular direction, this force must be opposed by an equal and opposite force, or else the body will move — that is, it will not be in equilibrium. This is Newton's third law, namely, that to every action there is always an equal and contrary reaction ; or, the mutual actions of any two bodies are always equal and oppositely directed in the same straight line. Or, putting it another way, If a body is in equilibrium under the action of two forces, it is self-evident that the two forces must be equal and oppositely directed. 1 See Table I. i6 NOTES ON BUILDING CONSTRUCTION Taking, for instance, the. case of two men pulling at a rope (Fig. 1 2). Fig. 12. Supposing the man A pulls with a force of 30 lbs., B must pull in the opposite direction also with a force of 30 lbs. or the rope would move to- ward A. Now, supposing B ties his end of the rope to a ring in the wall (Fig. 1 3) and goes away, while A goes on pulling with a force of 30 lbs. The wall will now supply a force of 30 lbs. to oppose the pull of 30 lbs. The force supplied by the wall is called the reaction of the wall. The next case to consider is that of a body in equilibrium under the action of parallel forces, as in Fig. 14, showing two men carrying a boy on a ladder. 'lbs. The boy's weight and the weight of the ladder act vertically downwards (Fig. 1 4) ; the men therefore have to supply vertical forces act- ing upwards to maintain equilibrium. Supposing the boy to weigh 60 lbs. and the ladder 90 lbs., the whole weight to ^'^S- be carried by the men is 150 lbs. That is, there are forces amounting to 150 lbs. acting vertically downwards, and the men have to supply the reactions to resist these forces, and prevent the ladder from moving downwards. If the boy is in the middle, it is clear that each man will have to bear half the total weight, amounting for each to 75 lbs. In the same way, when a horizontal beam supporting a single load placed at its centre (Fig. i 5), or loaded symmetrically throughout its length (Fig. 16), rests upon two walls, half the total weight is borne by each wall, or, in other words, each wall has to supply a reaction equal to half the weight. When the loads to be carried are symmetrically placed with regard to the supporting bodies, and are symmetrical in them- selves, the reaction afforded by each of these supporting bodies is equal, as in the illustrations given above. EXTERNAL FORCES ACTING ON A STRUCTURE 17 When, however, the weights with regard to the supporting b w A ^ 2 ' ' 2 Fig. 15. afforded hy one support differs from that to be afforded by the other support. Eeverting to the illustration of the boy carried on the ladder, if the boy does not sit in the centre of the ladder (Fig. 17), then the man A, nearer to whom he sits, will bear more of his weight than the other man — that is, he has to supply a greater reaction. In order to ascertain the 1- so zz* amount of weight each man will have to bear, a simple calculation is necessary, founded upon the principle of the lever, which no doubt is familiar to the student. By the principle of the lever we know that to pro- duce equilibrium the up- ward force or reaction act- ing at A, multiplied by the leverage AL, must equal the reaction at B multiplied by BL. We have, therefore, considering now only the weight of the boy, Eeaction at A x AL = reaction B x BL, Eeaction A x 5 = reaction B x 1 5, Eeaction A = ^ x reaction B, = 3 x reaction B. Eeaction A -H reaction B = 4 x reaction B. But we know that the two reactions together must be equal to the weight.^ .'. 4 X Eeaction B = 60 lbs. .•. Eeaction B = 1 5 lbs. And Eeaction A = 45 lbs. Therefore we see that, because the boy is three times as far from B as from A, A has three times as much of his weight to carry as B. ^ From the conditions of equilibrium (see Appendix IL) it appears that — by taking moments about A, Wx5-Ebx20 = 0. B.C. IV. C i8 NOTES ON BUILDING CONSTRUCTION We have, however, ignored the weight of the ladder, which, being a uniform load of 90 lbs., produces a reaction of 45 lbs. at A and 45 lbs, at B. Eeaction caused by Boy. Ladder. The total reaction at A= 45 + 45 = 90 B= 15 + 45 = 60 Total reactions caused by weight of boy and ladder = 150 The general rule for finding the reaction at either support, caused by a load, placed anywhere on a beam supported at both ends, may be stated as follows : — Rules for finding the Reactions at Supports. jluU I. — Tlie reaction at either support, caused hy a single load (W) placed anywhere upon a heam supported at both ends,is equal to the load multiplied by the distance from its centre of gravity to the other support, and divided by the length of the beam between the supports. Single Load. — Thus with refer- ence to Fig, 1 8 WxDB Eeaction at A = R, AB Fig. 18. Wliere W is in the centre, we have DB = |AB, and Reaction at A = = — ^ — = W- AB General Case of Single Load. tance of W from A = a, we have -If the load = W, the span = I, the dis- R. I — a W (1). Rb = 7W (1 A). R^t?W^ If TF he in the centre, then I (1 B). I 2 R. = R, Fig. 19. Two OR MORE Loads. — When a number of weights rest upon a beam the reac- tions can be found either by calculating the reactions produced by each weight separately, and adding them together, or by considering that they produce the same effects as would be produced by a weight equal to all of them put together, acting through their common centre of gravity. The second method is to be adopted in preference when the position of the centre of gravity is self-evident. Thus in Fig. 20 it is clear that the four weights, each of 100 lbs., produce the same reactions that would be produced 1 n A i I tOOtbi f 00 lbs < — 5>--* tOOtb, 100 Ibt P ^ iOlo'. >f R=|x4 io;,o - EXTERNAL FORCES ACTING ON A STRUCTURE 19 one weight of 400 lbs. placed midway between them, that is, at the centre of gravity of the weights (Fig. 21). ^ Uniform Load. — Again the reactions produced by the uniformly distributed load Sw (Fig. 16) will be equal to those of a con- centrated load W = ^10 act- ing through the centre of gravity of 8 w. When the load is uni- formly distributed over a por- tion or the whole of the beam, the reactions produced by it will be the same as those ! produced by a single weight equal to the load and placed at the centre of gravity of the load. Thus in Fig. 2 1 the reactions produced by the load of 400 lbs. distri- buted along EF would be ; exactly the same as those R=^*>loo I produced by a single weight of 400 lbs. acting at D, the e.g. of EF (see Fig. 21). JVhen the weights are unequal and placed unsymmetricaUy, their reactions are best found by the first method, that is, by taking each in turn and finding the reactions produced by it. The reactions produced at either support will be the sum of the reactions produced by each weight at that support. W2 Fig. 21. w •s:,o-->< io'„(fi- ItOO lbs Fig. 22. I t—5'„ O'—Xk — 5;, 0-->U 10'„0"-- ^100 lbs ^300 lbs Fig. 23, Thus in Fig. 23 the reactions produced would be as follows : — At A. At B. Reactions produced by Wj .... Reactions produced by W2 .... Total reactions caused by Wi + "VV, Or substituting the values of Wj and "Wg, we have Total reactions caused by Wj, + Wj i.e. by 300 lbs. +100 lbs. 20 20 ^2 i^^ + 2>^ = 150 + 75 = 225 20 20 "2 20^1 + 20^2 = 150 + 25 = 175 20 NOTES ON BUILDING CONSTRUCTION The general case of two weights unsymmetrically placed, as shown in Fig. 24. W2 ::!:dr^::t::::^!:::z^ -I- riff. 24. Suppose a to be the distance from the abutment A to the point of applica- tion of W^, 6 the similar distance for "Wg, and I the span or distance between the supports. Then I - a I — h. Keaction at A = — ^ — x Wj -\ j- Eeaction at B = ? + jW^ Wo I Several loads placed ^msymrfietrically.—^'h2i't&\&i: the number of weights, the reactions are found in a similar manner. Thus in Fig. 25 there are five weights, Wg, Wg, W^, W5, the distances Fig. 25. of which from A are respectively a, h, c, d, e, and I being the span or distance between the supports. l-a l-h ^„ ^ - C ,Tr ^ - ^ -.TT ^ ~ ^ -rtr The reaction at A = ^ + + W3 + + W,. Q/ 1) c d ^ The reaction at B = - + ^ + ^ W3 + + ^-^5. The weights W3, W^, W5 being known, as also the lengths of a, b, c, d, e, and I also known, it is easy to substitute the values and calculate the reactions. We arrive then at the following rule : — Bule II. — The reaction at either support caused hy any numher of loads placed upon a beam supported at each end is equal to the sum of the reactions at that support caused hy each load taken independently. EXTERNAL FORCES ACTING ON A STRUCTURE 21 Or, stated in another way, the reactions at the supports A and B caused by a number of loads Wp Wg, Wg, W^, W5, Wg, etc., will be as follows : — Total reaction at A = reaction at A of 4- reaction at A of Wg + reaction at A of W3 + reaction at A of "W4 + reaction at A of W5 + reaction at A of Wg, etc. Similarly, the total reaction at B = reaction at B of W^ + reaction at B of Wg + reaction at B of W3 + reaction at B of W^ + reaction at B of W5 + re- action at B of Wg, etc. The reactions of the individual weights at A and B are of course found by Rule 1. Load on the Supports. — It is evident that the load on a support is equal and opposite to the reaction the support exerts on the beam. Thus in the case of the men carrying the boy on a ladder, when the boy is sitting in the centre of the ladder the downward pressure on each man is equal to the reaction, or to 75 lbs. ; when, however, the boy is not sitting at the centre of the ladder, but at the position shown in Fig. 1 7, the man A has to bear a downward pressure of 90 lbs. and the man B of 60 lbs. In order to familiarise the student with the method of finding the reactions caused by various forms of load, it will be well to take as examples a few cases such as occur in practice. External Forces acting on a Girder. Example 1.— A girder AB of 30 feet span (Fig. 26) supports a c A B ^ 2o'o" > SO'O- M so'o"- — ■ g Fig. 26. beam C at a point 10 feet from the column at A. The load on the beam C is uniformly distributed, and amounts, including the weight of this beam, to 46 tons. The girder AB itself weighs 2'6 tons. Find the proportion of weight to be borne by the column and the wall. This weight will be equal to the reaction at each point of support. Since 46 the beam C is uniformly loaded, it will impose a load of — = 23 tons on the girder AB at the point C. The girder AB is therefore subject to two loads, namely — • 1 . Its own weight = 2 '6 tons. 2. The load upon it = 23 tons. The former is a uniformly distributed dead load, the reactions of which NOTES ON BUILDING CONSTRUCTION can be found as at p. 17. The latter is a single load placed unsymmetric- ally iipori the girder, and its reactions can be found by Rule I.^ At A ; tons. At B ; tons. The reaction of the weight of girder itself will be 1-3 15-3 1-3 7-7 Total reactions 16-6 9-0 If the reactions have been correctly worked out, their sum must be equal to the sum of the loads. In this case the sum of the weights is 23 + 2-6 = 25-6 tons. The sum of the reactions as found above is 16-6 + 9*0 = 25-6 tons. The result is the same in both cases — which shows the working to have been correct. If in this example the beam C had been supported in the centre of AB, 23 the reactions due to it at A and B would have been equal each to — = 11-5 tons, and the total reactions would have been altered accordingly — each being equal to 1-3 + 11*5 = 12-8 tons. The column A would thus have to bear less, and the wall B more, than in the former case. External Forces acting on a Cantilever. Example 2.— A wrouglit-iron cantilever, as shown in Fig. 27, is bolted to a wall and sustains a weight of 1 cwt. Find the other external forces acting on the cantilever. To oppose the downward tendency of W, the two bolts AB must supply an upward reaction W, but the proportion in E, which is divided between the two bolts depends upon the method of fixing. For in- stance, if the lower bolt were working in a slot, the top bolt would have to supply the whole ■^^S" of R and vice versa. "W also tends to turn the cantilever around the point B in the direction of the hands of a watch, and to oppose this the bolt at A must supply a horizontal reaction R^. From the principle of the lever, or, as it is termed, by taking moments about B (see Appendix II.) — 6xW-4xRi = 0, or R^ = ^W = l-5 cwt. 1 Reaction at A Reaction at 15-3, 7-7. STRESSES PRODUCED BY THE EXTERNAL FORCES 23 The bolt at A must therefore be of sufficient strength to withstand a pull of 1'5 ton, or in other words 1-5 ton is the stress in the top bolt. The reaction Eg ^ clearly equal to the reaction at A^^, but acting in the opposite direction. External Forces acting on a Boof. Example 3.— The roof shown in Fig. 28 is loaded with equal w / ^\ v*^-— — ; 1 \ A\ ...8>- Fig. 28. loads. W at the points D, C, and E. Find the reactions at the supports A and B. It can be seen at once from symmetry that the reactions must each be W equal to W + — , However, applying Rule II., p. 20, 24 W 8 „^ ^^ = 32^ + ¥+r2^^ 8 W 24 w :W + w The preceding must suffice to give a general idea of the manner of finding the reactions at the supports, and of thus determining all the external forces acting on the body or structure. Other cases, such for example as the re- actions due to the pressure of the wind, acting on one side of a roof, will be considered in subsequent chapters. The next step (see p. 14) is to find the internal stresses pro- duced hy the external forces, and for this we must consider the various structures separately, commencing with beams (Chap. III.) Chapter III. BEAMS/ EIGS. 3 1 and 3 2 show the manner in which a rectangular wooden beam, supported at the ends, breaks when subjected to a concentrated load greater than it can bear. The beam bends. Fig. 31. sinking most at the point under the weight, and the fibres oi the upper portion of the beam are crushed, and those of the lower portion torn asunder, as shown on a larger scale in Eig. 3 2. Fig. 32, There are therefore certain external forces at work tending to break the beam across, and certain internal forces which are resist- ing this tendency to rupture. In order to ascertain the strength of the beam, as compared with any load it may be called upon to bear, it is necessary to ascertain the nature and extent — 1. Of the external forces tending to break the beam across. 2. Of the internal forces tending to resist rupture, ^ The formulce for practical use are given in Equation 30, p. 54, and in Appendices VII. and XXL BEAMS A uniform load of sufficient magnitude would also produce rupture in the same way, but with a slight difference as to the form assumed by the beam just before the time of rupture. When a beam is supported at both ends and subject to a safe load, it will bend to a certain extent, and the upper fibres will be in compression and the lower fibres in tension. Now if we examine the external forces acting upon the beam in I'ig- 3 3j we see that the weight acting downwards causes a reaction at each support acting upwards, just as in Fig. 1 7 the weight of the boy on the ladder rendered it necessary for an upward force to be exerted by the man at each end. The downward vertical force caused by the weight acting to- wards the centre of the earth is balanced by the upward forces ^ caused by the reaction at each end, and the whole structure is in a state of equilibrium {i.e. the beam supports the weight). Conversely, we may say that the reactions acting vertically upwards are balanced by the weight acting vertically downwards. They are in the position of vertical forces each acting at the end of a lever under the weight as a fulcrum, and the action is the same as in bending a stick across one's knee. When we consider the effect of these reactions upon the beam we see that each acts with a leverage equal to the distance from the support to the point immediately under the weight, about which the beam tends to break. The power of these reactions to bend or break the beam, there- fore, consists of -D /.-U ^ A\ Kf^ (the reaction at A) x AC, or of Ej, (the reaction at B) x BC. Bending moment. — This is called the lending moment or, in some books, the moment of flexure. Since the bending moment of the reaction at C (the section of the beam under the weight) is proportional not only to the magnitude of R^ but also to the lever arm AC with which it acts, it is therefore equal. to AC X R^, therefore the bending moment due to R^ at C CB AB r\x> AC.-^.W .... (2). 26 NOTES ON BUILDING CONSTRUCTION In the same way the bending moment due to the reaction R3 at C is CB X R3 AO = CB.^.W .... (3), which shows that the bending moment at C due to is equal to the bending moment due to R^, a manifestly correct result.^ Let it now be required to find the. bending moment at any point P. Considering the left-hand part of the beam, the bend- ing moment is seen to be ^ AP.CB ^P^^^ = -AB-^- Now considering the right-hand part of the beam ; the reaction Rj, tends to turn that part of the beam against the direction of the hands of a watch, but the weight itself opposes and tends to turn it in the contrary direction. Thus the bending moment will be equal to the bending power of R3, less that of W, that is = PB . R3 - PC . W, AO = (PB. — -PC)W, PB.AC-PC.AB „^ - . W, AB (PO + CB)AC - PC(AC + CB) AB W, CB.AC-PC.CB (AC-PC)CB AP.CB AB AB AB The same result as before. The bending moment at any point of a beam can therefore be found by the following Eule for finding the Bending Moment at any given Point. Consider either of the portions into which the beam is divided at the given point, and multiply each force acting on that portion by the distance of its point of application from the given point ; these products can be taken as positive when the force tends to turn the portion of the beam under consideration in the direction of the hands of a watch, and negative when in the opposite direction. Distributed loads are to be reckoned as acting at their centre of gravity (see p. 19). The algebraic^ sum of these products is the bending moment required. 1 Strictly, the distances AC and CB should be measured horizontally, but in practice the bending of a beam is so small that these distances can be measured without any appreciable error along the beam, or in other words may be taken to be, after bending, equal to what they were before bending. 2 That is, the positive products are to be added and the negative products deducted. BEAMS— BENDING MOMENT 27 In practice that portion of the beam would be chosen on which, the fewest forces are acting. Example 4. — As a simple numerical example let W= 100 lbs. AC = 6 feet CB;= 14 feet |ab = 20 feet. 4R8 '^^^^ K. = iixlOO = 70lbs. ^ 20 E3 = — X 100 = 30 lbs. ° 20 And tlie bending moment at C = CBxE, It is also equal to ACxE. ^100 Ihs Fie. 34. : 14 feet X 30 lbs., 420 foot-lbs. 6 feet X 70 lbs., 420 foot-lbs. As will be seen subsequently, it is sometimes better to express lengths in inclies, and so expressed the bending moment at C would be = 420 X 12 = 5040 inch-lbs. Further, if the loads are expressed in cwts. the bending moments will be expressed in inch-cwts., and in inch-tons if the loads are expressed in tons. BENDING MOMENT UNDEE VAEIOUS CONDITIONS.^ We proceed now to consider the Bending Moment or Moment of Flexure caused in a beam fixed or supported in various ways, by the different distributions of load that are most likely to occur in practice. The following will be the notation used throughout : — M = Bending moment or moment of flexure in inch-pounds, inch-cwts., or inch-tons, according to the denomination in which W and w are expressed. Mg = Bending moment at centre. My = Moment at point P. = Moment at point S, and so on. A and B are used for the points of support. 'W = Total weight on the beam in lbs., cwts., or tons. w = Weight per inch run of the beam in lbs., cwts., or tons. I = length \ of the beam, 6 = breadth > all in d = depth ) inches. = Reaction at support A. = Reaction at support B. ^ The formulae for practical use will be found numbered 4 to 27 on pp. 28 to 40, also in a condensed form in Appendix VII. 28 NOTES ON BUILDING CONSTRUCTION Case 1. BEAM FIXED AT ONE END AND LOADED AT THE OTHER END. To find the bending moment Mp at any point P, consider the portion PE of the beam as shown in Fig. 35, then by the rule (p. 26) the bending moment is equal to the weight W multiplied by the leverage with which it acts, namely PE. Now if X be the distance of the point P from the wall, the leverage with which W acts at P is - x), therefore Mp-W(Z-a;) (4). <— X— »k- •< l-x ^ Ficr. 35. Fig. 35a. At the wall A the weight will be acting with the leverage of the full length of the beam Z, hence Mj^ = WZ .... (5). At the point E the leverage is nothing, and the bending moment M^ = 0 Graphic Representation. — The above can be shown graphically thus : — Draw AE (Fig. 36) as before.^ Plot A/ vertically to represent WZ (the bend- ing moment at A) to some convenient scale of inch-tons to an inch.^ Join E/ and draw P^ parallel to A/; then PK l-MC M Fig. 37. P El ! t< l-X >1 Fig. 38. = w(Z — £c], and it acts with a leverage equal to the distance of its centre of l — x gravity from P, i.e. — — -, therefore (I - X)2 (7). Fig. 39. A 1 1 1 1 1 1 1 1 1 1 1 / i r Fig. 40. \ Graphic Representation. — This case may be shown graphically, as in Fig, 39. 30 , NOTES ON BUILDING CONSTRUCTION Draw AE as before, and A/ — — to represent the bending moment at A. Draw /E a parabola with its vertex at E ; then the ordinate dropped w(l — cc)^ from any point P, distant x from A, will be equal to — , which is the bending moment at that point. ^ Simpler Construction. — To be of ready practical use, a simpler construc- AE tion is needed. Now if A/ is made less than — — a circular arc drawn through /, and tangent to AE at E, will practically coincide with the required parabola, and can be used instead of it. This is shown in Fig. 40, the circle being in full and the parabola in dotted lines. The value of the above limit as to the proportion of A/ to AE AE is shown on the same figure, where A/ is made = and the divergence between the circle and parabola is seen to be not sufficiently great as to lead to erroneous results, but in the case of A/' which is equal to — the error is seen to be considerable. An easy method of finding the centre of the required circle in such cases is also shown on this figure. EC is drawn perpendicular to AE, E/^ is joined and bisected at b, and 60 is drawn perpendicular to E/^^. The point of its intersection with EC is obviously the required centre. Case 3. — beam fixed at one end and loaded with a unifoemly DISTRIBUTED LOAD, AND ALSO WITH A CONCENTRATED LOAD AT ITS EXTREMITY. This case (Fig. 41) is a combination of Cases 1 and 2. Fig. 41. Fig. 42. Graphic Representation. — This case is shown graphically in Fig. 42 ; the ^ For a proof of this, reference may be made to any work on conic sections. BEAMS— BENDING MOMENT 31 diagram from Fig. 36 being plotted above, and that from Fig. 39 below the beam AE.^ The bending moment at A will be the sum of the bending moments caused by W and by wZ, so that, combining Equation 5 with Equation 6, M, = WZ+— (8), = A/n-A/, = /i/(less the thickness of the beam in Fig. 42). Combining Equation 4 with Equation 7, M, = W(Z-.) + ^fc^' . . . (9), = Pi + = (less the thickness of the beam). It is not absolutely correct to say that ln,f— the bending moment at A, and that ig = the bending moment at P, for it is obvious that the depth of the beam must be deducted from these ordinates, or the beam must be represented by a single line when drawing the diagrams. This case would apply to a balcony crowded with people and having a very heavy railing. GcLSt 4 (Fig. 43). BEAM FIXED AT ONE END AND LOADED WITH SEVERAL CONCENTRATED WEIGHTS, W^, Wg, Wg, ETC. The bending moment at any point produced by all the weights is equal to the sum of the bending moments produced by each at that point. W3 3 — /, — ^ Fiff. 43. Fig. 44. Graphic Representation (Fig. 44). Draw Ae = 'Wjl and join Ee, ef =^¥"2^2 and join F (the point on Ee under W^) with/, fg = WgZg and join G (the point on F/ under Wg) with g. It is evident that M, = W,l + W,l, + W,l„ .... (10), = Ae + ef+fg = Ag. At any point P distant x from A M, = W^{1 -x) + W,il, -x) + W,(l, - X), . (11), = 'Pi + ij+jk = 'Pk. ^ Both diagrams must be plotted to the same scale— that is, the same number of inch-tons or inch-lbs. to an inch must be taken for both. NOTES ON BUILDING CONSTRUCTION Case 5 (Fig. 45). — beam fixed at one end and suppoeting a LOAD UNIFOKMLY DISTKIBUTED OYER A PAET OF ITS LENGTH. Tf Z being the loaded portion of the heam At the point A M.^ = wz{\z + y) .... (12). At any point Q in AB distant x' from A llL^ = wz{lz + y-x') . . . (12 A). At any point P in BD distant x from A 'M.^ = w{z + ij-x)y.\{z + y- x), = lw{z + y-x)^ . . . (12 B). At the points = loa + 13 = |w22 . . . (12 C). At the point D M„ = 0. At any point in DE M = 0. Fig. 46. Graphic Representation (Fig. 46). Draw AG = = + ^/) and join G and the middle point between B and D. From B draw the vertical line BF,^ cutting the line from G in F. Between the points F and D (the points immediately under the extremi- ties of the load) draw a semi-parabola as for a beam BD uniformly loaded (Case 2). The vertical distance between GFD and ABD at any point gives the bending moment at that point. Case 6 (Fig. 47). — beam supported at both ends and loaded IN THE centre. Fig. 47. Fig. 48. ^ If BF be made less than -r--, the parabola can be drawn as a circle. BEAMS— BENDING MOMENT Mo = Reaction at A or B x leverage, 33 Reaction at A or B = Hence At any point P leverage = ,r W I Wl W 2' I W/l ■■ — I ; 2 \2 (13) . (14) , M, = 0, Graphic Representation. — Fig. 48 shows this Case graphically. Draw CD to represent to some convenient scale M(,= Join AD and BD, then the vertical distance from AB to AD or DB at any point is the bending moment at that point. Thus PF is the bending moment at P and is equal V2 ~ ^)* Case 7 (Fig. 49). — beam suppoeted at both ends and UNIFOKMLY LOADED, According to the rule given at p. 26 the reaction R^^^ produces a positive bending moment at C = ^ x ^, and the load on AC produces a nega- id I tive bending moment in the opposite direction at C = — x -. 2i 4 Hence the bending moment at C is wl I wl I M„ = — X X- ' 2 2 2 4 (15). To find the bending moment at any point P distant x from the centre of the beam, consider the portion AP of the beam, then by the rule — I I /I 2 \2 w/l + x 2V2 w/l ^2(2- + ^ %u /P 2V4 B.C.- -IV. (16). D NOTES ON BUILDING CONSTRUCTION Graphic Representation (Fig. 50). — As in Case 2 the bending moments are graphically represented by a portion of a parabola ADB as shown in Fig. 50, and if a suitable scale is chosen so that CD is less than ^ AC or -J AB, the arc of a circle can be used instead of the parabola without appreciable error. The method of drawing a parabola is given in Appendix III. To obtain, therefore, the bending moment at any point in AB, plot CD to represent — to some convenient scale, so that CD is less than ^ AB. Draw a circular arc through the points A, D, and B, then PQ will represent the bending moment at the point P; P^Qi the bending moment at P^, and so on. The bending moment at P can also be obtained by considering the portion PB of the beam. It will then be found that i The same value as before but with a negative sign. This difference in signs expresses the fact that the forces acting on the portion AP of the beam tend to turn that portion in the positive direction, i.e. in the same direction as the hands ot a watch, whereas the forces acting on the portion PB tend to turn that portion in the negative direction. The maximmn bending moment occurs at the centre of the beam, and is — . Now wl is the total load, which is written W, so that M„ = — . Comparison between Central and Distributed Load.— On referring to^Case 6 it will be seen that a central load W produces a bending moment — at the centre of the beam, that is, double the bending moment produced by a uni- formly distributed load of the same amount. In other words, the safe dis- tributed load on a beam is double the safe concentrated load at the centre. Case 8 (Fig. 51). — beam suppokted at both ends, loaded in CENTRE AND ALSO UNIFORMLY. This is a combination of Cases 6 and 7. Graphic Representation.— In the graphic representation. Fig. 52, the --a; A B A' J Pig. 51. Fig. 52. diagram of bending moments for the distributed load wl is below, and that for BEAMS— BENDING MOMENT 35 the single load W is above the line AB. It will be easily seen that the total bending moment at the centre = for the central load + for the distributed load. Therefore, combining Equation 13 with Equation 15, M, = M„ forW + M„fori(;/, And further, by combining Equation 14 with Equation 16, = for W + Mp for wl. = 1W -f A^hr (17). . . (18). It will be observed that the bending moment at any point of the beam can be obtained by adding together the ordinate of the triangle and of the curve at that point. Case 9 (Fig. 53). — beam suppoeted at both ends loaded AT any point. In this case the maximum bending moment is at D, the point of applica- tion of the load. With regard to the reactions at A and B see p. 17. l-m ^ m ^ -l-m-x — Fig. 53. Here the bending moment immediately under W is : Bending moment at D = M„ = reaction at A x leverage, AD, m = y W x{l- m), m{l — m) I W (19). Or M^, = - reaction at B x leverage, BD, 'I — m\ -J— jW xm, m{l — in) I W. At any point P in AD to the left of and at a distance x from W Mp = X leverage, m = J W X (Z - m - a;) . . , (20). At any point Q in DB to the right of and at a distance y from W = - X leverage, BQ, l — m = ^ ' • • (21)- At A and B and each = 0. 36 NOTES ON BUILDING CONSTRUCTION Graphic Representation. — The bending moments are represented hj a tri- angle, as shown in Fig. 54. Case 10 (Fig. 55). — beam supported at the ends and PARTIALLY LOADED WriH A UNIFORM LOAD. Fig. 55 shows the manner in which the beam is loaded, and the values of the reactions and R^j are as follows : I — m + n 1 (l — m — n){l — m + n) R^ = (i — m — u)w X I l + m — n 21 2 (l — m- n)(l + m-n) R^ = (Z - m - n)io X ^ = ^ . w. 1+ m-n -M-^QQQQ)^ ~ 2 ■ '(l-in -n) w Fig. 55. The distance of any point (P, P', or P") from A will be called x, and from B the distance will be called y. The bending moment at any point P situated between A and E is Mj. = R^ X X, (Z - m + n)(Z — m — n) 21 (22). Similarly the bending moment at any point P" situated between F and B at a distance y from B will be found to be (considering the portion P"B of the beam and neglecting the negative sign) (l + m — n)(l — m — n) ll,>, = li,xy = ^-^ ^\ \wy (23). Lastly, the bending moment at any point P' situated between E and F is equal to Mj., = 'Rj^x X - {x- m)w X — ^ , (I — m + n)(l - m — n) (x-m)"^ = ^ '.'wx-'^—^.w (24). It can also be shown by the application of the differential calculus that Mp' is a maximum when ^ = — SZ— • • • • (25). Graphic Representation. — It will be found that if the distributed load over BEAMS— BENDING MOMENT 37 EF were concentrated, one half at E and the other half at F, that neither nor R3 would be altered in value, and hence M^. and Mp" would also have the same value. But the bending moment at P' would be changed and would be equal to •,r/ (^-m + 7i)(Z-m-TC) (l-m-n)w M . ivx - (x - m), therefore 21 (x — nn)w I — m — n — {x — m) Fig. 56 represents the part EF of the beam by itself, and it is clear that he— z I — m — n = l^; and if z is the distance of P' from the centre of EF, I. Hence Mp' - M' ' = + 2 + z It will therefore be seen that the bending moment at P' is greater when the load is distributed than when it is concentrated at E and F. And further, Ly referring to Equation 16, it will be seen that the increase can be represented graphically by the ordinate of a parabola as shown in Fig. 57, or practically wl ^ I by the ordinate of an arc of a circle, if — ^ is represented by less than J. Now the diagram of bending moments, when the loads are concentrated at E and F, is as shown in Fig 58, where and EQ = R^ X m = FQi = E3xn = (l-m + n)(l — m ■ n (Z + m - n){l — in — n) 21 ■.vm, 38 NOTES ON BUILDING CONSTRUCTION so that the diagram, when the load is distributed, can he found by adding the diagrams in Figs. 57 and 58 together, as shown in Fig. 59. Instructive lessons can be learnt from this case by varying the values of m and n. Supposing, for instance, that m=n, then the maximum bending moment will be at the point for which (Equation 25) that is, at the centre of the beam, and the bending moment at that point, from E(,;[uation 24, will be n ■ '2' V7 ^ \ -\l-^-\-vx\w, (26). If now m = 0, so that n is also 0, that is, the load is distributed uniformly over the whole beam, then Mc = -g • the result previously obtained when considering Case 7 (Equation 15). Again, if m = ^, that is, the central half of the beam is uniformly loaded, M -4 16. w. BEAMS— BENDING MOMENT 39 from which it appears that if the two outer quarters of the beam were loaded and the central half unloaded, 32 So that the load on the central half produces a three times greater bending moment than the remainder of the load. Equation 26 can be written M^=|^^ + m^(Z- 2m)M;. Now (Z - 1m)w is the total load on the beam, that is W, so that Let m = \ and also m=hi the load will therefore be concentrated at the centre of the Z 2 beam (Case 1), then M — "W 0 4 * ' the result previously obtained (Equation 13). The case of a beam loaded with a concentrated load placed anywhere upon it can be deduced in a similar manner to the above from Equation 24, an exercise which is, however, left to the student. It will thus be seen that the case under consideration includes all the previous cases. Case 11 (Fig. 6o). — beam suppoeted at both ends, loaded WITH ANY NUMBER OF LOADS. The total reaction at each abutment is equal to the sum of the reactions w, N 0 1^ \0i) V osy \ \ \ \ \ \ 1 / / '^3 / t h 1 ' Z = the span OT = the distance of "Wi from A ,, "W2 from A mz— „ "W3 from A Fig. 60. caused by each load separately, thus (Fig. 60) I I I 40 NOTES ON BUILDING CONSTRUCTION The moment at any point P distant x from A is (if P be between aad O3) Mp = Reaction A x leverage — weights between P and A x leverage, f^Jl-m.) y^Jl-m^ Wo(Z-m„)\ , , , , = {-^l +-i + I ) xa;-W^(x-m)-W2(x-m2) (27). Graphic Representation^ (Fig. 60). — In the graphic representation the cotted lines show the triangle of bending moments due to each load separately (see ^^^^ Tlius Ar^B is the triangle for W^^, Ap.^B is the triangle for Wg, AggB is the triangle for "Wg. The bending moments caused by all the loads are represented graphically by a polygon, which can be obtained as follows : — The bending moment caused by immediately under itself is o^r,, the bending moment at this same point caused by the weight W2 is o^p^, and the bending moment caused by Wg is o^g^ ; the total bending moment at this point caused by W^, W^, and Wg is equal to o^r^ + o-^^p-^ + o-^q^, and o^t^ is set off equal to this. In the same way the total bending moment at Wg is found and set off as o^t^, and the total bending moment at Wg as Og^g. Join A^j^, t^t^, t^t^, and ^gB, and the polygon of moments is obtained. MOMENT OF EESISTANCE. The application of the preceding rules enables us to ascertain the effect produced by external forces acting transversely upon a beam — that is, in other words, to calculate the bending moments produced by loads or other external forces. The next step is to investigate the nature of the resistance offered to these forces. This resistance is afforded by the internal structure of the beam, which, for instance, consists in timber and wrought iron of a number of fibres lying closely together and running in the direction of the length of the beam, but in cast iron of closely adhering grains. There are two methods by which the value of this resistance can be ascertained — 1. By reasoning upon certain assumptions, 2. By experiment. Of these two methods the first is not practically used in the calculation of beams of rectangular cross section, but its applica- tion to them is described somewhat fully in detail, in order to clear the ground for the better understanding of the calculation of beams of more complex section. Method of Ascertaining the Moment of Resistance by Reasoning. ■ — The effects produced upon the fibres of a beam by external forces acting upon the beam may be shown as follows : — ^ See Appendix VI. where a similar ease is worked entirely by a graphic method. BEAMS— MOMENT OF RESISTANCE 41 Let Fig. 61 represent a straight rectangular beam supported at A and B and Fig. 6 1 a a cross section of the same beam. Mark upon tlie beam two vertical straight lines op, qr, and also a line NL drawn parallel to the length of the beam and midway of its depth, and intersecting op in n and qr in L c d e i / Fig. 61a. P Fiff. 61. The beam can be considered as made up of horizontal layers of fibres, so that oq will represent a portion of the top layer and pr of the bottom layer ; NL will represent the central layer, nl a portion of it, and cd and ef portions of intermediate layers. If the beam be now slightly bent as in Fig. 62 (where, however, the bend- C b- e' 1 f' P r' Fig. 62. Coin .pressioit' Tension ing is exaggerated to make the case clearer), it will be found that the lines op and qr, originally vertical, are inclined inwards, and that nl remains unchanged in length. It therefore follows at once that o'q' and c'd' are shorter than oq and cd, but that p'r' and e'f are longer than pr and ef. Fig. 63 is an enlargement of the central part of Fig. 62, and o^p^ has been drawn through n parallel to q'r'; clearly 0^0' and c^c are the amounts by which oq and cd have been shortened by the bending, and similarly e'e-^ and p'p.^ are the amounts by which ef and pr have been lengthened. It is evident, therefore, that all the fibres above nl have been compressed, and those below nl extended, by the stress put upon them by the bending action. Only the fibres in the layer repre- sented by NL are uninfluenced by the bending action ; they are neutral, hav- ing neither compression nor tension to resist, the neutral layer. P' Fig. 63. This layer of fibres is called 42 NOTES ON BUILDING CONSTRUCTION If a cross section of the beam be made, the neutral layer ^ will be intersected in a straight line (00, Fig. 6ia), and this straight line is called the neutral axis. As already mentioned at p. 24, if the bending of the beam were carried far enough to produce cross breaking, it would be found that the fibres at the cross section of maximum stress above the neutral layer were crushed, and the fibres below were torn asunder. It will therefore be seen that the resistance of the beam to bending is afforded by the resistance of the fibres above NL to compression and of those below NL to tension, and the question is. How can we arrive at the value of this resistance ? Figs. 64 and 65 may make us understand more clearly exactly what it is we wish to find out. Fig. €4. Taking a simple case of a beam supported at both ends and loaded in the centre. Fig. 64 shows the general direction of the forces which produce the bending moment and of those which offer resistance at the cross section CD under the load. and are the reactions at A and B respectively which act with a leverage equal to half the span of the opening, i.e. AC or BC, and R^ X AC or R3 X BC Supposing now that the beam is cut across at CD and separated, the load being also divided equally ; and that by some means the stress on each fibre is maintained exactly at what it was before the beam was cut ; the two halves of the beam will clearly remain in equilibrium, and the state of things is represented in Fig. 65. Considering the left-hand half of the beam, it will be seen that the forces W R^ and — tend to turn that portion of the beam in the positive direction, and that the internal stresses tend to resist it in the negative direction. The forces R^ : the bending moment. 1 The neutral layer is sometimes called the neutral surface and in some works the neutral axis. BEAMS— MOMENT OF RESISTANCE 43 and — , being equal and parallel, form what is called a coufle, and the pro- duct of one of the forces into the distance between them (i.e. x AC) is called the moment of the couple. It will be seen that the moment of this couple is equal to the bending moment at the section CD. To ensure equi- librium it is clear that the resisting stresses must have a. turning moment equal and opposite to the bending moment. This moment is called the moment of resistance of the section. The same reasoning applies to the right half of the beam, and can also be extended to any section of the beam. A A B W W Fig. 65. We have therefore established the important relation that at any section of a beam Bending Moment = Moment of Resistance. The moment of resistance at a section P will be denoted by Mp, and the next step is to find an expression for its value. From Figs. 64 and 65 we gain a general idea of the nature of the resistance which is offered by the fibres of the cross section and of the way in which this resistance counteracts the effects of the bending moment. We wish now to ascertain more exactly the amount of the resistance offered by the fibres of a cross section, and how the resistance is distributed among those fibres ; whether they all resist equally, or whether more resist- ance is offered by some than by others. In order to do this, certain assumptions must be made. These assumptions are based on experiment, and are practically true when the stresses do not exceed the elastic limit ; and as the working stress should not even approach, far less exceed, the elastic limit, these assumptions «iiui be accepted as giving practically reliable results. They are, however, not true so soon as the elastic limit is overstepped, and the divergence becomes greater and greater as the point of rupture is approached. It is clear, therefore, that the breaking load cannot be deduced by working on these assumptions, which are as follows : — 1. That the stretching or shortening in any fibre is proportional to the stress of tension or compression that comes upon it. 44 NOTES ON BUILDING CONSTRUCTION For example — If a compression of 1 ton per square iiicli shortens a fibre by xttVtt^^ i*"^ length, a compression of 2 tons per square inch on the fibre will shorten it yo^o o^-'^^ ^'^ length, 3 tons Y^^QQ-ths, and so on, up to the elastic limit ; and in the same way, if a tension of 1 ton per sqnare inch stretches a fibre j^Vo^^j ^ ^^^^^ '^'^ stretch it yQ^Q-yths, 3 tons y-^^^-^ths, and 5 tons y-Q^Q-g-ths. 2. That the amoixnt of shortening produced by a given stress causing compression will be equal to the amount of stretching produced by an equal stress causing tension. Thus if a compression of 1 ton per square inch ujion a fibre shortens it y^^jyth of its length, a tension of 1 ton per square inch will stretch it j qVo^-'^ of its length, and so on. 3. That the amount of stretching or shortening in a fibre is proportional to its distance from the neutral axis of the section under consideration.^ It follows from assumptions 2 and 3 (see Appendix IV.) that the neutral axis passes through the centre of gravity of the section. Thus if CDEF (Fig. 66) represents the cross section of a beam 12" deep, then 00, the neutral axis, is to be drawn through the centre of gravity of the cross section, that is in this case midway between CE and DF. p o s — Q-^r i If PQ be 1 inch from 00, RS „ 4 inches ,, TU „ 4 inches ,, VX ,, 5 inches X— F— Fig. 66. then the shortening of each fibre in the layer ES will be four times that of the fibres at PQ ; the shortening at CE six times that at PQ. The stretching at VX will be |- of that at TU, and the stretching at DF = |- that at TU. In the case under consideration, it is to be observed that, owing to the neutral axis being midway between CE and DF, the amotmt of shortening of the fibres at CE is equal to the lengthening of the fibres at DF. It will be seen that in Fig. 63, o^^ti^j, and o'np' are drawn as straight lines, which expresses graphically assumption 3. Combining assumptions 1 and 3, it will be seen that the stress in any fibre is proportional to its distance from the neutral axis, and further, from assumption 2, that the numerical value of the tension in a fibre is equal to the numerical value of the compression in a fibre at the same distance on the other side of the neutral axis. Thus in Fig. 66, if the tension on the fibres at TU is 8 cwts. per square inch, the compression on the fibres at RS is also 8 cwts. per square inch. The amount of stress is shown graphically in Fig. 67 ; that is, if CE represents the compression at CE, will represent the compression at PQ, and so on. ^ This assumption has been proved by experiment. See A Manual of Civil Engineering, by W. T. M. Rankine, F.R.S.S., etc., p. 250, 10th edition. — O BEAMS— MOMENT OF RESISTANCE 45 Stresses in beam 1" wide. — Applying these assumptions in the first instance to a simple case. Suppose A^B (Fig. 68) to be a rectangular beam 1" wide and 12" deep, supported at the ends and loaded in the centre, then each inch in depth of the beam will contain one square inch of fibres. Suppose also that the " elastic limit " of the material of which the beam is composed is 1 0 tons ^ per square inch. We know that the fibres will bear that stress without injury, but being determined to be well under that limit, we wish to arrange that the stress upon the fibres shall nowhere exceed 6 tons per square inch. The limiting stress then is to be 6 tons per square inch. On reference to Case 6, p. 32, it will be seen that the maximum bending moment occurs at the centre of the beam. If, therefore, the stress on the extreme fibres at the centre of the length of the beam does not exceed 6 tons per square inch, we can be quite sure that it will not exceed it anywhere else. Again, by assumptions 1 and 3, the stress upon the fibres is in proportion to their distance from the neutral axis. The stress at the neutral axis is nothing, but the compression upon the fibres above the neutral axis becomes w greater and greater as they approach the extreme fibre A^Bj, and the tension upon the fibres below the neutral axis becomes greater and greater as they approach the extreme fibre AB. Considering, therefore, the central cross section CD of the beam ; if c'd' be drawn so that Cc' = Bd' = 6 tons on a scale of 10 tons to one inch (Figs. 68 and 69a), the stress at any other point will be represented by the line drawn across the triangles of stress BOd' and COc' at that point parallel to the neutral layer. Thus mm' represents the stress at the point m =' 5 tons per square inch, as measured on the scale, nn' represents the compression at the point n = 2 tons per square inch, and pp' represents the tension at ^ = on the scale 4 J tons per square inch. ^ These are taken merely as simple convenient figures, not with reference to any particular material. NOTES ON BUILDING CONSTRUCTION Neutral D Fig. 69a. C tons 6 tons The mean compression on the fibres above the neutral axis will be that represented by 22 ( = 3 tons) at a point q, half-way between O and the extreme fibre. In the same way the mean tension on the fibres below the neutral axis is rr' = 3 tons. This mean compression acts over a surface 6" x 1", the total compression is therefore 6" X 1" X 3 tons = 18 tons. In the same way, below the neutral axis the total ten- sion is = 3 tons X 6" x 1" = 1 8 tons. We have then in Fig. 69 a cross section of the beam, and in Fig. 6ga a diagram which shows the resistance offered by the fibres (equal to the stress upon them) at any point in the cross section, and also the total resistance of the cross section both in compression and tension. The diagram also shows us how this resistance is distributed, i.e. that the fibres at the neutral axis contribute nothing to the resistance, but that the resistance of the fibres increases in proportion to their distance from the neutral axis until it reaches a maximum in the extreme layers of fibres CE and DF, where each undergoes a stress equal to the limiting stress of 6 tons per square inch. Graphic Method of Finding the Moment of Eesistance.^ — It is, however, convenient to combine the cross section of the beam with the diagram as follows : — Draw the cross section of a beam CEDF as before (Fig. 70), and make a new scale of tons in which the length CE = 6 tons, the limiting stress per square inch, and draw the stress dia- gram (Fig. 71) similar to Fig 69a above. Fig. 7 2 is merely a modification of Fig. 7 i and clearly gives the same results, and can be inserted in Fig. 70 as shown. Now the area of the triangle CEn is half the area of the rectangle CEOO, and, therefore, the total compression is — Tig- 70. = area of rect. CEOO x 3 tons, = area of triangle CEn x 6 tons, 6x1 ^ ^ == X 6 = 1 8 tons, the same value as before. In other words, a compression of 6 tons uniformly distributed over the triangle CEu is equivalent to the actual compression existing at the section of the beam. And in the same way a tension of 6 tons uniformly distributed over the triangle nDF is equivalent to the actual tension. The areas of the triangles CEn and nI>F can therefore be called the equivalent areas of resistance. In the case of beams more than 1 inch wide, the application of this method of showing the quantity and distribution of the resistances is very simple. Stresses in Beam 4 inches wide. — Thus a beam 4 inches wide and 12 ^ This method is described as it illustrates the principles and for the sake of those who prefer drawing to calculatious, but it is easier to use the formulae, p. 54. Fig. 71. Fig. 72. BEAMS— MOMENT OF RESISTANCE 47 inches deep is the same as four beams each an inch wide and 12 inches deep (such as that we have been considering) placed side by side. The diagram showing the intensity of stress on the fibres of such a beam will be the same as before (Fig. 69a). In the diagram (Fig. 73) showing the quantity and distribution of stress, as there are 4 inches in CE, each acted upon by a stress of 6 tons per square inch, CE will represent 4x6 = 24 tons. CEw will be the equivalent area for compression, and nDF the equivalent area for tension. The area CEw = ^ x 4" x 6", = 12 square inches. The resistance to compression of CEr!,= 6 tons x 12 square inches, = 72 tons. The tensile resistance of ?iDF = x 4" x 6" x 6 tons, = 72 tons : or in each case four times as much as the resistance (18 tons) of the beam 1" wide and 12" deep. In the same way the resistance of any beam may be found, whatever its breadth, depth, and strength of fibre. c In all these diagrams it will be noticed that the equivalent area representing tension is equal to that representing compression. It is evident that this must be the case, for if the total compression above the neutral axis were not equal to the tension below, there ^ would not be equilibrium. Value of Moment of Resistance. — Having thus drawn the triangles showing the amount and distribution of the resistance of the cross section, it will be easy to ^ f ' ascertain the value of the moment of resistance, which, as explained at p. 43, has to be made equal to the " bending moment produced by the load. It will make the general rule more clear if we deal first with a particular case. Let us take the case we have already been considering, that of a beam 4" wide and 1 2" deep, having a limiting stress upon the extreme fibres of 6 tons per square inch. We know from p. 46 that the amount and distribution of the resistance in compression is shown by the triangle CEn, that in tension by nDF, Fig. 73. This diagram tells us then that if 6 tons per square inch according to the scale were acting uniformly all over the triangle CEn, it would ofi"er in quantity and distribution exactly the same resistance as is offered by the vary- ing stress which actually occurs over the rectangle CEOO. Again, the actual varying tension on CODE is represented in quantity and distribution by 6 tons (on the scale) per square inch acting uniformly all over the triangle nDF. Now if 6 tons were acting on each square inch of CE?i, we know by the elementary rules of mechanics that it will be the same thing if we consider the total number of tons acting over CE», as being concentrated at the centre of gravit}^ and acting there. The centre of gravity of CEu is at the point c, and that of iiDF at (/.^ It has been shown above that the total compression is 72 tons, and that f the total tension is the same. ^ c is 2 inches or J of cji from e, ^ is 2 inches from/. 48 NOTES ON BUILDING CONSTRUCTION The resistance of the cross section is therefore equivalent to two forces, each equal to 72 tons, and acting at c and g. On reference to p. 43 it will be seen that these forces form a cou'ple, whose moment can be found by multiplying one of the forces by the dis- tance between the forces, or, as it is called, by the arm of the couple, in this case eg. The moment of this couple is clearly the moment of resistance required. The moment of resistance of the cross section under consideration is therefore = 7 2 x c£f. These centres of gravity are found by bisecting CE at e and DF at/ — joining ef and taking ec and gf equal to 1 of en and nf respectively. Now eg is evidently equal to cn + ng Fig. 74, and cn = fen = -I x 6 = 4" ng- or the moment of resistance .•• cg = X 6 = 4 Fig. 74.2 Thus, as in Fig. 73, so that the distance eg = 72 tons X 8 inches, = 576 inch-tons.^ Safe resistance Area — General Case. — Now to take a general case (Fig. 74). Suppose the breadth of the beam in inches is called b, the depth in inches d, and the limiting stress to be allowed on the outside fibres tons per square inch. As in the previous case, the triangles CEw and nD¥ are the equivalent areas of resistance each ^bd, the stress being ?•„ tons per square inch. The total compres- sion is therefore r^ x ^bd, and the total tension has the same value. cn = ng = i-~, or the length of the arm of the couple, will be ld + ^d = |d Eesulting Formula. — -The moment of the couple, i.e. the moment of resistance required, is therefore r^x^bdx^d = ^ rj)d^. We have now foiind a general formula for the moment of resistance of a rectangular beam, in terms of its breadth, depth, and of the limiting stress allowed upon the extreme fibres. The method by which this formula has been arrived at is shown step by step in the preceding paragraphs, and as the formula has been deduced from ^ Another way of looking at this is : The fibres in CEn act with an average leverage e(iual to tlie distance from their centre of gravity to the neutral axis -on— 4 inches ; in the same way the fibres of ?iDF act with an average leverage of 4". Thus we have total resistance tons tons CEnx6x4" + 7iDFx6x4", = 72x4 +72x4, = 576 inch-tons. 2 The shaded portions of this figure represent in magnitude and distribution the stresses over the section of the beam— just when the extreme fibres on each side are subjected to the limiting stress. BEAMS— MOMENT OF RESISTANCE 49 the geometrical demonstration, it is in agreement with it, and may be used instead of working out each case graphically. Two conclusions are self-evident from the formula, viz. : (1) That the moment of resistance in a rectangular beam varies directly as the breadth of the beam, i.e. that a beam 12x4 will have a moment of resistance twice as great as one 12x2; and (2) That it varies as the square of the depth of the beam, i.e. that a beam 12x2 will have a moment of resistance four times as great as one 6x2. It will be understood that the limiting stress should not be greater than the worlcing stress that can safely be allowed upon the particular material used and never greater than the elastic limit. When such a stress is used, the result gives the worlcing resistance of the beam ; but it should be remembered that if the breaking stress were used, the resistance found would not be the ultimate resistance, or actual resistance to breaking, because when the breaking point is approached the assumptions made (p. 43) no longer hold good. The value of the moment of resistance of any beam of rectangular section with any limiting stress can be obtained either by drawing the resistance diagrams and calculating the amount of the moment from them, or by substituting the values of h, d, and r„ in the formula. Effect of depth upon strength. — To illustrate the self-evident conclusions mentioned above, suppose we wish to ascer- tain the relative strength of three beams A, B, 0, of the same length, but differing ^ in depth — A being 4" deep, B 6" deep, C 8" deep, and the width of each being 2". Sup- pose also that the limiting stress to be D allowed upon the extreme fibres is to be ^ taken in each case at 3 tons. We will calculate these beams first from the diagrams showing quantity and distri- bution of resistance, then by the formula. The diagrams are as in Fig. 75. Calculation from Diagrams or Resistance. Calculation erom Formula To 1 Beam. 2 Dimen- sions. 3 Resistance of section = area of GE?!, or DwP X worlcing stress. 4 Lengtli of arm of couple, i.e. from c to g. = id 5 Moment of Resistance = Resistance of section x arm of couple. 6 Value of To, h, d. 7 Value of Moment of Resistance TO-. A ^ p. 2" X 4" tons tons ix2x2x3=: 6 fx 4"= 1" inch- tons tons 6x f" =16 To = 3 tons b = 2" c^ = 4" inch- tons 0 2x4x4 , 3 X =lb 6 B 2" X 6" ix2x3x3= 9 |x6" = V-" 9xi/"=36 ro = 3 tons 6 = 2" d = 6" „ 2x6x6 3x — - — =36 0 2" X 8" 4 X 2 X 4 X 3 = 12 I X 8"=-V-" 12x J/"=64 ro = S tons 6 = 2" d=8" „ 2x8x8 3 X — - — = 64 6 We see from the above that the B.C. IV. moments of resistance found in column E 5° NOTES ON BUILDING CONSTRUCTION 5 by calculation from the diagrams are tlie same as the moments of resistance found in column 7 from the formula. Moreover, we see that the width, etc., being the same, the moment of resistance, i.e. the strength of the beams, varies as : 16 -.36 : 64, i.e. as 4^ : 6^ : 8^, i.e. as the squares of the depths of the beams. Effect of width upon strength. — Let us see the effect of varying the width upon the strength of a beam. Take beam C, and instead of 2", give it first a width of 3", then of 4", then of 6", calculating by the formula we have : Size of Beam. 8" X 2" 8" X 3" 8" X 4" 8" X 6" Value of Moment of Besistance 3 X 3 X 3 X 3 X 2x8x8 3x8x8 6 4x8x8 6 6x8x8 = 64 = 96 = 128 = 192 We see from the above that when the depth remains the same the resist- ance or strength of the beam varies as the width. Thus the resistance of a beam 3" wide is one and a half times that of a beam 4" wide, twice that of a beam 6" wide, three times that of a beam 2" wide. General rule. — We see then, as already mentioned at p. 49, that the strength of rectangular teams of the same material vary directly as their breadth and as the square of their depth. Tliis indeed is apparent from an examination of the formula, for the resistance of the beam is shown by that formula to vary as bd^. General Formula.— This formula, Moment of resistance = r^, — ■ . (28), 6 is therefore, as has been mentioned above, a general ex- pression for the value of the moment of resistance of a beam of rectangular section, when it is subjected to such a load that the limiting stress on the fibres nowhere exceeds the elastic limit of the material.^ If the assumptions mentioned at p. 43 were true for all conditions of stress, ^ This formula is practically the same as that given by Kankine and other writers, who have approached the subject from the side of advanced mathematics. They r I . . » show the moment of resistance to be equal to where I is the moment of inertia of BEAMS— MOMENT OF RESISTANCE 51 ;]iat is, if they were true for every degree of stress from very small ones up to ;he breaking point, then would be that value of the stress either of tension )r compression which was known to cause rupture by tearing or crushing in ;he fibres of the material, and could be ascertained by direct experiments ipon the tensile strength and resistance to compression of the material. Experiment has shown, however, that these assumptions are true only for stresses such as are well within the elastic limit of the material, the shorten- ing or lengthening of the fibres varying regularly with the compressive or tensile stresses (see p. 11) so long as the elastic limit is not exceeded. Beyond that, the ratios of the shortening and lengthening to the corresponding stresses producing them advance more and more rapidly until, on approaching fracture, they become irregular, but always considerably in excess of what they were under the lower stresses. The effect of stresses in tension often differs also from the effect of the same stresses in compression, and the result of all this is that the position of the neutral axis is altered — it no longer passes through the centre of gravity of the cross section ; e.g. if the material is at the point of rupture stronger to resist compression than tension, the neutral axis will then be nearer the outer- most fibre in compression than to the outermost fibre in tension. Thus in a beam of such material, supported at both ends and loaded, the neutral axis will be nearer the top than the bottom edge. This difference between facts and the assumptions, at the time when the breaking stress is approached, renders the theory founded upon them inappli- cable for the purpose of ascertaining the resistance of beams to actual 'breaking. It should be noticed that in these examples the weight of the beam itself is left out of the consideration. When it is large enough to require attention it must, of course, be treated as a distributed load, and its effect be added to those of the load proper. METHOD OF ASCERTAINING THE MOMENT OF RESISTANCE BY EXPERIMENT AND CALCULATION. When we liave to deal with heams loaded so that they are more than slightly bent, or when the intensity of stress upon their fibres exceeds the elastic limit, the theory and the method just described no longer apply ; there is no sufficiently simple theory which is applicable, and we must have recourse to experiment. Careful experiments have accordingly been made, from which it has been found that the resistance of rectangular beams of the same length and of similar section similarly loaded up to the ireahing point varies directly as the breadth and as the square (/' the depth, i.e. as hd'^. This seems to agree with the result arrived at by reasoning in the case of beams only slightly bent by a load. the cross section, and yo is the distance from the neutral axis to the extreme fibre subject to a stress To. In the case of a rectangular timber beam 2/o=^ and hd? r I IxJP' 1 = — , therefore — =ro — ; and the formula reduces to that given above. NOTES ON BUILDING CONSTRUCTION If tlie agreement were complete, tlie formiila given at p. 50, i.e. M = r„ — , 0 g , might be applied even to cases of beams loaded up to the breaking point, by- taking r„ as the ultimate resistance of the fibres per square incli to tension or compression. This formula may be and is used for beams so loaded, when an important alteration has been made in it. The coefficient r„ no longer represents the resistance of the material to tension or compression up to the elastic limit ; but experiment shows that it has an entirely new value — depending upon the material, the form of beam, and disposition of the load— and that this new value can only be ascertained by experiment, and will be denoted by /„. Value of f„ for timber. — The method in which this value was ascertained was as follows : — Small experimental beams each 1" square and 1 2" long between the sup- ports were supported at the ends and loaded in the centre with a weight W. From Equation 13 we know that in such beams the bending moment = . Substituting/, for r„ in Equation 28, Ave have the moment of resistance = f!^ (29). Hence = f — , r 6 or /o=-— -Xy-, 4 0(1- The weight W was gradually increased until the beam broke, and then the value of W, I, h, and d in the above equation being known, / could be found. Modulus of rupture of timber. — The value of /„ thus found is called the modulus of rupture ^ of the material. It varies of course for different materials — the stronger the material the higher the modulus. Thus a beam of oak would require a greater weight W to break it than a beam of pine, and /„ would therefore be greater. It is not accurately known why the modulus of rupture thus found by experiment differs so widely, as it does in many instances, from the resistance of the fibres to either tension or compression. It must be remembered, however, that the fibres of beams are not in- 1 The modulus of rupture thus found is always equal to 18 times the weight W that will break the beam across. The reason for this is easily seen if the values of I, b, and d for the experimental beam are substituted in Equation 29 above — 1 = 12", 6 = 1", d = l". 4 ^bd^' ^Wxl_2 6 4 1x12 BEAMS— MOMENT OF RESISTANCE 53 dependent filaments, but are united together laterally with, an adhesion less or greater according to the nature of the material. This lateral adhesion has been shown by Mr. Barlow to lead to stresses of a very complicated character, which so far have not been dealt with by any simple mathematical theory, and its effect is practically to alter the value of the coefficient/,. There is also another consideration. The strength of the individual fibres cannot be uniformly the same, as it is assumed to be. So long as the limit of elasticity is not passed, this difference of strength is so small as not to affect the result ; but after that point the weaker fibres begin to yield more than the others, and finally, just before rupture, these weaker fibres break first and leave the whole strain to come successively upon fewer and fewer fibres, uutil complete rupture takes place. Under these conditions, /„ cannot be a constant coefficient. The value of /„ obtained by the experiments above referred to is gener- ally used in practice, though it is in most cases too high, for the following reason. The small beams (speaking of timber) experimented upon were as a rule selected specimens perfectly dry, free from defects, such as large or dead knots, etc., which tend to interrupt the continuity of the fibres and thus weaken the beam. Large scantlings cannot be obtained so well seasoned or free from defects as the small experimental beams, and therefore in calculating the strengtli of large beams the value of /, should be taken considerably lower than the value found for the experimental selected beam of the same material. For example, in the case of Red pine, the modulus of rupture found by experiments upon small selected beams, varies from 7100 to 9540 Ibs.^ ; whereas the average modulus found by breaking large beams is 45 cwts. or 5040 lbs. -per square inch.^ It would be expensive to ascertain the modulus by breaking very large beams, and after all, even if this were done, the timber of the beams used Avould probably not be exactly similar to that experimented upon, and would therefore have a different modulus. It will be seen, then, that the method of ascertaining and applying the modulus of rupture is only a rough way of correcting the imperfections of the theory upon which the formula is based. The method answers, however, sufficiently well in practice, and it has the advantage of being very simple. An engineer in a distant country having to deal with an entirely new material can at once ascertain its modulus of rupture by experimenting upon small bars as above described ; and having found the modulus he can use the ordinary formula given above. It is far easier for him to find the modulus of rupture in this way than to find the amount of extension and compression of the material under different stresses — which would be necessary if the perfect mathematical theory could be worked upon. Modulus of rupture for different sections of beam. — In using the modulus of rupture thus obtained by experiment, it must be remembered that it is only apphcable in the calculation of beams of similar section to the experimental beam from which the modulus was deduced. In cases where the modulus obtained from a rectangular beam of cast iron placed thus D was applied to the calculation of the resistance of a beam with a circular sec- ^ Rankine's Useful Rules and Tables, p. 199. ^ Seddon's Builders' Work. NOTES ON BUILDING CONSTRUCTION tion, the error caused was found to be 174 per cent, and in a case where it was applied to a square beam placed on edge, thus, 0> the error was 190 per cent. In practice, however, beams of these sections are not used for building construc- tion in any material, and they need not be further considered. Even a difference in the method of loading the beam used, as compared with the method in which the experimental beam was loaded to find the modulus, has been found to cause errors in results ; but these errors are comparatively small, and may be neglected in using a method which after all is merely a rough and ready way of arriving at the result. The moduli of rupture given in Table I. are those obtained by breaking beams by means of weights placed on their centres, but these moduli may, without material error, be used in calculating beams with any distribution of load. ^ Practical Formula. — In practice, the rough, and ready way of using the formula (Equation 29) in conjunction with the modulus of rupture is as follows : — To find the weight that will break a beam. — Find from pp. 27 to 40 the bending moment, M, for the given conditions of load — equate this with the moment of resistance, M (Equation 29, p. 52), and we have M='^-^ .... (30). Take the case of a beam supported at the ends and loaded at the centre,^ W"^ = breaking weight in centre I = length j h = breadth I all in the same dimensions, either feet or inches, d = depth J = modulus of rupture for the material (see Table I.) M = — (see Equation 13, p. 33) To find the load that a learn can safely hear. — is, however, the hreaJang load. To find the safe load the factor of safety F must be introduced. Taking W as the safe load with that factor. FW: W = " 61 fo^. To find the section of a beam to hear safely a given weight. — Take the same case as before — a beam supported at ends and loaded in centre. fjbd^ From this find hd^, and assuming a value for either b or d, find the other, Avhich will give the sectional dimensions of the beam. Examples of the calculation of beams are given at p. 75 and seq. ^ For moduli of rupture of other materials, see Part III. It must be clearly understood that M will vary according to the arrangement of the beam and loads. Its different values are given at pp. 27 to 40, also in a concise form in Appendix VII. (see also Appendix XXI.) BEAMS— SHEARING STRESS 55 SHEAEING STEESS. We have hitherto considered only the effect of the direct stresses upon a beam, that is, the stresses which eventually act upon the fibres by extending or compressing them in the direction of their length. There are, however, other stresses acting upon a beam the existence of which will be rendered evident upon considering the next three diagrams. Vertical Shearing Stress. — If a beam AB, Fig. 76, supported at the ends, be bent, the fibres above the neutral layer are evidently com- g pressed and those below it extended, but no other stresses in the beam are apparent to the observer. If, however, the continuity of the fibres be interrupted by cutting the beam into vertical slices, these slices, when the beam is loaded, will slip one past the other, as shown in Eig. 77. In the uncut beam there must be the same tendency for each section to slide vertically upon the next, although actual sliding is prevented by the continuity of the fibre or cohesion of the Fig. 77. Fig. 78. particles. The force causing this tendency is called the vertical shearing stress. Horizontal Shearing Stress. — In the same way, if the beam be divided into different layers of fibres by cutting it into planks, the planks or layers of fibres will slide upon each other as in Fig. 78, when loaded. In the uncut beam there must be a tendency to slide in the same way, and the force causing this tendency is called the horizontal shearing stress. Bules for finding Amount of Vertical Shearing Stress.^ The amount of the vertical shearing stress is easily ascertained by the application of two simple rules — ^ The practical formulce for finding the shearing stresses are on pp. 57 to 60, numbered (31) to (43c), each being in connection with figures showing the distribution of loads, diagrams of shearing stresses, etc. S6 NOTES ON BUILDING CONSTRUCTION Fig. 79 is a cantilever, or a beam / k h / 1 L til A t S Fig. 79. e B 1. Vertical Shearing Stress in Cantilevers. — The shearing stress at any section is equal to the weight on the beam between that section and the outer end. 2. Vertical Shearing Stress in Beams supported at each End. — The shearing stress at any section is equal to the difference between the reaction at either support and the weight between that support and the section in question. These rules hardly require any proof, but it may make them more clear to examine a case of each. SheariTU) stress in cantilevers. — AB, loaded at the end with a weight W. We may divide the beam into ima- ginary vertical sections at ki, hg, fe; these are only a few out of an infinite number of sections infinitely near to one another. Now "W acting downwards tends to shear off the piece B/, with a force W, acting, as shown by arrow thus, on the section fe ; this is resisted by the opposite surface of the section fe, the resisting force acting as shown by the arrow -f- upwards. Supposing then that the resistance prevents the shearing at surface fe, consider the next surface hg. Here the load W still acts with a force W downwards, tending to shear off the piece Hh. Again in the same way at hi, W the load acts downwards, and a resistance = W acts upwards, as shown by the arrows. Nothing has intervened to increase or diminish the force W caused by the load.^ The same takes place at any intermediate section, right back to Al. The shearing stress at any section from B to A is therefore always the same, i.e. = W as shown graphically in Fig. 82. Shearing stress in supported learns. — As regards the second rule, take the case of a beam (Fig 80) supported at the ends and uniformly loaded. SupjDosing the beam to be 8' span and loaded with tu lbs. per foot run : and each = 4u\ At the section marked the reaction 'Rj^ = 4w acts upwards ^, and is W/ lUj Wg Fiir. 80, Tm. 81. resisted by the four weights 4w acting downwards, so that the shearing force at section 1 = R.. ^ The weight of the beam itself does really increase it, but we have arranged to neglect it as insignificant. If it is not insignificant it would cause a shearing stress similar to that caused by a distributed load, as described on the next page. BEAMS— SHEARING STRESS 57 At the section marked one-fourth of the upward force of the reaction = 4i« is cancelled by the weight acting downwards, so that the upward force at the section 2 is Zw, which is resisted by Zw acting down- wards between it and the centre. Similarly at section 3 the shearing force is (R_^ - 2w) = 4w-2w = 2iu, and at section 4 it is Iw. At section 5 (the centre) the upward force of the reaction is quite neutralised by the four weights iw^, w^, w^, acting downwards and the shear- ing force = 0. Thus at each section the shearing force = — weights between the section and support A, as stated in Rule 2 above. The same reasoning applies to the part CB of the beam. The shearing stress is evidently decreasing from Aiv at the supports to 0 at the centre, and may be graphically shown as in Fig. 8 1 . The truth of the rules is so apparent, and their application to different cases so simple, that it will be sufficient to show a few of the most useful cases graphically, giving at the same time the value of the shearing stress in terms of the weights and lengths of the beams. The shearing stress at any point P is denoted by Sp, and at any other points A, B, C by S^, S^, S^, etc., respectively. Graphic Representations ^ of Vertical Shearing Stress, and Values for the Same. Case 1 (Fig. 82). — beam fixed at one end and loaded at the OTHEK. Shearing stress S = W throughout . . (31). / _>1 --/-AT ^— / /-X n ip )) w Fig. 83. AND LOADED Fig. 82. Case 2 (Fig. 83). — beam fixed at one end TJNIFOEMLY THROUGHOUT ITS LENGTH. Sp = wx. The point P being distant x from B . (32). 8^ = wl (32 A). Case 3 (Fig. 84). — beam fixed at one end, loaded at the OTHER END AND ALSO UNIFORMLY. A combination of Cases 1 and 2. The loads are omitted in the figure for clearness, but are as in Figs. 82 and 83. ^ These are merely graphic representations of results obtained algebraically, but in Appendix VI. will be seen a graphic method of obtaining the shearing stresses. -rmi,'. j„^X..m ..4ta:>fl«1^ 58 NOTES ON BUILDING CONSTRUCTION Sj. = W + wx (33). (33 A). Fig. 84. Case 4 (Fig. 85). — beam fixed at one end and loaded with SEVERAL concentrated WEIGHTS W^, Wg, Wg. S, = Wi + W2 + W3 . . . (34). S at any section = the sum of the loads between that section and the outer end of the beam. ....... (34 A). Case 5 (Fig. 86). — beam fixed at one end and loaded UNIFORMLY OVER A PART OF ITS LENGTH. The shearing stress at any point P under the load distant x from A is ^^ = w{z + y -x), . . . (35), and Sj^ = 'wz, . . . . . (35 A). S^^wz (35 B). w n lut it It 1 wlliilllll Fig. 87. Fig. 86 Case 6 (Fig. 87). — beam supported at both ends and loaded IN the centre.^ w S throughout = (36), except at C where the stress changes direction and S, = 0. . . .(36 A). The arrows indicate the different directions of the shearing stress 011 the opposite side of the centre. Thus S anywhere in CA the stress is thus tending to shear the ^^S- 88. Fig. 89. ^g^j^ -pig. 88, whereas in the half BO the stress is thus tending to shear the beam as in Fig. 89. At C the stresses in opposite directions cancel one another, and there is no shear. ^ See Appendix V. BEAMS— SHEARING STRESS 59 Case 7 (Fig. 90)! — beam suppokted at the ends and uniformly LOADED. See the remarks on p. 56 with, regard to this case. Case 8 (Fig. 91). — beam supported at both ends, loaded in the CENTRE AND ALSO UNIFORMLY. This case is a comhination, of 6 and 7. The load is oraitted for clearness. wl W ^ wl W (38). . (38 A). S(,= 0 (stress changing direction) (38 B). Case 9 (Fig. 92). — beam supported at both ends and loaded at Between A and D Between B and D ANY POINT. (39). (40). At D Sj, = 0 (stresses changing direction) (40 A). 6o NOTES ON BUILDING CONSTRUCTION Case 10 (Fig. 93). — beam supported at both ends and PARTIALLY LOADED WITH A UNIFORM LOAD. Between A and E Between B and F (I — mf — At any section P under the partial load distant x from A (I — lllf — TC^ 21 (x-m) > V) (41) . (41 A). (42) . Referring to p. 38 it will be a good exercise for the student to deduce the previous cases of beams supported at both ends from this case. Case 11 (Fig. 94). — beam supported at both ends and loaded "WITH ANY NUMBER OF CONCENTRATED LOADS. Between A and D Between D and E Between E and F Between F and B ^ — — I — + I S = R,-W S = - W - W, S = R = Wm, W„m„ Worn, I + 3"''3 I (43). (43 A). (43 B). (43 C). U— - t<- U— .i.vt2 A .f/ia !- Horizontal Shearing Stress. — We have hitherto considered only the vertical shearing stress, which tends to make sections of a loaded beam slide vertically npon one another, as in Fig. 77. We have now to consider the horizontal shearing stress, which tends to make the layers slide horizontally upon one another, as in Fig. 78. BEAMS— SHEARING STRESS 6i Fig. 95. This horizontal shearing stress is at every point in the learn equal to the vertical shearing stress at that point. Tliis is proved thus : — On or near the neutral layer of an unloaded rect- angular beam draw a little square, Fig. 95. If the beam is then loaded until it bends slightly, this square will be distorted into a rhombus, Fig. 96. Now we know that at the neutral layer of a rectangular beam there are no direct stresses. The distortion must therefore have been produced by the vertical shearing stresses V, V acting as shown by the arrows, and it is opposed by the horizontal shearing stresses H, H. Now since there is equilibrium, H and H must evi- dently be equal to V and Y, or the horizontal shearing stress must be equal to the vertical shearing stress at the point. ^ We know then the total amount of shearing stress at each section — the vertical is found as before explained and the horizontal is equal to it. Distribution of Shearing Stress. — It is necessary to know further how these shearing stresses are distributed over the section. It can be proved mathematically that the shearing stress is distributed over the section, as shown in Tig. 97. The curve is a parabola, the relative proportion of whose ordinates is shown in the diagram. The area^ of the parabola represents to scale the total shearing stress at the section. From this the horizontal shearing stress at any point can easily be ascertained. Thus suppose the total shearing stress S at any section of a beam 1 2" deep is 2 tons. Then I NL X CD = S, f X NL X 12" = 2 tons, NL == 2 tons X 1^ X y 5 _ 6 — 2 4' = i ton. Having thus found the stress on NL to be ^ ton, we make a scale on which the length NL = ^ ton, and we can find the stress at any other point by measuring the ordinate of the parabola at that point. It is unimportant in most cases to know the exact distribution of the shearing stress over the section, but we can see from the diagram that it is greatest at the centre and vanishes at the top and bottom layers of the beams. ^ Cunningham, p. 244. - The area of a parabola = | the circumscribed rectangle. The area = | x CD x NL. 62 NOTES ON BUILDING CONSTRUCTION This is exactly the converse of the distribution of the direct stresses. The difference is clearly shown by the diagrams, Figs, 98 and 99. Direct Stresses Neutral u iMyer U F Fig. 99. We see from these diagrams that the shearing stress is greatest along the neutral layer where the direct stress is the least, and least at the upper and lower fibres where the direct stress is greatest. In Fig. 99 the amount of the shearing stress is exaggerated in order that it may be clearly shown, but its amount is generally very small in com- parison with, the direct stresses ; so small that in rectangular beams whicli have so much, more substance near the centre than is required to meet the direct stresses, there is sure to be plenty to meet the small shearing stress, and its consideration may therefore safely be neglected. In built-up iron beams, however, it is different ; in those the shearing stress plays an important part (see p. 157). BEAMS OF UNIFOEM STEENGTH. If we glance at the graphic representation of the bending 3 moments at the different points of a beam supported at the ends and uniformly loaded throughout its length. Fig. 100, we see that the bending moment is greatest at the ^^g- centre and gradually diminishes to- wards the ends, and that in fact exactly at the points of support there is no bending stress whatever. If, therefore, the beam be made of the same section through- out, it is evident that it is unnecessarily strong at every point except the centre, and that it may be reduced in section as it approaches the points of support, because the bending moment becomes less and less as those points are approached. There are three ways in which the beam may be reduced so as to make it of equal strength throughout its length — 1. By varying the depth, keeping the breadth the same ; as in Figs. 102-105. 2. By varying the breadth, keeping the depth the same ; as in Figs. 106-109. BEAMS OF UNIFORM STRENGTH 63 3. By varying both breadth and depth. The same observations apply to other cases of loading. Thus, for example, in a cantilever of rectangular section, with either a single or a uniform load, there is superfluous material except near the point of fixing (see Figs. 102, 103), and in a supported beam carrying an isolated load there is superfluous material except under the load. In all such beams then a certain portion of the material is superfluous, and can be cut away without injuring their strength. In timber beams, as a rule, it does not pay to cut away this superfluous material, the cost of labour being greater than the value of the material saved ; but in some cases for the sake of appearance, or in large beams to reduce the weight, it may be desirable. It can, however, never be necessary to cut timbers to the exact curves shown. These curves are the forms which are theo- retically correct, but practically they are valuable only as dividing the parts of the beam which have no excess of strength from those parts which are unnecessarily strong and can therefore be cut away, thus making a beam of approximate uniform strength. The following figures show the shapes of beams of equal strength through- out their length so far as the direct stresses only are concerned, and in each case ignoring the shearing stress and the weight of the beam itself. The shearing stresses would necessitate some substance being left on at the ends of the supported beams. It is not difficult to prove mathematically that beams of equal strength throughout their length are of the forms shown, but such mathematical proof would here take i;p more space than can be afforded for it, and the question is not one of much practical importance. Case 1 (p. 28) is, however, worked out as follows, and the other cases can be worked out in a similar manner. Calculation to ascertain the form of a cantilever of equal strength throughout its length, loaded at one end and of the same breadth throughout. Bending moment at any point P (Fig. loi) distant x from the end W = Wx. Moment of resistance at P = ^fy^. Fig. 101. 64 NOTES ON BUILDING CONSTRUCTION Wl' X V Then since at any point bending moment = moment of resistance, we have — At P : Wx = i/;%2. At A : = ^M^- t. which is the equation to a parabola with its vertex at B ; the under side of the beam would therefore be shaped to this curve. Shapes of Beams of uniform Strength. a. WHEN THE BREADTH IS CONSTANT THROUGHOUT THE LENGTH OF THE BEAM, BUT THE DEPTH VARIED TO SUIT THE VARYING STRESS. Figs. 102-105 show the disposition of load and the shape of the beam in elevation, the dotted line being the beam of constant depth and breadth as calculated. In Fig. 102, AB is a parabola with the vertex at B. In Fig. 103, AB is a straight line. Fig. 102. K ' Fig. 103. In Fig. 104, AB and PB are parabolas with their vertices at A and B. Fig. 104. In Fig. 105, ABC is a semi-ellipse, AB being the major axis. BEAMS OF UNIFORM STRENGTH 65 A '^-^-^^"^ ---.-.^c^-- - 1 ✓ ^ — ■ — ^ ^ A < 1 Jf- .X. f ^ \ r Fig. 105. I. WHEN THE DEPTH IS KEPT CONSTANT THROUGHOUT THE LENGTH OF THE BEAM, AND THE BREADTH VARIED TO SUIT THE VARYING STRESS. Figs. 106-109 show the disposition of load and the shape of the beam in plan. In Fig. 106, ABD is a triangle. In Fig. 107, AB, DB are both parabolas with vertices at B. In Fig. 108, DAC and CBD are two triangles with their bases at DC. In Fig. 109, ACB and ADB are parabolas with their vertices at the centre points C, D. 66 NOTES ON BUILDING CONSTRUCTION DErLECTION.i In many structures it is necessary that the beams, whether of wood or iron, forming part of the structure, should be not only strong enough, but stiff enough ; that is, not only must rupture be prevented, but the beams must not bend too much. For example, it would never do if tlie joists supporting a ceiling were to bend beyond a certain small amount under the load, although they might be strong enough, because a very slight bending is sufficient to cause cracks in the ceiling below. Such joists must therefore be stiff as well as strong. It is necessary, therefore, to be able to calculate how stiff a given beam is, that is, how much it wiU bend under its load, in order to be able to judge whether the amount of bending is likely to cause inconvenience ; and if so, to increase the size of the beam in such a manner as to prevent such excessive and inconvenient bending. The amount of deflection that may be expected in a beam of known form and material, with a given amount and distribution of load, can easily be ascertained by the use of the formulae given below. The investigation of the formulae, however, involves the use of the Calculus ; their proof will therefore not be attempted in this work. The student must be content to accept these formulas, which are based upon those given by Professor Eankine, after investiga- tions made by himself and other mathematicians. Formulae for ascertaining the Deflection of Beams of any Kind. General Formula. — Th e general formula applicable to all kinds of beams, of any material and with various distributions of load, is Fig. 110. In this formula A = the maximum deflection in inches. Thus in Fig. iio, if the beam AB deflect as shown by the arc Ai/B, the line xy is denoted in the above formula by A. W = the total load on the beam. ^ The formulae for practical use are Epilations 44, 45, 46, 47, pp. 66, 67, 68, and in Appendices VIII., IX., XXI. Examples are given, pp. 76 to 80. BEAMS— DEFLECTION 67 I = the length of the beam in inches. E = the value of the modulus of elasticity of the material expressed in the same units as W. I = the moment of inertia about the neutral axis of that section of the beam where the greatest stress occurs with the given distribution of load. n IS a coefficient, the value of which varies for each class of beam and for each distribution of the load. The following Table shows the different values of n for three classes of beams and for four distributions of load. TABLE C. Arrangement of Beam and Load. Class of Beam. 1 Uniform cross section throughout its length. See Figs. 61 and 127. 2 Uniform strength. Same depth through- out its length. See Figs. 106 to T09. 3 Uniform strength. Same breadth throughout its length. See Figs. 102 to 105. Fixed at one end, loaded at the other . j; loaded uniformly Supported at both ends, loaded in the middle ) ) , . loaded uniformly . - 1 a 1 ■S" 6 1 \ 1 1 1 1 1 1 Maximum deflection under any load.— To find the maximum deflection of any beani under a given load it is merely necessary to substitute the values ot the different letters, and the result gives the value of A. The values of W and I are of course known ; E is given in Table I • and I, the moment of inertia, can be found by means of the information given m Appendix XIV. The deflection under proof load, or under loads which produce a known stress, may be ascertained from the following formulae n'r^l^ A = - for all beams, and ^_ n'.(rc + rt)l^ (45) for beams with cross sections of equal strength. 1 In these formulae r,, is the greatest stress on the weakest side of the beam ; ijo is the distance of the neutral axis from the extreme fibre of the beam on that weakest side at the section of greatest stress ; r^, are the limiting stresses in compression and tension respectively ; dg the depth of the beam at the section of greatest stress. The other letters have the same signi- fication as in Equation 44. The following Table shows the different values of n' for three classes of beams and for four distributions of load. 1 These are beams of such a form that the limiting stress is reached on the upper- most and lowermost fibre of the beam at the same instant. 68 NOTES ON BUILDING CONSTRUCTION TABLE D. Arrangement of Beam and Load. Class of Beam. 1 Uniform cross section throughout its length. See Figs. 6ia and 126. 2 Uniform strength and same depth throughout its length. See Figs. 106 to 109. 3 Uniform strength and same breadth throughout its length. See Figs. 102 to 105. Fixed at one end, loaded at the other . ,, uniformly loaded . Supported at both ends, loaded in the middle ,, uniformly loaded . 1 4 1 5 1 8 1 ? 1 2 ! Simpler Formula. — For ordinary rectangular beams of the same section throughout their length, such as wooden beams nearly always are, the for- mula 44 can be much simplified by su.bstituting for the moment of inertia I its value for a rectangular section, namely y--. (See App. XIV. and IX.) 1 'in'Wl^ Equation 44 becomes ^~~E6d^ .... (47). Comparison hetiveen strength and stiffness. — We found that the strength of bd^ rectangular timber beams varied as fo-f clii'ectly as the breadth, as the square of the depth, and as the modulus of rupture of the material, but inversely as the length. By examining Equation 47 we see that the amount of deflection of P rectangular beams varies as ^-^3 : that is, the greater the length the greater the deflection ; the greater the modulus of elasticity, the breadth, or the depth, the less the deflection. The deflection arises from want of stiffness, therefore the stiff7iess or resist- ance to deflection will vary according to an exactly opposite set of condi- EM^ tions— in fact, it varies as —^3- . It will be greater directly as E is greater (thus an oak beam, for which E is greater than for a similar fir beam, will be stiffer than the fir beam), also directly as the breadth and cube of the depth are greater ; but, on the other hand, it will be less as the cube of the length is greater : that is, of two beams of the same section and material, the longer will be less stift" than the other, not in proportion to its length, but of the cube of its length. Recapitulating, we see that for beams of rectangular cross section The strength of beams varies as — ^ — , „ stiffness „ „ Further, since the strength varies as bd^ and the stiffness as bd^, we see that by reducing the breadth and increasing the depth in such a Avay that the strength remains constant, we can obtain stiffer beams. Thus a beam 4 inches broad and 6 inches deep has the same strength as one 1 inch broad and 1 2 inches deep FIXED BEAMS 69 (4 X 62 = 1 X 122), ijut 4 X 6^ = 864 and 1 x 12^ = 1728, so that the second beam is twice as stiff as the first, though it only contains half the material. Amount of deflection allowable. — This depends greatly upon the nature of the structure. It is generally considered that for floors, or roof-timbers sup- porting ceilings, the maximum deflection allowable is -^jy" per foot of span. For roofs or floors without ceilings a somewhat greater deflection, say per foot run, and for rough or temporary structures per foot of span may be allowed in the case of supported beams. In cantilevers supporting ceilings the deflection allowable should be per foot of length, without ceilings and for rough structures per foot of length. It will be seen that these allowances make the depression at the end of the cantilever the same as that in the centre of the supported beam. FIXED BEAMS.i The general investigation of the formulse for the calculation of beams rigidly fixed instead of being merely supported, at one or both ends, involves mathematical knowledge quite beyond the province of these Notes. The important points of difference between fixed and sup- ported beams can, however, be explained in an elementary manner, and the student will be enabled intelligently to use the formulae connected with the calculation of such beams, and indeed to make the calculations in an approximate manner without having recourse to the formulae at all. Nature of stresses upon a beam fixed at the ends. — The stresses produced by loads upon a beam, whose ends are rigidly fixed, are different both in distribu- tion and amount from those produced by the same loads upon a beam whose ends are merely supported. Thus in a beam supported at the ends and loaded in the centre, as in Fig. 1 1 1, we know that the beam deflects, forming an approximately circular YW Fig. 111. arc,2 the part of the beam above the neutral layer is under compression (ccc), the part below in tension (ttt). If, however, the ends be rigidly fixed and prevented from tilting up, as in Fig. 112, the curve of deflection caused by the weight is quite different. 1 Theformulce for practical use are given in Equations 48 to 52, pp. 71, 72, and in Appendix VIII. 2 For beams of uniform section the curve of flexure is nearer an hyperbola when the load is at the centre. 70 NOTES ON BUILDING CONSTRUCTION From A to / and from B to i tlie bottom surface of tlie beam is concave ; from / to i it is convex. Fig. 112. Moreover, the stresses upon the beam are different from those o; the supported beam. From A to /and from B to t the upper surface is in teision (itt), the lower surface is in compression {ccc). From f to i the upper surface is in compression (ccc), and the lower surface is in tension (tU). It will be seen that at the points / and i, where the curvature changes^ the nature of the stress also changes ; the upper surface, which from A to / and from B to i was in tension, changes to compression, and the lower sur- face, which was under compression, to tension. These points, where the curvature and the stress change, are called the points of contra-flexure. Their distance from A and B varies according to the shape of the beam and the nature of the load. When that distance is known, the calculation of the bending moments, etc., becomes a simple matter. In fact, the beam thus fixed and loaded is exactly in the condition of two cantilevers A/ and Bi, carrying a beam fi between them, which is supported at its ends / and i by hanging from the ends / and i of the cantilevers. The cantilevers may be calculated as shown in Case 1, p. 28, and the supported portion as in Case 6, p. 32. The only difficulty, therefore, is to ascertain the exact distances of the points of contra-flexure from A and B. Position of points of contra-flexure in fixed teams with different distributions of loads. — Without making any attempt to describe the mathematical investigations by which the positions of the points of contra-flexure are ascertained, we will give their positions for different classes of beams with various distributions of loads. TABLE E. Distance or Points of Contra-Flexure. Arrangement of Beam and Load. Class of Beam. No. of Fig. Uniform cross section. Uniform strength, with uniform depth. Distance of points of contra-flexure. Fixed at both ends. Load in centre . „ ,, Loaded uniformly Fixed at one end andl ^oad in centre . supported at the other ) Fixed at one end and ^ r j j , , , , , . , \ Loaded uniformly supported at the other 1 0-25 I from A and B 0-211 I „ 0-273 I from A 0-267 I from A 0-25 Z from A and B 0-25 I i, 0-33 I from A 0-33 I from A Fig. 113 Fig. 115 Fig. 117 FIXED BEAMS 71 It will be seen that in beams fixed at both ends the points of contra-fiexure vary from '2111 to -25 Z from the ends; and in beams fixed at one end and supported at the other the point of contra-fiexure is from -267 ^ to -83 ^ from the fixed end; so that the distance of the point of contra-fiexure from the fixed end of any beam may, for approximate calculations, be taken without much error at ^ Z. Various Cases of Fixed Beams. One or two cases may now be calculated as examples : — Case 1. BEAM OF UNIFOKM CEOSS SECTION FIXED AT BOTH ENDS AND LOADED AT THE CENTKE. The points of contra-fiexure will be 0-25 I from the walls A and B (see Table E). It is evident that A/ and Bi, Fig. 1 1 3, are in the condition of canti- W levers ; each loaded at the end with a weight besides the weight of tlie beam fi which may be ignored. . •. = weight X leverage, = 17x7=^ ■ . • ■ (48). W I Wl (49). I The portion fi is like an ordinary supported beam of a length equal to - and loaded with W in the centre. W The reaction at / and at i will be — . . •. = Keaction x leverage, (50) Graphically the moments of flexure may be shown as in Fig. 1 1 4. 72 NOTES ON BUILDING CONSTRUCTION Shearing stresses. — In this case the shearing stresses are the same as in a beam supported at both ends and loaded in the centre (see Appendix VIII..) Deflection.— It can be shown that the maximum deflection is ^th of tlhe maximum of a similar beam supported at both ends (see Appendix VIII.) Case 2. — beam of uniform cross section fixed at both ends AND UNIFORMLY LOADED. In this case the points of contra-flexure are distant -211? from A and IE Fig. 115. 1 — 1 '■Unifoj'vi load " ■>■ C — 0-5781- ->T<— 0-311 /—>.^ ■■0-083wl^ = Fig. 115. = weight distributed on cantilever x leverage + weight at end X length of cantilever, = wx-2lUx—^ + X 21H, = 0-022w;Z2 + 0-061^^2^ 12 (51)- = Reaction x leverage - weight between C and / x leverage, •5781^1 -5781 -blSwl -5781 ~-~2~''—2 _ 0-334?/7Z2 wl^ 8 -24 • • (52). 12 24 values of and M„ as can be proved bpy another method; the fraction 0-211 is only approximate. FIXED BEAMS 73 Graphic Representation. — Graphically the tending moments may be shown in the shaded portions of Fig. 1 1 6. Fig. 116. The curve is a parabola whose centre ordinate ^d^ wl^ rvl'^ As in the preceding case, the shearing stresses are the same as in a beam supported at both ends and uniformly loaded (see Appendix VIII.) Deflection. — It can be shown that the maximum deflection is i-th of the maximum deflection of a similar beam supported at both ends (see Ap>]3endix VIII.) Case 3. — beam of uniform cross section fixed at one end, SUPPORTED AT THE OTHER, AND LOADED UNIFORMLY. In this case the beam is like a cantilever A/ and a supported beam /B, as shown in the portion Ai of Fig. 115. The point of contra-flexure / is at -211 the span ^ from A in Fig. 1 1 5 and Ai =-789Z, therefore in this case Af=^ll = •261l. The upper portion of A/ is evidently in tension and the lower portion in coimpression ; the upper portion of B/ is in compression, the lower portion in tension. In the figure the portions in compression are shown in thick lines, those in tension in thin lines. They will be practically the same for a beam loaded in the centre. Fig. 117. Bending moments for fixed beams with, different distribution of loads. — It is unnecessary to work out any more cases. The bending moments at dilfferent points for three classes of beams, with one or both ends fixed and wiith different distributions of load, are given i n the following Table. ^ Approximately (see p. 70). 74 NOTES ON BUILDING CONSTRUCTION TABLE F. Class of Beam. Arrangement of Beam and Load. Uniform cross section. Bending moment Uniform strength, with uniform depth. Bending moment At fixed ends. In centre. At fixed ends. In centre. Fixed both ends. Load in centre !, ,, Loaded uniformly » Loaded uniformly and in centre Fixed one end and "1 t i • supported at other /^"^^^ centre if", IUZ2 J W/ IS In centre of supported portion Loaded uniformly J wl- In centre of supported portion 6 — ""^'-'v^ >JV.^>-- J.i-pp^jJ.±>_liA Y XXJ.. Oljections to fixed learns in practice. — Fixing the ends of beams is generally objectionable in practice. In the case of timber, con- fining the ends keeps them from the air and causes them to rot ; in the case of iron beams it prevents them from expanding and contracting freely. ^ ^ There are other reasons against fixing special forms of beams, which will be mentioned in treating of those forms. As fixed beams are rarely required in practice, it is not necessary to go farther into the subject to consider their deflection; shearing stresses, etc. Enough has been said to show the student that the stresses in a beam fixed at the ends are very different from those in a supported beam, not only in amount but in distribution. Continuous Girders. When a beam or girder extends withouit break m itself over two spans or more it is said to be continuouts. Fig. 1 1 8 shows to an exaggerated degree the curves formed by Fig. 118. a uniformly loaded girder supported over two spans, AB, BC, ancd fixed at the ends. It will be seen that the general arrangement resembles twfo pairs of cantilevers, A/, Bt and B/, Ck, and between them two beams?, fi, jk, supported at the ends. ^ The fornudce for practical use are given in Apjiendix XI. BEAMS 75 The upper portions of A/, Bi, B/, and C/c are evidently in tension, the lower in compression; whereas in fi, jh the upper P'ortions are in compression and the lower in tension. The points of contra-flexure are at /, i, j, and k. Again, if the ends of the girder are not fixed, but merely sup- p.orted, the curves will be as shown in an exaggerated form in Fig. f tt B t I Fig. 119. 1. 1 9, and the portions in compression and tension respectively a.re shown by the thick and thin lines as before. It will be seen that the general curves of each span in Fig. 1 1 8 r'esemble those of a beam fixed at both ends (see Fig. 1 1 2), and tihe curves of each span in Fig. 1 1 9 resemble those of a beam fiixed at one end and loaded at the other, as shown in Fig. 1 1 7. The distances of the points of contra-flexure from the aibutments, and the value of the bending moments, vary according t'-o the section of the girder, the distribution of the load, etc. The calculations connected with continuous girders are very cjomplicated, and not suited for a work of this land. Enough imformation has been given above to indicate which portions are i:n compression and which in tension, and that is all that is r-equired by the syllabus ; but as cases occur of bressummers con- tinuous over two spans and of rafters continuous over many spans, s5ome further information on the subject is given in Appendix XI. EXAMPLES. In order to illustrate the application of the rules given in t:he previous pages, it will be well to give in some detail the c3alculations necessary for a few examples of such timber beams ais are used in practice. Before making calculations it is always desirable to state the ^preliminaries with great care. These include all known par- tticulars relating to the case, such as the span, the nature and dis- tbribution of the load, also the moduli of rupture and elasticity of tthe materials, etc. etc. Great care must be taken also that the dimensions, weights, eetc, are all expressed in the same units, i.e. that feet and inches, 76 NOTES ON BUILDING CONSTRUCTION tons and pounds, etc., are not mixed up in the formulae ; for if so, grave errors will ensue. Timber Cantilever for Balcony (uniform cross section). Example 5.- — To find the dimensions for oak cantilevers to support a balcony to carry a dead load. Conditions. — A temporary balcony 30 feet long and 6 feet wide is sup- ported by 11 cantilevers of English oak, built into a wall and projecting 6 feet from it. The weight of the balcony platform is about 20 lbs. per foot superficial, and it is liable to a dead load of about 100 lbs. per foot superficial. Find the scantling necessary for the cantilevers if they are made of uniform rectangular section throughout : (1) for strength ; (2) for stiffness ; the d'eflection being limited to -^fj" per foot. Preliminaries. — Here each cantilever, except the one at each end of the balcony, has to support an area of 6' x 3' = 1 8 feet superficial. The load on this area is Platform 20x18= 360 lbs. Load 100 X 18 = 1800 „ Total = 2160 lbs. Let h = the breadth of the beam. [A breadth convenient for fixing the platform to, 3" or 4", may be assumed, say 4".] Z = the length of the beam = 72". 2160 w = load per inch run of the beam = ■ = 30 lbs. / ^ /o = an ordinary modulus of rupture for English oak= 10,000 Ibs.^ E = an average modulus of elasticity for English oak= 1,200,000 lbs. F = the factor of safety for a temporary timber structure with a dead load = 5. Calculation for strength. — By referring to Fig. 39, Case 2, p. 29, we see that the bending moment is greatest at A — the point of fixing. As the cantilever is to be of uniform section throughout its length, if it is strong enough to resist this bending moment at A, it will be more than strong enough in all other parts. Now we know that at any point the bending moment = the moment of resistance when the forces are in equilibrium. Introducing the factor of safety F, by multiplying it into iv, we have (Equations 6, p. 29, and 29, p. 52) (Fw)P_ bd^ — a Jo ' rf2 = 6 ' _6 X 5 X 30 lbs. X 722 ~ 2 X 10,000 lbs. X 4"' = 58-3, (Z= 7'6 inches. 2 ^ It is to be understood that moduli of rupture, elasticity, resistances to tension and compression, etc. etc., used in this and all other examples, are taken from Table I. 2 If & had been taken at 3 inches, d would have been 8 '8 inches, a stiffer beam with less timber in it. BEAMS 77 This is the depth necessary when strength only is considered. We must now see if it is sufficient for stiffness. Calculation for stiffness— To ascertain the deflection of this cantilever UnWl^ when fully loaded, the general formula (equation 47) is A= -gyg • -l-iie beam is of uniform cross section fixed at one end and uniformly loaded, there- fore w=i (see Table C, p. 67, col. 1, line 2). 12 x|x 2160x 723 •'• ' 1,200,000 lbs. X 4" X 7-6^' = 0-57 inch. Since the deflection is to be limited to ^" per foot run of the beam (see p. 69), le. for the whole, it is clear that a beam 4" x 7-6" is not stiff enough. To find the necessary depth put A = We have g „ 12nWP Transposing, , 40 X 12 X i X 2160 X 723 6 X 1,200,000 X 4 = 1680, fZ= 11-9, say 12". This depth would be adopted, being greater than the depth (7-6") required for strength only. Therefore, to obtain the necessary stiffness, a beam 4"x 12" is required, supposing that it is decided the breadth shall be 4". If a breadth of 3" is sufficient, then it will be found by repeating the above calculations that the dimensions are (1) For strength, 3" x 8f", (2) For stiffness, 3" x 13". Timber Cantilever for Balcony (uniform strength, uniform breadth). Example 5 A. — To find the dimensions for oak cantilevers to support a balcony to carry a dead load. Conditions and Preliminaries. — Same as in Example 5; but supposing that for some reason it is desirable to have the cantilevers for the balcony, described in Example 5, as light as possible, or that it is thought that their appearance would be improved by making them of uniform strength, keeping the breadth the same but varying the depth. For strength the depth of the beam at the wall would be as before 7 -6". As regards stiffness, to find the depth for the allowed deflection ^V', we use the same formula as before, but the value of » is ^ instead of | (see Table C, p. 67, col. 3, line 2). We have ^ „_12nW;3 ^ Ebd^ ' 40 X 12 x-|x 2160 x 723 6 x 1,200,000 x 4 ' = 6718, d=l8-9, = 19 inches nearly. 78 NOTES ON BUILDING CONSTRUCTION Referring to Fig. 103 we see that the theoretical form of the beam would be as shown in Fig. 120. The calculated deptJi is given at AC, where the beam is built into the wall, and the line joining CB is the under side of the beam. In practice the end would not* be brought to a feather edge as shown at B, but would have a thickness there of an inch or two, as shown in Fig. 121. It will be seen that in most cases there would be no advan- tage in using a cantilever of uniform strength, for though the theoretical form would contain less timber, still, practically, it would contain nearly as much, and would involve more labour than the beam of uniform depth throughout. It would be better, Fig, 121. therefore, to iise the latter in ordinary cases. Balcony with live load. — It will be noticed that the load on the balcony has been taken as a dead load. If, however, the balcony is liable to be occupied by a crowd of excited people moving about, it would be safer to consider them as causing a live load. This live load should be reduced to a corresponding dead load by doubling it. The total weight would then be as follows Weight of platform as before = 360 lbs. Weight of people 2 X 1800 =3600 „ Total weight = 3960 lbs. Taking this value for the weight instead of 2160 lbs., the calculation would proceed as before with the following results : — Uniform cross section (1) For strength, 4" x 10-4", (2) For stiffness, 4" x 14-5". Uniform strength (1) For strength, 4" x 10-4", (2) For stiffness, 4" x 23-1". In practice it would be preferable to adopt a framed cantilever either of wood or iron. See also Examples 2 and 10. Example 6 -Timber Beam loaded in Centre. — To find the load a given baulk will carry. A pitch-pine baulk of 20 feet span 12" wide 12" deep is BEAMS 79 to be loaded in the centre with a dead load. It is to be used for temporary- purposes. What weight will it safely carry ? Preliminaries. — Z = 120" 6=12" d=12", /„= 1100, F = 4. Factor of safety. Introducing F we have from Equation 13, p. 33, and Equation 29, p. 52, or from Appendix VII. _ 11,000 X 12 X 122 X 4 ~ 6 X 120 X 4 ' = 3 tons (about). Fir Joist to carry given Load. Example l.—To find the scantling for a fir joist in a single floor. Conditions. — A joist of Baltic fir has a span of 20 feet, and is to carry a uniformly distributed dead load of 2400 lbs. Find the scantling it should have for (1) strength and (2) stiffness. This case will not be worked out in full, as it is a very simple one. Preliminaries. ■ — 1=120 inches for Baltic fir = 6600 lbs. E „ =1,444,000 lbs. 2400 lbs. w = — — = 10 lbs. per inch. F = factor of safety = 8. Calculation for strength. — Calculating for a beam of uniform rectangular section throughout its length (Fw)P _ hd^ 8 6 X (Fw)l^ = 524. Taking 6 = 3. d^=ll5, and d=l 3", as 13" would be inconvenient. Take 6 = 4. rf^ = 131, and d = 1 1-4, Calculation for stiffness. — To calculate deflection of the beam if 4" wide and 11-4" deep. By Equation 47, p. 68 . , l2n'Wf we have A = . Ebd^ Here n = -^l^ (see Table C, col. 1 , line 4). 12 X 5 X 2400 X 2403 A = 384 X 1,444,000 x 4 x 1 1*43' = -1 inch. 8o NOTES ON BUILDING CONSTRUCTION If, however, the beam is to carry a ceiling below, the deflection should not exceed -^q" per foot of span, or = |" altogether. To find the depth that the beam should have, so that the deflection shall not exceed |", we have by Equation 47 1" = 2 d-. Eb 1800, : 1 2 inches (about). Fig. 122. TJseful Notes. Strongest rectangular beam that can he cut from a round log (Fig. 122). Trisect a diameter "AB of the log. From the points of tri- section C, D raise perpendiculars CE and DF, cutting the cir- cumference at E and F. Join AE, AF, FB, and BE ; then AEBF is the strongest beam that can be cut out of the log. It can be shown that FB 1 „ , AF = V2~'^"^ """^"^y- Stiffest rectangular beam that can be cut out of a round log (Fig. 123). Divide a diameter AB of the log into four equal parts at the points C, O, and D. From C and D draw perpendiculars cutting the circumference at E and F. Join AE, AF, FB, and BE. Then AEBF is the stiffest beam that can be cut out of the log, and it can be shown that FB^ J_^4 AF-V3 7- Proportion of breadth to depth in a beam. — We have seen above that the ratio of breadth to depth is : For the strongest rectangular beam cut out of a round log, 1 to fj^. For the stiffest rectangular beam cut out of a round log, 1 to J2>. When, therefore, a single rectangular beam has to be cut from a round log, its proportions will be regulated by one of the above rules, according to whether its strength or stiffness is more important. In practice, however, the beams used are not generally cut simply out of round logs, and the proportions are governed by other considerations. The width is made sufficient for lateral stiffness and for fixing the superstructure, and the depth as great as convenient under the circumstances, taking care not to exceed the market size of the timber. Chaptek IV WROUGHT IRON ROLLED BEAMS.^ rriHE best section for a beam made of wrought iron is very different from that for a timber beam. We have seen that the direct stresses produced upon the cross section of a timber beam may be graphically shown as in Fig. 1 24. They are greatest on the layers most remote c e from the neutral axis, gradually decrease as they p approach the axis, and dwindle to nothing at all at the axis itself ~' In a rectangular solid section, however, just as much material is provided to resist the stress d f at PQ or VX, where this stress is smaU com- -^^S- i^i. paratively, as at CE or DE where it is largest, and indeed just as much at the neutral axis where there is no direct stress at all, but merely the maximum part of a comparatively small shearing stress (see p. 61), It is evident then that, so far as the direct stresses are con- cerned, the portions COD and EOE are useless, and might be removed. To resist the shearing stress a small portion must be left near the neutral axis. Nevertheless it would be undesirable to cut away these por- tions in the case of timber beams, as in most cases it would destroy the continuity of the fibres, and would lessen the resistance of the beam to lateral forces, that is, to those forces acting hori- zontally, which, although not as a rule calculated for (their amount being uncertain), should be allowed for in practice. Moreover, the wood obtained by thus cutting the beam would not pay for the labour of cutting. But in iron, which has in any case to be formed to a particular shape, it is evidently desirable for the sake of economy of material and weight to make beams of the shape best adapted to resist the stresses that will come upon them. ^ The. formulce for practical use are Equations 54, 55, 56, p. 87, 88, and in Appendix XXI. Examples are given at pp. 88 to 91. B.C. — IV. a utral 82 NOTES ON BUILDING CONSTRUCTION Wrought iron beams are therefore rolled of the form shown in rig. 125. c E CE being called the upper flange Ki DF „ lower I KM „ web iv, I By this arrangement, as will be seen, the bulk of ^emmmx^ the metal is placed in the flanges where the greatest Fig. 125. direct stresses exist, and these flanges are connected by the web KM. If the thickness of the web were regulated by the shearing stress only, the amount of which is comparatively small, it would be made very thm. Practically, however, for reasons which will soon be men- tioned, the web is made thicker than is necessary to resist the shearing stress. Let us now take a section of an ordinary rolled iron beam (Fig. 126), such as is kept in stock by merchants, and ascertain the nature and distribution of the direct stress upon it. Resistance of Wrought Iron. Ultimate Resistance. — We know from Parts II. and III. that the breaking stresses for average wrought iron may be taken as follows : — Tension . . . 25 tons per square inch. Compression . . . 16 to 20 „ ,, Shearing ... 20 „ But as rolled beams are often of inferior iron the breaking stress in tension should not be taken higher than 20 tons. Working Resistance. — The working or limiting stresses that may with safety be allowed in the case of rolled beams ^ are In Tension ... 5 tons per square inch. Compression . . . 4 „ „ Shearing . . . 4 „ „ These are lower than those sometimes allowed in the case of iron for roof trusses. To find the Moment of Resistance of an I Beam. There are several different ways, more or less accurate, of finding the moment of resistance of a rolled iron beam. A simple, and at the same time accurate method, is founded upon exactly 1 In consequence of the iron in rolled beams being free from joints, welds, and other sources of defective workmanship, it is the practice of some engineers to take the working stresses upon them as follows : — In Tension . . . . 6 tons per square inch. Compression . . . 5 ,, ,, Shearing . . . 5 ,, ,, As such beams are, however, often made from inferior iron, it is better and safer to adhere to the limiting stresses given above. WROUGHT IRON ROLLED BEAMS 83 the same principles, and worked out in almost the same manner, as the cal- culation described at p. 43 et seg. for a timber beam. Graphic Method, inchiding Web. — The internal direct stresses produced 111 a wrought iron beam, slightly bent by a central load, are similar to those produced m a timber beam. The fibres above the neutral axis are in compression ; those below it in tension. _ The intensity of stress in the different layers of fibres is greatest in those which are most remote from the neutral axis, and becomes smaller and smaller as they approach the neutral axis, until, at the axis itself, there is no direct stress at all. The amount and distribution of tlie stresses may be graphically shown, just as they are for a timber beam in Fig. 74. Fig. 126 is the section of a rolled iron beam 10" deep, with flanges 4" wide and |" thick, the web being also J" thick. I t... 1 .- k^'l- / . V— j-A ions per sq.in. h 3.G tons per sq.in. Fig. 126. Linear scale J inch= 1 foot. M' P' F 7/ / Fig. 128. Fig. 127. Linear scale I inch = 1 foot. Scale of stresses Scale of stresses 1 inch— 16 to7is. 1 inch—l& tons. The centre of gravity of this section will be at the central point n, and the neutral axis will pass through it (see p. 44). The limiting or working stress for wrought iron in compression being 4 tons, and in tension 5 tons per square inch, it is evident that the limit will be first reached in the flange which is under compression, i.e. in the present case (a beam supported at the ends and loaded) the upper flange. To put this in other words : When the stress upon each flange is 4 tons per square inch, the compression flange will be undergoing the limiting stress allowed for it, but the tension flange will be undergoing only four-fifths of the stress that it could rightly bear. Fig. 1 27 is a diagram showing the intensity of stress on the different layers NOTES ON BUILDING CONSTRUCTION of fibres at the moment when the extreme fibres under compression at CE are subjected to the limiting stress of 4 tons per square inch at ce. When this intensity is called out, a similar stress df of 4 tons per square inch occurs on the lowermost layer DF. This layer would safely withstand 5 tons per inch, but that stress cannot be called out without exciting a similar stress of 5 tons per square inch on the uppermost compression layer CE, i.e. a stress in excess of the limit of 4 tons per square inch. A cross section of the rolled iron beam is shown in Fig. 128, with the angles square for the sake of simplicity, and having marked upon it the amount and distribution of the stress upon the section. At CE the compression flange is 4" wide, the stress at CE is therefore four times ce = 4 X 4 tons =16 tons. At GH the flange is also 4" wide, and the stress four times gh = 4" x 3-6 tons = 14-4 tons. At KL the web is | inch wide and the stress upon it at that point is = ^xgh = ^ ■xZ-Q = 1-8 ton. Two points k and I are thus obtained on the diagram of stress, as shown in Fig. 129, !^ which is an enlargement of part of Fig. 128. |\ ;! The diagram of stress is completed (the \\ l\ web being rectangular) by joining kn and '■ (Fig. 128). A moment's consideration will show that the same two points can be obtained by dropping perpendiculars KK' and LL' on to CE, and joining K'n and L'ti, the inter- sections with KL are the points required. Thus the hatched portion above the neutral axis represents the amount and dis- tribution of the compression on the fibres of Fig- 129. the beam. In the same manner may be constructed the hatched diagram of tensile stresses below the neutral axis. It will be seen at once that the result we have obtained amounts to drawing Cn, En as far as their intersection with GH, and Bn, Fn as far as their intersection with I, J, and then completing the diagram by joining K'n, L'n, M.'n, and T'n. So that in practice Fig. 127 is not required. Thus CoknlpE is the equivalent area of compressive stress. This area, multiplied by 4 tons per square inch, will give the actual amount of resist- ance to compressive stress. Similarly the hatched figure DrmnpsF below the neutral axis, multiplied by the intensity of stress 4 tons per square inch, gives the amount of resist- ance to tension. Now we know that in the case of the timber beam the moment of resistance is equal to the stress of either kind multiplied by the distance between the centres of gravity of the equivalent areas shown on the diagram. It is, however, simpler in this case, and quicker, to take the parts of the rolled beam separately. Thus, in Fig. 128, the moment of resistance of the flanges is equal to stress area CoknlpE x distance between the centre of gravity of CoknlpE and of DrmnpsF. The moment of resistance of the web = stress kin x distance between the centres of gravity of kin and mpn. WROUGHT IRON ROLLED BEAMS Putting tliis into figures we have — M of Flanges = x 85 Limiting Stress Distance between Centres of Gravity of Areas CofcnZpE and DmnpsF tons X 4 X :|x||-x4x 9-53, = 72-43. 9 "5 3 inches, M of Web = Stress Area kin ^ Limiting Distance between Centres of Stress Gravity of Areas ^ I 215"" ^ ^2" ^ i I X 4 tons X 6 inches, = |ix4x6, _ =24-3. Therefore, M of Flanges and Web = 72-4 + 24-3, = 96*7 inch- tons. STRESS DIAGRAMS. In cases where the flanges of the beam diff'er very greatly from rect- Fig. 130. Fig. 131. angles, the nearest equivalent rectangles may be sketched in and the cal- culations carried out as if the flanges were of this shape. Thus in Fig. 1 30 abed may be taken as the shape of the flange ; and in Fig. 131 the flange may be considered as made up of opqj- and stuv. The moment of resistance of each rectangle should then be found, as previously explained, and the sum of these moments will be the moment of resistance required. Alternative Method. — The following is another method, which may occasionally be employed with advantage. In Fig. 132 the stress diagram for the shallow rectangle sv is obtained as explained for Fig. 128. Now it is evident that the depth su of the rectangle may be diminished until it becomes nothing, without altering the position of the points u-^ and The rectangle su would then coincide with the line itv. It thus appears that u-^v^ represents the stress in the layer of fibres uv ; and in the same way cc^i/^ represents the stress in the layer xy. Hence u-^, x^, y^, and are points in the stress diagram of the flange. And similarly any number of points can be found on the stress diagram, as shown in Fig. 132. Area and centre of gravity of stress. — The area of the stress diagram and the distance of its centre of gravity from n must now be found. The former can be obtained by reducing in the usual way to a triangle, or better still, by means of a planimeter, should one be available ; the latter can also be determined by a graphic method, which is too complicated for these Notes. 86 NOTES ON BUILDING CONSTRUCTION Both, however, can be found by the following simple artifice with ample accuracy for all practical purposes. n Draw the stress dia- gram on a thick piece of Bristol board, and cut it carefully out with a sharp penknife. To find the area weigh the cut-out dia- gram, and then weigh one square inch of the same Bristol board ; the first weight divided by the second will give the area required in square inches. To find the centre of gravity. — Make a small hole at each of the corners D and F, so that when a needle is passed through, the piece of cardboard will swing perfectly freely. Attach a small weight to a fine thread and secure the thread to the needle as shown in Fig. 133. Mark the position of the thread on the diagram, both when sus- pended from D and from F ; the intersection of these two lines is the centre of gravity required. In the particular case under consideration the centre of gravity lies on the centre line through n, owing to the symmetry of the diagram. Mathematical Method of finding the moment of resistance of a wrought iron I beam. — The graphic method described, pp. 83 to 85, gives exactly the same result as the accurate mathematical formula given in Rankine's and other works, and is simpler to understand. The mathematical formula is — ■ (53), where ro = limiting stress per square inch, I = moment of inertia (see Appendix XIV.), d I is worked out for the present example in App. Fig. 133. XIV., and found to be 120*9. Substituting in Equation 53, we have To—i tons for compression, 2/0 = 5 inches, M: yo 4x120-9 M = 96-7 as before. When the beam with its load is in equilibrium, the moment of resistance thus found is of course equal to the bending moment found in the manner alreaidy explained. WROUGHT IRON ROLLED BEAMS 87 The calculations necessary to ascertain the weight a given rolled iron beam will carry, or to find the dimensions of such a beam to carry a load of particular weight and distribution, are similar to those already described for timber beams. Examples of some of the cases likely to occur in practice are given as illustrations at p. 88 et seq. Approximate and Practical Pormulse for Rolled Iron Beams. Approximate Method, No. 1 {ignoring web), — In most books of formulas, and in many other works describing the calculation of rolled beams, the resist- ance to direct stresses afforded by the web is ignored ; the resistance of the flanges alone is calculated. Thus in Fig. 128, p. 83, the resistance indicated by the triangle Mn would be omitted. Instead, however, of taking the stress area CoklplS, as representing the varying resistance of the flange, the whole area of the flange is taken. Again, instead of taking the distance between the centres of gravity of the stress areas as the length of the arm of the couple, the full depth of the beam is taken. The resulting formula stands thus — M = area of flange x limiting stress x depth, = AxroXd . . . . . . (54). This formula gives defective results, and the thicker the web (in propor- tion) the greater the error. The error will therefore be greater in small rolled beams (as in the above instance) than in large ones. Taking the beam just dealt with, and ro = 4: tons M = 4x|x4xl0, = 80 inch-tons. We know from p. 85 that its actual value, as there calculated, is 96*7 inch-tons. This formula should therefore be employed only as a trial-method to pick out a suitable section, as shown in Example 9, and to be followed up by the accurate method already explained. Approximate Method, No. 2 {including web). — In the method just described the flanges only are considered, the web ignored. From a glance at Fig. 128, p. 83, it will be seen that the moment of the triangle of stress thus ignored, i.e. Jcln, is approximately = area Mn x §dl x r„, = \ area of web x x r^, = ^ area of web x ^ x r„. If, therefore, we add ^ area of web to the area of the flange and multiply by d, we shall have another formula for the moment of resistance, namely — M = (A + 1- area web) xr^y.d . . . (55). Taking the same beam as before, M = (4 X 1 4- ^ X 10 X I) X 4 X 10, = 113-inch tons. This formula gives too large a result, and is therefore not a safe one to use; the true value of M lies between the values found by the two approxi- mate formulae. 88 NOTES ON BUILDING CONSTRUCTION Practical Formula for Rolled Iron Beams, whose Length, Breadth, Depth, and Weight are given. The student should observe that in the case of rolled iron beams the choice is practically limited to the sizes which are usually rolled by manufacturers, and in many cases to what happens to be in stock. Most manufacturers publish a list of sections, and many also supply illustrations of full-sized sections. The process is therefore to select from such a list that section which best fulfils the requirements of the case. For such a purpose the approxi- mate formulae given above are very useful, but it sometimes happens that only the depth, the width, and the weight per foot are obtainable, and it is to be observed that this is the information usually given when specifying for a rolled iron beam. The follow- ing formula based on these data is therefore useful, namely — Approximate Method, No. 3.— Safe distributed load in tons = 0-55(^^- 0-3&(^)j^. . . (56), where w — weight of rolled iron beam in lbs. per foot, & = breadth in inches, d — depth in inches, L=rspan in feet. This formula is constructed on the supposition that the maximum safe stress is 5 tons per square inch, and it will be found on trial to give fairly good results. Distributed Load that can be safely borne by a Boiled Iron Beam of given Dimensions. Example 8. — Conditions. — Suppose an X rolled iron beam to be of the section sliown in Fig. 126 and to be 1 2 feet between supports, what load uni- formly distributed over its length will it safely carry ? Calculation by mathematical Method. — Preliminaries. — Taking the safe limit of stress in compression at 4 tons per square inch, we know from p. 86 that the moment of resistance of this beam = 96'7 inch-tons. The maximum bending moment (see Case 7, p. 33) Mg = — . But Mc = M, wl^ = 96-7 inch-tons, or since , , ^ ^ , -, . w/ 1=12 feet = 144 , 96-7 X 8 ■••^ = «*=-T44-' = 5-37 tons. Calculation by approximate Method No. 1. — The value of M found by the first approximate method which ignores the web was 80 inch-tous (see p. 87). WROUGHT IRON ROLLED BEAMS 89 Hence ^^=80, 8 , 80 X 8 wl= 144 ' = 4 '44 tons. Thus this method gives as the distributed weight that can safely be carried by the beam 4*44 tons, whereas by the more accurate method it was found to be 5*37 tons. Distributed Load that can safely be borne by a Rolled Iron Beam of given Length, Width, Depth, and Weight. Example 9.— A rolled iron beam 12 inclies deep and 6 inches broad is found from a Table of Sections to weigh 56 lbs. to the foot. Hence, from the a;pproximate formula No. 3 (56), for a span of 18 feet we have Safe distributed load = 0-55(56 - 0-3 x 6 x 12)i|, = 12-6 tons. The average thickness of the flange of such an iron beam is 1 inch. Hence, using approximate formula No. 1, — = A + r„ + c?. OP-,. .-, , , 8x6x1x5x12 Saie distributed load wl = — — — , 18 X 12 ' = 13-3 tons. The moment of inertia of this section will be found to be 400. Hence by the mathematical formula (see Equation 53, p. 86) wl^ rj. Safe distributed load wl = ^ ^ , ^ — 6x18x12' = 12-3 tons. Example 9 A.— Again, take the case of the rolled iron beam 16" x 6" of which the moment of inertia is 804, and let us find the safe distributed load for a span of 20 feet, when the beam is supported at the ends. On referring to a Table of Sections such as is published by manufacturers it will be seen that this section weighs 63 lbs. to the foot. Hence by approximate formula No. 3 Safe distributed load = 0-55 (63 - 0-3 x 6 x 16)i|, = 15 tons ; and by the accurate mathematical formula 5 X 804 X 8 Safe distributed load : 8 X 20 X 12' 16-8 tons. Section for a Joist of a specified Span to carry a given Distributed Load. Example 10. — Conditions. — A rolled iron floor girder of 20 feet span and fixed at the ends has to carry a uniformly distributed dead load of 1 ton per foot run of its length. Find the dimensions required (1) For strength, (2) For stiffness. NOTES ON BUILDING CONSTRUCTION Preliminaries. — 10 = \ ton per foot rim = -J^ ton per inch =186 lbs. 1 = 20 feet = 240 inches. i;' = 29,000,000 lbs. for wrought iron bars. Resistance to direct Stresses. — This being a case of a beam fixed at the ends, the greatest bending moment is at the point of fixing (see Case 2, p. 72), and is (Equation 51, p. 72) M.=A-(210)^ Therefore 12 = 400 inch-tons, and hence the rolled beam must have a moment of resistance of 400 inch- tons. It is practically necessary to select a section from those kept by manu- facturers. Supposing, therefore, we have a sheet of sections before us, and, as a first trial, select a section 12" deep, with flanges 6" wide, and 1" thick. Using the approximate formula No 1 given at p. 87 (Equation 54) we find M=6xlx4xl2, = 288 inch- tons, w^hich is a great deal too small. Trying next a section 17" x 6" x we find M = 6xl|x4x 17, = 510 inch-tons, which is more than necessary. A section 1 6" x 6" X 1" gives M = 384 inch- tons. This section will therefore probably do, since the value of M found by the approximate formula is too low. The moment of resistance of this section may be found by the graphic method thus : The upper and lower flanges may be taken as having a mean thickness of 1 inch, and the web ^ ^ inch thick. The equivalent areas are drawn as described at p. 83, and we have the following moments of resistance (Fig. 134) : — For flanges For web tons inch-tons 1 (6" + |6") X 1" X 15" X 4 = 337-5 TV'x|xix7"xf Xl4"x4 = 64-3 Total moment of resistance = 401 "8 This section will therefore be suitable in point of strength. ^ 15", the distance taken between the centres of gravity of the stress diagrams for the flanges, is a little less than the true distance, because the stress diagram for the flange is a trapezium, not a rectangle. The true distance is + 3 6-1-16 V' = 2(7-F0-51), = 15-02, and the moment of resistance for the flanges becomes 342 inch-tons. WROUGHT IRON ROLLED BEAMS 91 It must of course be understood that no holes are to be made in either flange. If any are necessary, the breadth or thickness of the top and bottom flanges must be increased accordingly. Besistayice to Shearing.- — We know then that the beam selected is able to bear the direct stresses that will come upon it ; the next question is whether the web will bear the shearing stresses. The greatest shearing stress is at the points of fixing, and it there amounts ^ wl . . 20 tons to — (see Case 7, p. 59) = — ^ — =10 tons. This has to be resisted by the sectional area of the web. This area is 14" X Y^g-" = nearly 8 square inches, . 10 tons the shearing stress will therefore be ^ = Ij ton per square inch. Now the safe working shearing stress may be taken as 4 tons per sqiiare inch, so that the web is amply strong enough. Deflection. — From Case 2, p. 72, and Appendix VIII., we see that the maxi- mum deflection of a beam of uniform section fixed at both ends and uniformly loaded throughout its length is at the centre, and is wP I ^= El'' 384- We know (Equation 53, p. 86) that - d I = Mx— , 16 = ^^^X2-74' A = - = 804. 186 x 2403 1 29,000,000 x 804 384' = 0-06 inch, or much less than the specified limit for floor girders (see p. 69) of per foot of span, viz. in this case, = 1". The section of girder chosen is therefore both strong enough and stiff enough for the purpose, if care is taken in setting it that the ends are properly fixed. Remarks on Rolled Beams. Large rolled beams not theoretically economical. Rolled heams much used because cheap and useful. Can be fixed at ends. EoUed iron beams when more than 12 or 14 inches deep are not theoreti- cally economical, because their section is uniform throughout their length, and their flanges are equal to one another. They are therefore not of uniform strength throughout their length, and thus more material is used in them than is theoretically required. A glance at pp. 34 and 59 will show that in a section of uniform strength both web and flanges would vary in dimensions at different parts according to the stress that comes upon them. NOTES ON BUILDING CONSTRUCTION Thus for a beam with a uniformly distributed load and supported at the ends the area of the flanges required at the centre to meet the maximum direct stress becomes unnecessarily great as the ends are approached, because the direct stress diminishes (see Fig. 50) towards the ends. This is still more the case with a load at the centre (Fig. 48). The web, on the contrary, would theoretically require to be greater at the supported ends to resist the shearing stress, and would gradually diminish towards the centre. The flange in tension might be smaller than that in compression, as it can be safely subjected to a greater stress per square inch. In the same way, beams of uniform strength for other dispositions of load would theoretically require to have flanges and web with varying sections to suit the stresses at diff'erent points. The method in which rolled beams are manufactured (see Part III.) renders it impossible that their section can be varied to suit the stress that comes upon them at different points. It results, therefore, that these beams rolled of uniform section throughout must contain more material than is theoretically required. This is not of much importance in small beams ; when, however, the beam is large, the waste of material becomes important, and it is better to build up such a beam, or girder, with plates and angle irons so disposed as to have at each point, as nearly as may be, the exact amount of material which is required to meet the stress at that point. The method of doing this is described in Chapter VIII. EoUed beams are cheap and convenient, and are therefore used for a great many useful purposes in spite of the fact that they contain more metal than is theoretically necessary to meet the stresses upon them. One advantage of rolled iron beams is that they can be placed with either flange uppermost, thus preventing all chance of mis- takes which occur with other girders put up by ignorant men. Moreover, as the flanges are equally strong, these beams can be fixed at the ends, thus greatly adding to their strength — giving them a great advantage over cast iron girders, which cannot so readily be fixed. Of course as compared with cast iron girders they possess a still greater advantage in being made of a sound and reliable material which will not give way suddenly. Steel rolled joists or beams are similar in form to those of wrought iron, and can be calculated in exactly the same way, using the following working stresses : tension 6-|- tons, compres- sion 6-|- tons per square inch. Market Sections. — EoUed I joists can readily be obtained in either iron or steel, from 3 to 30 inches in depth, with flanges up to 8" wide ; and in iron they may be obtained of much larger WROUGHT IRON ROLLED BEAMS 93 dimensions. The largest sizes are not economical to use, as explained at p. 92. Girders huilt up with the aid of I iron beams are sometimes used. The following are some of the forms : — Fig. 135. Fig. 136. Fig. 137. rig. 135 may be useful in some cases, but a beam built up with plate and angle iron would probably be better. Figs. 136 and 137 are distinctly faulty owing to the waste of material at the centre of the section. Chapter V. CAST IRON GIRDERS.^ THE calculation for cast iron girders may be based upon the same simple methods that have already been explained for timber beams and wrought iron I girders. There is, however, one principal point of difference, which makes the calculation at first sight not quite so straightforward and simple as in the cases previously explained. This difference is caused by the nature of the material. In the timber and wrought iron beams the resistance to tension per square inch was the same or not very different from the resistance to compression ; but in cast iron the resistance to tension is very small compared with the resistance to compression, as will be seen at once by a glance at the following figures. Resistance of Cast Iron. Ultimate Resistance. — The ultimate resistance, i.e. the resistance to actual rupture of cast iron obtained under prosper specifications for girder -work, may- be taken as follows : — Tension ... 9 tons per square inch. Compression. . . 48 „ „ ^ Shearing . . . 8^ „ „ TForking Resistance. — Working or safe limiting stress to be put upon cast iron per square inch : — Tension . . . ij tons per square inch. Compression. . .8 ,, ,, Shearing . . . 2"4 „ It will be seen from the above that the ultimate resistance to tension is about one-sixth of the resistance to compression. The limiting stress to be allowed in tension is therefore one-sixth of that allowed in compression. Manges. — If the tension flange is not made equal in strength to the compression flange it will tear across before the compressive 1 The formula for practical use will be found at p. 95, Equation 57, and in App. XXI. 2 The above figures are rather higher than those given at p. 315, Part III., for the average ultimate strength of ordinary varieties found in the market. CAST IRON GIRDERS 95 stress upon the other flange is equal to what the flange is able to bear. It will be seen that approximately, if the web be ignored altogether, the tension flange should, in order to be of equal strength with the compression flange, have an area six times as large. Approximate Formula for Calculation of Cast Iron Girders. The formulse ordinarily used for the calculation of the strength of cast iron girders are based upon the valuable experiments of the late Mr. Eaton Hodgkinson. Those experiments were made with girders having webs so thin that the webs may practically be ignored, because they afforded very little assistance to the flanges in resisting the direct stress. Mr. Hodgkinson arrived at the conclusion that Fig. 138 is the best form of section for a beam required to bear an ultimate or breaking strain : — "The section of the flanges being in the ratio of 6 to 1, or nearly in that of the mean crushing and tensile strength of cast iron. " He also inferred from his experiments that " when the length, depth, and top flange in different cast iron beams with very large flanges are the same, and the thickness of the vertical part between the flanges is small and invariable, the strength is nearly in proportion to the size of the bottom flange." Also that "in beams which vary only in depth, every other dimension being ^^^^^ ^^MWMX Mm ~d'-'nft the same, the strength is nearly as the depth." From these data^ he found the moment of Fio-. 138. resistance = C x a x c^. a being the area of the tension flange in inches. d being the depth of the girder. C being a constant or modulus found, by breaking several experimental beams, to be 6|- tons. This value of C gives the resistance to actual breaking; the working modulus may be taken at one-fifth of this, i.e. 1^ ton. As the bending moment equals the moment of resistance we have M = Oxaxd . . . . (57). For instance, in the case of a beam supported at the ends and loaded in the centre = Cad, W = 4Cad "IT' 4 X tons X ad 26 tons X ad ' I ■ ^ These results are not here stated in the exact form given by Mr. Hodgkinson, but are modified to make them in accordance with the formulse given above for timber and wrought iron beams. 96 NOTES ON BUILDING CONSTRUCTION If I be reduced from inches to feet W = ^ ; and this is tlie shape in which the formula is generally given in books, where W is the breaking weight in tons. It will be seen that this formula roughly gives the moment of resistance of the tension flange. In practice it is found that when cast iron girders fail, it is generally by the rupture of the flange which is in tension — the flange in compression is generally more than strong enough. If, therefore, the tension flange is strong enough, the girder is likely to be safe. The strength of the compression flange may, however, be approximately found by the same formula by substituting 48 tons, the breaking stress in compression, for the value of C. In recent practice the area of the top flange is raised from one-sixth of the bottom flange to one-third or one-fourth, so that by increased breadth sufficient resistance shall be offered to lateral bending. This formula is very simply applied as shown in Examples 11 and 11 A, and though it is only approximate, it is quite accurate enough for the calculation of cast iron girders. The material of which they are made is so treacherous and uncertain that it is always necessary to use a large factor of safety, or (what amounts to the same thing) a small limiting stress in calculating them. Any minute accuracy in the method used for ascertaining their resistance would therefore be useless. Graphic Method of ascertaining the Section for a Cast Iron Girder.^ The distribution of stress over a cast iron girder under a safe load, and subject only to a small limiting stress, can be shown in a manner similar to that already described for timber and for wrought iron beams, and its strength determined from the diagram thus obtained. Such a diagram will show the actual distribution of the stresses over both flanges and the web at the moment when the extreme fibres under tension are subjected to the working or limiting stress, and can easily be constructed as follows. ^ This is described as it illustrates principles, and for the sake of those who prefer drawing to calculations. CAST IRON GIRDERS 97 Take the case of a girder of the section shown in Fig. 139, supported at the ends and loaded in the centre. The lower flange will be in tension, and the centre of gravity will be at a point 7" from the bottom edge.^ The neutral axis NA will pass through the centre of gravity. The diagram of intensity of stress (Fig. 140) and the diagram of distribution of stress (Fig. 141) are drawn in the same way as at p. 83. If we limit the tensile stress on the lower- most fibre to li ton (the working tensile stress), we see that the maximum compressive stress per square inch d'a' upon the uppermost fibre will be (in proportion to its distance from the neutral . , 13 ^, , 13 3 39 axis) ~ X Vc' = y X - tons = = tons. Now the fibres in compression could safely bear 8 tons per square inch, so that in this girder they are not called upon to bear more than about one- third of the safe working stress. —^''/itons—->\ Neutral Axis 'i-^l'Atonpi 1-6 KV Fig. 140. Fig. 141. Scale of Stresses 1 inch = 4 tons. Linear scale ^ inch = 1 foot. Scale of stresses 1 inch= 12 tons. So far as mere compressive stress goes, the upper flange is therefore un- necessarily large, though it may nevertheless be desirable to make it of the dimensions shown, for the reason mentioned at p. 96. The figure showing the distribution of stress is obtained as before by multiplying the number of inches of fibres at each point by the intensity of stress at that point obtained from Fig. 140. See Appendix XII. B.C. -IV. H 98 NOTES ON BUILDING CONSTRUCTION Thus at CD there are 4" of fibres, each acted upon by a stress equal to 13 13 d'a' or ^ X the limiting stress, so that C-^D^ = x 4 = 7-| inches. At gh there 12 is 1 inch of fibres acted upon by — of the limiting stress, so that the width 12 of the stress area at this point is = -;^ inch and reduces gradually until at the neutral axis the direct stress is nothing. The equivalent shaded area in tension below the neutral axis is of course equal to the equivalent shaded area in compression above the neutral axis, being in each case about 17y square inches. The centres of gravity of these two equivalent areas are 15*97 inches apart, and the moment of the couple, i.e. the moment of resistance of the girder, is equal to the equivalent stress area of either kind in square inches x the limiting stress per square inch x the distance between centres of gravity = 17y square inches x 1|- ton x 15-97 inches = 417*5 inch-tons. Comparison with Result from approximate Formula. — Let us compare this result with that obtained from the approximate formula given at p. 95, namely — Taking C — ton, as at p. 95, we find Mr=lx(16xlJ)x20, — 400 inch-tons. But the limiting stress has been taken above at 1^ ton per square inch. If C be taken =1^ ton, M = 480 tons. Fig. 142. Fig. 143. Linear scale |" = 1 foot. Scale of stresses Scale of stresses 1 inch— 12 tons. 1 inch—^ tons. CAST IRON GIRDERS 99 Section with tapering Weh. — In a girder of tlie section shown in Fig. 142, and at p. 169, Part I., where the web tapers in thickness from the upper to the lower flange, supj)orted at the ends and loaded, the stresses would be as shown in Fig. 143. It will also be noticed that the maximum stress in compression, i.e. that on the extreme upper fibre, is only 3|- tons. Mathematical Method for calculating a Cast Iron Girder. The mathematical method may be described, though the graphic method just described is equally accurate, and safer to use. The formula used for M is — I, or — I, in which Vc yt rc = limiting stress in compression. ,, tension. 2/c = distance from NA to extreme fibre of compression flange. i/j= ,, tension flange. I = moment of inertia. The former of these two formulse is used when the compression flange is weaker than the other ; the latter formula when (as is usually the case) the tension flange is the ^'eaker. A case worked out by this method is shown at p. 104. Figs. 141 and 142 give a clear idea of the nature and distribution of the direct stresses on a cast iron girder, but for the reasons given at p. 96 it is seldom necessary to go into any accurate calculations for such girders. The shearing stress is distributed in the same way as described at p. 61 for timber beams. It is nearly all borne by the web, being a maximum at the neutral axis and diminishing as the flanges are approached, until at the flanges themselves the shearing stress is almost nothing. Practical Points connected with the rorm of Cast Iron Girders.^ Proportions of Flanges. — In both the sections given it will be seen that the compression flange is never called upon to undergo anything like the working stress that it can safely bear. Thus in Fig. 140 the stress on the extreme fibre in compression is only 2^1 tons instead of 8 tons per sc[uare inch, and in Fig. 143 it is only 3^ tons per square inch. It appears, therefore, that so far as resistance to crushing is concerned, the compression flange might be made smaller ; practi- cally, however, other points have to be considered. The flange which undergoes compression must not be made too narrow, or (though safe against crushing) it will fail by buck- ling sideways. The greater the span the wider the flange should be. No rule has been laid down for this, but it is well not to make the com- pression flange narrower than one-sixtieth of the span. ^ See also Part II. NOTES ON BUILDING CONSTRUCTION It will be further seen that in some cases, when from construction of the whole building it is impossible for the cast girder to buckle, the compression flange need not have an area exceeding one-sixth of the area of the tension flange, and this proportion is sometimes taken as the universal rule. When, however, the compression flange has to be widened to resist cross- buckling, or when it is required to be wide to support a wide form of load such as a wall above, then it is often necessary to make it of greater area than one-sixth of that of the tension flange. The consequence is that some writers propose one-fourth and others one-third as the proportion (see p. 96). Each case must, however, depend upon circumstances. As a rule a compression flange having one-sixth the area of the tension flange is large enough, if it is wide enough to resist cross-buckling, and to carry the load upon it. The web must of course contain sufficient area to withstand the shearing stress that comes upon it. In order to prevent it from "buckling" sideways it is customary to strengthen it by feathers or stiffeners, as shown at /in Fig. 144 and in Part 1. Figs. 247, 248. These stiffeners should, however, be avoided as far as possible, because the angles formed by their junction with the web and flanges tend to produce weak points in the castings, for the reasons explained in Part III. The depth of a cast iron girder must of course in most cases be governed by circumstances, but when possible it is desirable to give it a depth of at least one-twelfth the span, or one-tenth when considerable stiffness is required. The depth is, however, very often made much less than this in practice. Camber. — Cast iron girders should be constructed with a rise in the centre equal to -j-Itj to -2-^(5^ of their span, so that they may not when loaded sag or droop below the horizontal line. Points connected loith casting. — So far as the requirements of good casting go,^ for flanges 2' wide it is not wise to have a less thickness than 1 1", and for flanges 18" wide not less than 1", but as a rule 2 the thickness of any part should not exceed of the width of the part. When one flange is to be thinner than the other, as in Fig. 142, the web may taper from one to the other, each end being equal in thickness to the flange it joins. Some founders prefer that the thickness of the metal should be the same throughout web and flanges, as in Fig. 141. In any case, there should be no sudden variation in the thickness at any part, and no sharp angles. Girders of uniform Strength. — The method by which cast iron girders are made is such that they can without any difficulty be cast so as to be of uniform strength, the form varying according to the load they have to carry. Thus for a girder with a uniformly distributed load the shape would be 1 Wray. ^ Adams. CAST IRON GIRDERS lOI approximately that shown in Figs. 144 or 145, the material being reduced where the stress is less, so as to form curves.^ The alteration in section may be obtained by altering the depth, as in Fig. 144, and leaving the flanges of uniform width throughout, or by altering the width of the flanges at different points according to the stress 'upon them, keeping the depth of the girder constant throughout, as in Figs 146 147. ELEVA TION. f— I] TO Fig. 146. PLAN Fig. 147. A cast iron girder should never be fixed at the ends, because of the change from compression to tension which takes place (as shown in Fig. 112) in the top flange as the fixed end is approached. This is not desirable even when the girder is designed accordingly, and is unsafe if the ordinary design is followed, for then the top flange at the point of fixture will be unduly strained in tension. EXAMPLES OF CAST IRON BEAMS. Cast Iron Cantilever. Example 11. — Conditions. — A cast iron cantilever for a balcony (Fig. 148) has a projection of 6 feet, and has to carry a distributed dead load (including its own weight) of I ton per foot run. The depth at the wall must not exceed 3 feet. Find s'o'l the dimensions for the cantilever. ' Preliminaries. w = ^ ton = weight per foot run on the cantilever including its own weight, Fig. 148. ^ Theoretically the curve for Fig. 144 should be something between a parabola and an ellipse. In practice an arc of a circle with ends rounded is used, the ends being half the depth of the centre. The curves on plan Fig. 147 are parabolas. ^ The cantilevers are assumed to be 5 feet apart. I02 NOTES ON BUILDING CONSTRUCTION I = length. = 72 inches. C = 1 ton per square inch = coefficient for Hodgkinson's formula. a = area of top (or tension) flange in inches (to he found). d = depth of cantilever at wall =36 inches. E = modulus elasticity of cast iron = 17,000,000 Ihs. per square inch. Now since M = M, hence (Case 2, p. 28) '^^'^^ that is, ^ijx 72 X 36 = 1 xax 36. Hence area of tension flange is a = 3 square inches, and the flange may be 4" x f " area of compression flange = ^ x 3, = i square inch. Thus the web is thick enough to act as a compression flange, or an enlargement may be made for appearance, as shown in Fig. 149. The shearing stress at the wall will be = 'wZ = 3 tons (Fig. 83, p. 57). The thickness of web t required to resist shearing shearing stress ~ depth of web x working resistance to shearing per square inch' 3 tons t = , 36 x 2' the actual thickness f" is therefore 18 times as strong as it need be. Great waste of metal and sacrifice of appearance would result from making the cantilever of uniform depth throughout. The depth required by theory at different points may be found by equating the moment of flexure at each point with the moment of resistance. At any point P distant x from the wall W{1 - £C)2 2 We have then — — = Cad, , ^v(l — 1 2 Ga Taking Z = 6 feet, When £c=l A=^^^^ ^ I^' = |.2 X 1 = II = 25 inches 2 cZ = |2xi= i|=16 „ ■ 3 = X ^ = =9 „ .4 _ 2 2 y i _ JL = 4 : 5 cZ = i2 X i = • T-V = 1 inch X X X x = Q d = 0. Setting off these values as ordinates at the different points, and drawing the curve of the bottom flange through the extremities of the ordinates, we have Fig. 1 50. CAST IRON GIRDERS 103 The curve might have been ascertained by drawing a parabola with its apex at B through CB. It has also been shown that area of the web is largely in excess of the require- ments. It is evident, therefore, that large holes may be made in the web, as in Fig. 148, which will lighten the cantilever and improve its appearance without making it too weak. The end of the cantilever may with advantage be thickened as shown by the dotted line, Fig. 148. The deflection of the cantilever cannot be accurately ascertained, because it has been designed by an approximate formula, in which the resistance afforded by the web has not been taken into account. Assuming, however, that the stress on the tension flange will not exceed the limit of 1 ton per square inch, the probable deflection may be ascertained by Equation 45, p. 67. A: in which n' ■- Hence 1-. A = 1 (for a cantilever of uniform strength and breadth), ■ 1 ton, :72", : 1 6 inches nearly. 1 x 2240 x 722 17,000,000x 16' = 0'04 inch = inch. The allowable deflection, even if there were to be a ceiling below, is per foot run of cantilever = -^V', so that this cantilever is amply stiff enough. If any holes are mads in the upper flange to aid in securing the platform, a sufiicient width must be added to the upper flange to make up for the loss in area caused by the holes. Clearly, also, it is not strictly accurate to take the load (including weight of cantilever) uniformly at ^ ton per foot, as the weight of the cantilever itself varies from 8 lbs. per inch run at the wall to almost nil at the end. In this small cantilever this is of no consequence, but in larger work it would become important. Cast Iron Girder. Example 11 A. — To find the weight that a given girder will bear. Conditions. — A number of old cast iron girders in stock have their cross section throughout of the form and dimensions shown in Fig. 1 39. Span = 20 feet. What distributed load per foot run will they safely bear when supported at the ends ? 1. By the approximate rule (Equation 57), taking limiting stress in tension at 1|- ton per square inch, — = Qad, 8 Cad X 8 NOTES ON BUILDING CONSTRUCTION l|x 16x20x8 ^ 2402 ' = 0-067 ton per inch, = 0"8 ton per foot run. If C is taken at 1^ ton, as recommended at p. 95, the safe distributed load' ■would be |- 0*8 = 0*66 ton per foot run. 2. By the graphic method, taking the limiting stress in tension as 1^ ton. From p. 98 we know M to be 417*5 inch-tons. Hence -==417-5. =0-058 ton per inch, = 0-70 ton per foot. 3. By the mathematical method. — Finding I as in Appendix XIV. to be 1950, and knowing the tension flange to be the weakest, we have, since wl^ 1950 "8"" 7 ' = 417*8 inch-tons, and i<; = 0-058 ton per inch, as by the graphic method. Deflection. — To find the deflection of this girder when loaded with a distributed load of *7 ton per foot as found above by the graphic and mathe- matical methods. From Equation 44, p. 66, and Table C, we have A= -^v-, 5 n = for a girder of uniform cross section supported at the ends and loaded uniformly. I = 1950, see Appendix XIV. E = 17,000,000 lbs. per square inch. 5 14 X 2240 X (20 x 12)3 A = V i iripli 384 17,000,000 x 1950 6 The deflection allowable if there is to be a ceiling below would be per foot of span, or = The girders would therefore be amply stiff enough. Chapter VI. TENSION AND COMPRESSION BARS. "Tlsr the previous chapters we have considered various methods of J- calculating the strength of beams made of one piece of material — either wood, wrought or cast iron. Such beams can only be made of comparatively small dimensions, and in the case of wood or wrought iron it is, as already pointed out, uneconomical to make them beyond a certain size, because, in a uniform section, which does not accommodate itself to the varying stresses, much material is thrown away. Large beams or girders are therefore built up of smaller pieces suitably connected together, thus forming either plate girders or open-work girders in wrought iron, or trusses in timber. More- over, roof trusses are composed of a number of pieces or members jointed together (see Parts I.-II.) The various members of such structures will be subject to tension or compression or shear according to the position of the member in the girder or roof truss. The first step, after designing such a structure, is to ascertain the amount and nature of the stress each member has to bear ; and the next to design each member so as to safely resist the stress in it, as will be explained in the present chapter. Lastly, the various connections must be of adequate strength to transmit the stress from one member to another. This forms the subject of "joints" and will be considered in the next chapter. It will, however, be best to leave for the present the question of finding the stresses in the various members ; we will therefore deal first with the design of separate members, assuming the stress upon each as known. Tension Bars. Symmetrical Stress. — Let AB, Fig. 151, represent a tension bar stretched by the forces E^, E^. The dotted line shows the line of action of these forces, and we will suppose that it is exactly in the centre of the bar. By cutting the bar across at C, as io6 NOTES ON BUILDING CONSTRUCTION shown in Fig. 152, the tension forces excited in the fibres of the bar are, so to speak, brought to view ; and since the line of Fig. 151. Fig. 152. action of the force is central, these forces are equal, or, in other words, the stress is uniformly distributed over the cross section of the bar. Clearly therefore : Safe resistance to tension of the bar = safe resistance to tension of the material per unit of area x area of cross section. Thus a wrought iron bar 2'5" x 1" could resist safely 5 tons per square inch or 5x2'5=:12-5 tons. Or, again, a round iron bar 1" diameter could resist safely 5 X 0-78 = 3-9 tons. Effective Area.— Let us now inquire into the effect of making holes in the bar, such as would be made for rivets. In the first place let there be one hole at the centre line of the bar, as at D, Fig. 153. The cross section at D is clearly the Fig. 153. R; Fig. 154. weakest in the bar, and it is there where rupture would take place. The stress at the section is shown in Fig. 154, and it will be noticed that, as before, the stress is uniformly distributed, but that the area over which it acts is reduced. This reduced area is called the effective area of the tension bar and : Safe resistance to tension of bar = safe resistance to tension of the material X effective area. Or in symbols R = nx A (58). The same formula would clearly be applicable if two holes TENSION BARS 107 symmetrically placed were made in the bar, as shown in Figs. 155 and 156. 0 0 5- Fig. 155. Fig. 156. If now a third hole be made, as shown in Fig. 1 5 7, it becomes a question whether the bar will break as shown in Fig. i 5 8, or as -^5- Fig. 157. -ei^ ; Eg Fig. 158. Fig. 159. shown in Fig. 159. It has been found by experiment on ordinary fibrous iron that a bar is most likely to break along such a line that the area of the section along the line of fracture is a mini- mum. Therefore, if a + c + c + a is less than a + & + a, the bar will tend to break, as shown in Fig. 159, and the effective area will be 2(a, + c) X thickness of the bar. Unsymmetrical Stress. — So far it has been supposed that the holes are placed symmetrically with regard to the centre line of R 0 >- Fig. 160. Fig. 161. the bar. Fig, 160 shows a tension bar with a hole made in it nearer to one edge than to the other, and Fig. 161 exhibits the stresses in the fibres. It is clear that the intensity of stress will no longer be uniform, as in the previous cases, but will be greater at the edge near the hole than elsewhere ; the distribution of the stress can be found by calculation, but by a method rather beyond the scope of these Notes. The student should, however, observe ?o8 NOTES ON BUILDING CONSTRUCTION that an unsymmetrically placed hole weakens a tension bar more than a symmetrically placed hole does, and in the same way it will be seen that if the tension in the bar is not transmitted along the centre line, the bar will be weakened as shown in Tigs. 162, Fig. 162. Fig. 163. 163; for the intensity of stress being greatest at one edge, the bar will begin to yield first at that edge, and its whole area will never be opposed to the full working stress. It should be an aim, therefore, in designing tension bars to arrange the holes in them symmetrically, so that the stress shall be transmitted along the centre line. When this cannot be done, it is necessary to slightly increase the cross section accordingly. The formula Rj = x A is applicable to tension rods. The ends of such rods are often screwed with a minus thread, in which case the effective area is the area at the bottom of the threads (see p. 143). In fact, this formula is of general application when the stress is uniformly distributed over the cross section, and to use it the weakest section of the tension bar or rod must be found, and the area of that section is the effective area. Example 12. — A rectangular iron tension bar is required to resist a stress of 1 0 tons. The thickness of the metal is 1", and allowance is to be made for one f ' rivet hole. Find the width. If a is the total width the effective width is a — f , and the effective area (a — -I)!" square inches. Assuming a safe resistance to tension of 5 tons per square inch, we have 5(a-f)i=10, whence a = 4| inches. Example 13. — The stress in a wrought iron tension rod is V'SS tons. The ends are formed into links for pin joints (see Example 26, p. 140). Find the necessary diameter. The links at the ends of the rod w^ould be made of such dimensions as to be of at least equal strength with the rod. The strength of the rod is there- fore not reduced in any way, and consequently the effective area is the full area of the cross-section. Allowing a safe stress of 5 tons per -square inch, we have clearly 7-83 = 5xA, orA=l-56 square inch, which corresponds to a diameter of 1'4 inch, or a rod if" diameter would do. COMPRESSION BARS 109 Compression Bars or Struts.^ It might, at first sight, be thought that compression bars or struts could be treated in the same manner as tension bars, and that the resistance of a compression bar could be obtained by multiplying the area of the cross section by the resist- ance per square inch. But a moment's consideration will show that this cannot be the case, for in a tension bar the forces acting on it tend to straighten it, whereas C the forces acting on a compression bar practically tend to bend it,^ as shown in Fig. 1 64, thus causing a bend- ing moment which has to be resisted in addition to the compression proper. Fig. 165 represents the portion of -< \ - 1 a ■3— ^~ Fig. 165. the bar at AB to an enlarged scale, the arrows of length a show the compression proper, and the other arrows the bending stresses, which are distributed as in the case of a beam (see p. 43). The maximum intensity of compression will be at B and equal to a + h. At A it will be a minimum and equal a — h, and if & > a (as Yig. 164. in the figure), the fibres at A will be in tension. It is clear that a + h must not exceed the safe resistance to compression of the material. Fidler's rules for Struts.- — Unfortunately, however, there are many difficulties in the way of finding the value of b. In the first place it depends on the amount to which the resultant stress deviates from the centre of gravity of the cross-section of the strut, and it further also depends, as shown in a paper ^ by Professor Fidler, M. Inst. C.E,, on inequalities in the modulus ^ The formulae, for practical use will be found for iron struts in Table V. , in Equation 60, p. 113, and for timber in p. 115, and Table VI. As the question of long struts has hitherto been in an unsatisfactory condition three or four methods of dealing with them are given. Prof. Fidler's being preferred. ^ Theoretically, with the stress passing absolutely through the centre line of the strut, there is no tendency to bend, but practically this condition is never fulfilled, and the slightest divergence causes a bending stress ; so that, at best, the state of the strut is one of unstable equilibrium. ^ Published in Minutes of Proceedings, Inst. C.E., vol. Ixxxvi., 1886. I lO NOTES ON BUILDING CONSTRUCTION of elasticity whidi may exist in different parts of the column ; for instance, a variation of 2 per cent in the modulus of elasticity in a solid cylindrical column will reduce the strength by 26 per cent. Struts of rectangular section. — It can be shown, by an investigation too difficult for these Notes, that the average ultimate stress per square inch for an ideal column of rectangular section, in which there is no variation of the modulus of elasticity, is given by the formula r, = 7r2E.^V^' .... (59), where d is the shortest side of the rectangular cross section and I the length of 1,0,000 A 20,000 10,000 ^ JVROUGHT IRON. Q 100 200 300 Length divided by radius of gyration. Fig. 166. the column. This formula is given vnBer Gonstructeur, by Eouleaux; it is based on Euler's theory, and is plotted graphically in Fig. i66 in the curve BHC, on the assumption that the material is wrought iron, that is, taking E = 1200 tons. In this figure the abscissae measured along OP are the ratios ^ Vl2 • ^) the ordinates are the corresponding breaking weights in lbs. per square inch. AB is drawn at the ultimate resistance to crushing of the material. Now Prof. Fidler shows in the paper above alluded to, that under the most unfavourable conditions which are all likely to occur in practice, the column may be weakened to the extent shown by the curve ATC. This curve therefore gives the minimum strength of wrought iron columns of rectangular cross- _ I section for varying ratios of . ^. The actual strength of any particular column may therefore lie anywhere between these two curves, and Prof Fidler comes to this conclusion : " that the strength of columns cannot be defined by any ^ This ratio is length divided by radius of gyration (see top of next page). If /c be the ,. , ,. . ,■ ,^ o moment of inertia of the cross section radius of gyration oi a cross section then = — , -. area oi the cross section = for rectangular sections (see Appendix XIV.) ^ ^ " ^^12 AOO P COMPRESSION BARS III hard, and fast line, even when the modulus for the whole column and the ultimate strength of the material are accurately known ; but, on the contrary, the strength may have any value less than that of the ideal column within certain limits. The strength of columns must, therefore, be represented by an area, as shown in Fig. i66, within which the results of individual experiments may be expected to place themselves at haphazard," The advanced student is strongly recommended to read Prof. Fidler's paper,! in which will be found the formula he arrives at — a formula which is, however, too complicated for these Notes. Struts of any cross section. — "We have confined ourselves so far to columns of rectangular cross section ; the same reasoning, however, applies, whatever may be the cross section, but we must substitute for — =d (which is the radius of gyration for a rectangular section) (see Equation 59) the expression for the radius of gyration of any cross section, which can be found from ,r, J. p X- \o Moment of inertia of the cross section (Eadius of gyration)^ — ■ r . Area oi cross section The value of the moment of inertia of the cross section varies according to the axis about which it is taken, but when taken about an axis through the CG of the cross section, perpendicular to the direction in which the column tends to bend, it is a minimum, and the radius of gyration found from this moment of inertia is the one to insert in the formula. Practical Value of d. — It is, however, somewhat laborious to obtain this radius of gyration, and for practical purposes, at any rate for building con- y'^h^ ,\ i*--d — >! 1 •{"■ Fig. 167. struction, the value of can be taken as shown in Fig. 167, multiplying by a factor n as shown in Table V., which has been compiled from Fidler's formula and from similar tables given in the paper, and is recommended for use. Many other formulae have been proposed for finding the strength of columns, and some of them will be mentioned later on. Short and Long Compression Bars. It is often the practice to divide compression bars into two classes, viz. short compression bars and long compression bars. Short compression bars are those which fail by direct crushing, and their strengths are calculated as if the pressure were uniformly distributed over the cross section. Long compression bars are those ^ Also Bridge Construction, by T. Claxton Fidler. 112 NOTES ON BUILDING CONSTRUCTION which fail by bending, so that some of the fibres are torn. A sudden change in the manner of failing has therefore been assumed. Such an assumption is contrary to all experience and common sense, and it is evident that these two causes of failure lie at the extreme ends of a series, and that between them failure arises from a gradually varying combination of the two causes. Eeferring to Fig. 165 a compression bar may be said to commence to fail by cross-breaking when & = a ; but in fact the bending moment has begun to act from the first, and has been gradu- ally increasing the compression on one side of the bar and lessening it at the other, and it is therefore not right to assume that the pressure is uniformly distributed. It may be mentioned that some experiments by Mr. Kirkaldy point to the same conclusion. The following Table gives the proportion between length and least dimension of the cross section at which compression bars have been assumed to cease to become short and become long.^ Wood .... 20 Wrought Iron 10 Cast Iron .... 5 Of course there is in reality no abrupt transition from short to long com- pression bars as assumed in this Table, but the figures are given, as they are much used with the ordinary formulae. Securing ends of long struts. — The manner in which the ends of a long compression bar are secured has a great influence on its powers of resistance. If the ends are hinged or rounded the bar will tend to bend as indicated in Fig. 168, but if they are fixed or flat the bar will be stiffer and will tend to bend as shown in Figs. 169 and 170. Manifestly a compression bar with the Fig. 168. Fig. 169. Fig. 170. Fig. 171. 1 That is, it lias been assumed that a wooden strut whose length is more than 20 times the least dimension of its cross section will fail by bending ; if less than 20 times that it will fail by crushing. COMPRESSION BARS 113 ends fixed is considerably stronger than one with the ends rounded. Again, one end may be rounded and the other fixed, in which case the strength will have an intermediate value (see Fig. 171). In any particular case, before the ends of a compression bar are assumed to be fixed, the matter should be carefully considered, because it often happens that what may appear at first sight to be fixed is not fixed at all. The criterion is whether the ends are so connected as to constrain the bar to bend as shown in Fig. 169. Thus the end of a bar ^ fixed to a T iron as in Fig. 1 7 2 is not " fixed," because the bar can bend as shown in Fig. 1 7 3 by twisting the T iron. There is, however, a certain amount of con- straint due to the resistance of the T iron to bending, so that the bar is stronger than if its ends were purely hinged. These last remarks apply to the case when the ends are flat (see Fig. 174). Gordon's Formula. — The formula principally used in this country for long compression bars is that known as Gor- don's formula, namely — Ar. Fiff. 172. Fig. 173. 1 +a (60), where a is a constant determined by experiment, d is the least dimension of W ROUGH T IRON. L Red. colui ins- ends 'ounded 20 30 UO 50 00 Length divided by lesser side of section. Fig. 175. B.C. IV. 114 NOTES ON BUILDING CONSTRUCTION the cross section, as indicated in Fig. 167, A the area of the cross section, and the safe resistance to compression of the material per square inch. The value of the constant a depends not only on the material but also orv the form of the compression bar. The following Table (G) gives some oi these values. To compare this formula with Fidler's formula, when the material is wrought iron, the curves shown in Fig. 175 have been dra\^'n. It will be seen that the agreement is fairly good, but that Gordon's formula gives lower results for short columns and higher results for the longer columns. TABLE G. Material. Timber AVrought iron Cast iron ■ Form of Cross-section. Values of a when Rectangular )_ Circular J Rectangular Circular (solid) Circular (hollow) LT+n ]XIi-i Circular (solid) Circular (hollow) Rectangular Cross-shaped Both ends One end is are Both ends rounded, rounded or are fixed. the other hinged. fixed. 4 1 1 250 250 100 4 1 1 2500 2500 1000 \ * 1 1 ( 900 900 360 1 1 1 100 400 160 1 1 1 2"00 800 320 3 3 3 400 1(500 640 3 3 3 200 800 320 13 cwts. }- 4 tons. >■ 8 tons. N.B. — In using this Table it must be remembered that it is calculated iviih a factor of safety of 4- In many structures, such as those of cast iron or timber, a factor of safety of 6 to 8 or 10 is necessary for safety, and care must be taken in such cases to modify the value of r^ accordingly. Formulse for Wooden Struts — The formulae in use for wooden struts; are more at variance than those for iron, and it will be useful to compare som.e of them. For instance : — Professor Rouleaux gives this formula in Der Constructei ,IE /2 ■ • factor of safety (61)), in which I is the moment of inertia, E the modulus of elasticity which redmces to the following, taking E= 1,300,000 Ibs.^ and a factor of safety betwee;n 4 and 5, ^ bd^ R, = 2000-3- cwts. ^ The letters have been changed to agree with the notation in this book. See also Equation 59. ^ This modulus must be converted into cwts. before reducing the formula, because the safe load is to be expressed in cwts. COMPRESSION BARS 115 Professor Ritter gives the following — 1 + 0-0027 . . . (62), whicli is the same as Gordon's, but with a different constant. In both these formulee the ends are supposed to be rounded. Example 14, — To compare the formulae let us find by each of them the safe resistance to compression of a wooden strut 10 feet long and 4" x 3" cross section, ends rounded. By Gordon's formula : By Rouleaux's formula : 4x3x13 _ 4 /120\ ^+25oV"3-; 5 '8 cwts. 4 X 33 R, = 2000 By Ritter's formula : R. 1202' 15 cwts. 4 X 3 X 13 1 + 0-0027 X 402' 29-3 cwts. These results vary so, that it is difficult to make use of them, and we must resort to some other method. Table for Wooden Struts. — Table VI. was deduced from some experiments by Mr. Kirkaldy,^ but modified to agree with Mr. Fidler's views (see p. 110). The above example works out as follows by means of this Table : — I Ratio = ^ = 40 (col. R in Table), Safe intensity of compression per square inch= 1-06 cwt.. Total safe resistance = 4 x 3 x 1-06 = 12-7 cwts. Again, from Table VII. (Stoney's Table) we find for a ratio of 40, 80 tons as the breaking load per square foot, or taking a factor of safety of 4 the safe load = 20 tons per square foot. Hence in the case of the above example 20 X 20 X 4 X 3 ^ 12 X 12 ' = 33-3 cwts. This Table is, however, deduced from experiments with large timbers with flat ends adjusted as in ordinary practice, so that the ends might be con- sidered as partially fixed. Working out the example again for fixed ends by means of Table VI. we find Safe intensity of compression for ratio 40 = 2-75 cwts. per sqiiare inch, . •. R<. = 1 2 X 3-60 = 43-2 cwts. Table for practical use. — It is clear from the above tbat the subject of 1 Corps Papers, Royal Engineers (Vols. XXII. and XXIII.) ii6 NOTES ON BUILDING CONSTRUCTION long wooden struts is in an unsatisfactory condition. On the whole, Table VI. is recommended for use, and it moreover saves a considerable amount of numerical work. As a further comparison this Table is shown graphic- ally in Fig. 176 by the curve ATC ; likewise Rouleaux's formula, which, as already mentioned at p. 11 0, is the formula for an ideal column, is shown by the curve BHC. It will be observed that the curve ATC agrees well with the corresponding curve ATC in Fig. 166, which can be looked upon as an indication of the reliability of the Table. A B \ \h WOOD. - Red, Coht 77ms- ends rounded. % \ V — 0 10 so so UO 50 GO 70 80 Length divided by lesser side of section. Fig. 176. We now proceed to work out some examples. Cast Iron Column. Example 15. — Find the dimensions of the cast iron column supporting the end A of the rolled iron girder of Example 1, p. 21. The cross section to be circular and hollow. The load on the top of the pillar is equal to the reaction on the girder, which has already been found to be 16"6 tons. The length of the pillar is 10 feet. The practice with cast iron pillars is to make the ends large and flat, and such pillars can be considered as fixed. In this example it will be assumed that the ends of the pillar are " fixed." ^ Assume, as a first trial, that the external diameter is 4 inches, and tliat the thickness (i) of metal is ^ inch. Then Z = 120 inches, ^ It is, however, though usual, not good practice to have a large bearing surface on the top of a column, because the deflection of the girder causes nearly all the stress to pass down one side of the column, thus producing a bending moment. This may be avoided by having a small bearing surface at the top of the column, but then it would be in the condition of being fixed at the lower end and roundec at the top, and the value of a in Gordon's formula would be 3^^. COMPRESSION BARS 117 d = 4 inches, t = 0"5 inch, = 8 tons. 4 4 = 7ri(cZ-0-7ri(4-i), = 5"5 square inches. Or using a table of areas of circles A = area of circle 4" diameter - area of circle 3" diameter, = 12-58 - 7-08 = 5-5. Gordon's Formula. — Hence, using Gordon's formula, since (1= (for fixed ends), 8 X 5'5 I^c=i , 1 ./i20N2=19-6 tons. ^ "f" 'Sou Fidler's Formula. — Again, referring to Table V., p. 332, we find the value of n for a circular hollow column to be 3-1. The ratio of ^ to is I therefore - = 3-lxi|^ = 93 (see p. 306). From Table V. we find the I safe stress in tons per square inch. When — = 95 (the nearest to 93) the safe K stress is 3'66 tons per square inch. Hence Safe load = 5-5 X 3-66, = 20 tons. This is rather more strength than required, but looking to the uncertain nature of cast iron, and the possibility of flaws, this section is not too large. Since the actual load is 16*6, and according to Gordon's formula the safe load is 19"6 tons with a maximum intensity of stress of 8 tons per square inch, the actual maximum intensity of stress will be 16-6 X 8 = 6*8 tons per square inch. Cast Iron Column. Example 16. — Find the safe load that can be placed on a cast iron column of M section of the following dimensions : Length 1 5 feet, width of flanges 6", width across the flanges 8", thickness of metal (flanges and web) 1". It will be found that the area of the web is half the area of the flanges together. Hence, referring to Table V. and p. 333, the proper value to take for n is 4-2, so that I 15x12 Ratio - = X 4-2 = 126. K 6 If, therefore, the ends are supposed to be rounded, the safe stress per square inch is 0-86 ton, and the safe load = (2x6 + 6xl)x0-86 = 15-5 tons. But if the ends are supposed to be fixed, the safe stress would be 2-19 tons per square inch, Table V., and the safe load 39-4 tons. In practice it would not, however, be safe to rely on the ends being fixed unless special precautions are taken. ii8 NOTES ON BUILDING CONSTRUCTION Timber Strut. Example 17. — Find the cross section of a rectangular strut of Riga fir 13' 6" long, and capable of withstanding safely a stress of 25 cwts., the ends being considered rounded. There are clearly any number of cross sections that will fulfil the given conditions. If it were required that the strut should contain the minimum quantity of wood, then the cross section would be a square and the side of the square could be found by means of Table VI. by a series of approxima- tions. But in practice one side of the scantling is generally determined to enable the strut to fit in with other work. In this example let it be assumed that one side of the cross section is to be 4" ; then ^ . 13' 6" Ratio = — TTT- = 40, 4 and, referring to Table VI., the safe stress per square inch = 1-06 cwt. Hence the other side required _ 25 _ „ ~ 4^1^ ~ ^'^ • Or a cross section of 4" x 6" would be suitable. Timber Strut. Example 18. — Find the cross section of a rectangular strut of Riga fir 13' 6" long, and capable of withstanding safely a stress of 12 cwts. One side of the cross section is fixed at 4", the ratio ^ will be 40, and the ends are supposed to be rounded. If the side to be found is greater than 4", the safe stress per square inch will be as in the last example = 1*06 cwt., the side required is 12 ~4 X 1-06 ""^"^ ■ The side to be found is therefore smaller than the given side, and the ratio is therefore greater than 40, so that the safe stress must be less than 1"06 cwt. per square inch. In other words, the required side must be greater than 2'8" and must also be less than 4". In a case like the present we must proceed by a series of approximations. Assume that the required side is 3|", then ^ . 13' 6" Ratio = , „ =50, safe stress = 0'66 cwt. per square inch, and therefore the safe stress the strut can withstand = 0-66 X 3| X 4 = 8-6 cwts. The side is therefore still too small, therefore assume 3-|" as the required side, then ^ . 13' 6" Ratio = —-3 — = 43, and safe stress = 0'92 cwt. per square inch. Hence total safe stress 0-92 X 3| X 4 = 13-8 cwts. Therefore, according to the calculations, a cross section of 3|" x 4" is rather stronger than necessary, but in practice it would be a suitable se^ction. COMPRESSION BARS 119 Wrought Iron Strut. Example 19, — Design a wrought iron strut of tlie form shown in Fig. 177 to bear a stress of 1'69 ton, the length being 2' 11". In struts of this form it can always be arranged to make the tendency of the strut to bend as a whole less than the tendency of the sides to bend between the distance pieces,^ or in other words, such a strut will tend to fail as indicated in Fig. 178. The strength of such struts will therefore to a large extent depend on the number of distance pieces used. Clearly the portions of each side bar between two distance pieces (such as ah, Fig. 178) can be looked upon as a long compression bar hinged at the ends. A simple way of designing such struts is to assume some convenient cross section for the flat bar iron forming the sides, and then to calculate how far apart the distance pieces can safely be placed. Fig. 178. In this example the stress is small, and therefore very small side bars will probably suffice. For instance, assume x |-" as the cross section of each side bar. Then, as each has to bear half the stress or 0'85 ton, the intensity of stress will be 0-85 . , ~ 1| X ^ ~ ■'■'^ square inch. On referring to Table V. it will be seen that this intensity of stress corresponds to a ratio of 135. Hence, since « = 3'5 for a rectangular cross section, maximum safe distance between distance pieces -3 X — = 14-5" -8 X 3-5 -^^^ • One distance piece is therefore hardly sufficient, and it would be better to use two, as shown in Fig. 179. ^ The student is recommended to test this statement in the present case by taking the value of 7i = 3"0 from Table V. NOTES ON BUILDING CONSTRUCTION 2. ■11% ^ uVs'-- 11% ■> I Fig. 179. The cross section of the side bars might, however, with advantage be increased being rather narrow) say to 2", then intensity of stress A.Q K — = 1'13 ton per square inch. 0-85 2^ I Whence ratio = 152, and maximum distance between distance pieces 1S2 = |x-^ = 16-5". T Iron Principal of a Roof. Example 20.— The T iron principal of a roof is subject to direct compres- sion only. The portions between the supports are 1 0-8 feet long, and the stress is T-Sa tons. Find a suitable section for the T iron (see Example 42, 209). The purlins are supposed to be placed at the joints of the roof ; the principal is therefore only subject to direct compression and not to any bend- ing stress, as would be the case were the purlins distributed along its length. In this case, therefore, a section of J iron in which the stem is about equal to the table will be the best. Now in a roof of this size the principal would be made of a single T iron from wall plate to ridge, and on referring to Table H, p. 213, it will be seen that the upper part (DC) of the principal has less stress to bear than the lower part AD (Fig. 373), so that the upper part assists the lower part to resist buckling. Moreover, the purlin tends to " fix " the principal at D. The lower end (A) of the principal can, however, only be regarded as "rounded." On the whole, therefore, the principal from A to D can be regarded as a long column with one end "fixed" and the other "rounded," and the safe stress taken as the mean. Try in the first instance a section 4y x 4" x From p. 333 obtain 4-9 as a suflaciently accurate value for n. Then from Table V. we Ratio 7 k I 10-8x12 X 4-9 = 162. Safe stress = 1(0-98 x 2*14) = Total safe stress = 1-56 x (4 + 4)|, = 6-3 tons. Next try the section 4^" x 41" x Then I 1-56 ton per square inch. Ratio - Jc 10-8 X 12 — ^ — X 4-9 = 141 nearly. Safe stress = 1-86 ton per square inch. Total safe stress = 1-62 x (4 + 4 1)1, = 7-9 tons. which will just do Comparison with Molesworth's Tables On referring to the Table for trussed rafter roofs in Molesworth's Pocket-Booh of Engineering Formulce., it COMPRESSION BARS 121 will be found that tlie section of the principal is given as 3i" x 3 x \'. The roofs referred to in the above Table are of the same kind and shape as the one in which the stresses are found in Example 42, and the rise span) and camber of tie-rod are the same as assumed in the example. On the other hand, however, as stated in Molesworth's Pocket-Booh, in the Table the trusses are supposed to be 6 feet 8 inches apart, whereas in the example they were taken as being 8 feet apart ; and although not stated, it is probable that the covering on the roofs in the Table is corrugated iron or zinc, whereas in the example the roof-covering is countess slates on |" boarding. It will be interesting to investigate this a little more closely. Now, on referring to Example 42, Table H, it will be seen that the stress in the principal (AD) due to roof-covering, etc., and snow is 4-45 tons, and on referring to p. 209 it will be further seen that this stress is due to a load of 20-2 lbs. per square foot covered by the roof. If the roof-covering is altered to corrugated iron the load will be reduced by about 8 lbs. per square foot (see Table XIII.), or say to 12-2 lbs. per square foot. The stress of 4-45 tons will therefore be reduced to 1 2*2 4-45 X ^q:^ = 2"68 tons, and accordingly the total stress in the principal will be reduced from 7 "7 8 tons to 7-78 - 4-45 -1- 2-68 = 6-01 tons, and diminishing the distance apart of the trusses from 8 feet to 6 feet 8 inches will further reduce the stress to 6'8" 6-01 X -7-77= 5-01 tons. 8 0 We have now to find what stress a T iron principal 10-8 feet long and X 3 X ^ cross section can bear. As before, 10-8x12 Ratio = ^x 4-9 = 212. Safe stress per square inch = 1-01 ton. Total safe stress = I'Ol x (3 -i- 3) J, = 3 tons. From the above it appears that it would be wiser to use the stronger section obtained by the calculations on p. 120, at any rate for important roofs and in exposed situations, where full allowance for wind pressure ought to be made. Comparison with Hurst's Table. — In Hurst's Pocket-Booh 4i" x 3|-" x i" are given as the dimensions for the principals for a trussed rafter roof 40 feet span, rise | span, and camber of tie-rod span, covering countess slates on boards. Distance apart of trusses, 6 feet. Now in this case the length of the principal rafter from A to D (Fig. 360) is 11-7 feet, so that 11-7x12 „ Eatio = — ^ X 4-9 = 196. Safe stress per square inch= 1'12. Total safe stress = (4-5 + 3)i x 1-12, = 4-8 tons nearly, which would be almost enough for a corrugated iron covering. Chapter VII. JOINTS AND CONNECTIONS. A GEEAT variety of joints for connecting the various members of wooden and iron structures have been described in Parts I. and II. It is proposed to show in this Part how to determine the dimensions of the various portions of a joint when the stress to be borne is known. Eiveted joints are principally used for built-up wrought iron beams and roofs, but occasionally pin joints are employed. In roofs, screw connections, cotter joints, and knuckle joints are used, and these will also be described in this chapter. Points to be observed in making Joints. These are as follows ; — 1. The members of the structure connected together should be weakened as little as possible. They are liable to be weakened, as will be seen in the sequel, either by cutting away material (for instance, a minus screw thread), or by the design of the joint being such as not to transmit the stress uniformly over the cross section. 2. Each part of the joint should be proportioned to the stre;ss it has to bear. 3. Abutting surfaces in compression should be placed jas nearly as possible perpendicular to the stress. EIVETED JOINTS. The manner in which these joints are made and the technic;al names given to them were dealt with in Part I., and it ther'e- fore only remains to show how the strength of such joints can be calculated. We will first consider riveted joints in tension. Riveted Joints in Tension. Lap Joint. — Starting with a single lap joint with only ome RIVETED JOINTS 123 Fig. 181. rivet, as shown in Fig. 180, it is clear that this joint is liable to fail — 1. By one or both bars tearing across, Tig. 181. 2. By the rivet cutting into the metal of the bar, or by the bar cutting into the rivet, according to which metal is the softer. Fig. 182. 3. By the rivet "being sheared, Fig. i 8 3 (single shear in this case). 4. By the rivet shearing a piece out of the bar, Fig. 184. Let t be the thickness, h the breadth of the bar, and cl the diameter of the clenched rivet and of the rivet-hole,^ also the tension in the bar, the working resistance per square inch in tension = I— 5 tons. 1. To guard against the first manner of failing Y., = Q)-d)tr, (63). 2. The resistance to bearing Tj, has been found by experiment to vary as the thickness and as the diameter of the rivet, that is, as td ; so that if is the safe resistance to bearing per square inch Fig. 182. Fig. 183. (64). The value of of course depends on the quality of the iron employed, but it is usually safe to take it as 8 tons per square inch. 3. The resistance of the rivet to shearing T^ varies as the area of the cross section of the rivet, namely 0-l8d^, therefore T^= 0-7 8d^ XT, . . . (65). Tg can be taken at 4 tons per square inch. 4. To guard against the fourth manner of failing, the lap should be sufficient to prevent the piece of the bar shearing out. 1 See p. 136. 124 NOTES ON; BUILDING CONSTRUCTION Two surfaces have to be sheared, each havin^ an area oi tx ~ 2 (where I is the lap) and the resistance to shearing is therefore = ilr^ .... (66). The joint should be so arranged that T^,, T„ and T/ are each greater than T, so that the metal of the tension bar may be sub- jected to the full, safe, tensile stress it is capable of resisting ; that is, the joint itself must not be less strong but a little stronger than the tension bar to be connected. Example 21. — As an example assume 6=1-4", t = l", d=f. Then Ef = (1-4 - 0-75)i X 5 = 1 -62 ton. T6 = ix fx 8 = 2-99 tons. T, = 0-78 x (1)2 x 4 = 1-76 ton. By putting T/= 1-62 ton we get the least value of /, namely- l-62_ 1-62 '~^~~2~' = 0-81 inch. Practically, however, a lap of only 0-81 inch is quite insufficient, because the rivet holes would be much too close to the end of the bar, from a manufac- turing point of view. In the present case the lap should be 2 X IlxcZ 2^ inches (see p. 136). Single riveted lap joint. — The equations already deduced can be easily applied to the case of the joint shown in Fig. 185, thus R, = (6-3(^)ir^, Ts = 3(0-78tZ2)r,. Fig. 185. Double riveted lap joint- Fig. 186. -And similarly for the joints shown in Fig. 186 Rt = ib-3d)trt, Tb = t{6d)rb, T,= 6(078d2)r,. Butt joint with single cover plate. — The joint shown in Fig. I 87, with a single cover plate, can be looked upon as two joints like that shown in Fig. 185 put close together, and the equations RIVETED JOINTS 125 for that joint will therefore be applicable. Clearly the cover plate should have at least the same thickness as the plates to be joined.'- The double-riveted single cover joint shown in Fig. 1 88 can be calciilated from the equations for the joint shown in Fig. I 86. In the above it has been assumed, in finding the resist- ance to tearing of the bars, that the tensile stress was uniformly distributed over the cross section, A glance at Figs. o ■ 1 Fig. 188. 189 and 190 will, however, show that this assumption is not Fig. 189. warranted, and that the strength of the bar is less than cal- culated, the tendency of the stress being to draw the bars into the position shown. This is one of the cases mentioned at p. 108, in which the stress is not transmitted down the centre of the bar, and an allowance should be made ; but the values of T^ and Tj are not affected. Butt joint with double cover plate. — This reduction in the strength of a tie-bar can be avoided by using two cover plates, as shown in Fig. 191; which arrangement clearly allows of the stress being uniformly distributed. As regards the thickness of the cover plates, each should have at least half ^ the thickness of the plates ^ Prof. Unwin says that a single cover plate should be 1| the thickness of the pLates joined, at least in boiler work. — Machine Design. " Prof. Unwin says §, at least in boiler work. 126 NOTES ON BUILDING CONSTRUCTION to be joined ; the resistances of the covers to tearing and to bear- ing will then be at least equal to those of the plates. It is to be ob- served that the shearing- resistance of the rivets is doubled by the use of two cover plates, because each rivet has to shear along tivo sections instead of only along one, as shown in Tig. double shear." ^ 191. 192. These rivets are said to be in Joint, Fig. 191.- GO Q Q 00 QQ QO Fig. 193. Fig. 192. -The equations to find the strength of this joint are therefore -Rt = (b~Sd)trt, Tg=6(0-78(^2)r«. Comparing the joint shown in Fig. 193 with that shown in Fig. 1 94 it will be seen that, other things being equal, the first weakens the plates more than the second does, but that they have the same bearing resistance and the same resistance to shearing. The joint shown in Fig. 195 has the same beariujg and shearing resistance a.s the two above, but weaken.s the plates by only one rive t hole. Considering a sectiom AA, the plate is weakened by two rivet holes he, but be- fore it can tear across thiis section, the first or leading rivet a must fail by shearing or bearing, which practically makes up the difference : so that thits 1 Some experiments show that the resistance to sliearing on two sections is abouit § less than double the resistance on one section. Fig. 195. RIVETED JOINTS 127 section is as strong as the one through the leading rivet. Similarly, the section through the three rivets c?e/is stronger than the two first, there being three rivets ahc to fail as above, to set against the weakening due to two rivet holes. It should, however, be noticed that the cover plates should be made thicker than half the thick- ness of the plates, since their section is reduced by three rivet holes. Joint, Fig, 195. — The equations for this joint are therefore = (6 - d)trt, Ts = 2 X 6{0-78dys, and the thickness t' of each of the cover plates can be found from or Rt * ~ 2(6 - 3d)rt Joint, Fig. 196. — These equations also apply to the joint shown in Fig. \^\^ — vyc \y; O Fig. 196. 1 96 except as regards the cover plates. The upper plate, in which there is no joint, acts simply as a distance piece, and does not affect the resistance to shearing, or to bearing, of the rivets. But the lower cover plate now trans- mits more stress than the upper one, as shown by the arrows, and should therefore be made thicker ; in the present case very nearly in the proportion 1 2 j = -, so that and ^t' are the proper thicknesses to give to the top and bot/tom cover plates respectively. There is, however, a danger, when carry- ing out the work, of transposing the cover plates, so that in practice both are maide of equal thickness, and usually of the same thickness as the plates to be jointed, so as to be on the side of safety. Gfeneral Formulas for riveted Joints other than grouped Joints. The student ought to have no difficulty in deducing from the foregoing the following general formula for riveted joints (other than grouped joints). Let Ej be the tensile stress to be borne by the tension bar (denoted by Ej in the previous equations). a the number of rivets required for resistance to hearing. ^ the number of rivets required for resistance to shearing. 128 NOTES ON BUILDING CONSTRUCTION s the number of sections at which each rivet tends to shear (i.e. 5=1 for lap and single cover joints, and s = 2 for double cover joints). 7c the number of rivets in the first or outermost row, that is, the number of rivet holes by which the tension bar is weakened. h the breadth of the bar. t the thickness of the bar and also of the cover plates. d the diameter of rivet holes. '^^"'^ E, = r,.(6-M)< .... (67). rt b — M = n-~d- • ' ' ■ (68). _ X sx 0-78^2' rt {b - kd) t (69). Tg ' 0-78sd'^ ■ Whichever is greater, a or /3, must be adopted as the number of rivets in the joint. The calculations connected with the above are much shortened by the use of Table VIII. Double-cover Riveted Joint. Example 22.— Design a double-cover riveted joint similar to Fig. 195, the plates to be joined being 9" broad and ^" thick. We have s = 2, 1=1, b = 9, / _ 5" — 8 ' and from the practical rules for Eiveted Joints, p. 136, 9 - 1 X i- Hence from (68) a = f • ts"^' 16" = 6 nearly. (9-il)f (0-78 X 2 x(i|)2' = 4 nearly. Each side of the joint must therefore contain six rivets. On reference to p. 136, it will be seen that the minimum pitch ought to be 2|-" ; thus only three rivets can be placed in one row across the width of the bar. Hence the required joint will be as shown in Fig. 197. With the aid of Table VIII. the result can be obtained more quickly as follows : — The resistance to bearing of one rivet in |" plates is 4*7 tons, and RIVETED JOINTS 129 the resistance to double shear is 2 x 3"45 = 6'9 tons. The number required for bearing must therefore be adopted. This number is ^R, ^ 5(9-1 xlpl 4.7 4.7 = 6, nearly, as before. 1 L U/^ 1 0® ) Fig. 197. Formulae for Grouped Joints. We have next to consider " grouped " joints (see p. 9 3, Part I.) Equations for Joints with three Plates. For Shearing. — Taking a 1 — 1 1 — !• '1 — r — 1 r-H M 1 1 1 1 i M 1 1 1 1 i ' i 1 ' ! ' i j j i i ' A BCD Fig. 198. ^ ^ r\ r\ 00 ? ■ VLV O O O KJKJ Fig. 199. simple case of three plates joined together as shown in Fig. 198, we see that the joint can fail by the shearing of all the rivets, as indicated in Fig. 199. Let TCj be the number of rivets in each end group (i.e. between A and B and D and E) ; be the number of rivets between the joints (i.e. between B and C and C and D) ; t the thickness of the plates, and t.^ the thickness of the covers. Then it is clear that the number of rivets sheared through, if the joint were to fail as above, is 27ij + 2n.2, and therefore R, = (2ni + 2?i2)0-78cZ2rs . . . (70). For bearing. — The joint might also fail by the rivets cutting into the plates, as shown in Fig. 200. From A to B the resistance to bearing is tt-^ X dtdr^, and from B to C it is and from C to D So that altogether B.C. — IV. n^ X 2tdrfi, n^ X tdr;,. (71). K NOTES ON BUILDING CONSTRUCTION From Equations 70 and 71 the total number of rivets required can be obtained. Two values will be found for 2 {n^ + n^, one being the number of y Ongtnally tn ! contact with rivet. Separation between End of plates. Fig. 200, rivets required for shearing, and the other the number required for bearing ; the larger value is to be taken. So far we have only found the sum of + In^, and to obtain n-^ and separately we must consider two other ways in which the joint may fail. ^ ^ ^ The value of n._ — CT-cr Fig. 201. Failing as in Fig. 201. — In the first place it may fail as shown in Fig. 201, by the rivets between B and D shearing and the cover plates tearing across at B and D'. The resistance to tearing of the cover plates is, if is the number of rivets to be deducted, 2 X (6 - n^d)t^rt. depends on the width of the plates and on the pitch of the rivets ; for instance, with 10" plates and 3" pitch, ng would be 3, as shown in Fig. 202. The resistance to shearing of the rivets between B and D' is 27).2 X 0-78(/2rs, so that = 2(6 - n^d)t^rt + x 0-7 Sfi^r, . (72). It must now be determined wliat value should be given to t^. On referring to p. 127 it will be seen that when two plates have to be joined, the cover plate nearest the joint should be made f i thick, and by the same reasoning it will appear that for three plates We can therefore find the value of from equation 72, and hence the value of riy Practically, as already mentioned, it is usual to make t^ = t, but this will not affect the above calculations. Gutting into covers and plates. — The joint can also fail by the rivets cutting into the covers between A and B, and into the plates between B and D. The resistance to bearing of the covers between A and B is 2n.^t^dri), 1. Fig. 202. .-it. RIVETED JOINTS and the resistance of the plates to bearing between B and D is 2n2*(^rj, + ^^2^<^r^, (see p. 129). So that, assuming as before = t, % = i2n^ + 37i^)tdri, . . . (73). It is only necessary to see that the values already obtained for and 71^ are sufficient to give the bearing resistance required by this equation. On the whole, therefore, the strength of a grouped joint with three plates can be calculated from the four equations 70, 71, 72, and 73. Similar equations would be obtained whatever the number of plates to be joined, and the following are the general equations which the student should check by working out several cases. Ge7ieral Equations. — Let N be the number of plates on each side of the joint which are to be connected, then for shearing (see 70) Rt= (271^ + (N- 1)»2} 0-7 8(^2 . . (74), for bearing (see 71) N r ) R,= -| 2«i + (N-l)n2 j-rfirs ■ . . . (75), for tearing of covers and shearing (see 72) R, = 2(6-'H3(Z)i2?-t + (N-l)n2xO-78rf2rs . (76), for bearing on plates and covers (see 73) R, = I 2??,^ + |(N - 1)^2 I dtrt, . . (77). The total number of rivets is evidently + (N - l)n2 (77«). These equations can be solved very simply by the use of Table VIII., as will appear from the following example. Double-cover Gro7iped Joint. Example 23. — Design a double-cover grouped joint to connect four plates, each 9" x |", the end group of rivets being so arranged as to only weaken the plates by one rivet hole. Preliminaries. — Let cZ= l|- inch, which is the usual size of rivet for |" plates, and let r^ = 5 tons, r& = 8 tons, and = 4 tons. Then R, = 4(9-li)f x5, = 118 tons. Now from Table VIII. 0-78(^2^^ = 3-98 tons, dtrjj = Q-l4: „ .-. -(Z^ Fig. 228. It will be noticed that one end of the coupling is cut with a left-handed thread and the other with a right-handed thread ; thus when the coupling is turned in one direction the tie-rod is drawn together, but is slackened off when the coupling is turned in the other direction. We must first find the effective diameter of the tie-rod. Referring to Equation 58, p. 106, it will be seen that 3-88 = 5 -{- , 4 ■n2 4 X 3-88 5 X 77 or D = 1 -0 inch. Hence from Equation 80 d=\-\ X l-O-f-0-07, = X-I7 = l3-V' nearly. Therefore following the rule l = d, the female screw in the coupling should be made inch long. The length of screw thread to be cut on each half of the tie-rod depends on the amount of play desired ; under ordinary circumstances about 3 or 4 inches of screw on each will do, and consequently the least length of the coupling over all would be 6 or 8 inches. To find the outside diameter of the coupling theoretically required for strength, make the cross section of the coupling at A B equal to the effective cross section of the tie-rod. If d^ is the outside diameter of the coupling, then d^^ = T)'^ + d\ Or in this instance (^^2 = (1-0)2 + (1.1 7)2. Whence c?^ = 1 -5 4 = 1 Practically this diameter is quite insufficient, for a clearance must be allowed beyond the outer diameter of the thread, or the female screw could not be cut in the coupling ; the inner diameter of the coupling would there- fore have to be 1-|. The thickness of metal would also be, say, ^ inch, making the outer diameter of the coupling 2|-, or 2 x d — a very proper proportion. NOTES ON BUILDING CONSTRUCTION A usual form of COTTEK JOINTS. these joints is shown in Fig. 229 (see also Part I.) Such a joint tends to fail as follows — 1. By shearing of the gib G and cotter C (double shear). 2. By the cotter cutting into the tie-rod T. 3. By the gib G cutting into the sleeve ss. Fig. 229, Let h be the thickness of the cotter or of the gib bearing against the rod or the sleeve, and c the width ^ of the gib and cOtter ; t the thickness of the flattened part of the rod, and the thickness of each side of the sleeve, then Es = 2kc X r^. For the rod Vv^^htx r^. For the sleeve R'b = '^^k ^n- Moreover, the effective section of the flat part of the tie and of the sleeve must be proportioned to the resistance to tension of the tie-rod, namely K^. For a theoretical joint these resistances should each be equal to E^, hence 2A;cr, = E^ . . . (83), yt^rj = E^ . . . (84), and clearly t=2t^ . . ■ (85). If the value of k is assumed, the remaining dimensions for a given value of E^ can be found. Example 28. — The tie -rod of a roof is connected to the principal, and to the shoe at the abutments, by means of a cotter joint, as shown in Fig. 229. The tension in the tie-rod is 7-83 tons (see Example 42, p. 209, and Table H, p. 213). Find suitable dimensions for the cotter joint. Let k = ^", then from Equation 84, taking = 5 tons per square inch ^xix 5 = 7-83, whence i = 3*1 inches. This is too large a value for t. It will therefore be necessary to take a larger value for 7c, for instance 1 inch, then 7-83 X 1 Practically 1|" will do. t = 0 Hence (85) t 1-56. And 2 X 1 X c X 4 = 7-83, c = 1 inch nearly. A taper of 1 in 24 to 1 in 48 is usually given to the cotter. JOINTS IN WOODEN STRUCTURES 149 This is the minimum value of c. Practically, more room is required to fit in the wedges and cotter ; not less than 2 or 2^ inches. Lastly, to find the depth a of the sleeve. The effective section of the 7*83 sleeve is 2 x (a — = — - 5 2(a-l)f = ^, a-\ = 1-04, or a = 2'04 inches. Some allowance ought to be made for possible errors in manufacture, so that practically it would be well to make a = 2i", The depth of the fiat part of the tie-rod can also be made 2^ inches. The length of the sleeve should be made sufl&cient to prevent the cotter shearing a piece out of each side of it, and the same is true of the flat part of the tie-rod ; the calculations are, however, left to the student. The dimen- sions given in the figure are slightly more than those theoretically necessary. JOINTS m WOODEN STEUCTUEES. Numerous kinds of joints are employed by the carpenter in framing wooden structures, and a large number of these were described in Part I. The proper proportions for such joints have been determined by long experience, and the necessity to make any calculations in connection with them is never likely to arise. It is, however, thought that the consideration of the principles on which the strength of such joints could be calculated will be a useful exercise for the student, and a few of the joints given will therefore be examined with this object in view. "Fished" Joint for Wooden Beam. Example 29. — For instance, let us inquire into the joint shown in Fig. 230, which is shown in most books on Carpentry, and in Part I. If the bolts fit tightly into the ^ wood it can be considered that part uuuu juuu of the stress is transmitted by them, and the remainder will therefore have to be transmitted by the tabling. Clearly only the resistance to bearing Fig. 230. of the bolts need be considered. If the fish-plate is made of the same wood 6 — 2i as that to be joined, the thinner part t should be made equal to — - — , i being the depth of the indent. Hence 2t=='b-2i . . . . . (86). The resistance to tearing of the fish-plates will then be equal to the resistance to tearing of the beams to be joined. It will also be seen that then the resistance to bearing of each bolt of diameter d is 2i X d X rj). ft ft f, rt.A f. ^ A_ 'LI I I ]i 1 1 . y I I II I I' I □ i NOTES ON BUILDING CONSTRUCTION So that if there are n bolts on each side of the joint the total resistance Therefore the resistance to be demanded of the tabling is — 2ntdr]). But (remembering that we must deduct one bolt hole), Therefore the resistance of the tabling must be equal to 2t{a - dy^ — 2ntdrij. Now the indent must be of sufficient depth to prevent the fibres being crushed ; this condition is fulfilled by 2t{a - d)rt - 2ntdrf, = 2iarc . . . (87). Again, the length of the indent and the distance AB must each be sufficient to prevent shearing off of the wood. The resistance to shearing of either is ears, therefore t{a - d)rt — ntdr^) = eaVg . . ■ (88). From these equations the various dimensions can be found. As an example take the following data, which assume that the wood is of fair quality Baltic fir. a =12", b - 8", n = 4", rf = 12 cwts. per square inch, Table I a, p. 328. rg = 1 0 cwts. „ „ rs= 1-3 cwt. „ „ rft = 12 cwts. „ „ Substituting these values in Equations 86 and 87, we get 2t=8-2i, or t = 4~ i. And «(12-|)12-4i!x|x 12='ix 12x 10, or 99i=120i. Therefore 99(4 - i) = 1 20i, 219i=396, 4 = 1"8 inch, and < = 2*2 inches. Again, from Equation 88 we have 2-2 X (12 - f) X 12 - 4 X 2-2 X f X 12 = ex 12 X 1-3, whence e = 1 4 inches, or say llf inches. It would probably be better in practice to neglect the resistance offered by the bolts, and take them as only holding the joint together. To adapt the above equations to this case it is only necessary to omit the term 2ntdri). The equations then become 2t = b- 2i, t{a — d)rt = iavc, t(a — d)ri = eavg. Taking the same data as before, t = 4-i, ^(12 -1)12=1 X 12 X 10, JOINTS IN WOODEN STRUCTURES whence Hence and from Equation 88 (4 -i)135 . 135x4 255 i = 4-2-l = 120i, = 2'1 incliea. = 1-9, 1-9(12 -1)12 = ex 12 X 1-5, 1-9 X 11-25 X 12 or e= , 12 X 1-5 ' = 14^ inches nearly. It appears strange that the value obtained for e in this case should be so little greater than in the first case, but the reason is that the beam is much more weakened than in the first case owing to the greater depth of the indent, as will be seen by the following. In the first case the resistance to tension of the joint is 2<(a-d)rt = 2 X 2-2 X 11-25 X 12, = 594 cwts. In the second case it is 2i'(a-d!)re = 2 X 1-9 X 11-25 X 12, = 513 cwts. Further, the safe resistance to tension of the full section of the timber, namely 8" x 1 2", is 8 X 12 X 12 = 1152 cwts., from which it is evident that this is a very uneconomical form of joint, more than half the strength of the timber being lost. Scarfed Joint for Wooden Beam. Example 30. — Calculate the safe strength of the joint shown in Fig. 231, which is a bad form, as pointed out in Part I. Fig. 231. Preliminaries. — The various dimensions are given in the figure, taking as before = 1 2 cwts. per square inch, »c=12 ,, „ Vg—l'S ,, „ '•6=12 „ „ This joint can fail in three ways : (1) by tearing across da or be ; (2) by shearing along ca; (3) by the fibres at ab or a'b' being crushed. For the first we have E^ = 9 X 2f X 12 = 297 cwts. For the second R« = 24 X 9 X 1 -3 = 280 cwts. 152 NOTES ON BUILDING CONSTRUCTION For the third R, = 3i-x 9 X 10 = 315, Fiff. 232. It thus appears that this joint is most liable to fail by shearing the fibres along ca, and that 280 cwts. is the safe stress that it could be subjected to. The strength of the full beam = 9 x 9 x 12 = 972 cwts., so that this joint is still less economical than the last. Joint at Foot of Principal Rafter (Wood). Example 31. — To ascertain whether the joint shown in Fig. 232 is of sufl&cient strength to resist a compression of 61 cwts. in the principal. Preliminaries. — We will as- sume that this joint is made with inferior timber, and therefore take ?•( = 10 cwts. per square inch. — 7 5, Tg — 1'3 „ ^•6=7 „ „ The joint may fail in two ways : (l) By crushing the fibres at ad ; (2) by shearing the tie-beam along cd. As regards (1), the bearing surface is ade, allowing nothing for the tenon, which would not be fitted to bear ; de is inclined to the pressure, and only its equivalent at right angles to the pressure must be taken, therefore the bearing surface is the whole cross section of the principal, less the tenon, that is 4x2 = 8 square inches. So that R^=:8 X 7 = 56-5 cwts., which is not quite sufficient. As regards (2), R^ = 6x4x1 '3 = 31-2 cwts. But it will be observed that it is only the horizontal component of the stress in the principal that tends to shear the tie-beam along cd. This horizontal ^ component can be very simply found by a graphic ^" " method, which will be more fully explained in the sequel (see Chap. VIII.) Draw a triangle ABC (Fig. 233); AB being vertical, BC horizontal, and AC parallel to the principal, then CB will represent the horizontal component required, to the same scale that AC represents the compression in the prin- cipal. In Fig. 233 the scale chosen is 50 cwts. to one inch, AC is plotted as 61 cwts., and the horizontal component CB is found to scale 53 cwts. It thus appears that the resistance to shearing of the tie-beam along cd (31 '2 cwts.) is quite insufficient. The additional strength required can be obtained by means of a strap as shown in Fig. 234. To find the stress in this strap we must resolve the por- tion of the horizontal component it has to bear parallel to the strap. The }< 53 cwts ScaU 50(fi to 1 inch. Fig. 233. JOINTS IN WOODEN STRUCTURES 153 portion of the horizontal component which the strap has to bear is 53 — 31-2 = 21-8 cwts., and the resolved part of the stress can be found by means of a triangle as before, and as shown in Fig. 235. Thus it is found that the stress in the strap is 43-6 cwts. = 2-18 tons. u- 21-8 cwts >1 I Fig. 234. Fig. 235. Supposing the strap is made of |-" iron, then the necessary breadth h can be found from 26x;|x5 = 2-18, whence 6 = 0"87inch. A strap 1" wide would be used. By a calculation similar to that given in Example 27, it will be found that a -I" screw and nut are required. A bearing plate is required to prevent the strap cutting into the upper side of the principal at e ; its length will be 4", the breadth of the principal rafter, and to find the least breadth we have 4 X 6' X 7 = 43-6 cwts., whence h' = 1^ inch nearly. The bearing plate ^ at/ would be made of the same width. The above calculation supposes that the horizontal component is partly borne by the strap, and partly by the wood. Any shrinkage of the wood would, however, throw the whole stress on the strap, and it is therefore safer to consider that the whole of the horizontal component is resisted by the strap. The stress in the strap would then be (referring to Tig. 235) 53 43-6 X —106 cwts. = 5-3 tons, and making the strap of |" iron, 21-8 ^ 2& X f X 5 = 5-3 tons, 6=1-4. . So that a strap 1|-" x f " would be used. The bearing plates at e and / would be made about 4" wide, and the screw bolt of |-" diameter. 1 Unless there is any objection to weakening the tie-beam at this point, /would be better notched in to prevent slipping (see Part I. ) Chapter VIII. PLATE GIRDERS. 4 THE subject of plate girders was discussed to a certain extent in Part I., and it is now proposed to enter more into detail and to show how the various calculations can be effected. General form. — Fig. 236 shows in outline the general form of the cross section of a plate girder with a single web. The flanges consist of one or more plates (the thickness and number depending on the stress to be borne) riveted to a couple of angle irons ; the web consists of a thin plate, stiffened, however, against crumpling up, by means of vertical T or L irons. On referring to the distribution of the stresses in a rolled iron girder, p. 83, it will be seen that the \ K web of a plate girder being very thin can bear but a ^^^^^ very small part of the direct stresses. Practically the Fig. 236. whole of the direct stresses may be taken as borne by the flanges, and the whole of the shearing stresses by the web. Practical points to he attended to. — The following practical points should be borne in mind in designing a girder. The depth of plate girders varies from ^ to —g- the span ; ~ is said to be the most economical proportion. In floors of buildings much shallower girders are often used to save height. For calculations the " effective span" that is, the span between the centres of the bearings of the girder upon the abutments, should be taken, and the " effective depth," that is, the distance between the centres of gravity of the flanges (see Plates A and B). The width of the flange under compression should not be less than 3^ to ^ of the span, or it will be liable to buckle sideways. No plates of less than ^ inch thickness should be used, or they will soon be destroyed by corrosion. There should be as few joints as possible, especially in the tension fla.nge and web. Plate girders should be constructed with a camber or rise in the centre of about -j-g-g- to their span, or at any rate in excess of their calculated PLATE GIRDERS 155 deflection, so that if they should droop beneath their loads they may not sag below the horizontal line. Care should be taken to use market sizes of plates and angle irons ^ (see Part III.) as far as possible. Expense is incurred by using extra sizes. It may, however, be cheaper in some cases to use extra lengths, in order to reduce the number of joints. All parts should be so arranged as to be got at easily for riveting. The ends of girders should be bedded on lead or felt. FLANGES. As regards the direct stresses, we need therefore only consider the part of the girder shown in Fig. 237, and from ^^a^^^ Equation 54, p. 87, it will be seen that the moment "n I — of resistance of this section is M = r,A,DGr rAT) • (89), according to which is least, D being the " effective depth " for the whole cross section.^ Algebraic method of finding lengths of plates, — By equating the moments of flexure to the moments of resistance at differ- ent points along the girder, the number of plates required at those different points can be found. Such a method is, how- ever, cumbersome and tedious, and the required results can be obtained more easily and rapidly by the following graphic method. Graphic method of finding lengths for plates in tension flanges. — It has been shown (p. 33 seq^ that the moments of flexure at the various points along a beam can be represented graphically by a curve ; for instance, by a parabola when the beam is uni- formly loaded. Thus, in Fig. 238, akc represents the moments of flexure at every point of a uniformly loaded beam of span ac. JL Resistance of angle iron. „ „ 1st Plate. „ „ 2d Plate. „ „ 3d Plate. a a' a" < Fig. 238. Now considering the tension side of the girder, the moment of resistance offered by the two angle irons will be the same at every point, since the depth of the girder does not vary. Let ab represent this moment of resistance and draw M parallel to ac, thus we obtain by the intersection with the parabola two points a' and c' ; and the portions aa' and c'c require only the two angle irons so far as strength is concerned. Practi- cally, however, the first plate is also generally continued from one end of the girder to the other. Let now ef represent the moment of resistance of the first plate. As before, we obtain points a" and c" where the second plate ought to commence, and ^ The weights of angle and tee irons can be found by means of Table X. 2 At is effective area of tension flange, Ac is effective area of compression flange. 156 NOTES ON BUILDING CONSTRUCTION ill like manner the point a'" and c'", where the third plate ought to commence, can be found, and so on for as many plates as may be required to afford the necessary resistance. The same process is clearly applicable to the compression flange of the girder. Flanges for Plate G-irder of 20 feet span. Example 32. — The matter will be probably made clearer by an example. Let the effective span be 20 feet and the load 50 tons, equally distributed so that the moment of flexure at tbe centre is 1500 inch-tons ^ (see Fig. 240). Then we can draw tbe parabola akc as explained in Appendix III.^ Assume further that the depth of the girder to the backs of the L irons is 1' 7", and that 3" x 3" x ^" L irons and i" plates 10" wide are used. Tension, or lower flange. — It will be seen from Plate A that the cross section contains one rivet (|") hole in each angle iron, and two rivet holes in each plate.^ It will be found that the " effective depth," i.e. distance between centres of gravity for the L irons, is 17*1 inches.^ Therefore the moment of resistance of both angle irons in the tension flange (taking r^=b tons) is 2(3 + 2|--|)xix 17-1 X 5, = 395 inch-tons. Hence we can draw Id by making a6 = 395 inch-tons to the scale chosen for the figure, namely 3000 inch-tons to an inch, and we then find by measure- ment that aa' and c'c= 1*4 foot. Again, the effective depth of the first plate is 19"5", hence its moment of resistance is [10-(2 x|)]xlx 19-5 X 5, = 402 inch-tons. Of the second plate [l0-(2 x|)]xlx 20-5 X 5, = 423. Of the third [10 - (2 x |)] x 1 x 21-5 x 5, = 443. We can therefore draw a series of parallel straight lines at distances representing these several values in inch-tons, until we just pass beyond the apex of the parabola. In this manner we obtain the points /, g, etc., and by dropping perpendiculars the points a', a", a'", etc., so that the number of plates and length of each are shown in Fig. 239, and can be found by measurement.^ Compression, or upper flange. — On the compression side the moment of resistance of the two angle irons (taking = 4 tons) is 1 M,. = ^ (see p. 33)= g = 1500 inch-tons. ^ A circular arc can be substituted for the parabola if the rise of the arc does not exceed |th of the span (see p. 30). ^ In many cases, when the plates are wider, they are riveted together also at their edges, so that there would be four rivets to deduct. * It is as easy and more accurate to take " effective depths " separately for the pair of L irons and for each plate. ^ As the sum of the above moments is 1663 inch-tons, and by the data only 1500 are required, it is evident that the third plate need not be more than yV' thick. It will be found that the moment of resistance of such a plate is 275 inch-tons. PLATE GIRDERS 157 2(3 + 2-l-)xix 17-14x4, = 377 inch-tons.i And of the first plate lOxlx 191 X 4, = 390 inch.-tons. Of the second =410. Of the third = 430. Eepeating the ahove process as shown in Fig. 239 we obtain the number and lengths of plates given in Fig. 240. Resistance in inch-tons, 3d Plate, 430. 2d Plate, 410. 1st Plate, 390. Angle irons, 377 Total . 1607 Incli-tons. 1 '^\^ ^1 is 11 a' \a" a'" c'" c' c\ / 1 U-M-JS Fig. 240. Fig. 241. Fiff. 239. Angle irons, 395 1st Plate, 402 2d Plate, 423 3d Plate, 448 Total . 1603 inch-tons. As already observed, the first plate is continued in both the tension and com- pression flanges for the whole length of the girder, and is so shown in Figs. 239-241. It is to be observed that in practice the plates (after the first) will be somewhat longer than shown on the diagram, for they must include the next row of rivets beyond the point indicated in theory. WEB. Calculations for the Web — Shearing stress. — As already mentioned, the web can be considered as taking the whole of the shearing stress, and no great error will be involved by further assuming that the stress is uniformly distributed over the cross section. The methods of finding the amount of the shearing stress have already been explained in Chap. III., p. 55 et seq. The thin webs of plate girders tend to fail, much in the same way as long compression bars do, by bending or buckling, as shown in Fig. 242, and, to understand this, we must consider for a moment the stresses acting on a small portion of the web. Let Fig. 243 represent a small square marked on the web. We know from p. 61 that vertical shearing stresses V act along ^ The rivet holes are not deducted, as the rivets are supposed to fill them, so that the pressure is transmitted over the whole section. Some engineers would take rc=5 tons over the section, deducting the rivet holes. M of the angle irons would then be the same as for the tension flange 395 inch-tons. 158 NOTES ON BUILDING CONSTRUCTION AB and CD, and that horizontal shearing stresses H act along CB and AD. Now it is clear that these forces will distort the square and make it take the shape of a lozenge, as shown in an exaggerated manner in Fig. 244. This Fig. 243. Fig. 244. distortion clearly extends the material in the direction BD, and compresses it in the direction AC; and the tensions and compressions thus excited in the material will just balance the shearing stresses. This may perhaps be more clearly understood by imagining the square AC formed of four rods jointed B C VN^^-^"^H Fig. 246. at A, B, C, and D, and two springs introduced as shown in Fig. 245. When the forces H, H, V, V commence to act, the square will be deformed as shown in Fig. 246 ; the spring BD will be stretched and the spring AC compressed ; and (supposing the spring AC is not allowed to bend) the extension and com- pression of the springs will continue until they balance the applied forces. Now Professor Rankine has shown that the intensity of tension and compression excited as described above is equal to the intensity of the shearing stresses which produce them. We may therefore replace the vertical and horizontal shearing stresses, acting on the web of a plate girder, by a tensile and compressive stress of equal intensity, and acting in directions inclined at an angle of 45° (see Fig. 247). Now it is the compressive stress CC which tends to cause the buckling of the web, and for purposes of calculation we can consider the web as made up of a series of strips, in- clined downwards towards the Fig. 247. Fig. 248. supports at an angle of 45°, nr. JET PLATE GIRDERS 159 and each, acting as a long column. The length of these long columns is clearly^ \/2 X S, where S is the unsupported depth of the web between the angle irons (see Fig. 248), and they can be considered as "fixed" at the ends. The thickness (ij of the web is the least diameter of the long columns in question. We can therefore apply Gordon's formula (see p. 113, or Tables V. and XI.) to calculate the necessary thickness to be given to the web. It, however, frequently happens that an undue thickness would have to be given to the web to make it stiff enough. In such a case the additional resistance to buckling can be obtained "by riveting vertical T or L iron stiffeners to the web as shown in Fig. 249, © © © © —s — -/-^ ,/ © Q) © © © © 0 ® © © © © © © © © © 0 0 0 © 0 © © Fig. 249. Fig. 250. and if S is the clear space between the stiffeners, the length of the long columns into which the web is supposed to be divided is Sx s/\ and Gordon's formula or Table V. can be applied as before. As stated below,^ the base and perpendicular of the triangle, of which the column is the hypothenuse, are equal. Therefore, when the distance S between the stiffeners is less than S will represent the base and perpen- dicular of the triangle, and the length of the column will be shortened. From this it will be seen that stiffeners are theoretically useless unless they shorten the length of the assumed columns, and to be effective must be closer together than the depth. Example 33. — The depth between centres of rivets of a web is 21" and the unsupported depth (S) is 18", the thickness j-^^, and the shearing stress to be resisted is 9-3" tons. Find whether any stiffeners are necessary, and if so, at what distance apart they ought to be placed. Find also what thickness ought to be given to the web to obviate the necessity of stiffeners.^ ^ Length of the long column = \/2 x 5. sm 45 2 The length of the columns is represented by the length of a line drawn across the web in the direction that the web will fail, and as this direction is at an angle of 45°, its length can be represented by the hypothenuse of a right-angled triangle whose perpendicular is equal to the depth of the web. The formula V2 x 5 which expresses the length of the hypothenuse in terms of the depth of the web is arrived at from Euclid, Book I., Proposition 47. 2 The web being very thin the shearing stress may practically be considered as imiformly distributed, not as in Fig. 97. The depth of the web to resist shearing being taken as equal to the distance between the centres of the rivets connecting it to the flanges, 21" in the present example. 1 60 NO TES ON B UILDING CONS TR UCTION Substituting in Gordon's formula (see p. 113) as follows — rc = 4 tons, 1 2500' Z=18x ^/2, A = - we obtain — Safe intensity of compression in tbe web = 1 + 1 (^2x18)2 ,2500 " {-i-^f = 1"1 tons per square inch. Using Table V., n being 3-5, the ratio is 3'5 whence the safe intensity of compression is O'BB ton. Now the intensity of shearing stress per square inch is -^-~ = l-42 ton. 21Xt^ Evidently, therefore, the web is not quite stiff enough. To find the distance apart of the stiifeners we have 1-42. ^2 X 18 X 16 , or 285, 1+: (v/2.S)2' 2500 • Whence we find by reduction S2 = 221, S= 14-9 inches. The distance apart of the stiffeners can also be found by means of Table The ratio corresponding to a stress of 1'42 ton is 215. Hence V. 3-5- S = 215, where 3-5 in Table V. 13 "5 inches. To find the thickness of web required to obviate use of stiffeners we have V'218 from Table V. as above 3-5- = 215, = 0-425 = —g inch. We can also use Table XI. for this purpose. The stress per foot of depth of 9"3 X 12 the girder is — — = 5-3 tons, which for a net unsupported depth of 18" corresponds to a thickness of web of |". The shearing stress we have been dealing with is the maximum stress at the ends. The intermediate stresses diminish towards the centre (see Fig. 90), and the thickness of web may accordingly be reduced ; but it is never advisable, for practical reasons, to use less than Y' plates even at the centre of the girder. When the load on the girder is concentrated at one or more points it is usual to add stiffeners to the web at such points, to transmit the load equally to both flanges. PLATE GIRDERS i6i o o O Q Q © O 0 © Fig. 251. Joints in the Web. — These joints are arranged as shown in Fig. 251. One or two covers can be used as shown in the sections, but it is better work to use two covers. The thickness of one cover should be the same as that of the web. When two covers are used their combined thickness should be at least equal to that of the web, so long as neither of them is made less than 1 inch thick. As regards the number of rivets required to make the joint, it is to be observed that they must be suffi- cient on each side of the joint to take up the" shearing stress at the section of the girder where the joint is made. Example 34. — For instance, taking the same data as in the two previotis examples, namely — Shearing stress = 9-3 tons. Thickness of web = |- inch, Diameter of rivets = |- „ the bearing resistance of one rivet is 2-62 tons ; so that the number of rivets required on each side of the joint is = 3-5 nearly. These rivets are arranged as shown in Fig. 251, and the pitch would be 18 . , — = 5 inches nearly, o'o which is too great, being 13 times the thickness of the plates riveted. The pitch may be taken at 12 times that thickness,^ or practically at 4". Width of cover plates. — As regards the width of the cover plates, follow- ing the rule, p. 136, that the centres of rivets are not to be nearer the edge of a plate than li times their diameter, it will be seen that the width must not be less than Ax d = 6 d. In the above example, therefore, the least width of cover plate is 6 X I = 5^ inches, say 6 inches, and this would be a suitable width to take. Connection of the Web with the Flanges. The web is connected to the flanges by being riveted to the angle irons, and the number of rivets should be sufficient to with- stand the horizontal shearing stress. The easiest way is to calcu- late the amount of the shearing stress per foot, and then to find the number of rivets required for bearing and for shearing. Example 35. — In the above example, for instance, the shearing stress at each end per foot B.C. — IV, ^ See Stoney's Rule, p. 136. M i62 NOTES ON BUILDING CONSTRUCTION 9-3x12 = — — — = 5*3 tons. Now if 1^" rivets are used, tlie resistance of one rivet to bearing in web (taken as thick) is = 2-62 tons. {The resistance to shearing need not be considered, as it is much greater.) Hence the number of rivets required per foot 5-3 = =2, 2-62 ' so that a 6-inch pitch would be sufficient as regards bearing, but this would be too great in proportion to the thickness of the plates, and a pitch of not more than 12xf = 4|" should be adopted. In practice I" rivets with 4" pitch would probably be used in both this and Example 34. We have now considered the main points in the calculations required for plate girders. In the two following examples some further matters of detail will be considered. Example of Plate Girder -with Single Web.^ Example 36. — Design the girder shown in outline in Fig, 26, p. 21. Load. — The load to be borne by the girder is 23 tons placed 10 feet from the left support, and from Fig. 54 it appears that the greatest bending moment occurs at this point. The weight of the girder itself must, however, be taken into consideration, and this weight can be found by using either of the formulae given in Appendix XV. These formulaj can only be expected to give approximate results, but the error is, as a rule, too small to be of any practical importance ; they are based on the supposition that the load is uniformly distributed. To apply them in the present case we must therefore first find the " equivalent " uniformly distributed load, that is the uniformly distributed load that would produce the same bending moment at the centre of the girder that the actual loading does. Now if W is this load, we know by Case 7, p. 33, that the bending moment in the centre is WZ W. 30 , = -r- = — - — foot-tons, o y And from Equation 2, p. 25, the bending moment due to the actual loading is R^x 10'= 15-33 X 10 = 153-3 foot-tons, W= — x 153-3 = 40-9 tons. 30 This does not include what would be required to carry the weight of the girder itself, but the additional material would be small (about -16 ton), so that we can take W = 41 tons. A'p'proxi'mate, weight of girder. — Using Unwin's formula (see Appendix XV.) and taking the data given there, and taking D = 30 as below, we obtain 41 X 302 Weight of girder =. ^^^^^^^ 30 _ 302 » = 2-62 tons. As a check we may use Anderson's formula, and thus we obtain Weight of girder = X 41 X 30, O D (J = 2-2 tons^ 1 See Plate A, p. 352. PLA TE GIRDER WITH SINGLE WEB 163 The former result agrees witli tlie weight of the girder given at p. 21 (2-6 tons), and we will assume it to be the weight of the girder. Be^pth of girder. — The depth of girders is usually taken at from -|- to of the span. It should be observed that the deeper the beam the more liable is the web to buckle, and therefore a thicker web or a greater number of stiffeners will be required. On the other hand, a shallow girder requires more metal in the flanges, and is less stiff. We will assume a depth of 30", that is jlg- of the span. The distance between the centres of gravity of the flanges may be taken as 30". Deflection. — It remains to be seen whether the depth chosen (30") is sufficient to ensure that the deflection does not exceed a certain amount, say -^-q" per foot of span. The total admissible deflection would be = J^x 30 = 0-75 inch. Now, as the beam is to be subjected to certain limiting stress, we can use Equation 46, p. 67, namely — ED ' As regards the value to be given to n', it is to be observed that the beam will be stifl'er than a beam of uniform strength, but less stiff than a beam of uniform section, for the former w' = |-, and for the latter (witli central load) »' = yV- will therefore be a fair approximation to take w' = TV- Farther, let = 4 tons, Tt = 5 tons, Z = 360 inches. As regards the value of E it should be observed that the full value (29,000,000 lbs.) for wrought iron cannot be reckoned on in built-up girders, owing to the unavoidable imperfections of riveting. A value between 16,000,000 and 18,700,000 is generally assumed. We will take 17,920,000 lbs., or 8000 tons. Hence ^_ tVx 9 X 3602 ~ 8000 X 30 ' ■ A = 0-49 inch. So that the girder is amply stiff enough. Width of flanges. — The width of the compression flanges should be sufficient to prevent buckling sideways. It is found by experience that for this purpose the compression flange should be made about to of the span, that is in this case 12 to 9 inches. We will take 9 inches. The width of the tension flange is not subject to the same condition, but for uniformity we will assume it to be also 9 inches wide. Thickness of flange plates. — The thinner the plates used, the more nearly can the girder be made to approximate to a beam of uniform strength. On the other hand, corrosion affects thin plates to a serious extent, and the number of plates to be joined together is increased. It should also be 6bserved that very thin plates are liable to buckle along the edges when placed in the compression flange. Professor Unwin says that the thickness l64 NOTES ON BUILDING CONSTRUCTION should not be less than of the span. In practice the thickness is limited between ^' and |"; usually |-" or ^' is found to be suitable. We will in this example take -I" plates. Angle irons for connecting the flanges to the web. — The thickness will be taken the same as that of the flange plates, namely |". A table 3 inches wide gives sufficient room for riveting, but in the present case we will use angle irons 3 1" x 3^" x f", as they will afford more room for riveting. Calculation of flanges. — We are now in a position to proceed with the calculation of the flanges in the manner explained at p. 156. From Fig. 54, p. 35, it will be seen that the diagram for the moment of flexure due to the single load consists of a triangle, the height of which is the maximum moment of flexure. This maximum moment of flexure due to the load is in the present case equal to X 10 X 12 = 15-33 X 10 X 12, = 1840 inch-tons. We thus get the triangle AcZB shown in Fig. 252. The weight of the girder itself produces a different diagram of bending moments, namely the arc of a circle (see p. 34) ; and the height of the arc is the bending moment at the centre, that is 2-6 X 30 X 12 3 = 117 inch-tons. We thus obtain the arc AcB shown in Fig 252. Fig. 252. The diagram for the total bending moments can of course be found by adding the ordinates of the arc to those of the triangle, and thus, finally, we get AeB as the diagram of bending moments. Tension flange (see Plate A). — The rivet holes must be deducted from the cross section to obtain the effective section. Deductinsr one f ' rivet hole from each angle iron, and taking the depth to the backs of the angle irons at 29", i.e. 1" less than the distance be- tween the centres of gravity of the flanges, whence the "effective" PLA TE GIRDER WITH SINGLE WEB 165 depth ^ of the angle irons is 2 7" we obtain the moment of resist- ance of both angle irons = 2(3^ + 31-1-1)1x27x5, = 595 inch- tons. Two rivet holes must be deducted from each plate, so that moment of resistance of the first plate is = (9-2 x|)|x 29|x 5, = 413 inch-tons nearly. In the same way it will be found that the moment of resistance of the second plate is = 423 inch-tons, and of the third plate 434 inch-tons. The total moment of resistance of the L irons and the three plates is therefore 1865 inch-tons, which is less than the maximum bending moment of 1944 inch-tons, but practically suflScient. Following out the process explained at p. 155, we obtain the completed diagram given in Fig. 253, which shows the number of plates required and the length of each. On referring to the Tables in Part III., giving the market sizes of iron, it will be seen that plates of the full length can readily be obtained, and that therefore no joints are needed ; but as an exercise for the student the joints i Plate - Ji \ex tended ; to Jorin I cover Fig. 253. have been so arranged in Fig. 253 that no plate will exceed 13 feet in length, and, moreover, the joints have been so placed that " grouped " joints can be formed. Each of these joints requires to be designed and calculated on the principles explained in Chap. VII., and in fact one of the joints has been worked out in Example 24, p. 132. Compression flange. — The diagram of bending moments just ^ i.e. distance between centres of gravity of angle irons. 166 NOTES ON BUILDING CONSTRUCTION obtained for the tension flange holds good for the compression flange, and is re-drawn in Fig. 254, but upside down, so as to show the plates in their proper relative positions. Fig. 254. As already remarked at p. 136, it is not necessary in this flange to deduct the rivet holes to find the effective section, but the safe stress is reduced to 4 tons per square inch. The moment of resistance of both angle irons will therefore be 2(3| + 3i -Dfx 27x4, = 537 inch-tons. The moment of resistance of the first plate will be 9 xf X 29|x 4 = 396, and so on. The total moment of resistance of the L irons and of the three plates will be found to be 1754 inch-tons, but the maximum bending moment is 1944 inch-tons. It will be therefore seen that three f" plates do not give sufficient resistance at the point of maximum moments of flexure, but practically they are quite safe. By making the outside plate Y thick the moment of resist- ance will be increased to 1897 inch-tons, which is near enough. Repeating the operation we can obtain the number and length of plates required, and also ascertain in what position the joints ought to be placed if full-length plates cannot be obtained. The flanges can now be considered as designed, with the exception of the joints, for which the student is referred to Example 24, p. 132. Weh. — To design the web the first step is to find the shearing stress in each portion of it. The simplest way is to make use of the diagrams given in Chap. III. As the girder is subjected to two loads, we must combine the diagram for a uniform load (Fig. 90, p. 59) with that for concentrated load placed at any point (Fig. 92). PLATE GIRDER WITH SINGLE WEB 167 Thus, in Fig. 255, Ae/B represents tlie shearing stress due to the weight of the girder, and AghklB the ^ shearing stress due to the load of 23 tons, so that AlmnoB repre- ^ sents the shearing stress to which e, the web is subjected. j It will be observed that owing ! to the weight of the girder being ^ small in comparison with the load to be carried, the shearing stress is almost constant from A to D and again from D to B. It will therefore be theo- retically best to make the web of a certain uniform thickness from A to D, and of another less thickness from D to B ; these thicknesses must now be found. For the part AD we have a maximum shearing stress of 16 "6 tons to deal with (seep. 22), and the depth of the web (S) between the angle irons may be taken as 22" and the distance between the centres of rivets as 25 '1 (see Plate A). Assuming the web to be ^ thick, the intensity of shearing stress is 16-6 ^ . ' — — =r= 1"32 ton per square men. 25-1 X 1^ r 1 Now referring to Table V., we find the value of - for the equivalent long column to be, taking n = 3'5, _ 3.5X^^=218, 2 and hence the safe intensity of stress (for a column fixed at both ends) to be nearly 1-40 ton ; so that a i" web will do very well. We can also find the thickness of the web by means of Table XI. The net unsupported width of the plate is 22" and the shearing stress per foot is ^^25^1^^ ^ ''''^^ tons, and it will be seen on referring to the Table that ^" is the thickness required. It should be noticed that the f rivets, which have been selected, are a little small for a ^" plate, but this difficulty can be overcome by drilling the holes, or by using a f " plate for this part of the web, supported by stiffeners. Pitch of rivets to connect the web to the flanges. — The stress per foot is 1^ X 12 = 7 "94 tons nearly. Taking rg = 4 tons and = 8 tons the resistance to bearing of a f ' rivet in a ^" plate is 3-0 tons (Table VIII.), and the resistance to double shear is 1*8 X 2 = 3'6 tons ; so that the number of rivets per foot 7-9 = ¥-o = '-'- Practically a pitch of 4" would do. For the part DB of the web we have a maximum shearing stress of 9-0 tons (see p. 22). Assuming the web to be f " thick, the intensity of shearing stress is 9-0 " =0-95 ton. I . 25-1 X I But the value of - is 1 See Appendix V., p. 298. i68 NOTES ON BUILDING CONSTRUCTION f And from Table V. the safe intensity of stress is 0-86 ton, so that a |" plate is somewhat thin. It should be observed, however, that the maximum shear- ing stress occurs only at the abutment B, and it will also be found that -Z^" is thicker than necessary according to the Table. There can, therefore, be no doubt that f" is thick enough. To use Table XI. we have : net unsupported width of plate =22", and shearing stress per foot = 4-4 tons, which correspond to a thickness of |". Fitcln, of rivets to connect the web to the flanges. — The stress per foot is 9-0 X 12 = 4-3 tons, 25-1 and the resistance to bearing of a number of rivets per foot rivet in a |" plate is 2-2 tons. Hence the 4-3 or a 6" pitch will do as regards bearing ; but in order to keep the plates together the pitch should not exceed 12 x |" = 4|-", and may practically, for the sake of uniformity, be made 4" as in Plate A. Joints in the web. — Two joints will be required in the web, so as to keep the plates within the ordinary market sizes. One of these joints will be best placed immediately under the load, and the joint can be made by means of two _L irons, which will also act as a stiffener to transmit the load to both flanges of the girder. Keferring to p. 160 we see that the least width of the table of the ± iron should be 8 x | = 6 inches. If the stem of the JL iron is I" thick, there will be enough room for riveting, as shown in Fig. 256. It will benoticed that this joint is placedat the point where there is noshear- ing stress in the web, so that theoretically there is no need for a cover ; but prac- U'l pitch U; pitch s/i rivet Packing c:7 rig. 256. Q @ © ''© 0 \ Q 0 Q © h"j'itch 1 Q <«'- Q Q © Q Q Q Q © 0 © © © I'ig. 257. tically it would not do to dispense with it, and, moreover, in this before said, it fulfils the important duty of assisting to transmit the load of 23 tons. Theoretically, no rivets are required to connect the X ^^"^ to the web, but practically they would be placed at about 4" pitch to put the girder together. Fig. 257 represents the joint. The second joint will be best placed midway between D and B. The shearing stress at this point found from Fig. 2 5 5 is 8'4 tons.^ The cover plates ^ The student is recommended to draw Fig. 255 to a large scale to obtain this value. PLATE GIRDER WITH SINGLE WEB can be made 6" wide, and or -j-y is a suitable thickness. To number of rivets, the bearing resistance of |" rivets in a |" plate is 2 therefore number of rivets 169 find the '25 tons. 2-2 nearly, or a 4" pitch will do. End pillars. — The ends of the girder should be stiffened to resist the shearing stress at those points. The most economical way is to stop the longi- tudinal angle irons at the ends of the girders, and to rivet two vertical angle irons to the web, using packing pieces as shown in Fig. 258. The ends of the girder are generally bedded on sheet lead. The rivet heads should be countersunk to allow of this, and four of the rivets would be left out at one end to enable the girder to be bolted down to the top of the column. oJ Fig. 259 and Plate ^ A ^ show the girder as arranged to comply S with the calculations made above. The total length of the girder is 31' 0", 1' 0" more than the distance between centres of bearings, or 2' 0" more than the clear span, thus allowing 12" at each end to rest on the cap of the C.I. column and on the wall. The bearing area is there- 9 fore X 1 = 0-75 square foot, so that the 9"0 pressure at the wall end is =12-0 tons 075 per square foot, and this is not more than a hard stone can bear. If, however, the stone is soft, a casting ought to be placed under the end of the girder to distribute the pressure. <2> [■^ ! •'J I ll^ i 1-^ J 11^ L__i_ i ^ In Plate A the joggles in the angle iron are omitted for the sake of simplicity and economy in construction. NOTES ON BUILDING CONSTRUCTION Example of Box Girder. Example 37. — Calculate and design a box plate girder to act as a bressummer, as shown in Tig. 260 (see Plate B, p. 353). Load. — The first step is to find tlie load to be borne by the girder. The portion of the wall supported by the girder is shown in dotted lines ^ in Fig. 260, and the cubic content of this portion is Fig. 260, ^ It might be objected that the load due to the wall ought to be taken as the weight of the wall contained between vertical lines drawn from the ends of the BOX GIRDER 171 (20 + 14 164-14 ■) 14 f 14 + 10 16 + 14 ) 9 + |-^x2 + 8x8 + -^xl j- = 229 cubic feet, A cubic foot of brickwork weighs 1 cwt. (see Table XVII.),-^ so that weight of brickwork to be supported = 229 cwts. The first floor is supported over a length of 1 8 feet, and since the floor is 20 feet wide, area supported = 18x10 = 1 80 square feet. The second floor is supported for a length of 14 feet by the girder, hence floor area supported = 14 x 10 = 1 40 feet. The total for both floors is therefore 320 square feet. Assuming that these are warehouse floors, and that the load is Z\ cwts. per square foot, the total load produced by the two floors is 320 X 3-5 = 1120 cwts. Lastly, the length of the roof supported is 14 feet, that is, a horizontal area ofl4xl0 = 140 square feet is supported. Allowing 40 lbs. per square foot (Tredgold), the load due to the roof is 140x40 _ . = 50 cwts. 112 The total load on the girder (except its own weight) is therefore 229 + 1120 + 50 = 1399 cwts., = 70 tons very nearly. A'pfroxi'maU weight of girder. — Clearly this load can be taken as uniformly distributed. Using Anderson's formula (Appendix XV.) to obtain the weight of the girder we get the weight per foot run of girder 2240 x 70 . c,or^^■U =4 X 70 lbs. = 280 lbs. 560 Hence total weight of girder ■ * " = 280 lbs. X 20 = 5600 lbs. = 2"5 tons. Therefore the total distributed load the girder has to sustain is 70 + 2-5 = 72-5 tons. Deflection. — As the load is considerable we will take the depth at of the span, that is 24". It remains to be seen whether this depth will give sufiicient stiffness. In the formula (Equation 46) n'(rc + r()Z2 A = - ED We have in this case rc = 4 tons, rt = 5 tons, 1 = 240 inches, E = 8000 tons (see p. 163), D = 24 inches, girder ; it is considered, however, that the bond of the brickwork causes it to take part of the weight. 1 Under some circumstances it would be better to take 130 lbs, per cubic foot as the weight of the brickwork, to allow for the bricks being saturated with water. 172 NOTES ON BUILDING CONSTRUCTION and for n we must give some value between that for a beam of uniform strength and depth (i), and for a beam of uniform cross section loaded uniformly = ■^-^). We will take n' = i- Hence ^_ ^x9x240^ 8000 X 24 ' = 0-3 inch, 60 that the girder is amply stiff enough. It should be noticed that the value thus found for A is only approxi- mate, because it depends on the value of n' and of E, both of which can only be approximately estimated ; but the error would not be sufficient to affect the result practically, and does not, probably, amount to more than 10 per cent. Calculation of flanges. — As the girder has to support a 14" wall, the flanges might also be given a width of 14" (see Plate B). This width, it will be observed, is much more than is required to prevent side buckling ; but if, on calculation, a less width be found to give a better arrangement of the plates it might be adopted so long as the width is not less than 9", which is too narrow for a box-section, and would require very large corbels for the brick- work. The size of the L irons used for connecting the webs to the flanges must first, however, be determined ; there are two L irons to each flange. The shearing stress is considerable, and therefore the thickness of the metal must be such as to give sufficient bearing area to the rivets without having to place them too close. Referring to Fig. 263, p. 173, it will be seen that the vertical shearing stress at the abutments is 36-25 tons, which is distributed over a length of 24 inches, and at a foot from the abutments it is reduced to 36-25 X 9 — = 32-6 tons. Hence the average horizontal shearing stress for the first foot close to the abutments is 12 36-25 -f- 32-6 ^ X ■ ^ = 1 7-2 tons, and since there are two webs the stress the rivets have to resist per foot is 8-6 tons. If I" rivets are used, it will be seen on referring to Table VIII. that if the resistance to bearing is taken at 8 tons per square inch, the resistance of one rivet in a plate whose thickness is t will be 7-0 X t, and if the pitch of the rivets is 4", that is 3 rivets per foot, the total bearing resistance of the rivets per foot will be 7 X i X 3 = 21 X i nearly. 8*6 ' Hence = — = 0-41 inch, a thickness of will therefore do, and L irons 3" x 3" x would be a suitable size to use. It will be found that the centres of gravity of these L irons are 22" apart. Hence their moment of resistance on the compression side is BOX GIRDER 173 2(3 + 3-T-V)/^x22x4, = 428 inch-tons. Bat the moment of flexure at the centre of the girder is 2175 inch-tons. Hence the plates have to supply a resistance of 2175 - 428 = 1747 inch-tons. We can assume as a sufficiently near approximation for the present inquiry that 25 inches is the mean effective depth for the plates. If is the total thickness of the plates at the centre and 6 the breadth, we have on the compression side Hence if and for this, one y If 6x e X 25 X 4 = 1747, .-. 6<= 17-47. 6 = 14 inches, i = 1*25 inch, -V' and two -g-" plates would do. 6=12 inches, i = 1 '45 inch, and three y' plates would do. If 6=10 inches, t = 1*74 inch, and one ^ and three -I" plates would be required. Either of these three arrangements would require about the same weight of iron, but the first two are preferable in having fewer plates, and it does not appear to be a matter of much moment which of them is chosen. We will adopt the first arrangement, so that a width of 14 inches will be chosen. Fig. 261. Compression boom ^ — ^ >k plate ^-'---^^ \ ^^x-^^ opiate ^^-^ L jrons 20' 0" - - >i [-<■ 1 Scale UOOO inch-tons to one inch L iro7is \ 'Is plate ; iece Section m.n. Kg. 264. Rivets to connect the web to the flanges. — The calculations for these are also left to the student, being similar to those already explained. Vic ■Web ^weh , I, 11 8-- Joint riff. 265. 176 NOTES ON BUILDING CONSTRUCTION End pillars. — The clear span is 20 feet, but th^ girder must be made longer at each end so as to bear on the stone sills, say 9 inches ; and as the shearing stress is considerable, it would be advisable to adopt the arrangement shown in Fig. 264. The completed girder is shown in Fig. 265, the thickness of the plates being considerably exaggerated, also in Plate B. Intermediate diaphragms. — In a large girder, two or three of these should be inserted. They are pieces of plate iron riveted to both webs to prevent racking, that is, to Fig. 266. prevent the girder bending over as shown in Fig. 266. Saddle-backed or Hog-backed Girders have a curved upper flange such as that of the Cast-hon Girder, Fig. 144. Both flanges can be made of nearly uniform section throughout instead of as shown in Fig. 238. Such girders are however very seldom used in connection with buildings, aud need not be further described. Chapter IX. BRACED OR FRAMED STRUCTURES. General Remarks. — Before proceeding to the consideration of lattice girders and roofs, which are braced structures, we must investigate a few general propositions in connection with such structures, and we will also give two methods of finding the stresses in them. Let us consider in the first place the simplest case of a weight W supported by two rods, as shown in Fig. 267, A and B being Fig. 267. Fig. 268. Fig. 269. Fig. 270. fixed points. The two rods are evidently in compression, and as they are equally inclined to the direction of the weight they will be subject to equal stresses. The same remarks apply to the rods shown in Fig. 268, with the exception that they are in tension instead of in compression. Fig. 271. Fig. 272. If the inclination of the rods to the vertical is diminished, as in Figs. 269 and 270, the stress in them will also be diminished; and on the other hand, if the inclination is increased, as in Figs. 271 and 272, the stress will be increased. B.C. — IV. N 178 NOTES ON BUILDING CONSTRUCTION In Figs. 273 and 274 the rod AD is shown at a greater inclina- tion to the vertical than the other rod. Clearly the stress is less Fig. 273. in AD than in BD. Fig. 274. If, as in Figs. 275 and 276, BD is vertical, D w Fig. 275. Fig. 276. that is, the inclination to the vertical is nothing, the stress in BD will be equal to "W, and there will be no stress in AD. Having thus obtained an idea of the manner in which the position of the rods relatively to the weight affects the stresses in them, we proceed to show how the actual value of the stresses (in terms of W) can be found. Eeturning to Fig. 267, draw la (Fig. 277) parallel to W, Id parallel to BD, and ad parallel to AD, then by a theorem called the " triangle of forces," ^ if la is drawn to scale to represent W, dh 3 will represent C2 and ad will represent to the same scale. Supposing, for instance, that W= 1*5 ton, and that la is made 1"5 inch, then it will be found that = 0-93 inch, a^^= 0*93 inch, or 02 = 0-93 ton, Ci = 0-93 ton. Further, la represents the direction of W, and ad will give the direction of Ci and dl that of Cg, as indicated by the arrow- heads in Fig. 277. Transferring these directions to Fig. 267 we see that the rods AD and BD are in compression. It is very important to observe that in Fig. 277 the arrows follow each other, as it were, in the same direction of rotation. Again, applying the theorem to Fig. 268, we obtain Fig. 278. As before, if W= 1'5 ton, and 1*5 inch, dl = 0-93 inch, ad = 0-93 inch, 0-93 ton, Ti= 0-93 ton. ^ A proof of this theorem will be found in any work on Statics. Fig. 277. Fig. 278. we obtain FRAMED STRUCTURES 179 In this case the arrows follow each other in the reverse direction of rotation as compared with the previous case, and transferring to Tig. 268 we find that the rods are in tension. Applying the theorem in succession to Figs. 267 to 270 and Figs. 273 to 276, we obtain Figs. 279-286,^ and assuming in each case that W=l-5 ton as before, we find the stresses marked on the figures. The arrows showing the directions of the forces are also marked. The student is recommended to compare these results with the preliminary remarks. The, theorem of the triangle of forces can be thus stated : — If three forces acting on a point are in equilibrium, then, if a triangle be drawn whose sides are parallel to the directions of the three forces, the length of the sides will represent the magnitude of the forces, and the directions of the forces will follow each other round the triangle. Thus, taking a general case, if F, G, H are three forces acting on the point P as in Fig. 287, then Fig. 288 will be the triangle of Fig. 287. Fig. 288. Fig. 289. forces ; and if the magnitude of one of the forces be known, that of the other two can be found. Fig. 290. Fig. 291. Fig. 292. Now supposing we replace the force H (Fig. 289) by two forces K and M, which together have the same effect as H (in other words, H is the resultant of K and M), then the triangle cda (Fig. 290) would be the triangle of forces for K, M, and H (strictly H reversed), so that we obtain the polygon abed to repre- sent the forces Gr, F, M, and K. In the same way, if F were re- placed by two forces IST and 0 (Fig. 291), we should obtain the polygon aelcd (Fig. 292), and so on. 1 Figs. 279-286. See Plate I. at end of volume. i8o NOTES ON BUILDING CONSTRUCTION This result is called the " polygon of forces," and it can be stated as follows : — If any number of forces acting at a point are in equilibrium, then, if a polygon be drawn having its sides parallel to the foices, the sides of this polygon will represent or be proportional to the magnitude of the forces ; and the directions of the forces will follow each other in rotation round the. polygon. In the case of the ^polygon of forces it is only possible to find the magnitude of two of the forces when the magnitude of all the others is known. Thus in Fig. 290, if only the magnitude of G were known, all that could be done would be to draw cb and the direction of cd, but the point d could not be fixed, as the magnitude of K is not known ; but if the magnitude of K is known, then d can be fixed, and drawing da and la we see that the magnitudes of M and F are found. The student is recommended to study the above very care- fully, as these results will be found of the utmost importance in the sequel. Some different forms of framed structures. — A framed structure is composed of a num- ber of bars jointed to- gether. It is usual to assume that these joints are hinged, and offer therefore no resistance to the movement of the bars. Such an as- sumption, unless pin joints are used, is not true, but it errs on the side of safety. The simplest frame is a triangle, as in Figs. 293 and 294, and such a frame is perfectly braced; that is, it is rigid or stiff, and can- not change its form.^ This being the case, the reactions and can be found by the 1 A very minute change takes place in consequence of the elongation and shorten- ing of the members produced by the stresses, but this may practically be disregarded. FRAMED STRUCTURES i8i Fig. 295. Fig. 296. Fisr. 299. rule given at p. 18 for beams, and thus, taking the numerical data given in Fig. 294, we find = 3:0 X 7 = 5 cwts., = T4 X = 2 cwts. A frame composed of four bars, as in Fig. 295, is, however, not perfectly braced, and would collapse under the action of the load W. But if an addi- tional bar DB be added (Fig. 296), it will be- come perfectly braced. A moment's considera- tion will show that this additional bar, called a "brace," is in tension. The frame could also have been braced by adding a bar AC (Fig. 2 9 7), which would, how- n ever, have been in com- pression, and C3 would then be unneces- sary. If the bars AD and CB are equally inclined, and another weight W be added at D (Fig. 298), the frame would retain its shape, but the least deviation would cause it to collapse; it is, in fact, in unstable equilibrium, and practi- cally a brace would have to be added. If only one brace is added, it must be capable of resisting both com- pression and tension, that is, a simple rod would not do ; or two rods capable of resisting tension only may be added, as in Fig. 299. In this case the frame is over-hracecl, and it is possible, by drawing up the rods by means of coupling screws, to introduce stresses in l82 NOTES ON BUILDING CONSTRUCTION the frame in addition to those produced by the weight. In such a case the frame is said to be self -strained — a state of things which is of course to be avoided as much as possible. In Fig. 300 is shown a frame composed of five bars. If a brace EC be added (Fig. 301), the frame is still in- B completely braced, and another brace, either AC or EB, must be intro- duced. Figs. 302 and 303 show different ways in which a five -sided frame could be perfectly braced, and Fig. 304 shows the frame over- braced. From the above it will appear that to per- fectly brace a frame it must be divided up into triangles. Fig. 302. Fig. 304. Maxwell's Diagrams. Eeturning to Fig. 294, p. 180, which is redrawn in outline in Fig. 305,^ and applying the triangle of forces to the joint A, we obtain the triangle shown in Fig. 306, and therefrom the stresses given. The triangle also shows that the bar CA is in compression, as re- presented by the arrow pointing towards A, and that AB is in tension, as represented by the arrow pointing away from A. 1 Figs. 305-311. See Plate II. at end of volume. MAXWELL'S DIAGRAMS For the joint B we obtain in the same manner the triangle shown in Fig. 307. And lastly, at C we obtain the triangle given in Fig. 308, and we see that, as should be the case, the values obtained for and Cg are the same as before. Method of drawing a stress diagram} — The above three triangles can be combined together in one diagram, Fig. 309, thus: Commence by drawing the triangle for joint A. We now have C^, and on this side of the triangle we can draw the triangle for the joint C, as in Fig. 308, taking care to reverse the direction of C^. Lastly, the triangle for joint B can be drawn on C^, Fig. 309, and if the diagram has been correctly drawn, it will finish off or " close " where it com- menced. Sucli a diagram is called a stress force diagram,aTid is one of the methods of finding the stresses in a framed structure. These diagrams were originally proposed by Professor Clerk-Maxwell. It will be noticed that in this example we could have com- menced by drawing the triangle for the joint C, and then obtain the other two triangles, thus finding the values of and with- out previous calculation ; but it is not always possible to do this, as will be seen by the following example : — In Fig. 310, which is an outline copy of Fig. 297, but loaded with two weights W and W^, if we were to commence at the joint C we should have four forces to deal with, namely, the weight W and the stresses in the three bars connected together at that joint. Therefore, as already seen at p. 180, we cannot determine the stresses. Finding, however, E^ and Eg as usual, we can draw the triangle for joint B, as shown in Fig. 311. Now proceeding to joint 0 we form a polygon and obtain the stresses C3 and C^. This polygon is drawn by reversing the direction of C^, drawing W vertically to scale, then drawing C3 to meet C^, drawn parallel to their respective bars. Taking D as the next joint, we find C^, and the polygon for the joint A can then be added. This diagram could, however, have been drawn by commenc- ing at the joint D, without determining E^ or E3. The student is recommended to see what effect altering the values of W and has on the shape of the diagram. He will find, for instance, that when W = W^, then Cg = T and C^ = 0 ; or, again, when = 0, then C3 = 0 and C^ = 0, which is, more- over, evident on inspection. ^ In Appendix XYI. will be found more detailed rules for drawing a Maxwell's diagram. i84 NOTES ON BUILDING CONSTRUCTION The above method will be applied in the sequel to the determination of the stresses in roofs. Bow's system of notation has not been used for the simple roof trusses here dealt with, but it is admirable for complicated structures. It is explained in Ap- pendix XVII. and used for the braced girders at p. 354. METHOD OF SECTIONS. We can now proceed to the consideration of the second method of de- termining the stresses in a framed structure. Imagine the frame shown in Fig. 3 1 2 cut in two parts along the line PP ; then, considering the left portion, it is clear that this portion could he kept in equilibrium by applying to each bar that is cut, a force of the same mag- \ 1 A ^ Fig. 312. Fig. 313. nitude and direction as the stress existing in the bar before the section is made. Thus the forces R^, W, C3, Cg, and T, acting on the portion of the frame shown in Fig. 3 1 3, are in equilibrium. Hence the sum of the moments of these forces about any point is zero (see App. II.) Now it is quite inmiaterial about which point moments are taken. If, therefore, we wish to find Cg, we can select the point A, which is the intersection of Cg and T (the other two unknown forces), and so we obtain an equation containing Cg in terms of W, and the perpen- dicular distances from A on to the directions of Cg and W, which are called the lever arms (R^ does not appear in the equation, as it also passes through A). If I is the lever arm for W, then IW is the moment of W ; and if is the lever arm for Cg, Z^Cg is the moment of Cg — the sign being negative, because Cg tends to turn the portion of the frame in what we have assumed to be the negative direction (see p. 26). Hence IW -\G^ = 0, or 03 = ^ W. I and l-^ could be found by calculation, when the dimensions of the frame are known, but the process is rather complicated, and, moreover, they can be obtained graphically very simply and rapidly, and with sufficient accuracy for all practical purposes, by drawing perpendiculars from A (which can be called the " turning point ") to the directions of "W and Cg, and measuring the lengths of these perpendiculars to the same scale that the drawing of the frame is made to. Carrying this out in the present example we find (Fig. 314) Z = 6-70 feet, = 3 -93 feet, METHOD OF SECTIONS 185 and consequently 6-70 T93' W-1-7W. To find T we must take moments about the point D, the intersection of Co and Cg. Measuring the lever arms we find ^ -9-5T-8-3W + 15R^=0. Hence T = l-58R^-0-87W. And lastly, to find Cg we must produce T and Cg to meet at 0 (Fig. 315); and dropping perpendiculars from O on to the directions of W and R^ we find -6-7C5+19-3W- 12-6R^ = 0. Hence C5= - (r88R^ - 2-9W). E A <—-/-—» Fia;. 314. Fig. 315. This method is called the Method of Sections, and is a modification of that originally proposed by Professor Rankine. In the present form it was, it is believed, originally suggested by Professor Ritter.^ Rule fok Method of Sections. — This method can be stated as follows : — • Sever the structure in two parts by means of an imaginary line (which need not be a straight line), cutting if ^possible through three bars of the structure. Select one of the parts, and apply to the severed ends of the bars forces equal in magnitude and direction to the stresses in those bars. This part of the structure will then be in equilibrium. To find the stress in any one of the severed bars, take moments about the point of intersection of the other two bars. An equation will thus be obtained containing the unknown stress it is required to* find, any loads that may be acting on the part of structure selected, any reaction that may be acting on it, and the lever arms of the various forces involved. The following terms are used in connection with this method, and are tabulated as follows : — Turning ]}oint is the point about which moments are taken, and is generally the intersection of two of the unknown stresses. Lever arm of a force is the perpendicular distance from the turning point 1 It will be observed that the forces applied to the ends of the several rods in Figs. 313, 314, and 315 are all tensions. Hence clearly tensions are positive but compressions are negative, if the forces are thus drawn. 2 See Iron, Bridges and Eoofs, by Professor Ritter, translated by Lieut. H. R. Sankey, R.E. 1 86 NOTES ON BUILDING CONSTRUCTION to the direction of the force. Practically these lever arms are obtained graphically. It was mentioned above that the section should if possible be taken through three bars. This is not always possible, and the method must then be slightly modified. The above is only an outline of the method, which is a very powerful one. The student is referred to Professor Eitter's work above mentioned, where he will find the method very fully described. We have now considered two methods of determining the stresses in a braced structure, and it will be found that in some cases the iirst, and in other cases the second method, gives the result in the quickest and simplest manner ; and again in other cases the methods are equal in this respect. Experience will soon show which method to apply, and it does not appear to be of any use laying down general rules on the subject. It should be observed that both methods require a drawing to sc^le of the structure, and to obtain accurate results this draw- ing should be made to as large a scale as possible. We now proceed to apply these methods to some simple cases of framed structures. Chapter X. BRACED OR OPEN WEBBED GIRDERS. "DEACED or open webbed girders are much used in engineering works, but have only a limited application in building con- struction. It will therefore only be necessary to consider one or two simple cases. Moreover, the general theory of these girders is difficult, and beyond the scope of this book. Warren Girders. — A very usual form of braced girder is shown in Fig. Fig. 316. 316, and names are given to the different parts as follows : — AB is the compression boom or flange. *io*i9 tension boom or flange. a^ciii, «2^i2' ^3*^13' ^^°-5 called braces, and are tension braces or compression braces according to the stress in them, a^, a^, etc., are the apices. A modification of this girder would be obtained by turning it upside Fig. 317. down, as in Fig. 317, when the nature of the stresses in the diagonals is of course reversed. These forms of braced girders are known as "Warren girders. Fig. 318. Lattice Girder. — Fig. 318 shows another form of girder, which is called a lattice girder. Vertical braces are added at the ends, and are called end pillars. i88 NOTES ON BUILDING CONSTRUCTION It is most usual to incline the braces of a Warren girder at 60° and those of a lattice girder at 45°. M" Girder.^ — Fig. 319 shows another form of braced girder, called the N girder. In this the upright braces are called verticals. Mr A Fig. 319. X X X X X X X X X X Fig. 320. N (braced) girder. — Sometimes additional braces are added, as in Fig. 320 ; but, as explained at p. 181, the girder is then over-braced and may be self- strained, so that these braces should not be added unless the girder is subject to a moving load, and then only to a certain number of bays next the centre, depending on the proportion of the moving load to the weight of the girder. Application of tlie Load; — The load is supposed to be concentrated at the apices, either at the top or at the bottom ; occasionally, but unusually, at both top and bottom. To do this, arrangements must be made to transmit the load to the apices by means of secondary bearers. The weight of the braces and booms can also be considered as applied at the apices, and practi- cally the weight of the girder can be taken as uniformly distributed. Proportion between span and depth of girder. — As in former cases, the depth of the girder must be sufficient for stiffness, and it is found that a ratio of about J^- gives the best results. The exact depth must be arranged so as to get an exact number of triangles, and will therefore depend on the angle of inclination chosen for the braces. Stresses in Warren Girder. Example 38. — We will take as an example the girder shown in Fig. 321 supporting loads of W-^, Wg, Wg, etc., tons on the upper apices. w, w, W3 w w, w, w, W3 w, ^ la, iflj ^a^ lil y la^ w •11 "-12 <.'l6 "17 '^IS ^1! a, 4 dy. Fig. 321. To find the number of triangles we have, d being the depth, = tan 60 = ^3, _ M 1" V^' ^ Also called WhiiJple-Murphy girder. STJiESSES IN WARREN GIRDER If n is number of triangles, 189 2c3! n(Aa^) = span = 1 2d = n -y^, n=10-4. whence .. - - . -, , , x ^-u Therefore ten triangles will do, and will give a depth rather greater than -jL- span, as shown in the figure. K- a. Fig. 322. To find the stresses, take a section and select the left portion of the girder for calculation (Fig. 322). We will adopt the following notation :— C4 5, the compression in the bar a^a^ of the top boom. B^,;^^, the stress in the brace a^a-^^. Ti3,;l4, the tension in the bar a-^^a^^ of the lower boom, a, the distance between the apices. And similarly for the otlier bars. To find the str obtain the equation To find the stress €^,5 we must take moments about a^^, and we thus Supposing that = W, = W, as is most usually the case, then R^=4-5W, and Or and But Therefore + 2cZxC4,5-a(W + 3W + 5W + 7W-9x4-5W) = 0. + 2dxC4,5 + 24-5Wxa = 0, C, , = - 24-5W 4.5 2d ^ = cotan 60' a 1 C 24-5 — — W. The - sign shows that 0^,5 is in compression. _ Ao-ain, to find T^g^^^ we must take moments about a^, and we obtain _dxTi3,i4-W3xa-W2X 2a-WiX3a + R,x4ct = 0. As before, when the apices are equally loaded, we get - d X T, o ,4 - a(W + 2W + 3W - 4 X 4-5 x W) = 0. igo NOTES ON BUILDING CONSTRUCTION a ^ 5: 5 — '•"lb ^ ^ v« 5 — ^ — 5 — a" 5 < That is, „ 12a ia According to the rule, to find B^^^^ we ought to take moments about the intersection of a a. and a^^a^^. But these lines do not meet, or, in other words, their intersection is at infinity. All the lever arms would therefore be infinite. Although the result can be obtained by following the rule, it is simpler in this case to resolve vertically (see Apjoen- dix II.) the forces acting on the portion of the girder shown in Fig. 322. It will be observed that both ^4.5 ^'^^ '^\^,\'^ ^^"^^ resolved part vertically, since they are both horizontal, and that the resolved part of B x^in 60°, Hence ^4.14 Or, if the apices are equally loaded, _ w In the same manner the stresses in all the other bars can be found. These have been worked out, and are marked on Fig. 323. In the above example the loads are supposed to be applied to the upper apices. If they were applied to the lower apices, the stresses would be slightly modified. The student is recommended to re-work the example with the loads applied to the lower apices. Stresses in Whipple-Murphy Girder. Example 39. — Find the stresses in the girder shown in Fig. 324, the lower apices being loaded with 3-0 tons each. To find the stresses in the bars a^a^^ a^^^g^ and ayy^Yii take an oblique section along the line 00, as shown in Fig. 325. The stress in a^a^ can be found by taking moments about ct^o, thus (the depth being 4 feet) — + 403^^-4 X 3-0-8 X 3-0 + 12 X 13-5 = 0. WHIPPLE-MURPHY GIRDER 191 Whence €3^^= — 31 "5 tons. a^a^ is therefore in compression. To find T^i,i2, take moments about dg, so that o.Y)P^Y2. tension. - 4Tii,i2-4 X 3-0 - 8 X 3-0 + 12 x 13-5 = 0, '^11.12 ~ +31'5 tons, cli as Oi as «6 ^ Ob 9t=13-5ions Fig. 324. Fig. 326. Fiff. 327. thus- The stress in the vertical be found by resolving vertically, '^3,12 - - ^'^ + 13-5 = 0, 19 = + 7'5 tons. Another section must be taken to find the stress in the brace cu^f'ix, shown in Fig. 326. Kesolving vertically, Bg^j^ X cos 45° + 13-5 - 3-0 - 3-0 = 0, V-5 X v/2, Bo = - 10-6 tons. To find the stresses in the bars a^a^ and (t^a-^^^, we can take a section NOTES ON BUILDING CONSTRUCTION cutting through only two bars (Fig. 32 7).< — Resolving vertically, Bo^;^ X COS 45° + 13-5 = 0, or ■'^o,!" -13'5x a/2) = - 19-1 tons. To find Tq^jq Ave can take moments about any point in aga^, for instance, about P, vi^hose distance from is c, then - Tq^^q X c sin 45° + 13-5 x c cos 45° = 0. But sin 45° = cos 45°, so that Tq,;^q= +13-5 tons. The stresses Bq_^ and T^^^q can also be found very simply, by diagram, as shown in Fig. 328. Fig. 328. / i \ J 5- 'S ions. Fig. 329. It will be observed that a straight section taken to obtain the stress in a^a-^^ will cut through four bars, but by taking a circular sec- tion, as shown in Fig. 329, we need only cut through three bars, and by resolving vertically we obtain at once ^5,14 = + 3 tons, a result which is self-evidently correct. The stresses in the remaining bars of the girder can be obtained in precisely the same manner to the above, and as an exercise the student is recommended to carry out the cal- culations for each bar and check the results he obtains by those given in Fig. 330. He is also advised to find the stresses when the girder is turned upside down, as shown in Fig. 331- PRACTICAL FORMULA 193 General Eemarks. — It is to be noticed that in both the above examples all the top horizontal bars are in compression, and all the lower horizontal bars in tension. Further, all the braces sloping downwards away from the centre are in com- pression, and all the braces sloping downwards towards the centre in the Warren girder and the verticals in the Whipple-Murphy girder, supported as in Fig. 330, are in tension. The stress in both top and lower booms is greatest at the centre, and gradually diminishes towards the ends ; the stress in the braces is least at the centre, and increases towards the ends. The student is recommended to compare carefully these stresses with those in a plate girder. Dimensions of the various bars. — The stresses being obtained, the dimensions of the various bars can be found as explained in Chap. VI. For the small braced girders used in building construction, the booms are generally made of horizontal plates riveted to- gether as in plate girders, and connected by L irons with a vertical web plate, to which the braces are secured. The ties are generally made of flat bar iron, and the struts of L, T, or LJ iron. Short practical Pormula for all open webbed Girders. Tlie stresses in the flanges Bg may practically be taken as in a plate girder. Applying this to Example 38, p. 188, for a plate girder (uni- formly loaded) we have from Equations 15 and 89 o .:BsXd==~ and Bs= — , w= — in Example 38: d= , a r 7 2 ' W (10a)2 2 a 8 ^3 25aW J3 ' aW liere = W in Example 38, 25W V3 B.C. IV. 194 NOTES ON BUILDING CONSTRUCTION wHch is practically near enough to the result in Example 38, viz. 24-5W V3 The stress on any har is equal to the shearing stress at the point multiplied by the cosecant of the angle of inclination of the bar to the horizontal, and divided by the number of triangulations. Thus in Example 38 the shearing stress at 'B^,ji= — W^ — Wg _W2-Wi + E^= -4W + 4-5W=r-5W. By above rule the stress in B4,i4 = '5W cosecant 60° 2 W which agrees with the result by the other method (see p. 190). In vertical bars (the cosecant of 90° being 1) the stress is equal to the shearing stress at the point (see Vgjig, p. 191). Graphic Method. Figs. 1 and 2, Plate C, show Example 38 worked out by a Max- well's diagram, and Figs. 3 and 4 show the same for Example 39. In Example 38, 2 tons per upper apex has been assumed as the load. The notation is by Bow's method, which is explained in Appendix XVII. By measuring the different lines of the diagrams on the scales given it will be seen that the stresses found graphically agree with those found by calculation. The thick lines denote compressions, the thin lines tensions. Lattice girders may be calculated by the method given for Warren girders at p. 188, or by the graphic method. A separate calculation or diagram should be made for each triangulation. Bowstring Girders have a curved upper flange or low, wliich for an uniformly- distributed load should theoretically be a parabola, but is practically made an arc of a circle (see p. 34) ; the lower flange or string is horizontal. For an uniformly distributed dead load only vertical bars are required, suspending the string (on which is the load) from the bow. Where Z = span, w = weight of bow and load per unit of length, (i = depth of girder at centre, = distance between the vertical bars. wP The horizontal thrust (T) throughout bow = tension in string =g-^ The compression along the bow at any point distant x horizontally from the centre = ^w^x^ + T". Thrust on bow at spriuging=T x secant of angle of tangent of bow at this point with horizontal. Tension on each vertical bar = mw. Advantages. — The stress on the string and verticals is uniform, that on the bow varies only from 8 to 12 per cent, so that the bars and plates may be of uniform dimensions. As the girder can be made of a depth equal to 4 its span, the sections of the flanges will be light and economical in comparison with those of shallower girders. Bowstring girders are very seldom used in connection with buildings and need not be further described. Chaptee XI. TRUSSED BEAMS O EVEEAL kinds of trussed beams were described in Part I., and ^ it will be noticed that the construction of these beams is simple ; the calculations to obtain the stresses accurately are, how- ever, difficult, and beyond the scope of these Notes. It is clear that the beam itself bears a certain proportion of the load, and the trussing makes up the difference. The part borne by the beam depends on the comparative stiffness of the beam and of the trussing, and this varies with the temperature, which affects the length of the tie-rods. Moreover, the beam is not in a simple state of bending stress, but is also subject to direct compression arising from the trussing, and liable at any time to alteration by tightening or slackening the nuts of the tie-rods. These con- siderations will show that the accurate calculations are compli- cated and practically useless, as they would have to be based on assumptions which might at any time cease to be correct. It is therefore usual in practice to assume that the beam does not bear any of the load ^ as a beam, and this as- sumption amounts to the same thing as to suppose that the beam is jointed at the point or points i^ — where the struts are con- * FLfr. 332. nected to it. Thus the trussed beam shown in Eig. 332 is assumed to be jointed as in Fig. 333. This assump- tion clearly errs on the safe side, and makes a rough allowance for the compression produced in the beam by the tension in the tie-rod. With this assumption there is no difficulty in finding the 196 NOTES ON BUILDING CONSTRUCTION stresses by either of the methods given in the preceding chapter. For instance, employing Maxwell's diagrams it will be seen that Fig. 334 is the diagram of the stresses in the trussed beam shown in Fig. 332. Having thus obtained the stresses, the dimensions of the '« various parts can be found as ex- plained in Chaps. III. and VI. ■^^S" Trussed beams of the form shown in Fig. 335 are, as explained at p. 181, incompletely braced, and are therefore unsuitable if the load is not uniformly distri- njPlui thread Fig. 335. buted. Fig. 336 shows the change of form which would take place if the beam were jointed and one point only were loaded • Fig. 336. this figure exhibits, therefore, the tendency when the beam is con- tinuous, although the stiffness of the beam controls the deforma- tion. For uniform loads, however, such beams are suitable, as the stiffness of the beam would be quite sufficient to counteract any small unevenness in the load that might occur. Example 40. — Design a trussed beam 1 8 feet span to carry a load of 8 cwts. per foot run. The first step is to find how many struts to employ. If we place one ptrut at the centre, the beam will be divided into two portions of 9 feet each, and there will be a load of 9x8 = 72 cwts. on each of these portions. Considering each, portion as a beam loaded uniformly and supported at both ends, it will be found on referring to Tables II. and III. that for strength 6" X 9" deep, and for stiff'ness 3|" x 1 1" deep, is required. If, therefore, timber 3f X 11, or its equivalent in section, is available, one central strut will be suffi- cient, and the trussed beam would be of the shape shown in Fig. 332. We will, however, suppose that timbers of this section are not obtainable ; it will therefore be necessary to use more than one strut. If two struts are used, as shown in Fig. 335, the total span will be divided into three portions of 6 feet each, and if each of these portions is looked upon as a beam uniformly TRUSSED BEAMS 197 loaded and supported at both ends, it will be found tbat for strength and for stiffness a beam 4" x 7" deep is needed. We will decide on this section. The depth of the trussing must next be fixed. If it be made deep, the stresses will be much reduced, but in many cases a deep trussing is incon- venient. A rule applicable in general cases is to make the depth of the trussing from centre of beam to tie-rod about of the span. In the present Ra=AS cwts IS avis. Fig. 337. case, therefore, a depth of l' 6" would do very well, so that the struts will be 1' 2" long. Fig. 337 shows the trussed beam in outline. Distribution of the Load. — We may consider that half the load on AC and half the load on CD is concentrated at C ; that is, the load at C is 1(6 X 8 -1- 6 X 8) = 48 cwts. The remaining half of the load on AC is transmitted direct to the point of support, and it therefore does not affect the stresses in the trussing. At D there will clearly be the same load as at C, hence the reactions E.^ and Eg are each equal to 48 cwts. We wiU now find the stresses both by the method of sections and by Clerk-Maxwell diagram. Stresses by Method of Sections. — To find the stresses in AE and CE take the section shown in Fig. 338, and for the stress in AE take moments about C. - l-44T^-i-6 X 48 = 0, 6 X 48 = 200 cwts.. 1-44 = +10 tons, the stress in CE take moments To find about A. + 6 X 48 -f- 6 X Cj^ = 0, C-^ = — 48 cwts., = — 2'4 tons. The stress in EF can be found by taking the section shown in Fig. 339, and taking moments about C, then - 1-5 X Tg-f-e X 48 = 0, Yf T„ = 6 X 48 cwts. Fiff. 339. 1-5 = 9-6 tons. The stress in FB will clearly be equal to that in AE, and the stress in DF to that in CE. We do not require the direct compression in AC, CD, or DB for the approximate method we are employing. 198 NOTES ON BUILDING CONSTRUCTION Stkesses by Clerk-Maxwell Diagram. — Commencing at the abutment A we obtain the triangle fgh (Fig. 340), from which we find T-^ = 10 tons as before. Proceeding to E we obtain the triangle 1dm (Fig. 341), from which we obtain Tg = 9-6 tons, Cj^ = 2*4 tons. h Fig. 340. Scale 100 cwt. — 1 inch. Having obtained the stresses, the diameter of the tie-rod can be found as Fig. 341. Scale 100 cwt. = 1 inch explained in Example 13, and is given in Fig. 335. The stress in the struts is so small that no calculation of dimensions is needed. Chapter XII. ROOFS. MANY different kinds of roofs were described in Parts I. and II., and Tables of the dimensions used in practice were given. Cases may, however, occur in which the conditions are different from those assumed in these Tables, and it then becomes necessary to find the stresses in the various members of the roof, and there- from deduce the dimensions. The general method of finding the stresses in roofs will therefore be described in an elementary manner, and wiU be illustrated by examples. Loads to be borne by Roofs. Roofs are subject to several kinds of loads, but these loads can be classed under two heads — 1. Permanent Loads. — These consist of the weight of the roof truss itself, of the purlins, rafters, and roof covering; frequently also the weight of a ceiling has to be borne by a roof, and sometimes the weight of a lantern. The weight of the truss itself must be estimated much in the same way as the weight of a girder, but there does not appear to be any formula available corresponding to that of Unwin's for girders. Table XII., however, gives the information required for several different kinds of roofs, and even in a special case the weight required can be deduced therefrom with sufficient accuracy for practical purposes, care being taken in large structures to check the assumed weight by that deduced from the completed design, modifying the dimensions if found necessary. The weights of purlins, rafters, roof covering, etc., are also given in Tables XII. and XIII. The use of these Tables will be better shown when working out the examples. 200 NO TES ON B UILDING CONSTR UCTIO N 2. Occasional Loads.— The occasional loads consist of the weight of snow and of the wind pressure. Bnow.~T\i% weight of snow a roof is Hkely to have to bear depends on the locahty in which it is to be erected, and at best only an approximate estimate can be given. It is usual to assume m England that a roof of very flat pitch may have a depth of 6 inches of snow on it, and the depth will of course diminish as the pitch increases. It can be further assumed that the weight of a cubic foot of snow varies from 6^ to 11 lbs., according to the amount of consolidation, so that an allowance of 5 lbs. per square foot of horizontal surface covered by the roof is ample. Wind pressure.~The wind pressure is more difficult to allow for, and the experimental data on the subject are at present very deficient. It is, in fact, only since the Tay Bridge disaster that Enghsh engineers have inquired at all closely into the subject, and the estimated allowance to be made for wind pressure has vlried very considerably. In Unwin's Iron Bridges and Hoofs, however although published long before the Tay Bridge disaster, the subject IS treated at some length, and the following is principaUy derived from that work. Assums that the wind blows horizontally. When it meets the inclined surface of a roof it will produce a pressure, normal (i.e. perpendicular) to the surface of the roof, and there will also be a force exerted along the surface of the roof. Now as air is almost a perfect fluid, this force along the surface of the root will be extremely minute and can be neglected. Therefore we can state that the wind pressure acts perpendicularly to the surface of the roof Let P be the horizontal wind pressure per square foot, that is P is the force exerted by the wind on a vertical surface one square foot in'area and let P, be the pressure normal to the surface of the roof per square 'foot then accordmg to Hutton's experiments ' Pj, = P(sin i)i-84cosi-i . . C90) where i is the angle of pitch of the roof. ' ' ^ ^' _ Amount of wind pressure to he calculated for.— As regards the value to be given to P, the data at our disposal are not so numerous or so trustworthv as might be wished ; the general practice, however, is to give P some value between 40 and 60 lbs. per square foot.i For ordinary roofs, unless in positions of exceptional exposure, 50 lbs. per square foot will be sufficient to take. Table XIV. is derived from Equation 90, which it must be remembered is only empirical, on the assumption that P= 50 lbs. per square foot. ' A wind pressureof 41 lbs. per square foot has been registered at Greenwich. Tlie gale which destroyed the Tay Bridge was calculated to have a force of 42 lbs per square foot. The Board of Trade require 56 lbs. per square foot to be taken for railway bridges which are often much exposed (Wray, Seddon). LOADS ON ROOFS 201 Stresses in roofs of special shape.— It may occur in large roofs of special shape that the wind pressure produces (say) compression in some member of the roof, in which the permanent load and the snow produce tension, or vice versa. In such a case the wei4t of the snow actually assists the member in question, but as the maximum wind pressure may occur when there is no snow on the roof, this assistance from the snow should not be reckoned on.^ Hence the stresses due (1) to the permanent load (2) to the snow, (3) to the wind pressure, should be determined separately, and that' combination of (1), (2), and (3) which produces the greatest stress in any member must be adopted in designing it. However, in the comparatively small roofs used in ordinary building construction, such as those illustrated m Parts I. and II., the wind pressure does not reverse the stresses in any of the members. Stresses in ordinary roofs. — In these it is only necessary to obtain the stresses produced — (1) By the permanent load and the snow combined, (2) by the wind pressure ; and from a consideration of these stresses the maximum stress to be borne by each member can be found, as will be shown in the sequel. Tredgold assumed that the wind pressure acted vertically and uniformly over the whole surface of the roof. Such an assump- tion is clearly erroneous, nevertheless the scantlings given on his authority in the Tables, Parts I. and II., are dependent on this assumption. The lighter scantlings for wooden roofs given in Parts I. and II., and in Table XV., were obtained on the assump- tion that the wind pressure acts perpendicularly to the surface of the roof, and only on one side, and are recommended for use. Distribution of the Loads. The weight of the roof covering of the snow and of the wind pressure is borne in the first instance by the common rafters, and by them transmitted to the purlins. Taking a simple case, in which there is only one purlin on each side of the roof, each common rafter will be supported in the centre by the purlin, at the top by the ridge piece and at the bottom by the wall plate. If the rafter is continuous, and these three points are accurately in a straight line, it can be shown (see Appendix XL) that f ths of the load on the rafter will be supported by the purlin, y^gths by the ridge piece, and j-^^ths by the wall plate. If, however, the three points of support are not in a straight line, this distribution would be changed. Supposing the rafters were not continuous, but (as it were) jointed just over the purlin, then clearly the dis- 1 " It is suggested that, except perhaps in very cold climates, the snow need not be considered at° all. Since long before such a wind force as 50 lbs. per foot square could take effect upon a roof all the snow would have been blown offit."— Wray, Instruction in Construction (Seddon). 202 NOTES ON BUILDING CONSTRUCTION tribution would be as follows : — half the load on the purlin, quarter on the ridge piece, and quarter on the wall plate. This latter distribution causes greater stresses in the roof, and it will therefore be better to accept it. Now let a, I, c (Fig. 342) be three purlins (and we will suppose, as is usually the case, that ab = he and that w is the load ^ on the portion db of the rafter and also on the portion he), then, if the rafter is considered to be jointed over each purlin, half the load on the portion ab ^or ^ will be transmitted to the purlin b, and half the load on the portion he [ov I) will be transmitted to the same purlin. On the whole, therefore, the purlin h will have a load = w transmitted to it from the rafter in question. Applying this to the case where there are Tig. 343. •w s two purlins on each side of a roof, as shown in Fig. 343, the two purlins will each support a load w and the ridge piece and wall plate — . The total load on a purlin will be that due to the sum of the loads transmitted by all the common rafters supported by that purlin. Eeferring back to Fig. 342 it will be seen that purlin 1 The loads are shown acting vertically, but the argument holds equally for the wind pressure. LOADS ON ROOFS 203 6) supports the load on a strip of the roof — ^ — ^^'^ toending the whole length of the purlin, that is, from one truss s^upporting the purlin to the next. This is indicated in Fig. 344. The purlins transmit the load to the principal truss, and, Hooking upon the purlin as a beam supported at each end, half tbhe load on the purlin will be borne by each truss supporting it. Mg. 344. Fig. 345. TThus, referring to Fig. 345, at each point where a purlin is fixed tto the principal rafter of the truss there will be a load = \ load on purlin AB + -J- load on purlin BC, tthat is the load due to the hatched portion of the roof. Fig. 345. Usually the trusses are equidistant. In such a case the lload at every point of the principal rafter where a purlin is fixed iis equal to the load on a strip of the roof equal to distance apart of purlins X distance apart of trusses. So far we have tacitly assumed that the purlins are placed i immediately over the joints of tthe truss as shown in Fig. 346. ^Sometimes, however, smaller pur- ]lins are used, and they are dis- Itributed along the principal rafter fas shown in Fig. 347. In such fa case we can consider ab as a Ibeam supported at I, and Ic as f another beam supported also at II, so that half the load on each ( of these beams will be transmitted Ito the joint at &. The load on Ithe joint at I will therefore be 1the same as if there were only ( one purlin at I. The above Fig. 347 refers strictly only to the loads due to roof covering, rafters and 204 NOTES ON BUILDING CONSTRUCTION purlins, wind and snow. The weight of the truss itself should strictly be considered as applied at all the joints, but it is sufficiently accurate to consider this part of the load as applied to the same joints {i.e. the upper joints) as the other loads. In every case, therefore, the load acting on a joint of the principal rafter (and in almost every case these are the only joints that have to be considered as loaded) is the load on a strip of roof whose width extends half-way from the joint in question to the joints on either side of it, and whose length is the distance between the trusses. Example 41. — To take a numerical example, let it be required to find the loads on the joints of the iron roof shown in Fig. 348, the trusses beino- supposed to be 10 feet apart. ° Permanent Load and Snoio. — First as regards the permanent load and the snow, we want to know the load per square foot of roof surface. Referring to Table XIII. we find Per square foot of roof surface. Eoof covering (countess slates) Slate boarding (1" thick) From Table XII. Common rafters and purlins Principal Snow .... 8 lbs. 3-5 „ 11-5 lbs. Per square foot of covered area. . 2-0 lbs. . 3-5 „ . 5-0 10-5 lbs. Fig. 348. Fig. 349. This last part of the load must be reduced to the load per square foot of roof surface. Referring to Fig. 349 it will be seen that it is the load on 1x1-15 square foot of roof surface, so that this part of the load is, per square foot of roof surface, 10-5 r— -r=9-l lbs. I'lo Therefore the total load per square foot due to permanent load and suow = 11-5 + 9-1 = 20-6 lbs. The load at joint A is due to a strip of roof 5 feet wide as indicated by the arrows in Fig. 348, and 10 feet long, but as this load is supported directly by the abutment, it does not cause any stresses in the roof, and need not therefore be considered. LOADS ON ROOFS 205 The load on joint D is due to a strip of roof 5 + 5 = 10 feet wide by 10 feet long. Hence load at D = 10 X 10 X 20-6 = 2060 lbs., = 0-92 ton. The load at C is due to two strips, each 5 feet x 10 feet, so that the load at C is the same as at D. Clearly the load at E is also the same. Load due to Wind. — Next, to find the load due to the wind. Supposing that the wind is blowing from left to right, then the left side only of the roof will have any wind pressure to bear. As already mentioned, the direction of the wind pressure is taken normal to the surface of the roof, and from Table XIV. we see that for a pitch of 30° the wind pressure to be allowed for is 33-0 lbs. per square foot. The load at joint A is due, as before, to a strip of roof 5 x 10, so that the load equals 5 x 10x 33 = 1650 lbs., = 0-73 ton. The load at joint D is 10 X 10 X 33 = 3300 lbs., = 1-46 ton. And the load at joint C is 5 X 10 X 33 = 1650 lbs., = 0-73 ton. The load on any roof can be found in a similar manner. Reactions at the Abutments. As regards the permanent load and the snow, the reactions at the abutments can be found as explained at p. 23 (Example 3). The reactions due to the wind pressure cannot, however, be found in so simple a manner. Let Fig. 350 represent a roof (the arrangementof the bracing, ^y, being immaterial, is omitted), il* then, regarding the roof as a I whole, the total wind pressure Pj, can be considered as concentrated at D, the centre point of the side of the roof. Further, P^, can be replaced by two forces — one a vertical force, P^, and the other a horizontal force, Pg. The reaction at each abutment can also be replaced by two forces, one vertical and the other horizontal, as shown in the figure. and Vg can be found at once, as being the reactions due to a vertical load P^, and thus (Eule I., p. 18) V = 4^ X P = ^P 2o6 NOTES ON BUILDING CONSTRUCTION .'A IS V — 4 V P — IP B — s~ ~ As regards and H^, however, all we can say is, that H, + H3 = P„ . . . (91), but we cannot, from the conditions of equilibrium, determine the value of or H^. This will be easily understood by considering that, if the support B were quite smooth, and therefore unable to offer any horizontal resistance, the support A would have to furnish all the horizontal reaction. "We must therefore endeavour to find some other condition to determine the relative values of and H3, or else make some assumption. ISTow in the case of wooden roofs a fair assumption to make would be, that ^ T.. ^ each abutment will iust afford suffi- Jbig. 351. • . 1 • 1 . cient horizontal reaction to make the total reaction parallel to the normal wind pressure, as is shown in Fig. 351. In this case EB(m + = P^ X m, Eb = -^P. . . (92), m + n and K = -^'P. . • (93). And m and n can be found by measurement. In iron roofs, however, the case is rather different. To pro- vide for the expansion and contraction due to variations of temperature, one end of the roof is fixed and the other end left free to move ; and so to enable this end to move freely it is usual in large roofs to place steel roUers under the shoe. In small roofs, where the shoe at the free end simply rests on a stone template, the horizontal resistance of the free end cannot exceed the frictional resistance. Now the coefficient of friction for iron upon stone can be taken at about 0--i5, and to find the resistance to friction we have only to multiply the vertical pressure by this number, so that, supposing B to be the free end, we have, if Vg is the vertical reaction due to the wind pressure, roof covering, roof truss, and snow, H3 = 0-45V3. . . . (94), REACTIONS DUE TO WIND PRESSURE ON ROOFS 207 and this is the greatest value H3 can have, and = - O^SV^, so that, knowing H^, we can easily find H^. The assumption made in the case of wooden roofs is therefore only tenahle in the case of small iron roofs, so long as it does not require a greater value of than the above. Now, for ordinary pitches, the assumption that the reactions are parallel to the wind pressure holds good, and equations 92 and 93 are applicable, but for high pitches it does not hold good. Let us inqiTire at what pitch the change tales place. Fig. 352 is the outline of a roof, the reac- tions at the abutments being parallel to P,,. It will be seen that the inclination of I^B to Vb is a, the angle of pitch. Hence Bb cos a = Vb, E, sin a = H„. A fixed end Fis. 352. Free end Fig. 353. also Eesults which are also evident from the triangle of forces, Fig. 353. But by taking moments about A, |P^ = 2cos2axEB. Hence H =■ sin a and V = 4 cos a 4 cos ^a' Now if L denote the permanent load, _ P, L Y = — - — h- ^ 4 cos a 2 Therefore the maximum value of Hb, from equation 94, = 0-45 —^ + - V4 cos a 2 and the value of a required can be obtained from the equation P^sina ^ ^ / P L -r — ^ = 0-45 +-Z 4 cos \4 cos a 2 It will be seen that a depends on L, and it will be found on trial that for very light roof coverings (such as zinc) a = 36°, and for heavy roof cover- ings, such as in Example 41, a = 45°. When an iron roof expands by an increase of temperature, the abutments will resist the expansion by exerting horizontal forces and h^, which will be equal , ^ and opposite to each other ; but ^>v!l as soon as these forces exceed the frictional resistance of the -p. y- -^/^ free end, that end will begin to ^dedend ^'ig- ^54. r^eeend move. Now supposing that a strong wind is blowing from right to left, whilst the roof is expanding, the wind causes a horizontal reaction TL'^, and the increase of temperature a horizontal reaction 2o8 NOTES ON BUILDING CONSTRUCTION in the opposite direction, as shown in Fig. 354. Supposing, as may well happen, that then clearly there will be no horizontal reaction at the abutment B, or, in other words, abutment A will have to supply the whole of the horizontal reaction, that is h; + a, = p,. A similar case may occur when the wind is blowing from left to right, and the roof is at P»^JJ^ ^ the same time contracting ; f then, as in Fig. 355, if H, + A, = P^, and the horizontal reaction at the free end will be zero. Cases to be Considered. Eeviewing what has been said with regard to small iron roofs, supported on substantial abutments and with the free end not resting on rollers, it will be seen that there are four cases to be considered, as follows : — The free end being on the right. ■ Eeactions parallel to normal wind press- ure for pitches from 36° to 45°, ac- cording to the roof covering, or hori- Case 2. Wind on right j zontal reaction at free end equal to frictional resistance for pitches above 45°. Case 3. Wind on left 1 tt • ^ Case 4. Wind on right | horizontal reaction at free end zero. It will be necessary to find the stresses in each case, and then pick out the maximum stresses to which the various bars may be subjected. On the other hand, in large iron roofs, where the free end of the trusses are supported on rollers, or, again, in small iron roofs having one end supported by a column, which can offer little or no horizontal resistance, the horizontal reaction at the free end can be taken as zero. In such roofs, therefore, only two cases need be considered, namely — Case 1. Wind on left 1 tt • ^ i o TTtr- J • 1 i. r -Horizontal reaction at iree end zero. Case 2. Wind on right j Case 1. Wind on left EXAMPLE OF IRON ROOF 209 We have now shown how to find the loads and reactions acting on roof trusses, and the next step is to show how the stresses can be obtained. This we proceed to do by means of a numerical example. Stresses in an Iron Boof Truss. Methods of finding the Stresses.— The stresses can be found either by the method of sections or by the graphic method (Clerk-Maxwell diagrams) If the stresses m all the bars are required, then the latter method will be the better one to use ; but if it is only required to find the stresses in one or two of the bars, then the method of sections will give the result, as a rule more quickly. As an illustration, however, the stresses will be found by both methods, by the graphic method below, and by the method of sections m Appendix XVIII. Example 42.— Find the stresses in the iron roof truss shown in Fig 373, P- 215, and 356,1 Plate III, assuming the following data :— Span 40 feet. Distance apart of trusses 8 feet. Purlins placed over joints of truss. Eoof covering — Countess slates on |" boarding. Pitch 21|° or i. span. Rise of tie-rod -^^ of span. Stresses due to Permanent Load and Snow. Preliminaries.— To find the permanent load and snow on the roof we refer to Table XIII., whence Per square foot of roof surface. Eoof covering and boarding . .10-5 lbs. Per square foot of covered area. Common rafters and purlins . .2-0 lbs. Principal . . . . _ .3-5 Snow 5-0 10-5 lbs. 1 1^'^ ^ r08 ^^^^^y P^^ square foot of roof surface. Hence total load per square foot of roof surface = 10-5 + 9-7 = 20-2 lbs. Now the length of principal rafter is V202-}-(i/)2 = 21-6 feet. Hence the loads are (see p. 203) at joint D 21-6 -—X 8x20-2 = l745 lbs., = 0-78 ton. And joints C and E will be loaded to the same amount. The reactions at A and B will be each equal to I X 0-78 = 1-17 ton. 1 Figs. 356-359. See Plate III. at end of volume. B.C. IV. 2IO NOTES ON BUILDING CONSTRUCTION The external forces due to the permanent load and the snow are therefore as drawn in black in Fig. 356, Plate III. Maxwell's Diagrams. — Commencing at joint A (Fig. 356), Plate III., by drawing ah parallel to AC, he to AF, etc., we get the triangle ahc (Fig. 357), which, measured on the scale of forces, gives the stresses in AD and AF, and shows that AD is in compression and AF in tension. We next proceed to joint D : there three bars meet, and there is also one external force, hence we obtain the quadrilateral ahdea, and we thus find the stresses in DC and DF. Next pro- ceeding to joint F we get the quadrilateral chdfc ; as this quadrilateral is of peculiar shape it is shown separately in Fig. 3 $8, and from it we find that both FC and FG are in tension. The student is strongly recommended thus to draw separately the diagram for each joint until he has made himself completely master of the subject. We have thus practically found the stresses in the left half of the truss, and the stresses in the right half can be deduced at once by symmetry ; for clearly the stress in CE will be the same as that in DC, and so on. The diagram has, however, been completed for the whole truss. At joint C we get the five-sided figure eghfd (Fig. 359). At joint E the quadrilateral ghlk (Fig. 357). At joint G the quadrilateral fhlc. To draw this quadrilateral all we have to do is to draw from Z a line parallel to GB, the other sides being already determined. This line ought, if the diagram has been correctly drawn, to pass through c, or, as it is expressed, the diagram ought to " close." We thus have a very useful and simple check on the accuracy of the work, for if any mistake has been made the diagram will not close. Finally the triangle ckl gives the stresses at joint B, and the numerical values of the stresses can be obtained by measuring the lines representing them on the scale given (| inch = 1 ton), the values being then filled in to column 2, Table H., p. 213. Stresses due to Wind Pressure. 1 We now proceed to find the stresses ''^K \Pc due to the wind. On referring to Table XIV. we find that, for a roof of \ p ^^sT^"'^ -^ 2lf° pitch, the wind pressure normal G^^^=~^**jB to the roof should be taken at 25*2 -11-5 -X- , -^^^^ g^y. 25 ii3g_ per square foot. Fig. 360. Following the method explained a,t p. 205 we then obtain the load caused by the wind on the joints D and C, as shown in Fig. 360, namely— P _!l:^x 8x 25 lbs., = 0-96 ton. P, = iP, = 0-48 ton. P^=0-48 ton, and Pj,, the total wind pressure, is P^=2P„=l-92 ton. We will suppose that A is the fixed end of the truss, and that the end B is free to move, but not placed on rollers. Case I (see p. 208). Wind on left. Reactions parallel to wind pressure.'^ — 1 Since the pitch is less than 45° (see p. 207). EXAMPLE OF IRON ROOF 211 Eeferring to Equations 92 and 93, p. 206, and to Figs. 360 and 361 we see tliat 1 ' and ^ 40 ^ 28-4 X 1-92 = 1-36 ton, Rb = -^x 1-92 = 0-56 ton. The external forces due to the wind from the left are therefore as shown in Fig. 361.2 ^ Maxwell's Diagrams. — Commencing at joint A we find two external forces, and ; as, however, these two forces are acting in exactly opposite directions, we need only plot their difference, namely — 1-36 - 0-48 = 0-88 ton. "We thus obtain the triangle ahc, which gives us and S^^ (Fig. 362). Proceeding now to joint D we get the quadrilateral cade, in which ad represents P^ and is drawn to scale equal to 0-96 ton. For joint F we obtain the quadrilateral hcef. This quadrilateral is also shown separately in Fig. 363. Next taking joint C we get the five-sided ^guTe fedgh (see also Fig. 364). Then proceeding to joint E we get the straight line gh as the diagram. This shows that S^^ = and = 0. For joint G we obtain the triangle fhh, and not a quadrilateral, since = 0- This triangle will be a check on the accuracy of the drawing, for the diagram ought to close at h. Lastly for joint B we get the triangle hhg, and we have thus obtained the stresses in all the bars. They can be measured to scale on the diagram and are then inserted in Table H., p. 213. Case 2, Wind on right. Reactions 'parallel to normal wind pressure. — A moment's considerati6n of Fig. 365 will show that this case is exactly the Fig. 365 opposite of Case 1, and we can immediately write down the values of the various external forces, as has been done in the figure, and it is further manifestly unnecessary to go through the calculations for the stresses, as we can deduce them from those obtained under Case 1. Thus the stress in CE in the present case will be equal to the stress in DC in Case 1, and so on. Column 4 has thus been filled in in Table H. Case 3 (p. 208). Wind on left. Reaction at free end vertical. — In this case _28-4 m + 7i ~ 40 ■ 2 Yigs. 361-364. See Plate IV. at end of volume. 212 NOTES ON BUILDING CONSTRUCTION (Fig. 366) it is a little more difficult to find the values of and R^, because the external forces acting on the roof are not parallel to each other, and we ft^O-52 Fig. 366. know neither the direction nor the magnitude of R^. Now when three forces acting on a body are in equilibrium, their directions meet in a point,^ so that, as we know the direction of and E3, we can find the direction of by joining A with the point of intersection of Pj, and R^. But with flat roofs, as in the roof under consideration, this intersection is a long way below the roof, and the intersection is acute, and therefore cannot be obtained with great accuracy. In such a case, therefore, a better way is to find the value of R^, which can be done by taking moments about A. We obtain + 10-8xP^-40xR3=0, 10-8 X 1-92 ^3 = - ^ = 0-52 ton. Now draw ah (Fig. 367) parallel to P^ and measure along it 1'92 ton to scale ; next be parallel to R^, and measure along it 0-52 ton a to scale ; then ac will be the direction of R, and we also find by measuring ac that R^=l-45 ton. We can therefore now proceed to find the stresses. Maxwell's Diagrams. — Commencing at joint A (Fig. 368)2 -we find two exter- ^,\>.K nal forces, P^^^ and R^. Let ad parallel to P^ and ac parallel to R^ represent these forces (Fig. 369), then adec is the complete diagram for joint A. It will be found that the remainder of the diagram for the roof can be drawn as in Case 2 ; the description is not therefore repeated, but the diagram is given in Fig. 369. Case 4. Wind on right. Reaction at free end vertical. — In this case we can obtain the reactions very simply by dia- gram. Thus produce Pj, and R3 to meet at K (Fig. 370) and join AK; then AK is the direction of R^ (see Fig. 367). Now, plotting KA to scale to represent Fig. 370. ^ This can be easily proved by taking moments about the intersection of two of the forces. The moments of these two forces will be nothing, but for equilibrium the sum of the moments of three forces is nothing, therefore cleai-ly the moments of the third force about this point must be nothing ; that is, its direction must pass through the point, or, in other words, the three forces meet in a point. 2 Figs. 368, 369. See Plate V. at end of volume. EXAMPLE OF IRON ROOF 213 (1-92 ton), and drawing ah parallel to E^, we obtain the triangle of forces for the three forces P„, E^, and E.„ and by measurement we find E^ =0-87 ton, E3 = 1-26 ton. As a check, taking moments about A, we have -4OXE3 +26-3 X 1-92 = 0, E3 = 1-26 ton. Maxwell's Diagrams. — Fig. 37 ji shows the loads for this case, and the corre- sponding diagram is given in Fig. 372, and the stresses obtained therefrom are entered in Table H. Tabulation of Stresses. We have thus obtained the stresses in the various members of the roof under various conditions. These stresses must now be tabulated, so that we may be able to pick out the maximum stress to which each member of the roof truss may be subjected. This has been done in Table H below, and the stresses^ that are to be added together, to obtain the maximum stress, are printed in italics. The seventh column gives the maximum stresses, and it will be seen that the maximum stress to be borne by corresponding members on either side of the roof is not quite the same. This is due to the fact that one end of the roof truss is fixed and the other end is free to move. But as these differences are very small, and it would be practically unwise, at any rate in so small a roof truss, to make corresponding members on either side of different scantlings, these differences will be ignored and the scantlings will be calculated to meet the stresses given in column 8. It should be noticed also that the stress in the principal rafter is greater in the lower part than in the upper part ; but, in a small roof truss like the one under consideration, no difference would practically be made in the scantling of the principal from A to C. In column 8, therefore, the stress in the upper part of the principal rafter has been put equal to that in the lower part. TABLE H. Compressions — Tensions + ? Stresses. 2. Perma- nent load and snow. Wind. 7. Maximum stresses. 8. Stresses to be used in calcu- lating scant- lings. Reactions parallel Reaction free end vertical 3. On left. 4. On right. 5. On left. 6. On right. -3-27 -1-89 -3-38 -1-62 -7-83 -7 -83 + 3-38 + 1-57 + 3-67 + 0-83 + 7-83 -f7-83 -0-73 -0-96 0-00 -0-96 0-00 -1-69 -1-69 -If- 16 -3-28 -1-89 - 3-33 -1-62 -7-49 -7-83 + 1-93 + 2-17 + 0-32 + 2-20 + 0-16 + 4-13 + 4-13 + 2-34 + 1-30 + 1-31 + 1-54 + 0-69 + 3-88 -1-3 -88 + 1-93 + 0-32 ^2-17 + 0-38 + 2-02 + 4-10 -f 4-13 -4-16 -1-89 -3-28 -1'98 -3-0 -7-44 -7-83 -0-73 0-00 -0-96 0-00 -0'96 -1-69 -1-69 -V4-16 + 1-57 + 3-38 + 1-88 + 2-70 + 7-54 + 7-83 -4-45 -1-90 -3-27 -1-98 -3-00 -7-72 -7-83 ^ Figs. 371, 372. See Plate V. at end of volume. 2 See p. 185. Many writers take compressions as -|- and tensions as - NOTES ON BUILDING CONSTRUCTION Calculation of the Scantlings. Having obtained the stresses in the various members of the roof truss the next step is to decide on the form, and to calculate the scantlings of the vari- ous members, so that they may be able to safely resist these stresses. Suitable forms, such as are used in practice, have been described in Parts I. and II., and the student is therefore referred back for information on that part of the subject. As regards the calculations of the different members and joints, several examples have already been worked out in Chap. VI. of this Part, and the student is specially referred to Examples 13, 19, 20, 26, 26a, 27, and 28. The roof truss can now be considered as completely designed, and the result is given in Fig. 373. The same truss is shown in Fig. 374, when the purlins are distributed along the length of the principal rafter, in which case the principal rafter is increased in section but no alteration is made in the other members of the truss. The calculations to find the size of the principal rafter in the case where a number of small purlins are distributed along its LENGTH would be as follows : — We must first find the load on the principal. The permanent load and snow consists of Per square foot of roof surface. Koof covering and boarding , . .10*5 Per square foot covered. Purlins 1"5 Snow 5-0 Or 6"5 — 6 lbs. per square foot of roof surface. Total=10-5 + 6 = 16-5 lbs. This load is vertical and must be resolved perpendicularly to the principal by multiplying by cos 2lf° thus — 16-5 X cos 2lf° = 15 lbs. The normal wind pressure is 25 lbs. per square foot (see Table XIV.), so that the total load per square foot is 15 + 25 = 40 lbs. Hence total load on one half of the principal rafter is Distance Length, apart of trusses. 40 X 10-8 X 8 = 3456 lbs., = 1-54 ton. This load produces a bending stress in the principal, the compressive element of which must be added to the direct compression. The moment of flexure is wV^ ^ 1-54 x 10-8 x 12 ~8"~"8"~ 8 ' = 25 inch-tons, whether we consider the principal as jointed at the strut, in which case the lower half of it is a beam supported at both ends, or whether we suppose it to be continuous over the strut, when the lower half is a beam supported at one end and fixed at the other. 2 1 6 NO TES ON B UILDING CONS TR UCTION We must now assume some section of T iron, and see if it is strong enough. Assume, for instance, a section 4" x 5|-" x y, the table being 4" wide. According to the rule given in Appendix IV. the distance of tbe centre of gravity from the edge of the table is - 4x1 + 5 X 3 , „ . , X = f — =1-8 inch. 4 + 5 And according to Appendix XIV. the moment of inertia about an axis passing through the centre of gravity is I - H 4 (1 -83 - 1 -33) + 1 (1 -33 + 3-73)}, = 13-65. Treating 'principal as jointed. — Hence the greatest compression due to bending (at the edge of the table), on the supposition that the lower half of the principal is supported at both ends. -r. ^Vo 25x1-8 . , Irom Equation 53 r„ = — = i^.q^ = tons per square inch. The direct compression produces a stress per square inch (see note below) 7-83 7-83 = ^=(ST5)l=i'^^°^^^^^^^- . Hence total maximum compression = 3'3 + 1*7 = 5 tons per square inch. Practically this section would do. Treating principal as continued. — If, however, the principal is considered as continuous over the strut, and is supposed to be supported at one end and fixed at the other, the maximum compression due to bending at the edge of the stem becomes 25 X (5-5 -1-8) So that on this supposition the section is too small. Next, try a section 3" x 5^", the table being i" thick and the stem It will be found that the distance of the CG from the edge of the stem is 3""92, and that 1= 15-9. Hence maximum compression (at edge of stem) 25 x 3-29 / = =5'l7tons. 15-9 And the total compression is 5-17 + 1-49 = 6-66 tons. This section is therefore apparently too small, on the supposition that the principal is continuous over the strut. This supposition, however, re- quires that the three points A, D, and C D^'^^^^^^^"^^ ""^^^t"*^ shall remain in a straight line, but the maximum compression occurs in Fig. 375. t]je principal AC when the wind is blowing from the left, and this wind pressure bends the roof as shown in Fig. 375 in a very exaggerated manner, and it is evident that this bending of the principal will considerably relieve the compressive stress on EXAMPLE OF IRON ROOF 217 the stem of the T iron, so that it can be taken that this section is quite strong enough. iyofe.— It has been assumed that the direct compression is uniformly distributed over the cross section of the T iron. The detiection due to the distributed load tends, however, to increase the intensity of the stress towards the edge of the table. The increase is, however small If for instance, the deflection were i", the moment of flexure due to the direct compression would be 3-2x^8 ^ ^^^^'^'^ ^^'^^'^ "^"^"^ "^'^^ ^ ''^^^^ 4"x5|"xi" a maximum compression of ton per square inch, to be added to the I'T ton already found. Diagrams for various Roof Trusses. In Figs. 376-395 ^ diagrams for different kinds of roof trusses are given, with the hope that they may be of assist- ance to the student. It is strongly recommended that, as an exercise, some span be as- sumed, and that the stresses be obtained by following these diagrams ; the results obtained being occasionally checked by the method of moments. In these diagrams only one case for the wind pressure has been shown, since the other cases pro- duce very similar diagrams. It shouki be observed that the diagram for the roof truss shown in Pig. 392 cannot be drawn by the graphic method alone, because either at the points E or M the stresses in three of the bars are unknown. But the stress in MN can easily be found by taking a section as shown in Fig. 396 for permanent load and snow, and in Fig. 397 for wind. The value thus found for S diagram completed. WOODEN EOOFS. It is not worth while to give examples of the calculations for timber roofs. These can very seldom he necessary, and, if required, ivould he similar to those for iron roofs. Tables of Scantlings for Wooden and Iron Roofs are given in Tahle XV. p. 340, and in Parts L and II. 1 See Plates VI., VII., VIII., IX., and X., at end of volume. 2 For Figs. 398-430, see Appendix XVIII. Fig. 397.2 can then be inserted, and the Chapter XIII. STABILITY OF BRICKWORK AND MASONRY STRUCTURES. TT is proposed in the following chapter to show, in an elementary manner, how the dimensions for various kinds of brickwork and masonry structures can be calculated. The mortar with which the stones or bricks of a wall are bedded assists the wall by its resistance to tension ; but, on the other hand, the resistance to compression of mortar being less than that of stone or brick, it reduces the resistance to crushing of the wall. On the whole, the effect of the mortar is to increase the strength of the wall as regards overturning. In important structures, however, the tenacity of the mortar is not taken into account, because settlements are liable to occur which may dislocate the joints, and such walls are treated as if built up of uncemented blocks. We will first consider the following problem. Single Block. — A rectangular block is supported by a flat surface, and is pressed against it by an inclined force F (Fig. 431). Find in wbat manner tbe pressure between the block and the flat surface is distributed. It will be observed on looking at the plan that, for simplicity, the force F is assumed to act in the central vertical plane of the block, and is supposed to include the weight of the block. The point C, where the direction of the force intersects the supporting surface, is Fig. 431. called the centre of pressure. Now F may be resolved into two components — one perpendicular to the supporting surface or joint, and the other parallel to it. Y and X are these components. The component X tends to make the block slide along the surface and is resisted by the friction ; but at present we are not concerned with this part of the question. A K R G c. H " F E D Y A STABILITY OF BRICKWORK AND MASONRY The normal component Y is the one that causes the pressures between the block and the surface. Since C is not in the centre of GH it is clear that these pressures will not be uniformly distributed, but will be greater towards AE than towards BD. In the first place, since GH is midway between AB and ED, and C lies on GH, the pressure at all points of any line KL drawn parallel to AE will be the same, and therefore we can take the pressures along GH to represent the pressures over the whole surface of the joint. Secondly, on the supposition that the joint is a perfect one (and in a masonry structure the principal function of the mortar is to make the joint prac- tically perfect), we can say that the pressure along GH diminishes xmiformly from G towards H. The pressures along GH may therefore be represented by a triangle GMO as shown in Fig. 432, so that the pressure p at any point V is given by the ordinate VN, and this or- dinate will also give the pressure at any point on KVL (Fig. 431). First. Let 0 be situated between G and H ; this means that there is no pressure on the joint from 0 to H. Now the system of parallel forces acting over the whole joint and repre- sented by the triangle GMO have a resultant K acting through the centre of gravity of the triangle, and for equilibrium R must be equal and opposite to Y, that is R = Y. Again, R is represented by the area of the triangle GMO, and is equal to the area x AE. That is ^xGOxAE = Y. 2 Fiff. 432. R: Therefore MG = 2Y GO X AE But since Y passes through the centre of gravity of the triangle GMO we have GO = 3GC, and for MG we can write X'{mas. > 2Y Hence (95), SGCxAE • and since Y, GO, and AE are known, f^-oisx.) can easily be found. Example 43. — As a numerical example let Y = 4 tons, GO = 0-5 foot, AE = 1 foot, 2 4 then = = 5*3 tons per square foot. Second. Let O and H coincide, as in Fig. 433. In this case we have GC = i-GH, and this is the condition which must exist in order that the pressure may just vanish at H, or more strictly along DB. NOTES ON BUILDING CONSTRUCTION Let GH = 3 feet, and the remaining data be as before, then Fig- 433. Now if 0 were situated at the middle point of GH, the pressure would clearly be uniformly distributed over the joint, and its intensity would be Y , ^ GH X AE from which we see that when the centre of pressure is so situated that the pressure vanishes along one edge of the joint, then the pressure along the opposite edge of the joint will be double what it woiild have been had the centre of pressure been at the middle point. This is an important result, and should be carefully remembered. ^ Thus far we have not considered the horizontal component X of the force F. This component is balanced by the resistance to friction of the joint so long as slipping does not occur. Now the resistance to friction is where fx is called the coefficient of friction, and is simply a number (always less than 1) depending on the nature of the surfaces. Thus for stone against stone If, however, in a masonry or brickwork joint the strength of the mortar is taken into account, the resistance opposed to X will be the resistance to shearing of the joint. In practice, however, it is not usual to take account of the strength of the mortar. It will be observed that the horizontal compon- ent X and its effect on the joint are independent of the position of the centre of pressure. Uncemented Block Structures. — If we consider any masonry or brickwork structure, such as a retaining wall for earth or water, a boundary wall or an arch, we see that such a structure can be dissected as it were into a series of blocks separated by joints, such as the one we have just been considering. There will be a resultant external force acting at each joint through the centre of pressure of the joint, and this force will be resisted by the com- pressions (and tensions in some cases) acting over the surface of the joint. ^ See Appendix XIX. STABILITY OF BRICKWORK AND MASONRY 221 Bef. — The line joining the centres of pressure is called the " line of pressures," or " Moseley's line of resistance." Thus in Fig. 434, representing a portion of an arch, the broken line dbcdef is Moseley's line of resistance; a, h, c, d, e, and / being the centres of press- ure at the respective joints. Conditions of Stability. I. For important struc- tures, or for structures the failure of which might lead to serious consequences, such as loss of life : {a) No portion of any joint should be brought into tension. Eef erring to Fig. 433, p. 220, we see that, to conform to this condition, the centre of pressure must not be nearer the edge of any joint than one-third the width of the joint (GH); or, in other words, Moseley's line of resistance must lie within the middle third of the structure.-^ (6) The intensity of pressure should nowhere exceed what the material is capable of safely resisting. That is,p(^ax.) must not exceed the safe resistance to crushing or cracking of the material and mortar (see Table Ia). (c) The resistance to sliding must be greater than the horizontal component of the external force at each joint.^ II. For unimportant structures (such as boundary walls) : {d) The intensity of pressure should nowhere exceed what the material or mortar is capable of safely resisting. (e) The intensity of tension should nowhere exceed what the mortar can resist. (/) The safe resistance to shearing of any joint must be greater than the horizontal component of the resultant external force acting at the joint. These conditions will now be applied to some numerical examples. The middle half is sometimes taken for unimportant structures, as in Example 51, p. 254. The middle third should be taken for all important arches, as in the Example, Appendix XX. 2 See p. 248. 222 NOTES ON BUILDING CONSTRUCTION Brick Pier. Example 44. — Find whether a brick pier 14 inches square and 8 feet high can resist a thrust of 3 cwts. inclined at an angle of 55° and applied at the top of the pier, as shown in Fig. 435. If the pier is not p strong enough, find what additional thickness of brickwork sfnvts. is required, and how high it ought to be carried up. The \ tenacity of the mortar not to be taken into account. Preliminaries. — We will assume that one cubic foot of brickwork weighs 1 cwt. The total weight of the pier will therefore be „ 14x14 11 cwts. nearly, and acts vertically through the centre of gravity of the pier, which is easily found. The direction of the weight W intersects that of the force P in the point A, and these are the only two external forces acting on the pier. Their resultant can be found in the usual way by means of the triangle A5c. To find the centre of pressure, produce A6 to intersect the plane of the base of the pier. It will be observed that this point falls outside the actual base of the pier, and, since the tenacity of the mortar is not to be taken into account, it follows that the pier is not strong enough to withstand the pressure of 3 cwts.^ It will therefore be necessary to increase the thickness of the pier. Suppose the thickness is increased to 18 inches, which would appear to be about what is required, as the centre of pressure will be brought nearer to the wall owing to the additional weight. Fig. 436. Fiff. 437. ^ The student will also observe what a very small thrust is sufficient to overturn a pier of the given dimensions. STABILITY OF BRICKWORK AND MASONRY 223 then the first question is how far up is it necessary to make the pier this extra thickness. Joint KL. — As a trial, suppose the increased thickness commences 3 feet from the top of the pier, then it must be ascertained whether the joint KL at the point of change is safe (Fig, 436). The external forces acting on the upper three feet are the thrust of 3 cwts., and the weight of the portion of the pier EL acting through its centre of gravity. Determining the resultant of these two forces in the usual way, it will be seen that its direction passes very near to K, so that the portion EL of the pier would be on the point of overturning. Next try if increasing the thickness of the pier 2' 6" from the top will do. Carrying out the same construction as before, we find that the centre of pressure now lies 0-9 inch within the edge K^. Crushing. — We must now find the maximum intensity of compression, that is,P(max.) The vertical component (Y) of the total pressure on the joint K^L^ can be found by dropping a perpendicular M on to ScZ the direction of W (Fig. 437), and it is thus found by measurement that Y = 5-85 cwts. Hence applying Equation 95 we have 5-85 P(max.) = 2 X 3^0-9^ ^4 = cwt. per square inch. And since i5,max.) can safely be as much as 0-5 cwt. per square inch (Table Ia.) there is no fear of crushing, and condition (&) is therefore fulfilled. Sliding. — As regards condition (c), the force tending to cause sliding is the horizontal component of the thrust of 3 cwts. By measurement as above, it is found that X= 1-7 cwt. The resistance to sliding is found by multiplying the vertical component (Y) of the pressure on the joint by the coeflicient of resistance ^ to sliding, which can be taken at 0-47. Hence the resistance to sliding is 0-47 X 5-85 = 2-7 cwts., jcrictuts. Fig. 438. Fig. 439, Fig. 440. ^ 0-47 is the co-efficient of friction for masonry and brickwork with wet mortar (see Table XVa.) NOTES ON BUILDING CONSTRUCTION so that tlie joint is safe against sliding.^ Finally, therefore, it is safe to commence thickening the pier at 2' 6" from the top. Joint GH. — It remains to be seen whether the conditions (b) and (c) are fulfilled at joint G^H, Fig. 438. We have now to consider the portion L^G-^ of the pier. The external forces acting on this portion of the pier are the resultant pressure acting on joint K^L-^, the value of which is found by- measurement from Fig. 437 to be 6"1 cwts., and the weight of the portion L^G-^ of the pier, namely 9*6 cwts., acting through the centre of gravity of that portion, as shown in Fig. '438. The centre of pressure of joint G-^H is found to be quite close to G^^, so that the pier is still not thick enough at the bottom. We have therefore to find how far below the joint K-^L^ the thickness of 18 inches can be carried. Assume that it is carried down 4' 6" below K^L^ to joint MN (Fig. 439), then, proceeding as before, the centre of pressure of joint MN is found to be 0'3 inches within the edge M, and from Equation 95, p. 219, 13-7 i'(max.) = 2X 3 ^ 0.3 ^ 14. = 2'18 cwt. per square inch. Orushing. — The maximum intensity of compression is therefore more than the safe intensity, and consequently MN must be taken nearer to K-^L^, say at 3' 6" below K^L^i (Fig. 440). We then find 11-9 ^'dnax.l-S X 3 ^ Q.g ^ = 0-63 cwts. per square inch. M,N,, 'U-U cwts. Fig. 441. Fig. 442. 1 According to Kanldne, there is no danger of sliding if the angle S (Fig. 437) between the normal to the bed joint and the direction of the resultant pressure is less than from 25° to 36° the angle of repose of fresh masonry. In this case, the angle at S is only 16°. CHIMNEY 225 Below K^L^, therefore, the pier must again be thickened to the next brick dimensions, that is to 23 inches. Lowest portion of pier. — We now have to consider the portion of the pier shown in Fig. 441. The external forces acting on this portion are its weight (4-4 cwts.) and the resultant pressure of 12-1 cwts. acting at Cg as found by measurement from Fig. 441. Crushing and sliding. — It is then found that the centre of pressure is Co, which is 3*5 inches from the edge, and the vertical component of the resultant" pressure on the joint is also found to be 16-4 cwts., so that 16-4 P(m^^.) = 2 X 3 X 3-5 X 14' = 0-22 cwt. Condition (b) is therefore fulfilled. As regards condition (c), the force tending to cause sliding is found by measurement to be 1-7 cwt. ; and since the vertical component of the pressure is 16-4 cwts. the resistance to sliding is 0-7 X 16-4 = 11-5 cwts. The pier has now been designed, and is shown in Fig. 442. once most Chimney. Example 45. — Find what wind pressure the chimney shown in Fig. 443 can safely resist. It will be seen at that AB is the joint likely to fail. Now the fall of a chimney may lead to serious conse- quences, and it is therefore well to lay down the rule that the centre of pressure shall be kept so much inside of one edge that no tension will be excited at the other edge. Refer- ring to Appendix XIX. we see that for a hollow square chim- ney, AC must not be less than ^ AB in order that the above condition maybe fulfilled ; and this value can be applied with sufficient accuracy to the case iinder consideration. But AB is 22'5", so that least value of AC = 3-75". It will be found that there are 8 7 cubic feet of brickwork above the joint AB. There- fore (allowing 1 cwt. per cubic foot for weight of brickwork) the weight of the chimney above the joint ABis 87 cwts. B.C. IV. Fiff. 443. 226 NOTES ON BUILDING CONSTRUCTION Let P be the normal wind pressure ; the total wind pressure on the side of the chimney will therefore be ll'3"x6xP = 67-5P cwts., and this force is applied at the centre of gravity of the side of the chimney. The part of the chimney above the joint AB is therefore under the action of three external forces — namely, 87 cwts. acting vertically through C.G., 67-5P cwts. acting horizontally at E, and R the reaction to the resultant pressure acting upwards at C. Taking moments about C we have /22-5 V 2 - 3-75 187 11-25 X 12 X 67-5P = 0, whence Pr= 0-1428 cwt.= 16 lbs. It therefore appears that this chimney does not fulfil the condition laid down, and although the tenacity of the mortar would materially increase the powers of resistance, this chimney cannot be considered as safe. Crushing. — The maximum intensity of compression with 16 lbs. wind press- ure should also be found. In this case the vertical component of the resultant pressure on the joint AB is equal to the weight of the chimney, namely, 87 cwts., and this pressure is distributed, but not uniformly, over an area of 9-9 square feet, or 1425 square inches. Following the rule given at p. 220 we have , = 2 X = 0-122 cwt. per square inch nearly, f'dnax.) — " '- -J 425 so that there is no fear of crushing Avith a wind pressure of 16 lbs. per square foot. Sliding— As regards sliding, the force tending to cause sliding is the wind pressure equal to 16 x 67-5 = 9-6 cwts.. 112 : 61 cwts. and the maximum resistance of the joint AB to sliding is 0-7 x 87 = so that there is no fear of sliding. Stay. — This chimney can, however, be made perfectly safe by fixing an iron stay as shown in Fig. 444. Let it be ascertained what the cross section of this bar ought to be. We have seen that the chimney itself can safely resist 16 lbs. wind pressure. Taking 50 lbs. as the maximum wind press- ure, it follows that the iron will have to resist 50-16 = 34 lbs. per square foot. So that the wind pressure the stay has to resist is 67-5 X 34 ^ ^ = 20-5 cwts. 112 take To find S, the stress in the stay, moments about A (Fig. 443) 7 x 8- 5-62 x 20-5 = 0, S= 16-5 cwts. = 0-82 ton. When the wind is blowing in the direction shown in Fig. 443, the stay has to act as a strut, and it must therefore be calculated as a long column ENCLOSURE WALLS 227 with ends fixed. By the help of Table V. it will be found that an L iron 2" X 2" X f " can safely bear 0-9 ton (ends rounded), and will therefore be strong enough either as strut or tie. ■■'/i/m////mm/"' Enclosure Wall. Example 46. — Find the wind pressure per square foot a long enclosure wall of the section and plan shown in Fig. 445 can safely resist. Ignoring tenacity of the mortar. — If the tenacity of the mortar is neglected, it can be shown that, with a wind pressure of 16 '2 lbs. per square foot, the centre of pressure would be at A, so that, to say nothing of crushing the bricks, the wall would be on the point of overturning. This investigation is, however, left as an exercise for the student. Long enclosure walls should therefore be built in good tenacious mortar, and it is well to build the lower courses in cement, especially if the situation is exposed. By a long enclosure wall is meant one whose unsupported length is more than about ten times its height. Shorter walls than these receive considerable support from each end. Considering tena- city of the mortar. — We will now suppose that this wall is built in a good hydraulic lime mortar, and that the adhesion of this mortar to the bricks can be safely taken at 0-06 cwt. per square inch (see Table Ia). The tenacity of the mortar itself is greater, but of course we can only reckon on the adhesion, as it is less. Now, looking at the plan, we see that we can take a portion of the wall from E to F as a unit. The weight of this portion of the wall (allowing, as before, 1 cwt. per cubic foot of brickwork) is found to be 59 cvvts. This weight is distributed uniformly over the area of the joint AB in section and portion EF on plan, which is found to be 1215 square inches. The pressure due to the weight of the wall is therefore 59 j^j-g = 0*05 cwt. per square inch nearly. Let P be the wind pressure per square foot, then the total wind pressure acting at the centre of gravity of the unit of the wall is 10' 6"x 6' 7-|"x P = 69-5P. The full height of the wall is not taken on account of the inclination of the coping. The wind pressure will clearly produce tension on the windward side, that is at B, and compression at A, and the intensities of these stresses can be —0 0 > Fig. 445. 228 NOTES ON BUILDING CONSTRUCTION found by considering the portion of the wall as a beam fixed at one end, namely, at the joint AB. Referring to p. 50 (footnote) we see that we must first find the moment of inertia of the section, and this we can do by adding together the moments of inertia of two rectangles thus — I = _i_ X 1 8" X (13-53) + 3-^9 X 1 2) X 9^, = 10251. The distance of the extreme fibre from the "mean" fibre is sensibly 13-5 . , = — — 6 ■ 2 5 inches. So that moment of resistance 10251 inch-cwts. But moment of flexure = 69-5P X Hence P = 6-25 (6' 7i") X 12 2 10251 X 2 6-25 X 69-5 X 79-5 " To find P we must therefore decide what value to give to r^. Now we have seen that the mfe adhesion is O'l cwt. per square inch ; but can be given a greater value than this, because the compression produced by the weight of wall (viz. 0'05 cwt. per square inch) relieves the tension. Hence we can take = 0-1 + 0-05, = 0-15 cwt. per square inch. Hence P = -089 cwt. = 10 lbs. Therefore even if the wall is built in good hydraulic lime it cannot mfehj resist more than 10 lbs. per square foot of wind pressure. It would of course take a greater wind pressure than this to cause the wall to fail. Thus, taking the ultimate adhesion of best hydraulic lime to hard stock bricks as 36 lbs. per square inch or 0"32 cwt., we can put r(, = 0-32 + 0-05, = 0-37 cwts. And 10251 X 2 X 0-37 "'6-25 X 69-5 X 79-5' = 0-22 cwt., = 24"5 lbs. per square foot. In an exposed situation, therefore, this wall would undoubtedly in course of time be blown down. In an unexposed situation, however, the wall miglit be safe enough, especially as the friction of the wind against the ground is known to lessen considerably its force near the surface. The wall can be materially strengthened by building the first ten or twelve courses in cement. In this case the safe adhesion can be taken at about 0'6 cwt. per square inch, and hence 10251 X 2 X 0-6 ~6-25 X 69-5 X 79-5' = 0-35 cwt., 39'5 lbs. per square foot, so that the wall is now strong enough. ENCLOSURE WALLS 229 It is clearly unnecessary in this case to find the maximum intensity of compression, neither need any calculations be made as regards the resistance to sliding, as the resistance we can now reckon on is not merely the friction of brick against brick, or stone against stone, but the shearing strength of the mortar, and this is much more than enough. Chapter XIV. RETAINING WALLS.^ A EETAHSTING- wall is a wall built for the purpose of " retain- ing " or holding up earth or water. In engineering practice such walls attain frequently enormous proportions, being used in the construction of railways, docks, water-works, etc. In building construction, however, their dimensions are never very large, and the difficulties attending their design and construction, which in engineering works are often very great, are therefore much reduced. Usual crqss Sections for Eetaining Walls. — The form of cross section given to such walls varies considerably according to circumstances, and often according to the fancy of the designer. A few of these forms, such as might be useful in building con- struction, are given in Figs. 446-456. Fig. 446 shows a simple rectangular wall. Such a section is suitable only for very low walls (say not exceeding 5 feet), as it is wasteful of material. Figs. 447 and 448 show a better section, in which the back of the wall is sloped. Except in the case of a wall for retaining water, the alterations in the thickness of the wall would be carried out by means of offsets, as in Fig. 448. 1 For Practical Formula see pp. 235, 239, 244, and App. XXL RETAINING WALLS 231 rig. 449 shows a wall with vertical back and with a " battered" face. In point of stability a wall with a battered face may or may not be, according to circumstances, more stable than a wall with a straight face and sloped back. It has the appearance of being more stable, and there is the advantage that a small altera- tion in the batter, caused by rotation or inclining forward, is not noticeable, whereas the eye is offended if a wall with a vertical face rotates ever so little. To prevent lodgment of water, how- ever, the batter should not exceed 1 In Figs. 450 and 451 walls with battered face and sloping Fig. 450. Fig. 451. backs are shown. It will also be noticed that the footings are inclined, thus giving greater resistance against sliding forward. The section shown in Fig. 450 is, however, clearly not suitable for retaining water. In the case of walls for retaining earth, counterforts are some- times added, as in Figs. 452 and 453, and Professor Eankine has shown that a slight saving in masonry is thereby effected. In the preceding cases it has been supposed that the top- surface of the earth to be retained is horizontal, and also level with the top of the wall. Frequently, however, a bank has to be sup- 232 NOTES ON BUILDING CONSTRUCTION ported, as sliown in Fig. 454 or 455. Such walls are called " surcharged " walls and clearly require greater strength. Fig. 452. Fig. 453. Another case in which greater strength is required is when a building is placed close to the wall as shown in Fig. 456, because Fig. 454. Fig. 456. Fig. 455. the weight of the building tends to overturn the wall by increasing the natural thrust of the earth. RETAINING WALLS FOE WATER. The calculations for finding the dimensions of walls for retain- ing water, so long as the depth of water to be retained is not great, are comparatively simple. Such a wall is acted on by two external forces, namely, the pressure of the water and the weight of the wall. As regards the pressure of the water, it is known from hydro- statics that it is : (1) Directly proportional to the depth below the surface. (2) Normal to the surface exposed to the pressure. Applying this to the case of a wall with a vertical back (Fig. 457) ; at the lowest point B the pressure will be proportional to the depth of water AB, and will be horizontal, since it is normal to the surface pressed. The pressure of the water at B RETAINING WALLS FOR WATER 233 can therefore be represented by BD, which is drawn horizontally equal to the depth of water at B. At A the pressure is clearly zero ; hence ordinates from AB to AD represent the pressures at all points along AB. Fig. 457. The area of the triangle ABD therefore represents the total water pressure on the wall (see p. 219), and the resultant pressure will pass through the centre of gravity of this triangle. The centre of pressure E is therefore two-thirds of the way down, that is AE = f AB . . . (96). And if we consider 1 foot of length of the wall, the resultant pressure (P) on this portion of the wall will be equal to the weight of a prism of water whose cross section is the triangle ABD and length 1 foot (Eig. 458). Fig. 458. Fig. 459. The volume of the prism is ABxBD , AB2 where H is the height of water above the point B. And, since a cubic foot of water weighs 0*557 cwt., we have NOTES ON BUILDING CONSTRUCTION P= 0-557 x-—cwts. . . (97). A We therefore know the magnitude, point of application, and direction of P, which is all we require to know. In the case of a wall with sloping back some modifica- tions must be introduced. Thus in Fig. 459 the pressure at B will be represented by BD drawn normally to AB and equal to H. AD will then represent the pressure along AB, so that the triangle ABD will represent the total pressure, and AE = f AB. Also w AB X H P = 0-557 X — 2 — • • P is therefore completely determined. As regards the second external force, namely, the weight of the wall, if we assume a trial section the weight can easily be obtained, and we can then ascertain whether this trial section is suitable by applying the rules given at p. 221. In the case of a wall for retaining water it is of the utmost importance that no cracks should be formed, and for this it is necessary that no tension shoiild be excited. Hence the centre of pressure must not approach the outer edge nearer than one- third the width of the joint. The various points that have been considered will now be illustrated by means of an example. Reservoir Wall. Example 47. — A wall to retain water to a depth of 6 feet is required ; the top of the wall is to be 1 foot above the surface of the water and the batter of the face i?-- The wall to be built in brickwork and cement. Preliminaries. — So far as stability is concerned, the top of the wall might be built to a feather edge, but if the wall were thus built the top would soon be in want of repair ; moreover, the saving in brickwork would be exceedingly small, and would not compensate for the additional difficulty in building. A width of 9" at the top, although somewhat narrow, will be assumed. Since the batter of the face of the wall is given, it remains only to find the batter of the back of the wall, or, what comes to the same thing, to find the width of the wall at the footings. Now it can be shown that if the wall were of triangular section, i.e. with a feather edge at the top, that all joints would be of equal strength, and the effect of giving a certain width to the top of the wall is to make the lowest joint the weakest ; we need therefore only consider this joint. The courses will be built at right angles to the face of the wall, so that the joints will have a slope of -^q. RETAINING WALLS FOR WATER 235 As a first trial, assume the thickness of the wall at the footings to be 4 feet. The weight of one foot of length of the wall will be 0-75 + 4 ^ — X 6-6 X 1 cwt. = 15-6 cwts., and this force will be applied at the centre of gravity marked C.G. in Fig. 460, which can be found graphically by a method explained in Molesworth's Pochet-Book} The water pressure is equal to 6 4 6 0-557 X ■ — ^ — , = 10-7 cwts. It is perpendicular to AB, and is applied at the point E. ii-S cwts. Fig. 460. Fig. 461. The resultant of these two forces intersects the joint DB in the point C, DB and as DC is 1-05 foot and is less than — the wall must be made thicker. Next try 5 feet ; the weight will be 19 cwts. and the water pressure 11-2 cwts. It will be found that DC = 1-7 foot, which is within the middle one-third of the width of the joint. This thickness is therefore sufficient, and the section of the wall will be as shown in Fig. 461. Strictly it should be ascertained that con- ditions (5) and (c), p. 221, are fulfiUed. This is left as an exercise for the student. As a further exercise, the student can find the proper dimensions to give to this wall supposing the back to be vertical. Practical Kule for thickness of walls to retain water. — The following is an old rule for retaining walls for water — "Width at bottom = height x 0*7, Width at middle — height x 0-5, Width at top = height x 0-3, 1 Page 386, 22d edition. 236 NOTES ON BUILDING CONSTRUCTION and a wall designed accordingly would be as shown in Fig. 462. It will be seen that this section has a smaller base but somewhat greater area than the one just designed. Its weight per foot run is 19-6 cwts. The resultant in this case also falls just at the one-third of the width of the base within the toe. On comparing Figs. 461 and 462 it will be seen that the width at the top has a considerable effect on the cross section of the wall. EETAINING WALLS FOE EAETH. The section of a wall for retaining earth could be at once ascer- tained in the manner just described for walls for retaining water if we knew the direction and magnitude of the pressure exerted by the earth on the back of the wall. We saw in the last example that the water pressure on the back of a wall can be easily found, but it is not so with regard to earth pressure, which varies with every different kind of soil, and also with the conditions in which the soil is ; for instance, whether loose or compressed, wet or dry. Various theories on the subject have been propounded by Professor Eankine, Boussinesq, and others, but they do not appear to be very satisfactory, and in any case are far too complicated for these Notes. Under these circumstances it will only be possible to give a mere outline of the subject. If a steep bank of earth is left to itself it will, under the action of the weather, gradually crumble down until it has taken up a certain slope as shown in Fig. 463. The angle of inclination of the slope at which crumbling ceases is called the " angle of repose," and this angle varies with each different ^ kind of earth. This angle is gener- -^'S- ^63. ally denoted by In Table XVI. will be found the angle of repose for several of the usual kinds of earths, but this angle can be found experimentally in any particular instance by shovelling the earth up into a heap and measuring the angle of the slope so formed. Now suppose that the earth behind a re- ; taining wall is loose, ^ and that it would im- Fig. 464. mediately fall down to RETAINING WALLS FOR EARTH 237 the angle of repose were the wall removed, then it will be seen from Fig. 464 that the wall, aided by the friction of the earth, supports the wedge of earth represented in the figure by the triangle ABF, which is tending to slide down BF. The resultant pressure of this wedge of earth will be some force such as E, and it is to find this force E that the various theories already alluded to have been proposed. Once found, the dimensions of the wall can be ascertained precisely as in the case of the reservoir wall (Example 47). If, however, the earth is at all consolidated, so that it would not fall to the natural slope immediately on removal of the wall, the pressure on the back of the wall is much less than that due to the wedge ABC. The earth may even be sufficiently consolidated to stand vertically on removal of the wall, in which case there is practically no pressure on the back of the wall, and if it can be ensured that the earth will always remain in this condition, the wall need only be built thick enough to resist its own weight. It then acts simply as a covering to the earth to prevent its being degraded by the action of the weather. Such walls are called " breast walls." On the other hand, however, infiltration of water will increase the pressure of the earth against the back of the wall, beyond what it is when the earth is quite loose. The student will at once see that to frame a theory to meet all these different conditions must be a difficult matter. "The presence of moisture in earth to an extent just sufficient to expel the air from its crevices, seems to increase its coefficient of friction slightly ; but any additional moisture acts like an unguent in diminishing friction, and tends to reduce the earth to a semi-fluid condition, or to the state of mud. In this state, although it has some cohesion, or viscidity, which resists rapid alterations of form, it has no frictional stability; and its coefficient of friction, and angle of rejDose, are each of them null. " Hence it is obvious that the frictional stability of earth depends, to a great extent, on the ease with which the water that it occasionally absorbs can be drained away. The safest materials for earthwork are shivers of rock, shingle, gravel, and clean sharp sand, whether consisting wholly of small hard crystals, or containing a mixture of fragments of shells ; for those materials allow water to pass through without retaining more than is bene- ficial. The cleanest sand, however, may be made completely unstable, and reduced to the state of quicksand, if it is contained in a basin of water-holding materials, so that the water mixed among its particles cannot be drained off. " The property of retaining water, and forming a paste with it, belongs specially to clay, and to earths of which clay is an ingredient. Such earths, how hard and firm soever they may be, when first excavated, are gradually softened, and have both their frictional stability and their adhesion diminished by exposure to the air. In this respect, mixtures of sand and clay are the worst ; for the sand favours the access of water, and the clay prevents its escape. 238 NOTES ON BUILDING CONSTRUCTION " The properties of eartli with respect to adhesion and friction are so vari- able that the engineer should never trust to tables or to information obtained from books to guide him in designing earthworks when he has it in his power to obtain the necessary data either by observation of existing earth- works in the same stratum or by experiment." ^ Mathematical Formulse. — Eankine's Theoey — Earth, liori- zontal top. — It lias been shown by Eankine that in the case of a wall retaining loose granular earth (i.e. not consolidated), the top surface of the earth being horizontal and the back of the wall vertical, the pressure P of the earth acts horizontally, at a point ^d of the height from the base, and that its magnitude is P = ivIL^ 1 — sin ^ (99), 2 1 + sin \ Ki=0"7 tan ( — z — ), where (j) is the angle of repose (see Table XVI.) Then Case (1) Wall with vertical sides and hacldng horizontal at the top (Fig. 468). _ T = . • (101). (2) Sloping walls with vertical bach (Fig. 469). Batter of face Tf T^ = 0-86T. 8 1\ = 0-80T. fi T^ = 0-74T. Case (3) Wall with vertical face and sloping hack (Fig. 470), or (Case 3a) with offsets (Fig. 471). Ti=0-85T. Fig. 472. Case 4. Fig. 473. Case 5. RETAINING WALLS FOR EARTH 241 Fig. 474. Case 6. Case (4) Suecharged Eevetments. — Substitute for H in the formula the vertical height H' measured to the point G, found by setting off the distance H along the earth slope, as shown in Eig. 472. Case (5) When the lanh is of less height above the wall than the distance IT would give, take the actual height of the bank, as shown in Fig. 473. Case (6) When the earth slopes from the front edge of the loall the point G- should be found, as shown in Fig. 474. "Values for w and W will be found in Table XVII. Retaining Wall for Light Vegetable Earth. Example 48. — Design a wall to retain light vegetable earth, consolidated and dry, to a depth of 8 feet, the backing being horizontal at the top and flush with the top of the wall. The back of the wall to be vertical and the face to battel- A. The wall to be built in ordinary brickwork. Preliminaries. — From Table XVII. we find ^y = 90 lbs. per cubic foot for vegetable earth. W = 1 1 2 „ „ for brickwork in mortar. Also from Table XVI. K = 0-26 for moist vegetable earth. B.C. IV. NOTES ON BUILDING CONSTRUCTION Calculation by practical formula. — Hence Equation 101, p. 240, = 1-86 foot ; and = 0-80 X 1-86, = 1-49, or say 1' 6" ; and as the batter is y the thickness at the top will be 1' 0", and at the bottom 2' 0". The foundations can, witli advantage, be inclined so as to resist the pressure which, tends to make the wall slip forward. We thus get a wall as shown in Fig. 475. Graphic method. — To compare the formula given above with Kankine's and Scheffler's methods, draw a cross section of the wall, and apply the graphic construction as shown in Fig. 476, taking the angle of repose as 49°. It will be found that BP is 2-96 feet, so that 90 lw(BP)2= —(2-96)2 = 394 lbs. Also the weight of one foot length of the wall= 1345 lbs. Taking Rankine's method first, we find R as the resultant of the weight and the thrust, and it will be seen that R intersects the plane of the base sufficiently inside the wall. the resultant found according to Scheffler's method, intersects the base still more inside the wall. It is left as an exercise for the student to find the maximum intensity of compression at the edge. RETAINING WALLS FOR EARTH 243 The student should observe that an extreme case has been chosen for this example ; the earth is very light and the angle of repose large. If the earth were not consolidated, the angle of repose would be 29°, and it will be found that the mean thickness (T^) would then be 2-4 feet. Eetaining Wall for London Clay. Example 49. — Same wall as in the previous example, but designed to retain London clay recently excavated and saturated with water (see Fig. 477). In this case we find, on referring to Table XVII., i«= 120 lbs. per cubic foot, and K = 0-53. Hence Equation 101, p. 240, T = 0-53 x 8 \/If| = 4-4 feet, Tj = 0-80 X 4-4 = 3-5 feet. Therefore a wall of the section shown in Fig. 477 is required. ^London clay aturated 'with water Tis. 478. Surcharged Retaining Wall (Loamy Earth). Example 50. — Find the dimensions of the surcharged retaining wall shown in Fig. 478. The earth is supposed to be of a loamy nature ; the iace of the wall is to be vertical, and the back to have a batter of about |, the wall to be built of rubble masonry. Preliminaries. — From Table XVII. we find w= 80 lbs. per cubic foot, W = 140 „ and K = 0-33. On measuring H along the slope we find = 9-3 feet, and therefore T = 0-33 X 9-3 ^^ = 2 -3 feet. Hence from (3), p. 240, Tj = 2-3 X 0-85, = 2 feet nearly. The dimensions in Fig. 478 give a mean thickness of 2 feet and a batter of f at the back. 244 NOTES ON BUILDING CONSTRUCTION Baker's practical Rules for Thickness of Retaining Walls. Proportion of thickness to height. — "As a result of his owu experience " Sir Benjamin Baker " makes the thickness of retain- ing walls in ground of an average character equal to of the height from the top of the footings." He also says that "A wall quarter of the height in thickness, and battering 1 inch or 2 inches per foot on the face, possesses sufficient stability when the backing and foundation are both favourable." Also that "under no ordinary conditions of surcharge or heavy back- ing is it necessary to make a retaining wall on a solid foundation more than double the above, or ^ of the height in thickness." Eqiiivalent fluid pressure. — He says further that " Experiment has shown the actual lateral thrust of good filling to be equivalent to that of a fluid weighing about 10 lbs. per cubic foot; and allow- ing, for variations in the ground, vibration, and contingencies, a factor of safety of 2, the wall should be able to sustain at least 20 lbs. fluid pressure, which will be the case if ^ of the height in thickness." These rules may be usefully applied to check the calculations made by other methods. Foundations. Ordinary firm earth will safely bear a pressure of about 1 to 1-^ ton per square foot, while moderately hard rock will bear as much as 9 tons. It is therefore not worth while to make any calculations for the foundations of ordinary walls because the pressures they cause are so small. The foundations of retaining walls are subject, however, to considerable pressures, which moreover are not uniformly distri- buted. They should therefore be of such a width that the maximum intensity of pressure is not greater than the soil can safely bear, and the centre of pressure should not be nearer the outside edge than ^ the width of the foundation. Example 50a. — Pressure. — As an example suppose Nc, Fig. 478a. (the normal constituent of the resultant of the weight of the wall for 1 foot in length and the pressure of the earth upon it) to be equal to 2 tons. Assume foundations such as those shown in Fig. 477, the centre of pressure c being at J the width ef of the foundation from the outer edge, then the maximum intensity of pressure will be at e (see p. 219), and it will be equal to 2 x -I = f tons per square foot. It is evident, therefore, that on firm earth, clay, or harder soils the maxi- mum intensity of pressure would not be too great. If the soil were very loose and unfit to bear weight then the concrete RETAINING WALLS FOR EARTH 245 foundation would have to be extended so that the maximum pressure upon it should not exceed what the soil could bear, and in extreme cases the founda- tion would have to be prepared by driving piles. If it is very important for any reason that the pressure on the earth should be uniform, the toe of the concrete must be extended so that c should be the centre of the earth pressed upon. Concrete, base. — Care must be taken that the concrete is made so thick that there is no danger of its breaking across. In Fig. 478a ahde is in the condition of a can- tilever nearly imiformly loaded and tending to break off at Id. Thus see Appendix VII. 6 ' taking /o= 100 lbs. per square inch,^ ' wl- 2 X 12(x 18)2 X 100 6 X 24 = 5400 lbs. = 2|^ tons nearly. Therefore the cantilever will bear 2|- tons distributed or ij ton per foot superficial without breaking. Whereas the most intense pressure at the end is only ^ ton per foot 2^ (giving a factor of safety of = nearly 4), and the mean pressure distri- buted only ^^n. The concrete therefore is amply thick enough. If made 12" thick it would bear about i ton per foot-superficial. Sliding on concrete. — To prevent danger of the wall sliding forward on its concrete base, the tangent of the angle MQE (between the normal NQ to the base and the resultant pressure EQ) must not be greater than -J x the coefficient of friction of brickwork with damp mortar, Table XVcc, i.e. not greater than ^ x '74 = -59. The angle EQN measures 21°, the tangent of which is '38, so that the condition is amply fulfilled and the wall quite safe against sliding. Sliding on clay. — In the same way by using the proper co- efficients from Table XVa it will be found that the wall would slide on wet clay but would be safe against sliding on dry clay. I Baker. Chapter XV. ARCHES. IN Parts I. and II. some preliminary remarks were made in connection with arches ; the names of the different parts were given, and also some practical details as to the construction of brick arches. It is proposed to show in this chapter how to ascertain the thickness of the arch ring in order that the arch may safely resist the load it has to bear. The calculations in connection with the stability of the arch ring are amongst the most difficult that occur in engineering, but there is a simple graphic method by means of which the desired result may be obtained by a series of approximations, and it is this method which will be described. It should, however, be observed that in the large majority of cases the arches used in building construction have but a small span, and are, moreover, only repetitions of what have been built before, so that there is ample experience to fall back upon. There is therefore no necessity, in such cases, to make any calculations, and the dimensions can be obtained by referring to a table such as that given in Molesworth's Pocket-Booh^ or Table XVIIa. Graphic Method of determining the Stability of an Arch. Let I, J, K, L be any voussoir^ of an arch. The load on this voussoir consists of its own weight and of the portion of the weight carried by the arch included between the dotted lines (Fig. 479). Let w be the resultant of both these weights acting through their common C.G. The voussoir is also subjected to the pressure P of the voussoir above it, and to the reaction E of the voussoir below it. These are the only three forces acting on the voussoir, and if we know w and P we can find E and also its 1 Page 108, 22(i edition. 2 The student is referred to Parts I. and II. for the meaning of the various technical terms employed. ARCHES 247 point of application C, which is the centre of pressure for the joint I, J. Now we know (see p. 221) that for stability the point C must lie inside I, J, and that, moreover, to prevent crushing the material of the voussoir, C must not approach I or J nearer than a certain distance, depending on the value of K, and on the resistance to crush- ing of the material. On this point Dr. Scheffler has laid down a general rule that the point C should lie within the middle, hcdf of the joint, ^'ig- but other authorities recommend that C should lie within the inner third of the joint,-^ in order that no tension be excited (see p. 221). Dr. Scheffler's assumption is amply safe for all ordinary cases, and will therefore be accepted for these Notes. Now, for purposes of calculation, the arch can be considered as divided up into a number of imaginary voussoirs, as shown in Fig. 480, and we can suppose that the centre of pressure at each joint has been found, and that these centres of pressure are then joined together by a broken line. If the arch had been divided up into a greater number of imaginary voussoirs, as in Tig. 481, the line joining the centres of pressure would be almost a continuous curve, and if a still greater number of voussoirs had been assumed, the line would hardly be distinguishable from a curve. This line is called the line of resistance, and from it we can obtain the point of application (centre of pressure) and direction of the pressure at any section of the arch. Applying, therefore, the results arrived at in connection with one voussoir, we see that for safety the line of resistance must lie within the middle half of the arch ring, as shown in Fig. 485. For mere stability, if the material of the arch ring were un- crushable, the line of resistance might just touch the outside of the arch ring. The line of resistance also gives the direction of the resultant ^ For important arches the line of resistance should always be kept within the middle third of the arch ring, as in the Example, Appendix XX. 248 NOTES ON BUILDING CONSTRUCTION pressure on any joint of the arch ring, this direction being a tangent to the curve ; and should this direction be too much inclined to the joint the arch would fail by sliding.^ As a matter of fact, how- ever, the inclination seldom, probably never, is sufficient to produce failure in this manner, except in straight arches (see Fig. 1 4, Part I.) To find the line of resistance. — The problem therefore is to find the line of resistance, and herein lies all the difficulty. On looking at Tig. 481 it will be seen that the line of resistance is hori- zontal at the point D ; at this point therefore the pressure is horizontal. \^ I n Suppose the portion DB of the arch is re- ^ ot moved (Fig. 482), then we have the left-hand // j \ portion of the arch kept in equilibrium by — // Y (1) The portion of the load it has to sup- ll-an \ P«^t (Wj). L^jr__n_.:^ y. ^2) The horizontal pressure H at D. / ^ (3) The pressure at the abutment A. / pjg_ 482. Now, a force is determined when we know (1) its direction, (2) its point of application, and (3) its magnitude. is, or can be, so determined. Of H we only know the direction ; and as to we know nothing except that it is equal and op- posite to the resultant of "Wj^ and H. We therefore cannot solve the problem until we know the point of application and the magnitude of H. To obtain these we must resort to what is known as Moseley's Principle of Least Resistance. Professor Eankine states this principle as follows : — " If the forces, which balance each other in or upon a given body or structure, be distinguished into two systems, called respectively active and passive, which stand to each other in the relation of cause and effect, then will the passive forces be the least which are capable of balancing the active forces, consistently with the physical condition of the body or structure." In the case under consideration W^^ represents the active forces and H and the passive forces. It therefore follows that, in the actual arch, the magnitude of H will be the least possible consistently with the physical condition of the arch. And it will also be seen that to each value of H there is a corresponding line of resistance. Combining this with the condition relative to the line of resistance (p. 247) we arrive at the following result. The value of H will be such that the corresponding line of resistance is just included ivithin the inner half of the arch ring. This line of resistance is called the line of least resistance. Theory of grafhic metJiod. — To determine the true line of least resistance mathematically is most difficult, even in simple cases ; but by the graphic method of trial and error already referred to, the line of least resistance can be found very expeditiously — only ^ " To insure stability of friction the normal to each joint must not make an angle greater than the angle of repose with a tangent to the line of pressures drawn through the centre of resistance of that joint" (Rankine, Civil Engineering). ARCHES 249 approximately, it is true, but with ample accuracy for practical purposes. This method is as follows : — ■ Referring to Fig. 482, let be the lever arm of H from the centre of pressure of the joint at A (viz. the point of application of Pj^), and a the lever arm of from the same point. Then taking moments about this point we have ■WjX(i-Hx/!- = 0, or H = rWi .... (102). Hence H vs^ill be least when a is least and It is greatest, and that is the case when the point of application of H is as high up as possible (that is at E, Fig. 483), and the point of application of as low down as possible (that is at K). Fig. 483.1 / " ' Pig. 484.1 Fig. 485.1 On the other hand, the greatest admissible value of H will be when a is great- est and In, least, that is when H is applied at F, and P^^ at G, as in Fig. 484. The true value of H lies somewhere between the least and greatest values thus found, and is the least value of H that allows of the line of resistance being just included within the inner half of the arch ring. Thus, supposing that it has been found that a certain value gives the line of resistance shown in Fig. 485, then any less value of H would cause the line Fig. 487. 1 In Figs. 483, 484, 485 and others, the curved dotted lines represent the centre half of the arch. 250 NOTES ON BUILDING CONSTRUCTION of resistance to pass below K, is therefore the least admissible value of H, and the corresponding line of resistance is the line of least resistance. Clearly, therefore, the next step is to establish a method of finding the line of resistance corresponding to any assumed value of H. Suppose that the portion AD of the arch ring is divided up into a number of imaginary voussoirs, as shown in Fig. 486, and let the load on each of these voussoirs be found, namely w^, w^, w^, etc. Now the first voussoir is kept in equilibrium by three forces, as shown in Fig. 487. We know H and w^, but P^^ has to be found. To do this, draw ah (Fig. 488) to represent H, and aa-^ to represent ; then (by the triangle of forces, p. 179) will represent P^. We therefore have only to draw P^ parallel to through the intersection of H and to find Cy Proceeding to the second voussoir we have forces as shown in Fig. 489 ; and Fig. 490 gives the triangle of forces. Cg therefore can be found as before. The same process can be repeated for each succeeding voussoir, and by this means the various centres of pressure can be found, and by joining them the line of resistance can at once be obtained (Fig. 491). It will be observed that Figs. 488 and 490 can be combined to form one H Fig. 492. ARCHES— LINE OF LEAST RESISTANCE 251 figure, as shown in Fig. 492, and similarly the triangles of forces for all the succeeding joints can be combined into one figure (Fig. 493), thus saving a considerable amount of drawing. It will also be observed that the broken line formed by the forces H, P^, Pg, etc. (Fig. 491), approximates to the line of resistance, and if double the number of voussoirs had been taken, the approximation would be still closer, as shown in Figs. 494 and 495. With a very large number of voussoirs, the line of resistance and the broken line formed by the direction of the pressures at the joints practically coincide. Practical Graphic Method of finding the Line of Resistance in an Arch with a Symmetrical Load. — The line of resistance corresponding to any particular value of H can therefore be found by the following very simple graphic method. Trial line. — Find the loads w^, w^, w^, etc., acting on the arch 252 NOTES ON BUILDING CONSTRUCTION (Fig. 496), and draw the diagram of forces, Tig. 497, thus obtaining the forces V^, Pg, P3, P^, etc. Then find the intersection of H and w^, then the intersection of P^ and w^, and so on. Join these points of intersection, and the broken line so obtained will be approximately the line of resistance required. It will be observed that this line of resistance (Fig. 496) is not included within the inner half of the arch ring. The reason is that the point of application of H has been chosen too high. Fig. 498. j,y Fig. 499. True line. — Lowering the point of application, which will also increase H to (Fig. 498), we obtain a new diagram of forces (Fig, 499) and a new line of resistance, which is now found to just fit in within the inner half of the arch ring, and is there- fore the line of least resistance required. To find where the line of resistance is horizontal. — On refer- ring to p. 248 it will be seen that the section D was taken through the point where the line of resistance is horizontal, and we must now show how this point can be obtained. 7V Fig. 500. Fig. 501. Taking the simplest possible case, when the load is uniformly distributed, as in Fig. 500, it is clear from symmetry that D will be situated at the centre of the arch ring. ARCHES— LINE OF LEAST RESISTANCE 253 Or, again, if the load is symmetrically distributed D will still be at the centre of the arch ring (Fig. 501). In both these cases the line of least resistance will be symmetrical about the point D. Now let us investigate the effect a concentrated load,^ placed at any point on the arch ring, has on the position of the point D, as shown in Fig. 502. Let W be this concentrated load, and let it be applied at the point P. If we take any section of the arch, as in Fig. 503, the thrust at the section can be represented by the horizontal thrust H and a vertical force V. What we want to find is the section at which V is nothing. If we regard V in the light of a shearing stress we see on referring to p. 5 6 that since, as a consequence of Eule 2, the shearing stress is nothing under the resultant of all the loads, therefore the section of the arch we require will be under the resultant of W and w^, . . . w^, w^. In any case, therefore, the point D can be found by determining the resultant load on the arch, and this can be done very simply, by finding the centre of gravity of all the loads, as explained in Appendix VI. Thickness of the arch ring. — The thickness of the arch ring should be sufficient to allow of the intensity of pressure being not greater than what the material is able to bear safely. Moles worth gives a rule, based on Eankine's, for finding the thickness of the arch at the crown, or, in other words, the depth of the keystone. It is as follows : — Depth of keystone = 71 ^radius at crown . . (103), where n= OS for blbckstone, = 0"4 for brickwork, = 0'45 for rubble stonework. The depth thus found is expressed in feet. 1 A different method of treating an arch with an nnsymmetrical load will be found in Appendix XX. 254 NOTES ON BUILDING CONSTRUCTION The actual intensity of pressure at any section of the arch ring can, however, be found, as soon as the line of least resistance is known, by applying Equation 95. Semicircular Brick Arch. Example 51. — Find the line of least resistance in the symmetrically loaded semicircular brick arch shown in Fig. 504 when loaded with a uni- form load of 1 ton per foot run, including its own weight. Fig. 504. Preliminaries. — It will be found, by applying Equation 103, that the thickness of the arch ring at the crown is = 0-4 v^5 = 0-9 foot nearly. 14" is therefore the nearest brick dimension, and we will call this 1'2 foot. The arch being uniformly loaded, we need only consider one half of the archj and the thrust at the crown will be horizontal, as shown in Fig. 505. Fig. 505. The total load on the half arch is 6*2 tons, and its direction is at a distance ARCHES— LINE OF LEAST RESISTANCE 255 of r9 foot from tlie point A. Now, as seen at p. 249, H will be a minimum wlieu applied at the point E, and its lever arm will then be 5 + (fx 1-2) = 5-9 feet. The lever arm of the load is l-9 + (|x 1-2) = 2-2 feet. Hence 2-2 - 2-3 tons. Again, H will be a maximum when applied at F, and when moments are taken about G. In this case the lever arm of H is 5 + (^x 1-2) = 5-3 feet, and of the load l-9 + (fx 1-2) = 2-8 feet. Hence _2-8 = 3-26 tons. The true value of H, giving the line of least resistance, must lie between these two values, provided of course the arch ring is able to withstand the load put upon it. As a first trial draw a line of resistance through E. Following the pro- cess explained at p. 251 we obtain the line of resistance given in Fig. 506, "Iton that is, draw H = 2"3 tons horizontally through E to intersect the first load of 1 ton, from this intersection draw a line parallel to P-^ (Fig. 507) to meet the second load of 1 ton, then from this intersection a line parallel to Pg to meet the third load, and so on until a line parallel to P^ is drawn to meet the last load. From this point of intersection a line parallel to Pg is drawn and gives the direction of the pressure at the springing ; Pg in Fig. 507 being NOTES ON BUILDING CONSTRUCTION the magnitude. This line of resistance passes below the soffit of the arch, and therefore the value of H must be increased. It should be observed that the direction of Pg passes through K, the point about which moments were taken, and this forms a valuable check on the accuracy of the drawing. As a second trial increase H to H(max.), namely, 3-26 tons. The point of application of H will in this case be F, and the line of resistance shown in Fig. 508 will be obtained, which also cuts through the arch ring at the soffit. As a last trial apply H at E and take moments about G, so that the line of resistance will pass through G. We find TT 2-8 H = —x 6-2 = 2-95 tons, and the corresponding line of resistance shown in Fig. 510 is obtained. This line of resistance is within the arch ring at all points, but at the haunches it is outside the inner half of the arch ring, and therefore does not comply with Dr. Scheffler's rule. To raise the line of resistance at the haunches so that it will be within the inner half of the arch ring, we can either increase the load above the haunches or increase the thickness of the arch ring at the springing. The first alternative is left as an exercise for the student. Thickening arch. — Supposing that the thickness of the arch ring is in- creased to 1' 6" at the springing, as shown in Fig. 511, the point G is now at a distance of f x 1-5 = I'lS from the inner point of the springing, and the lever arm of the load becomes l-9xl-13 = 3-0 nearly. Hence, if H is applied at E, 3-0 , , H = X 6-2 = 3-15 tons, and the 5'9 True line of resistance shown in Fig. 5 1 1 is obtained. This line of resistance is practically everywhere included within the inner half of the arch ring, and is therefore the line of least resistance required. Practically, however, a brick J ARCHES— EXAMPLE 257 arch would not be built of varying depth owing to the expense of cutting the bricks, and the arch ring would therefore be made 1' 6" deep throughout. This expedient could, however, be adopted if the arch were built in masonry. Thickness to resist crushing. — The thickness of the arch ring must be suffi- cient to prevent the intensity of pressure being greater than the pressure the material can safely bear. Now on inspection it would appear that the greatest intensity of pressure occurs either at the point M (Fig. 513) or at Fis. 510 Fig. 510a. the abutment, and to find the required thickness it will be necessary to ascertain at which of these points the intensity of pressure is greatest. At the point M the line of least resistance is practically perpendicular to the joint between the voussoirs, and on referring to Fig. 5 1 2 it will be found that the pressure at the point M is 5-1 tons. We must now employ Equation 95, p. 219, and we can write Y = 5'1 tons, GC = J^ inches, AE = b (the width of the arch ring). _ 2x 5-1x4 0-97 Hence or 3 X 14 X 6 0-97 Now proceeding to the abutment we find from Fig. 5 1 2 that the pressure is 6-8 tons ; but this pressure is inclined to the joint, and we must resolve perpendicularly to the joint. "We thus find (Fig. 514) Y = 6 tons, also GC = i^ inches. 2x6x4 0-89 Hence 3x18x6 B.C. -IV. S 2s8 NOTES ON BUILDING CONSTRUCTION 0-89 K = . E F Fig. 512. Fig. 511. Therefore I is greater than \, or, in other words, the maximum intensity of pressure is greater at M than at the abutment. -.S' 0"- Fig. 514. Fig. 515. If the arch is built in ordinary brickwork in mortar, p can be taken at about 0"5 cwt. per square inch. Hence , 0-97 x 20 . , 6 = — ^-^ — = 38 inches. ARCHES— EXAMPLE 259 But if the arch were built in cement, f can be taken at 0"8 cwts. per sq^uare inch, and then ^ 0-97x20 0 = — — — =24 inches. U'o Arch with unsymmetrical Load. (See Appendix XX.) Abutment for Arch. Example 51a. — As an example it will be assumed that the arch is situated at one end of the wall of a building, as shown in Fig. 515. The forces acting on the abutment are the thrust of the arch, namely, 6*8 tons as found above, and the weight of the brick- work in the wall immediately over the wall, including the weight of the abutment itself. It will be assumed that this weight is 10 tons, acting as shown in Fig. 515. These two forces intersect at A, and completing the triangle of forces we obtain B as the centre of pressure at the joint DC. In order that no tension be excited at C we must make DB = -g-BC. It must also be ascertained that the intensity of pressure is not too great. The vertical 'pressure, on the joint DC is 1 6 tons. Hence taking the width of the arch as 2 3" we have from (95) _ 16x20 6'6"x23" = 0"36 cwt. per square inch nearly. It will be seen from Table I A that O'S cwt. is safe. The maximum pressure at the joint at A will, however, be somewhat greater; the vertical pressure will be reduced by about 1 ton (viz. the weight of the abutment below A) ; and as A is (according to the dimensions taken) at the width of the joint from the outer edge ^ ^ 15 x 20 = 0*48 cwt. per square inch nearly. Resistance to sliding can be found as before. The concrete foundations to the abutment can be designed on the same principles as above, taking care that the maximum intensity of pressure is not greater than the soil can withstand with safety. Table of thickness of arches.^ — The arches used in building construction are not generally of very great span, and they do not as a rule require to be calculated. Their thickness is often governed by appearance, or it may be found in Tables founded upon experience. One of these is given at p. 343, and others in Molesworth's and Hurst's Pocket-Books. ^ See Appendix XXI. Chapter XYI. HYDRAULICS' (As applied to Building Construction). HYDEAULICS is a subject which is related to building con- struction only to a limited extent, and in fact only two subdivisions of the subject need be considered, namely — 1. The motion of liquids through pipes in connection with water supply and disposal of sewage. 2. The delivery of water from a jet in connection with stand- pipes. We must first consider and define some of the terms in use. Pressure. — If we imagine a small cube immersed in a liquid as shown in Fig. 516, the pressure exerted by the liquid on each of the six faces of the cube will be normal to that face, and if the cube is sup- posed to be very small indeed, all these pressures will be almost equal. If the cube be now sup- posed to diminish without limit it will become a point ; the pressures will become exactly equal, and will then be the pressure at a point, as at P, Tig. 516. for instance, in Fig. 517. It is shown in works on Hydrostatics that the pressure at a point in a liquid is propor- tional to the depth below the surface of the liquid, and in fact that the intensity pressure, that is, the pressure per unit of area (a square inch, for instance), is equal to the weight of a column of liquid of a height equal to the depth below the surface and one unit of area in cross Fig. 517. section. Thus if D is 50 feet and the liquid is water, a cubic foot of which weighs almost exactly 6 2 "4 lbs. ' For Practical Formulce, see pp. 267, 281, and App. XXI. HYDRA ULICS— DEFINITIONS 261 at a temperature of 52°"3 Fahr., the pressure on a square inch at the point P would be 50 X -^x 62-4 = 21-67 lbs. 144 The intensity of pressure at the point P is therefore 21*67 lbs., or as it is usually expressed, 21*67 lbs. per square inch. Head of pressure. — The depth of the point P (Fig. 5 17) is also called the head of pressure at the point P, or simply the " head," and is generally expressed in feet. Head of elevation. — The height of the point P (Fig. 517) above some datum level is called the head of elevation of the point P. Loss of head. — "When a liquid is in motion, each particle is constantly moving from a place of greater head to a place of lesser head, and the difference between the two heads is called the loss of head. This loss of head may be entirely a loss of head of pressure, or entirely a loss of head of elevation, or, again, partly a loss of head of pressure and the remainder a loss of head of elevation. The following examples will assist to illustrate the above — AB (Fig. 5 1 8) is a pipe connected to a tank T ; at A there is a tap. When the tap A is closed, the head at the point P will be: Head of pressure, EP ; Head of elevation, FP. At A the head of pressure is G-A, and there is no head of elevation with reference to the datum chosen. Dahitn level. F ' " Fig. 519. When the tap is opened, however, the pressure at A is reduced 262 NOTES ON BUILDING CONSTRUCTION to zero, and if the resistance of the pipe to the flow of the water is uniform, the pressure at each point of the pipe can be represented by the straight line AL, as shown in Fig. 519. Therefore the head at P will now be : Head of pressure, KP ; Head of elevation, FP. It will be observed that the head of elevation is not altered, but that the head of pressure is reduced by the amount KE. With reference to a second point P', the head of pressure is K'P', and the head of elevation Y'^' ; but the " loss of head " between the two points is K'O. This loss of head is made up of K'M, the loss of the head of pressure (KM is drawn parallel to PP0> ^-nd of MO, equal to P^N, which is the loss of the head of elevation. If the pipe is horizontal, as in Fig. 520, it is clear that the ^ Fig. 520. loss of head between the two points P and P' is entirely loss of head of pressure, and there is no loss of head of elevation. On the other hand, if water is flowing in an open channel, as in Fig. 521, the loss of head is entirely loss of head of eleva- tion ; in fact, the water is not flowing under pressure. Another way of looking at the matter is as follows : — The water, when flowing in the pipe or in the channel, has to overcome the resistance of the pipe or of the channel, which means that the water has to do a certain amount of tvork, and this work is proportional to the loss of head. Supposing, for instance, that 1 lb. of water, in flowing through a pipe, loses 1 0 feet of head, then the work done by that pound of water is 10 foot-lbs. Wetted perimeter. — In an open channel, or in the case of a pipe not flowing full, the portion of the cross section of the channel, or of the pipe, wetted by the liquid (from E to F, Figs. 522 and 523) is called the wetted perimeter. It is also called the " harder." Hydraulic mean depth. — The quotient of the area of the HYDRA ULICS— DEFINITIONS 263 cross section of the liquid divided by the wetted perimeter is called the hydraulic mean depth, and will be denoted by E. Fig. 522. Fig. 523. ' MOTION OF LIQUIDS IN PIPES. The motion of a liquid in a pipe is due to gravity, and if the pipe were perfectly smooth, and offered no resistance, the velocity would increase without limit. But, as already mentioned, pipes do offer a resistance depending on their diameter and on the state of their inner surfaces, so that the velocity, instead of in- creasing continually, is limited. Moreover, the velocity of the * liquid is not uniform throughout the cross section of the pipe, but is greatest in the centre, diminishing gradually at first, and then rapidly towards the sides. Close to the sides the liquid does not flow freely, but is disturbed by numerous small eddies. It was at one time thought that neither the diameter nor the state of the surfaces had any influence on the resistance, and to account for this it was supposed that these eddies formed, as it were, a liquid lining to the pipe. Darcy has, however, shown by means of an elaborate series of experi- ments that this view, due to De Prony, is erroneous, and has also shown that the mistake arose from combining experiments on pipes of large diameter but with rough surfaces, and experiments on smooth pipes but of small diameter, and that the resistance offered by the rough surface in the one set happened to exactly equal the resistance due to the small diameter in the other set ; so that it appeared that neither the state of the surface nor the diameter had any influence on the resistance. Although this supposition is now known to be erroneous, yet in many of the formulae in general use for obtaining the discharge of pipes, the effect of the diameter on the resistance is not taken into account, nor is any notice taken of the degree of roughness of the surface ; practically this latter is of little moment, for whatever the surface may have been originally, it will speedily be covered with sediment. These formulae are empirical, and are 264 NOTES ON BUILDING CONSTRUCTION based on numerous experiments that have been made on the flow of liquids in pipes. It should, however, be observed that these experiments, made at various times by different observers, are neither numerous enough nor were they sufficiently system- atically carried out to obtain results of scientific accuracy, although they may be good enough for practical purposes. We have to consider two cases of the flow of liquids through pipes, namely — 1. When the pipe is flowing full and the liquid is therefore impelled to move by the pressure of the head of liquid. 2. When the pipe is flowing partially full. In this case the liquid is not under pressure, and simply flows down, owing to the slope of the pipe, as it would in an open channel. We will consider each of these cases separately. Discharge feom a Pipe flowing full under Pressure. As already mentioned, numerous formulae have been proposed to find the discharge from a pipe, but we will only consider four and then select for future use the one best adapted for our purpose. One of these, known as Eytelwein's formula, is where V= 108^^ X H_ oas . . (104), V is the nYiean velocity of the water in feet per second ; H the effective, head of water in feet, that is, the actual head reduced by the various losses of head due to bends, etc. (see p. 268); L is the total length of the pipe in feet ; D is the diameter of the pipe in feet. Neville's formula is said to be more accurate than Eytelwein's, but it is more complicated and is as follows : — Both in Hurst's and Molesworth's Pocket-Booh will be found tables calcu- lated from this formula, which are of great assistance in practical calculations. Thrupp's formula. — The latest, and probably the most accurate formula, is that recently elaborated by Mr. Edgar C. Thrupp, which is as follows : — „ , /z-R V = -^ ^ ^ x(S)» . . . (106), where K is the hydraulic mean depth (which is equal to ^ for pipes flowing full, see p. 280), S is the slope of the pipe fi.e. and x, y, z, and c are con- PIPES FLOWING FULL UNDER PRESSURE 265 stants depending on the material of which the pipe is made and on the sur- face of the pipe. For further information on this formula, and for values ot the constants, reference is made to the Proceedings of the Society of Engineers (Dec. 188Y). The following cases may, however, be useful :— For wrought iron pipes over 3 J" diameter, ■nO-65 1 0-004787 For wrought iron pipes l" diameter. For new cast iron pipes, V = 0-004787 X (S)i-8 x(S)^ (107) . (108) . (109) . "0-006752 Darcy's formula.— Th.Q fourth formula we will mention is Darcy's, namely — v.oy5,H . . . (no). where C is a coefacient to be found from the following table.^ TABLE J. Diameter of pipe in inches = 12xD. 1 2 3 4 5 6 7 8 Value of C. 65 80 93 99 102 103 105 106 107 Maximum value of C for very large pipes, 113-3. To compare these formula we will work out some examples. Example 52.— Find the velocity in a pipe 1 inch in diameter, 100 feet long, the effective head being 1 foot. We have By Equation 104 (Eytelwein' s formula), V = 108 sJl X X - 0-13, = 1-43 feet per second. By Equation 105 (Ne ville's formula ), V = 140 ^/i X jJ^^ X - 11 v/i >^ T2 X T^TT' = 2-02 - 0-65, = 1*37 feet per second. By Equation 110 (Darcy's form ula), taking C = 80 for this size of pipe, V = 80x/ix'^io<^, = 1-15 feet per second. By Thrupp's formula (for wrought iron 1" diameter). Equation 108, 1 From Molesworth's Pocket-Boole of Engineering Formula. 266 NOTES ON BUILDING CONSTRUCTION 1 y — V? ^ 12^ X lT(ny/) 0-004787 ' 0-0786 "0-004787 X 1-29' = 1-27 feet per second. Example 53.— Find the velocity in a pipe 1 inch in diameter, 100 feet long, the effective head being 10 feet. By Equation 104, ^= 108 ^f^^A^^- 0-13, = 4-8 feet per second. By Equation 105, 140^f^^^^- 1 1 ^i^T^^, = 4-99 feet per second. By Equation 110, V = 80 a/1 V 1 V 1 0 V V 4 X 3-2- X Yo^, = 3-65 feet per second. By Thrupp's formula (Equation 108), i 0-004787 ' ^ 0-0786 0-004787 X 3-594' = 4-59 feet per second. Example 54.— Find the velocity in a pipe 4 inches in diameter, 100 leet long, the effective head being 1 foot. By Equation 104, V= 108 v/ix 3^x^^0-0-13, = 2-99 feet per second. By Equation 105, = 6 01 teet per second. By Equation 110, taking C= 102, = 2-94 feet per second. By Thrupp's formula (for cast iron. Equation 109), V = k_^T2) , 1 a 0-006752 ^^^^ ' 0-209 0-006752 X 10' = 3-09 feet per second. Example 55.— Find the velocity in a pipe 4 inches in diameter, 100 feet long, the available head being 10 feet. By Equation 104, ■^=108x/|xAxtV^-0-13, = 9-73 feet per second. PIPES FLOWING FULL UNDER PRESSURE 267 By Equation 105, V = 1 40 ^/i X T-*^ X T-i^ - 11 -^i X t\ X = 10"53 feet per second. By Equation 110, V = 102v/ix i*2XtV% ==9-31 feet per second. By Thrupp's formula (for cast iron, equation 109), ^ ~ 0-006752 ^^100'' ' 0-209 0-006752 x 10^ = 9-79 feet per second. Practical Formula (Darcy's). It will be observed that in all cases Darcy's formula gives the lowest velocity, and practically it is better to under-estimate the velocity than to over-estimate it. It is also a simple formula to use, so that on the whole it is to be preferred to Thrupp's in cases where great accuracy is not essential, and it will therefore be used to work out any further examples in this book. The discharge of the pipe can of course easily be found as soon as the mean velocity is known ; thus Discharge in cubic feet per second = V x r = 0-785D2V . (111). ttD^ Discharge in gallons per minute = V x x 6-25 x 60, G = 294D2V . . . (112). In practice, however, the question is generally to find what diameter of pipe is required for a given discharge, the effective head and length of pipe being known. Combining equations 110 and 112 together we get /D H G = 294D2xC^^x whence, by squaring and simplifying, ==7:^(s-IT • • • D being expressed in feet. Or d=l'e3[^.-^,) . . • (114), when d is expressed in inches. or or 268 NOTES ON BUILDING CONSTRUCTION To obtain the fifth root of the expression in brackets, either logarithms can be used, or else a table of fifth roots,' such as published in Molesworth's Pocket-Booh Equation 114 can, however, be written H«' = CD gallons per minute, and the tap at B two gallons per minute, when both taps are open, allowing for incrustation. Find also the discharge from each tap when the other is closed. The available head in the main is 90 feet. The levels are shown thus +10 with reference to the main. To FIXD SIZE OF BC AND AC. — The best way to proceed is to assume the size of one of the branches. For instance, supposing BC is taken as ^" diameter. L Then, from Table XVIIL, H G2: Hence if we take L = 30 feet and G = from C to B, thus H: 2 gallons, H will be the loss of head 30x 22 = 24 feet. Now since B and C are at the same level the available head at C ought to be 24 feet. In the pipe AC the water has to rise 19 feet, hence the available head is 24-19 = 5 feet. WATER SUPPLY TO A HOUSE 275 Hence ^-02 = ^x32 = 36. n. 5 So that (Table XVIII.) a pipe is too small and a f" is too big. "We liave, however, made no allowance for the various minor losses of head. It will be well to see whether in this case they are of sufficient magnitude to take into account. Minor losses of head. — In the first place, if the pipe AC is connected to BD by means of a T piece, as shown in Fig. 531, the general flow of the water will be along DB, as shown by the arrows. Consequently in entering the pipe CA there will be a loss of head due to the change in the direction of the flow, a part of which may be taken as the loss of head required to impart the velocity in the pipe AC, but, in addition, eddies are formed (called gurgli- tation) which cause another loss of head, generally taken as Uvice that due to velo- city. On the whole it is usual to reckon the loss as three times that due to velocity.^ We must therefore first find the velo- city :— Supposing a |-" pipe is provision- ally decided upon, then we must reckon the velocity Avhen the diameter of the pipe has been diminished by incrustation. From Table XVIII. it will be seen that the diminished diameter is 0-64 inch. Hence from Equation 122 V = 0-49 X (0-64)2' = 3*6 feet per second, 3-g2 and = = 0-2 foot. ^ 64-4 So that the loss of head is 3 X 0-2 = 0-6 foot. Bends. — We will neglect the loss due to bends ; it must be small owing to the low velocity. Tap. — The tap, if smaller than the pipe, will caiise a slight loss of head. Supposing that the bore of the tap is |", or 0"48 inch, making a slight allowance for incrustation, then to discharge three gallons per minvite the issuing velocity must be = 6*4 feet per second TT 6-42 or H= ——= 0-63 foot. 64-4 But the velocity in the pipe is 3*6 feet per minute, which represents a head of 0'2 foot as already seen. Hence the additional head required to produce the issuing velocity, or, in other words, the loss of head due to issue^ is 0-63 -0-2 = 0-43 foot. Total minor Loss of Head. — Altogether, therefore, we must reckon on a loss of head of 0-6 + 0-43, See Hurst's Architectural and Surveyor's Handbook. 276 NOTES ON BUILDING CONSTRUCTION = r03 foot, or say 1 foot. The available head is therefore reduced to 4 feet we thus have 20 — X 32: 4 45. It thus appears (Table XVIII.) that a f " pipe is the proper size. To FIND Size of CD. — We now have to find the diameter of the pipe CD. We found that the available head at C ought to be 24 feet, and there is a loss of head in the pipe CD of 10 feet, due to difference of level. Hence the available head is 90 - 10 - 24 = 56 feet. Therefore - . G2 = — x (2 + 3)2 = 45, H 56 which corresponds to a |" pipe. Since the pipes AC and CD are both of the same diameter, the joint at C would be made as shown in Fig. 532, instead of as in Fig. 531, and this would prevent the loss of head of 0'6 foot, and the loss would occur in the ^" pipe. Referring back to p. 274 it will be seen that the diameter of the pipe BC was assumed as Had a larger diameter been assumed the loss of head in BC would have been less, so that the available head at C producing the discharge in AC would also have been less, and conse- quently this pipe would have to be made bigger. On the other hand, the available head in CD would have been greater, hence this pipe would have been smaller ; that is, the service pipe would be smaller than the branches, which is not advisable. Thus the arrangement worked out is practically the best. To find the discharge from one tap when the otlier is closed.- — First, when tap A is open, it is clear that the available head is 90-29 = 61 feet, or, allowing for the various minor losses of head, say 60 feet. Hence from Table XVIII. 100 + 20 Fig. 532. -.G2: H 60 X G2 = 48. Thus or G = 4-9 gallons per minute. When tap B is open and tap A closed, we have to deal with two sizes of pipe. Let H, be the head at the point C, then 90 — 10 available head in the portion CD of the pipe. L H and similarly for the pipe CD L 100 H Hj^ will be the Hence for the pipe BC 30 G2 = S^.G2 = ■^1 (a), G2: 80 -H, G2 = 48 (6). Now the discharge G is the same in both pipes, (a) by Equation (6), Hence, dividing Equation WATER SUPPLY TO A GROUP OF BOUSES 30 5H, 277 whence 100 48(80 -Hj)' 3x 48(80-Hi) = 50Hi, ^ 3 X 48 X 80 ill = 194 = 59-5 feet. Inserting this value in Equation {a), 5 X 59-5 G2 = 30 9-9, or G = 3'15 gallons per minute. Of course we ought to get the same result from Equation (b\ namely — ^,^48(80- 59^^ 100 Water Supply to Eight Houses in a Street. Example 63. — Eight houses, as shown on plan in Fig. 533, are to be supplied with water from a branch main connected to the main at the point Fig. 533. A, i 2U0 —2U0- ■r -SAO *h—r20—> A H, ,1 0 Main, Hr5 Vertical scale 10 times horizontal scale Fig. 534. A. The levels of the branch main are shown in Fig. 534, and each house is fitted up as in Example 62. Find the diameter of the branch main capable of supplying each house approximately with 5 gallons of water per minute. On referring to Example 62 it will be seen that the highest tap was 29 feet above the main, that it was connected to the main by 120 feet of f" pipe, and that the discharge from it was 4-9 gallons per minute when the head in the main was 90 feet. It will also be seen that the discharge is almost the same when both taps are running. W^e will therefore ignore the lower tap in the present example. Size of Main — Half the taps opened simultaneously. — It is very un- likely that all the houses will require water at the same time, and a very 278 NOTES ON BUILDING CONSTRUCTION safe assumption to make is that half the number of houses require the full amount of water at the same time. Supposing, therefore, that the upper taps in houses 1, 2, 1', 2' are opened simultaneously and that an average of 4-9 gallons is running out of each, then, from Example 62, the head at A2 must be 61 feet more than the height of the tap, that is H2 = 61 +31 = 92 feet. Hence the available head in the portion AgA of the branch main is 92-5 -92 = 0-5 foot. Again, the discharge at Ag must be * 4 X 4'9 = 19*6 gallons per minute. L 2 • Hence the value of the expression g . (jt is 600 , — - X 19-62=460,000 nearly. 0-5 ' And referring to Table XVIII. it will be seen that a 4" pipe is the nearest market size, allowing for incrustation. Theoretically the branch main ought to be made to diminish after the junction to each house, but the practice is to maintain the same diameter, for some distance at any rate. All tipper taps opened simultaneously. — As an exercise let us inquire what the discharge would be supposing all the upper taps were open. Although it may appear paradoxical, the discharge from each tap will not be very much diminished. The direct solution of the question, however, leads to several cumbersome quadratic equations, and the best way is to work by approximation. Thus as a first approximation let us suppose that the average discharge per tap is 4-0 gallons per minute when all the taps are open, then the dis- charge at A^ must be 8x4 = 32 gallons per minute. Now from Table XVIII. we have for a 4" pipe ^. G- = 420,000. Hence 120 X 32^ II: 420,000 ' = 0-29 feet, which is the loss of head between A and A^. Hence H^, the head at A^, 92-5 - 0-29 = 92-21 feet since there is no difference in level. The discharge from the tap in house No. 4 will therefore be 2 48 X (92-21 - 29) ^4 = 120 ' or . = 5 gallons per minute. It follows that the discharge at Ag will be 32 -2x5 = 22 gallons. Hence 240 x 22^ 420,000 =^'^^' or Hg = 92-21 + 5 -0-28, = 96-93. WATER SUPPLY TO A GROUP OF HOUSES 279 Therefore discliarge in house No. 3 is 48(96-93 - 24) or Gg = 5-4 gallons per minute. The discliarge at will therefore be 22 - 2 X 5-4 = 11-2. Hence „ 240 x 11-2^ ^ H= 420,000 or H2 = 96-93- 7 -0-07, " = 89-86. Hence discharge in house No. 2 will be 48(89-86 -31) ^2 - 120 = 4-8 nearly. We still have the discharge from house No. 1, so that clearly the original assumption of 4 gallons per minute is considerably too small. As a second trial assume 4-9 gallons as the average discharge of each tap per minute, then repeating the above calculations we will find H4 = 92-5 -0-44 = 92-06, = 5 gallons. H3 = 92-06 + 5 - 0-48 = 96-78, G3 = 5 -4 gallons. H2 = 96-78 - 7 - 0-19 = 89-59, G2 = 4-8 gallons. = 89-59 - 0-04 = 89-55, G^ = 4-8 gallons. Practically, therefore, we get the same values as we did before, when only half the number of taps were running, and the reason is, that a small alteration in the head produces a considerable alteration in the discharge of a 4" pipe, but no practical difference in the discharge of a f " pipe. As a further exercise let us find the discharge in the pipe connecting house No. 3, Avhen only the corresponding tap is opened. We have for the f'P^P' ^ 2_48(H,-24) '3 120 and for the 4" pipe ^ ^ 420,000(92-5 + 5 - H, ^ 240 + 120 Therefore 420,000(97-5 - H„) 48(H3 - 24) = \ „ / „ 420,000\ 420,000 x 97-5 H3U8 + '—J = '—^ + 48 x 24, whence H3=97-5 very nearly, that is to say, that the flow of water from A to H3 is so little, that there is no appreciable loss of head. Therefore 28o NOTES ON BUILDING CONSTRUCTION , 48(97 5-24) ^3 - 120 ' = 5 "4 2 gallons per minute. When the pipes are new, the discharge will of course be greater, for instance in the last case, since the corresponding value from Table XVIII. for a pipe 0-75 inch diameter is 117. ^, ^ __ 117(97-5 -24) 3 120 G'3 = 8-4 gallons per minute. DiSCHAEGE FROM PiPES FLOWING PAETIALLY FULL. It has been shown by experiment that the mean velocity of flow of a liquid in a pipe is roughly proportionate to Vis, where E is the mean radius or hydraulic mean depth (see p. 262), and S is the slope of the pipe (see p. 264). An approximate formula would therefore be V = CVlS . . . (123), the units of measurement being the foot and the second, and C being a constant. Different values have been ascribed to C by various experimenters as follows : — Beardmore . . . 95, Downing . , . . 100. But Darcy makes C vary according to the diameter of the pipe, as in the Table given at p. 265. Neville's formula — namely, V=140^RS-11^IS . . . (124) — is considered to be more accurate, but as in practice the choice is limited to certain sizes of pipes, the additional accuracy is not worth the considerable complication introduced into the calculations. The student will observe that these formulae are similar in form to those given for pipes running full under pressure. Now for a pipe running full 7rD2 1 _D and ^ = Li So that D H from which it will be seen that the formulae for pipes running full are a special case of those now given. A distinction must, however, be made PIPES FLOWING PARTIALLY FULL 281 H between j- and S ; the former is the head of water divided by the length of the pipe, whereas S is the slope of the pipe, and may vary at different points. Thus in ^'ig- 535 if the pipe were running full we would have to substitute Fig. 535. H H ^ — ^Y- for ^ in Equation 110, and the velocity of the water would be the same throughout the pipe ; but if the pipe were only partially full, then for the portion AB and for the portion BC and the velocity would be greater in BC than in AB. Darcy's formula. — For the same reasons as those given at p. 267, Darcy's formula is preferred, and will be used in working out examples. To find the discharge we must multiply the velocity by the area of discharge. Let A be this area, then F = A.C^ES . (125). To find A in circular pipes we must sub- tract the area of the triangle NKM (Fig. 536) from the area of the sector NKM, thus— ^ j)2 D2 ^ ^^^^--T sin 4> ^ = 360 ^'^4 whence R = . J / 7r<^ sin ~V360 ¥~ D2 ■7r(f> 360 (126), • (127). D These expressions are given to show how the discharge could be calculated if necessary, but in practice what we require to know in connection with drain pipes is whether, when conveying the average quantity of sewage, the velocity is sufficient to keep the pipe clean, without flushing.^ Of course the size of the drain pipe depends on the maximum quantity of liquid to be conveyed. Velocity. — The question therefore is, knowing the diameter of the pipe and the discharge, to find the velocity. We have frqm (123) 1 See p. 283. 282 NOTES ON BUILDING CONSTRUCTION A But R = p' ■where P is the wetted perimeter expressed in feet. Further, F ^ = V Hence = . ^^^^ . S, , c^r . s or Y^=—^ . . . (128). jSTow we do not know P, and we must therefore assume a trial value P^, and thus we obtain an approximation (V^) to V. But F = A V where is the corresponding value to P^ obtained from Table XXIII. If F^ is greater than F, then is also too great or P^ is too large, and vice versa. Two or three trials ought to give a suf- ficiently accurate value of V. Example 64. — A 15-inch drain pipe, laid at a slope of -g-^o, is discharg- ing 0'2 cubic foot per second. Find the velocity. Assume = 0-6 x D, = 0-75 foot. „ 1102x0-2 Then from Equation 128 v^"' = -— — — — or Vj^ = 2'5 feet per second. P But from Table XXIIl. the corresponding value of A^^ when ^ = 0'6 is 0-034 X D2, = 0-053 square foot. Hence 1^ = 0-053 x 2-5, = 0-132 cubic foot per second, so that clearly P^ is too small. Assume as a second trial P2 = 0-7 X D, = 0-87 foot. . o 1102x0-2 Then V 3 ^2 -0-87 x 200' or V2= 2-4 feet per second. But A2 = 0-052 xD2, = 0-0813 square foot. Hence == 0-195 cubic foot. Vo is therefore a sufficiently near approximation. We will, however, try the next value of P from Table XXIIL, namely — PIPES FLOWING PARTIALLY FULL 283 P3 = 0-8D, = 1-0 foot. , 1102x0-2 Then 3 1-0x200 = 2-3 feet per second. But A3 = 0-075 X D2, = 0-117 square foot. Hence F3 = 2-3 X 0-117, = 0-27 cubic foot. Drain Pipes. A drain pipe must be large enough, to convey the maximum discharge, due to heavy rains, without running quite full, so that there may be no danger of the pipe being under pressure, and the velocity of discharge should be ascertained to see whether a heavy rainfall will flush the drain. The next step is to find the velocity of discharge when the pipe is only conveying the average quantity of sewage ; it will then be seen whether arrangements must be made for periodical flushing during dry weather. In the case of small drains (6 to 9 inches diameter) the velocity ought to be not less than 3 feet per second, and for larger pipes, in which the sewage is generally well diluted with water, the velocity ought to be not less than 2 feet per second, in order to dispense with periodical flushing.^ The method of calculation will be best illustrated by an example. Simple System of Drains. Example 65. — Fig. 537 shows two branch drains BC and BD con- nected to a main drain AB. The lengths of the pipes and the levels of the bottom of the pipes are also shown on the figure. The pipes are laid at an even slope between these levels. The discharges to be reckoned upon are as follows : — Cubic feet per second. Maximum. Average. Discharge in BC Discharge in BD 4 3 0-2 0-03 ^ "These velocities should be increased for small pipes, say for house drains under 6 inches in diameter, and for mains under 12 inches in diameter " (Hurst). 284 NOTES ON BUILDING CONSTRUCTION Find what size of pipes should be used, and whether any arrangement for periodical flushing ought to be provided. We will work out this example by means of Tables XIX. and XXIII. Pipe BG. — For the maximum discharge we have F = 4 cubic feet per second. H = 4-7 - 1-6 = 3-1 feet. L = 320 feet. Hence H-^ ~ 3-1 = 1650. On referring to Table XIX. it will be seen that this number corresponds to a 12-inch pipe, and that this pipe will run very nearly full when dis- charging the maximum quantity. To find the velocity of discharge we may consider that the jjipe is running full. The discharge in that case can be found from L „ jj-F2= 1850, 1850 X 3-1 ~ 320 ' or F = 4'23 cubic feet per second. Hence V. =4-23, 4 or, since D = 1 foot, 4-23 X 4 TT X 1^ = 5"4 feet per second. The velocity is therefore well above the limit for flushing. Next, to find the velocity when the average quantity of sewage is being discharged. Proceeding as in Example 64, assume P = 0-8D, then, since (Table I.) C= 109*5 for a 12" pipe, f T? Trq 109-52 X 0-2 X 3-1 from Equation 128 V,^ = , ^ 1 0-8 x 1 x 320 ' = 3-07 feet per second. Hence, since (Table XXIII. j = O-OTSD^, Fj = 0-075 X l^x 3-07, = 0-23 cubic foot. This is a sufficiently near approximation, and since the velocity is a little over 3 feet per second, there is no necessity to make any arrangements for flushing. Pipe BD. — For the maximum discharge we have F = 3 cubic feet per second. H= 13-8- 1-6 = 12-2 feet. L = 416 feet. Hence L o 416 , On referring to Table XIX. it will be seen that a 9" pipe will do. SIMPLE S YSTEM OF DRAINS 285 Let US inquire how full this pipe will run when discharging the maxi- mum quantity. Assume then (Equation 128) 2-2 X D = 2-2 X 0-75, 3_ 108^ X 3 X 12-2 1 ~2-2 X 0-75 X 416' == 8"5 feet per second. Hence, using Table XXIII., Fj = 0-699 X 0-75^ X 8-5, = 3*2 cubic feet per second. For Pg = 2-0 X D we find V9 = 8"8 feet per second, = 2*95 cubic feet per second. It can therefore be taken that the wetted perimeter is 2-1 x D, or 18-9 inches. The un-wetted perimeter is therefore 28-3 - 18-9 = 9-4 inches, and the angle subtended at the centre by an arc of 9-4 inches is 9*4 — — X 360 = 120° nearly. 28-3 ^ Hence, when discharging 3 cubic feet per second, the pipe will be filled to the depth shown in Fig. 538. Check ly Neville's forimila. — As a further exercise let us test the above result by means of Neville's formula (p. 264). We have, using Table XXIII., • 0-633 X 0-75 Fig. 538. 2-1 - = 0-225 foot. V = 140 7' 0-225x^-11 4I0 7 0-225 12-2 416' 11-36-2-06 = 9-3 feet per second, so that there is a fair agreement between the results. To find the velocity when the average discharge is flowing. 0-6x0-75, „„ 108^x0-03x12-2 then Assume P 0-6 x 0-75 x 416 = 1-32 foot per second, Fi= 1-32 X 0-034 X 0-752 = 0-025. Since the average discharge is 0-03 cubic foot per minute, the velocity just found is a sufficiently near approximation, and as it is less than 2 feet a second, it is clear that this drain would require periodical flushing during dry weather. The depth of liquid in the pipe for the average discharge is shown in Fig. 539. The calculations to find this depth are left as an exercise for the student. Pi'pe AB. — For the maximum discharge we have F = 4 + 3 = 7 cubic feet per second. H = r6. L=1283. Fig. 539. 286 NOTES ON BUILDING CONSTRUCTION 1283 Hence — — x 7^ = 39,300 nearly. 1-6 And from Table XIX, it appears that a 24-incli pipe is the nearest market size that will do. Assume P^^ = 2-0 x D, 111-52 X 7 X 1-6 then 2-0 X 2 X 1283 ' V, = 3-0, F^ = 3-0x 22 X 0-595, = 7-1 cubic feet nearly. The average discharge is 0-2 + 0-03 = 0-23 Assume P^ = 0*6 x D, then Vj'^ lll-52x 0-23 X 1-6 0-6 x 2x 1283 ' = 1 -44 feet per second, r^= 1-44 X 22 X 0-034, = 0-196. As a second trial, if we assume = 0-65 x D we will find 1-40 feet per second, = 1-40 x 22 X 0-042 = 0-236, / n\ which is near enough. / / \ \ This drain will therefore require periodical flush- / L 21^". X \ ing, more especially because the velocity is low even I I ,A, I I when discharging the maximum quantity. When \ \ ^/~..'\ j I discharging the average quantity the drain will be \\ / \/ / filled to the extent shown in Fig. 540, and it will be seen that the wetted perimeter is large in comparison with the amount of liquid, and this is one cause of Fig. 540. ^i^e lo^^ velocity. Egg-shaped Sewer. The wetted perimeter can be reduced by using an egg-shaped sewer as shown in Fig. 542, and the velocity of discharge will therefore be somewhat increased. The j)rincipal advantage, how- ever, is that the depth of the liquid is increased, which is im- portant in the case of sewage. Such sewers are therefore suitable when the slope is small, the maximum discharge large, and the average discharge small. Example 66. — Design an egg-shaped sewer suitable for the pipe AB in the previous example. Egg-shaped sewers can be, and are made of various forms, but the most usual form is as follows : — ^ Let (Fig. 541) N = diameter of bottom of sewer, M = diameter of top of sewer, O = radius of sides, K = depth of sewer, 1 See also Part II. EGG-SHAPED SEWER 287 then N = |, M = ^, 0 = K. Fis. 5-11. Fig. 542. In the present case we must, in the first place, ensure that the sewer is large enough to carry the maximum discharge without running full. We have already seen that a circiilar pipe of 24 inches is required, and it will be near enough for practical purposes if we make the area of the egg-shaped sewer equal to that of the circular sewer. Xow the area of an egg-shaped sewer of the form given above can be found from the formula Area = 0-525 x . . . . (129). And the area of a circle 2 feet in diameter is 3-14 square feet. Hence 0-525 X K2 = 3-14, K = 2-45 feet. Therefore the depth can be taken as 2' 6". Hence N= 10", M= 1' 8", and applying these dimensions we obtain the sewer shown in Fig. 542. Let us now see to what extent the velocity is greater than with the 2-foot drain pipe when the average discharge is flowing. The quantity flowing being small, it will probably not rise above the invert, so that we may regard it as flowing in a 10" pipe. Supposing that 10 = 1-2 X then and 12' lOO^x 0-23 X 1-6 X 12 1-2 X 10 X 1283 10^ = 1-5 X 0-216 X = 0-225, which is a very close approximation, and thus we see that the velocity is only very slightly increased. Check hy Nevilles formula.— To check this result by using Neville's formula we have . 102 288 NOTES ON BUILDING CONSTRUCTION Hence V = 140^ 0-15 x^^l-li;/ 0-15 xA|, = 1-92 -0-63, = 1 "3 feet per second nearly. The results therefore agree fairly well. It appears therefore that, at any rate in this case, an egg- shaped sewer does not increase the velocity to any material extent. JETS. It is sometimes necessary to know, in connection with fount- ains and hydrants, the height to which a jet of water will rise. When a stone is thrown vertically into the air it rises to a height depending on the velocity with which it was thrown. If the stone is not thrown vertically, but at an angle, it will describe a curve depending on the velocity and on the angle of elevation, i.e. the angle at which the stone is thrown. In the same way a vertical jet of water will rise to a height depending on the velocity with which it issues from the orifice (or nozzle), and an inclined jet will describe a curve depending on the velocity and the angle of elevation. Now it is shown in books on dynamics that in the case of a stone thrown vertically, and neglecting the resistance of the air, the height to which it will rise is given by ya " = 6—4 If the stone is thrown at an angle, then the cnrve described by it is repre- sented by the equation ^2 1/ = xtan B - 16'1 5-7), cos*^ vr where V is the initial velocity in feet per second, and Q is the angle of eleva- tion, as shown in Fig. 543. Fig. 543. To jini the range, or the distance from 0 where the stone strikes the horizontal plane, that is OA = R, put y = 0, then JETS 289 tan (9= 16-1 R E = TTTT • sin ^ cos 1d"1 (131). R Further, to find the greatest height the stone rises to, or BD, put V = '^^ tlien BD = /(-(max.) = ta^l ^ sin 6 COS 16-1 V2cos2^" 32-22 V2 sin2 ^ (max.) — 64-4 sin- ^cos^^. (132). These formulae are applicable to jets projected in vacuo, and to make allowance for the resistance of the air the simplest way is to multiply the initial velocity by a reducing factor, or, in other words, to calculate the path for a less initial velocity than the actual. The result is not quite accurate (see Fig. 544), but is near enough for practical purposes. The reducing factor dimin- I Full velocity Z Reduced velocity S Actual path Fig. 544. ishes as the ratio of the diameter of the jet to the head increases, and can be found from Table XXIV., which is based on experiment. Example 67. — A jet of water is issuing from a |^-inch nozzle with a velocity of 60 feet per second. Find : (a) Tlie height to which the jet will rise when it is thrown vertically. (h) The distance to which the jet will reach, and the greatest height it will attain, when thrown at the angles of 15° and 55° respectively. A velocity of 60 feet per second corresponds to a head of 60^ 64^ 56 X 12 H = 56 feet. Hence d being diam. of nozzle •= 1350. We can therefore take (see Table XXIV.) J = 0-91. Consequently for case (a), from Equation 130, (0-91 X 60) 64-4 46 feet. B.C. IV. 290 NOTES ON BUILDING CONSTRUCTION — 1- It should be observed that if the jet is perfectly ver- tical, and there is no wind, the water falls back on the jet, and thus reduces the height as shown in Fig. 545. Again for case (b), when the angle of elevation is 15°, we have from Equation 131 R = (0-91 X 60)^ sin 15° X cos 15' 16-1 = 46 feet ; and from Equation 132 (0-91 X 60f sin^ 15° ^(max.) — 64-4 Fig. 545. = 3-1 feet. Lastly, when the angle inclination is 55°, we have from Equation 131 (0-91 X 60f R = 16-1 87 feet ; sin 55° X cos 55°, and from Equation 132 ^(max.) (0-91 X 60) 64^4 = 31 feet. X sin^ 55° Issuing velocity of a Jet. — We must now see how to obtain the issuing velocity of a jet. If the nozzle offered no obstruction to the issuing stream of water, it is clear that the velocity would be that due to the head at the nozzle ; but the nozzle does offer an obstruction, or, in other words, causes a loss of head, the amount of which depends upon the shape of the nozzle. The reduction in velocity caused by the nozzle has been ascertained by experiment as follows : — A well-shaped nozzle, such as the one shown in Fig. 546, reduces the velocity to 0-97 X velocity due to the head. Fig. 548. JETS 291 1 to \ 4 " \_ 13 >' 2 5 " 1 3 1_ 12 JL_ 2 4 1 A converging nozzle, as in Fig. 547, reduces the velocity to 0-9 4 X velocity due to head. And a cylindrical nozzle, as in Fig. 548, causes the following reductions : — When the diameter is length . . . . 0-81 \ 0-77 f Multiplied by n.^TO / velocity due ^ 16 i to head. 0-68 j The reduced velocity obtained by applying the above is of course the initial velocity of the jet to be used in finding the height or the range. Example 68. — A 1" nozzle for a vertical jet is connected to the end of a 1" pipe, 300 feet long. The head at the farther extremity of the one-inch pipe is 100 feet, and the nozzle is 10 feet higher than this point. Find the height of the jet and the quantity of water that will be discharged. Let "V feet per second be the mean velocity of the issuing jet of water ; then if G is the discharge in gallons per minute, we have from Equation 122 •49' = 2-04x jij-x V, = 0-126 X V. Now if the nozzle is of a good shape (see Fig. 546) the velocity due to the head at the nozzle will only be greater than V. Hence if H is the head at the nozzle / y Hence or 0-97; =^^-^H' V2 = 60-7H. {~^\ =60-7H, VO-126/ ' G2=0-96H nearly. Again, we have from Table XVIII. for a 1" pipe, allowing for incrusta- tion and remembering that the nozzle is 10 feet above the farther extremity of the 1" pipe, 100-10-H G2 = 250 x 300 Hence 0-96H = ^^90-2^.H, 300 300 or H = 41 -7 feet. The ratio of H to the diameter of the nozzle is therefore 41-7x12 = 2000. Hence by interpolation from Table XXIV. the factor J is 0-82. from the above equation And since NOTES ON BUILDING CONSTRUCTION we have V2=60-7 X 41-7, _ (0-82)2 X 60-7 ^ 41.7 64-4 = ^6-4 feet. Lastly for the discharge we have G' = 0-96 X 41-7, or G = 6'3 gallons per minute. Water Supply for House and Jet. Example 69. — A jet, to rise to a height of 30 feet above the nozzle, is required for an ornamental piece of water. The water is to be obtained from a reservoir which is also to supply the house, the arrangements for which are the same as in Example 62. The various lengths and levels are given in Fig. 549. Find the diameters of the pipes that will be required, and also how much higher the jet will rise when no water is being supplied to the house. We must first find the head required to produce a jet 30 feet high. To do this we require to know the issuing velocity. Datum level. Vertical scale exaggerated S Unus. Fig. 549. 40x12x8 T neaiiv, 3 The head required at the nozzle will be about 40 feet, hence the ratio of the head to the f " diameter of the nozzle is H 'd ' = 1280. So that, referring to Table XXIV., it will be seen that the factor J can be taken at 0*90. Hence if V is the issuing velocity by Equation 130 3^^(0;90^ 64-4 or V = 49 feet per second nearly. If the nozzle is well shaped, the velocity due to the head will be 49 = 50'5 feet per second. 0-97 ^ Hence the head required to produce a jet of 30 feet with a good nozzle is 50-5'^ 6^4 It is to be observed that the head assumed (namely, 40 feet) for the purpose of calculating the factor J is very nearly the calculated head. Had 39-6 feet. JE TS— EXAMPLE 293 there been any considerable difference the calculated head would have been looked upon as a first approximation, from which a more accurate value of J could have been found. The next step is to find the discharge from the jet, which will be, from Equation 122, 2"04 X 49 X 3^ G = — . — = 14 gallons per minute nearly. On referring to Example 62 it will be seen that a head of 90 feet is required at the point B to obtain the necessary water supply to the house. Hence the available head from B to A is 90 + 12-39-6, = 62-4 feet. Applying, therefore. Table XVIII. we have 62-4 ' and to this number corresponds a 1;^" pipe. To find the diameter of the portion BC of the pipe we have the available head : 112-90-12 = 10 feet, and the discharge is : 14 + 5 = 19 gallons per minute. Hence '-1^^^=206,000 nearly, a number which corresponds in Table XVIII. to a 3|^ inch pipe. The jet will not be exactly 30 feet high, because the available diameters of pipes are not quite those required to produce a jet of this height under the conditions given. Working backwards we find — Loss of head from C to B, '^:^^^^ = 9-9feet, 208,000 and the loss of head from B to A will be 2^^^.60-5 feet. 810 Hence the head at A will be 112 -9-9 -60-5 = 41-6. The velocity due to this head is \''41 -6 X 64'4 = 51-8 feet per second, hence the issuing velocity will be 51-8 X 0-97 = 50-3. Therefore height of jet = Q'^Q' x 50-3^ _ ^^.^ 64-4 When no water is being supplied to the house, the jet will be a little higher. In this case the loss of head from C to B 5700 X 142 208,000 Hence the head at A will be = 5-4. 112 - 5-4 - 60-5 = 46-l. The ratio of the head to the diameter of the nozzle is 294 NOTES ON BUILDING CONSTRUCTION Hence from Table XXIV. the value of J is about 0-89. Therefore the height of the jet will be 46-1 X (0-97 X 0-89)2 = 34-3 feet. Strictly a slight correction ought to be made, because the issuing velocity of the jet is slightly increased. Hence the discharge will be greater — con- sequently the loss of head somewhat greater than calculated above ; but as the formulae used are only approximate it is not worth while going into such niceties. APPENDICES. APPENDIX I. Factors of Safety. The following Table shows the Factors of Safety recommended by various eminent engineers for application in several cases that arise in practice. Authority. Nature of Structure. Nature of Load. Factor of Safety. Remarks. Cfl-st Iron, U Dead 4 U Live or varyin*^ (3 Stress of one kind only u " 10 Equal alternate stresses TT U 1. Varying with shocks 15 of different kinds " Dead 3 to 4 6 to 8 Dead 3 i> Live 6 s » • Live 6 s Dead s Pillars 6 8 ]) subject to vibr3,tiou . . Live 8 S y transverse shock 10 "Wrouglit Iron. u General ..... Dead 3 u Live or varying 5 Stress of one kind only u 8 Equal alternate stresses u Varying with shocks 12 of different kinds R Dead 3 R vjirciers ...... 4 to 6 s Dead 3 8 6 B Bridges Mixed 4 Riveted joints .... Dead 4 U Roofs 4 8 Tension and compression bars . Live with shocks 6 88 4i to 7 According to proportion 8 Compression bars Dead 4 of length to least dia- meter Steel. U Dead 3 U Live and varying 5 Stress of one kind only u 8 Equal alternate stresses u Varying and shocks 12 of different kinds R Dead 3 R Live 4 to 6 C Mixed 4 Timber. u Dead u Live and varying 10 Stress of one kind only u 15 Equal alternate stresses u Varying and shocks 20 of different kinds R 10 8 Exposed to weather Dead 10 8 8 8 For temporary purposes 4 Brickwork and Masonry. U Dead 20 U Live 30 R Dead 4 to 10 S 20 B. Board of Trade. C. Commissioners. R. Rankiae. 8. Stoney. 88. Shaler Smith. U. Unwin. 296 NOTES ON BUILDING CONSTRUCTION The Factor of Safety determined upon for any particular case will depend upon what is accurately known of the str,ength and quality of the particular material to be used, upon the importance of the structure, etc. APPENDIX II. Wote on Equilibrium. When the forces acting on a body lie in a plane, there are three conditions of equilibrium, namely — 1. The algebraic sum of the resolved parts of the forces along some straight Hne is equal to zero. 2. The algebraic sum of the resolved parts of the forces along another straight line perpendicular to the former is also equal to zero. 3. The algebraic sum of the moments of the forces about any point is equal to zero. By the resolved part of a force along a straight line, is meant the effect the force produces along that straight line. Thus if a force is parallel to a straight line, the resolved part is equal to the force ; but if the force is perpendicular to the straight line, the resolved part is zero. When the force is inclined, the resolved part is equal to the force multiplied by the cosine of the angle included between the direction of the force and the straight line. Thus the resolved part of X along AB (the direction A to B being taken as positive). Fig. 29, is + X cos 6 and the resolved part of Y wdl be - Y cos . By the moment of a force about a point, is meant the magnitude of the force multiplied by the length of the perpendicular drawn from the point to the direction of the force. The moment measures the power the force has of turning the body it is acting upon about the point ; and if one direction of rotation (say like the hands of a watch) be taken as positive, then the reverse direction will be negative. Thus in Fig. 29 the moments of the forces X and Y about the point P will be - Xa and + Yb. ■< a >- Fig. 29. Fig. 30. As an example, let the conditions of equilibrium be applied to the case of a ladder resting against a wall shown in Fig. 30. Resolving horizontally, Rg - Rj = 0, ,, vertically, Ro-W = 0, Moments about A, Rj x 0 - R3 x 0 + W x a - Rj x 6 = 0. APPENDICES 297 From which it appears, by solving these three equations, that R2 = W, ^3 = ^ . W. Sometimes a solution is arrived at sooner hy taking moments about more than one point, instead of resolving. APPENDIX III. To draw a Parabola. Suppose the object to be to find the curve of bending moments for a Span A C B 1^ ii D Fig. 550. c I B. (15 1S\ Fig. 551, 298 NOTES ON BUILDING CONSTRUCTION uniformly loaded beam. The data given are the two points of support AB (or, in other words, the span) and the maximum bending moment at the centre CD ; that is, the three points A, B, and D (Fig. 550) are given, and it is required to draw a parabola through these three points. Produce CD to F, making DF = CD ; join AF and BF. Divide AF and BF into an even number of parts by repeated bisections, and number the points thus obtained as shown in Fig. 551. Then join 1, 1 ; 2, 2 ; 3, 3 ; etc. etc. These lines are tangents to the required parabola, which can then be drawn in. (The parabola has not been drawn in the figure, as it would confuse the construction, and the tangents sufficiently indicate the shape.) APPENDIX IV. To show that the Neutral Axis in a Beam passes through the Centre of Gravity of the Cross-section (see p. 44). For a rectangular cross-section. Let To be the stress in the outside layers of fibres (o'g' or^'j-', Fig. 62). the depth of the beam. X, the distance from the neutral axis of any layer of fibres situated above the neutral layer, such as c'd', Fig. 62. Ax the thickness of the layer. 6Aa;, the area of this layer of fibres (h being the width of the cross-section), x' and 6A«' the same for a layer such as e'f. Then the total tension over the cross-section of layer c'd' is, by assump- tions 2 and 3, p. 44, X X rJ)Llx. Hence the sum of the compressions above the neutral axis = — ^ZjXIXx, a ' and similarly the sum of the tensions below the neutral axis = — f-2a;'Aa;'. • a But for equilibrium the tensions must be equal to the compressions. Hence 2xAcc — Sx'Acc' = 0. If now X be the distance of the centre of gravity from the neutral axis, we have, by taking moments about this axis, «(62Ax + b2Ax') = &2ccAx - h^x'Ax', = 0. that is, a; = 0) or the neutral axis passes through the centre of gravity. The same reasoning can be applied to a cross-section of any shape, with the difference that the breadth of each layer, instead of being constant, varies. APPENDIX V. Diagrams of Shearing Stress. These, for beams supported at the ends, are often differently arranged from those given at pages 58 to 60. APPENDICES 299 By many writers the shearing stresses acting in one direction, say tlius are called positive, and those acting in the other direction mgative, or vice versa. The diagram of positive shearing stress is shown on one side of the line representing the beam, and that for the negative shearing stress on the other. Thus in Fig. 263, p. 173, which represents the shearing stress caused by an uniformly distributed load, the diagram for the shearing stress of this direction is below the line, that for the shearing stress ■^y^- is above the line representing the beam, instead of the whole diagram of shearing stress being below the line as in Pig. 90, p. 59. Fig. 255, p. 167, is similarly arranged for a distributed and a concentrated load, both diagrams from A to D being below the line, and both from D to B above the line, instead of being as in Fig. 91, p. 59. Similarly in Fig. 87, p. 58, half the rectangle denoting the shearing stress from one abutment to C might be below the line and the other half above the line. In Fig. 4, Appendix VI., the diagi-am of shearing stress with direction may be called positive, and is above the line from A up to the point where the direction changes to from this to B it may be called negative, and is below the line. The area of the shearing stress diagram between any point and either abutment is equal to the bending moment at the point. This is evident in Fig. 82, p. 57, where (P being the point), the area of the shearing stress diagram at P (being the rectangle under PB)=:PB x W = ^Yx and the bending moment at PMp = "\Va;. In Fig. 83, P being the point, the area of shearing stress diagram at P (being the triangle under PB) = xx= and Mp = irax-=— • When the portions of shearing stress diagram between the point and the abutment chosen are of different signs or directions, the difference of these areas must be taken. Thus in Fig. 90, if the area be taken of the diagram under PB, we have to take the area of the triangle under BC minus the area of the triangle under PC _ivl I ^ _ wlx lOX^ ~~2 2" and wl X Mp= — xx-wxx- _wlx wr? ~~2 2"' By working out other cases the student will find that the rule given above always holds good, and it is sometimes very useful. APPENDIX VI. Graphic method of finding the Bending Moments and Shearing Stresses in a loaded Beam. In Fig. I AF is a beam carrying the loads W3, W^, at the points BCDE, through which draw vertical lines downwards as dotted. Draw another vertical line YZ (Fig. 4) at a convenient distance. Along 300 NOTES ON BUILDING CONSTRUCTION YZ lay off W^, Wg, W^, on any convenient scale of loads. Take any point 0, which is called the "j;o/e," and draw the lines «, 6, c, d, e, joining to it the extremities of the lines indicating the loads to the pole. Bending Moments. — Starting from any convenient point X (Fig. 2) verti- cally under A, draw in succession a, h, c, d, e, parallel to a, b, c, d, e of Fig. 4, respectively— intercepted between the verticals from A to F of Fig. i, and terminating at V on the vertical from F. Join XV, Fig. 2. The polygon thus formed is called the "Funicular pohjgon." In Fig. 4 draw the horizontal line oH, and the dotted line oN — the latter parallel to XV. Then YN will represent and NZ will represent Ep measured on the scale of loads. Moreover in Fig. 2 the ordinate UT of the "funicular polygon" vertically under any point P measured on a proper scale represents the bending moment at P. Take a foot as the unit of the linear scale, and a lb. as the unit of the scale of loads. Then if Ho is equal to 1 foot on the linear scale, UT on the scale of loads will represent the number of foot pounds of the bending moment at this point. If, however, Ho is more than one foot of the linear scale in length, then the length UT on the scale of loads must be multiplied by Ho to find the number of foot pounds of the bending moment. Of course any other units may be substituted for the foot and pound — for example, the inch and ton. Shearing Stresses. — These are shown in Fig. 3, the construction of which is obvious. The dotted horizontal lines are drawn through the extremities of the loads set off along YZ, and are intercepted between the verticals from Fig. i as shown. A dotted horizontal line is drawn also through the lower extremity of Ra on Fig. 3, and at this point the shearing stress changes from to or + to - . The shearing stress at any point of AF is equal to the portion of the vertical from that point intercepted between the upper and lower lines of the shearing stress diagram (Fig. 3) . Thus SQ is the shearing stress at the j)oint P. Example.— In Fig. 1 take AF = 20 feet. AB = 4 feet. 80 lbs. BC = 2 „ W2 = 120 CD = 6 20 „ DE = 4 „ 60 EF = 4 Take the linear scale to be 10 feet = l inch, and the scale of loads 120 lbs. =1 inch. Lay off AB, BC, etc., on Fig. i on the linear scale, and Wi, Wg, Ws, "W4, Fig. 4 on the scale of loads. Complete the figures as described above. Measure oH. It equals 10 feet on the linear scale. Measure YN on the scale of loads. It measures 168 lbs., which should be equal to Ra. NZ measures 114 lbs., which should equal Er. To check this by calculation (see page 60) we have ^ 80x16 120x14 20x8 60x4 Ra = 1 h - 1 20 20 20 20 3360 Similarly Rf = 114 lbs. =168 lbs., which agrees. APPENDICES 301 100 50 Scales. 10 Lifiear Scale 10 ft. = i inch. ) 100 Scale of moments 1200 ft. lbs,= \ inch. 20 feet Scale of loads 120 ids. = 1 inch. ICKDO 500 O 1000 200 pounds 2000 foot pounds 302 NOTES ON BUILDING CONSTRUCTION To find the scale of bending moments we have Ho in feet on the linear scale x number of lbs. 2}er inch on load scale = the number of foot pounds in one inch on the scale of moments. 10 ft. X 120 = 1200 foot pounds .•. on scale of moments 1200 ft. lbs. = 1 inch. Draw the scale. Then to find the bending moment at any point P, draw PUT a vertical inter- cepted by the upper and lower sides of the polygon. UT will be the bending moment. Measuring UT on the scale of moment we find it = 500 foot pounds. ^ Shearing 5' '» • • uniform ~ 32md^ Beam fixed both ends centre A ^ T^'^Ebd^ A )) )) ... uniform = s\Ebd^ A Beam supported one end, fixed other . centre A )) )) " uniform APPENDIX X. Relative Strength of Beams of different Sections. The following are given as they may be of interest to the student, but they are of no practical value. APPENDICES Square Beam and Cylindrical Pole of Diameter equal to ^ SIDE OF Beam. _ M of sqimre learn =fol _ 12, yo~ d 2 6 In a square beam i=c?.-. M = -:^ ... 6 M of cylindrical pole='^= — — 4 4 8x4 _ /oX3-14xc;g 32 = . . . n. from I. and II. M square beam : M circular pole : : \fod^ : i-^fod^ ° O : : Wod^ ■■ W : : 10 to 6. The Strength of a Square Beam side Vertical is to that of a Square Beam with Diagonal Vertical :7 :5 A Triangular Beam of the same area and depth as a Rectangular Beam is only one-half the strength of the latter. The strength of a cylindrical pole to that of a square beam of the same area is as 1 to 1-18. The strength of a cylindrical beam to that of the strongest beam that can be cut out of it is as 1'53 to 1. APPENDIX XI. Continuous Beams of Uniform Section loaded uniformly with Points of Support equidistant from one another and in the same Straight Line. On pp. 74, 75 are shown the forms assumed by a uniformly loaded girder of uniform strength extending over two spans. B.C. IV. X 3o6 NOTES ON BUILDING CONSTRUCTION In practice, however, small continuous beams are generally of uniforik section. When such beams are fixed at the ends and continuous over one or more supports, each portion may be treated as a beam fixed at both ends (see p. 72, Case 2), and the moments of flexure, etc., will be found accord- ingly. When, however, a beam is continuous over one or more supports, and Total load = 2iwZ. C Fig. 1. 3wl Ra = Ma = 0 , / lOwl Mc = wZ2 Mb = 0. supported only at the ends — which is sometimes the case in practice — the Fig. 2. Total loa.i=3wl. C D .„,./.„-± /—J Ma = 0 Mc = t^' ^iD = i~ Mb = 0. resulting stresses are different, and are very difficult to find entirely by calculation. Total load=4wZ. Fig. 3. — -/- RA=iiwl Rc= Ma = 0 Mc=?-^ Me = ^— Md = |— Mb = 0. Reactions and Bending Moments. — Figs. 1 to 3 show the reaction and the bending moment at each point of support for beams continuous over from 2 to 4 spans. Each span = I Load on each span = wZ Deflection. — This will be a maximum in the two end spans, and may be obtained for those from Case 3, p. 73, and Appendix VIII., though if the two end spans are not reduced their inner ends will not be perfectly fixed and the actual deflection will be in excess of that calculated. Points of Contrary Flexure. — These, if required to be known, may for each, of the intermediate spans be found as in the case of a beam fixed at both ends (p. 72), or for each of the end spans, as in the case of a beam fixed at one end and supported at the other. To find the Bending Moment from the Reactions. — When the reactions are known, the bending moment at any point may be found by the ordinary method described in Chapter III. APPENDICES 307 Thus in Fig. i The bending moment at a section infinitely close to C will be M. = R X AC - X - C A 2 = %wl X I - — « 2 = - \wl'^. Again in Fig. 3 Mo = \\wl Beams continuous over several Supports. The following table gives the reaction on each support caused by uniformly loaded beams of uniform section continuous over several supports. Uniformly Loaded Beams continuous over several Supports. Table of Distribution of Loads. No. of Span. Number of each Support, and Load borne by it in terms of W. or of load on each span. 1st 2d 3d 4th 5th 6th rth 8th 9th 10th 1 2 3 4 5 6 7 8 9 1 3 ? 4 \ts 1 1 1 5 4 1 TlTi 6 6 15 2 2 0 9 1 T 1 0 1 1 nr 3 2 4 3 lis TT5T 16 1 TTT 4 4 0 6 0 1 "5 "SIT 3 1 1 TiT 2 6 ■iTS- ■Si 10s TiTJ 13 7 TTT 374 TTS- S 1 1 4 TTT 3 2 IS 37 10 6 TTTT 14 3 TTT 3 9 2 Tg¥ 6 3 5 TTiT 1 1 Tg- 4 3 TS- 10 8 TtTf 1 4 3 TTT 3 8 6 ■j-rs- 5 2 9 1 5 118 TUT 13 7 TTT 3 9 2 Trs 5 2 9 4 1 TOT 161 TTT 3 7 4 TT? 5 3 5 5 6 TTT 44 0 5 11 15 2 6 0 1 200 Illustration. — The principal rafter of the roof dealt with in Chapter XII. is treated in j)age 214, first as if it were in two portions each supported at both ends, and next as a beam continuous over two spans. In both cases it is uniformly loaded. In this example the maximum bending moment is the same under either supposition, being in the case of each supported portion -— - in the 8 8" support. In the case, however, where a rafter extends over several points of support (see p. 202, Fig. 343) the maximum bending moment to which it is subjected will be very different if it is a continuous beam, compared with what it would be in a jointed and supported beam. centre, and m the continuous beam at the intermediate point of 3o8 NOTES ON BUILDING CONSTRUCTION Thus in Fig. 4, if it be assumed that there are joints at the points of Fig. 4. support, then AC, CD, DB are supported beams (see Case 7, p. 33), and the maximum bending moment is — at the centre of each. If AB is assumed to be continuous, then (see Fig. 2 above) the maximum bending moment is at C or D, i.e. -| x - being less than — , and therefore if the rafter is really continuous, all the points being accurately in a straight line a less scantling will do for it than if it were jointed and supported.^ APPENDIX XII. Method of finding the Centre of Gravity of an unsymmetrical Cross Section of a Girder by Calculation. Divide up the cross section into rectangles all parallel to each other and A to the bottom edge of the section. Rule. — Multiply the area of each rectangle by the distance of its centre of gravity from the bottom edge, and add these products together. Divide by the whole area of the section ; the quotient is the distance of the centre of gravity from the bottom edge. Example. — Thus the cross-section of the cast iron girder shown in the figure can be divided up into three rectangles, and the products are 4x 1 X 19-5 + 18 X 1 X 10+ 16 X 1 x 0-5 = 266. The whole area is 4x1 + 18x1 + 16x1= 38. Therefore distance of CG from bottom edge = — = 7 mches. Since the section is symmetrical vertically, the centre of gravity is determined as shown in the figure. Should, however, the figure be unsym- CG. ^ In reality the case lies between the two, because C and D are not rigid supports. APPENDICES 309 metrical in all directions, the above process must be repeated with reference to an axis perpendicular to the first one. Thus in the present case, to find the distance of the centre of gravity from the axis AB the products are 4x1x8 + 18x1x8 + 16x1 x 8 = (4 + 18 + 16)8. Hence distance of centre of gravity (4 + 18 + 16)8 ^. ^ = 77, — T77~ = 8 inches. 4+18 + 16 APPENDIX XIII. Graphic Method of finding Centre of Gravity of unsymmetrieal Cross Section of Girder. Draw the polygon OPQR having its sides parallel to o-pcp respectively, and intercepted between verticals drawn through the centres of gravity of ABC. From the intersection of R and 0 at Y draw a vertical. The point at which this cuts the centre line of the girder will be its centre of gravity. The figures show that this agrees with the calculations above (App. XII.) NOTES ON BUILDING CONSTRUCTION APPENDIX XIV. Moment of Inertia (I) of Various Sections.^ Section. (1) Rectangle (breadth h, deptli d) . . . . . (2) Isosceles triangle (base h, heiglit d) .... (3) Circle (radius r) ....... (4) Annidus (E, outer radius, r inner radius) (5) Semicircle either thus \^ or thus ^'^"^ Value of I. hd:^ T2 36 ^(E4-r^) 0-1 Ir* (6) T (see Fig. i). (7) I girder divided into rectangles as in Fig. 2 — neutral Fig. 1. Fig. 2. GENERAL RULE FOR ANY SECTION MADE UP OF A NUMBER OF SIMPLE FIGURES. a. Find the moment of inertia of each of the simple figures about an axis, traversing its centre of gravity parallel to the neutral axis of the complex figure. b. Multiply the area of each simple figure by the square of the distance be- ttveen its centre of gravity and centre of gravity of the whole figure. Add the results so found (by a and b above) for the moment of inertia of the whole figure. (Rankine.) I for Balled Beam. — Applying No. 7 formula to find I for the section of the rolled iron girder, p. 83, we have T-V{|x93 + 4(103-93)}, = tV{^ X 729 + 4(1000 - 729)}, = 120-7. ^ In each case the moment of inertia is taken about an axis parallel to the neutral axis passing through the centre of gravity of the section. TT is the ratio between the circumference and diameter of a circle, and its numerical value is 3'1416. APPENDICES I for Cast Iron Girder. — Applying Rankine's rule to the section of the „ ■< — 4 — > cast iron girder, Fig. 3, we have a. I of upper flange I of web I of lower flange 4 X 1^ 4 ~ 12 ^ 12' 1 X 183 5832 12 16 X 13 16 ~ 12 ~ 12' Total 5852 12 487-6. h. Area x square of distance of centre of gravity from the neutral axis Of upper flange = 4x1 2-5'^ = 625, Web = 18 x 32 = 162, Lower flange = 16 x 6"52 = 676, Total of a and & = 1950 = I. so ■16 Fig. 3. It should be observed that the thickness assumed for the lower flange, namely, 1 inch, is rather little for the width, but it simplifies the example. APPENDIX XV. rormulsB for finding approximately the Weight of Girders. The weight of a girder can be found with a fair degree of accuracy by either of the following formulae, the first being, however, the more accurate. Weight of girder in tons W X = ^^fz — (Prof. Unwin's formula), W X L ~~560~ (Anderson's formula), where W = distributed, or equivalent distributed, load to be borne by the girder, in tons. L = span in feet. C = a factor whose value is from 1400 to 1500 for small girders, and from 1500 to 1800 for large bridges. rc = stress in compression boom (generally about 4 tons per square inch). D = eft'ective depth of girder in feet. APPENDIX XVI. Bules for drawing Maxwell's Diagrams {see p. 183). The diagram of the truss itself is called the frame diagram. The diagram of the forces is called the stress diagram or "diagram offerees." 312 NOTES ON BUILDING CONSTRUCTION In the stress diagram lines are drawn parallel to the several lines, of the frame diagram, thus 1. Draw each hiown force in succession in the direction in which you know it acts. 2. ^ Then from the two " open ends " draw the two unknown forces parallel to their directions on the frame. The first unknown force working in this direction -> is at the end of the last known force, that is, at the end to which the arrow mentioned in next paragraph points. 3. Place arrows on the stress diagram beginning with the known forces, and make them all follow each other pointing in the same direction. 4. By transferring these arrows in imagination back to the frame, and considering them with reference to the point in q^uestion, you know whether each member is in compression or tension. The above rules are applied to each joint of the frame in succession. Example.— Thus in the example illustrated in Figs. 305 to 308, Plate II. Fig. 1. Frame Diagram. Fig. 2. Stress Diagram. we know (pp. 180, 181) that \V = 7 cwts. Ra = 5 cwts. Rb=2 cwts. • Commencing at joint A in the frame (Fig. i) draw for the stress diagram the known Ra = 5 cwts., and mark upon it the arrow head showing its direction, i.e. upwards. Then working in the direction draw from its end q, to which the arrow points, the APPENDICES 313 unknown T parallel to AB of the frame, and from the other end 0 draw Ci parallel to AC. These intersect at ^, and by measurement 22^ = 3-94 cwts. =T 0|i = 6-4 cwts. =Ci. ISTow place arroM's on the stress diagram (Fig. 2). We have marked the arrow on Ra pointing upwards — following round in the same direction the arrow on T points towards p and that on Ci towards O. Transferring these to the frame we see that Ci is in compression pushing towards A, and T in tension pulling from A. Proceeding to the next joint 0 on the frame, we have Ci and W known and C2 unknown. On the stress diagram we have Ci already drawn. Draw W vertically = 7 cwts. , and draw a line from the end s of W parallel to C2 ; this will be the line sO. We know the direction of the arrow for W is downwards. Following on in the same direction the arrow for will point toward O and that for Ci toward p. The latter is in a different direction from what it was before, because Ci is now in com- pression toward C. Transferring these arrows in imagination to the frame, we find Cj C2 both in compression. We come now to the last joint, B, and have Rb Co T all known. Draw Rb = 2 cwts. and a line T = 3 "94 cwts. from s parallel to AB of frame. This will be the line sr. Again the arrows are placed — on Rb pointing upwards, on C.2 toward s, and on T towards r. Transferring these to the frame we find Cj in compression toward B, and T in tension from B. If a line drawn through s parallel to AB and equal in length to 3 "94 cwts. did not come exactly to r, it would be certain that some mistake had been made in the drawing. In such a case the diagram is said not to "close" properly, and this property of not closing when there is anything wrong in the diagram is very valuable, because the error thus shown to exist can be investigated and corrected, whereas, in using formuloe there is often nothing to show the existence of an error, which remains undetected and makes the result incorrect. The non-closing may be due only to inaccurate drawing. If this is suspected, the stress upon one or more bars may be checked by Ritter's method (see p. 185). APPENDIX XVIL Bow's System of lettering Maxwell's Diagrams {seep. 184). The following words quoted are from Mr. Bow's own description of liis system.^ " This plan of lettering consists in assigning a particular letter to each enclosed area or space m, and also to each space (enclosed or not) around or bounding the truss, and attaching the same letter to the angle or point of concourse of lines which represents the area in the diagram of forces. Any linear part of the truss or any line of action of an external force applied to it is to be named from the two letters belonging to the two spaces it separates : and the corresponding line in the reciprocal diagram of forces which represents the force acting in that part or line, will have its extremities defined by the same two letters." Example."^ — Figs, i and 2 give an example of a truss and a diagram of forces so treated. ^ From Bow's Economics of Constructio^t. ^ Ibid. , but condensed. 314 NOTES ON BUILDING CONSTRUCTION The spaces in the truss are lettered ABC, those around the truss DEFGH. These letters are attached to the corresponding points of concourse in Fig. Fig. 1. Truss. Fig. 2. Diagram of Forces. Each line in the truss is lettered according to the letters of the spaces it separates. Thus the line separating spaces A and F is AF and is shown as AF on the force diagram below. Similarly the line separating spaces B and C is BC. If any triangle in the truss be taken, such as A, the stresses on its sides as shown on the force diagram will be found to radiate from A. Thus, in Fig. 2, AB, AF, AD radiate from A. Examples of the application of Bow's method to girders will be found at p. 194. APPENDIX XVIII. Calculation of the Stresses upon the Iron Roof (Example 42, p. 209) by the Method of Sections. Referring to Example 42, p. 209, we now proceed to find the stresses by means of the method of sections (see p. 185). Stresses DUE TO Permanent Loads. — Commencing as before at joint A (Fig. 3565 Plate III.), we take the section shown in Fig. 398, which cuts through only two bars ; therefere to find the stress in AF we can take moments about any point in AD, round m for instance. Ra = 1'17 tons (see p. 209). We find by measurement that the lever arm of Ra is 8 '53 feet, and of the stress in AF, 2-39 feet. Hence 8-53 X 1-17- 2-39 xS^F = 0, or Saf — 8 -53x1 -17 2-39 = +4-16 tons. Similarly to find Sad we can take moments about any point in AF. for instance, lever arm of Ra =10 "So feet, Sat>= 2-72 feet. We find. APPENDICES 315 Hence 10-35 xll7 + 2-72xS^=0, or Sad= ~ 4 "45 tons, so that AD is in compression. To find Sdf we can take the section shown in Fig. 399, and the lever arms are marked on the figure. Taking moments about A we have 10 X 078 + 10-78 xScF = 0, SDf= -0-72 ton. H=l-17 Fig. 399. Fig. 400. The same section will enable us to find Sdo ; the lever arms are, as shown in Fig. 400, for Ra, 11-07 feet, Wi, 1-04 „ Snc, 2-90 „ and therefore whence 11-07 X 1 -17 - 1-04 X 0 -78 + 2-90 x 8^,^,= 0, S„„= -4-20 tons. Lastly, to find Sfg and Sfc we take the section shown in Fig. 401, and obtain the lever arms as marked, whence 20 X 1-17 - 10 X 0-78 - 6-67 X 8^0 = 0, K- --20'0" Sf„= +2-34 tons. For Syo we must take moments about the intersection of FG and DC, thus 3-25 X 1-17 + 6-75 x 0-78 - 4-69 x S^^, Spo= +1-93 ton. On examination (see Table H) it will be found that the two methods do not in all cases give exactly the same values for the stress. The differences are due to slight errors in the drawing of the diagram and in measuring the lever arms. Stresses due to "Wind Pressure. — It does not appear necessary to describe the working of this method again in detail, and therefore the various sections to be taken are simply marked on the figures, together \vith the turning points and corresponding lever arms ; and the calculations are given below for each stress. Case 1. Wind on left — Reactions ■parallel to loind pressure. — S.^} : Section 1, 1, Fig. 402. Turning point a. 2-69 x 8^0 + 10-0x0-88 = 0, Sad= -3-27 tons. Fig. 401. 2-69 n'O-' Fig. 402. Fig. 403 31 6 NOTES ON BUILDING CONSTRUCTION Saf: Section 1, 1, Fig. 403. Turning point I. -2-61xSaf + 1O-0x0-88 = 0, Saf = +3-38 tons. Sdf : Section 2, 2, Fig. 404. Turning point A. AD xSpF + ADx 0-96 = 0, ScF= '0-96 ton. That is, the compression in DF is equal to the wind pressure acting at D. Fig. 404. Fig. 405. Sdc : Section 2, 2, Fig. 405. Turning point F. 2-9 xSdc + IO-Sx 0-88 = 0, Si,c= -3-28 tons. Thus we see (allowing for errors of measurement) that Sad — Sec- Fig. 406. Sfo : Section 3, 3, Fig. 406. Turning point c. - 4-65 xSfo + 3-62 X 0-88 + 7-18 X 0-96^0, Sfc= +2-17 tons. Fig. 407. APPENDICES 317 Sfo : Section 3, 3, Fig. 407. Turning point C. - 6 -67 X Sro - 10 -8 X 0-96 + 21 '6 X 0 -88 = 0, Sf.o= +1'30 ton. Fig. 408. Section 4, 4, Fig. 408. Turning point G. - 2-9 xScE- 9-81 X 0-56 = 0, Scj.= -1-89 ton. Fig. 409. ScG : Section 4, 4, Fig. 409. Turning point d. 4-65 xSco- 2-62 X 0-56 = 0, SeQ= +0-32 ton. 0-56 Fig. 410. Fig. 411. SoE : Section 5, 5, Fig. 410. Turning point B. -10-8xS<,E-0x0-56 = 0, ScE-0. Sob : Section 5, 5, Fig. 410. Turning point E. + 2-78 X Sob- 7-8 X 0-56 = 0, S(,3=+l-57 ton. Seb: Section 6, 6, Fig. 411. Turning point G. - 2-9 xSeb- 9-81 X 0-56 = 0, Seb = - 1 -9 ton. 0-56 31 8 NOTES ON BUILDING CONSTRUCTION Again we find slight discrepancies between the vahies found by the two methods, and, as before, these discrepancies are due to errors in measurement. Case 2. Wind on right — Beactions parallel to normal wind pressure. — The stresses can be deduced from Case 1. Case 3. Wind on left — Reaction at free end vertical. — It should be noted that Pa and Ra being no longer exactly opposite in direction (as they were in Case 1), we cannot now use their difference only, but must introduce them separately into the equation. 1 Fig. 412. Fig. 413. Sad '• Section 1, 1, Fig. 412. Turning point a. + 2-69 X Sab -10-0 X 0-48 + 9-58 X 1-45 = 0, Sad= -3-38 tons. Saf : Section 1, 1, Fig. 413. Turning point h. - 2 -ei X Saf - 10 -0 X 0-48 + 9 -92 x 1 -45 = 0, Saf= +3 "67 tons. 2 Fig. 414. Sdf: Section 2, 2, Fig. 414. Turning point A. AD xSdf + ADx 0-96 = 0, Sdf= — 0 96 ton, which is the same value as already obtained under Case 1. In fact, since this stress does not depend on the reaction at the abutment, but only on the load at the joint (as is evident from the equation of moments), it follows that it is unnecessary to recalculate it each time. Sdc : as in Case 2. Scd = Sai,= -3-33 tons. s APPENDICES Syc : Section 3, 3, Fig. 415. Turning point c. - 4 -65 X S,c+ 7-18x0 -96 + 3-51 x 1 -45 - 3 -62 x 0 -48 = 0, Spo= +2-2 tons. Sfg : Section 3, 3, Fig. 416. Turning point C, - 6 -67 x Sfo - 10 -8 X 0 -96 - 21 -6 x 0 -48 + 21 -4 X 1 -45 = 0, S,o=+l-54. Fig. 417. Fig. 418. ScB : Section 4, 4, Fig. 417. Turning point G. - 2-9 xScE- 11-04 X 0-52=^0, ScK = -1-98 ton. Sco : Section 4, 4, Fig. 418. Turning point d. + 4-65 X Sco - 3-40 X 0-52 1:^0, Sca = +0-38 ton, Sro could also be found very easily from this section. Fig. 419. Fig. 420. Sgk : Section 5, 5, Fig. 419. Turning point B. - 10-8 xSqe- Ox 0-52 = 0, Sge=0. Sgb : Section 5, 5, Fig. 420. Turning point E. + 2-75 xSgb- 10-0x0-52 = 0, S<,B= +1-88 ton. Fig. 421. S^E : Section 6, 6, Fig. 421. Turning point G. -2-9xSbb-11-04xO-52 = 0, Seb= - 1 -98 ton. 320 NOTES ON BUILDING CONSTRUCTION Case 4. W ind on right — Reaction at free end vertical. Sad : Section 1, 1, Fig. 422. Turning point F. 2-9 xS^„ + 5-40 X 0-87 = 0, S^„=: -1-62 ton. Fig. 423. Section 1, 1, Fig. 423. Turning point D. - 2-75 X Saf + 2-65 X 0-87 = 0, Sai.= +0-84 ton. Sdj. ; the stress in DF is clearly 0. Sot '• Similarly, as in Case 3, Sbc = Sai,= -1-62 ton. Fig. 425. Section 3, 3, Fig. 424. Turning point c. -4-65 xSi,o + 0-88x 0-87 = 0, Si,c= +0-16 ton. Section 3, 3, Fig. 425. Turning point C. -6-67x8^0 + 5-26x0-87 = 0, S™= +0-69 ton. Fig. 426. Fig. 427. ScE : Section 4, 4, Fig. 426. Turning point G. - 2 -9 X ScK - 0 X 0 -96 - 11 -04 X 1 -26 + 10-8x0 -48 = 0, ScE= - 3-0 tons. Sco : Section 4, 4, Fig. 427. Turning point d. + 4-65 xSco- 7 -18 x0-96 + 3-62x0-48 -3-37 x 1-26 = 0, ScG= +2-02 tons. APPENDICES 321 Fig. 428. • Fig. 429. : Section 5, 5, Fig. 428. Turning point B. - 10-8 xSoE- 10-8 X 0-96 = 0, Sge= ^ 0'96 ton. Sqb : Section 5, 5, Fig. 429. Turning point E. + 2 75 X Sob + 10 -8 X 0 -48 - 10 -0 X 1 -26 = 0, Sgb= +2-7 tons. I Fig. 430. Sbb : Section 6, 6, Fig. 430. Turning point G. - 2 -9 X - 11 -04 X 1 -26 + 10-8x0 -48 = 0, Seb= - 3'0 tons. APPENDIX XIX. Centre of Pressure in Masonry Joints [see p. 219). It can be shown by an investigation, similar to, but more complicated than that given at p. 219, that the position of the centre of pressure for a joint of any section -svhatever, when the pressure vanishes along one edge, depends on the moment of inertia of the section. Rankine ^ gives the result as follows :— Let A denote the area of the joint. y, the distance from the centre of gravity of the figure of the joint to the edge farthest from the centre of pressure. h, the total breadth of tlie joint in the same direction. I, the moment of inertia of that figure, computed as for the cross section of a beam relatively to a neutral axis traversing the centre of gravity at right angles to the direction of the deviation to be found. S, the deviation to be found. Then S = — . Ay 1 A Manual of Civil Engineering, by W. T. M. Rankine, F.R.S., etc v 378 10th Edition. ' B.C. — IV. V 322 NOTES ON BUILDING CONSTRUCTION If the joint is rectangular, as shown in Fig. 43 1, p. 218, and the centre of pressure is situated in GH, then : A = BD.GH, GH y = ^, h = BD, I = — .BD.GH3, 12 ' 8 = ^ -GO. 2 GH 2 . BD . GH3 Hence or GC = 2 12.BD.GH2' GC = IGH, as before. The following values of GC for a few sections will be found useful in practice : — Descri2:)tion of Section. Value of GC. Eectangular ..... Circular ..... Hollow rectangular ) , . 1 > chimneys ,, circular j IGH |GH IGH ) . ^ *GH j ^PP^o^^™^te APPENDIX XX. Arch with Unsymmetrioal Load. The following memorandum by Mr. Henry Fidler describes how the line of least resistance in an arch with an unsymmetrical load may be drawn, by the method devised by Professor Fuller, and modified by Professor Perry. Let ahcd be a segmental arch divided into any convenient number of imaginary voussoirs (in this case 9), and loaded unequally. The positions and intensities of the loads are indicated by the vertical arrows 1, 2, 3, 4, etc., and the direction of each vertical force is supposed to pass through the centre of gravity of each voussoir. The centre third of the depth of the arch ring is indicated by the dotted arcs, and it is required to draw the polygon of forces corresponding to the load and contained within the boundaries of the central third. With any pole e, draw the force diagram. Fig. 2, shown below the arch, plotting the vertical loads 1, 2, 3, etc. to any convenient scale. Draw the polygon of forces JKLMNORSTUV in the usual way, terminating with the closing line JV, to which ef is parallel, giving the vertical components of abutment reactions fg, fh, for the right and left hand abutments respectively. For the sake of convenience repeat the polygon of forces jhlmnorstuv above the arch to an increased vertical scale if necessary, plotting the vertical ordinates from a horizontal base AB. On AB take any two convenient points 2^, q, and join pr, qr. 324 NOTES ON BUILDING CONSTRUCTION From the points I, m, n, o, etc. of tlie polygon draw horizontal lines intersecting the straight lines pr, qr. From the points of intersection let fall the perpendiculars I, II, III, IV, V, etc. From the points of intersection of the perpendiculars 1, 2, 3, 4, 5, etc. with the upper and lower boundaries of the central third draw horizontal lines intersecting the corresponding perpendiculars I, II, III, IV, V, etc. Join the points thus obtained and complete the irregular figure WXYZ. Within the area W, X, Y, Z, select the pair of straight lines, Xx ; Zx ; meet- ing at X (on the vertical through r) containing the smallest possible angle at x, and contained wholly within the boundaries of the area W, X, Y, Z. The straight lines Xx and Zx bear the same relations to the polygon required to be drawn within the central third of the arch ring as the straight lines pr, qr, do to the polygon jklmno, etc. From the points of intersection of the lines Xx and Zx with the verticals I, II, III, etc. draw horizontal lines intersecting the verticals 1, 2, 3, etc. By joining the latter points thus obtained the required polygon lying within the central third of the arch ring is drawn. A second force diagram, Fig. 3, may now be drawn from the polygon last obtained giving a polar distance T equal to the minimum horizontal thrust of the arch required to meet all the conditions. If it be not found possible to draw any pair of straight lines within the boundaries of the area W, X, Y, Z, and meeting at x, the depth of the arch ring must be increased until these conditions are met. The above method is of course equally applicable to an arch with a symmetrical load : and it avoids the necessity of drawing repeated trial lines of least resistance, which is very tedious. APPENDIX XXI Short Formulae to be remembered.^ It is dangerous to trust to the memory for formulae that are not in habitual use, and the invaluable Pocket-books of Molesworth and Hurst are not always at hand. A student who has grasped the principles iipon which the different calculations are founded can generally build up the formulas he requires, but in everyday out-of-door life there is not always time for tliis, and it is well to have a few simple formulfe in one's memory to aid in rapidly forming an opinion when reference to books of any kind is impossible. The following are good formulte to remember. In those for supported beams the safe distributed load is given, in cwts. for timber, in tons for iron. From this the safe weight for other distributions of the load can easily be ascertained (see Appendix VII.), ^ = depth*^^}°^ beam or girder in inches. S = span of beam or girder in feet. ^ These short formulae are derived from those given in this volume at the pages mentioned in connection with each. They have been made shorter by substituting in some cases feet for inches, cwts. for tons. APPENDICES 325 Timber Beams, supported at ends and uniformly loaded. Strength. — For Fir. The safe distributed dead load in cwts. = J^" inches. S m feet. * ' ^ '■' (See Case 7, p. 59.) This assumes the modulus of rupture/,, =6500 lbs., or about 60 cwts., and a factor of safety of 5. For English Oak and Pitch Pine multiply the result of (I.) byl^. For Teak multiply by 2, and for Greenheart by 3. Stiffness.— When a beam supported at the ends is loaded with the dis- tributed safe weight found by (I.) then (see Equation 45, p. 67) n . , S^ in feet. JJenection at centre = — — — ; — • n\ \ •^^ dm. inches. " ' ' • \ •/ For _PVr, Oa\ Teak, Greenheart. The distributed dead load in cwts. ) • • i to cause a deflection of per \ = ^^^hes foot of span. J in feet. (See Equation 47, p. 68.) For dead load in centre to cause the same deflection, multiply the result of III. by i. In II. and III. the modulus of elasticity is taken at 1,210,000 for large scantlings. Cast-iron Girders supported at ends and uniformly loaded. The safe distributed dead load in tons = ^ . . . . (IV.) S a being the effective area of tension flange in inches— the depth of girder in inches — S the span in feet. (See Equation 57, p. 95.) This assumes c = %\ tons and factor of safety = 4. Wrought iron, rolled, or Plate Girders, supported at ends and uni- formly loaded. The safe distributed dead load in tons = , . , (V.) - (See Equation 54, p. 87.) a being the effective area of compression flange in inches. The ultimate resistance to compression is taken at 18 tons per square inch, the factor of safety as 4. The limiting stress per square inch is therefore 4 4 tons. Open Webbed Girders. Load is as in (V.) above. Stress on any bar in web _ shearing stress at the point x length of bar depth of girder. ' (^^'^ (See p. 194.) Retaining Walls. For water (see p. 235) mean thickness =| of height . (VII.) For average earth (see p. 244) „ =1 . . (VIII.) Arches, thickness of, from 10' to 25' span. Masonry — Block, 1 inch for every foot of span . . . (IX.) Eubble, 1^ inch „ „ . . . . (X.) Brick, ^ brick for every 5 feet of span . . (XI.) ^ H. E. Aide Memoire. 2 Seddon. 326 NOTES ON BUILDING CONSTRUCTION. Water Supply. — Pipes flowing full (see p. 267). G = Gallons supplied per minute. H = Head in feet. L = Length of pipe in feet. d = diameter of pipe in inches. For i" pipes take f, for 1" pipes take |, and for pipes of over 12" diameter take {h, of the result found by XII. For i pipes take d J larger than as found by XIII. (XIII.) (XII.) a o Hi «H o 03 d I a hi cS I— ( O II «' ^ 00 fj a 5. ' -"^ Ji « ^ S CO ■§) J3 •S -3 .2 a la II II < a il go II II ■< a m 03 + + II II + ^ II II M So II II 02 00 -< tQ OQOQ II II ■t a 03 02 II II II ■0 Ma 02 OQ OQ 1 1 3 C 3 o" + + H ll^^li^ o QJ 0) (D II II II 03 •< o o 03 02 03 11^ II the sub- of W, ;e the : II o - O ^ CO ( 1^ Tj^ OiCOGOt— C0C01>-C COiMCOGSCOCO'^rcO 00000000 00000000 o o 0^0^00 0^0^ o-^oooooo (M-^OOCOCOCQrH 00000000 00000000 00 o o 00 00 o" o" cT Tt'*" T*r ^'^jrcM' 00000000 00000000 000000000 o't^icTtM lo i-TcTcyr 00000000 OOlOOOOOCO 00_O O^OO^O O O CO cT i>r (m" of r-TcT 00000000 O'^OO'MiCClQO OOCMCqCSOCDCOiOCi .3 t=D tC >^ tJD ^ bO tH !> P !-. ^ ^ ^ elm 2 « CD 53 >s O ffl o -If c 3 g o g OJ O) 8 o 'St- a s o ^ S g 3 rti-H 2-2 o «5 0 g S M 2 ^ " ^ fc< 03 « 2 ^ t*,' C4 £ 0 328 NO TES ON B UILDING CONS TR UCTION TABLE Ia. Safe Eesistanee of Materials. This Table only includes the materials most generally in use, and the values given can be depended upon as quite safe for the ordinary quality of materials. ^ Cast iron Wrought iron built-up girders "Wrought iron rolled joists Steel, mild . Cast iron Wrought iron built-up girders Wrought iron rolled joists Steel, mild . Oak, English Fir, such as Kiga or Dantzic Pine, American yellow Cast iron Wrought iron Steel, mild . Cast iron Wrought iron Cast iron Wrought iron 1"5 ton per sq. inch 5 tons do. 6 do., do. 8 do., do. Safe Resistance to Tension. Oak, English Fir, Baltic . Pine, American yellow Pine, pitch . Teak , Moulmein . 16cwts. per sq. inch 12 do., do. 3 do., do. 10 do., do. 10 do., do. Portland cement, neat, after 9 men. B'kwork in cement^ Do. in mortar (hy- draulic lime) 2 Concrete, Portland cement, 5 to 1 Safe Resistance to Compression. 8 tons per sq. inch 4 do., do. 5 do., do. 6-5 do., do. 13 cwts., do. 10 do., do. 6 do., do. Pine, pitch . Teak, Moulmein Portland cement, neat, after 9 months Mortar, common . Concrete, lime ,, Portland cement, 5 to 1 Concrete.Portland cement, 10 to 1 lOcwts. per sq. inch 12 do., do. 9 do., do. O'o cwt. do. O'o do., do. 2 cwts. do. 1 cwt. do. Bricks, ordinary stock Brickwork in mor- tar (good) 3 Brickwork in ce- ment 3 Granite, Aberdeen Limestone, granu- lar Sandstone, ordin- ary Masonry, rubble Safe Resistance to Shearing. 2*4 tons per sq. inch 5 do., do. 5 '5 do., do. Fir, such as Riga or Dantzic, along the grain 1 '3 cwt. per sq. inch Oak English, across grain II 11 along the grain I'O cwt., per sq. inch 0'5 cwt., do. 0-1 do., do. 0-3 do., do. 0"S cwt. per sq. inch 0-5 do., do. 0-8 do., do, 10 cwts. do. 9 do., da 5 do., do 0-4 cwt. do. 5 cwts. per sq. inch 2 do., do. Safe Resistance to Bearing steel, mild . 10 tons per sq. inch 5 do., do. I Oak, English 8 tons per il Fir, such as ' sq. inch or Dantzic 25 cwts. do. 3 "5 tons per sq. inch 4 '5 do'., do. Safe Modulus of Rupture. Oak, English Fir, such as Riga or Dantzic 16 cwts. per sq. inch 11 do., do. Pine, American yellow Teak, Moulmein . 1 12 cwts. per sq. inch 10 cwts. per sq. inch 20 do., do. 1 For more detailed information see Part III. 2 These figures represent the adhesion of fresh mortar to bricks The best hydraulic mortar cannot be safely taken as adhering to the best stock bricks after six months with a greater ultimate force than 36 lbs. per square inch. The adhesion to soft place bricks IS only about 18 lbs. per square inch. Cement mortars after a short time exert a tensile and adhesive resistance nearly equal to the tensile strength of good brick, i.e. about 280 to 300 lbs. per square inch. 3 The resistance of brickwork to cracking or crushing is much less than that of the bricks alone. Alter allowing three to six months (according to the mortar) for setting, good stock brickwork will begin cracking at a pressure of 200, 400, or 700 lbs. per square inch, according as it is laid in gray Chalk lime, Lias lime, or Portland cement mortar. For ultimate crushing from one and a half time to twice these pressures would be required. {Wray and Seddon.) TABLES 329 >5* ■Ho ;± 00 CO oCi^^C>o^■iO^O'^3^oc>(^^so CO .CoCO^COOi0005COS<3 I:^':O00^00OiCOOi(NOiO5>-lt^?^ CO 05 10 T-H v^Oi OO >-( (M (M e-HCOltilOOO»OtoCOS^-*i^i»<^ 0O!NClf5CO<»-* i-H ^ oi5CO>SOOOC30<5-* vi-CO l:^'ooi^l>.ij:5iO~^COSO(NS'> 1-Hl-SO'^'»?^t7HC>CO<>0-^9>-jhCi i-H '^S i-H i 00 jf^ i> lii 10 -i^-M ©5 -i 95 tO^COS^Cq^(N'~ii-l'Mr-i'~H CO (M 1-1 '-S n3 PI s ^ r^M CO ^ I ^ ^ ^ rcJ ■£ O OJ o rt g-i o o js ^ tzj CO § Hcq 9 -5 E 1 :3 o bD Pi ft a 330 NOTES ON BUILDING CONSTRUCTION pq 0) ^ O CD o O ^00 OO >~H «0 ®0 CO "^^ ■ • T' ^ 7^ "^lo 05 >-i : . . . ^ . . Q) rH C— I .<2>i-i'~H-i : : 6 6 o eH . . rH 00 Co : : ^ S xn o S pi c3 cn .5 (U CO COC00tlCOCM(MoiH«co<*.t--t--CO«D«Dkr5lOVO-^Tt<-^COCOCO(M(M o >a o 10 oiooioeinoiaoiooiooiootooio rHiHeiNMe<5**10IO«05OI>t-00000»09OOr-j TABLES I 333 For practical purposes the value of - cau be found with sufficient accuracy from I I - = nr, K 0 where n and b have the following values for various sections in general use. — Section. m Holloiu Metal i/w'-b Solid ITolloiij Oi Metal '/w'^b 3-5 2-5 3-0 4-0 3-1 Section. Ti 1 ylrea of IVeb = l/i area of flanges. ^ ^ I Area o/lVei^'/s "f flanzes. 6-0 4-9 4-2 3-9 4-2 Section. I — I +1 4-9 3-9 3-8 4-3 4-9 Example. — Find the safe stress for an L wrought iron strut, 3" x 3" x i", 10' 0" long, the ends being considered rounded. We have 6 = 3, » = 4-9, Z=120; hence I 120 - = 4-9 —-=196. K 3 From the Table the safe stress per square inch is 0"70 ton nearly. Hence Safe stress = 0*70 x (3 x + 2l x |) = 1-9 ton nearly. 334 NOTES ON BUILDING CONSTRUCTION TABLE VI. Strength of Wooden Struts. Rounded Ends. Wood supposed to be average quality fir, such as is used in roof work. R = ratio of length least dimension of cross section, rc = safe stress per square inch in ciots. If both ends are fixed take value of corresponding to ^. If one end is fixed and the other rounded take mean of values of r R corresponding to — and R. R R R R 5 9-55 19 3-85 33 1-56 54 0-58 6 9-25 20 3-60 34 1-43 66 0-55 7 8-85 21 3-88 36 1-35 58 0-52 8 8-30 22 3-18 36 1-28 60 0-49 9 7-75 23 2-92 37 1-22 62 0-46 10 7-20 24 2-75 38 1-16 64 0-43 11 6-70 25 2-60 39 1-11 66 0-41 12 6-25 26 2-41 40 1-06 68 0-39 13 5-80 27 2-25 42 0-96 70 0-37 14 5-40 28 2-11 44 0-87 72 0-35 15 5-05 29 1-96 46 0-79 74 0-33 16 4-75 30 1-82 48 0-72 76 0-31 17 4-45 31 1-71 50 0-66 78 0-29 18 4-15 32 1-60 52 0-62 80 0-27 Example.— Find, the safe stress that can be applied to a wooden strut 3" X 2", and 10 feet long. 10x12 - 2 — ' '•c = 0-49 ; safe stress = 3 x 2 x 0-49 = 2-9 cwts. TABLE VIL Strength of Timber Columns. The following Table has been deduced by Mr. Stoney, C.E., from a series of experiments made by Mr. Brereton, C.E., on square columns of American yellow pine {Pinus strohus). Ends adjusted as in ordinary practice. Ratio of length to side of column 10 15 20 25 30 35 40 45 50 Breaking weight in tons per sq. foot of section 120 118 115 100 90 84 80 77 75 TABLES 335 TABLE VIII. Strength of Rivets (Iron Rivets in Iron Plates). Safe resistance to shearing, 4 tons per square inch. Safe resistance to bearing, 8 tons per square inch. Diam. of rivet. Resistance to single, shear — tons. Resistance to bearing of one rivet in plates of various thicknesses — tons. A" i" 1" 1" 1" r 1" i" 0-78 0-25 1-00 1-50 2-00 2-50 3-00 3-50 4-00 "1 1-23 0-31 1-25 1-87 2-50 3-12 3-75 4-37 5-00 3'' I 1-77 0-37 1-50 2-25 3-00 3-75 4-50 5-25 6-00 111 8 2-40 0-44 1-75 2-62 3-50 4-37 5-25 6-12 7-00 15" 1 « 2-76 0-47 1-87 2-81 3-75 4-69 5-62 6-56 7-50 1" 3-14 0-50 2-00 3-00 4-00 5-00 6-00 7-00 8-00 U" 3-98 0-56 2-25 3-37 4-50 5-62 6-75 7-87 9-00 li" 4-91 0-62 2-50 3-75 5-00 6-25 7-50 8-75 10-00 TABLE IX.1 Dimensions of Eyes of Wrought Iron Tension Bars. Ratio of area of metal at side of eye to Ratio of diameter of area of bar. Ratio of maximum pin to width of bar, or diameter of bar thickness of bar to Hammered eye. Hydraulic forged width. (Pin in (if round) (Metal section at eye. (Same section single shear.) (not to be < 0-67). back of eye = = tliat at back as at sides of bar.) of eye.) 0-67 0-66 0-74 0-21 0-75 0-67 0-75 0-25 1-00 0-75 0-75 0-38 1-25 0-76 0-80 0-54 1-33 0-79 0-85 0-61 1-50 0-83 0-93 0-70 1-75 0-84 1-00 0-88 2-00 1 0-88 1-13 1-08 ^ This Table is taken from Instruction in Construction (revised edition), and is tbunded on that given by Mr. Shaler Smith. 336 NOTES ON BUILDING CONSTRUCTION TABLE X.1 Weight of Angle Iron and Tee Iron. 1 foot ill length. Sum of the 'Width and Depth in Inches. Thick- ness. 2 2i 03 2i 21 2| Cliches. lbs. lbs. lbs. lbs. lbs. lbs. lbs. lbs. lbs. lbs. lbs. 1 •57 •62 •68 •73 •78 •83 •88 -94 -99 1^04 1-09 ■i T<5" i •81 -89 •97 1-05 1-13 1^21 1-29 1-37 1-45 1^52 1-60 1-04 1-15 1^25 1-36 1-46 1^56 1-67 1-77 1-88 1^98 2-08 5 TS" 1-24 1-37 1-50 1-63 1-76 1^89 2^02 2-15 2-28 2^41 2-54 3 3* 3i 31 3* 3i 33 3^ 4 4i 1 8 1-14 1-20 1-25 1-30 1-35 1^41 1^46 1-51 1-56 1-62 1-72 3 i r68 1-76 1-84 1-91 1-99 2-07 2^15 2-23 2-30 2-38 2-54 2^19 2-29 2-40 2-50 2-60 2^71 2^81 2-92 3-02 3-13 3-33 5 TS- 2^67 2-80 2-93 3-06 3-19 3-32 3^45 3-58 3-71 3-84 4-10 3-13 3-28 3-44 3-59 3-75 3-91 4-06 4-22 4-38 4-53 4^84 7 TB' 3^57 3-75 3-93 4-11 4-29 4-48 4-66 4'84 5-02 5^20 5-56 4i 4| 5 H 5i 5| 6 6i 6| 7 i 2^70 2-85 3-01 3-16 3-32 3-48 3-63 3-79 3-95 4-10 4-26 3-54 3-75 3-96 4-17 4-38 4-58 4^79 5-00 5-21 5-42 5-63 6 4^36 4-62 4-88 5-14 5-40 5-66 5-92 6-18 6-45 6-71 6-97 5^16 5-47 5-78 6^09 6-41 6-72 7-03 7-34 7-66 7-97 8-28 7 5^92 6-29 6-65 7-02 7^38 7-75 8-11 8-48 8-84 9-21 9-57 i 6-67 7-08 7-50 7-92 8^33 8-75 9^17 9-58 10-00 10-42 10-83 9 TB' 7^38 7-85 8-32 8-79 9^26 9-73 10-20 10-66 11-13 11-60 12^07 n 71 8 8i 8i 8| 9 9i 9i 9| i 5^83 6-04 6-25 6-46 6^67 6-88 7-08 7^29 7-50 7-71 7-92 5 TB- 7^23 7-49 7-75 8-01 8-27 8^53 8-79 9^05 9-31 9-57 9-83 3 5 8-59 8^91 9-22 9-53 9-84 10-16 10-47 10-78 11-09 11-41 11-72 7 TB' 9^93 10^30 10-66 11-03 11-39 11-76 12-12 12^49 12-85 13-22 13-58 1 2 11^25 11-671 12-08 12-50 12-92 13-33 13-75 14^17 14-58 15-00 15-42 9 12-54 13-01 13-48 13-94 14-41 14-88 15-35 15-82 16-29 16-76 17-23 13-80 14-32 14-84 15-36 15-89 16-41 16-93 17-45 17^97 18-49 19-01 10 11 Hi 12 12i 13 13^ 14 14J 15 § 12-03 12-66 13-28 13-91 14-53 7 TB' 13-95 14-67 15-40 16-13 16^86 17-59 18-31 19^04 19^77 20-50 21-22 4 15-83 16-67 17-50 18-33 19^17 20-00 20-84 21^67 22^50 23-34 24-17 9 T¥ 17-70 18-63 19-57 20 51 21^44 22-38 23-31 24^25 25^19 26-12 27-06 19-53 20-57 21-61 22-66 23^70 24-74 25-78 26-83 27^87 28-91 29-95 3 4 23-13 24-38 25-63 26-88 28^13 29-37 30-63 31^88 33^13 34-38 35-63 12 124 13 13i 14 15 16 17 18 19 20 23-70 24-74 25-78 26-83 27-87 29-95 32-03 34-12 36-20 38^28 40-36 3 4 28-1329-37 30-63 31-88 33^13 35-63 38-13 40-63 4M3 43-63 46-13 7 8 32-45,33-91 35-36 36-82 38-28 41-19 44-12 47-02 49^95 52-87 55-78 1 36-67,38-33 40-00 41-67 43^33 46-67 50-00 53-33 56^67 60-00 63-33 Note. — When the base or the •web tapers in section, the mean thickness is to be measured. ^ From D. K. Clark's Rules and Tables. TABLES 337 TABLE XI. Web Plates— thickness of. TahU 1 /or determining the thickness of loeb -plate, suitable for a given shearing stress per foot depth of girder, at varying widths of plate between supports (whether L irons of flanges or vertical stiffeners). The Table is founded on the formula 48< - - (Unwin's /rort Bridges and Roofs), where T is the stress in tons ; t the thickness of web plate ; h the unsupported width of plate in feet. To use the Tahle.—Find the least unsupported width of plate in the top row of figures, and underneath it find the number nearest in excess of the given shearing stress in tons per foot depth of girder. The corresponding thickness of plate will be that required. For example, a girder 3 feet deep has a maximum shearing stress of 18 tons = 6 tons per foot of depth : the distance between the L irons is 30 inches Under the number 30 we find 6-3 tons, and the corresponding thickness is f , which is to be the thickness at abutments. Tliick- ness of web in inches. Net unsupported width of plate in Inches, whether between flange L irons or vertical stiffeners. 12 15 3-1 5-5 8-0 10-9 14-2 17-4 207 24-1 27-4 18 21 24 27 1- 2 2- 2 3- 5 5-3 7-4 9-8 12-3 15-0 17-9 30 33 36 1-0 1-8 3- 0 4- 5 6-3 8-0 10-8 13-4 16-1 39 42 45 •45 •9 1- 5 2- 3 3- 3 4- 6 6- 0 7- 6 9-5 48 1- 3 2- 0 3- 0 4- 2 5- 4 6- 8 8-6 51 36 7 ■2 8 7 8 9 ■3 Obviously the Table can be used for determining the thickness of web plate at any part of the girder, as well as at the abutments, when the shearing stress is known ; and it is easy to find thicknesses for unsupported widths not given in the Table by estimation between the widths above and below. The safe compression shearing stress used in the formula is 4 tons per square inch, and as the assumed diagonal pillar, 12 inches wide, is not independent but continuous with the adjacent parts, the stresses given in the Table are well on the safe side. If, therefore, the given stress is but little over a tabular number it will be safe to use the corresponding thickness instead of the next greater. If, for example, the stress is 1 1 tons per foot of depth, and the unsupported width 30 inches, it will be quite safe to use a web plate |" thick. TABLE XIL2 "Weight of Roof rraming, Ceilings, etc. Ceilings. Lath and plaster ceiling . . . • 8 ) lbs. per square foot hori- g!j:l^ jo^'^ts 3 I zontal s urface covered. 1 Kindly given to the wiiter by Mr. C. Light. 2 The weights of framing of iron roofs are taken from Unwin's Bridges and Hoofs ; those for wooden roofs from Instruction in Construction, by Col. Wray, R.E. B.C. IV. 2 338 NOTES ON BUILDING CONSTRUCTION Framing, etc., Wooden Roofs. Common rafters 3 ) 1^^- ^'l^^^^ ^^^^ ^""^ Collar beams • • ^ ■' suriace. Framing' for wooden roofs, including " purlins and ridge boards, but exclusive of tie beams — 20 feet span, king post, rise \ span 30 ,, )5 " 40 „ queen post „ 50 5, )5 " 60 ,, )) " Tie beams — 20 feet span, king post 30 40 50 60 queen post lbs. per square foot of roof surface. 2 5) " 2 J> " 3 )5 " 4 iy » 11 lbs. per foot run of tie. 20 j> " 18 )) " 20 )) " 30 J) » Framing of Iron Roofs. Trussed Roofs. Pent . Common truss Bowstring Eoofs. Manchester . Lime Street Birmingham Arched Roofs. Small Corr. Iron. )> " Strasburg Rail Paris Ex. large Dublin Derby Sydenham . )) St. Pancras Clear Span. Distance apart of Prin- cipals. Feet. 15 37 40 54 55 72 84 50 100 130 140 50 154 211 40 60 97 153 41 8lh 120 72 240 Of Pur- lins, etc. Feet. 5 12 14 20 9 10 14 26 12 11 26 24 13 26 16 24 Weight per square foot of covered area in lbs. Of Prin- cipal. 11 2-0 6-5 4-6 4-2 2-6 0-8 291 9-5 3-4 10-8 7- 9 8- 4 7-4 3-5 3-5 3-0 7-0 2-8 5-9 5-6 4-5 4-9 7-3 5- 5 7-3 6- 0 3-9 2-9 17-1 Total Iron- work. 3- 5 4- 6 5- 5 9-5 11-6 7- 0 8- 5 3-0 7-0 6- 4 9-6 11-0 12-0 15- 0 10- 7 16- 8 11- 8 11-3 24-5 Total with covering. 6-9 5-2 9-0 8-0 2- 5 3- 5 TABLES 339 TABLE XIII.i Weight of Roof Coverings. Lead covering, including laps, but not boarding or rolls Zinc covering „ 14 to 16 zinc gauge Corrugated iron, galvanised, 1 6 W.G. . 18W.G. . 20 W.G. . Sheet iron, 16 W.G. „ „ 20 W.G. ... Slating laid with a 3-inch lap including nails, but not battens or iron laths — Slates, Doubles, 13 inches x 9 inches, at 18 cwts. per 1200 . . „ Ladies, 16 inches x 8 inches, at 31-5 cwts, per 1200 „ Countesses, 20 inches x 10 inches, at 50 cwts. per 1200 „ Duchesses, 24 inches x 12 inches, at 77 cwts. per 1200 ....... Tiles, plain, 1 1 inches x 7 inches, laid with a 3-inch lap and pointed with mortar, including laths and absorbed rain ...... „ pan, 13|- inches x 9 J inches, laid with a 3-inch lap and pointed with mortar, including laths and absorbed rain ...... „ Italian (ridge and furrow), not including the board- ing, but including mortar and absorbed rain . Slate battens, Z\ inches x 1 inch — For Doubles For Countesses Boarding, | inch thick . )5 1 )) Wrought Iron Laths, angle irons- For Duchess Slates For Countess „ Cast Iron plates, | inch thick Thatch, including battens Per square foot. h\ lbs. to 8| lbs. to If „ 3i 2 2i 8l °2 12 14 2 44 6J TABLE XIV. Wind Pressure. Wind pressure normal to a roof- surface from Hutton's formula, viz. PN-P(sin i)i-84cosi-i^ taking P=50 lbs. per square foot. Pitch of Roof (i) l6° 15° 20° 21° for \ span 25° 26°Jor \ span 30° 33° J or J span 35° 40° 45° 50° Normal wind pres- sure in lbs. per 8q. foot (Pn) 12-1 18 22-6 25-2 28 -8 30-2 33-0 36-6 37-8 41 '6 43 47-6 ^ This table is taken from Instruction in Construction, by Col. Wray, R.E. 340 NOTES ON BUILDING CONSTRUCTION TABLE XV. Scantlings for "Wooden Roofs. The roofs are supposed to be of Baltic fir covered with Countess slates laid on inch boards ; the maximum horizontal wind force is taken at 45 lbs. per foot super acting only on one side of the roof at a time, equivalent to a normal wind pressure of 30 lbs. per square foot for a pitch of 30°, and 40 lbs. per square foot for a pitch of 45". The common rafters to be 1 ft. from centre to centre, but in sheltered positions they may be placed 1 ft. apart in the clear. A EooFS WITHOUT CEILINGS. —Pitch up to 30°. Nature of Roof. Span in feet. Common Rafters. Collar. In tixing the collar to the rafter the latter should not be cut into. Remarks. "Walls cap- able of resisting thrust.* Collar placed J- way up. Walls not capable of resisting thrust, t Collar placed \- way up. * The following solid walls are strong enough, when built in Lias lime mortar 1 to 2, to resist the thrust ol roof; allowance must be made for door and wimlow openings. Height to be taken from level of the tlooi below roof. Stone Walls. Brick Walls. Couple. 8 10 12 Bdth.Depth. 2" X 3" 2"X3J" 2"x4" 16" thick, not over 15 ft. high. o e -irx ( 9" thick, not over 7' high Span of roof 10 -j ' ° Span of roof 18' 1^^^' " " jq' " t When the walls are not capable of resisting the tlrnis', of the roof, place the collar low down ; but if th( collar is required half-way up, the scantlings musi be increased as follows :— Rafters, add one-fourth to both breadth and depth Collar, add J" to depth ; but it would be better to us( the scantlings for walls capable of taking the thrust and make some arrangement to prevent the walls fi on spreading, such as tying the wall plates together ai intervals. Collar Beam. 8 10 12 14 16 18 irx2i" irx2r li"x2i" l|"x3" 2" X 3J" 2"X3|" 2"x3J" 2"x4" 2"x4j" 2J"x5" 2i"x5i" 2i"x6" 2"x2J" 2"x2A" 2"x2|" 2"x3" 2"x3i" 2" X 4" King Post. Trusses 10' centre to centre. Tie Beam.{ Depth in- cludes 3" for joints. Principal Rafters. King Post. Struts. Straining beam. . Purlins.f 10 ft. bear- ing. Common Rafters, j 20 22 24 26 28 30 3"x4i" 3"x4|" 3i"x4r 3i" X 4|" 4"X4}" 4"X4J" 3"x5" 3" X 5i" 3J"x5i" 3i" x 5f " 4"x5^" 4"X6" 3"x2i" 3"x2f" 3i"x2|" 3f' x2|" 4" X 2f " 4"x2}" 3" X 3" 3" X SJ" SV'xSi" i3|"x4" 4"x4" ' 4"x4i" 5"x7i" 5"x7|" 5" x 8" 5" X 8i" 5"x8J" 6" X 8|" 2"x3i" 2"x3^" 2"x4" 2"x45" 2"xU" 2"x4f' Queen Post. Trusses 10' centre to centre. 32 34 36 38 40 42 44 46 4 J" X 4r 4i"x4|" 4f" X 4|" 4|"x4i" 4|"X6" 5" X 5" 5"x5i" 5i"x5i" 41"X4|" 4!"X5" 4|"x5" 4J"x5i" 4|" X bV 5" X bl" 5"x5|" 5J"X5J" Queen Posts. 4*"X2|" 4|"x2i" 4J"x2i" 4|"x2i" 4|"x2f" 5"x2|" 5"x2r 5i"x2r 4l"x2h" 4l"x2i" 4|" X 3" 4J"x3i" 4f"x3J" 5" X 3J" 5"X3J" 5i"x3r 4J"x5i" 4i"x6" 4|"X6J" 4J"x6|" 4|"x7i" 5"x7i" 5"x8" 5i"x8i" 5"x7r 5"x7f' 5"x8" 5"x8" 5"X8J" 5"x8J" 5"x8i" 5"xS|" 2"x?A" 2"x3-l" 2" X 4" 2" X 4" 2"x4i'' 2"x4L" 2"x4.';'' 2" X 5" TABLES 341 TABLE ILY.— Continued. B Roofs with Ceilings.— Pitch up to 30°. Nature of Roof. Span in feet. Common Rafters. Collar.* * In fixing the collar to the rafter, the latter should not be cut into. As regards walls capable or not cap- able of resisting the thrust of roof see remarks under A. Collar Beam. 8 to IS. Add J" to depths given in A. Add f" to depths given in A. King Post. Trusses 10' centre to centre. Tie-beam. J Principal Rafters. King Post. Struts. Straining beam. Purlins.§ 10 ft. bearing. Common Rafters. 20 22 24 26 28 30 4" X 7" 4"x7r 4i"x8" 4J" X 8i" 4^X9" 4J"x9r 4" X 4" 4"x4r 4i" X 4J" il" X 5" 4J" X 5J" 4|" X 5|" 4" X 3 ' 4" X 3" 4J" X 3" 4J" X 3" 4J" X 3" 4J" X 3" 4" X 2i" 4"x3" 4i" X 3" 4^" X 3" 4i"x3r 4|"X3|" 5" X 7*" 5" X 7|" 5"x8" 5" X 8J" 5" X 8|" 5" X 8|" 2" X 3J" 2" X 3i" 2" X 4" 2"X4J" 2" X 44" 2" X 41" Queen Post. Trusses 10' centre to centre. 32 34 36 38 40 42 44 46 4rx7i'' 4|" X 7|" 4|"x8i" 5"x8J" 5"x9" 5i"x9" 5rx9J" 5i" X 10" 4|" X 51" 4J"x5|" 4rx6i" 6"x6" 5" x 6J" s^xer 6i"x6r 6rx7i" Queen Posts. 4|"x3" 4i" X 3" 4J"x3" 5" X 3" 5"x3r 5rx3r 5i"x3r 5i"x4" 4|"x2r 4f" X 2|" 4|" X 3" 5"x3" 5"X3J" 5J" X 3" 5i"x3i" 5i"x3|" 4|"x6r 4|"x7i" 4|"x8i" 5"x8i" 5"X9" 5J"X9" 5i"x9J" 5J"xlO" 5"x7r 5" X 7J" 5" X 8" 5" X 8" 5" X 8i" 5" X 8*" 5" X 8|" 5" X 8}" 2"x3i" 2" X 3|" 2" X 4" 2" X 4" 2" X 4i" 2"x4i" 2"x4r 2"x5" For Roofs of 45° Pitch.— Add 1" to the depth of common rafters, purlins, and struts, and I" to the depth of the principal rafters, as given in A and B. t The joint of the tie beam with the principal rafter should be placed immediately over the supporting wall. If this cannot be conveniently done, the depth of the tie beam should be increased one or two inches. § If the purlins, instead of being placed immediately over the joints, are placed at intervals along the principal rafter, increase the depth of the latter, given in the Tables, as follows : ^4. f r '>^ii'^out ceiling 2", without ceiling U", King post roof I ^ .^^ Queen post roof | ^ .^^ ^ |; The purlins if placed 2 feet apart and with 10 feet bearing may be made 3"x 6". The scantlings of the principal rafters, struts, and straining beam can be slightly modiiied by means of the following rough rule : " For every J" deducted from the lesser dimension of the scantling, add 4" to the other dimension, and vice versd." For the tie beam, purlins, and common rafters, so long as the depth is about double the breadth, I" deducted from the breadth requires J" to be added to the depth. This Table is derived from the War Office practice. TABLE XVa. CoeflS-cients of Friction. 4> = Angle of Repose. /= tan (p = coefficient of friction. 9 / / Masonry and brickwork (dry) ,, with wet mortar ,, with slightly damp mortar ,, on dry clay „ on moist clay 31° to 35° 25^° 27° isi° •6 to -7 •47 •74 •51 ■33 Wood on stone Iron OD stone Wood on wood (dry) Metals on metals (dry) Smoothest and best greased surfaces 22° 35° to 163° 14° to 26J° 8^° to lli° If to 2° •4 •7 to -3 •25 to -5 •15 to -2 •03 to -036 ^ Rankine's values. NOTES ON BUILDING CONSTRUCTION TABLE XVI. Retaining Walls.^ Angle of repose (<^) of various earths, and value of K = 0-7 tan Description of Earth.

), ii between the joints. ftg, number of rivet holes to be deducted from the cross section of cover plates. Tft, total safe resistance to bearing. Ts, ,, ,, shearing. <, thickness of plates. fg, ,, „ cover plates. a, number of riTOts required for bearing. „ „ shearing. Plate girders. A(,, effective area of compression flange. Aj, „ tension „ D, „ depth, i.e. distance between centres of gravity of flanges. tyj, thickness of web. Braced or framed girders. 0^,5, compression in the bar a^,a^ of the top boom. d, depth, i.e. distance between centres of gravity of flanges. ^4,14, stress (tension or compression) in brace a^,a.^^. n, number of triangles. T^3_^^, tension in the bar a^2,a^^ of the lower boom. Roofs. H^, horizontal reaction at abutment A due to wind. P. P. K A B B horizontal wind pressure, normal „ vertical „ reaction at abutment A. stress (tension or compression) in member AD. vertical reaction at abutment A due to wind. B temperature, wind. temperature. Masonry structures. p, intensity of pressure at any point of a bed joint. 'P(max)i greatest intensity of compression on a bed joint. Tq see Beams. Retaining walls. H, height of wall. K, constant depending on <^ (angle of repose). T, thickness of wall with vertical sides. Tj, mean thickness. W, weight of a cubic foot of the material of the wall. ^, angle of repose of the earth to be retained. w, weight of a cubic foot of the earth to be retained. 350 NOTES ON BUILDING CONSTRUCTION Arches. H, horizontal thrust. P^jPg, etc., total pressure on joints between voussoirs. N V, vertical component of pressures. w-^^w^, etc., loads on each voussoir. W, total load on an arch. Hydraulics. A, area of cross section of flow. D, diameter of a pipe in feet. d, „ „ inches. F, discharge in cubic feet per second. G, discharge in gallons per minute. H, effective head of water in feed, i.e. the actual head reduced by the various losses. Hb, loss of head due to bends. H^, „ „ elbows. H„, „ „ orifice of entry. H^, „ _ „ velocity. J, factor for jets. L, length of a pipe in feet. R, hydraulic mean depth. S, slope of a pipe. V, mean velocity of flow in feet per second. TABLE XXVI. WORKING STEESSES. The undermentioned are the working stresses that have been used in this volume : — Cast Iron. Girders Wrought Iron. Rolled beams. Built-up girders Roofs . Tension tars and rods Riveted joints Tension l| tons per square inch (see p. 94). Compression 8 Shearing 2-4 ( Tension 5 tons per square inch (see p. 82). Compression 4 Shearing 4 Tension 5 „ „ (see p. Compression 4 „ » ( » Tension 5 „ ,, ( Compression (depends on ratio of ( ,, length of struts to least diameter) Tension 5 tons per square inch ( „ Shearing 4 „ „ ( „ Bearing 8 „ » ( » ). ). 156). 156). 108). 119). 106). 123). » )• Pin joints Screws . Timber. Fir (superior) Fir (inferior) Oak Brickwork. In Mortar In Cement TABLES 351 Shearing 4 tons per square inch (see p. 140). Bearing 5 „ „ ( „ 141). Shearing 2 „ „ ( „ 146). Tension 12 cwts. per square inch (see p. 150). Compression 10 53 35 ,3 )■ Shearing I'S J> 53 55 ). Bearing 1 2 » 53 35 ,5 ). Tension 10 3> 5J 55 152). Compression 7 )> 33 33 „ ). Shearing 1-3 5> )3 33 5, ). Bearing 7 5! 33 33 ,5 ). Tension 1 6 55 33 33 328). Compression 13 )) 53 35 55 )• Shearing 5 J) 35 35 33 )• Bearing 2 5 S) 53 33 ,5 )• Compression 0*5 cwts. per square inch (see p. 223). Adhesion 0-06 ?> 35 ( 55 227). Compression 0"8 33 ( 33 259). Adhesion 0'5 55 35 ( 55 5, ). 352 NOTES ON BUILDING CONSTRUCTION NOTES ON BUILDING CONSTRUCTION INDEX A Abutment for arch, 259. Adams, reference to, 100. Anderson's formula for the weight of girders, App. XV. Angle irons, resistance of. Table I., weight of. Table X. Angle of repose of earths. Table XVI. Annulus, moment of inertia, App. XIV. Appendices : {see List of, p. xv.) Area, effective, of tension bar, 106. Area of resistance, equivalent, 47 ; safe, 48 ; mechanical method of finding, for timber beam, 48 ; for rolled iron beam, 83 ; for cast iron beam, 97. Arches, 246 ; abutment for, 259 ; example of brick arch, 254 ; graphic method of determining stability of, 248, 251 ; do. of drawing line of least resistance for symmetrical load, 251 ; do. for unsymmetrical load, App. XX. ; line of least resistance, 248. Arch ring, thickness of, 253, Table XVIIa. ; table of thickness, 259. Axis, neutral, 42 ; note on, App. IV. B Baker's practical rales for retaining walls, 244. Bars {see Compression bars, Tension bars) ; wrought iron, resistance of. Table I. Basalt, weight of, Table XVII. Bath-stone, weight of. Table XVII. Beams, 24 ; bending moment, 25 ; cantilevers, 28 ; cast iron, 94 ; comparison of strength, stiffness, and of fixed and supported, App. VIII. ; continuous, 74, App. XI. ; deflection, 66; fixed, 69; moment of resistance, 40 ; reactions at supports, 1 8 ; strength of rectangu- lar beams, 54; rolled iron, 81; rolled steel, 92 ; shearing stress, 55 ; strength and stiffness, com- parison between, 66, App. VIII. ; supported, 32 ; timber beams, 79 ; trussed, 195 ; uniform strength, 62. {See above headings in Index.) Bearing stress, 7 ; resistance of rivets to, 123, Table VIII. ; resistance of pins to, 141. Beech, ultimate resistance, moduli, and weight, Table I. Bending moment, 25 ; graphic method of finding, App. VI. ; various cases for cantilevers and supported beams, 28-40 ; for continuous beams, App. XI. ; and for fixed beams, App. VIII. Bends, loss of head due to, 270, Table XXI. Block structures, uncemented, 220. Booms of girders, 154 {see Flanges). Box girder, 170 {see Plate girder) ; ap- proximate weight, 171 ; deflection, 171 ; diaphragm, 176 ; effective span, 170; end pillars, 176; flanges, 172 ; section of, 175, Plate B. Bow's notation, 184, 194, App. XVII. Bowstring girder, 194. Braced frameSjperfect, 182 ; overbraced, 181. 356 INDEX Braced girders, 187 ; application of load, 188 ; different forms of, 187 ; dimensions of bars, 193 ; graphic method for, 194 ; lattice 187, 194 ; N, 188; N (braced), 188; prac- tical formula for, 193; pro- portion between span and depth, 188; Whipple - Murphy, 188 ; stresses in, 188, 190. „ structures, 177 ; different forms of, 180. Breaking load, 5. „ stress, 8, 49 (see Ultimate stress). Brick chimney, calculations for, 225. „ pier, calcidations for, 222 ; arch, 254. Bricks, brickwork, coefficient of friction. Table XVa. ; safe resistance of, Tablela. ; weightof, Table XVII. ; Brickwork structures (see Masonry). Built-up X beams, 93. C Cantilever, cast iron, 101 ; timber, 76-79 ; wrought iron, 22. Cast iron, resistance of, 94 ; moduli, weight, iiltimate resistance, Table I. ; safe resistance, 94, Table la., XXVI. ,, cantilevers, example, 101. „ columns, examples of, IIG, 117. ,, girders, 94, 103 ; approximate formulse for, 95 ; camber for, 100 ; casting, 100 ; centre of gravity of, App. XII., XIII. ; depth, 100 ; example of, 103 ; flanges, 99 ; graphic method for finding section, 96 ; mathe- matical method, 99 ; mo- ment of inertia for Apja. XIV. ; practical points, 99 ; resistance of, 94 ; stress areas, 97 ; uniform strength, 100 ; web, 100. „ pipes, thickness of, Table XX. Cement, Portland, safe resistance of. Table la.; weight of. Table XVII. Centre of gravity ; calculation for cross section of girder, Apji. XII. ; graphic method of finding, for parallel forces, App. VI. ; graphic method for cross section, App. XIII. ; mechanical method of determining, 86. „ pressure, 218, App. XIX. Chimney, centre of pressure, App. XIX. ; example of, 225. Circle, moment of inertia for, App. XIV. Circular arc, substituted for parabola, 3 4 . Classification of stresses, 8. Clay, angle of repose. Table XVI. ; weight. Table XVII. ; coefficient of friction. Table XVa. Clerk-Maxwell (see Maxwell). Coefficients of friction. Table XVa. Columns, cast iron, exanqjles of, 116, 117 ; timber, strength of. Table VII. Comi^arative effect of dead and live loads, 5. Compression, 6 ; resistance to, of cast iron,94 ;wrought iron, 82 ; steel, 92 ; other materials. Table I., la. bars, 105, 109 ; ex- amples, 115, 116- 121; Euler's formula, 110 ; Fidler's views, 110 ; Gordon's for- mula, 113 ; long and short. 111 ; Eitter's and Rouleaux's form- ula, 114 ; securing ends of, 112 ; strengtli of. Tables V-, VI. ; Table for wooden, 115. „ riveted joints, 136. Concentrated load, 3. Concrete, safe resistance of. Table la. ; weight of. Table XVII. Conditions of equilibrium, 14. Connections, 122. Continuous beams or girders, 74, App. XI. ; example, App. XI. INDEX 357 Contra-flexure, points of, 70-75, App. XL Cotterjointjformulseand example, 148. Couple, 43. Cover,plates, 124, 125, 128, 131, 132. Covering (see Eoof). D Darcy's formula, pipes flowing full, 267 ; pipes partially full, 281. Dead load, 4. ,, and live loads, comparative effect of, 5. Deflection, 12, 66 ; amount allowed, 69 ; of beams and cantilevers, 66 ; of beams of fir, Table III. ; of continuous beams, Ai)p. XI. ; of fixed beams, Ajsp. VIII. ; of rect- angular beams, A]3p. IX. ; of rolled beams. Table IV. ; formulae, 66, 68 ; of supported beams, 66, App. VIII. Depth of girders, plate, 154; braced, 188 ; bowstring, 194 ; cast-iron, 100 ; rolled, 91. Diagrams, Bow's system of lettering, App. XVII.; Clerk - Maxwell's, 182 ; method of drawing, 183 ; riiles for drawing, App. XVI. ; for various roof trusses, 217; for Warren and N girders, 194 ; for trussed beams, 196, 198. Dimensions for riveted joints, 136. Distributed load, 4 ; compared with concentrated load, App. VII. Double-cover grouped joint, 131 ; for plate girders, 132. Drain pipes, 283 ; egg-shaped sewer, 286. Drains, example, 283. E Earths, angle of repose. Table XVI. ; weight of. Table XVII. Effective area of tension bars, 106 ; effective depth and span of girder, 154. Egg-shaped sewer, example, 287. Elastic limit, 10 ; of iron and steel. Table I. Elasticity, 10; modulus of, 12; modulus for various materials, Table I. Elbows, loss of head due to, 270, Table XXII. Elm, ultimate resistance, moduli, and weight. Table I. Elongation, 11. Enclosure wall, calculation for, 227. Equilibrium, 13 ; conditions of, 14 ; note on, App. II. Equivalent area of resistance, 47. Euler's theory, for compression bars, 110. Examples (.see List, p. xvii.) External forces acting on a structure, 15. Eyes of wrought -iron tension bars, 139, Table IX. Eytelwein's formula, flow of liquids in pipes flowing full, 264. F Factor of safety, 9, App. I. Fidler's rules for struts, 109 ; views on compression bars, 110; Table for strength of struts. Table V. ; drawing of line of least resistance for arch with unsymmetrical load, App. XX. Fir, ultimate resistance, moduli, and weight, Table I. ; safe resistance. Table la. Fished joint, 149. Fixed beams, 69 ; bending moment, 71-74; comparison of strength, etc., with supported beams, App. VIII.; objections to, 74 ; stresses on, 69 ; various cases, 71-74. Flanges of plate girders, 155, examples, 156, 164; width of, 163, 172; of cast iron girders, 99. Flexure, moment of 25 (see Bending moment) ; points of contra-flexure, 70-75, App. XL Flow of liquids in pipes, 263 ; pipes running full, 264 ; pipes running partially full, 280. Forces, polygon of, 180 ; triangle of, 179. 358 INDEX Formulse, arches, thickness of, 253, App. XXI. ; beams, 54, App. VII., XXI. ; beams, different sec- tions, App. X. ; beams, continuous, App. XL ; beams, fixed, 71, 72 ; rolled, 87 ; bending moments for different distributions of load, 28- 40, App. VII. ; App. VIII. ; cast iron girders, 95, 99 ; centre of gravity, App. XII. ; compression bars, 113-115 ; continuous girders, App. XI. ; cotter joint, 148 ; de- flection, 66-68, App. VIII., IX., XXI. ; egg-shaped sewer, area of, 287 ; equilibrium, App. II.; fished (wood) joint, 149, 150; fixed beams, 71,72, App. VIII. ; girder, approximate weight of, App. XV. ; head, loss of, 268-271 ; jets, 288 ; masonry joints, 219, App. XIX. ; moment of inertia, App. XIV. ; open webbed girders, 193 ; pipes flowing full, 264-268 ; pipes partially full, 280-282 ; plate girders, 155 ; retaining walls, 234, 238,240,244; rivets, 136; riveted joints, 123-131 ; rolled iron beams, 87, 88 ; roof, reaction at supports, 206; screw, 146; shearing stress, 56-60, App. VII.; short, for mem- ory, App. XXI. ; struts, 113-115. Foundations of retaining walls, 244. Fracture, 9. Framed structures, 177 ; different forms of, 180. Frames, 180 ; overbraced, 181 ; per- fectly braced, 180, 182. Friction, coefficients of. Table XVa. Fuller, Prof., reference to, App. XX. Funicular polygon, App. VI. G Gib, 148. Girders, bowstring, 194; box, 170; braced, 187; built-up, with rolled iron beams, 93 ; cast iron, 94, 103 ; continuous, 74, App. XI. ; hog-backed, 176 ; lattice, 187 ; N 188 ; N (braced), 188 ; open- webbed, 187; plate, 153 ; uniform strength, 64, 101 ; Warren, 187 ; weight, formula for, App. XV. ; Whipple-Murphy, 188. {See above headings in Index.) Gordon's formula for compression bars, 113. Granite, safe resistance of. Table la. ; weight of. Table XVII. Graphic method of finding moment of resistanceinbeams,46; inwrought iron rolled beams, 83 ; in cast iron girders, 96 ; bending moments and shearing stresses^ App. VI. ; centre of gravity of cross section of girder, App. XIII. ; resultant and centre of gravity of parallel truss, App. VI. Gravel, angle of repose, Table XVI. ; weight of. Table XVII. Gravity, centre of {see Centre of gravity). Greenheart, moduli, weight, ultimate resistance. Table I. Grouped joint, double-cover riveted, formulse for, 129 ; examples, 131, 132. H Head (hydraulic), loss of, 261, 268 ; due to bends, 270, Table XXI. ; to elbows, 270, Table XXII. ; to orifice of entry, 269 ; to velocity, 269. Hodgkinson's experiments on cast iron girders, 95. Horizontal shearing stress, 55, 60. House chimney, calculation for, 225. Hydraulics, 260-294; definitions: head of elevation, 261 ; head of pressure, 261 ; hydraulic mean depth, 262 ; intensity of pressure at a point, 261 ; loss of head 261 ; pressure, 260 ; wetted peri- meter, 262 {see Head, Pipes, etc.) Hurst, references to, 121, 239, 259, 264, 270, 275, 283. I X section, moment of inertia for, App. XIV. {see EoUed beams). Inertia, moment of, 50, 86, 104 ; of various sections, App. XIV. INDEX 359 Intensity of stress, 8. Iron, angle and tee, weight of, Table X. „ rolled, 81 ; girders (see Girders). „ cast, resistance of, 94 ; moduli, weight, ultimate resistance, etc.. Table I. ; safe resistance, 94, Table la., XXVI. „ columns (see Columns). „ pipes, strength, thickness, weight. Table XX. „ roofs, example of, 209 ; principal for, 120; struts for, 119. „ struts (see Struts). wrought, resistance of 82 ; moduli, weight, ultimate re- sistance, Table I. ; safe resist- ance, 82, Tables la., XXVI. J Jets, 288; example, 289, 291, 292 ; factors to find velocity, Table XXIV. ; height, 289 ; issuing velocity, 290 ; nozzles, 290 ; path described by, 289 ; range, 288. Joints, 122 ; at apex and feet of truss, 144 ; cotter, 148 ; in webs of plate girders, 161, 168; pin, 138, 140 ; riveted, 122 (see Riveted joints) ; screw, 145, 147. „ in wooden structures, fished, 149 ; scarfed, 151 ; ut foot of principal rafter, 152. „ (see Masonry). Joist, fir, 79 ; rolled iron, 89 ; steel, 92. L Lattice girders, 194. Layer, neutral, 41. Least resistance, Moseley's principle of, 248. Lever arm, 185. Light, C, reference to. Table XI. Lime, quick, weight of, Table XVII. Limestone, safe resistance of. Table la. ; weight of, Table, XVII. Limiting stress, 8. Line of resistance, 221. Links, various forms, 139. Live load, 4. „ and dead loads, comparative effect of, 5. Load, 2 ; breaking, 5 ; concentrated, 3 ; concentrated and distributed loads compared, App. VII. ; dead, 4 ; distributed, 4 ; live, 4 ; safe, 5 ; safe for fir beams. Table II. ; on supports, 18 ; working, 5. Long compression bars (see Compres- sion bars). Loss of head (hydraulic), 268 (see Head). M Masonry, safe resistance of. Table la. ; coefficient of friction. Table XVa. ; weight. Table XVII. „ joints, centre of pressure, 218, App. XIX. „ structures, important, 221, unimportant, 221 ; con- ditions of stability, 221; line of resistance, 221; resistance to sliding, 221, 224; resistance to crush- ing, 221, Table la.; re- sistance to shearing, 221 ; single block, 218 ; un- cemented 220. Materials, factors of safety for, 9, App. I. ; ultimate resistance, moduli, weight in, Table I. ; safe resistance. Table la. ; weight, Table XVII. Maxwell's diagrams, 182 ; for braced girders, 194 ; for roof trusses, 217; for trussed beams, 196, 198; Bow's system of lettering, App. XVII. ; method of drawing, 183 ; rules for drawing, App. XVI. Mean fibre, 135. Method of sections (see Sections) ; graphic (see Graphic). Modulus of elasticity, 1 2 ; of rupture, 52 ; of rupture for different sec- tions, 53 ; of rupture and elas- ticity for different materials, Table I. Molesworth, references to, 120, 247, 253, 259, 264, 265, 268, 270. 36o INDEX Moment, bending, 25 ; of flexure, 25 ; of inertia, 50, App. XIV. ; of re- sistance, 40 (see Bending Mo- ment, Resistance, etc.) Mortar, safe resistance of. Table la. ; weight. Table XVII. Moseley's line of resistance, 221 ; principle of least resistance, 248. N N Girder, 188 ; examples, stresses on, 190; Maxwell's diagrams for, 194 (see Braced girders). „ (braced) girder, 188. Neutral axis, 42 ; note on, Ap]3. IV, „ layer, 41. Neville's formula, flow of liquids in pipes, 264, 280. Newton's third law, 15. Notation, Bow's, 184, App. XVII. ; no- tation in this volume. Table XXV. Nozzles for jets, 290. 0 Oak, ultimate resistance, moduli, weight. Table I. ; safe resistance, Table la. Open-webbed girders, 187 (see Braced girder). Orifice of entry, loss of head due to, 269. Overbraced frames, 181. P Parabola, to draw a, App. III., to substitute circular arc for, 30, 34, 156. Perfectly braced frames, 180, 182. Perimeter, wetted, 262. Perry, Prof., reference to, App. XX. Pier, brick, calculation for, 222. Pillars, cast iron, examples of, 116, 117 ; at end of girders, 169. Pin joints, 138 ; example of, for roof truss, 140, 144. Pine, American yellow, and Pitch, ultimate resistance, moduli, and weight. Table I. ; safe resistance, Table la. Pipes, flow of liquids in, when run- ning full, 264, Tables XVIII., XIX. : formula;— Darcy's,265,267 ; Neville's, Eytelwein's, Thrupp's, 264 ; when running partially full, 280, Table XXIII.: formulae — Beardmore's, Downing's, Neville's, 280 ; Darcy's, 280, 281 ; thickness, weight, and strength of, Table XX. „ drain, 283. Pitch of rivets, 136. Plate girders, 154 ; booms or flanges, 155, 163, 166; connection of web, 161, 168 ; cover plates, 133; depth of, 163 ; flange plates, thick- ness of, 156, 163 ; flanges, width of, 163; hog-backed, 176 ; joints in flanges, 132, 165 ; joints in web, 161 ; proportion between depth and span, 154 ; web, 157, 166; web plates, thickness of. Table XI. (see Box girder). Plates, wrought iron, resistance of. Table I. Pole of diagram, App. VI. ; cylindri- cal, strength of, App. X. Polygon of forces, 180 ; funicular, App. VI. Portland cement, safe resistance, Table la. ; weight. Table XVII. ; stone, weight. Table XVII. Pressure, distribution of, on masonry joint, 219, App. XIX. ; hydraulic, 261 ; wind, 200, Table XIV. Principal rafter, iron, calculations for, examples, 120, 144 ; as jointed and as continuous, 216 ; loads on, 203 ; joint at head and foot, ex- ample, 144 ; wood joint at foot, example, 152. Punched holes in iron plates, 137 ; in tension joints, 137. R Rafter, common, wood, loads on, 201, continuous, App. XI. „ principal (see Principal rafter). Range of jet, 288. Rankine, reference to, 5, 9, 44, 50, INDEX 361 53, 66, 86, 158, 185, 224, 238, 242, 248, 353, App. I., XIV., XIX., Table XV«. Eeaction at supports of loaded beams, 15-20 ; of a cantilever, 22 ; of continuous beams, App. XI. ; of a girder, 21 ; of a roof, 23 ; rules for finding, 18, 20. Rectangle, moment of inertia, App. XIV. Eectangular beams, strength of, 54 ; deflection of, App. IX. Repose, angle of, 236, Table XVI. Resistance, 1 4 ; of masonry to crush- ing, 221; equivalent area of, 47 ; safe area of, 48 ; of masonry to sliding, 221; least (Moseley's prin- ciple), 248 ; ultimate, 49 ; Avorking, 49 {see Working stresses). „ to tension, compression, and shearing, ultimate and safe — of cast iron, 94 ; wrought iron, 82 ; of vari- ous materials. Tables I., la (see Working stresses). ,, to find moment of — beam, by experiment, 51 ; by graphic method, 46 ; by reasoning, 40; rolled beam, by graphic method, 83 ; by mathematical method, 86 ; cast iron beam, by graphic method, 96 ; by mathematical method, 99. Retaining walls, sections, 230 ; for earth, 236, 239 ; examples of, 241 - 243 ; foundations, 244 ; formula5, 238, 239, 244 ; graphic method, 238 ; for water, 232 ; example of, 234. Revetments, surcharged, 241. Ritter's formula for compression bars, 115 ; method of sections, 184, App. XVIII. Rivet holes, diameter and pitch of, 136 ; punched in iron plates, 137 ; rimered out, 138. Riveted joints, 122 ; butt, with single cover plate, 124; butt, with double cover plates, 125, 128 ; dimensions of, 136 ; double cover grouped, 131 ; examples of, 128, 131-134 ; formulae for grouped, 129 ; general formulae for, 127 ; in compression, 135 ; ia tension, 122 ; lap, 123 ; oblique, 134. Rivets, iron, 136 ; in drilled holes, 138 ; in punched holes, in iron plates, 137 ; strength of, Table VIIL „ steel, 138. Rolled wrought iron beams or joists, 81 ; advantages of, 92 ; de- flection, 91, Table IV. ; depth, 91; examples of, 88- 9 1 ; example, load borne by given beam, 88, 89 ; girders built up with, 93 ; market sections of, 92 ; moment of inertia, App. XIV. ; practical formulae for, 87, 88, 90 ; re- marks on, 91 ; section re- quired for given load, 89 ; stress diagrams, 83, 85. „ steel joints or beams, 92. Roofs, iron, 196; calculation by method of sections, App. XVIII. ; calculation of principal, 120, 214 ; diagrams for various trusses, 217 ; distribution of loads, 201 ; ex- ample, 209 ; joint at apex and sj)ringings, 144 ; loads, 199 ; oc- casional loads, 200 ; permanent loads, 199 ; pin joint, 140 ; reac- tions at supports, 22, 23, 205 ; snow, 200 ; stresses on, 213 ; strut, 119 ; weight of framing, Table XII. ; weight of roof cover- ing, Table XIII. ; wind pressure, 200, 208, 210, Table XIV. Rouleaux, formulae for wooden struts, 115. Rules for drawing Maxwell's dia- grams, App. XVI. ; finding reac- tions, 18, 20 ; shearing stress, 55 (.see Formulae). Rupture, modulus of, 52 ; for dif- ferent sections, 53 ; for various materials. Table I. 362 INDEX S Safe load, 5. „ stress 8 (see Working stresses). Safety, factor of, 9, App. I. Sand, angle of repose, Table XVI.; weight, Table XVII. Sandstone, safe resistance to compres- sion, Table la. ; weight, Table XVII. Scantlings for wooden roofs. Table XV. Scarf joint, in wooden beam, 151. Scheffler's theory, retaining wall, 239; arch, 247. Screw of tie-rod of roof, 147. Screws, 145. Sections, method of, 184 ; applied to iron roof, App. XVIII. ; to Warren girder, 188 ; to Whipple-Murphy girder, 190 ; lever arm, 185 ; turning point, 185. „ of box girder, 175, Plate B ; cast-iron girder, 97 ; plate girder, 154, Plate A ; rolled beams, 83 ; market, of rolled beams, 92. Seddon, reference to, 53, 100, 200, 201 ; Tables I, la., IX., XII., XIII., XV. (see Wray). Semicircle, moment of inertia for, App. XIV. Sewer, egg-shaped, 287. Shaler Smith, references to, 139, Table IX. Shearing, resistance to, of cast iron, 95 ; wrought iron, 82 ; of iron and steel. Table I. ; of rivets, 123, Table Villi ; safe resistance to, of iron, steel, and timber. Table la. „ stress, 7 ; beams (various cases), 57-60 ; diagrams, 55, 60, 157, 166, App. V. ; distribution of, 60 ; graphic method of finding, App. VI.; horizontal, 55, 60 ; resistance to, in rolled beam, 9 1 ; rule for finding, in beams, 55 ; vertical, 55 ; in web of plate girders, 157, 166, 173, Table XI. Shingle, angle of repose. Table XVI. Short compression bars, 112 (see Com- pression bars). Shortening, 11. Slate, weight of. Table XVII. Sleeve, 148. Sliding, resistance to, 221, 224, 245. Stability of brickwork and masonry structures, 218. Steel, ultimate resistance, moduli, weight. Table I. ; safe resist- ance, Table la. ,, rolled joists, market sections, 92 ; resistance of, to compres- sion and tension, 92. Stiffeners, 159. Stififest rectangular beam cut out of a round log, 80. Stiffness, 12 ; of beams compared, App. VIII. Stoney, references to, 136, 137, 138, 141, 161, Table VII. Strain, 5, Table of, 9. Strength of materials. Table I., la. ; of beams of fir. Table II. ; of rectangular beams com- pared, App. VIII. ; of beams of different sections, App. X. ; of rivets, Table VIII. ; of struts. Table V. ; of wooden struts. Table VI. „ uniform, 64. Stresses, 6 ; in beams, 41 ; bearing, 7 ; breaking, 8 ; classification of, 8 ; compressile, 6 ; in fixed beams, 69 ; limiting, 8 ; intensity of, 8 ; repeated, 1 1 ; safe work- ing, 8, Table XXVI.; shearing, 7 ; Table of, 9 ; tensile, 6 ; torsional, 7 ; transverse, 6 ; variation of, 1 1 (see Stress diagrams, Working stress, etc.) Strongest rectangular beam that can be cut out of a round log, 80. Structures, braced or framed, 177. Struts (see Compression bars), 109 ; examples — cast - iron columns, 116, 117 ; timber, 115, 118 ; wrought-iron, 119, 120; Fidler's rules for, 109 ; formula? for, 113, INDEX 363 114, 115 ; Tables for, Tables V., VI., VII. Supports, loads on, 18 ; I'eactions at Rule I., 18 ; Rule II., 20. Surcharged revetments, 241 ; example, 243, T T iron, moment of inertia for, App. XIV. ; weight of. Table X. Tables, various (see List, p. xvi.) Taps, flow of water from, 272, 278. Teak, ultimate resistance, moduli, and weight. Table I. ; safe resistance, Table la. Tension, 6 ; resistance to, of cast iron, 94 ; of various materials. Tables I., la. ; of wrought iron, 82 ; riveted joints in, 122. „ bars, 105 ; effective area of, 106 ; examples of, 108 ; eyes of wrought iron, 139, Table IX. Terms in use, 2 Thickness of arch I'ing, 253, Table XVIIa. „ of pipes. Table XX. Third law, Newton's, 15. Thrupp's formula for the flow of liquids in pipes, 264. Tie-rod, screw of, 147. Timber beams, 24 ; examples, of, 78, 7 9 ; modulus of rupture, 52; of different sections comj)ared, App. X. ; for- mula? for, 54, App. XXI. ; stiffest beam from log, 80 ; strength of, 54 ; strength of beams of fir, Table II. ; strongest beam from log, 80 ; struts, 115, 118. „ ultimate resistance, moduli, and weight. Table I. ; safe resistance. Table la. „ cantilever, 76-79. „ roofs, note, 217 ; scantlings for. Table XV. „ structures, joints in, 149. Torsion, 7. Transverse stress, 7. Triangle of forces, 179. Trussed beams, 195 ; example, 196. Turning point, 185. U Ultimate resistance, 49 ; of cast iron, 94 ; of wrought iron, 82 ; of vari- ous materials. Table I. Uniform strength, beams of, 62 ; constant breadth, 64 ; constant depth, 65 ; cast-iron girders, 100. Un win's formulas for weight of girders, App. XV. ; references to, 1 25, 1 36, 140, 163, 200, Apps. I., XV., Table XVII. V Variation of stress, 11. Velocity, loss of head due to, 269. Vertical shearing stress, cases of, 55- 60 ; rules for finding, 55. W Walls, enclosure, 227 ; retaining, 230; retaining for earth, 236; for water, 232 (see Retaining walls). Warren girder, 187 ; Maxwell's dia- grams for, 194 (see Braced girder); stresses on, 188. Water supply, examples of, for a house, constant, 274 ; intermittent, 271 ; for a street, 277 (see Hydraulics, Pipes, etc.) Web of plate girders, 157, 166, 173 ; connection to flanges, 161, 167 ; joints in, 161, 168, 175 ; thick- ness of, 160, 168, 174, Table XI. Weight of earths, stones, etc., Table XVII. ; of girders (Anderson's and Unwin's formulae), App. XV. ; of metals and timber. Table I. ; of roof covering. Table XIII. ; of roof framing. Table XII. ; of T and angle iron. Table X. Whipple-Murphy girder, 187 ; ex- ample, 190 ; Maxwell's diagram for, 194 (see Braced girder). 364 INDEX Wind pressure, on chimney, 225 ; on roof, 200, Table XIV. ; on wall, 227. Wohler's experiments, 11. Wooden roofs, scantlings for, Table XV., note, 217. ,, structures, joints in, 149 {see Joints). „ beams {see Timber beams). Working load, 5. „ resistance, 49. „ stresses or resistance, general. Tables la., XXVI.; Brick- work, Table la. ; cast iron, 94 ; concrete, Table la. ; pin joints, 141 ; rolled beams, 82 ; riveted joints 123 ; screws, 146 ; tim- ber, Table la. Wray (Seddon), references to, 100, 200, 201, Table la., IX., XIL, XIII., XV. Wrought iron, resistance of, 82 ; mod- uli, weight, ultimate resistance. Table I. ; safe resistance, 82, Table la, XXVI. „ cantilever, 22 ; strut for roof, 119; rolled beams, 81; principal for roof, calculations, 120, 214, 216. END OF PART IV. Printed by R. & R. Clark, Edinburgh. PLATE I. Compressions Red Tensions Blue Scale of Forces 1 Inch- = 1 Ton. O'lS , , , . ^ 1 2 aTans D PIG 283, PIG 284. FIG 285. PIG 236. PLATE 2. Compressions Red Tei\sions Blue ScaLe of forces 1 Inch -3 Owt J ^ 0 7 2 3 4 S 6 .7 B 9 OvtS C FIG 811 T PLATE 4. Compressions Red Tensions Blue Scale of Poroes % - 1 Ton. FIG 361. FIG 362. FIG 363 PLATE 5. Ccst^jressions Red Tensions Blue ■ Scale cf Forces -J Ind\-1 Ton. OS o 1 J 5 ficms PLATE 7. Compressions Red Tensions , .Blue PLATE 8. Compressions Red Tensions Blue