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REESE LIBRARY UNIVERSITY OF CALIFORNIA. Received.. ^-/tfai*^ ,i8?/ Accessions No. Shelf No.- / ARCHITECTURE. BUILDING CARPENTEY STAIRS/ ETC. THE ARCHITECT'S AND BUILDER'S POCKET BOOK KIDDER. Of Mensuration, Geometry, Trigonometry, Rules, Tables, and Formulas relating to the strength and stability of Foundations, Walls, Buttresses, Piers, Arches, Posts, Ties, Beams, Girders, Trusses, Floors, Roofs, etc., etc. Statistics and Tables Relating to Carpentry, Masonry, Drainage, Painting and Glazing, Plumbing, Plastering, Roofing, Heating and Ventilation, Weights of Materials, Capacity and Dimensions of Churches, Theatres, Domes, Towers, Spires, etc., etc. By F. E. Kidder. C.E., Consulting Architect. Upwards of 600 pages and over 400 plates. Fourth edition, enlarged (1887) morocco flaps, $3 50 "The book admirably fulfills its purpose of becoming an indispensable companion in the work of every architect, young or old." American Architect. " We do not hesitate to recommend this new pocket book." Building. THE AMERICAN HOUSE CARPENTER. HATFIELD. A Treatise on the Art of Building, comprising Styles of Archi- tecture, Strength of Materials, and tLe Theory and Practice of the Construction of Floors, Framed Girders, Roof Trusses, Rolled Iron Beams, Tubular Iron Girders, Cast Iron Girders, Stairs, Doors, Windows, Mouldings and Cornices, together with a Compend of Mathematics. A Manual for the Practical use of Architects, Carpenters, and Stair Builders. By R. G. Hatfield. Re-written and enlarged. Numerous*fine wood en- gravings 8vo, cloth, 5 00 NOTICES OF FORMER EDITIONS OF THE WORK. "The clearest and most thoroughly practical work on the subject." "It is a valuable addition to the library of the architect, and almost indispensable to every scientific master-mechanic." .ft. R. Journal. THE THEORY OP TRANSVERSE STRAINS. HATFIELD. And its Application to the Construction of Buildings, includ- ing a full discussion of the Theory and Construction of Floor Beams, Girders, Headers, Carriage) Beams, Bridging, Rolled Iron Beams, Tubular Iron Girders, Cast Iron Girders, Framed Girders and Roof Tresses, with Tables calculated expressly for the work, etc., etc. By R. G. Hatfield, Architect. Fully illus- trated. Second edition, with additions 8vo, cloth, 5 00 "The 'Theory of Transverse Strains' is undoubtedly the most important contribution to the Science of Building that has yet appeared." London, Architect. "This work i* a genuine surprise. It is so seldom one finds an architect who is thoroughly posted in what may be termed the 'Engineering' of Architecture, that we may well be pardoned the above expression. * ~* * We lay down the book with the feeling i hat the author has filled a gap in architectural literature, and done it well." Engineering News, STAIR-BUILDING MONCKTON. In its various forms and the One Plane Method of Hand Railing as applied to drawing Face Moulds, unfolding the center line of Wreaths, giving lengths of Balusters under all Wreaths. Numerous Designs of Stairs, Newels, and Balusters for the use of Architects, Stair Builders, and Carpenters. By James H. Monckton. Illustrated by 81 full-page plates df working drawings, etc . . 4to, cloth extra, 6 00 ''It may certainly be considered the champion work of its kind, and not only stair builders, but architects, owe you a debt of gratitude for making their labors lighter in this regard." O. P. HATFIELD, Architect. COTTAGE RESIDENCES. DOWNING. A Series of Designs for Rural Cottages and Cottage Villas, SARGENT. and their Garden Grounds. By A. J. Downing. Containing HARNEY. a revised list of Trees, Shrubs, and Plants, and the most recent and best selected Fruit, with some account of the newer style of Gardens. By Henry Winthrop Sargent and Charles Downing. 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Containing a Complete Treatise on Framing Hip and Valley Roofs, together with much valuable instruction for all Me- chanics and Amateurs, useful Rules, Tables never before pub- lished. By H. W. Holly. New ed., with additions. 18mo, cloth, 75 SAW FILING. The Art of Saw Filing Scientifically Treated and Explained on Philosophical Principles. With explicit directions for putting in order all kinds of Saws, from a Jeweler's Saw to a Steam Saw Mill. Illustrated by 44 engravings. Third edition. By H. W. Holly 18mo, cloth, 75 * SEVEN LAMPS OF ARCHITECTURE. New and beautiful edition, containing very fine copies of all the plates of the original London edition. Lamp of Sacrifice. Lamp of Truth. Lamp of Power. Lamp of Beauty. Lamp of Life. Lamp of Memory. Lamp of Obedience. By John Ruskin 8vo, extra cloth, 6 00 DITTO. With all the plates 12mo, extra cloth, 1 25 DITTO. " " ....12mo, russet cloth, 075 LECTURES ON ARCHITECTURE AND PAINTING. Delivered at Edinburgh. By John Ruskin. 12mo, russei cloth, 050 DITTO. 15 plates, full page, printed separately. 12mo, russet cloth, 1 00 * STONES OF VENICE. New and beautiful edition. Containing fine copies of all the plates and engravings of the original London edition. Colored and plain. Vol 1. Foundations. Vol.2. Sea Stories. Vol.3. The Fall. 54 full page plates 3 vols., 8vo, extra cloth, 18 00 DITTO. With engravings only. 3 vols. in two, 12mo, russet cloth, 1 50 DITTO. With 54 plates and engravings 3 vols. in two, cloth. 3 00 DITTO. With 54 plates and engravings. 3 vols. in box, 12ino, ex. cloth, 4 50 DITTO. With 54 plates and engravings. 3 vols., 12mo, half calf, 7 50 DITTO. People's edition 3 vols. in one. Neat blue cloth. 125 THE POETRY OF ARCHITECTURE. Cottage, Villa, etc., to which is added Suggestions on Works of Art. With numerous illustrations. By John Ruskin. 12mo, russet cloth, 50 ST. MARK'S REST. THE HISTORY OF VENICE. Written for the help of the Few Travelers who still care for her Monuments. Parts I., II. and III., with two Supplements. By John Ruskin 12mo, russet cloth, 050 AN INQUIRY INTO SOME OF THE CONDITIONS AFFECTING "THE STUDY OF ARCHITEC- TURE" IN OUR SCHOOLS. By John Ruskin. 12mo, paper, 10 THEORY OF TRANSVERSE STRAINS AND ITS APPLICATION IN THE CONSTRUCTION OF BUILDINGS, INCLUDING A FULL DISCUSSION OF THE THEORY AND CONSTRUCTION OF FLOOR BEAMS, GIRIJERS, HEADERS, CARRIAGE BEAMS, BRIDGING, ROLLED-IRON BEAMS, TUBULAR IRON GIRDERS, CAST-IRON GIRDERS, FRAMED GIRDERS, AND ROOF TRUSSES ; WITH TABLES, Calculated and prepared expressly for this Work, OF THE DIMENSIONS OF FLOOR BEAMS, HEADERS AND ROLLED-IRON BEAMS ; AND TABLES SHOWING RESULTS OF ORIGINAL EXPERIMENTS ON THE TENSILE, TRANS- VERSE, AND COMPRESSIVE STRENGTHS OF AMERICAN WOODS. BY R. G. HATFIELD, ARCHITECT. FELLOW AM. INST. OF ARCHITECTS ; MEM. AM. SOC. OF CIVIL ENGINEERS; AUTHOR OF "AMERICAN HOUSE CARPENTER." THIRD EDITION, REVISED AND ENLARGED. JOHN WILEY & SONS, 15 ASTOR PLACE. 1889. COPYRIGHT, 1877. JOHN WILEY & SONS. PREFACE. THIS work is intended for architects- and students of architec- ture. Within the last ten years, many books have been written upon the mathematics of construction. Among them are several of particular excellence. Few, however, are of a character adapted to the specific wants of the architect. The subject is treated, by some, in the abstract, and in a manner so diffuse and 'general as to be useful only to instructors. In other works, where a prac- tical application is made, the wants of the civil engineer rather than of the architect are consulte.4. Writers of scientific books, as well as the -public at large, have failed to appreciate the wants of the architect. Indeed, many architects are content to forego a knowledge of construction ; following precedent as far as pre- cedent will lead, and, for the rest, trusting to the chances of mere guess-work. For such, all scientific works are alike useless ; but there is a class of architects who, through a faulty system of edu- cation, have failed to obtain, while students, the knowledge they need ; and who now have little time and less inclination to apply themseh^es to abstract or inappropriate works, although feeling keenly the need of some knowledge which will help them in their daily duties. For this class, and for students in architecture, this book is written. In fitting it for its purpose, the course adopted has been to present an idea at first in concrete form, and then to lead the mind gradually to the abstract truth or first principles upon which the idea is based. This method, or the manner in which it is executed, may not meet the approval of all. Nevertheless, it is hoped that those for whom the work is written may, by its help, acquire the knowledge they need, and be enabled to solve readily the problems arising in their professional practice. 4 PREFACE. To adapt the work to the attainments of younger students, the attempt has been made to present the ideas, especially in the first chapters, in a simple manner, elaborating them to a greater extent than is usual. The graphical method of illustration has been employed largely, and by its help some of the more abstruse parts of the science of construction, it is thought, have been made plain. Results obtained by this method have been analyzed and shown to accord with the analytical formulas heretofore employed. In a discussion of the relation between strength and stiffness, a method has been developed for determining the factor of safety in the rules for strength. Rules for carriage beams with two and three headers are given. The subject of bridging has been dis- cussed, and the value of this system of stiffening floors defined. Especial attention has been given to the chapters on tubular iron girders, rolled-iron beams, framed girders and roofs; and these chapters, it is hoped, will be particularly acceptable to architects. The rules for the various timbers of floors, trussed girders, and roof trusses, are all accompanied by practical examples worked out in detail. Tables are given containing the dimen- sions of floor beams and headers for all floors. These tables are in two classes ; one for dwellings and assembly rooms, the other for first-class stores; and give dimensions for beams of Georgia pine, spruce, white pine and hemlock, and for rolled-iron beams. Immediately following the tables will be found a directory, or digest, by which the more important formulas are so classified that the proper one for any particular use may be discerned at a glance. The occurrence recently of conflagrations, resulting in serious loss of life, has shown the necessity of using every expedient cal- culated to render at least our public buildings less liable to destruction by fire. To this end it is proposed to construct timber floors solid, laying the beams in contact, so as to close the usual spaces between the beams, and thus prevent the passage of air, and thereby retard the flames. The strength of these solid floors has been discussed in Article 702, and a rule been obtained for the depth of beam or thickness of floor. By this rule the depths for floors of various spans have been computed, and the results re- corded in table XXI. PREFACE. 5 Tables XXIII. to XLVI. contain a record of experiments made, expressly for this work, upon six of our American woods. In these experiments and in computations, the author has been as- sisted by his son, Mr. R. F. Hatfield. In the preparation of the work, he has had recourse to the works of numerous writers on the strength of materials, to whom he is under obligation, and here makes his acknowledg- ments. The following are the works which were more particu- larly consulted : Baker on Beams, Columns, and Arches. Barlow on Materials and on Construction. Bow on Bracing. Bow's Economics of Construction. Campin on Iron Roofs. Cargill's Strains upon Bridge Girders and Roof Trusses. Clark on the Britannia and Conway Tubular Bridges. Emerson's Principles of Mechanics. Fairbairn on Cast and Wrought Iron. Fenwick on the Mechanics of Construction. Francis on the Strength of Cast-Iron Pillars. Haswell's Engineers' and Mechanics' Pocket-Book. Haupt on Bridge Construction. Hodgkinson's Tredgold on the Strength of Cast-Iron. Humber on. Strains in Girders. Hurst's Tredgold on Carpentry. Kirkaldy's Experiments on Wrought-Iron and Steel. Mahan's Civil Engineering. Mahan's Moseley's Engineering and Architecture. Moseley's Engineering and Architecture. Poisson's Traiie de Mecanique. Ranken on Strains in Trusses. Rankine's Applied Mechanics. Robison's Mechanical Philosophy. Rondelet sur le Dome du Pantheon Franais. Sheilds' Strains on Structures of Ironwork. Styffe on Iron and Steel. Tarn on the Science of Building. Tate on the Strength of Materials. Tredgold's Carpentry. Unwin on Iron Bridges and Roofs. Weisbach's Mechanics and Engineering. Wood on the Resistance of Materials. GENERAL CONTENTS. INTRODUCTION. CHAPTER I. THE LAW OF RESISTANCE. CHAPTER. II. APPLICATION OF THE LEVER PRINCIPLE. CHAPTER III. DESTRUCTIVE ENERGY AND RESISTANCE. CHAPTER IV. EFFECT OF WEIGHT AS REGARDS ITS POSITION. CHAPTER V. COMPARISON OF CONDITIONS SAFE LOAD. CHAPTER VI. APPLICATION OF RULES FLOORS. CHAPTER VII. GIRDERS, HEADERS AND CARRIAGE BEAMS. CHAPTER VIII. GRAPHICAL REPRESENTATIONS. CHAPTER IX. STRAINS REPRESENTED GRAPHICALLY. CHAPTER X. STRAINS FROM UNIFORMLY DISTRIBUTED LOADS. CHAPTER XI. STRAINS IN LEVERS, GRAPHICALLY EXPRESSED. GENERAL CONTENTS. CHAPTER XII. COMPOUND STRAINS IN BEAMS, GRAPHICALLY EXPRESSED. CHAPTER XIII. DEFLECTING ENERGY. CHAPTER XIV. RESISTANCE TO FLEXURE. CHAPTER XV. RESISTANCE TO FLEXURE LIMIT OF ELASTICITY. CHAPTER XVI. RESISTANCE TO FLEXURE RULES. CHAPTER XVII. RESISTANCE TO FLEXURE FLOOR BEAMS. CHAPTER XVIII. BRIDGING FLOOR BEAMS. CHAPTER XIX. ROLLED-IRON BEAMS. CHAPTER XX. TUBULAR IRON GIRDERS. CHAPTER XXI. CAST-IRON GIRDERS. CHAPTER XXII. FRAMED GIRDERS. CHAPTER XXIII. ROOF TRUSSES. CHAPTER XXIV. TABLES. DIGEST OR DIRECTORY INDEX. ANSWERS TO QUESTIONS. CONTENTS. INTRODUCTION. ART. 1. Construction Defined 2. Stability Indispensable. 3. Laws Governing the Force of Gravity. 4. Science of Construction, for Architect rather than Builder. 5. Parts of Buildings requiring Special Attention. 6. This Work Limited to the Transverse Strain. 7. In Construction Safety Indispensable. 8. Some Floors are Deficient in Strength. 9. Precedents not always Accessible. 10. An Experimental Floor, an Expensive Test. 11. Requisite Knowledge through Specimen Tests. 12. Unit of Material Its Dimensions. 13. Value of the Unit for Seven Kinds of Material. CHAPTER I. THE LAW OF RESISTANCE. 14. Relation between Size and Strength. 15. Strength not always in Proportion to Area of Cross-section. 16. Resistance in Proportion to Area of Cross-section. 17. Units may be Taken of any Given Dimensions. 18. Experience Shows a Beam Stronger when Set on Edge. 19. Strength Directly in Proportion to Breadth. 20. By Experiment Strength Increases more Rapidly than the Depth. 2 1. Comparison of a Solid Beam with a Laminated one. 22. Strength due to Resistance of Fibres to Extension and Compression. 23, Power Extending Fibres in Proportion to Depth of Beam. CHAPTER II. APPLICATION OF THE LEVER PRINCIPLE. 24. The Law of the Lever. 25. Equilibrium Direction of Pressures. CONTENTS. 26 Conditions of Pressure in a Loaded Beam. 27. The Principle of the Lever. 28. A Loaded Beam Supported at Each End. 29. A Bent Lever. 30. Horizontal Strains Illustrated by the Bent Lever. 31. Resistance of Fibres in Proportion to the Depth of Beam. CHAPTER III. DESTRUCTIVE ENERGY AND RESISTANCE. 32. Resistance to Compression Neutral Line. 33. Elements of Resistance to Rupture. 34. Destructive Energies. 35. Rule for Transverse Strength of Beams. 36. Formulas Derived from this Rule. 37 to 51. Questions for Practice. CHAPTER IV. THE EFFECT OF WEIGHT AS REGARDS ITS POSITION. 52. Relation between Destructive Energy and Resistance. 53. Dimensions and Weights to be of Like Denominations with those of the Unit Adopted. 54. Position of the Weight upon the Beam. 55. Formula Modified to Apply to a Lever. 56. Effect of a Load at Any Point in a Beam. 57. Rule for a Beam Loaded at Any Point. 58. Effect of an Equally Distributed Load. 59. Effect at Middle from an Equally Distributed Load. 60. Example of Effect of an Equally Distributed Load. 61. Result also Obtained by the Lever Principle. 62 to 65. Questions for Practice. CHAPTER V. COMPARISON OF CONDITIONS SAFE LOAD. 66. Relation between Lengths, Weights and Effects. 67. Equal Effects. 10 CONTENTS. 68. Comparison of Lengths and Weights Producing Equal Effects. 69. Tne Effects from Equal Weights and Lengths. 70. Rules for Gases in which the Weights and Lengths are Equal. 71. Breaking and Safe Loads. 72. The above Rules Useful Only in Experiments. 73 Value of a, the Symbol of Safety. 74. Value of a, the Symbol of Safety. 75 Rules for Safe Loads. 76. Applications of the Rules. 77. Example of Load at End of Lever. 78. Arithmetical Exemplification of the Rule. 79. Caution in Regard to a, the Symbol of Safety. 80. Various Methods of Solving a Problem. 81. Example of Uniformly Distributed Load on Lever. 82. Load Concentrated at Middle of Beam. 83. Load Uniformly Distributed on Beam Supported at Both Ends. 84 to 87. Questions for Practice. CHAPTER VI. APPLICATION OF RULES FLOORS. 88. Application of Rules to Construction of Floors. 89. Proper Rule for Floors. 90. The Load on Ordinary Floors, Equally Distributed. 9 I . Floors of Warehouses, Factories and Mills. 92. Rule for Load upon a Fioor Beam. 93. Nature of the Load upon a Floor Beam. 94. Weight of Wooden Beams. 95. Weight in Stores, Factories and Mills to be Estimated. 96. Weight of Floor Plank. 97. Weight of Plastering. 98. Weight of Beams in Dwellings. 99. Weight of Floors in Dwellings. 100. Superimposed Load. 101. Greatest Load upon a Floor. 102. Tredgold's Estimate of Weight on a Floor. 103. Tredgold's Estimate not Substantiated by Proof. 104. Weight of People Sundry Authorities. 105. Estimated Weight of People per Square Foot of Floor. 106. Weight of People, Estimated as a Live Load. 107. Weight of Military. 103. Actual Weights of Men at Jackson's and at Hoes' Foundries. 109. Actual Measure of Live Load. II 0. More Space Required for Live Load. III . No Addition to Strain by Live Load. CONTENTS. 1 1 1 1 2. Margin of Safety Ample for Momentary Extra Strain in Extreme Cases. 1 1 3. Weight Reduced by Furniture Reducing Standing Room. 1 1 4. The Greatest Load to be provided for is 70 Pounds per Super- ficial Foot. 1 1 5. Rule for Floors of Dwellings. 116. Distinguishing Between Known and Unknown Quantities. 117' Practical Example. 118. Eliminating Unknown Quantities. 1 1 9. Isolating the Required Unknown Quantity. 120. Distance from Centres at Given Breadth and Depth. 1 21. Distance from Centres at Another Breadth and Depth. 1 22. Distance from Centres at a Third Breadth and Depth. 123. Breadth, the Depth and Distance from Centres being Given. 124. Depth, the Breadth and Distance from Centres being Given. 125. General Rules for Strength of Beams. 126 to 135. Questions for Practice. CHAPTER VII. GIRDERS, HEADERS AND CARRIAGE BEAMS. 136 A Girder Denned. 137 Rule for Girders. 138. Distance between Centres of Girders. 139. Example of Distance from Centres. 140. Size of Girder Required in above Example. 141. Framing for Fireplaces, Stairs and Light-wells. 142. Definition of Carriage Beams, Headers and Tail Beams. 143. Formula for Headers General Considerations. 144. Allowance for Damage by Mortising. 145. Rule for Headers. 146. Example. 147. Carriage Beams and Bridle Irons. 148,-Rule for Bridle Irons. 149. Example. 150. Rule for Carriage Beam with One Header. 151. Example. 152. Carriage Beam with Two Headers. 153. Effect of Two Weights at the Location of One of Them. 154. Example. 155. Rule for Carriage Beam with Two Headers and Two Sets of Tail Beams. 156. Example. 157. Rule for Carriage Beam with Two Headers and One Set of Tail Beams. 158. Example. 159 to 166. Questions for Practice. 12 CONTENTS. CHAPTER VIII. GRAPHICAL REPRESENTATIONS. 167. Advantages of Graphical Representations. 168. Strains in a Lever Measured by Scale. 169. Example Rule for Dimensions. 170. Graphical Strains in a Double Lever. 171. Graphical Strains in a Beam. 172. Nature of the Shearing Strain. 173. Transverse and Shearing Strains Compared. 174. Rule for Shearing Strain at Ends of Beams. 175. Resistance to Side Pressure. 176. Bearing Surface of Beams upon Walls. 177. Example to Find Bearing Surface. 178. Shape of Side of Beam, Graphically Expressed. 179 to 187. Questions for Practice. CHAPTER IX. STRAINS REPRESENTED GRAPHICALLY. 188. Graphic Method Extended to Other Cases. 189. Application to Double Lever with Unequal Arms. 190. Applicati6n to Beam with Weight at Any Point. 191. Example. 192. Graphical Strains by Two Weights. 193. Demonstration. 194. Demonstration Rule for the Varying Depths. 195. Graphical Strains by Three Weights. 196. Graphical Strains by Three Equal Weights Equably Disposed. 197. Graphical Strains by Four Equal Weights Equably Disposed. 198.^Graphical Strains by Five Equal Weights Equably Disposed. 199._General Results from Equal Weights Equably Disposed. 200. General Expression for Full Strain at First Weight. 201. General Expression for Full Strain at Second Weight. 202. General Expression for Full Strain at Any Weight. 203. Example. 204 to 209. Questions for Practics. CONTENTS. 1 3 CHAPTER X. STRAINS FROM UNIFORMLY DISTRIBUTED LOADS. 210. Distinction Between a Series of Concentrated Weights and a Thoroughly Distributed Load. 211 1 Demonstration. 212. Demonstration by the Calculus. 213. Distinction Shown by Scales of Strains. 214. Effect at Any Point by an Equally Distributed Load. 215. Shape of Side of Beam for an Equably Distributed Load. 216. The Form of Side of Beam a Semi-ellipse. 217 to 220. Questions for Practice. CHAPTER XL STRAINS IN LEVERS, GRAPHICALLY EXPRESSED. 221. Scale of Strains for Promiscuously Loaded Lever. 222. Strains and Sizes of Lever Uniformly Loaded. 223. The Form of Side of Lever a Triangle. 224. Combinations of Conditions. 225. Strains and Dimensions for Compound Load. 226. Scale of Strains for Compound Loads. 227. Scale of Strains for Promiscuous Load. 228 to 233. Questions for Practice. CHAPTER XII. COMPOUND STRAINS IN BEAMS, GRAPHICALLY EXPRESSED. 234. Equably Distributed and Concentrated Loads on a Beam. 235. Greatest Strain Graphically Represented. 236. Location of Greatest Strain Analytically Defined. 237. Location of Greatest Strain Differentially Defined. 238. Greatest Strain Analytically Defined. 239. Example. 240. Dimensions of Beam for Distributed and Concentrated Loads. 241. Comparison of Formulas, Here and in Art. 150. 242. Location of Greatest Strain Differentially Defined, 243. Greatest Strain and Dimensions. 244. Assigning the Symbols. 245. Example Strain and Size at a Given Point. 246. Example Greatest Strain. 247. Example Dimensions. 14 CONTENTS. 248, Dimensions for Greatest Strain when h Equals n. 249. Dimensions for Greatest Strain when // is Greater than n. 250i Rule for Carriage Beams with Two Headers and Two Sets of Tail Beams. 251. Example. 2.52. Carriage Beam with Three Headers 253, Three Headers Strains of the First Class. 254. Graphical Representation. 255. Greatest Strain. 256. General Rule for Equably Distributed and Three Concentrated Loads. 257. Example. 258. Rule for Carriage Beams with Three Headers and Two Sets of Tail Beams. 259. Example. 260. Three Headers Strains of the Second Class. 261. Greatest Strain. 262. General Rule for Equally Distributed and Three Concentrated Loads. 263. Example. 264. Assigning the Symbols. 265. Reassigning the Symbols. 266. Example. 267. Rule for Carriage Beam with Three Headers and Two Sets of Tail Beams. 268. Example. 269 and 270. Questions for Practice. CHAPTER XIII. DEFLECTING ENERGY. 271. Previously Given Rules are for Rupture. 272. Beam not only to Be Safe, but to Appear Safe. 273. All Materials Possess Elasticity. 274. Limits of Elasticity Denned. 275. A Knowledge of the Limits of Elasticity Requisite. 276. Extension Directly as the Force. 277. Extension Directly as the Length. 278, Amount of Deflection. 279. The First Step. 280. Deflection to be Obtained from the Extension. 281. Deflection Directly as the Extension. 282. Deflection Directly as the Force, and as the Length. 283. Deflection Directly as the Length. 284. Deflection Directly as the Length. 285. Total Deflection Directly as the Cube of the Length. 286. Deflecting Energy Directly as the Weight and Cube of the Length. 287 to 291. Questions for Practice. CONTENTS. 1 5 i CHAPTER XIV. RESISTANCE TO FLEXURE. 292, Resistance to Rupture, Directly as the Square of the Depth. 293. Resistance to Extension Graphically Shown. 294. Resistance to Extension in Proportion to the Number of Fibres and their Distance from Neutral Line. 295. Illustration. 296. Summing up the Resistances of the Fibres. 297. True Value to which these Results Approximate. 298. True Value Denned by the Calculus. 299. Sum of the Two Resistances, to Extension and to Compression. 300. Formula for Deflection in Levers. 301. Formica, for Deflection in Beams. 302. Value of F, the Symbol for Resistance to Flexure. 303. Comparison of F with , the Modulus of Elasticity. 304. Relative Value of F and E. 305. Comparison of F with E common, and with the E of Barlow. 306. Example under the Rule for Flexure. 307 to 310. Questions for Practice. CHAPTER XV. RESISTANCE TO FLEXURE LIMIT OF ELASTICITY. 311. Rules for Rupture and for Flexure Compared. 312. The Value of a, the Symbol for Safe Weight. 313. Rate of Deflection per Foot Length of Beam. 314. Rate of Deflection in Floors. 315 to 319. Questions for Practice. CHAPTER XVI. RESISTANCE TO FLEXURE RULES. 320. Deflection of a Beam, with Example. 321. Precautions as to Values of Constants F and <?. 322. Values of Constants F and e to be Derived from Actual Experiment in Certain Cases. 1 6 CONTENTS. 323. Deflection of a Lever. 324. Example. 325. Test by Rule for Elastic Limit in a Lever. 326. Load Producing a Given Deflection in a Beam. 327. Example. 328. LoaJ at the Limit of Elasticity in a Beam. 329, Load Producing a Given Deflection in a Lever Example 330.--Deflection in a Lever at the Limit of Elasticity. 331. Load on Lever at the Limit of Elasticity. 332. Values of W, /, b, d and c5 in a Beam. 333. Example Value of / in a Beam. 334. Example Value of b in a Beam. 335. Example Value of d in a Beam. 336. Values of P, n, l>, d and 6 in a Lever. 337. Example Value of in a Lever. 338. Example Value of b in a Lever. 339. Example Value of d in a Lever. 340. Deflection Uniformly Distributed Load on a Beam. 341. Values of U, I, b, d and <5 in a Beam, 342. Example Value of U, the Weight, in a Beam. 343. Example Value of /, the Length, in a Beam. 344. Example Value of l>, the Breadth, in a Beam. 345. Example Value^of d, the Depth, in a Beam. 346. Example Value of (5, the Deflection, in a Beam. 347. Deflection Uniformly Distributed Load on a Lever. 348. Values of U, n, (>, d and d in a Lever. 349. Example Value of U, the Weight, in a Lever. 350. Example Value of , the Length, in a Lever. 351. Example Value of b, the Breadth, in a Lever, 352. Example Value of </, the Depth, in a Lever, 353. Example Value of (5, the Deflection, in a Lever. 354 to 357. Questions for Practice. CHAPTER XVII. RESISTANCE TO FLEXURE FLOOR BEAMS. 358. Stiffness a Requisite in Floor Beams. 359. General Rule for Floor Beams. 360. The Rule Modified. 361. Rule for Dwellings and Assembly Rooms. 362. Rules giving the Values of c, /. b and &. 363. Example Distance from Centres, 364. Example Length, 365. Example Breadth. 366. Example Depth. CONTENTS. I 367. Floor Beams for Stores. 368. Floor Beams of First-class Stores. 369. Rule for Beams of First-class Stores. 370. Values of c, I, b and d. 371. Example Distance from Centres. 372. Example Length. 373. Example Breadth. 374. Example Depth. 375. Headers and Trimmers. 376. Strength and Stiffness Relation of Formulas. 377. Strength and Stiffness Value of a, in Terms of B and F. 378. Example. 379. Test of the Rule. 380. Rules for Strength and Stiffness Resolvable. 381. Rule for the Breadth of a Header. 382. Example of a Header for a Dwelling. 383. Example of a Header in a First class Store. 384, Carriage Beam with One Header. 385. Carriage Beam with One Header, for Dwellings. 386. Example. 387. Carriage Beam with One Header, for First-class Stores. 388. Example. 389. Carriage Beam with One Header, for Dwellings More Precise Rule. 390. Example. 391. Carriage Beam with One Header, for First-class Stores More Pre- cise Rule. 392. Example. 393. Carriage Beam with Two Headers and Two Sets of Tail Beams, for Dwellings, etc. 394. Example. 395. Carriage Beam with Two Headers and Two Sets of Tail Beams, for Firsl-clasj Stores. 396. Example. 397. Carriage Beam with Two Headers and One Set of Tail Beams. 398, Carriage Beam with Two Headers and One Set of Tail Beams, for Dwellings. 399. Example. 400. Carriage Beam with Two Headers and One Set of Tail Beams, for First-class Stores. 401. Example. 402. Carriage Beam with Two Headers and Two Sets of Tail Beams More Precise Rules. 403. Example h less than . 404. Example // greater than n. 405. Carriage Beam with Two Headers and Two Sets of Tail Beams, for Dwellings More Precise Rule. 406. Example. 407. Carriage Beam with Two Headers and Two Sets of Tail Beams, for First-class Stores More Precise Rule. 1 8 CONTENTS. 408. Example. 409. Carriage Beam with Two Headers and One Set of Tail Beams More Precise Rule. 410. Example. 411. Carriage Beam with Two Headers and One Set of Tail Beams, for Dwellings More Precise Rule. 412. Example. 413. Carriage Beam with Two Headers and One Set of Tail Beams, for First-class Stores More Precise Rule. 414. Example. 415. Carriage Beam with Two Headers, Equidistant from Centre, and Two Sets of Tail Beams Precise Rule. 416. Example. 417i Carriage Beams with Two Headers, Equidistant from Centre, and Two sets of Tail Beams, for Dwellings and for First-class Stores Precise Rules. 418. Examples 419. Carriage Beam with Two Headers, Equidistant from Centre, and One Set of Tail Beams Precise Rule. 420. Example. 421. Carriage Beams with Two Headers, Equidistant from Centre, and One Set of Tail Beams, for Dwellings and for First-class Stores Precise Rules. 422. Example. 423. Beam with Uniformly Distributed and Three Concentrated Loads, the Greatest Strain being Outside. 424. Example. 425. Carriage Beam with Three Headers, the Greatest Strain being at Outside Header. 426. Example. 427. Carriage Beam with Three Headers, the Greatest Strain being at Outside Header, for Dwellings. 428. Carriage Beam with Three Headers, the Greatest Strain being at Outside Header, for First-class Stores. 429. Examples. 430. Beams with Uniformly Distributed and Three Concentrated Loads, the Greatest Strain being at Middle Load. 431. Example. 432. Carriage Beam with Three Headers, the Greatest Strain being at Middle Header. 433. Example. 434. Carriage Beam with Three Headers, the Greatest Strain being at Middle Header, for Dwellings. 435. Carriage Beam with Three Headers, the Greatest Strain being at Middle Header, for First-class Stores. 436. Example. 437 to 442. Questions for Practice. CONTENTS. CHAPTER XVIII. BRIDGING FLOOR BEAMS. 443, Bridging Defined. 444. Experimental Test. 445. Bridging Principles of Resistance. 446. Resistance of a Bridged Beam 447. Summing the Resistances. 448. Example. 449. Assistance Derived from Cross-bridging. 450. Number of Beams Affording Assistance. 451. Bridging Useful in Sustaining Concentrated Weights. 452. Increased Resistance Due to Bridging. CHAPTER XIX. ROLLED-IRON BEAMS. 453. Iron a Substitute for Wood. 454. Iron Beam Its Progressive Development. 455. Rolled-Iron Beam Its Introduction. 456. Proportions between Flanges and Web. 457. The Moment of Inertia Arithmetically Considered. 458. Example A. 459. Example B. 460. Example C. 461. Comparison of Results. 462. Moment of Inertia, by the Calculus Preliminary Statement. 463. Moment of Inertia, by the Calculus. 464. Application and Comparison. 465. Moment of Inertia Graphically Represented. 466. Parabolic Curve Area of Figure. 467. Example. 468. Moment of Inertia General Rule. 469. Application. 470. Rolled-Iron Beam Moment of Inertia Top Flange. 471. Rolled-Iron Beam Moment of Inertia Web. 472. Rolled-Iron Beam Moment of Inertia Flange and Web. 473. Rolled-Iron Beam Moment of Inertia Whole Section. 474. Rolled-Iron Beam Moment of Inertia Comparison v/ith other Formulas. 475. Rolled-Iron Beam Moment of Inertia Comparison of Results. 476. Rolled-Iron Beam Moment of Inertia Remarks. 477. Reduction of Formula Load at Middle. 2O CONTENTS. 478. Rules Values of W, /, <5 and 7. 479. Example Weight. 480. Example Length. 481. Example Deflection. 482. Example Moment of Inertia. 483. Load at Any Point General Rule. 484. Load at Any Point on Rolled-Iron Beams. 485. Load at Any Point on Rolled-Iron Beams of Table XVII. 486. Example. 487. Load at End of Rolled-Iron Lever. 488. Example. 489. Uniformly Distributed Load on Rolled-Iron Beam. 490. Example. 491. Uniformly Distributed Load on Rolled-Iron Lever. 492. Example. 493. Components of Load on Floor. 494. The Superincumbent Load. 495. The Materials of Construction Their Weight. 496. The Rolled-Iron Beam Its Weight. 497. Total Load on Floors. 498. Floor Beams Distance from Centres. 499. Example. 500. Floor Beams Distance from Centres Dwellings, etc. 501. Example. 502. Floor Beams Distance from-Centres. 503. Example. 504. Floor Beams Distance from Centres First-class Stores. 505. -Example. 506. Floor Arches General Considerations. 507. Floor Arches Tie Rods. 508. Example. 509. Headers. 510. Headers for Dwellings, etc. 511. Example. 512. Headers for First-class Stores. 513. Carriage Beam with One Header. 514. Carriage Beam with One Header, for Dwellings, etc. 515. Example. 516. Carriage Beam with One Header, for First-class Stores. 517. Example. 518. Carriage Beam with Two Headers and Two Sets of Tail Beams. 519. Caniage Beam with Two Headers and Two Sets of Tail Beams, for Dwellings, etc 520. Example. 521. Carriage Beam with Two Headers and Two Sets of Tail Beams, for First-class Stores. 522. Example. 523. Carriage Beam with Two Headers, Equidistant from Centre, and Two Sets of Tail Beams, for Dwellings, etc. CONTENTS. 21 524. Example. 525. Carriage Beam with Two Headers, Equidistant from Centre, and Two Sets of Tail Beams, for First-class Stores. 526. Example. 527. Carriage Beam with Two Headers and One Set of Tail Beams, for Dwellings, etc. 528. Example. 529. Carriage Beam with Two Headers anal One Set of Tail Beams, for First-class Stores. 530. Example. 531. Carriage Beam with Three Headers, the Greatest Strain being at Outside Header, for Dwellings, etc. 532. Example. 533. Carriage Beam with Three Headers, the Greatest Strain being at Outside Header, for First-class Stores. 534. Carriage Beam with Three Headers, the Greatest Strain being at Middle Header, for Dwellings, etc. 535. Example. 536. Carriage Beam with Three Headers, the Greatest Strain being at Middle Header, for First-class Stores. 537 to 545. Questions for Practice. CHAPTER XX. TUBULAR IRON GIRDERS. 546. Introduction of the Tubular Girder. 547. Load at Middle Rule Essentially the Same as that for Rolled-Iron Beams. 548. Load at Any Point Load Uniformly Distributed. 549. Load at Middle Common Rule. 550. Capacity by the Principle of Moments. 551. Load at Middle Moments. 552. Example. 553. Load at Any Point. 554. Example. 555. Load Uniformly Distributed. 556. Example. 557. Thickness of Flanges. 558. Construction of Flanges. 559. Shearing Strain. 560. Thickness of Web. 561. Example. 562. Construction of Web. 563. Floor Girder Area of Flange. 564. Weight of the Girder. 22 CONTENTS. 565. Weight of Girder per Foot Superficial of Floor. 566. Example. 567. Total Weight of Floor per Foot Superficial, including Girder. 568. Girders for Floors of Dwellings, etc. 569. Example. 570. Girders for Floors of First-class Stores. 571. Ratio of Depth to Length, in Iron Girders. 572. Economical Depth. 573. Example. 574 to 579. Questions for Practice. CHAPTER XXI. CAST-IRON GIRDERS. 580. Cast Iron Superseded by Wrought Iron. 581. Flanges Their Relative Proportion. 582. Flanges and Web Relative Proportion. 583. Load at Middle. 584. Example. 585. Load Uniformly Distributed. 586. Load at Any Point Rupture. 587. Safe Load at Any Point. 588. Example. 589. Safe Load Uniformly Distributed Effect at Any Point. 590. Form of Web. 591. Two Concentrated Weights Safe Load. 592. Examples. 593. A relied Girder. 594. Tie-Rod of Arched Girder. 595. Example. 596. Substitute for Arched Girder. 597 to 602. Questions for Practice. CHAPTER XXII. FRAMED GIRDERS. 603.^-Transverse Strains in Framed Girders. 604. Device for Increasing the Strength of a Beam. 605. Horizontal Thrust. 606. Parallelogram of Forces Triangle of Forces. 607. Lines and Forces in Proportion. CONTENTS. 23 608, Horizontal Strain Measured Graphically. 609, Measure of Any Number of Forces in Equilibrium. 610. Strains in an Equilibrated Truss. 611, From Given Weights to Construct a Scale of Strains. 612, Example. 613, Horizontal Strain Measured Arithmetically. 614. Vertical Pressure upon the Two Points of Support. 615. Strains Measured Arithmetically. 616. Curve of Equilibrium Stable and Unstable. 617. Trussing a Frame. 618. Forces in a Truss Graphically Measured. 619. Example. 620. Another Example. 621. Diagram of Forces. 622. Diagram of Forces Order of Development. 623. Reciprocal Figures. 624. Proportions'in a Framed Girder. 625. Example. 626. Trussing, in a Framed Girder. 627. Planning a Framed Girder. 628. Example. 629. Example. 630. Number of Bays in a Framed Girder. 631. Forces in a Framed Girder. 632. Diagram for the above Framed Girder. 633. Gradation of Strains in Chords and Diagonals. 634. Framed Girder with Loads on Each Chord. 635. Gradation of Strains in Chords and Diagonals. 636. Strains Measured Arithmetically. 637. Strains in the Diagonals. 638. Example. 639. Strains in the Lower Chord. 640. Strains in the Upper Chord. 641. Example. 642. Resistance to Tension. 643. Resistance to Compression. 644. Top Chord and Diagonals Dimensions. 645. Example. 646. Derangement from Shrinkage of Timbers. 647. Framed Girder with Unequal Loads, Irregularly Placed. 648. Load upon Each Support Graphical Representation. 649. Girder Irregularly Loaded Force Diagram. 650. Load upon Each Support, Arithmetically Obtained. 651 to 656, Questions for Practice. 24 CONTENTS. CHAPTER XXIII. ROOF TRUSSES. 657. Roof Trusses considered as Framed Girders. 658. Comparison of Roof Trusses. 659. Force Diagram Load upon Each Support. 660. Force Diagram for Truss in Fig. 98. 661. Force Diagram for Truss in Fig. 99. 662. Force Diagram for Truss in Fig. zoo- 663. Force Diagram for Truss in Fig. 101. 664. Force Diagram for Truss in Fig. 102. 665. Force Diagram for Truss in Fig. 103. 666. Force Diagram for Truss in Fig. 104. 667. Force Diagram for Truss in Fig. 105. 668. Force Diagram for Truss in Fig. 106. 669. Strains in Horizontal and Inclined Ties Compared. 670. Vertical Strain in Truss with Inclined Tie. 671. Illustrations. 672. Planning a Roof. 673. Load upon Roof Truss. 674. Load on Roof per Foot Horizontal. 675. Load upon Tie-Beam. 676. Selection of Design for Roof Truss. 677. Load on Each Supported Point in Truss. 678. Load on Each Supported Point in Tie-Beam. 679. Constructing the Force Diagram. 680. Measuring the Force Diagram. 681. Strains Computed Arithmetically. 682. Dimensions of Parts Subject to Tension. 683. Dimensions of Parts Subject to Compression. 684. Dimensions of Mid-Rafter. 685. Dimensions of Upper Rafter. 686. Dimensions of Brace. 687. Dimensions of Straining-Beam. 688 to 692. Questions for Practice. CHAPTER XXIV. TABLES. 693. Tables I. to XXL Their Utility. 694 Floor Beams of Wood and Iron (I. to XIX.). 695 Floor Beams of Wood (I. to VIII.). 696 Headers of Wood (IX. to XVI.). 697 Elements of Rolled-Iron Beams (XVII.). 638. Rolled-Iron Beams for Office Buildings, etc. (XVIIL). CONTENTS. 25 699. -Rolled-Iron Beams for First-class Stores (XIX.). 700. Example. 701. Constants for Use in the Rules (XX.). 702. -Solid Timber Floors (XXI.). 703. Weights of Building Materials (XXII.). 704. Experiments on American Woods (XXIII. to XLVI.). 705. Experiments by Transverse Strain (XXIII. to XXXV., XLU. and XLIII.). 706. Experiments by Tensile and Sliding Strains (XXXVI. to XXXIX., XLIV. and XLV.). 707. Experiments by Crushing Strain (XL., XLI. and XLVI.). TABLES. I. Hemlock Floor Beams for Dwellings, Office Buildings, etc. II. White Pine III. Spruce IV. Georgia Pine " V. Hemlock Floor beams for First-class Stores. VI. White Pine VII. Spruce VIII. Georgia Pine " IX. Hemlock Headers for Dwellings, Office Buildings, etc. X. White Pine " XI. Spruce XII. Georgia Pine " XIII. Hemlock Headers for First-class Stores. XIV. White Pine " XV. Spruce XVI. Georgia Pine " " " XVII. Elements of Rolled-Iron Beams. XVIII. Rolled Iron Beams for Dwellings, Office Buildings, etc. XIX. " " " First-class Stores. XX. Values of Constants Used in the Rules. XXL Solid Timber Floors Thickness. XXII. Weights of Materials of Construction and Loading. XXIII. Transverse Strains in Georgia Pine. XXIV. " " " Locust. XXV. ' " " White Oak. XXVI. " " " Spruce. XXVII. " XXVIIL " " " XXIX. " " " White Pine. XXX. " " " XXXI. " " " XXXII. " " " 26 CONTENTS. XXX III. Transverse Strains in Hemlock. XXXIV. " " " XXXV. " " " XXXVI. Tensile Strains in Georgia Pine, Locust and White Oak. XXXVII. " " " Spruce, White Pine and Hemlock. XXXVIII. Sliding Strains in Georgia Pine, Locust and White Oak. XXXIX. " " " Spruce, White Pine and Hemlock. XL. Crushing Strains in Georgia Pine, Locust and White Oak. XLI. " " " Spruce, White Pine and Hemlock. XLII. Rupture by Transverse Strain Values of B. XLIII. Resistance to Deflection Values of F, XLIV. Rupture by Tensile Strain Values of 7". XLV. " " Sliding " " " G. XLVI. " " Compressive Strain Values of C. DIGEST OR DIRECTORY. INDEX. ANSWERS TO QUESTIONS. INTRODUCTION. ART. I. The science of Construction, as the term is used in architecture, comprehends a knowledge of the forces tending to destroy the materials constituting a building, and of the capacities of resistance of the materials to these forces. 2. One of the requisites of good architecture is Sta- bility. Without this the beautiful designs of the architect can have no lasting existence beyond the paper upon which they are delineated. 3. The force of Gravity is inherent not only in the contents of a building, but also in the materials of which the building itself is constructed ; and unless these materials have an adequate power of resistance to this force, the safety of the building is endangered. Hence the necessity of a knowledge of the laws governing the force of gravity in its action upon the several parts of a building, and of the expe- dients to be resorted to in order to resist its action effect- ually. 4-. It may be objected by some that this knowledge pertains rather to building than to architecture, and that the architect is required merely to indicate the outlines of his plans, leaving to the builder the work of deter- mining the arrangement and dimensions of the materials. This objection is not well founded. Between the duties of 28 INTRODUCTION. the architect and those of the builder there is a well-defined line. This may be shown by a consideration of the operation of building as it is usually conducted. The builder is selected generally from among those who compete for the work. Each builder competing fixes the amount for which he is willing to erect the building, after an examination of the plans and specifications and an estimate of the cost of the work. To arrive at this cost the arrangement and dimensions of the materials must be fixed ; and if not fixed by the plans and specifications, in what way shall they be determined ? Shall it be by the builder ? The builder has not yet been selected. Shall each builder estimating be permitted to assign such di- mensions as his caprice or cupidity shall dictate ? The evil effect of such a course is apparent. The only proper method is to have the arrangement and dimensions of the materials all definitely settled by the architect in his plans and specifi- cations. Moreover, the necessity for a knowledge of this subject by the architect is manifest in this, that he is constantly liable, without this knowledge, to include in his plans such features as the action of gravity would render impossible of produc- tion in solid material, or which, if executed, would not pos- sess the requisite degree of stability. 5. In considering the requisites for stability in a build- ing, the various parts need to be taken in detail : such as Walls, Piers, Columns, Buttresses, Foundations, Arches, Lintels, Floors, Partitions, Posts, Girders and Roofs. 6. It is the purpose of the present work to treat principally of those parts which are subjected to trans- verse strains. 7. In the construction of a floor, the safety of those who are to trust themselves upon it is the first consideration. TO OBTAIN A RULE FOR FLOOR TIMBERS. 29 8. Floors are not always made sufficiently strong. Scarcely a year passes without its record of deaths conse- quent upon the failure of floors upon which people should have assembled with safety. Many floors now existing, and not a few of those annually constructed, are deficient in material, or have an improper arrangement of it. 9. The strength of a floor consists in the strength of its timbers. The dimensions of the timbers for any given floor may be ascertained, practically, by an examination of other similar floors which have been tried and found sufficiently strong. But if no similar floor is found, how is the problem to be solved ? 10. The amount of material required may be found by constructing one or more experimental floors, and testing them with proper weights ; but this w r ould be attended with great expense, and probably with the loss of more time than could be spared for the purpose. II. There is a simple method, which is- quite ascertain and less expensive. The chemist, from a small specimen, makes an analysis sufficient to determine the character of whole mines of ore or quarries of rock. So we, by proper tests of a small piece of any building material, may deter- mine the characteristics of ail material of that kind. 12. To obtain, then, the requisite knowledge of the strength of floor timbers, let us adopt a piece of convenient size as the unit of material. Let it be a piece one inch square and one foot long in the clear between the bearings. This we will submit to a transverse force, applied at the middle of its length, sufficient to break it crosswise, and learn from the result the power of resistance it possesses. Numerous experiments of this nature have been made 30 INTRODUCTION. upon all the ordinary kinds of timber, stone and iron, and the average results collected in tabular form. (See Table XX.) A few results are here given. 13. The unit of material, when of Hemlock, breaks with 450 pounds : White Pine, " " 500 Spruce, " " 550 White Oak, " " 650 Georgia Pine, " " 850 Locust, " " 1200 " Cast-Iron " " 2100 " These figures give the average unit of strength for these several kinds of material, when exposed to a transverse strain at the middle of their length. CHAPTER I. THE LAW OF RESISTANCE. ART. 14. Relation between Size and Strength. Having ascertained, by careful experiment, the power of resistance in a unit of any given material, the next question is : What is the existing relation between size and strength ? Is the increase of one proportionate to that of the other ? In two square beams of equal length, but of different sectional area, the larger one will bear more than the smaller. From this it appears that the resistance is, to a certain degree at least, in proportion to the quantity of material, or to the area of cross-section. There is an element of strength, however, other than this, and one which modifies the proportion very materially. IS. Strength not aBway in Proportion to Area of Cross-eeiioii. That the strength of any two pieces of equal length is not always in proportion to the area of cross-sec- tion, is shown by attempts to break two given pieces. For example, take two beams of equal length, but of differing area of cross-section; the one being 3 x 8, and the other 5x6 inches. The former has 24 and the latter 30 inches of sectional area. If the strength be in proportion to the sectional area, the weights required to break these two pieces will be in the proportion of 24 to 30 their relative areas of cross-section ; but they will be found (the pieces being placed upon edge) to be in the proportion of 24 to the smaller piece being actually stronger than the larger ! 32 THE LAW OF RESISTANCE. CHAP. I. 16. Resistance in Proportion to Area of Preliminary to seeking the cause of this apparent want of proportion, it will be well to show first that, under certain conditions, the resistance of beams is directly proportional to their area of cross-section. Let there be twenty pieces of smooth white pine, each one inch square, and one foot long in the clear between the bearings. The resistance of anyone of these pieces is limited to 500 pounds. This has been ascertained by experiment as before stated in Art. 13. Let four of these pieces be placed side by side upon the bearings. The resistance of the four is evidently just four times the resistance of one piece ; or 4 x 500 = 2000 pounds. Let four more pieces be placed upon the first four : the strength of the eight amounts to 2 x 2000 = 4000 pounds. Add four more, and the combined resistance of the twelve pieces will be 3 x 2000 = 6000 pounds. The resistance of four tiers of four each, or of sixteen pieces, will be 16 x 500 = Soco pounds. The total strength of the twenty pieces, piled up five tiers high, will be 20 x 500 =10,000 pounds. Thus we see that the resistance is exactly in proportion to the amount of material used.* 17. Units may be Taken of any Given Dimensions. In this trial we have taken as the unit of material a bar one inch square. W v e might have taken this unit of any other dimension, as a half, a quarter, or even a tenth of an inch square, and, after finding by trial the strength of one of * The truth of this proposition depends upon obtaining, in the experiment, pieces of wood so smooth that, in being deflected by the weight, they will move upon each other without friction ; a condition not quite possible in practice to obtain. This friction restrains free action, and, as a consequence, the weight required to effect the rupture will be somewhat greater than is stated. RELATION BETWEEN AREA AND STRENGTH. 33 these units, could have as readily known the strength of the whole pile by merely multiplying the number of units by the strength of one of them. We will now consider the relation between breadth and depth. 18. Experience Snows a Beam Stronger when Set on Edge. One of the first lessons of experience with timber of greater breadth than thickness, is the fact of its pos- sessing greater strength when placed on edge than when laid on the flat. As an example : a beam of white pine, 3x8 inches, and 10 feet long between bearings, will require 9600 pounds to break it when set on edge ; while three eighths of this amount, or 3600 pounds, will break it if it be laid upon the flat. Here again, as in Art. 15, we have a fact seemingly at variance with the one but just previously established namely, that of the resistance being in propor- tion to the area of cross-section. We will now investigate the apparent anomaly. (9. Strength IMrectly in Proportion to Breadth. First, as to the breadth of a beam. If two beams of like size are placed side by side, the two will resist just twice that which one of them alone would. Three beams will resist three times as much as one beam would. So of any number of beams, the resistance will be in proportion directly as the breadth. This is found by trial to be true, whether the beams are separate or together, solid ; for a 6 x 8 inch beam will bear as much, and only as much, as three beams 2x8 inches set side by side, and, in both cases, on the edge. In other words, when the depths and lengths are equal, a beam of six 34 THE LAW OF RESISTANCE. CHAP. L inches breadth will bear just three times as much as a beam of two inches breadth, or twice as much as one of three inches breadth. So this fact appears established, that the resistance of beams is directly in proportion to their breadth. 20. By Experiment Strength Iiicrcasc more Rapidly than the Depth. In regarding the depth of beams, another law of proportion is found. Having two beams of the same breadth and length, but differing in depth, we find the strength greater than in proportion to the depth. If it were in this proportion, a beam nine inches high would bear just three times as much as one three inches high, whereas experiment shows it to bear much more than this. 21. Comparison of a Solid Beam with Jt Laminated one. To test this, let there be two beams of equal length, breadth and depth, one of them being in one solid piece FIG. i. FIG. 2. (Fig. 2), while the other is made up of horizontal layers or veneers, laid together loosely (Fig. i). Placing weights upon these two beams, it is seen that, although they contain a like quantity of material in cross-section, and are of equal height, the solid beam will sustain much more weight than the laminated one. Let the several parts of the latter beam be connected together by glue, or other cementing material, CHANGE IN LENGTH OF FIBRES. 35 and again applying weights, it will be found that it has be- come nearly, if not quite, as strong as the beam naturally solid. From these results we infer that the increased strength is due to the union of the fibres at each juncture of the hori- zontal layers. But why does this result follow? If the simple knitting together of the fibres is the cause, then why, in considering the breadth, is a solid beam no stronger than two beams, each of half the breadth, as has been shown ? 22. Strength clue to Resistance of Fibres to Extension and Compression. An examination of the action of the beams under pressure in Figs, i and 2 may explain this. The weights bending the beams make them concave on top. In Fig. i the ends of the veneers or layers remain in vertical planes, while, in the other case, the end of the solid beam is inclined, and normal to the curve. It is also seen that the upper surface in Fig. 2 is shorter than the lower one, although the two surfaces were of the same length before bending. This change in length has occurred during the process of bending, and could only happen through a change in length of the fibres constituting the beam. In the operation of bending, one of two things must of necessity take place : either the fibres must slide upon each other, as in Fig. i, or else the length of the fibres must be changed, as in Fig. 2 ; and since in practice it is found that the fibres are so firmly knit as effectually to prevent sliding, we have only to consider the effects of a change in the length of the fibres. The resistance to this change is an element of strength other than that due to quantity of material, and its nature will now be examined. 3^ THE LAW OF RESISTANCE. CHAP. I. 23. Power Extending Fibres in Proportion to Depth of Beam. If a beam be made of four equal pieces, as in Fig- 3> an d be held together by an elastic strap firmly attached to the under side of the beam, and by two cross pieces let into the horizontal joint and closely fitted ; and if upon this beam FlG - 3- a weight be laid at the middle sufficient to elongate the strap and open the vertical joint at the bottom a given distance say an eighth of an inch ; then, if the weight and the two upper quarters of the beam be removed, and a weight laid on at the middle suffi- cient to open the joint, to the like distance as before, it will be found that this weight is just one half of that before used. In this experiment, the strap may be taken to represent the fibres at the lower edge of the beam. We here find a relation between the weight and the height of the beam. The greater the height, the greater must be the weight to produce a like effect upon the fibres of the lower edge. Double the height requires double the weight. Three times the height requires three times the weight. Therefore we decide that, in elongating- the fibres at the bottom, the weight and the height are directly in pro- portion. It must be observed that Fig. 3 and its explanation arc not to be taken as a representation of the full effect of a transverse strain upon a beam. The scope of the experi- ment is limited to the action of the fibres at the lower edge. The other fibres, all contributing more or less to the resistance, are, for the moment, neglected, in order to show this one feature of the strain namely, the manner in which fibres at any point contribute to the general resistance. Galileo, of Italy, who, two hundred and fifty years since, NEUTRAL LINE. 37 was the first to show the connection of the theory of trans- verse strains with mathematics, not recognizing in his theory the compressibility of the fibres at the concave side of the beam, supposed that in a rupture by cross strain all the fibres were separated by pulling apart ; as might be shown in Fig. 3, in case the rubber were extended up each side to the top, instead of being confined to the lower edge. We are greatly indebted to Galileo for his studies in this direc- tion ; but Hooke, Mariotte, and Leibnitz, about 1680, found the theory of Galileo to be defective, and showed that the fibres were elastic ; that only those fibres at the convex side of the beam suffer extension ; that those at the concave side suffer compression and are shortened ; and that, at the line separating the fibres which are extended from those which are compressed, they are neither lengthened nor shortened, but remain at their natural length. This line is denominated the neutral line or surface. It will here be observed that the amount of extension or compression in any fibre is proportional to its distance from the neutral line. CHAPTER II. APPLICATION OF THE LEVER PRINCIPLE. ART. 24. The Law of the L,ever. The deduction drawn from the experiment named in Art. 23 depends for its truth upon what is known as the law of the lever. This law, in so far as it applies to transverse strains, will now be con- sidered. 25. Equilibrium Direction of Presures. When equal weights, suspended from the ends of a beam supported upon a fulcrum, as at W in Fig. 4, are in equilibrium, it is found that the point of support 'is just midway between the two weights, provided that the beam be of equal size and weight throughout its length. It will be observed that the directions of the strains upon the beam are vertical, those at the ends being down- w ward, while that at the middle is upward ; also that the strains are evidently equal, the upward pres- sure at the middle being just equal to the sum of the two weights at FIG. 4. the ends; for if unequal, there would be no equilibrium, but a movement in the direction of the greater power. We decide, then, that the pressure upon the fulcrum is equal to the sum of the two weights.* * In ascertaining the pressure at the fulcrum, the weight of the beam itself should be added to the sum of the two weights, but to simplify the ques- tion, the beam, or lever, is supposed to be without weight. REACTION EQUAL TO THE PRESSURE. 39 25. Conditions of Freure in a Loaded Beam. In Fig. 5 we have a beam supported at each end, and a weight W laid upon the middle of its length. Comparing this with Fig. 4 we see that the strains here ..j^ ^J^ are also vertical but in re- versed order, the one at the middle being downwards, FIG. 5. while those at the ends are upwards. In other respects we have here the same conditions as in Fig. 4. The downward pressure at the middle is equal to the upward pressures or reactions at the ends ; and, since the weight is placed midway between the points of support, the reactions at these points are equal, and each is equal to one half the weight at the middle. 27. The Principle of the Lever. In Fig. 6 is shown a lever resting upon a fulcrum W t and carrying at its ends the weights R and P. w Here, the fulcrum W is not at the middle as in Fig. 4, but at a point which divides the lever into two unequal parts, m and n. In accordance with the prin- FIG. 6. ciple of the lever, the two parts m and n, when there is an equilibrium, are in proportion to the two weights P and R; or, the shorter arm is to the longer as the lesser weight is to the greater ;* or, m : n : : P : R * For a demonstration of the lever principle see an article, by the author, in the Mathematical Monthly, published at Cambridge, U. S. f vol. I, 1858, page 77. 4O APPLICATION OF THE LEVER PRINCIPLE. CHAP. II. from which we have Rm = Pn ~ n (1.) m v ' and. ^m As an example: suppose the lever to be 12 feet long, and so placed upon the fulcrum as to make the two arms, m and , 4 and 8 feet respectively. Then, if the shorter arm have suspended from its end a weight, R, of 500 pounds, what weight, P, will be required at the end of the longer arm to produce equilibrium ? Formula (#.) is appropriate to this case. Therefore P=R 500 x ~= 250 pounds; equals the weight required on the longer arm. From Art. 25 it is evident that the sum of the weights R and P is equal to the upward force or reaction at W. Therefore, we have, W= + P and W-R=P Substituting this value for />in formula (^.),we have n _ m n W = R ' and, multiplying by PRESSURE ON SUPPORTS MEASURED. 41 and, since n + m is equal to the whole length of the beam, or to /, therefore R=W^ (3.) In a similar manner, it is found that P= w (4.) 28- A Loaded Beam Supported at Eaeli End. In Fig. 7 a weight W, is carried by a beam resting at its ends upon two supports. Here we have, with the pressures in reversed order, similar con- ditions with those shown in Fig. 6. Here, also, it will be observed that the weight W is equal to the sum of the upward resistances R and P (Arts. 25 and 26) neglecting for the present the weight of the beam itself and that the upward resistance at R may be found by formula (3.) ; while that at P is found by formula (4.). For example : suppose the weight W, Fig. 7, to be 800 pounds ; and that it be located five feet from one end of the beam and eight feet from the other end. Here W 800, m 5, n = 8 and / = 13. To find the pressure at R, we have, by formula (#.), R W = 800 x = 4Q2 T 4 T pounds. I 13 To find the pressure at P, we have, by formula (4.\ P W~ 800 x -- = 307^ pounds. To verify the rule, we find that + 37A = 800 pounds = W. APPLICATION OF THE LEVER PRINCIPLE. CHAP. II. Either one of these upward pressures, or reactions, being- found, the other may be determined by subtracting 1 the first from W. From the above, we see that the portion of a weight borne by one support is equal to the product of the weight into its distance from the other support, divided by the length between the two supports. 29. A Bent L.evcr. In Fig. 8 let P C G be a rigid bar, shaped to a right angle at C, and free to revolve on C as a centre. Let R and H be two weights at- tached by cords to the points P and G, the cords passing over the pulleys D and E. Let the weights be so proportioned as FIG. 8. to produce an equilibrium. Here P C G is what is termed a bent lever, and the arms a and b are in proportion to the weights R and H ; or, a \ b \\ R \ H and 30. Horizontal Strains Illustrated by the Bent Lever. To apply the principle of the bent lever let a beam R E (Fig. 9) be laid upon two points of support, R and E, and be loaded at the middle with the weight W. The action -of this weight upon the beam is similar in its effect to that taking place in the bent lever of Fig. 8, producing horizon- FIG. 9. tal strains, which compress the fibres at the top of the beam and extend those at the bottom. (Art. 23). HORIZONTAL STRAIN MEASURED. 43 Let the line P C represent the line of division between the compressed and the extended fibres. Then P C G may be taken to represent the bent lever of Fig. 8 ; for the upward pressure or reaction at R, moving the arm of lever P,C, which turns on the point (7, as a centre, acts upon the point G, through the arm of lever *C G, moving the point G horizontally from E, and thus extending the fibres in the line G E. Now, if H represents this horizontal strain along the bottom of the beam, and -R the vertical strain at P both being due to the action of the weight W; if the arm P C be called b, and C G called a, then, as before, a : b : : R : H from which H= R t a For an application: let b in a given case equal 10 feet, a equal 6 inches, or 0-5 of a foot, and R equal 1200 pounds; what will be the horizontal strain in the fibres at the lower edge of the beam ? From the above formula, b 10 H -- R = 1200 x 1200x20.= 24,000 or the horizontal strain equals 24,000 pounds. 31. Reitaiicc of Fibres in Proportion to the Depth of Beam. From the proportion in the last article, a : b : : R : H we have Ha = Rb and dividing by a b we have J^-^-jK b a For any given material, the power of the fibres to resist tension is limited, and, since this power is represented by H, 44 APPLICATION OF THE LEVER PRINCIPLE. CHAP. II. therefore H is limited. In any given length of beam, b, which is dependent upon the length, is also given ; hence TT TT -D r> -T- becomes a fixed* quantity ; and since -7- = , therefore is a fixed quantity. But R and a may vary individually, provided that the quotient of R divided by a be not changed. So, then, if R be increased, a must also be increased, and in a like proportion ; if R be doubled, a must be doubled ; if one be trebled, the other must be trebled ; or, in whatever proportion one is increased or diminished, the other must be increased or diminished in like proportion. Therefore R and a are in direct proportion. Take a as equal to one half of the depth of the beam, or , and R as equal to one half the weight at the W middle of the beam, or . Then, since a is in proportion to R, d is in proportion to W, or the depth of the beam must be in proportion to the weight. This result is the same as that arrived at in Art. 23 ; that the power of the fibres at the bottom to resist extension is in proportion to the depth of the beam. CHAPTER III. DESTRUCTIVE ENERGY AND RESISTANCE. ART. 32. Reistance to Comprcion Neutral Line. We have shown the manner in which the fibres at the convex side of a beam contribute to its strength by their resistance to extension. It may now be observed that the resistance to compression of the fibres at the concave side is but a counter- part of the resistance to extension of the fibres at the convex side. Whatever resistance may be given out in one way at one side of the beam, a like amount of resistance will be called up in the other way at the other side. The one balances the other, like two weights at the ends of a lever (Figs. 4 and 6). If the powers of resistance to compression and extension be equal, as is the case in some kinds of wood, then one half of the fibres will be compressed while the other half are extend- ed ; and, should the beam be of rectangular section, the neu- tral line will occur at the middle of the height of the beam, and the condition of equilibrium will be as shown in Fig. 4- If the capability to resist compression exceeds the resist- ance to extension, as in cast-iron, then the greater portion of the fibres will be employed in resisting tension, and the neu- tral line will be nearer to the concave side ; an equilibrium represented by Fig. 6, in which the shorter arm of the lever may represent the portion of the fibres subjected to compres- sion, and the longer arm those suffering tension, and where R, the heavier weight, may represent the power of any given number of fibres to resist compression, while P, the lesser 46 DESTRUCTIVE ENERGY AND RESISTANCE. CHAP. III. weight, represents the power of an equal number of fibres to resist tension. In Art. 31 the power of the fibres at the convex side of a beam to resist extension was shown to be in proportion to the depth of the beam. This result was obtained by taking the position of the neutral line at the middle of the depth. The like result will be obtained even when the neutral line occurs at a point other than the middle. For, whatever be the proportionate distance of this line from the lower edge, that distance, for the same material, will always bear the same proportion to the depth of the beam. 33. Elements of Resistance to Rupture. Having now sufficient data for the purpose, the several elements of strength which have been developed may be brought to- gether, and their sum taken as the total resistance to rup- ture. First. We have the rate of strength, or the weight in pounds required to break a unit of the given material one inch square and one foot long, when supported at each end (Arts. 12 and 13). Let ^represent this weight. Second. We have the strength in proportion to the area of cross-section, or to the product of the breadth into the depth (Arts. 16 and 17). If b be put to represent the breadth, and d the depth, both in inches, then this element of strength may be represented by b x d or bd. Third and last, we have the strength due to the resist- ance of the fibres to a change in length, which has been shown to be in proportion to the depth (Arts. 22, 23 and 31), and may therefore be represented by d. Putting these three elements of strength together, and representing by R the total resistance, we have, DESTRUCTIVE ENERGIES. 47 R = Bxbdxd or R = Bbd 3 (5.)* and this is the total power of resistance to a cross strain. 34. a Destructive Energies. It is requisite now to con- sider the destructive energies. It has been shown (Art. 27) that the power of a weight, acting at the end of a lever, is in proportion to the length of the lever. This is seen in Fig. 6, where a small weight acting at the end of the longer arm produces as great an effect as the larger weight upon the shorter arm. This principle may be stated thus : The mo- ment of a weight is equal to the product of the weight into the length of the arm of leverage at which it acts. If n (Fig. 6) be the arm of leverage, and P the weight act- ing at its end, then the moment of P is equal to the weight P multiolied by the length of the lever n ; or, Moment = Pn. Let 5 represent the weight which it is found on trial is required to break a lever or rod of given material, one inch square, and projecting one foot from a wall into which it is firmly imbedded ; the weight being suspended from the free end of the lever. Then, since the moment equals the weight into its arm of leverage, as above stated, which arm in this case equals unity, we have 5 x i = Pn * Strictly speaking, the whole power of abeam to resist rupture is due to the resistance of the fibres to compression and extension, as will be shown in speak- ing of the resistance to bending and it is usual to obtain the amount of this power by a more direct method ; arriving at the total resistance by one opera- tion, and this based upon a consideration of the resistance offered by each fibre to a change of length, and taking the sum of these resistances ; but it is thought that the method here pursued is better adapted to securing the object had in view in writing this work. 48 DESTRUCTIVE ENERGY AND RESISTANCE. GHAP. III. or the power of resistance of such a rod equals S, the weight required to break it. Having this index of strength, S, and knowing (Art. 33) that the resistance to breaking is in proportion to the breadth and the square of the depth, then for levers larger than one inch square, and longer than one foot, when the destructive energy equals the resistance, we have Pn = Sbd* (0.) that is, for the moment, or destructive energy, we have P, the weight in pounds, multiplied by ;/, the length in feet ; and for the resistance, we have 5, the index of strength for the sectional area of one inch square, multiplied by the breadth of the lever, and by the square of its depth ; the breadth and depth both being in inches. 35. Rule for Tranvere Strength of Beams. This for- mula, (6.), gives a rule for the transverse strength of lever?. From it we may derive a rule for the transverse strength of beams supported at both ends. We know, for example, from Arts. 25 and 26, that the strains in a lever are the same as in a beam which is twice the length of, and loaded at the middle with twice the weight supported at the end of the lever. Therefore, when P is equal to the half of W, the weight at the middle of a beam (Fig. 5), and n is equal to the half of /, the length of the beam, we have W I WL Pn x - = - and since, (form. 0), Pn = Sbd* by substitution we have =Sbd* (7.) or Wl= 4Sbd a (8.) in which Wl equals the moment or destructive energy of a weight at the middle of a beam, and ^Sbd* equals the resist- RULE FOR STRENGTH OF BEAMS. 49 ance of the beam. But this resistance was found (Art. 33) to be equal to Bbd 2 ; therefore, 4$bd* = Bbd* hence Wl = Bbd 2 (9.) This is the required rule for the strength of beams sup- ported at each end. In it W equals the pounds laid on at the middle of the beam, / the length of the beam in feet, b and d the breadth and depth respectively of the beam in inches, and B the weight in pounds at the middle required to break a unit of material (Art. 12) of like kind with that in the beam, when strained in a similar manner. It may be observed here that from Bbd 2 ^Sbd 2 as above, we have B = 4S or, the weight at the middle required to break a unit of material, when supported at each end, is equal to four times the weight required to break it when fixed at one end only, and the weight suspended from the other.* * Professor Moseley, in his "Engineering and Architecture," puts S to rep- resent the index of strength, but his definition of this index shows it to be not the same as that for which S is put in this work. While, with us, S represents the resistance to rupture of a unit of material (one inch square and one foot long), fixed at one end and loaded at the other ; in his work (Art. 408, p. 521, Mahan's Moseley, New York, 1856), S is placed to represent the "resistance in pounds opposed to the rupture of each square inch at the surface exposed to a tensile strain" To compare the two, let M be put for the 6* of Prof. Moseley. Then his ex- pression (Art. 414, p. 528) for rectangular beams, I be 3 P = - S becomes 6 a P M -r- in which da P is the weight at one end of a beam, which is fixed at the other end, and c is the depth and a the length, both in inches. If for c we put */and for a we put , representing feet instead of inches, so that a = 12 n, then 50 DESTRUCTIVE ENERGY AND RESISTANCE. CHAP. III. 36. Formulas Derived from t!ii Rule. From the gene- ral formula, (9.), of Art. 35, any one of the five quantities named may be found, the other four being given. bd a P - M - and 72 Pn = Mbd* Now we have found (form. 6\ that Pn - Sbd* Multiplying this by 72 gives 72 Pn = TZ Comparing this value of 72 Pn with that from Prof. Moseley, as above, we have from which M=-]2S or M, the S of Prof. Moseley, is equal to 72 times the S of this work. We also find that Prof. Rankine (Applied Mechanics, Arts. 294 and 296) similarly designates the index of strength ; or, as he and Prof. M. both term it, "the modulus of rupture." Prof. R. defines it the same as Prof. M. ; except, that instead of limiting it to the tensile strain, he applies it equally to that ele- ment, tension or compression, which first overcomes the strength of the beam. Prof. Rankine further defines it (p. 634) to be " eighteen times the load ivhicli is required to break a bar of one inch square, supported at two points one foot apart, and loaded in the middle between the points of support" Now the bar here de- scribed is identical with the unit of material adopted in this work (Arts. 12 and 13) ; to designate the strength of which we have used the symbol B. To compare the two, we have, as above found, M= 725 and also, (Art. 35) B - 4$ Multiplying the latter equation by 18, we have 18.5 = 726" or iB = M or as defined by Prof. Rankine, M, the S of Prof. Moseley, is equal to 18 times the value of B, the index of strength as used in this work. Hence the values of S, as given for various materials by Profs. Moseley and Rankine, are 18 times the values of B in this work for the same materials. Owing, however, to a con- siderable variation in materials of the same name, this relation will be found only approximate. RULES FOR BREAKING WEIGHT. 51 For examplej Bkd' j- (13.) Bbd* In these formulas B is the breaking weight in pounds ap- plied at the middle. The value of B (Arts. 33 and 35) is given for the length in feet, and the breadth and depth in inches. QUESTIONS FOR PRACTICE. 37. What kind of strain is a floor beam subjected to? 38. In a beam subjected to a transverse strain, how- does the breadth contribute to its strength ? 39. How does the depth contribute to its strength? 4-0. What are the elements of resistance, and what is the expression for this resistance ? 4-1. When a beam supported at each end carries a load at its middle, what is the amount of pressure sustained by the two points of support, taken together? 52 DESTRUCTIVE ENERGY AND RESISTANCE. CHAP. III. 42. What portion of the load is upheld .by each sup- port? 43 B If the load be not at the middle, what is the sum of the pressures upon the two points of support ? . In the latter case, what proportions do the parts borne at the two points of support bear to each other? 45. What expression represents that borne by the near support. 46. What expression represents the pressure upon the remote support ? 47. If a beam, 12 feet long between bearings, carries a load of 15,000 pounds, at a point 4 feet from one bearing, what portion of this load is borne by the near support ? And what is the pressure upon the remote support? 48. When a beam is subjected to transverse strain at its middle, what constitutes the destructive energy tending to rupture? 49. When the destructive energy and the resistance are in equilibrium, what expression represents the conditions of the case ? 50. What is the breaking load of a Georgia pine beam, 15 feet long between the bearings; the breadth being 4 inches, the depth 10, and the load at the middle ? 51. How many times as strong as when laid on the flat is a beam when set on edge ? CHAPTER IV THE EFFECT OF WEIGHT AS REGARDS ITS POSITION. ART. 52. Kelation between Destructive Energy and Resistance. In a beam, laid upon two bearings, and sustain- ing a load at the middle, we have discovered certain relations between the load and the beam. The load has a tendency to destroy the beam, while the beam has certain elements of resistance to this destructive power. The destructive energy exerted by the load is equal to the product of half the load multiplied by half the length of the beam. The power of resistance of the beam is equal to the product of the area of cross-section of the beam, multi- plied by its depth and by the strength of the unit of mate- rial. At the moment of rupture, the destructive energy and the power of resistance are equal ; or, as modified in Art. 35, Wl = Bbdd or, as in formula (9.), WL = Bbd* 53. Dimensions and Weights to be of Like Dciiomiua- tioiis with Those of the Unit Adopted. In applying the above formula it is to be observed, that the length', breadth and depth, in any given case, are to be taken in like denominations with those of the unit of material adopted (Art. 33). For ex- ample : if the unit of material be that of this work, then, in the application of the formula, the breadth and depth are to be taken in inches, and the length between bearings in feet. 54 EFFECT OF WEIGHT AS REGARDS ITS POSITION. CHAP. IV. It is also requisite that the weight be taken in like denomi- nation with that by which the resistance of the unit of mate- rial was ascertained. If the one is in ounces, the other is also to be in ounces ; if one is in pounds, the other must be in pounds ; or, if in tons, then in tons. The strength of the unit of material adopted for this work is given in pounds ; therefore, in applying the rule, the weight given, or to be found, must necessarily be in pounds. 54. Position of the Weight upon the Beam. The loca- tion of the weight upon the beam now requires considera- tion. Upon our unit of material, which is supported at each end, the load is understood to have been located at the mid- dle of the length ; so, in using formula (P.), the weight given, or sought, must be located at the middle of the length of the given beam. 55. Formula Modified to Apply to a L.ever. By pro- per modifications this formula may also be applied to the case of a weight suspended from one end of a lever or pro- jecting beam. To show the application, we proceed as fol- lows : In Fig. 10 one half of the load W is borne on each one of the supports A and B. w w FIG. 10. FIG. ii. In Fig. ii we have a beam of the same length, and sub- jected to the same forces, but in reversed order (Art. 26). BEAM AND LEVER COMPARED. 55 While Fig. 10 represents a beam supported at both ends and loaded in the middle, one half of Fig. n may be taken to represent a lever projecting from a wall and loaded at the free end. In these two cases the moment or destructive energy tending to break the beam is the same in each, and yet it is produced in Fig. n with only one half the weight, acting at the end of a lever only one half the length of the beam. We have, therefore, or, in a lever, it requires but a quarter of the weight to pro- duce a given destructive energy, that is required in a beam of equal length, laid upon two supports that is to say, if two beams of like material, and of the same cross-section, be subjected to transverse strains, in like positions as to breadth and depth, one beam being supported at both ends and load- ed in the middle, and the other one firmly fixed in a wall at one end and loaded at the other ; and if the distance between the wall and the weight in this latter beam be equal to the distance between the bearings in the former ; then but one quarter of the weight requisite to break the beam supported at both ends will be required to break the projecting one. If the former requires 10,000 pounds to break it, then the latter will be broken by 2500 pounds. The proportion between the weights is as 4 to I. But suppose the weights upon the two beams are equal.' In this case the lever will have to be made stronger, and its sec- tional area enlarged sufficiently to carry 4 times the weight. Hence we have, for beams fixed at one end and loaded at the other, = Bbd* in which W is the weight suspended from the end of the lever, and / is the length of the lever ; or, to correspond with 56 EFFECT OF WEIGHT AS REGARDS ITS POSITION. CHAP. IV. the symbols used in Art. 34, where P equals the weight and ;/ equals the length of the lever, we have 4/fc = Bbd* (15.) 56. Effect of a Load at Any Point in a Beam. The next case for consideration is that of the effect of a weight located at any point in the length of a beam, the beam being supported at both ends. In Arts. 27 and 28 it was shown, in cases of this kind, that R, the portion of the whole weight borne at the nearer end, is (form. 3.) equal to Wj-\ an d that P, the portion resting upon the more remote end, is (form. 4-) equal to W,- ; where W 7 equals the weight on the beam, R the portion of the weight carried to the near support, P the portion carried to the remote support, / the length of the beam, m the distance from the weight to the near support, and n the distance to the remote support. As shown in Art. 34, the effective power or moment of a weight is equal to the product of the weight into the arm of the lever, at the end of which it acts. In Fig. 6 the weight R may be taken to represent the reaction of the point of support R in Fig. 7; and the destructive effect at the point of the fulcrum W in Fig. 6, taken to be the same as that at the location of the weight W in Fig. 7, as the strains in the two pieces are equal ; and hence, the moment of R, Fig. 6, is equal to the product of R into its arm of lever ;;/, or equal to Rm. Taking the value of R in formula (3.), and multiplying it by its arm of lever, /#, we have LOAD AT ANY POINT RULE TESTED. 57 Again, taking the value of P in formula (^.), and multi- plying it by its arm of lever, ;/, we have p n =w- n = w m f The two results agree, as they should. 57. Rule for a Beam Loaded at Any Point. These formulas may be tested by taking the two extreme condi- tions, the load at the middle and at the end. First : When the load is at the middle m n \l the destructive energy, as above, will be D = W 1 ^ = W ^f=W^- the same value as obtained in Art. 35. Second : When the weight is moved towards the nearer end, m becomes gradually shorter, and when the weight in its movement reaches the point of support, m becomes zero, and n equals /. The destructive energy will then be as it ought to be, for the weight no longer exerts any cross strain upon the beam. The destructive energy therefore of a weight, W, when laid at any point upon a beam, is When laid at the middle, it is as above shown, 58 EFFECT OF WEIGHT AS REGARDS ITS POSITION. CHAP. IV. In formula (7.) we have therefore, by substitution, W?f = Sbd* Multiplying by 4, we have and since, by A rt. 35, we have = Bbd* (16.) a rule for the resistance of a beam when the weight is located at any point in its length. 58. Effect of an Equally Distributed Load. Let the effect of an equally distributed weight now be considered. Formula (16.) gives the effect of a weight at any point of a beam that is, the effect of the weight at the point where it is located ; but what effect at the middle of the beam is produced by a weight out of the middle ? When a weight is hung at the end of a projecting lever, its effective energy, at any given point of the length of the lever, is equal to the product of the weight multiplied into the distance of that point from the weight (Art. 340. In Arts. 27 and 28 we have the effect of the weight W upon its points of support. For the remote end, in Fig. 7, this is P= W -j. This is the reaction, or power acting upward LOAD AT ANY POINT EFFECT AT MIDDLE. 59 at the point of support P. We have, Arts. 56 and 57, the moment or destructive energy due to this reaction equal to but if, instead of the whole distance ;/, we take only a part of it, or say to the middle of the beam, or /, we have, instead of/X Px$t = W~ x \l = $W^-=$Wm or, we have, for M, the effect at the middle due to a weight placed at any point, This result may be tested as in Art. 57; for let m |-/, then M |- Wm becomes which is a quarter of the weight at the middle into the whole length, as shown in Art. 55. Again, taking the other extreme ; when m becomes zero, then M = j- Wm becomes which is evidently correct, for when the weight is moved from over the clear bearing on to the point of support it ceases to exert any cross strain whatever upon any point of the beam. From the above, we conclude that the effect produced at the middle of a beam, by a weight located at any point of its length, is equal to the product of half the weight into its distance from its nearest point of support. This result would be true of a second weight, and a third, and of any number of weights. If the weights R, P, Q, etc. (Fig. 12), be located on a beam, at distances from their near- 60 EFFECT OF WEIGHT AS REGARDS ITS POSITION. CHAP. IV. cst point of support equal to ;;/, r, s, etc., their joint effect at the middle of the beam will be m + \Pr + \Qs + etc. or 59. Effect at middle from an Equally Distributed I,oad. - We may now ascertain the effect produced at the middle of a beam by an equally distributed load. Let a beam, A B (Fig. 12), of homologous material, and of equal sectional area throughout its length, be divided into any number of equal parts. T~|s The weight of any one of these parts will equal that of any other part, and therefore we have in this beam a case of an equally distributed load. Now, suppose the weight of each of these parts to be concentrated at its centre of gravity, and represented by a ball, as R, P, or Q, suspended from that centre of gravity. Let / equal the length of each of the parts into which the beam is divided, then m = - /, r = - / and s = - /, and, since M - Win, we have for 22 2 the effect of the weight R, at the middle of the beam, = ~R-t\ for the effect of P, M=- 22 2 - - -f\ and for the 2 effect of Q, M = - Q - /; etc., for all the weights on one half of the beam. If these results be doubled (for the effects of the weights on the other half would equal these), we shall have the total effect at the middle of the beam of all the weights. When, LOAD, CONCENTRATED AND DIFFUSED. 6 1 as in this case, the beam is divided into six parts, we have for the total effect at the middle, - tQ 222 Now if we put the symbol U to represent a uniformly distributed load, we have ^ R = P=Q = ~ therefore In this case t equals , therefore M=^U~= l - Ul 468 In which U equals the whole weight uniformly distributed over the beam. We have seen (Art. 35) that \Wl is the destructive ener- gy of a weight concentrated at the centre of the beam. We now see, as above, that this same effect is produced by \UL We therefore have i-7/ ^Wl or, multiplying by 4, $u=w or, when the effects of the two loads upon a beam are equal, one half of /, the distributed load, will equal the load W, concentrated at the middle. 60. Example of Effect of an Equally Distributed Load. Let R, P, Q, etc., each equal 20 pounds; or the whole load U equal 6x 20 = 120 pounds. Let the whole length, 12 feet, be divided into six equal parts, and the equal loads be sus- pended from the centre of each of these parts. Then from the nearer point of support, A, the distance m to R is one foot ; the distance r to P is three feet ; and the distance s to 62 EFFECT OF WEIGHT AS REGARDS ITS POSITION. CHAP. IV. Q is five feet ; and, since R, P y and Q are each equal to 20, and (Art. 58) M = Wm therefore M = -J- x 20 M 10 (i +3 + 5) 10x9 = 90 ' The like effect, 90 pounds, is had from the three weights upon the other half of the beam. Adding these, we have 180 pounds. This is the destructive energy exerted at the mid- dle of the beam by the six weights, or by U, the 120 pounds equally distributed along the beam. As a test of this, let it now be shown what weight concentrated at the middle of the- beam would produce the like effect. In Art. 35 we have for the destructive energy, D = j- Wl, from which W= and ^ since, as above, D = 180 and I = 12, we have W =. --- = 60 J pounds. This is the weight concentrated at the middle. Above, we had /, the equally distributed Aveight, equal to 1 20 pounds, or twice 60. Therefore 2W = U. Thus, as before, it is seen that an equally distributed weight produces an effect at the middle equal to that produced by one half the weight if concentrated at the middle. 6L Rcult alo Obtained toy the Lever Principle. This result may also be obtained by an application of the lever principle. In Fig. 13 a double le- I ver is loaded with weights, pro- ducing strains similar to those in a beam such as Fig. 12. Here the arm of lever at which R acts is five feet, that of P three feet, and Q one foot ; therefore, LOAD EQUALLY DISTRIBUTED. 63 Rx$ = 2OX$ = 100 Px 3 = 20 X 3 = 60 Q X I =: 20 X I = 20 1 80 pounds. This- is the whole energy, because the weights on the other side of the fulcrum do not add to the strain at W\ they only balance the weights R, P, and Q. The full effect, therefore, at the middle of the beam is 180 pounds, as before shown, and this effect is produced by 3 x 20 = 60 pounds equally distributed. Now, what concentrated weight at the end of the lever would produce an equal effect ? Since the weight P, at the end of a lever, multiplied by n, the length of the lever, is the moment or destructive energy of the weight, therefore Pn = 1 80 the moment as above, or = n 6 and this is one half of 60, the distributed weight which pro- duced a like effect. Hence we find that a given load, if concentrated at the middle of a beam, will have a destructive energy there equal to that of twice said load equally distributed over the length of the beam ; or, in other words, an equally distributed load will need to be double the weight of a concentrated load to produce like effects upon any given beam. In formula (9.) W represents the concentrated weight at the middle. If for W we substitute its equivalent %U, we have (17.) 64 EFFECT OF WEIGHT AS REGARDS ITS POSITION. CHAP. IV. QUESTIONS FOR PRACTICE. 62. A white pine beam, 6x9 inches, supported at each end, and set upon edge, is 12 feet long. What weight laid at 4 feet from one end would break it ? 63. What weight equally distributed over the length of the above beam would break it ? 64. What weight concentrated at the middle of the length of the same beam would break it ? 65. What weight would break this beam if suspended from one end of it, the other end being fixed in a wall ? CHAPTER V. COMPARISON OF CONDITIONS SAFE LOAD. ART. 66. Relation between Lengths, Weights and Ef- fects. In the consideration of the effect of weights upon beams, we have deduced certain formulas applicable under various conditions. These rules Avill now be presented in such manner as to show by comparison : first, what relation the lengths and weights bear to each other when the effects are equal ; and, second, the resulting effects when the lengths and weights are equal. 67. Equal Effects. Take the four Figs., 14, 15, 16 and 17. w FIG. 14. FIG. 16. RRRRRRRR FIG. 15. FIG. 17. The lengths of the beams and the amounts of the weights with which they are loaded, are such as to produce equal 66 COMPARISON OF CONDITIONS SAFE LOAD. CHAP. V. effects. For example, the dimensions are such that in all of the figures, I 2n and s ^ ? ; and the weights 8 10 are so proportioned that W = 2 P 4 R. By comparison, we find that in Fig. 14 the destructive energy is In Fig. 15 the destructive energy is equal to the sum of the products of the several weights R, into their respective distances from the point of support ; or, Jfr(i + 3+ 5 + 7) = i6Rs = i6 In Fig. 16 the destructive energy is In Fig. 17 the destructive energy equals the sum of the products of the several weights R, into one half their respec- tive distances from the nearest point of support (Art. 58), or, 2 2 [i^ (1 + 3 + 5 4-7)] = 16)= i6Rs= i When the load is at any point upon the beam, the destruc- . - J7 mil tive energy is W =--. This case is a modification of Fig. 16, for, when m n = %l we have, 68. Comparison of I^en^tlis and Weiglit Producing Equal Effects. We now see that, in order to produce equal effects, we must have the length and weight in Fig. 16 twice EQUAL WEIGHTS AND LENGTHS. 67 those in Fig. 14 ; and the length and weights of Fig. 17 twice those of Fig. 15. Again, we see that, while the lengths of Figs. 14 and 15 are the same, the weights of the latter are equal in amount to twice that of the former ; and that the same proportions exist in Figs. 16 and 17. 69. Tlie Effects from Equal Weights and Lengths. In regard to the second relation, as expressed in Art. 66. We have, in Figs. 18, 19, 20, and 21, examples showing the w FIG. 18. FIG. 20. FIG. 19. FIG. 21. difference of effect when the load upon each beam is equal to the load upon either of the other beams, and the lengths of the beams are equal. The destructive energy is in Fig. 18, D = Pn 19, D " 20, D " 21, > 68 COMPARISON OF CONDITIONS SAFE LOAD. CHAP. V. 70. Rules for Cases in wliieli the Weights and Lengths are Equal. Putting these equal to the resistance for levers, we have (Art. 35) for the case shown in Fig. 18, Pn = Sbd 2 " 19, $Un - Sbd 2 " 20, \Wl= Sbd 2 21, \Ul = Sbd s and, since (Art. 35) ^S = B y S = J/?. If in the above we substitute this value for 5, we shall have the following rules : For case i, ^Pn Bbd 2 (15.) " 2, 2 Un = Bbd 9 (18.) " 3, Wl = Bbd 9 (9.) " 4, $Ul= Bbd 2 (17.) and in case 5, W ~ = ^^ (.Z0.) this last being that of a load located at any point in the length of a beam (Art. 57). 71. Breaking and Safe Loads. These rules show the relation of the load to the resistance. Before showing their applications, the proportion which exists between the break- ing load and what is called the safe load will be considered. 72. The above Rules Useful Only in Experiments. The rules thus far shown have all been based upon the con- dition of equilibrium between the destructive power of the load and the resistance of the material ; or, in other words, an equilibrium at the point of rupture. Hence they are chiefly useful in testing materials to their breaking point. MARGIN FOR SAFETY. 69 73. Value of <r ? the Symbol of Safety. To make the rules useful to the architect, it is requisite to know what por- tion of the breaking load should be trusted upon a beam. It is evident that the permanent load should not be so great as to injure the fibres of the beam. The proportion between the safe and the breaking weights differs in different materials. The breaking load on a unit of material being represented by B, as before, let T represent the safe load, and a the proportion between the r) r) two ; or, T : B : : i : a =- then T-. The values of a, 1 a for several kinds of building materials, have been found and recorded in Table XX., an examination of which will show that a, for many kinds of materials, is nearly equal to 3, a number which is in general use.* 74. Value of , the Symbol of Safety. In the rules a may be taken as high as we please above the value given for a in the table ; but never lower than the value there given. T) If a be taken at 4, then, as above, T = = \B equals the safe power of the unit of material, and we have Wl = \Bbd 2 ; or, ^.Wl^Bbd*, as the proper rule for a beam supported at each end and loaded at the centre. In order, however, to * This is the value as fixed by taking the average of the results of the tests of several specimens of the same kind of material, or material of the same name. Owing to the large range in the results in any one material, it is not safe, in a general use of this symbol, to take it at the average given in the table. It should for ordinary use be taken higher. When the kind of material in any special and important work is known, and tests can be made of several fair specimens of it, and from the results com- putations made of the values of a, then an average of these would be safe to use. For the ordinary woods in general rules, it is prudent to take the value of a at not less than 4. 70 COMPARISON OF CONDITIONS SAFE LOAD. CHAP. V. make the rules general, we shall not adopt any definite num- ber, as 4, but use the symbol a. the value of which is to be taken from the table in accordance with the kind of material employed, increasing its value at discretion. (Sec note, Art. 73.) 75. Rules for Safe Loaris. The rules, with this factor a introduced, will then be as follows : Rule i, 4Pan = Bbd* (19.) " 3, Wai = Bbd 2 (21.) " 4, \Ual- Bbd 2 mn 5, 4,Wa-j-=Bbd' (23.) 76. Applications of tlie Rules. In this form the rules are ready for use applying them as below. Rule i is applicable to all cases where a load is suspended from the end of a lever (Fig. 18), said lever being fixed at the other end in a horizontal position. Rule 2 applies to cases where a load is equally distributed upon a lever fixed at one end (Fig. 19). Rule 3 is applicable to a load concentrated at the middle of a beam supported at both ends (Fig. 20). Rule 4 is applicable to equally distributed loads upon beams supported at both ends (Fig. 21). Rule 5 is applicable to a load concentrated at any point upon a beam supported at both ends (Fig. 7). 77. Example of Load at End of Lever. To show the practical working of these rules, take, first, an example coming under rule i, formula (19.), 4Pan = Bbd 9 LEVER EXAMPLE SYMBOL OF SAFETY. 71 Let it be required to find the requisite breadth and depth of a piece of Georgia pine timber, fixed at one end in a wall, and sustaining safely, at five feet from the wall, a weight of 1200 pounds ; the ratio between the safe and breaking weights being taken as i to 4, and the value of B for Georgia pine being 850 (Art. 13). 78. Arithmetical Exemplification of the Rule. The first thing, in applying a rule, is to distinguish between the known and the unknown factors of an equation, by so trans- posing them that those which are known shall stand upon one side, and the unknown upon the other side of the equa- tion. In rule i, formula (19.), as above, the known factors are 4, P, a, n and B ; therefore we transpose, so that Substituting the known quantities for the symbols of the first member, we have 4 x 1200 x 4 x g 79. Caution in Regard to , the Symbol of Safety. The working of this problem is interrupted to remark that students are liable to err in estimating the value of a, making it a fraction instead of a whole number. Thus, if the pro- portion between the safe and the breaking weights be as i to 4, they, starting with the idea that the safe weight is to be one fourth of the breaking weight, make a equal to J, instead of 4. This is a serious error, as the result would be a destructive energy of only one sixteenth (for |- : 4 : : i : 16) of the true amount, and consequently the resultant resistance of the timber would be but one sixteenth of what it should 72 COMPARISON OF CONDITIONS SAFE LOAD. CHAP. V. be, and in practice it would be found that the beam would break down with only one fourth of the amount considered the safe weight. To farther explain the value of a, let W equal the break- ing weight, and T the safe weight ; the proportion being as 4 to i. Then T \W, or ^T W. Now, in formula (9.) (Wl Bbd 3 }, in order to preserve equality, it is requisite, in removing the symbol W denoting the breaking weight, that we substitute its equal, or ^T. So when, in the new for- mula for safe weight, W is understood to represent not the breaking but the safe weight, ^T becomes ^W, and we have ^Wl = Bbd 2 ; therefore the symbol a is to be not a fraction but a whole number. Returning from this digression to the expression at the end of Art. 78, and reducing it, we have 96000 - = 1 12-94 Here we have the value of the breadth multiplied by the square of the depth, but neither the one nor the other is as yet determined. 80. Various methods of Solving a Problem. There are at least three ways of procedure by which to determine the value of each of these factors. The breadth and depth may be required to be equal ; the breadth may be required to bear a certain proportion to the depth ; or, one of the factors may be fixed arbitrarily. First. If the timber is to be square, then b will equal d, bd 2 = d\ and d = ty 112-94 = 4-83 that is, the dimensions required are 4-83, or, say 5 inches square. MANNER OF WORKING A PROBLEM. 73 Second. Let the breadth be to the depth in the proportion of 6 to 10, then b : d\ : 6 : 10 or iob = 6d or b = o-6d Then 112-94 = bd 2 = o-6dx d* = o-6d s = d 3 = 188-23 =w that is ^=5-73, and b= 5.73x0-6 = 3-44 The timber should be therefore 3-44 inches broad, and 5.73 inches deep ; or, 3^ x 5} inches. Third. The breadth or depth may be determined arbitra- rily, or be controlled by circumstances. Let the breadth be fixed, say at 3 inches, then 112-94 = bd 2 3<af* ; -ii2i=,/. = 37-65 d 6-14 The dimensions should be 3 x 6-14, or, say, 3 x 6J inches. Again, let the depth be fixed, say at 6 inches, then 112-94 = bd* = bx 6* 112- thus giving as the dimensions of the beam 3- 14 x 6, or, say 3 J x 6 inches. We have now these four answers to the question of Art. 77, namely : If the beam be square, the side of the square must be 5 inches. If the breadth and depth be in the proportion of 6 to 10, the breadth must be 3! and the depth 5f inches. 74 COMPARISON OF CONDITIONS SAFE LOAD, CHAP. V. If the breadth be fixed at 3 inches, then the depth must be 6J inches. If the depth be fixed at 6 inches, then the breadth must be 3^ inches. 81. Example of Uniformly Distributed Load on Lever. Take an example coming under rule 2, formula (20.), Let the conditions be similar to those given in Art. 77, ex- cept that the weight is to be equally distributed, instead of being concentrated at the end. What are the required di- mensions of breadth and depth ? The formula transposed becomes, 2Uan _ B As the known factors are all the same as in the last ex- ample, except the numerical co-efficient, which here is only one half of its former value, it follows that bd 2 in this case must be equal to one half of bd* in the previous case ; or, - =' 56.47 = bd* Now to apply this result : First. If the timber be square, 56-47 = d 3 = ^S4 3 Second. If the breadth and depth are to be as 6 to 10, 56-47 = 0-6 d 3 and = 4.55x0.6 = VARIOUS LOADS COMPARED. 75 Third, If the breadth be fixed at 2 inches, then 56.47 = bd 3 = 2d a ^ = j* = 28 . 24 2 d= 5-31 Fourth. If the depth be fixed at 5 inches, then 56-47 = bd 2 b x 5 9 The four answers are, therefore, 3-^ square 2f x 4^ 2x5! and 2 J x 5 ; and the beam may be made of the dimen- sions named in either of these four cases and be equally strong. 82. Load Concentrated at Middle of Beam. In an example under rule 3, the value of bd 2 in the formula Wai = Bbd s , would be just one quarter of that required by rule i. 83. Load Uniformly (Distributed on Beam Supported at Both Ends. In cases under rule 4, the values of bd* would be only one eighth of those under rule i ; and, in general, the five rules given are so related that when the result of com- putations under any one of them has been obtained, the re- sult in any other one may be found by proportion, in compar- ing the two rules applicable. COMPARISON OF CONDITIONS SAFE LOAD. CHAP. V. QUESTIONS FOR PRACTICE. 84. What breadth and depth are required for a white pine beam, of sufficient strength to carry safely 3000 pounds equally distributed over its length, the beam being 12 feet long and supported at each end ? The breadth is to be one half of the depth, and the factor of safety a equals 4. 85. What would be the size if square ? 86. What would be the depth if the breadth be fixed at 3 inches ? 87. What would be the breadth if the depth were fixed at 6 inches ? CHAPTER VI. APPLICATION OF RULES FLOORS. ART. 88. Application of Rules to Construction of Floors. Having completed the investigation of the strength of beams to resist rupture so far as to obtain formulas or rules applicable to the five principal cases of strain, we will now show the application 1 of these rules to the solution of such problems as occur in the construction of floors. As these rules, however, are founded simply upon the resistance to rupture, the size of a beam determined by them will be found to be much less than by rules hereafter given ; and the beam, although perfectly safe, will yet be found so small as to be decidedly objectionable on account of its excessive deflection. Owing to this, floor beams in all cases should be computed by the rules founded upon the resistance to flexure, as in Chapter XVII. 89. Proper Rule for Floors. Floor beams are usually subjected to equally distributed loads. For this, formula (22.) is appropriate, as it " is applicable to equally distributed loads upon beams supported at both ends." It is = Bbd 2 90. The Load on Ordinary Floors, Equally Distributed. The load upon ordinary floors may be considered as being equally distributed ; at least when put to the severest test a densely crowded assemblage of people. For this load all floors should be prepared. 7 APPLICATION OF RULES FLOORS. CHAP. VI.- 91. Floors of Warehouses, Factories and Mills. The floors of stores and warehouses, factories and mills, are re- quired to sustain even greater loads than this, but in all the load may be treated as one equally distributed. 92 B Rule for Load upon a Floor Beam. Each beam in a floor is subjected to the strain arising from the load upon so much of the floor as extends on each side half way to the next adjoining beam ; or, that portion of the floor which is measured by the length of the beam and by the distance apart from centres at which the beams are laid. Denote the distance apart, in feet, at which the beams are placed (measuring from the centres' of the beams) by c. Then cl will equal the surface of the floor carried by one of the beams. If the load in pounds upon each superficial foot of the floor be expressed by /, then the total load upon a floor beam will be cfl. This is an equivalent for U, the load. By substituting for U its value cfl in the formula we have %acfl> = Bbd* (24.} which is a rule for the load upon a floor beam. 93. Nature of the Load upon a Floor Beam. Before this formula can be used, the value of /must be determined. This symbol represents a compound weight, comprising the weight of the materials of construction and that of the superimposed load. The weight of the materials of construction is also in itself a compound load. A part of this load the floor plank and ceiling (the latter being either.of boards or plastering) will be a constant quantity in all floors ; but the floor beam WEIGHTS OF MATERIALS OF CONSTRUCTION. 79 will vary in weight as the area of its cross-section. In all cases of wooden beams, however, the weight of the beam is so small, in proportion to the general load, that a sufficiently near approximation to its weight may be assigned in each case before the exact size of the beam be ascertained. 94-. Weight of Wooden Beams. For example, in floors for dwellings, the beams will vary from 3x8 to 3 x 12, ac- cording to the length of the beam. If the timber be white pine (the weight of which is about 30 pounds per cubic foot, or 2^ pounds per superficial foot, inch thick), the 3x8 beam will weigh 5 pounds, and the 3x 12 beam 7-^ pounds; or, as an average, say 6^ pounds per lineal foot for all white pine beams for dwellings. For spruce, the average weight is about the same. Hemlock, which is a little heavier, may be taken at 7 pounds ; and Georgia pine (seldom used in dwell- ings) should be put at about 9 pounds per lineal foot. 95. Weight in Stores, Factories and Ulilfls to be Esti- mated. For stores, factories and mills the weight is greater, and is to be estimated. 96. Weight of Floor Plank. The weight of the floor plank, if of white pine or spruce, is about 3 pounds; or, if of Georgia pine, about 4^ pounds per superficial foot. 97. Weight of PIaterSng. The weight of plastering varies from 7 to 1 1 pounds, and is, on the average, about 9 pounds, including the lathing and furring, per superficial foot. 98. Weight of Beams in Dwellings. The weights of beams, given in Art. 94, are for the lineal foot, but it is re- quisite that this be reduced so as to show the weight per square foot superficial of the floor. When the distance from 8O APPLICATION OF RULES FLOORS. CHAP. VI. centres at which the floor beams are placed is known, the weight per lineal foot divided by the distance between cen- tres in feet will give the desired result. Thus, let the distance from centres of white pine floor beams be 16 inches, or i-J- feet. Then 6f -4- i-J = 4^ pounds. As the average distance from centres in dwellings differs little from 16 inches, the weight of beams may be safely taken at 5 pounds per superficial foot for white pine and spruce. 99. Weight of Floors in Dwellings. In summing up we have, for the weight of the floor plank, 3 pounds ; for the plastering, 9 pounds, and for the beams, 5 pounds ; and the sum of these items, 17, or, in round numbers, say 20 pounds is the total weight of the materials of construction upon each superficial foot of the floor of ordinary dwellings ; and this is large enough to cover the weight per superficial foot, even when a heavier kind of timber, such as Georgia pine, is used. 100. Superimposed Load. We have now to consider the superimposed weight, or the load to be carried upon the floor. 101. Greatest Load upon a Floor. * Mr. Tredgold, in speaking of bridges, says (Treatise on Carpentry, Art. 273): " The greatest load that is likely to rest upon a bridge at one time would be that produced by its being covered with peo- ple." Again he says : " It is easily proved that it is about the greatest load a bridge can possibly have to sustain, as well as that which creates the most appalling horror in the case of failure." The floors of churches, theatres, and other * The substance of the following discussion of the load per foot upon a floor was read by the author before the American Institute of Architects, and published in the Architects' and Mechanics' Journal, New York, in April, 1860. TREDGOLD'S ESTIMATE OF LOAD ON FLOOR. 81 assembly rooms, and also those of dwellings, are all liable to be covered with people at some time (although not usually), to the same compactness as a bridge. Therefore, to find the greatest strain to which floor timbers of assembly rooms and dwellings are subjected, it will be requisite, simply, to weigh the people ; or, to find an answer to the question in the ex- periments of those who have weighed them. 102. Tredgold's Estimate of Weight on a Floor. Mr. Tredgold, in the article quoted, says : "Such a load is about 120 Ibs. per foot ;" and again, at page 283 of his Treatise on the Strength of Iron, he says: " The weight of a superficial foot of a floor is about 40 Ibs. when there is a ceiling, counter- floor, and iron girders. When a floor is covered with peo- ple, the load upon a superficial foot may be calculated at 120 Ibs. Therefore 120 + 40 = 160 Ibs. on a superficial foot is the least stress that ought to be taken in estimating the strength for the parts of a floor of a room." 103. Tredgold's Estimate not Substantiated by Proof. Mr. Tredgold's most excellent works on construction have deservedly become popular among civil engineers and archi- tects. With very few exceptions, the whole of the valuable information advanced by him has stood the test of the ex- perience of the last fifty years ; and notwithstanding that many other works, valuable to these professions, have since appeared, his works still remain as standards. Statements made by him, therefore, should not be dissented from except upon the clearest proof of their inaccuracy ; and only after obtaining ample proof is the statement here ventured that Mr. Tredgold was in error when he fixed upon 120 pounds per foot as the weight of a crowd of people. In the writings of Mr. Tredgold, his positions are gener- ally sustained by extensive quotations and references ; but 82 APPLICATION OF RULES FLOORS. CHAP. VI. in this case, so important, he gives neither reference, data from which he derives the result, nor proof of the correct- ness of his statement. This proof must be sought else- where. 104. Weight of People Sundry Authorities In the year 1848, an article appeared in the Civil Engineer and Architects Journal, containing information upon this subject. From this article we learn that upon the fall of the bridge at Yar- mouth, in May 1845, Mr. James Walker, who was employed by government to investigate the matter, stated in evidence before the coroner, that his estimate of the load upon the bridge was based upon taking the weight of people at an average of 7 stone (98 pounds) each ; and admitted that this was a large estimate, rather higher, perhaps, than it ought to be ; yet he did so because it was customary to estimate them at this weight ; and further, that he calculated that six people would require a square yard for standing room. At this rate there would be two persons in every three feet, and the weight would be 65 pounds per foot. Herr Von Mitis, who built a steel suspension bridge over the Danube, at Vienna, estimated 15 men, each weighing 115 Vienna pounds, to a square fathom of Vienna. This, in Eng- lish measurement and weight, would be equal to 39 men in every hundred square feet, and nearly 55 pounds per foot. Drury, in his work on suspension bridges, lays down an arbitrary standard of two square feet per man of 10 stone weight. This equals 70 pounds per superficial foot. In testing new bridges in France, it is usual for govern- ment to require that 200 kilogrammes per square metre of platform shall be laid on the bridge for 24 hours. This is equal to 41 pounds per foot. The result of combining the above four instances is an average of 57! pounds per foot. WEIGHT OF PEOPLE. 83 But we have a more accurate estimate, founded upon trustworthy data. Quetelet, in his Treatise on Man, gives the average weight of males and females of various ages as follows : Average weight of males at 5, 10 and 15 years, 61-53 " 20 " 25 " I35-59 " " 30, 40 " 50 " 140.21 Average weight of females at 5, 10 and 15 years, 57-50 20 " 25 " 116-33 " " " 30,40 " 50 " 121-80 6J 632-96 Total average weight in Ibs. 105-5 105. Estimated Weight of People per Square Foot of Floor. The weight of men, women and children, therefore, is 105.5 pounds each, on the average. This may be taken as quite reliable as to the weight of people. Now as to the space occupied by them. It is known among military men that a body of infantry closely packed will occupy, on the average, a space measur- ing 15 x 20 300 square inches each. At this rate, 48 men would occupy 100 square feet, and if a promiscuous assembly should require the same space each, then there would be a load of 50-64 pounds upon each square foot. In military ranks, however, the men would weigh more. Taking the weight of males from 20 to 50 years, in the above table this being the probable range of the ages of soldiers the average is found to be 137-9; a weight of 66 pounds upon each superficial foot of floor ; and this weight may be taken as the greatest which can arise from a crowd of people. 84 APPLICATION OF RULES FLOORS. CHAP. VI. 106. Weight of People, Estimated a a Uve Load. But this is simply the weight, no allowance being made for any increase of strain by reason of the movement of the people upon the floor. We will now consider the increase made in consequence of the agitation of the weight through walking and other movements. In walking, the body rises and falls, producing in its fall a strain additional to that due to its weight when quiet. The moving force of a falling body is known to be equal to the square root of 64^ times the space fallen through in feet, multiplied by the weight of the body in pounds. By this rule, knowing the weight and the height of fall, we may compute the force. The weight in the present case, 66 pounds, is known, but the height of fall is to be ascertained. This height is not that of the rise and fall of the foot, but of the body ; the latter being less than the former. The elevation of body varies considerably in different persons, as may be seen by observ- ing the motions of pedestrians. Some rise and fall as much as half an inch at each step, while others deviate from a right line but slightly. If, in the absence of accurate obser- vation, the rise be assumed at a quarter of an inch, as a fair average, then the moving force of the 66 pounds, computed by the above rule, would be 76.4 pounds. This would be the moving force at the moment of contact, and the effect pro- duced would be equal to this, provided that the falling body and the floor were both quite inelastic ; but owing to the presence of an elastic substance on the soles of the feet, and at the joints of the limbs, acting as so many cushions, the force of the blow upon the floor is much diminished. The elasticity of the floor also diminishes the effect of the force to a small degree. Hence the increase of over ten pounds, as found above, would be much diminished, probably one half, or, say to six pounds. ACTUAL WEIGHT OF MEN. 85 I07L Weight of Military. This six pounds would be the increase per foot superficial. To make this effect general over the whole surface of the floor, it is requisite that the weight over the whole surface fall at the same instant ; or, that the persons covering the floor should all step at once, or with regular military step. It will be found that this is the se- verest test to which a floor of a dwelling or place of assem- bly can be subjected. In promiscuous stepping the strain would be much less, scarcely more than the quiet weight of the people. 108. Actual Weights of men at Jackson's and at Hoes' Foundries. The above results, it must be admitted, are de- rived from data with reference to the height of fall, and to the lessening effect of the elastic intervening substances, which are in a measure assumed, and hence are not quite conclusive. They need the corroboration of experiment. To test them, I experimented, in April, 1860, at the foun- dry of Mr. James L. Jackson in this city. He kindly placed at my service his workmen and his large scale. The scale had a platform of 8^ x 14 feet. It was of the best construc- tion, and very accurate in its action. Eleven men, taken indiscriminately from among the workmen of the foundry, stood upon the platform. Their combined weight while standing quietly was 1535 pounds, being an average of 139-55 pounds per man. This is but a pound and a half more than was derived from the tables of Quetelet. It is quite satisfac- tory in substantiating the conclusion there drawn.* * In May, 1876, since the above was written, by the courtesy of Messrs. R. Hoe & Co., of this city, who placed at my disposal their platform scale and men, I was enabled, by a second experiment, to ascertain the weight of men and the space they occupy. Selecting twenty-six stalwart men from their smith shop, they were found to weigh 3955 pounds, and to occupy upon the platform a space 7 x *1\ = S 2 ^ square feet, or 753- pounds per superficial foot. This is a 86 APPLICATION OF RULES FLOORS. CHAP. VI. 109. Actual measure of Live Load. After ascertaining the quiet weight of the men, they commenced walking about the platform, stepping without order, and indiscriminately. The effect of this movement upon the scale was such as to make it register 1545 pounds ; an increase of only ten pounds, or less than one per cent. They were then formed in a circle and marched around the platform, stepping simultaneously or in military order. The effect upon the scale produced by this movement was equal to 1694 pounds, an increase of 159 pounds, or over ten per cent. This corroborated the results of the computation before made most satisfactorily ; ten per cent of the weight per foot, 66 pounds, being 6-6 pounds. As a final trial, the men were directed to use their utmost exertion in jumping, and were urged on in their movements by loud shouting. The greatest consequent effect produced was 2330 pounds, an increase of 795 pounds, or about 52 per cent. 110. More Space Required for Live Load. This seems a much more severe strain than the former, but we must consider that men engaged in the violent movements neces- sary to produce this increase of over 50 per cent need more standing room. Packed closely, occupying only 15 x 20 inches (the space allowed per man in computing the weight per foot to be 66 pounds), it would not be possible to move the limbs sufficiently for jumping. To do this, at least twice as much space would be required. But, to keep within the limits of safety, let only one half more space be allowed. In this case the 66 pounds would be the weight on a foot larger average than found at Mr. Jackson's, or than any previous weight on record, and is accounted for by the fact that these were muscular men, weighing about 12^ pounds each more than the heaviest hereinbefore noticed, and much heavier than it were reasonable to expect in assemblages generally. MEASURE OF LIVE LOAD. S/ and a half, or there would be but 44 pounds on each foot of surface. Add to this the 50 per cent for the effects of jumping, or 22 pounds, and the sum, 66 pounds, is the total effect of the most violent movements on each foot of the floor ; the same as for the weight of men standing quietly, but packed so much more closely. III. 3fo Addition to Strain by Live Load. The greatest effect, then, that it appears possible to produce by an assem- bly on a floor, is from the regular marching of a body of men, closely packed ; and amounts to 66 + 6-6 = 72-6 pounds per superficial foot. This result would show the necessity of providing for ten per cent additional to the weight of the people. This in general is not needed, for the conditions of the case generally preclude the possibility of obtaining this additional strain upon the floor. The strain of 66 pounds is only obtained by crowding the people closely together in the whole room. To obtain the ten per cent additional strain, they must be set to marching ; but there is no space in which to march, unless they march out of the room, and in doing this the strain is not increased, for the weight of those who pass out is fully equal to the stress caused by the act of marching. Were both ends of the room quite open, or were it a long hall, as a bridge, through which the people could march solid, the throng being sufficiently numerous to keep the floor constantly full, then the ten per cent would need to be added, but not in ordinary cases of floors of rooms. 112. Margin of Safety Ample for Momentary Extra Strain in Extreme Caes. It may be argued still, that, although the room be full and marching can only be effected by some of the people leaving the floor, yet this additional strain will be 88 APPLICATION OF RULES FLOORS. CHAP. VI. obtained in consequence of the exertion made in the act of taking the very first step, before any have left the room. To this we reply that the strain thus produced would not endanger the safety of the floor, because this strain, when compared with the ultimate strength of the beams sustaining it, would be quite small, and its existence be but momentary. Beams made so strong as not to break with less than from three to five times the permanent load would certainly not be endangered by the addition for a moment of only ten per cent of that load. 113. Weight Reduced by Furniture Reducing Standing Room. Hence, for all ordinary cases, no increase of strength need be made for the effects of motion in a crowd of people upon a floor, and therefore the amount before ascertained, 66 pounds, or, in round numbers, say 70 pounds, may be used in the computations as the full strain to which the beams may be subjected. Indeed, the cases are rare where the strain will even be as much as this. When we consider the space occupied in dwellings by furniture, and in assembly rooms by seats, the presence of these articles reducing the standing room, the average weight per foot superficial will be found to be very much less. 114. The Greatest Load to be provided for i ?O Pounds per Superficial Foot. As a conclusion, therefore, floor beams computed to safely sustain 70 pounds per superficial foot, or to break with not less than three or four times this, will be quite able to bear the greatest strain to which they may be subjected in the floors of assembly rooms or dwellings; and especially so when the precaution of attaching them to each other by bridging* is thoroughly performed, thereby ena- * The subjects of Floor Beams and of Bridging are farther treated in Chap- ters XVII. and XVIII. RULE FOR FLOORS OF DWELLINGS. 89 bling the connected series of beams to sustain the concen- trated weight of a few heavier persons or of some heavy article of furniture. MS. Rule for Floors of Dwellings. We now have, by including the weight of the materials of construction as shown in Art. 99, the total weight per superficial foot, as follows : / = 70 + 20 = 90 for the floors of dwellings. With this value of/, formula (&4-\ i- acfl 2 = Bbd* becomes J- ac 90 /* = Bbd 3 or, when a = 4 i$ocl'=Bbd* (25) 116. Distinguishing Between Known and Unknown Quan- tities. This formula may now be applied in determining problems of floor construction in dwellings, in which the safe strength is taken at one fourth of the breaking strength. In distinguishing between known and unknown quantities, we will find generally that B and / are known, while c, b and d are unknown. From formula (25.) therefore, we have, by grouping these quantities, _ L 8oT = ^ (M-) (17. Practical Example. Formula (26.) is a general rule for the strength of floor beams of dwellings. As an example under this rule, let it be required to find the sectional dimensions and the distance from centres of the beams of a floor of a dwelling; the span or length between bearings being 20 feet, and the material, spruce. 90 APPLICATION OF RULES FLOORS. CHAP. VI. Here B = 550 and /= 20; and from formula (26.) i8ox2o 2 bd* - = --- = 1300 550 c 118. Eliminating Unknown Quantities We have here the numerical value of a quotient, arising- from a division of the product of the breadth and square of the depth, by the distance from centres at which the beams are to be placed. Two of the three unknown quantities are now to be as- signed a value, before the third can be determined. Circum- stances will indicate which two may be thus eliminated. In some cases the breadth and depth of the timber arc fixed, and the distance from centres is the unknown quantity ; in others, the distance from centres and the depth may be the fixed quantities, and the breadth be the factor to be found ; or, the distance and the breadth be fixed upon, and the depth be the quantity sought for. Generally, the breadth and depth are assigned according to the requirements of the case, or simply as a trial to ascertain the scope of the question, and the distance from centres is the dimension left to be deter- mined by the formula. 119. Isolating the Required Unknown Quantity. In the solving of a question of either kind, the formula must first be transposed so as to remove all of the factors, except the one sought for, to the same side of the equation ; thus, bd* , .,, = 130-9 becomes either or c = b bd 2 130-9 Assuming the value of any two of the factors, we select the proper formula and proceed with the test for the third factor. RULE FOR DISTANCE FROM CENTRES. 9! (20. distance from Centres at Given Breadth and Depth. For example, fix the breadth and depth at 3 and 9 inches. Then to find c, the above expression, W* c = becomes 130-9 130-9 130-9 The value of c being in feet, this gives about i foot 10 inches, or 22 inches. 121. Distance from Centres at Another Breadth and Depth. The above result may be considered too great, and beams of less size and nearer together be more desirable. If so, assume a less size, say 3x8; we then have 3 x 8 2 102 c ^- -=i-47 130-9 130-9 This gives c equal to about \*j\ inches. 122. Distance from Centres at a Third Breadth and Depth. With the object in view of economy of material, let another trial be had, fixing the size at 2\ x 9. In this case , = 2*X9! = 202-5 = I . S5 130-9 130-9 This gives for c about i8J inches. The answers then to this problem are, for 3 x 9 inches, 22 inches from centres, "3x8 " \j\ and " 2j x 9 " i8 " " These trials may be extended to any other proportions thought desirable, fixing first the breadth and depth, and then determining the corresponding value of c. (See pre- caution, Art. 88.) 92 APPLICATION OF RULES FLOORS. CHAP. VI. 123. Breadth, the Depth and Distance from entre being Given. Again, it may be desirable to assume a value for c, and then to ascertain the proper corresponding breadth and depth. In this case, one of the two unknown factors, b and d, must also be assumed. Let us fix upon c = 1-5 and d 8, then the formula in Art. 119, c l 3O-9 x 1-5 becomes b - ^ = 3.07 j - VJ \~> V> \J i -L 1 \~> O t/ ^ or, say 3 inches for the breadth. 1 2 4-. Depth, the Breadth and Distance from Centres being Given. If the breadth be assumed, say at 2^, then, with <; = 1.5, to find the depth we have (Art. 119), d= 8-86 = 8| inches. Thus, when placed at 18 inches from centres, we have, in the one case 3x8 inches, and in the other 2^ x 8J inches. 125. General Rules for Strength of Beams. Any other case of wooden beams for dwellings may be treated in a similar manner, using formula Beams of any material for any building may be deter mined by the general formula in all cases regarding the caution given in Art. 88, QUESTIONS FOR PRACTICE. 126. In the floor of a dwelling, composed ot 3 x 9 inch beams 16 feet long, how far from centres should spruce beams be placed ? 127. How far if of hemlock? 128. How far if of white pine? 129. In the floor of a dwelling, composed of 2j x 10 inch beams 19 feet long, how far from centres should spruce beams be placed? 130. How far if of hemlock? 1 3 I. How far if of white pine ? 132. In a floor of 4 x 12 inch beams 23 feet long, and re- quired to carry 150 pounds per superficial foot (including material of construction), how far from centres should spruce beams be placed, the factor of safety being 4? 133. How far if of hemlock? 134. How far if of white pine ? 135. How far if of Georgia pine? CHAPTER VII. GIRDERS, HEADERS AND CARRIAGE BEAMS. ART. I36 A Girder Defined. By the term girder is meant a heavy timber set on posts or other supports, and serving, as a substitute for a wall, to carry a floor. 137. Rule for Girders. A girder sustaining a tier of floor beams carries an equally distributed load ; the same per superficial foot as that which is carried by the floor beams. In determining the size of the girder formula (24-) is appli- cable, namely, \acfl* = Bbd 2 138. Distance between Centres of Girders. In apply- ing this formula to girders, it is to be observed that c repre- sents the distance between centres of girders, Avhere there are two or more, set parallel ; or, the distance from centre of girder to one of the walls of the building, if the girder be located midway between the two walls ; or, an average of the two distances, if not midway. As an example of the latter case, in a building 30 feet wide, the centre of a girder is 12 feet from one wall and 18 feet from the other. Here GIRDERS DISTANCE BETWEEN. 95 139. Example of Distance from Centres. What is the required size of a Georgia pine girder placed upon posts set 15 feet apart, the centre of the girder being 12 feet from one wall and 18 feet from the other; the load per foot super- ficial of floor, including the weight of the materials of con- struction, being 100 pounds, and the value of a being taken at 4? 140. ize of Girder Required in above Example. By transposing formula (@4-) we have B and if the breadth be to the depth in the proportion of, say 7 to 10, then (Art. 80) *. = 0-5 XAX 15 x 100 x 15* .. 5 -- _ d* 1134-45 07x850 d = y U34-45 - 10-43 and b = 0-7 x 10-43 = 7'3- Therefore the girder should be 7-3 x 10-43 ; o r > to avoid fractions, say 8 x n inches. 14-1. Framing for Fireplaee, Stairs and Light-wells. We will now consider the subject of framing around open- ings in floors, for fireplaces, stairs and light-wells. 142. Definition of Carriage Beams, Headers and TaiB Beams. Fig. 22 may be taken for a representation of a stair- way opening in a floor ; AB and CD being the walls of the 96 GIRDERS, HEADERS AND CARRIAGE BEAMS. CHAP. VII, FIG. 22. building, A C and BD the carriage beams or trimmers, EF the header, and the beams which reach from the header to the wall CD, such as GH, the tail beams. 14-3. Formula for Headers General Considerations. First, the headers. The load upon the header EF is equally distributed, therefore formula (22.) is applicable. %Ual Bbd 2 The header carries half the load upon the tail beams, or the load upon a space equal to the length of the header by half the length of the tail beams. Let g represent the length of the header, n the length of the tail beams, and / the load per foot superficial ; then /, the load upon the header, equals and, as g here represents /, the length, therefore, HEADERS RULE PRECAUTION. 97 and formula (22.) becomes = Bbd* lafng 2 = Bbd 3 I44 B Allowance lor Damage by Mortising. This last formula should be modified so as to allow for the damage done to the header by the mortising- for the tenons of the tail beams. This cutting of the header ought to be confined as nearly as possible to the middle of its height, so that the injury to the wood may be at the place where the material is subject to the least strain. If this is properly attended to; it will be a sufficient modification to make the depth of the header one inch more than that required by the formula. Thus, when the depth by the formula is required to be 9 inches, make the actual depth 10; or, for d* substitute (dij, d being the actual depth. The rule, thus modified, will determine a header of the requisite strength with a depth one inch less than the actual depth. This will compensate for the damage caused by mortising. The expression in the last article then becomes \afng> =, Bb(d-lJ 14-5. Rule for Headers. Generally, the depth of a header is equal to the depth of the floor in which it occurs. Hence, when the depth of the floor beams has been deter- mined, that of the header is fixed. There remains then only the breadth to be found. We have, for the breadth of a header (from Art. 144) afng* ~- (See precaution in Art. 88.) 98 GIRDERS, HEADERS AND CARRIAGE BEAMS. CHAP. VII. Example. In a tier of nine inch beams, what is the required breadth of a white pine header at the stair- way of a dwelling, the header being- 12 feet long, and carry- ing tail beams 16 feet long; the factor of safety being 4? In formula (27.), making a = 4, /= 90, n= 16, g 12, B = 500 and d 9, the formula becomes 4 x 90 x 1 6 x 1 2 2 b = - TOT- = 6-48 4 x 500 x 8 The breadth of the header should be 6^, or say 7 inches, and its size 7x9 inches. 14-7. Carriage Beam and Bridle Irons. A carriage beam, or trimmer, in addition to the load of an ordinary beam, is required to carry half the load of the header which hangs upon it for support. As this is a concentrated load at the point of con- nection, all mortising at this point to re- ceive the header FIG. 23. should be carefully avoided, and the requisite support given with a bridle iron, as in Fig. 23. 148. Rule for Bridle Iron*. In considering the strain upon a bridle iron, we find that it has to bear half the load upon the header, and, as the iron has two straps, one on each side of the header, each strap has to bear only a quarter of the load upon the header. We have seen (Art. 14-3) that the load upon the header equals %fng, where g represents the length of the header, n the length of the tail beams, both in feet, and / the load per BRIDLE IRONS RULE. 99 superficial foot. The load upon each strap of the bridle iron will, therefore, be equal to Good refined iron will carry safely from 9000 to 15,000 pounds to the square inch of cross-section. Owing, however, to the contingencies in material and workmanship, it is pru- dent to rate its carrying power, for use in bridle irons, at not over 9000 pounds. If the rate be taken at this, and r be put to represent the number of inches in the cross-section of one strap of the bridle iron, then 9000^ equals the pounds weight which the strap will safely bear; and when there is an equilibrium be- tween the weight to be carried and the effectual resistance, we shall have \fng- from which r = 72000 (4-9. Example. For an example, let f = 100, n 16, and g= 12 ; then 100 x 16 x 12 r - - 0-266 72000 If the bridle iron were made of J by i^ inch iron (-J-x i = 0-375) tne SIZG would be ample. For such a header they are usually made heavier than this, yet this is all that is needed. It is well to have the bridle iron as broad as possible, in order to give a broad bearing to the wood, so that it shall not be crushed. ISO. Rule for Carriage Beam with One Header. To return to the carriage beam, or trimmer. The weight to be carried upon a carriage beam is compounded of two loads ; one the ordinary or distributed load upon a floor beam, as TOO GIRDERS, HEADERS AND CARRIAGE BEAMS. CHAP. VII. shown in formula (24)\ the other a concentrated load from the header. Of the former a carriage beam is required to carry one half as much as an ordinary beam ; or, the load which comes upon the space from its centre half way to the adjacent common beam. This is the half of that shown in formula (2A), or \acfl 2 = Bbd 2 The symbol W in formula (23.) represents the load from the header, and is equal (Art. 14-3) to \fng. The carriage beam carries half this load, or %fng ; hence \fng = W or, by formula (23. \ mn mn , mn* Combining this with the formula for the diffused load, we have \acfl 2 + afg "^- = Bbd 9 or This is a rule for the resistance to rupture in carriage beams having one header. (See Art. 241, and caution in Art. 88.) 151. Example. As an example, let it be required to show the breadth of a white pine carriage beam 20 feet long, car- rying a header 10 feet long, with tail beams 16 feet long, in a floor of lo-inch beams, which are placed 15 inches from centres ; and where the load per superficial foot is 100 pounds, and the factor of safety is 4. Transposing formula (29) we have _ b-af CARRIAGE BEAM TWO HEADERS. IOI in which a 4, f 100, c = 15 inches = ij feet, /= 20, g 10, n 16, m ln = 20 16 = 4, B 500 and d= 10. Therefore, 2 + .ioxi6 a x-A-) - - SLZ = 4 x loo x- -= 5oox 10 The breadth required is 5.096, or sa}" 5 inches. The trimmer should be 5 x 10 inches. 152. Carriage Beam with Two' Headers. For those cases in which the opening in the floor (Fig. 25) occurs at or near the middle (instead of being at one side, as in Fig. 22), two headers are required ; consequently the carriage beam, in addition to the load upon an ordinary beam, has to carry tivo concentrated loads. To obtain a rule for this case the effect produced upon a beam by two concentrated loads will first be considered. 153. Effect of Two Weights at the Location of One of Them. The moment of one weight upon a beam is (Art. 56) W '-j~. This is the effect at the point of location of the weight. A second weight, at another point, will produce a strain at the location of the first weight. To find this strain, let two weights, W and V (Fig. 24) be located upon a beam resting upon two supports, A and B. Let the distance from W to the support which may be reached without passing the other weight, be represented by ;;/, and the distance to the other support by ;/. From V let the distances to the supports be designated respectively by s and r ; s and n being distances from the same support. The letters W and V, representing the respective weights, are to be carefully assigned as follows : Multiply IO2 GIRDERS, HEADERS AND CARRIAGE BEAMS. CHAP. VII. one of the weights by its distance from one support, and the product by the distance from the other. Treat the other weight in the same manner ; and that weight which, when so multiplied, shall produce the greater product is to be called W. For example, in Fig. 24 let the two weights equal 8000 and w FIG. 24. 6000, /= 20, the distances from the 8000 weight to the sup- ports equal 4 and 16, and those from the 6000 weight equal 5 and 15. Then 8000 x 4 x 16 5 12000 and 6000 x 5 x 1 5 = 450000 The former result being the greater, the former weight, Sooo, is to be called W, and the latter V. The moment or effect of the weight W at its location is equal, as before stated, to W -=-. The effect of the weight V at the point W will (Art. 27) be equal to the portion of V borne at A, multiplied by the arm of lever m (Arts. 34 and 57). The portion of V sustained by A is (Arts. 27 and 28), Vj ; hence the effect of V at W will be Vj x m = V ^. Adding the two effects, we have This is the total effect produced at W by the two weights. EFFECT OF TWO CONCENTRATED LOADS. 103 In like manner it may be shown that the 'total effect at V is / / These are the moments or total effects at the two points of location. The first, when modified by the factor of safety a, gives a j (Wn + Vs) = Sbd 2 = -bd* (see Art. 35) from which we have 4 ^( Wn + Vs) = Bbd* (30) for the dimensions at W. Then, also, 4<t ( Vr + Wm) = Bbd> (31.) for the dimensions at V. (See caution in Art. 88.) Example. When the beam is to be of equal cross- section throughout its length, as is usually the case, then formula (30.), giving the larger of the two results, is to be used. For example, let a weight of 8000 pounds be placed at 3 feet from one end of a beam 12 feet long between bearings, and another weight of 3000 pounds at 5 feet from the other end. Then, as directed in Art. 153, 8000 x 3 x 9 = 216000 3000 x 5 x 7 = 105000 The weight of 8000 pounds having given the larger pro- duct, it is to be designated by W, and the other weight by V. 104 GIRDERS, HEADERS AND CARRIAGE BEAMS. CHAP. VII. Making a =. 4, we have for the greater effect (form. 30.\ 1M = Bbd* 4 x 4 x x ( 8ocx> x 9 + 3000 x 5 ) = Bbd~ = 348000 and with =$oo, and b=o-jd, we have B x o-jdx d 2 348000 348000 d s -^^ -=004-29 500 x 07 d = 9.98 b = 0-7 x 9-98 = 699 or the beam should be 6.99x9.98, or 7x10 inches. 155. Rule for Carriage Beam with Two Headers and Two Set of Tail Beam*. Let the rules of Art. 153 be applied to the case of a carriage beam with two concentrated loads, as in Fig. 25. FIG. 25. When the opening in the floor is midway between the walls, the two sets of tail beams are of equal length ; or, m=s ; and n=r ; therefore mn=sr. The weights are also equal ; therefore Wmn = Vrs ; or, the strains at the headers CARRIAGE BEAM WITH TWO HEADERS. 105 are equal. By moving the opening from the middle, the weight at the header carrying the longer tail beams is in- creased ; so also the product of the distances to the supports is increased ; therefore the letter W is to be put at that header which carries the longer tail beams, for then the pro- duct Wmn will exceed the product Vrs. The weight at W is equal to the load upon one end of the header which is lodged there for support. This is equal to (Arts. 14-3 and 150) ^fgm ( m being the length of the tail beams sustained by this header), or W= \fgrn. In like manner it may be shown that V= %fgs. By substituting these values of W and V in formula (30.) we have In addition to this load, the carriage beam is required to carry half the load upon a common beam, or half that shown at formula (24-), or \acfl*. The expression for the full effect at W therefore is Bbd* = Bbd 2 = af[m (mn + s ") f + \cl* ] In like manner we find for the full effect at Bbd* = a/[s (rs + m *) f (See caution in Art. 88.) 106 GIRDERS, HEADERS AND CARRIAGE BEAMS. CHAP. VII. These two formulas (32. and 33.) give the sizes of the carriage beam at W and V respectively, but when the beam is made equal in size throughout its length, as is usual, the larger expression (form. 32.) is to be used. 156. Example. What is the required breadth of a Georgia pine carriage beam 25 feet long, carrying two headers 12 feet long, so placed as to provide an opening between them 5 feet wide; the tail beams being 15 feet long on one side of the opening and 5 feet long on the other ; the floor beams being 14 inches deep and placed -18 inches from centres ; the load per superficial foot being 150 pounds, and the factor of safety being 4? Taking m to represent the longer tail beams, we have a = 4, 7=150, m=i$, n = 10, J=5, g = 12, /= 25, c = 1 8 inches = i| feet, B = 850 and d 14. Formula (32. \ now becomes 850x^x14' = 4 XI 5of I 5( I 5 XI o+5 2 )^7 + i XI i x2 5 2 *= 'S+' + * = 5-3* showing that the breadth should be 5.38. The beam may be made 5^ x 14 inches. (57. Rule for Carriage Beam wills Two Headers and One Set of Tail Beams. The preceding discussion, and the rules derived therefrom, are applicable to cases in which the two headers include an opening between them. When the headers include a series of tail beams between them, leaving an opening at each wall (Fig. 26), then the loads at W and V are equal ; for the total load is that which is upon the one series of tail beams, and is carried in equal portions at the ends of the two headers a quarter of the whole load at each CARRIAGE BEAM TWO HEADERS ONE SET TAIL BEAMS. IO/ FIG. 26. end of each header. If by j we represent the length of the tail beams, we have W= V=\jfg, and from formula (30.) we have, for the effect at W y Add to this half the load upon a common beam, \acfl* (Art. 92), and we have, as the full effect at W, and, for the size 01 the beam at W, = Bbd* ($4-) Similarly, we find for the size of the beam at V, af-~s (r + m) + \cl * = Bbd' 108 GIRDERS, HEADERS AND CARRIAGE BEAMS. CHAP. VII. These are identical, except that s (r+ m) in (35.) occupies the place of m(n + s) in (34*)- (See caution in Art. 88.) As in Art. I53, care must be taken to designate by the proper symbols the weights and their distances. In that article the proper designation was found by putting the letter W to that weight which when multiplied into its distances m and n would give the greater product. Here, as the weights are equal, the comparison may be made simply between the two rectangles mn and rs. Of these, that will give the greater product which appertains to the weight located nearer the middle of the beam ; this weight, therefore, is to be designated by W y and will be found at that header which is at the side of the wider opening. The distances m and n appertain to the weight W. The symbols being thus carefully arranged, formula (34-) gives the larger result, and is to be used when the beam is to be of equal sectional area throughout. I58. Example. To show the application of this rule, let it be required to find the size of a carriage beam in a tier of beams 12 inches deep and 16 inches from centres, with a weight per superficial foot of 100 pounds. In this case what should be the breadth of a white pine carriage beam 20 feet long between bearings, carrying two headers 12 feet long each, with one series of tail beams 10 feet long between them, so located as to leave an opening 6 feet wide at one wall and 4 feet at the other; the factor of safety being 4 ? Here we have the two distances m and s equal to 6 and 4, and putting ;;/ for the larger we have a = 4, /= 100, j 10, -=12, 1=20, m = 6, n 14, 5 = 4, c = i \, .5=500 and d=i2. Transposing formula (34) to find b, we obtain CARRIAGE BEAM QUESTIONS. 109 4X100 [~IOXI2 b = -i =x -x6(i4+4) + i><iVx20 2 4-34 500XI2 2 L 20 J The breadth is required to be 4-34 inches, and the size of carnage beam, say 4|- x 12 inches. (See caution, Art. 88.) QUESTIONS FOR PRACTICE. 159. A building, 26 feet wide between the walls, has a tier of floor beams 12 inches deep and 14 inches from cen- tres, supported at 16 feet from one of the walls by a gir- der resting upon posts set 15 feet apart. Upon that side of the building where the girder is 16 feet distant from the wall a stair opening occurs, extending 14 feet along the wall, and 6 feet wide. The floor is required to carry 150 pounds per foot superficial, including the weights of the materials of con- struction, with a factor of safety of 4. The girder, trim- mers and header all to be of Georgia pine. NOTE. The resulting answers to the following questions will be smaller than if obtained under rules in Chapter XVII. (See Art. 88.) 160. What must be the breadth and depth of the girder, the breadth being equal to 55 hundredths of the depth ? 161, What should be the breadth of the carriage beams? 162. What should be the breadth of the header? 163. What should be the area of cross-section of the bridle iron ? 110 GIRDERS, HEADERS AND CARRIAGE BEAMS. CHAP. VII. 164. Another opening 6 feet wide in the same tier of ^eams, has headers 10 feet long, with tail beams on one side 6 feet long and on the other side 4 feet long. What should be the breadth of the carriage beams ? 165. What ought the breadth of the floor beams of the aforesaid floor to be on the 16 feet side of the girder, if of white pine ? 166. In the same tier of beams there is still another pair of carriage beams. These carry two headers 16 feet long, and the two headers carry between them one series of tail beams 8 feet long, thus forming two openings, one at the girder 3 feet wide and the other at the wall 5 feet wide. What should be the breadth of these carriage beams ? CHAPTER VIII. GRAPHICAL REPRESENTATIONS. ART. 167. Advantages of Graphical Representations. In the discussion of the subject of rupture by cross-strains, rules have been given by which the effect in certain cases has been ascertained ; for example, that at the middle of a beam which rests upon two supports ; that at the wall in the case of a lever inserted in the wall ; and that at any given point in the length of a beam or lever. These rules are perhaps sufficiently manifest ; but when it becomes desirable to know the effect of the load in a new location, or under other change of conditions, an entirely new computation is needed. To obviate the necessity for this labor, and to fix more strongly upon the mind the rules already given, the method of representing strains graphically, or by diagrams, is useful, and will now be presented. 168. Strains in a L.ever measured by Scale. In Fig. 27 we have a lever AB, or half beam, in which the destructive energy or moment of the weight P, suspended from the free end B, is equal to the product of the weight into the arm of leverage at the end of which it acts (Art. 34) ; or FIG. 27. From A drop the vertical line AC = c, make it by any convenient scale equal to \IP, and join C and B. The tri- 112 GRAPHICAL REPRESENTATIONS. CHAP. VIII. angle ABC forms a scale upon which the strain produced at any point in AB may be obtained, simply by measurement ; for, at any point, D, the ordinate DE ( y), drawn parallel with the line AC, is equal (measured by the same scale) to the strain at the point D. In the two homologous triangles ABC and DBE, we have this proportion : I 7 CX */: c :: x : , == -^ By construction c = \IP, therefore UPx y = --f = px equals the weight into the arm of lever at the end of which it acts ; or Px = y is the destructive energy or moment of the weight P at the point D. In this equation (Px y) since P is constant, the value of y is dependent upon that of x, for however x may be varied, y will vary in like manner. If x be doubled, y will be doubled ; if x be multiplied or divided by any number, y will require to be multiplied or divided by the same number. We conclude then that we may assign any value to x desirable, or select any point in AB for the location of D, from D draw an ordinate DE, parallel with the line AC, and measuring the ordinate by the same scale by which c was projected, find the strain or destructive energy exerted upon the beam at the selected point D. 169. Example Rule for Dimensions. For example, let P 100 and / = 20, then AB \l = 10, and = IO X IOO = IOOO Now from a scale of equal parts (say tenths of an inch, or any other convenient dimensions), lay off c equal to ten of the divisions of the scale ; then each division represents 100 CORRESPONDING FORMULA. 113 pounds and c=iooo = %/P. Draw the line CB, and from any point D draw the ordinate y. Suppose that y, measured by the same scale, is found to equal 7^; then the strain at D equals J\ x 100 = 725 pounds. If y 6, then the strain at D equals 600 pounds ; and so of any other ordinate, its measure will indicate the strain in the beam at the end of that ordinate. We have, therefore [as in Art. 34, formula (#.)] Px = Sbd 2 and, with a the factor of safety, and putting for S its equivalent \B (Art. 35), or, 4/ter = Bbd* (36.) It is to be observed that the b and d of this formula are those required at D, the location of the ordinate y. When x equals the length of the lever AB, equals -J/, we have 2Pal = Bbd 2 and if P be taken as W, W being the load at the centre of a whole beam, we have 2 *%Wal=Bbd* Wai = Bbd 2 the same as formula 170. Graphical Strains in a Double L,ever. In Fig. 28 we have a beam AB resting . upon a point at the middle C y and carrying the two equal "Tx^ f loads R and P suspended from j[ \$ fi the ends. The half of this beam, or CB, is under the same conditions of FlG< * 8 ' strain as the beam AB in Fig. 27, and since the weights R and GRAPHICAL REPRESENTATIONS. CHAP. VIII. P are equal, and C is at the middle of AB, the one half of the beam, or AC, is strained alike with the other half CB. Therefore a strain at any point in the length of the beam is measured by an ordinate from that point to the line A WB, and formula (36.) is applicable to this case also, conditioned that x does not exceed ^/. 171. Graphical Strains in a Beam. In Fig. 29 we have a beam AB, resting upon two supports A and B, and loaded at middle with the weight W, one half of which, R, is borne upon A, and the other half, P, is supported by B. This beam has the same strains as that of Fig. 28, there- FlG - 2 9- fore (see Art. 26) the same formula (36.) is applicable, namely : tfax = Bbd a P = ^W, and by substitution (37.) 2 Wax = Bbd 3 a rule applicable to this case, conditioned that x shall not exceed /. When x = / then we have Wai = Bbd 2 the same as given in formula (21.). Again, if x be diminished until it shall reach zero, then 2 Wax = o or the strain is nothing. This is evidently correct, as the effect of the weight, in producing cross-strain, disappears at OF THE SHEARING STRAIN. the edge of the bearing. We are not to be permitted, how- ever, in shaping the beam to its exact requirements, to re- move all material at and upon the bearing wall, for there is another strain, known as the shearing strain, for which provision is to be made at the end of the beam. This strain we will now consider. (72. Mature of the Shearing Strain. The nature of the shearing strain, as well as of the cross-strain, is very clearly shown in Fig. 30, a diagram suggested by a similar one in "Unwin's Wrought-Iron Bridges and Roofs, London, 1869." In this figure a semi-beam, AB, fixed in a wall at A, is cut through at CD, and the severed piece, CB, is held in place by means of a strut at D and a link at C, which resist the compression and ten- sion due to the cross-strain arising from the weight P ; and by the weight R (equal to P ) suspended over a pulley E, FIG. 30. which prevents the severed beam from sinking, or resists the shearing strain. As the link C and strut D are both acting in a horizon- tal direction, they can have no effect in resisting a vertical strain, consequently the weight P must be entirely sustained by the counter-weight R, and as the action of the latter is directly opposite to that of the former, it must be equal to it in amount. In the above arrangement we may see that were the strut D removed, the beam CB, under the action of the weight Pj would revolve upon C as a centre, closing the gap at the bottom ; hence the strut D is compressed. In like manner, if the link at C were removed, the Il6 GRAPHICAL REPRESENTATIONS. CHAP. VIII. weight P would cause the beam to revolve on D, making wider the opening at the top, and showing that the link C is in tension. If the tension at C be represented by /, the compression at D by c, and the depth CD by d, then td = cd=Px CB Disregarding the weight of the beam, the shearing strain at CD equals the weight P. As this strain is wholly inde- pendent of the distance between C and B, the beam may be cut at any point in its length with a like result as to the amount of the shearing strain. At every point we shall have R = P, or the shearing strain equal to the weight. If the weight of the beam be included in the considera- tion, the shearing strain at any point will equal the weight P plus the weight of so much of the beam as extends beyond the point at which the shearing strain is considered. Let CD be the cross-section at which it is required to find the shearing strain ; let JT equal the distance from this cross-section to B, in feet ; and let e represent the weight per foot lineal of the beam ; then the weight of the piece CB will equal ex, and the shearing strain at CD will equal P+ ex y or the destructive energy is D == P + ex f73. Transverse and Shearing Strains Compared. Be- fore this formula can be available, it is needed to know the resistance of the different materials to this kind of force. Experiments have been made upon wrought-iron which show that its shearing resistance is about seventy-five per cent of its resistance to tension. If, in the absence of the ex- periments necessary to establish the resistance to shearing in materials generally, it be assumed that they bear the MEASURE OF SHEARING STRAIN. 117 same proportion to their tensile resistance as is found in wrought-iron, this shearing strength may be put equal to in which T equals the absolute resistance to tension per square inch of cross-section. The resistance of certain woods to tension may be found in Table XX. When D = R we have P+ex= \Tbd This gives bd, or the area of cross-section, equal only to the destructive energy. In this case rupture would ensue. We therefore introduce the factor of safety, a, and have a(P+cx)=\TV& (38} The portion of T considered safe is from one sixth to one ninth. We then have a 6 to a = 9. As an example: Suppose a semi-beam (as AB, Fig. 30) of white pine to be 10 feet long, and loaded at the end with P= 10,000 pounds ; what would be the required area of cross- section at the wall ? Here the weight of the beam is so small in comparison with the load P that it may be neglected in the computation. Throwing it out of the formula, we have (39.} Let a 9 and T = 12000 ; then loooo x 9 = {. x 1 2000 x bd 10000 x o -?-bdiQ Jxl2OOO Il8 GRAPHICAL REPRESENTATIONS. CHAP. VIII. To compare this requirement with that for the cross- strain, we make use of the formula for this strain, (19.), 4Pan = Bbd* and, making a = 4, have 4 x 10000 x 4 x 10 500 x bd a 4 x 10000 x 4x 10 . = W = = 3200 and, making d = 16, have b x i6 2 = 3200 therefore the area will be 12^ x 16 = 200 square inches. This is the area required at the wall, but at the end B, the point of attachment of the weight, we have seen (Fig. 27) that the destructive energy in cross-strain is zero. Were this the only effect produced by the weight P, the beam might be tapered here to a point. Owing, however, to the shearing effect of the weight, we find, as above, a requirement of material equal to 10 inches in area, or the beam 12^ inches wide would require to be eight tenths of an inch thick ; and the rope supporting the weight should be so attached as to have a bearing across the whole width of the piece. 174. Rule for Shearing Strain at Ends of Beams. The shearing strains at the two supports upon which a beam is laid are together equal to the weight of the beam and the load laid upon it. If the beam be of equal cross-section throughout its length, and the load upon the beam be located at the middle, or symmetrically about the middle, then the SIZE OF BEAM AT ENDS. IIQ weight of the beam and its load will be sustained half upon each support. In this case, the shearing strain at the two supports will be equal, and each equal to half the total load. Putting W for the load upon the beam, and el for the weight of the beam, then for the shearing strain at each end of the beam we have Putting this equal to the safe resistance [see formula (38, ,), Art. 173] we shall have (W + el) = \Tbd bd (40.) When the load is not at the middle nor symmetrically disposed about the middle, the portion borne upon each support may be found by formulas (3.) and (4>), Art. 27. The shearing strain at each support is equal to the reaction of the support or to the load it bears. 175. Resistance tio Side Pressure. Beyond the fore- going considerations, there is still another of some impor- tance. Care should be taken that the surfaces of contact of the wall and the beam are of sufficient area to be unyielding. Usually the wall composed of brick or stone is so firm that there need be no apprehension of its failure, and yet it is well to know that it is safe. It should, therefore, be carefully considered, to see that the given surface is sufficiently large for the given material to carry safely the weight proposed to be distributed over it. In calculations for heavy roof trusses this precaution is particularly necessary. The upper surface of the joint, or underside of the beam, 120 GRAPHICAL REPRESENTATIONS. CHAP. VIII. requires. especial attention. This is usually of timber, and parallel with the fibres of the material. The pressure upon the surface tends to compress these fibres more compact!) 7 together by closing the cells or pores which occur between the fibres. When pressed in this way, timber is much more easily crushed, as may readily be supposed, than when the pressure is applied at the ends of the fibres in a line parallel with their direction. The resistance to side pressure approaches the resist- ance to end pressure in proportion to the hardness of the material. By experiments made by the author some years since, to test the side resistance, results of which are recorded in the American House Carpenter, page 179, it appears that the hard- est woods, such as lignum-vitae and live oak, will resist about i times the pressure endwise that they will sidewise ; ash, if times ; St. Domingo mahogany, twice; Baywood mahog- any, oak, maple and hickory, about 3 times ; locust, black walnut, cherry and white oak, about 3^ times ; Georgia pine, Ohio pine and whitewood, about 4 times ; chestnut, 5 times ; spruce and white pine, 8 times ; and hemlock, 9 times. Their resistance to side pressure is in proportion to the solidity of the material, or inversely in proportion to the size of the pores of the wood. In the above classification, the comparison is not that of the absolute resistance of the several kinds of wood to side pressure. It is only a comparison of the results of the two pressures on the same wood. Whitewood, classed above with Georgia pine, resists sidewise only as much, absolutely, as white pine. Its power of resistance to end pressure is the lowest of any of the woods, being but one half that of white pine. The average effectual resistance to side pressure per square inch of surface, /, for BREADTH OF BEARING ON WALLS. 121 Spruce =250 pounds. White pine = 300 " Hemlock = 300 " Whitewood = 300 " Georgia pine 850 Oak = 950 Under these pressures only a slight impression is made, and the woods may be safely trusted with these respective amounts. 176. Bearing Surface of Beams upon Walls. The sur- face of , the beam in contact with the wall must be sufficient in extent to insure that it shall not be exposed to more pres- sure than is above shown to be safe. If b equal the breadth of the beam, h the length of the bearing surface, and p the resistance per inch, as above, then the total resistance equals R = bhp The destructive energy for one end of the beam is, as before (Art. 174), D = When there is equilibrium, then R = D, or l) = bhp Owing to the deflection of the beam by the load upon it, its extreme ends may be slightly raised from off the bearing surface, and in consequence the pressure be concentrated at the edge of the wall. No serious effect will ensue from this, for if the pressure be greater than the timber can resist at the edge, the fibres will be crushed there, but only suffi- ciently so to allow the surface of contact to extend towards 122 GRAPHICAL REPRESENTATIONS. CHAP. VIII. the end of the beam, until it is so enlarged as to effectually resist any further crushing. Beams which are likely to be depressed considerably should have their ends formed so that their under surface will coincide throughout with the wall surface when the greatest load shall have been put upon them. 177. Example to Find Bearing Surface. Let a white pine carriage beam 6 inches wide, 24 feet long between bearings, and weighing 15 pounds per lineal foot, be loaded with 12,000 pounds, equally distributed over its length. What should be the length of the bearing upon each wall ? By transposition, formula (41.) becomes W+el = k 2bp In this case, W 12,000, e 15, / = 24, b 6, and p = 300; then 12000 + 15x24 , ? = h = 3.43 2x6x300 or the end of the beam must extend upon the wall, say 3^ inches. The usual bearing for floor beams, which is 4 inches, would in this case be amply sufficient. Where the concentrated weight is so large in comparison with the weight of the beam, the latter Aveight may be neg- lected without any serious result ; for had we considered the 12,000 pounds only, in the above example, the value of h would have been 3.33, only a tenth of an inch shorter than the former result. 178. Shape of Side of Beam, Graphically Expresed. As will be observed, we have digressed from the principal subject. This became necessary in order to explain the apparently anomalous result of leaving the beam without any SHAPE OF SIDE OF LEVER. 123 support at the ends. For it was seen that in an application of the formula for cross-strains the requirement of material gradually lessened towards the ends of the beam, until at the very edge of the bearings it entirely disappeared. To prevent the beam, with its load, from falling as a dead weight between the bearings ; or, to provide against the shearing strain, as well as against the crushing of the material upon its bearings, we have turned aside so far as seemed to be needed. And before returning to the main subject, it may be well here to show that the line CB in Figs. 27 and 29, limit- ing the ordinates of cross-strain in the lever and beam, does not show, as might be supposed, the shape of the depth of a lever or beam having a cross-section of equal strength throughout its length. A short consideration of the relation between the strains at given points in the length will show the true shape. By construction, c, Fig. 27, is equal to %tP, and from this we have shown (Art. 168) that and when the destructive energy and the resistance are equal \lP^Sbd 2 and Px = Sbdf from which c\y\\ Sbd* : Sbdf and when S and b are constant c : y : : d 3 : df or, the ordinates are in proportion to the squares of the depths, and not directly as the depths themselves. From these ordinates, however, the shape of the side of the lever may be directly found by taking their square roots. For let AB in Fig. 3* be the upper edge of the lever, and 124 GRAPHICAL REPRESENTATIONS. CHAP. VIII. T t CB the line limiting the ordinates of cross strain. Then, if AD be made equal to the square root of AC, and, corresponding- ly, d t , d in d ilit etc., be each made respectively equal to the square root of the ordinate upon which it lies, and if a line be drawn through the ends of d n d :t , d llt , etc., this line, DEB, will limit the shape of the lever. This curve line is a semi-para- bola, with its vertex at B and its base vertical at AD. By con- struction, each ordinate y is in proportion to x, its dis- tance from B, or (since y equals d 2 ) d 2 is in proportion to x, a property of the parabola. Hence to obtain the shape of the lower edge of the lever, any method of describing a para- bola may be used, making AD, its base, equal to (form. 19.) FIG. 31. FIG. 32. Bb As a whole beam is in like condition with two semi-beams, as to the cross strains, there- fore the shape of a whole beam of equal strength throughout its length is that given by two semi-parabolas placed base to base, as in Fig. 32. QUESTIONS FOR PRACTICE. (79. In a semi-beam, or lever, 10 feet long, fixed in a wall, and loaded at the free end with 3672 pounds, what is the destructive energy at the wall ? 180. Make a graphic representation of the above by a horizontal scale of one foot to the inch, and a vertical scale of 1000 foot-pounds to the inch. What is the height CA of the triangle of cross-strains, in terms of the scale selected ? (81. Measuring horizontal distances from the free end, what are the lengths, by the scale, of the respective ordinates at the several distances of 5, 6, 7, 8 and 9 feet; and what the amount of cross-strain corresponding thereto at these several points in the beam ? 182. What will be the required depth at the wall, and at 9 and 8 feet respectively from the free end ; the lever being of Georgia pine, 6 inches broad, and the factor of safety 4? 183. In a white pine beam, 4 inches broad, 16 feet long between bearings, and loaded at the middle with 3250 pounds, what should be the respective depths at the several distances of 3, 5, 7 and 8 feet from one end, the factor of safety being 4? 184. A white pine semi-beam, 12 feet long and 4 inches broad, is loaded with 693 pounds at the free end, including the effect of the weight of the beam itself. The factor of safety is 4, the beam is of constant breadth and depth 126 GRAPHICAL REPRESENTATIONS. CHAP. VIII. throughout its length, and its weight is 30 pounds per cubic foot. What is its required depth at the wall ? What is the weight suspended from the end of the beam ? What is the shearing strain at the wall ? What is the shearing strain at 5 feet from the wall ? 185. A beam of Georgia pine, 4 inches broad and 20 feet long, is loaded at the middle with 9644! pounds. The beam is 17 inches high at the middle, and tapered in parabolic curves to each end. The material of the beam is estimated at 48 pounds per cubic foot. What is the weight of the beam? 186. What is the shearing strain at each wall ? With a factor of safety of 9, how high is the beam required to be at the ends to resist the shearing strain safely ? (87. How far upon each wall is the beam required to extend, in order to prevent crushing of the material ? CHAPTER IX. STRAINS REPRESENTED GRAPHICALLY. ART. 188. Graphic Method Extended to Other Cases. In Figs. 27, 28 and 29, with a given maximum strain upon a semi-beam, or upon a full beam, we have a ready method of finding the strain at any given point in the length. This simple method of ascertaining the strain at any point, graphically, is based upon a principle which is applic- able to strained beams under conditions other than those given, as will now be shown. 189. Application to Double Lever with Unequal Arms. In Figs. 28 and 29 the load upon the beam is at the middle. But it may be shown that the triangle of strains is applicable in cases where the load is not at the middle. Let R and P, Fig. 33, represent two unequal weights, FIG. 33. suspended from the ends of a balanced lever AB. From the law of the lever, we have (Art. 27) Rm Pn 123 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. If CD, called g, be made of a length to represent Pn, then will it also represent Rm ; for Rm = Pn. Hence, since the triangle BCD is the triangle of strains, in which an ordinate, y, showing the strain at any given point in DB, may be drawn, therefore the triangle ACD will give ordinates, y' , measuring the strains at the points in AD, from which they may be drawn ; or, since Pn : g : : Px : y >.* n so also Rm : g :: Rx' : / (4$-) f90. Application to Beam with Weight at Any Point. In Fig. 34, AB represents a beam supported at each end, carrying a load W at a point nearer to A than to B. This w FIG. 34. beam is strained in all respects like that in Fig. 33, except that the strains are in reversed order. Therefore an ordi- nate, y, drawn across the triangle BC W, will indicate the strain at the point of its location. So an ordinate, /, across the triangle ACW, will indicate the strain at its point of SCALE OF STRAINS WEIGHT AT ANY POINT. 1 29 location. Or, generally, the two triangles ACW and BCW limit the ordinates \vhich measure the strains at any point in the length of the beam. Thus when g =. Pn = Rm we have y Px and / = Rx' and since P= W and R = Wj (Art. 27) we have y = W ~, x (44-) y <=W U jx> (45) Now, since Rm Pn =g, equals the destructive energy of the weight at its location, therefore any ordinate across the triangles ACW and BCW equals, when measured by the same scale, the destructive energy at the location of that ordinate, and when the resistance is equal to the destructive energy we have for the strain at any point to the right of the weight Putting for 5 its equivalent \B (Arts. 35 and 57) to agree with the unit of dimensions, we have, for the safe weight, (46.) which, with x at its maximum equal to n, is identical with formula (23). For the safe weight at any point to the left of the weight we have 4Wajx'=zBbd* (47.) 191. Example. As an example in the application of these expressions, let it be required to find the strains at 130 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. various points in the length of a white pine beam, the maximum strain being given. Let the beam be 10 feet long and loaded with 2000 pounds at a point three feet from the left-hand end. What is the strain at the location of the weight? What are the several strains at 2, 4 and 6 feet from the right- hand end and at 2 feet from the left-hand end ? Take first the strains to the right. .m Here, by formula (44-), y = W jx, and with x at its maximum we have y 2000 x x 7 = 4200 In Fig. 35, make the length between the bearings A and B by any scale, equal to 10 feet, and CW, or g, equal to 42 units of any other scale. Then each of these units will \ s \ I >--"- FIG. 35. represent 100 pounds of strain. The number of units in the length of the ordinates, y, at the several distances, x equal to 2, 4 and 6 feet, and of x' 2 feet, will give, when multiplied by TOO, the strains at these several points. Thus it will be found that, DEPTH OF BEAM WEIGHT AT ANY POINT. 131 at 2 feet from B, y 12, and 12 x 100 1200; " 4 " " B, y = 24, " 24 x 100 2400 ; " 6 " " B, y 36, " 36 x loo = 3600 ; and " 2 " " A, y' 28, " 28 x 100 = 2800. Now, if it be required to find the proper depth of the beam at these several points, we take, for the right-hand end, formula in which W represents 2000 pounds, the weight upon the beam, and in which W-j-x will give the strain at each ordinate ; and by transposition have and if a = 4, B 500 and b = 3, we have 500 x 3 x 10 = 6 and therefore when x 2 then ^ 2 = 6-4x2=J2-8 and ^=3.58 4:^4 " d 2 = 6-4x4^25-6 " ^=5.06 ^r = 6 " ^' = 6.4x6 38-4 " ^=6.20 " x=n = ? " d* 6-4 x 7 =44-8 " ^=6-69 For the left-hand end we use formula (47-) 132 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. 4 X 2000 X 4 X 7 d 2 = -- ^x'iA.-^x 1 500 x 3 x 10 and hence, when x' = 2 then */' = 14-93 x 2 = 29-9 and </ 5-47 " y = i=3 " ;/*:=: 14-93 x 3 =44.8 " d = 6-6g This last result agrees with the last from the right-hand end, as it should, for they are both for the same location. The above results are all obtained by computations, but the value of d*, at as many points as may be desired, can be obtained by scale, in a similar way with the ordinates for the destructive energy ; but this scale, for the purpose of obtain- ing the depths, must be made with the principal ordinate, g t equal to the requirement (see form. #$.), and then the square root of each ordinate drawn across the scale will be the required depth at its location. For example : Make g, by any convenient scale, equal to 44-8 as above required ; then the several values of d* at 2, 4 and 6 feet may be found by measuring the ordinates drawn at these several distances from B. The square root of each ordinate will equal the depth of the beam there. The results obtained by measurements, although not exact to the last decimal, are yet sufficiently exact for all practical purposes. If it be required to find the exact dimension, this may be done by computation, as shown, and the diagram will then serve the very useful purpose of checking the result against any serious error in the calcula- tion. MEASURE OF STRAIN FROM TWO WEIGHTS. 133 192. Graphical Strains toy Two Weiglit. The value of graphic representations is manifest where two or more weights are carried at as many points upon a beam. In Fig, 36 we have a beam carrying two weights A' and B r . The destructive energy of the weight A', at its location, is equal to (Art. 56) and the destructive energy of the weight B' ', at its location, is equal to D" = B' ~ mn Make AE equal to A'-j- by any convenient scale. By the TS same scale make BF equal to B'-j-. Draw the lines CE and DE, CF and DF. Now, while AE represents the effect of the weight A 1 at the point A, so also AG measures (A rt. 190) the effect, at the same point, of the weight B' ; therefore make EJ equal to AG, then AJ is the total effect at A of both weights. 134 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. In like manner (FK being made equal to BH), BK measures the total effect at B. Draw the line CJKD. and by dropping a vertical ordinate from any point in the beam CD to this line, we have the total strain in the beam at that point. 193. Demonstration. The above may be proved, as follows : First. Let the ordinate occur between the two weights as LM, Fig. 37. Extend the lines CF, DE and JK, till they meet at R and 5, and draw CR and DS. FIG. 37. Now the effect of B' at B, is measured by BF, and at L by LP (Art.\B9). Also the effect of A' at A, is measured by AE, and at L by LN. The joint effect of A' and B' at L, is thus LP+LN, and if it can be shown that PM equals LN, then LP+LN=LP+PM = LM equals the joint effect of the two weights A' and ', at L. In two triangles of equal base and altitude, two lines drawn parallel to the respective bases, and at equal alti- tudes, are equal; from which, conversely, if two triangles of equal base have equal lines drawn parallel to the base, and SCALE OF STRAINS DEMONSTRATION. 135 at equal altitudes, then the altitudes of the two triangles are equal. In the present case we have AE = GJ\ for A G EJ by construction ; and if, to each of these equals we add the common quantity GE, the sums will be equal, or AE= GJ The two triangles ADE and GSJ are therefore standing upon equal bases, AE and GJ. Moreover, at equal distances, AB, from the line of bases AJ, and parallel with it, we have the two lines BH and FK, made equal by construction. Consequently, the two triangles have equal altitudes. Hence all lines drawn across them, parallel with and at equal distances from the base, are equal, and therefore LN and PM, having these properties, are equal, and LM=LP+LN equals the true measure of the strain induced at L by the weights A' and B' ; or, in general, any vertical ordinate drawn across AJKB will measure the total strain caused by the two weights at the location of the ordinate. Damonsf ration Rule for the Varying Depths. Second. Let the ordinate occur at one end, between B and D, as OQ, Fig. 37. Here we have OT for the strain caused by A', and O V for the strain caused by B' ; or the total strain equals OT+OV. Now if VQ can be proved equal to OT, we shall have equal to the total strain at O. We have the two triangles BDH and FDK, with bases 136 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. W BH and FK, made equal by construction, and with equal altitudes BD, and we have the two lines OT and VQ drawn parallel with, and at equal altitudes ( BO ) from the base; consequently OT and VQ are equal, and OQ meas- ures the total strain of the two weights at O\ or, in gene- ral, any vertical ordinate drawn across BDK will measure the total strain at the location of the ordinate. Since it may be shown in like manner that any vertical ordinate drawn across ACJ will measure the total strain at its location, therefore we conclude that a vertical ordinate from any point in the beam CD to the line CJKD will show the total strain in the beam at that point. In practice, the scale of strains CJKD may be con- structed as just shown, in detail, but more directly by obtaining the points J and K in the following manner : We have for the joint effect of the two weights at the location of one of them, A, (see Art. 153) which becomes, on changing W and V into A 1 and B f , **'"\'+ffs) (51.) equals the length ol the ordinate AJ. TWO WEIGHTS STRAINS AND DEPTHS. 137 In like manner we have (52.) for the length of the line BK. The points J and K are to be obtained by these expres- sions. The scale is then completed by connecting these points and the ends of the beam by the line CJKD. The strain at any point in the beam may then be readily meas- ured, sufficiently near for all practical purposes. If, however, the exact strain is desired, this may be obtained as follows : Putting g for AJ, p for BK, and // for AB, we have for the several ordinates s : p :: x : y y=tx (53.) S m : g : : x' : y r h p-g ::.*": y" -g h(y"-g) = x(p-g) hy"-hg = x"(p-g) hy" = x"(pg) + hg If it be required to know the depth of the beam at every point, to accord with the strain there, then, instead of mak- ing the two principal ordinates as above shown, find their lengths thus : 138 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. By formulas (30.) and (51.) make AJ equal to m d* = (56.) Bb and by formulas (31) and (52.) make BK equal to 4a-(B'r + Am) (57.) Draw the line CJKD, and then an ordinate drawn across this scale at any point will give the square of the depth at that point. The square root of this length will be the required depth there. 195. Graphical Strains by Three Weights. In Fig. 38 we have a graphical representation of the strains resulting from three weights. FIG. 38. This figure is constructed by making AJ equal to the moment of A' at A, BK equal to the moment of B' at B, and CL equal to the moment of C' at C, all by the same MEASURE OF STRAINS FROM THREE WEIGHTS. 139 scale. Connect J, K and L each with the ends of the beam E and D. Make JF equal to AM + AN, KG equal to BO + BP, and LH equal to CQ + CR. Join E, F, G, H and D, and this line will be the boun- dary of any vertical ordinate from any point in ED, which, by the same scale as used for AJ, etc., will measure the strain at the location of the ordinate. In this diagram, the points F, G and H may be found directly, as follows : To find F, we have (Art. 153) A'~ for the effect of A', B'J- for B f , and so, in like manner, we may have C' ~j- for that of C'. Added together, these will equal AF = (A'n + B's + C'v ) (58.) To find G t we have A'^ for A', B'-- for B f , and C-r for C ; which together give n ^ A'ms + B'rs + C'rv >(JT = To find H, we have A f ~ for A', B'~ for B f , and C~ for C ; which added, will equal CH -(A'm + B f r - If it be desirable, the strains may, as in the last figure, be computed ; for putting g for AF, p for BG, k for CH, h for AB, and q for BC, we have, for an ordinate between C and D, 140 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. FIG. 38. v : k : : x : y 9*& ^ For an ordinate between E and A we have m : g : : x' : y' m For an ordinate between A and B we have, as in Fig. 37, y" = h (63) and for ordinates occurring between B and C we have y = tJL x "> + k (64.) These expressions give the strains at any point, due to the three weights. In like manner, we may find the strain at any point in a beam, arising from any number of weights. To obtain the squares of the depths at various points by scale, make AF equal to THREE EQUAL WEIGHTS SYMMETRICALLY DISPOSED. 141 Make BG equal to A'ms'+ B'rs + Crv Make CH equal to 4a^(A'm + B'r + Ct) / fay \ d 2 = : ( 67 -) Bb The square roots of ordinates upon this scale will give the depths required at their several locations. 196. Graphical Strains by Three Equal Weights Equa- bly Disposed. Let us now consider the effect of equal weights, equably disposed. In Fig. 39 we have three equal weights, L, placed at equal distances apart upon a beam, ED, the distance from either wall to its nearest weight being one half that between any two of the weights ; or, EA - CD = The line EFGHD is obtained as directed for Fig. 38. It may also be obtained analytically, thus : First. The line AF, or the effect at A of the three weights, equals the sum of the three lines AJ, AO and AN. 142 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. FIG. 39. Let EA = CD = t, and AD h, then t + h = /, and (Art. 56) tx/t th as per Art. 195. CDxEA _ fr//x* t /A -L--J-- -L t - -^L t or th . th Second. The line BG, or the effect at B of the three weights, is equal to the sum of the line BK and twice the line BQ. Let EB = /, BD h, and / + // = /; then T BK = Z-- th and FOUR EQUAL WEIGHTS SYMMETRICALLY DISPOSED. 143 Third. The effect at C produced by the three weights is equal to that at A. We have, then, for the total effect at A, AF = B, tt tt th I th 197. Graphical Strains by Four Equal Weights Equably Disposed. When there are four equal weights, as in Fig. 40, similarly disposed as in Fig. 39, the effect at A is, \ 'M -.-v--- .. I FIG. 40. from load at A, hxt L ~ ~~ ht C, " 'JD, ht ~T ht 144 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. or the total effect at A, of the four weights, is The effect at B is, from load at A, L^-j^ = \L-r C, or the total effect at Z?, of the four weights, is The effect at C is equal to that at B, and the effect at D is equal to that at A. 198. Graphical Strains by Five Equal Weights Equably Disposed. When there are five equal weights, as in Fig. 41, similarly disposed as those in Fig. 39, the effect at A is, T hxt T ht from load at A, Lj- %L-j B, ^hxt-^lL 11 C, |A:X-/ 7 .= |y " M, FIVE EQUAL WEIGHTS SYMMETRICALLY DISPOSED. 145 L L L L L ABC M or the total effect at A, of all the weights, is The total effect at B is, from load at A, c. II tt " M, ht -j- ht - l - y -f or the total effect at B, of all the weights, is 146 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. The total effect at C is, from load at A, ty x h- \L^ / / , = or the total effect at C, of all the weights, is 7 The effects produced at D and J/ are, respectively, like those at B and A. 199. General Result* from Equal Weights Equably Dis- posed. In looking over the results here obtained, it will be seen that in each case the effect is equal to gLr, in which g is put for the numerical coefficient, L for any one of the equal weights with which the beam is loaded, / and h the respective distances from the point at which the strain is being measured to the ends of the beam, and / for the length of the beam. All of these are simple quantities except the coefficient g, and this it will be shown is subject to a certain law and may be stated in general terms. TOTAL STRAIN AT LOCATION OF FIRST WEIGHT. 147 2 00 a General Expression for Full Strain at First Weight. The coefficient g is a fraction, having its numerator and denominator both dependent upon the number of weights upon the beam. Let us first consider the value of the numerator in measuring the effect of the weights at A, the location of the first weight from the left. With three weights, g, the coefficient, was | + f + | = f, the numerators being 1 + 3 + 5=9. With four weights, g was equal to - = , the numerators being 1+3 + 5 + 7=16. I I ^ I I 7 I Q 25 With five weights, g was equal to - = , and the numerators 1+3 + 5 + 7 + 9 = 25. In general, we shall find that the numerator of the frac- tion g, is in all cases equal to the sum of an arithmetical progression comprising the odd numbers i, 3, 5, etc., to n terms ; ;/ being put to represent the number of weights upon the beam, the first term being unity, and the last being 2n\. To find the sum of this progression, we have in which S = the sum, a the first term, / = the last term, and n = the number of terms ; or _ I + (2n i)n n + 2n 2 n O > r^ n: ft' Hence, the numerator of the coefficient of the expression showing the effect of any number of weights at the location, A y of the first weight, is equal to the square of the number of weights ; thus, when there are 148 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. 2 weights, n = 2, and the numerator = 2 2 = 4 3 " = 3, " " = 3 2 - 9 4 = 4 " " = 4 2 = 16 5 " =5. " " = 5 2 = 25 6 = 6, " " = 6 2 = 36 and so for any number of weights. In considering the value of the denominator of g it will be observed that it is derived by taking the value of h in each case in terms of /. With three weights, // = 5^ ; with four weights, // = 7/ ; and with five weights, // = qt ; so that in general, h-=-(2n\)t. The denominator of the fraction generally, therefore, is 201. n a The value of the coefficient is, consequently, , and the full effect at A of any number of equal weights equably ' _ /// disposed upon a beam is - L .- . 2n\ I 201. General Expression for Full Strain at Second Weight For the effect at the location B we have the ex- pression pL.-_ ; in which the same quantities occur as before, except in the case of the coefficient /. This coefficient is composed of two classes of fractions. The first of these is based upon the relation between the dis- tances EA and EB, and since EA is in all cases equal to -J- of EB, therefore this part of the coefficient / will be equal to i In the second fraction of the coefficient, the numerator is, as in the case at A, equal to the sum of an arithmetical pro- gression, but extending one less in the number of the terms, so that in place of n s we put (n i) 2 . The denominator is found by taking ;/ i for , or 2(01)!, equal to 203, for 201. The value of TOTAL STRAIN AT LOCATION OF SECOND WEIGHT. 149 this fraction is therefore - -~ . To this, adding the first fraction, we have 2/2-3 and for the full effect at B, of all the weights, ( t+ -<!=%* V z 3/ / From the above, the value of the coefficient / is as follows (2 if with 2 weights, / = i + ^- x2) _ 3 =i + -f =| ' 3 " , = t + =- = i + i =V " 5 The numerators of these results are in the order of 2n, $n, Sn, nn and 14;?; the numerals differing by 3. The de- nominators are the products of i, 3, 5, 7 and 9, each by 3. We may continue therefore the values to any number of weights by following these laws, thus f i 17 X 7 IIQ for 7 weights, / = for 8 weights, / = HX3 33 20 x 8 __ 160 i3><3"~ ~39~ or, in general, the effect at B for any number of weights may be had directly from the previous expression. 150 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. 202. General Expression for Full Strain at Any Weight. For the sum of effects at C, it is seen that we have kL-r-, and it can be shown that the coefficient k is the sum i n 2\ a of two fractions namely, f and - _ or For the effect at D we have For the effect at E we have or, putting them in sequence, we have (n-o}' at A the effect g= f C " " k = D " " u= E , ., n - GENERAL EXPRESSION TOTAL STRAIN AT ANY WEIGHT. 15 1 and so for any number of weights upon one end of the beam. An examination of this series shows that in the first of the two fractions the numerator is equal to the square of the number of weights preceding the one under consideration ; for instance, at A, where there are no weights preceding, we have the numerator o ; at B there is one weight preced- ing, and hence the numerator is i 2 equals i ; at C there are two weights preceding, hence the numerator equals 2 2 equals 4 ; at D there are three weights, hence the numerator equals 3 2 equals 9 ; etc. For the denominator of the first fraction we have, for the several cases in consecutive order, the values J > 3> 5> 7> e tc. ; an arithmetical series of the odd numbers. In the second fraction we have a numerator equal to the square of the difference between n and the number of weights preceding the one at which the strain is being measured ; and a denominator of 2n minus the denominator of the first fraction. Let r represent in any case the number of weights pre- ceding the one at the location of which we wish to know the strain. Then we shall have, as the coefficient of the effect at that point, r* (n-r) a and for the full effect, or the destructive energy, D = L--( ^ + <*--. \ (68.) I \ 2r+ i 2n (2r+i) / in which L represents one of the equal weights with which the beam is loaded ; // the distance from the weight at which the strain in the beam is being measured to the right-hand end of the beam ; t the distance from the same point to the left-hand end ; / = h + / the length of the beam between sup- 152 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. ports ; n the number of equal weights equally disposed upon the beam, as in Fig. 41 ; and r the number of weights between the point where the strain is measured and the left-hand end of the beam, not including the one at the point where the strain is measured. 203. Example. What is the strain at the fifth weight from the left-hand end of a beam 22 feet long, loaded with 1 1 weights of 100 pounds each ; the weights placed at equal distances from centres, and the distance from each end of the beam to the centre of the nearest weight being equal to half the distance between the centres of any two adjoining weights? Here the distance between centres of weights will be 2 feet, t will equal 9 feet, and h will equal 13 feet, L = 100, n n, and r 4. From these the strain at the fifth weight will be (form. 68.) D = ioox- + -- = 2950 22 V 8+1 22 (8+ 1) QUESTIONS FOR PRACTICE. 204-. A beam 12 feet long is loaded at 4 feet from the left-hand end with 4000 pounds. What is the strain at that point ? 205. What are the strains, respectively, at 2, 4, and 6 feet from the right-hand end ? 206. A beam 14 feet long is loaded with two weights; one, A', weighing 3000 pounds, is located at 4 feet from the left-hand end ; the other, B' , weighing 5000 pounds, is at 6 feet from the right-hand end. What strain is caused by these two weights at the point A ? What strain is caused at Bl 207. In the above beam what strain is caused by the two weights at a point 2 feet from the left-hand end ? What strain is caused at a point 2 feet from the right- hand end? What strain is produced at the middle of the beam ? 208. Abeam 20 feet long is loaded with three weights; one, A', of 3000 pounds, at 3 feet from the left-hand end; one, B' 9 of 2000 pounds, at 1 1 feet from the same end ; and the third weight, C' , of 4000 pounds, at 4 feet from the right-hand end. 154 STRAINS REPRESENTED GRAPHICALLY. CHAP. IX. What is the full effect of the three weights at the location of each weight, at 2 feet from the left-hand end, at 2 feet from the right-hand end, at 6 feet from the same end, and at the middle of the beam ? 209. Abeam 16 feet long is loaded with 20 weights of zoo pounds each, the weights being equally distributed. What strain do these weights produce in the beam at the ninth weight from one end ? CHAPTER X. STRAINS FROM UNIFORMLY DISTRIBUTED LOADS. ART. 210. Extinction Between a Series of Concentrated Weights and a Thoroughly Distributed ILoad. The distribu- tion of the load upon a beam, as shown in Figs. 39, 40 and 41, is essentially that of a uniform distribution over the entire length of the beam. For if the beam be divided into as many parts as there are weights, by vertical lines located midway between each two weights, it is seen that the parts into which these lines divide the beam are all equal one with another, and the weight upon each part is located 'in a vertical line passing through the centre of gravity of that part. Hence this beam, taken with the loads upon it, is an apparently parallel case with a beam having an equally distributed load. An application of formula (68.), however, will show that the case is that of a beam loaded with a series of concentrated weights, and not with a thoroughly distributed load, although it closely approximates the latter. We find that the results of computations made with this formula differ according to the number of weights upon the beam, but approach a cer- tain limit as the number of weights is increased ; a limit which is that of a beam with an equally distributed load. 211. Demonstration. For example, let us find by for- mula (68.) the effects at the middle of the beam under differing numbers of weights. 1 56 STRAINS FROM UNIFORMLY DISTRIBUTED LOADS. CHAP. X. We may modify the formula to suit this case, for Lxn = U, when U equals the total weight upon the beam, or L = , and h = t = \l. By substituting these values, we have 2r+ (69.) To apply this modified formula to the question : First. Let there be five weights equally disposed, or n = 5 ; then r 2, and we have Second. Let there be nine weights or n 9, then r 4, and we have *=~ If = 25, then r 12, and Fourth. \i n 101, then r = 50, and + W) = SERIES OF CONCENTRATED LOADS. 157 Comparing the coefficients of these several results, we have when n 5, the coefficient = |~f =J-f T V " = 9> " " " = 25, " " = 101, " " = -AWr = 4 + The result in all cases is equal to a half, plus a fraction which decreases as n increases, or which has unity for its numerator, and a denominator equal to twice the square of n. The coefficient may be expressed then by \ + Now, when the number of weights is unlimited, or the load thoroughly and equally distributed over the whole length, then n is infinite, and the denominator of the last fraction becomes infinity. In this case, the fraction itself equals zero and consequently vanishes. Hence the coefficient tends towards , and with the loads subdivided to the last degree, and infinite in number, actual- ly becomes \ ; for, with these conditions fulfilled the case is actually that of an equally distributed load, and then x = \U- = i*7/. (See Art. 59.) This value of the coefficient may be concisely derived by the use of the calculus, as will now be shown. 212. Demonstration by the Calculus. To obtain a for- mula to represent the strain caused at any point by an equally distributed load, let RPTS, Fig. 42, represent graphically an equally distributed load, SR being equal to TP, and let it be required to find the ordinate EF, equal to the effect at any point E, caused by the whole load. 158 STRAINS FROM UNIFORMLY DISTRIBUTED LOADS. CHAP. X. F D '' C C B P A I FIG. 42. To do this we may proceed as follows : Let the ordinate AG represent by scale the strain caused at A by a small weight A', concentrated at A. Then will EJ represent (Art. 190) the effect of A' at E. Again, let the ordinate BH represent by scale the strain at B caused by a small weight B' , concentrated at B. Then will EK represent the effect of B' at E. The sum of these, EJ+EK, will equal the joint effect of the weights A' and B' at E. Or (Art. 190) A'hx B'tx, U ~ 'I ' I Let the loads A' and B' be very small ; equal to a small portion of the equally distributed load SRPT, and repre- sented graphically by the thin vertical slices at A and B respectively, and let these slices be reduced to the smallest possible thickness. By the rules of the calculus we may represent the thickness of the slices, when infinitely reduced, by dx, the differential of x, or rate of increase. If e be put to represent the weight per lineal foot of the equally dis- tributed load SRPT, then edx will represent the weight of the thin slice at A, or equal A'. So also edx t will represent the weight of the slice at B, or equal B'. STRAINS COMPUTED BY THE CALCULUS. 159 Substituting these values for A' and B' in the above expression, we obtain ~ ehxdx etx.dx. e , , , , D - + '- '- = j (hxdx + txpX This is the effect at E of the two loads at A and B, but these loads are infinitesimally small, therefore the expression is to be considered merely as the sum of the differential, or rates of increase of the strains produced by the two parts into which the whole of the equally distributed load RSTP is divided by the ordinate EF. The strain itself is to be had by the integral which is to be derived from the above differential of the strain. Therefore, by integration, we have (Arts. 4-62 and 463) ^ ( hxdx + tx t dx t ) = 4 (\hx* + \tx?} y By integrating between x o and x = /, also between x t =. o and x t /i, or making the integral definite, we have but h = I - t therefore ht = (I t) t and h 2 = (l-t) a therefore = (l-t)t+(l-t} s = ltt*+l*2lt+t* = I s -It l6o STRAINS FROM UNIFORMLY DISTRIBUTED LOADS. CHAP. X. and the formula et y = j (ht + k s ) becomes *=Ti (l '- K > y = \et(l-f) (70.) This result gives the value of the ordinate y, drawn at any point, and is comparable with the formula for the para- bola*, in which / equals the base, and the maximum ordi- nate, y, equals the height. Therefore, if the curve line RFDP be that of the parabola, it will limit all the ordi- nates, y, which may be drawn from the line RP. In the above discussion e was put for the weight of one foot lineal of the load, therefore the whole load U equals el, or e = . If in formula (70.) we substitute for e this value of it, we have and when h t \l we have, for the ordinate at its maximum or at the centre, y = (72.} * For here we have an ordinate to the curve from any point in the base, which is in proportion to the rectangle [t x (/ /)] of the two parts into which the base is divided by that point, a property of the parabola. (See Cape's Mathe- matics, 1850, Vol. II., p. 48.) COMPARISON OF RESULTS. l6l We thus see that the true value of the coefficient discussed in Art. 211 is equal to one half. This result (67) is the effect at the middle of the beam, and shows that an equally distributed load will need to be twice the weight of a concentrated load to produce like effects upon any given beam ; a like result with that which was obtained in another way at Art. 59. 213. Distinction Shown by Scales of Strains. By the calculus, the coefficient, as has just been shown, is equal to|, but those by formula (69.) exceed i by a certain fraction (Art. 211). A comparison of the scales of strains in 'Figs. 41 and 42 will show that the line limiting the ordinates is not a para- bola, but a polygonal line. In proportion to the increase in the number of the weights, and their consequent diminution in size and distance apart, this polygonal figure approximates the parabolic curve ; and in like proportion do the corre- sponding coefficients approach the coefficient obtained by the calculus; until finally, when the number of the weights becomes infinite, or the load is absolutely an equably distrib- uted one, then the coefficients are identical. The difference between the two expressions is that which is shown between the areas of the polygonal and parabolic figures. 214. Effect at Any Point by an Equally Distributed Load. One other lesson may be learned from this discus- sion. It has been shown (Arts. 59 and 61) that the effect at the middle of the beam, from an equably distributed weight, is, equal to that which would be produced by just one half of the weight if concentrated there ; and now we see (Arts. 2(1 and 212) that this proportion holds good, not only at the middle of the beam, but also at any point in its length. 162 STRAINS FROM UNIFORMLY DISTRIBUTED LOADS. CHAP. X. The expression (71.) just obtained, gives the effect produced by an equally distributed load at any point in the beam. It was shown (Art. 56) that the effect at any point of a load concentrated at that point, is equal to W - w ht I ~l Now when the effects in the two cases are equal, we have or, 4*7 = W ,i showing that when the effects at any point are equal, the concentrated load is equal to just half of the uniformly distributed load. 215. Shape of Side of Beam for an Equably Distributed Load. We have seen (form. 71.) that the effect at any point in a beam from an equably distributed load is and that the curve drawn through the ends of a series of ordinates obtained by this formula is a parabola (Art. 212, foot note). From this may easily be derived the form of the depth of a beam (the breadth being constant), which shall be equally strong throughout its length to bear safely an equably dis- ECONOMIC FORM OF BEAM. 163 tributed load. The formula (71.) gives the strain at any point, and when put equal to the resistance (Art. 35) is - Sbd* r> Substituting for 5 its value we have for the safe weight (Art. 73) f , . , 2UaAt from which a* ^^ This gives the square of the depth at any point, and when h t = l we have equals the square of the depth at the middle. Now make CD, Fig. 43, equal by formula (73.} to d* equals -r, and through D draw the parabolic curve RDFP, Across the figure draw a series of ordinates, as CD and EF. Then any one of these ordinates is equal to d* or the square of the required depth of the beam at the location of that ordinate. To find d, the depth, at each of these points, we have but to make CG equal to the square root of CD, and EH equal to the square root of EF, and in like manner find corresponding points to G and H on each ordinate, and draw the curve line RGHP through these points ; then this curve line will define the top edge of a beam (RP being the bottom edge), which shall be equally strong at all points to bear safely the equably distributed load. 164 STRAINS FROM UNIFORMLY DISTRIBUTED LOADS. CHAP. X. I T P f FIG. 43. 216. The Form of Side of Beam a Semi-el lipe. The form of the top edge of the beam as obtained in the last article is elliptical, as may be shown thus : The equation to the ellipse, the co-ordinates taken as in Fig. 43, is* 12 U 2 = - (2(IX X 2 ) in which x (= RE, Fig, 43) is the abscissa, u (= EH) is its ordinate, a(=RC=%f) is the semi-transverse diameter, and b ( CG \ / ~CJ}) is the semi-conjugate diameter: therefore If = CG* = CD and,' by formula (72.\ in which CD, the height of the parabola at the middle in Figs. 42 and 43, is represented by y, at its maximum we have y = \Ul. In the above value of u* substituting for a, and b, their values as here shown, we have and since Ix x' = x (I x) = th of Fig. 42, therefore By referring to formula (71.) it will be seen that this value of u a is identical with that given for y, the ordinate to the * Cape's Mathematics, Vol. II., p. 21, putting ;/ for;-. FORM OF BEAM AN ELLIPSE. I6 5 parabola, consequently y u*, and therefore the curve RGHP is elliptical. To obtain the shape of the beam, instead of drawing a series of ordinates in a parabola, and taking the square root of each ordinate, we may at once draw the semi-ellipse RGHP. Formula (73.) gives the value of d* at middle, therefore for d at middle make CG, Fig. 44, equal to -\/'~ Ual " V ~ (U) and through RGP draw a semi-ellipse, then RGPCR will be the shape of the beam. 7? FIG. 44. As an example : With a beam of white pine 10 feet long, 5 inches broad, and loaded with 10,000 pounds equably dis- tributed, and with a factor of safety a = 4, what should be the height at the middle? Formula (74-) becomes 10000 x 4 x 10 2 X 500 X 5 = 8-94 or the height of the beam is to be 9 inches, and the form of the side is to be that of a semi-ellipse, with 10 feet for its transverse diameter, and 9 inches for its semi-conjugate diameter. 166 STRAINS FROM UNIFORMLY DISTRIBUTED LOADS. CHAP. X. QUESTIONS FOR PRACTICE. 2(7. In a scale of strains for an equally distributed load, what curve forms the upper edge ? 218. In a beam, 10 feet long, having 1000 pounds equably distributed over its length, what are the strains at 2, 3, and 4 feet respectively, from one end ? 219. What should be the depth at the middle of this beam, if it be of white pine, if the breadth be made equal to fff of the depth, and if 4 be the value of the factor of safety ? 220. In order that the beam be of equal strength throughout its length, of what form should the upper edge be when the lower edge is straight, and the beam of parallel breadth throughout ? CHAPTER XL STRAINS IN LEVERS, GRAPHICALLY EXPRESSED. ART. 22 1 Scale of Strains for Promiscuously Loaded I^ever. In Fig. 45 we have a semi-beam loaded promiscu- ously with the concentrated weights A, B, C and D. FIG. 45. To construct a scale of strains for this case, make EF, by any convenient scale, equal to the product of the weight A into the distance EK\ make FG equal to BxEU\ make GH equal to CxEV', and HJ equal to DxET. From each weight erect a perpendicular, join K and F y L and G, M and //, and N and J\ then any vertical ordinate, as QP or ^5, drawn from the line EK to the line JNMLK, will, when measured by the same scale as that with which the points F, G, H and J were obtained, give, at the loca- tion of the ordinate, the effect produced by the four weights. 168 STRAINS IN LEVERS, GRAPHICALLY EXPRESSED. CHAP. XI. In the construction of this figure, each triangle of strains is made upon the principle shown in Art. 168, and the several triangles are successively added. An ordinate crossing all these triangles must necessarily be equal to the sum of the strains at its location caused by all the weights. The strain at any ordinate may also be found arithmeti- cally, by taking the sum of the products of each weight into its horizontal distance to the ordinate, measured from the weight towards the wall ; those weights which occur between the ordinate and the wall not being considered, as they add nothing to the strain at the ordinate. 222. Strains and Sizes of Lever Uniformly Loaded. When the weights are equably distributed over a semi-beam, the equation to the curve CFA, Fig. 46, limiting the ordi- II FIG. 46. nates of strains, may be found by the use of the calculus, as in Art. 212 ; for if ABHJ be taken to represent the equa- bly distributed load, then in considering the effect at the wall of a very thin slice of this load, as EG (reducing it infinitely) we obtain the differential of the strain. Let AE=y, then dy, its differential, may be taken as the thickness of the thin slice of the load at EG, when reduced to its smallest possible limits. Putting e for the weight of a lineal foot of the load, then edy will equal the weight of the thin slice. The effect or moment of this slice STRAINS IN LEVER COMPUTED BY CALCULUS. 169 at the wall, equals its weight into its distance from the wall, therefore we have for the differential of the moment edy x (ny) = du or, endyeydy = du The integral of this expression is (Arts. 462 and 4-63) / (endyeydy) = eny^ey* = u Applying this, or integrating between y equals zero and ^ equals n, we have ert%evt = \en* BC = u or for the strain at the wall, BC, u = \en* (75.) and for the strain at any point, E, x = \ey> (76.) From this latter, by transposing, we have which is the equation to the parabola * a proof that the curve CFA is that of a semi-parabola, in which A is the apex, and CD the base. These considerations pertain to the scale for strains. A scale for depths may be had by proceeding as follows : The value of e in formulas (75.) and (76.) is, from U = en (in which U equals the whole load upon the semi-beam) * For, putting - = /, then y 2 = -x becomes y 2 = 2/je, the equation to the parabola. See Cape's Mathematics, Vol. II., p. 47. STRAINS IN LEVERS, GRAPHICALLY EXPRESSED. CHAP. XI. e = . Substituting this value for e in formula (75.) we n have u = %--n 2 n Putting this equal to the resistance (Art. 35) gives us = Sbd* and substituting for 5 its equivalent \B, and inserting the symbol for safety (Art. 73), we have 4*4 / =5 Bbd* or, 2Uan = Bbd 2 [which agrees with formula (#0.)] for the size of the semi- beam at the wall. Again, subjecting formula (76.) to like changes, we have for the size of the semi-beam at any point 2U-y* = Bbd 2 (77.) in which y is the distance of that point from the free end of the semi-beam. 223. The Form of Side of L,ever a Triangle. If a semi- beam, subjected to an equally distributed load, be of rect- angular section throughout, and of constant breadth, then, in order that it may be equally strong at all points of its length, the form of its side must be a triangle. This may be shown as follows : Formula (77.) gives by transposition in which the coefficient -^r~ , for the case above cited, is FORM OF LEVER FOR EQUABLY DISTRIBUTED LOAD. I/I composed of constant factors ; hence d* will vary as y*, and therefore d will be in proportion to y. From this, formula (78.) is shown to be the equation to a straight line, and in such form that when y equals zero, d also becomes zero. From this, the side elevation of the semi-beam must be a tri- angle, with the depth at the wall (for then y becomes equal to n ) equal [from formula (78.) or (00.)] to d 2Uan (79.) As an example, let it be required to define the depth of a semi-beam of white pine, 10 feet long and 5 inches broad, carrying 5000 pounds equably distributed along its length, and with a factor of safety, a, equal to 4. Formula (79.) becomes d = 4/2 X 5000X4 X 10 y 500 x 5 This is the depth at the wall, as at AC, Fig. 47, in which AB is the length of the semi-beam. By joining B and C we have ABC for the shape of the side of the required semi-beam. FIG. 47. 224-. Combination of Conditions The forms of strain scales for loads under various simple conditions having been denned, we may now consider those arising from combina- tions of conditions. 225. Strains and I>imenioiis for Compound Load. Take the case of a semi-beam or lever, carrying an equably distributed load, and also a concentrated load at the free end. J/2 STRAINS IN LEVERS, GRAPHICALLY EXPRESSED. CHAP. XI Let the line AB, Fig. 48, represent the length of the FIG. 48. lever, R a weight suspended from its free end, and DC the face of the wall into which the lever is secured. In formula (75.) we have the strain at the wall, in which e equals the weight per lineal foot of the load, or e . Substituting this value in the formula, we have 21 = \Un as the strain at the wall; therefore make AD = %l7n, and by the same scale make AC RxAB = Rn. Join B and C, and describe a semi-parabola from B to D Avith the apex at B, and the base extended from D parallel with AB ; then any vertical ordinate drawn from the curve DB to the straight line CB will measure the strain at the point of intersection with the line AB. The scale here given is that for strains ; the scale for depths will now be shown. We have seen in Art. 223 that the form of the side of a lever required by a uniformly distributed load is that of a triangle, the vertical base of which is determined by formula (7#.) ; and it is shown at Art. 178, that the form, for a load concentrated at the end of a lever, is a semi-parabola, with its apex at the free end of the lever, and its base vertical at the fixed end and equal to FORM OF LEVER FOR COMPOUND LOAD. 173 Therefore let AB, Fig. 49, be the length of the lever FIG. 49. secured at A in the wall DC, and having suspended from its free end, B, the weight P, and also carrying an equa- bly distributed load ABEF. Make, by formula (79.), AD = and join B and D ; then ABD is the scale for the depths required by the equally distributed load U. Make, as above, AC=V^ Bb and upon AC as a base and AB for the height describe the semi-parabola ABC, which gives the scale for depths due to the concentrated load P. Now, an ordinate drawn at any point, as G, vertically across the combined scales of depths, as H to J, measures, by scale, the required depth for the lever at the point G. The length of any ordinate, as HJ, may be determined analytically thus. The portion of the ordinate representing the equably distributed load is, by formula (77.), 2Ua ~Bbn 174 STRAINS IN LEVERS, GRAPHICALLY EXPRESSED. CHAP. XI. For the remaining part of the ordinate we have formula (36.) (in which x is equivalent to the y of this case), Adding these we have for the full length of the ordinate HJ, or for the depth at the point G, d 2Ua Bbn Bb y (80.) in which U is the weight equably distributed over the length of the lever ; P, the weight concentrated at the end of the lever ; #, the length of the lever ; j/, the horizontal distance from the free end of the lever to the location of the ordinate at which the strain is being measured ; a, the factor of safety ; b, the breadth of the lever, and B the resistance to rupture as per Table XX. 226. Scale of Strains for Compound Loads. Fig. 5 represents the case of a semi-beam like the preceding, except that the concentrated load is located at some other point than the extreme end. FIG. 50. The curve DB is found as in Fig. 48, and the line CE in the same manner as there, except that, in finding AC, the distance m from the wall to the weight R is to be substi- tuted for n, the length of the lever. LEVER PROMISCUOUSLY LOADED. 175 227 Scale of Strains for Proniicuoiis Load. A semi- beam, equably loaded, may also have to carry two or more concentrated loads. In this case, for the scale of strains we combine the methods required for the two kinds of loads, as in Fig. 51. Here AB represents the length of the semi- beam ; the curve DB, for the equably distributed load, is obtained as in Art. 222 ; and the triangles for the concen- trated weights are found as in Art. 221. A vertical ordinate drawn anywhere across the figure, and terminated by the curve DB and the line KJHEB, will measure the strain at the location of that ordinate. The depth of the beam at that point may be found by putting the strain as above found equal to the resistance ; or. or (Art. 35), from which, D = Sbd* D = \Bbd'< Bb in which D represents the destructive energy or the strain as shown by the length of the vertical ordinate obtained as above directed ; a, the symbol for safety (Art. 73) ; E equals the resistance to rupture as per Table XX., and b and d are the breadth and depth, respectively the breadth being constant. STRAINS IN LEVERS, GRAPHICALLY EXPRESSED. CHAP. XL QUESTIONS FOR PRACTICE. 228. In a semi-beam 6 feet long, carrying 500 pounds at 2 feet from the wall, and 300 pounds at 5 feet from the wall, what are the respective strains at i, 2, 3, 4 and 5 feet from the free end ? What is the strain at the wall ? 229. In a scale of strains for a semi-beam equably loaded, what curve limits the upper edge ? 230. A semi-beam, 8 feet long, is equably loaded with 100 pounds per foot lineal. What is the strain produced at 5 feet from the free end ? 231. Of Avhat form is ; the side of the last-named semi- beam required to be, in order that the beam may be of equal strength at all points, the breadth being constant ? 232. In a semi-beam 7 feet long, carrying 1000 pounds at its free end, and 100 pounds per foot lineal, equably dis- tributed, what are the respective strains at 3, 5 and 7 feet from the free end ? 233. In a semi-beam 10 feet long, carrying an equably distributed load of 1000 pounds, and concentrated loads of 800, 500 and 700 pounds, at the several distances of 3, 6 and 8 feet from the free end, what are the respective strains at 2, 4, 7 and 9 feet from the free end ? CHAPTER XII. COMPOUND STRAINS IN BEAMS, GRAPHICALLY EXPRESSED. ART. 234. Equably Distributed and Concentrated Loads on a Beam. We have now to consider the effect ot com- pound weights upon whole beams. Of this class we shall take first the case of an equably distributed weight, together with a concentrated one, as in Fig. 52. In this figure the curve of strains RFDP for the equably distributed load is a parabola, with its apex at D. The FIG. 52. height CD is, by formula (70.), to be made equal to -//; and HJ, by the same scale, and by Art. 192, is to be made equal to A' -j- . Join J with R and with P. Then any vertical ordinate FG drawn across the figure, and termi- 1/8 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. nated by the curve RFDP at top, and by the line RJP at bottom, will measure the strain, j, at E, the point of inter- section of the ordinate with the line RP. To obtain this strain analytically, we have, for the ordi- nate EF, formula (71.), which is (putting u for y ) ujht u = \U-j- and, for the ordinate EG, formula (44*)> which is (putting b' for/, A' for W and // for x) Now, since b'+u EG+EF = y, therefore ^ *,, y = u+b' = \U-j + A'-h y = j(Ut + A'm) (81.) equals the strain at any point between H and P. To find the requisite depth of the beam at any point, the breadth being constant, we put the strain equal to the resistance, or (Art. 35) y = Sbd 2 = Bbd* or, for the safe weight, ^ay = Bbd* from which Bbl 235. Greatest Strain Graphically Represented. To find the longest ordinate, and consequently the greatest strain, arising from the compound loads of Fig. 52, draw the tangent KL parallel with JP\ then an ordinate FG drawn from GREATEST STRAIN LOCATION DETERMINED. 179 the point of contact, F, will be greater than any other which may be drawn across the figure. 236. Location of Greatest Strain Analytically Defined. The point of contact between a curve and its tangent is not easily found by mere inspection, but analytically its exact position may be defined. FIG. 52. To do this, let (Fig. 52) a' = HJ, V = EG, u = EF, h = EP and h + / = / = RP. We now have, from the similar triangles HJP and EGP, n : a' : : h : V = From formula (70.), in which y = u = \et(lf), we have u = %eh(l/i) = \ehl- \eh* therefore n (83.) This is the value of an ordiriate drawn at any point be- tween H and P. But it is required to find where this ISO COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. ordinate will be at its maximum. This may be done by the calculus. Obtain the differential of formula (83.\ and placing it equal to zero, derive its integral ; from which the value of h will be obtained. This represents the distance from P to the ordinate y, when at its maximum, and there- fore determines the point E, the location of the ordinate, as required. 237. Location of Greatest Strain Differentially Defined. First. For the value of h we are to find the differential of formula (83.) and put it equal to zero ; thus : dy = ( + \el\dh %e x 2hdh = o V;z / = ehdh Now, since el= U, therefore e = -,-, and Again, <i=ff?=A' therefore ~ (84.) GREATEST STRAIN DETERMINED ANALYTICALLY. l8l or the distance of the ordinate from the remote end of the beam is equal to half the length of the beam, plus a fraction which has for its numerator the product of the concentrated weight into its distance from the nearest bearing, and for its denominator the weight which is equably distributed along the beam. This formula of the value of h is limited in its applica- tion to those cases in which n exceeds h in value. When, on the contrary, k exceeds n , then the longest ordinate is at the location of the concentrated weight, and n is to be substituted for h. The reason for this may be seen by an inspection of the figure. 238. Greatet Strain Analytically Defined. Second. To find the length of the ordinate y, we have, by formula (83.), n and by substituting for / its value, h + t, y - a'h XT / Ai mn j" U r Now, a = A -j- , and e -r, therefore ' h y n 1 82 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. which gives the greatest strain resulting from both the concentrated and distributed loads. This formula is identical with formula (81.), obtained by another process. 239. Example. As an example, let it be required to find the location and length of the longest ordinate of strains produced by a load of 4000 pounds, concentrated at three feet from one end of a beam 16 feet long, together with a load of 3000 pounds, equably distributed over its length. First. The location of the ordinate, or the value of h. This, from formula (84-), is or the longest ordinate is situated within one foot of the location of the concentrated weight. Second. The amount of strain at this ordinate. This, by the above formula, is 12 3+ix 3000x4) = 13500 or the greatest resulting strain at any one point of the combined weights equals 13,500 pounds. 240. IMmeiiiions of Beam for Distributed and Concen- trated Loads. The amount of strain, just found is the actual moment of the loads. Putting this equal to the resistance (Art. 35), we have, for the safe weight, a^(A 'm +#) = Sbd 2 = \Bbd 2 or / 4* 7 (A 'm + Ut) = Bbd* (85.) DIMENSIONS OF BEAM FOR COMPOUND LOAD. 183 which is a rule for obtaining the dimensions requisite for re- sisting effectually the greatest strain arising from the com- bined action of a concentrated and an equably distributed load ; and in which A' equals the concentrated load, and U the equably distributed load, both in pounds ; / is the length of the beam between bearings ; m the distance from the con- centrated weight to the nearer end of the beam ; h the dis- tance from the location of the greatest strain to the more distant end of the beam ; and / equals / //. /, m, h and / are all to be taken in feet, and the value of h is to be had from formula (84-} ; care being exercised that when h ex- ceeds n in value, then n is to be used in place of h, and m in place of /. In the latter case formula (85.) becomes 4a ^(A'm + tUm) = Bbd 2 = Bbd* 24-1. Comparison of Formulas, Here and in. Art. ISO. Formula (29. \ given in Art. 150, for a carriage beam with one header, is for a case similar to that of the last article, but is not strictly accurate. Instead of the two strains being taken at the same point, E (the location of the longest ordinate), as in Fig. 52, they are taken, the one for the concentrated load, at the location of this load, and the other, that for the equably distributed load, at the middle of the beam ; or, the maximum strain for each load. Taken in this manner the result is in excess of the truth, as //7+ CD is greater than FG. The error is upon the safe side, the strains being estimated greater than they really are. In most cases this error would not be large, and the only objection to it would be that it requires a little more mate- rial in the beam. Formula (29.) may therefore be employed 1 84 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. in ordinary cases where a low priced material, such as wood for example, is used for the beams ; but where a more costly material is involved, economy would dictate that the strain be not over-estimated, and that it be correctly obtained by the use of formula (85.) in Art. 24-0. (See also caution in Art. 88.) 242. Location of Oreatet Strain Differentially De- fined. In Fig. 53 we have a scale of strains, RABPF, by which is found the effect arising at any point in the length of the beam from two concentrated loads, together with an equably distributed load. The curve RFP is a parabola (foot note, Art. 2(2) found as in Fig. 52, and the moment of the two concen- trated loads equals AH at H and BJ at J, and is found as in Art. 194- and Fig. 37. FG is the ordinate for strains occurring between H and J 1 and is defined thus : .L Let HJ = d' EJ x, EG = v = b'+fr EF = , AH=a f , BJ=V and a'b f = c. Then, from similar triangles, GREATEST STRAIN LOCATION BY CALCULUS. 185 and, since x = h s, v = b'+p and c a' b' t there fore d' a f -V and v b -\ 77 < Formula (70.), gives [putting /// for t(lt) and u for y\ u = %eht and since y u + v, consequently y =. ^eht + b' -\ -j, (h s) (87.) This is the value of the ordinate for the strain at any point between H and J. To obtain the longest ordinate which can be drawn here, proceed as in Arts. 235 to 237, and as follows: First reduce formula (87.) thus, a'-b' . a'-b' a'-V then v + u = y = \ehl^eh s + b' -i -r, h -- - s In this expression, rejecting the quantities unaffected by the variable h, we have, for the differential of y, 1 36 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. dy = (^el + ^^dh - ehdh = o or, ( \el + - }dk=. ehdh or, its integral gives * = #+= (**.) 243. Greatest Strain and Dimensions The above gives the value of h. To obtain the value of y at its maximum take formula (87.). In this, for the value of a' we have AH, equal to the joint effect at H of the two concentrated loads ; or, putting a' for the D of formula m and for the value of b' (form. 52.) b' = j(B'r + A'm) The value of e (from el U) is equal to 7- . By sub- stituting this value for e we have =U + y+k-s (89.) This equals the strain from the compound weights of Fig. S3, and is the same as (87.), for j == \e. Either formula will give the strain at any required point between H and J (Fig. 53) by putting h equal to the dis- DIMENSIONS OF BEAM FOR COMPOUND LOAD. 1 87 FIG. 53. tance between that point and P ; but when the greatest strain is required, h must be obtained from its value in formula (88.). To obtain the dimensions in this case, we put the strain equal to the resistance, and have, with a as the factor for safe weight (Arts. 35 and 73) ~ = Sbd * = = Bbd * and from this formula may be found the dimensions required for resisting effectually the greatest strain in the beam, the value of h being derived from formula (88.). 24-4. Aignlng the Symbol*. It is important to observe here that of the two moments a' and ', a' designates the larger of the two, while m and n represent the distances from a' to the two ends of the beam, m being the distance to that support which may be reached without passing the 1 88 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. other weight. Again r and s are to be regarded as the distances from b' to the two ends of the beam ; k and s dating from the same end of the beam as n ; and as n is the greatest possible value of //, it is to be substituted for it when by the formula for h its value is found equal to or greater than n. In order to ascertain which of the two moments a and b' is the greater, a trial must be had by the use of the expres- sions in the last article designating their respective values. When the two concentrated weights are equal, then the nearer weight to the middle of the beam will produce the greater moment, and may at once be designated as a. 245. Example Strain and Size at a Given Point. As an example, let a beam, 10 feet long, be required to carry an equably distributed load of 100 pounds per foot lineal, a con- centrated load of 2000 pounds at a point two feet from the left-hand end, and a second concentrated load of 800 pounds located at 3 feet from the right-hand end. What will be the resulting strain at 4 feet from the right-hand end ? Formula (87.) is equals the required strain. In designating m and s we find (Art. 243) for the larger weight (2000 x 8 + 800 x 3) = 3680 and for the smaller (8OO X 7 + 2000 X 2) = 288O and hence (Art. 153) m = 2 and s = 3. DIMENSIONS AT A GIVEN POINT. 189 We now have e = 100, h = 4, /= 10, t = lh = 6, 2, n = S, J=3, r = 7 and ^'=5. becomes x 100 X4 x 6 =1200 With ^' = 2000 and B' = 800, a' = (A'n+B's) = (as above) 3680, and b f = j (B r r+A' m) = (as above) 2880 and #' ' = 36802880 = 800. We therefore have, as a resulting 1 value of y in formula 1200 + 2880 + -(43) = 4240 y This equals the effect at 4 feet from the right-hand end pro- duced by the three weights. To find the dimensions of the beam at this point, make the strain just found equal to the resistance [see Art. 24-3 at formula (90.)], and we have 4a x 4240 = Bbd* and, if a = 4 and B = 500 (see Table XX.), we have 500 Let =3, then we have d = 6-j$\ or, the beam at 4 feet from the right-hand end should be 3 x 6- 73 inches in cross-section. 246. Example Greatest Strain. Again, let it be re- quired to show the greatest strain produced at any one point by the three weights of the last article. The first dimension required here is that of h. For this we have, as per formula (88.\ 190 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. from which h= $ -^ --- - =6-6 This result being less than n, since n equals 8, is there- fore the correct value of h, and from it we obtain (from /) / 3.4. Formula (89.} now gives '= which is the required greatest strain. 247. Example Dimensions. What sized beam of equal cross-section throughout would be required to carry safely the loads upon the beam of the last article, when B = 500 and a = 4? The greatest strain at any point was found to be 4578 pounds, therefore 4a x 4578 = Bbd 9 4.X4X4578 500 and with b taken equal to 3, then d=6>gQ. The beam must be 3x7 inches. 248. Dimensions for Greatest Strain when // Equals n. When, in formula (90.), h = n, or is greater than n t then t = m, hs = d f , and c*f = b'+a'-b' = a' also, DIMENSIONS OF BEAM COMPOUND LOAD. and the formula becomes or, supplying the value of a' (Art. 243), which is a rule for a beam carrying two concentrated loads and a uniformly distributed load, when h = n as above stated. 249. Dimensions for Greatest Strain when h is Greater than n. As an example under this rule, what are the breadth and depth of a Georgia pine beam 20 feet long, carrying 2000 pounds uniformly distributed over its whole length, 10,000 pounds at 7 feet from the left-hand end, and 8000 pounds at 5 feet from the right-hand end ; the factor of safety being 4 ? Here a = 4, [7= 2000, /=2O, ^=850, m = 7 and s = 5 (since 7 x 10,000 = 70,000 exceeds 5 x 8000 = 40,000 ), #=13, r=i5 and d' = 8. The value of // is to be tested, to know whether it is equal to or greater than n. By formula (88.), and Art. 243, a'b' a'-b r d r a' = -j-(A'n+B's) (10000x13 + 8000x5) = 59500 V = j(B'r+A'iri) (Sooox 1 5 + 10000x7) = 47500 192 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. a' b' = 59500 47500= 12000 12000 20 This gives a value to h greater than that of n and shows (Art. 24-4) that n must be substituted for //, and that the problem is a proper one for solving by formula (91.)\ therefore r 7X13 7 __ _ ~] 4x4 looo- -- + -(10000x13+8000x5) ' - ~~ =" = 1205. 65 If the breadth b be taken at 8 inches, then ^= that is, the beam should be 8 x i2| inches. 250. Rule for Carriage Beams with Two Headers and Two Sets of Tail Beams. By proper modifications, formula (90.) may be adapted to the requirements of a carriage beam with two headers, as in Fig. 25. These modifications are as follows : By Art. 150 we have hence U~, = \cfht also, from Arts. 153 and 243, a = and, from Art. (55, A' = \fgm and B' = \fgs therefore CARRIAGE BEAM WITH TWO HEADERS. 193 m Similarly we find b' To obtain the maximum strain, h is to be determined by formula (88.\ in which for e we have U cfl <-----[----Ti= and therefore a'-V In these deductions, f equals the weight per superficial foot of the floor, c the distance apart from centres at which the beams in the floor are placed, and g the length of the header. (For cautions in distinguishing between m and s, and between a' and b' t see Art. 244.) By formula (90.) and the modifications proposed, we therefore have Bbd* and as auxiliary thereto we have, as above, a' =~ and a'-b' IQ4 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. and thus we have in formula (92.) a rule for a carriage beam carrying two headers and two sets of tail beams. (See cau- tion in Art. 88). 251. Example. To show the application of this rule, let it be required to find the breadth of a white pine carriage beam, 20 feet long and 10 inches deep; the beam to carry two headers 10 feet long, one located 9 feet from the left- hand end, and the other 6 feet from the right-hand end. The floor beams are to be placed 15 inches from centres, and the floor is to carry 100 pounds per superficial foot, with the fac- tor of safety a 4. Here the header at the left-hand end is the nearer of the two to the middle of the carriage beam, and therefore (Art. 244) m 9. From formula (92.) we have, for the value of b, in which a = 4, B 500, d* io 2 , /= 100, c = ij, g 10, I 20, m = q, n \\, r = 14, s = 6 and d' = 5. From the auxiliary formulas- of Art. 250, a' = 100 x IQX - (9 x ii + 6 2 ) = 15187-5 4 x 20 b' = ico x io x ^ 2 - (14 x 6 + 9 2 ) == 12375 a'b 1 = 15187-5 12375 = 2812-5 2812-5 _ / '- =IO + "" RULE FOR CARRIAGE BEAMS. Here , since it is but n, is less in value than h, and must be used in its place ; we therefore have recourse to formula (91-), Art. 248. By this formula the value of b is This is a general rule. To make it conform to the re- quirements for a carriage beam, we have for U the equally distributed load \cfl (Art.\BO). A' = \fgrn (Art. 250), and B' = Ifgs. Hence = I ^ r + g SooxiooL 20 ^ J J or, the carriage beam should be 5^ or, say 6 inches broad. In this computation, no allowance is made for the weaken- ing effect of mortising, it being understood that no mortises are to be made ; the headers being hung in bridle irons (Art. 147). (See Art. 88). 252. Carriage Beam with Three Header*. It some- times occurs in the plan of a floor that two openings, the one a stairway at the wall, the other an opening for light at or near the middle of the floor, are opposite each other, as in Fig. 54. In this arrangement the carnage beam has three headers to carry, besides its load as an ordinary floor beam. 196 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. FIG. 54. Cases of this kind may be divided into two classes : one that in which the header causing- the greatest strain occurs between the other two ; the other, that in which it occurs next to one of the walls. We will first consider the latter case. 253. Three Headers Strains of the Firt Class. When the well hole for light occurs at the middle of the distance between the walls, its two headers will be equally near the centre of the length of the carriage beam ; and, were their loads alike, the headers would produce equal strains upon the carriage beam ; but the loads are not alike, for the tail beams carried by one header, those which reach to the wall, are longer than those carried by the other. Hence the header carrying the tail beams, one end of which rest on the wall, has the heavier load ; and, as it has the same leverage as the header on the other side of the well hole, it will therefore have the greater moment, and will produce the greater strain upon the carriage beam. CARRIAGE BEAM WITH THREE HEADERS. I 9 7 The stair header will add to the strains upon the carriage beam at the points of location of the other two headers, and this addition will be greater at the middle header than at the farther one, but still not so much greater as to cause the total strains at the one to preponderate over those at the other. 254. CJrapIiical Representation. Let Fig. 55, construct- ed similarly with Fig. 53, represent the strains in a carriage beam supporting three headers, one of the outside ones, as at A, producing the greatest strain. In this figure the curve DKE is a parabola (Art. 212) and is the curve of strains for the uniformly distributed load upon the carriage beam, FIG. 55. of which KL represents the strain at the middle of the beam ; and CF, BG and AH, vertical lines, by the s'ame scale, represent the strains caused by the three head- ers at the points C, B and A, respectively. Any or- dinate drawn, parallel to AH, across this figure, and ter- minated by the boundary line DFGHEKD, will measure the strain in the carriage beam at its location. Hence that point at which an ordinate thus drawn proves to be longest of any which may be drawn, is the point where the strain upon the carriage beam is the greatest, and the length of this ordinate measures the amount of this strain. IQ8 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. Draw the tangent ST parallel to GH. If its point of contact with the curve occurs between Q and , then HQ will be the longest possible ordinate ; but, if it occur between K and Q, then HQ will not be the longest. When AH and BG are equal, the point of contact will be at K. In the case under consideration (the well hole in the middle of the floor) the tangent will usually touch be- tween Q and , giving HQ as the longest ordinate. 255. Greateit Strain. With the loads A, B and C in position as in Fig. 55, the longest ordinate may be found FIG. 55. by formula (87.), where / ri y = \eht + b' -i -77 (hs) and in which m + n = r+s=/i + t = t (for the position of these letters see Art. 244), \eht represents the strain from distributed load, and a '~^-(h s) // v */ the uniformly stands for the length of an ordinate drawn from GH to BA at the distance h from D towards A, and repre- sents at the location of the ordinate the strain from the three concentrated loads. In all cases, except where b' is very nearly or quite equal to a', h will exceed , and, in GENERAL RULE FOR COMPOUND LOAD. 199 general, for all problems of the class of which we are treat- ing, it may be assumed, without material error, that h will always exceed n. Then m and n take the place of t and h in formula (87.), and it becomes (Art. 248) y = \ernn + a' mn or, The value of a' is (form. 58.) a' = ~ hence y = % u^ + j(A 'n + B's + Cv) (95.) In this formula y equals the greatest strain in the beam. 256. General Rule for Equably E>itributecl and Three Concentrated Loads. Putting the strain y of last article equal to the resistance (Art. 35) gives us % U ni ~ + (A 'n + B's + Cv) = Sbd* and with B ^S and a as the coefficient of safety, 'n + B's + C'v) = Bbd 2 (96.) which is a general rule for beams carrying a uniformly dis- tributed load and three concentrated loads similarly placed with those in Fig. 55. In this rule, U is the uniformly dis- tributed load, and A f , B' and C the three loads concen- trated at A, B and C in the figure. 257. Example. As an example, we will ascertain the required breadth of a Georgia pine beam of average quality, 200 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. 20 feet long and 14 inches deep, with a load of 2000 pounds equally distributed over its length, a concentrated load of 4000 pounds at 3 feet from the left-hand end, a like load at 7 feet from the same end, and one of 7000 pounds at 7 feet from the right-hand end. Take as the factor of safety a = 4. Then /= 20, ;;/ = 7, ;/ = 13, s = 7, r 13, v 3, u = 17, d 14, U 2000, A' 7000, ^ = 4000 = C' and B 850, and from formula (96.) / ---- -- - --- \ b = 85oxTo(* X200 X J 3 + 7oooxi 3 + 4000x7 +4000x3^=4.84 or the breadth should be 4^ inches. 258. Rule for Carriage Beam with Three Heatler and Two Sets of Tail Beams. To modify formula (96.) so as to make it applicable to a carriage beam, we have for 7, the uniformly distributed load, (Art. 150) U=%cfl; for the load at A, caused by the header carrying the tail beams, one end of which rests upon the wall, A ' \fgm ; for the load at B, B' \fg(sv) ; and for the load at C the same, C' = : kfg(sv). Formula (96.) now becomes b = b = TM ^ cnl + gmn + g ^~ b = -^ [cnl+g (mn + s*-v*)] (97.) which is a rule for carriage beams carrying three headers and two sets of tail beams, located, as in Fig. 55, with A, the heaviest strained header in an outside position relative to the other two headers. 259. Example. Under the above rule, what should be the breadth of a spruce carriage beam 20 feet long and 12 CARRIAGE BEAM WITH THREE HEADERS. 2O I inches deep, carrying three headers 15 feet long, located as in Fig. 54. The well-hole for light, in the middle of the width of the floor, is 6 feet wide, and the stairway opening, at one of the walls, 3 feet wide. The beams of the floor are placed 15 inches from centres, and are to carry 90 pounds per superficial foot, with 4 as the factor of safety. 16 Here / 20, m = s = - 7, n 13, v^g\^, d\2, <:=ii, /=90, a 4 and ^= By formula (97.) b = or the breadth should be 3f inches. -3 2 )] = 3-64 260. Three Headers Strains of the Second Class. We will now consider the other class named in Art. 252, that in which the header causing the greatest strain occurs between the other two. The conditions of this class of cases are represented in Fig. 56, in which AH=a', BG = b' and CF=c f , repre- senting by scale the combined concentrated strains at A, B and C respectively, and KL is the strain at the middle due to the uniformly distributed load. The parabolic curve 202 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. (Art. 212) EKD and the line DGHFE form the boun- daries of the scale of strains, as in Art. 254. For the proper assignment of the symbols ///, n, r, s, etc. see Art. 244, taking the two larger of the strains of Fig. 56 for the two given in that article. The longest vertical ordinate across the scale of strains will ordinarily be at QH] the exceptions being when the strain at B is nearly or quite equal to that at A. In the latter case, however, the diminution at QH will be so small that that ordinate may be assumed, without material error, to be the greatest. Taking it as the greatest, formula (87.) becomes, as in Art. 255, 261. Greatest Strain. The manner of obtaining the value of a', the strain produced by the three concentrated loads, will now be shown. The strain at A, produced by the load A', is (Art. .mn 56) A'-j-. The strain at *? B, produced by B' ', is of the strain at B may be had by the The effect at A proportions shown in the triangles BGE and ARE ; for the effect is proportional to the horizontal distance from E (see Art. 192); therefore, THREE LOADS GREATEST STRAIN. 203 equals the effect of the weight B' at the point A. Also, for the effect at A of the weight at C y the effect of C' at C being C y-, we have Cuv C'nuv C'nv u : n : : -, : , --- = =- I lu I equals the effect at A of the weight at C. The joint effect at A of the three weights is therefore or, a f = A f n Adding this to the effect of the uniformly distributed load, mn ~, gives " (P*.) This represents the greatest strain arising from the uni- formly distributed load and the three weights disposed as in Fig. 56] A' at the middle being the greatest strain and B' the next greatest. 262. General Rule for Equally Distributed and Three Concentrated Load. Putting the strain [form. (##)] in equilibrium with the resistance (Art. 35) we have *** tii) \'n + B's) + C'^- = Sbd' = 204 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. and with the symbol for safety added, b = Wi Un + A'n + B's) + C. 'wv\ (99.) which is a rule for beams loaded with an equally distributed load and with three loads relatively disposed as in Fig. 56 ; A' being the greatest strain, and B' the next greatest, and A' being at the middle. 263. Example. As an example under this rule : What should be the breadth of a Georgia pine beam of average quality, 20 feet long and 12 inches deep, carrying 4000 pounds uniformly distributed, 6000 pounds at 4 feet from the left-hand end, 6000 pounds at 9 feet from the same end, and 7000 pounds at 6 feet from the right hand end ; with the factor of safety a = 4? Assigning the symbols to the loads and spaces as in Fig. 56, we have # = 4, ^ = 850, d 12, /= 20, m = 9, nii, r 14, s = 6, v 4, U= 4000, A' 6000, B' 7000 and C' 6000. Substituting these values in formula (99.) gives b Tt -- 3 --- [0(^x4000x11 +6000x1 1 + 7000x6) + Ly t 850x12 (6oooxi 1x4)] =9- 37 or the breadth should be 9f inches. 264-. Aigning the Symbol*. In working a problem of the kind just given, it is of prime importance to have the symbols denoting the weights and distances properly located. In doing this, the first point to settle is as to which of the two classes (Fig. 55 or 56) the case in hand belongs. Make a sketch, such as Fig. 55 or 56, according to the probable position of the largest strain, letter the weights and COMPOUND LOAD DISTINGUISHING THE RULE. 205 distances as there shown, and then compute the three strains by the following formulas. For Fig 55 the strains will be as follows (Art. 195) : At A, the strain a = -r(A'n + B's+ Cv) " B, " b' = j(A'm + B'f) + C'^ (101) " C, " c' = j(A 'm -f B'r + Cu) (102.) In the diagram, AH is to be made, by any convenient scale, equal to a', BG to b', and CF to c , as found by these three formulas, and KL, the height of the para- bola, is, by the same scale, to be made equal to %U ' j . U is the load equably diffused over the beam ; A', B ! and C' are the loads concentrated at A t B and C respectively, and / is the span, or length of the beam between bearings. For Fig. 56 the strains will be as follows : At A, the strain a' = ~ (A 'n + B's) + C' H j (103) " B, " b' = j(A r m+B'r+Cv) (104) v n Q c > __ __ /^ 'ft i j^i s In the case of a carriage beam the loads A', B' and C' in the formulas (100.) to (105.) are those from the headers ; and equal %fgm, etc. In this, / and g are constant, as to the three loads in any given case, and m represents the length of one set of tail beams ; consequently the loads A' B' and C will vary as the length of the tail beams. Hence, in the preliminary work required to ascertain to which of the two classes any given case belongs, it will suf fice to use simply the length of the tail beams, instead of the full weights A', B' and C. 205 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. For example: Take the case given in Art. 259, where / = 20, m = 7, s = 7, n=\$, r = 13 and v = 3, the let- ters being assigned as required by Fig. 55. Here the tail beams carried by the header at A are 7 feet long, and those carried by the two other headers are 4 feet ; there- fore A = 7 and B C 4, and by formulas (100.) and (101) ij . ' 4x7 + 4x3) = 45 85 7 , . 4x13x3 b == ^(7 X 7 + 4X13) + 2^-- =43-15 The result here obtained, a/ being larger than ', shows that the case has been rightly assigned to the first class, that of Fig. 55. 265. Reassigning the Symbols. The result of a compu- tation of the strains may show that the arrangement of the symbols was erroneous ; instead of the greatest strain being in the middle it may be found at one side, or vice versa. Then the lettering of the loads and spaces must be changed, to agree with the proper diagram and formulas, before com- puting the dimensions of the beam ; using formula (96.) or (97.) for the class shown in Fig. 55, and formula (99.) for the class shown in Fig. 56. 266. Example. As an illustration of the above, take a case presumably belonging to the class first treated (Fig. 55), where' the greatest strain is an outside one. Let / = 20 ; and let the greatest load, 1750 pounds, be designated by A', with its distances in = 7 and ;/ = 13 ; the second load, 1250 pounds, be designated by B' y with its distances r = 12 and J = 8 ; and the third load, 1250 pounds, be called C, and its distances v = $ and u = 17. To find COMPOUND LOADS EXAMPLES. 2O/ the united effect at each station, we have, according to formulas (100.) and (101.), a' = f 1750x13 + 1250x8 + 1250x3 \ 12775 8 / -\ 1250x12x3 b'= --(1750x7 + 1250XI2J + - ^- =13150 Here b' exceeds a and shows that a mistake has been made as to the class to which the case belongs. We must change the symbols and arrange them for the second class (Fig. 56). The middle weight is to be called A' ; the weight be- fore called A', at 7 feet from one of the walls, is now to be B' ; and the third weight C ' . With these changes made, we have ^'=1250, v = 1750, ' = 1250, / = 20, m 8, Ji12, s = 7, r 13 and ^ = 3; and, from for- mulas (103.) and (104.), 8 / \ 1250x12x3 a' = (1250x12 + I750X7J + - 2_ =13150 b' = (1250x8+ 1750x13 + 1250x3^ = 12775 The result is now satisfactory, and shows that the prob- lem belongs to the second class, the one in which the great- est strain occurs at the middle, and this notwithstanding the fact that the greatest of the three weights is at the outside. It will be seen that the results of the two trials are the same, but reversed, that which was at first taken for a' being now taken for b' . 267. Rule for Carriage Beam with Three Headers and Two Sets of Tail Beams. Formula (99) may be trans- formed so as to make it specially applicable to carriage beams. If, in Fig.s^y we suppose the spaces EC and AB to be openings in the floor, then one set of tail beams will extend 208 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. s D from C to A, and another from B to D, giving- three headers, one each at A, B and C. The load on the header A will equal that upon C, and will equal one quarter of the load upon the space occupied by the tail beams AC, or -fg(inii). Similarly the load at B will be \fgs. Of the several factors composing formula (99.) we now have and since and the formula itself becomes Cnv = \fgnv (mv) U \cfl n = \cfln b = jntfcfln + tfS* ^^ + i/^) + \fgnv(m-v)\ b = t> = b = (cnl+gn mv +g/) + \fgnv (mv]\ v) +gms* +gnv(mv)\ +gn(mv) ( b = - (106). CARRIAGE BEAM WITH THREE HEADERS. 209 which is a rule for carriage beams carrying three headers and two sets of tail beams relatively placed as in Fig. 56, the header producing the greatest strain being between the other two. 268. Example. What should be the width of a carriage beam 20 feet long, 12 inches deep, of Georgia pine of average quality, carrying three headers 14 feet long ; the headers placed so as to afford a stair opening 4 feet wide at one wall, and a light well 5 feet wide, 6 feet from the other wall? The floor beams are 15 inches from centres and carry 200 pounds per foot superficial, with the factor of safety a = 4. In this case we have B = 850, /= 200, a = 4, c = ij, <^= 12, / = 2O, v = 4, ;# = 9, # = 11, s = 6, r = 14 and g = 14, and by formula (106.) 1 X2Q + 4H i4xu(9'-4 3 )] = 5-56 or the breadth should be, say 6 inches. 210 COMPOUND STRAINS, GRAPHICALLY EXPRESSED. CHAP. XII. QUESTIONS FOR PRACTICE. 269. In a beam 20 feet long, carrying an equably dis- tributed load of 2000 pounds, and, at 4 feet from one end, a concentrated load of 5000 pounds, what is the great- est strain produced, and where is it located ? 270. In a floor composed of beams 12 inches deep, and set 15 inches from centres, there is a Georgia pine carriage beam 22 feet long, carrying two headers with an opening between them. The headers are 14 feet long, and are placed at 5 and 12 feet respectively from the left-hand wall. The floor is required to carry 200 pounds per superficial foot, with the factor of safety a = 4. What must be the breadth of the carriage beam ? CHAPTER XIII. DEFLECTING ENERGY. ART. 271. Previously Given Rulc are for Rupture. In the discussion of the subject of transverse strains, the rules adduced thus far have all been based upon the resist- ance of the material to rupture, or the power of the material to resist the destructive effect produced by the load which the beam is required to carry. 272. Beam not only to Be Safe, but to Appear Safe. It is requisite in good construction that a loaded beam be not only safe, but that it also appear safe ; or, that the amount of deflection shall not appear to be excessive. In determining the pressure a beam may receive without injury, real or ap- parent, it is requisite to investigate the power of a beam to resist bending, rather than breaking that is, to ascertain the Laws of Deflection. 273. All Material Possess Elasticity. Any load, how- ever small, will bend a beam. If the load be not excessive, the beam will, upon the removal of the load, recover its straightness. The power of the beam by which it returns to its original shape upon the removal of its load, is due to the elasticity of the material. All materials possess elasticity, though some, as lead and clay, have but little, while others, as india- rubber and whalebone, have a large measure of it. 212 DEFLECTING ENERGY. CHAP. XIII. 274. Limits of Elasticity Defined. When a beam is bent, some of its fibres are extended and some compressed, as was shown at Art. 22 ; and when the pressure by which the bending was effected is removed, the fibres resume their original length. Should the pressure, however, have been excessive, then the resumption will not be complete, but the extended fibres will remain a trifle longer than they were before the pressure, and the compressed fibres a trifle shorter. When this occurs, the elasticity is said to be in- jured ; or, the pressure has exceeded the limits of elasticity. When the fibres are thus injured, they are not only in- capable of recovering their original length, but (the pressure being renewed and continued) they are not able to maintain even their present length, and therefore the deflection must gradually increase, and the fibres continue to alter in length, until finally rupture will ensue. 275. A Knowledge of the Limits of Elasticity Requisite. To secure durability, it is evident that a beam subject to transverse strain should not be loaded beyond its limit of elasticity. Hence the desirability of ascertaining this limit. 276. Extension Directly as the Force. Let the effect offeree in producing extension be first considered. Suspend a weight of one pound, by a strip of india-rubber one foot long, and measure the increase in the length of the rubber. Then, double the weight, and it will be found that the in- crease in length will be double. If the extension caused by one pound be one inch, then that caused by two pounds will be two inches. Three pounds will increase the length by three inches ; or, whatever weight be suspended, it will be found that the extensions will be directly in proportion to the forces producing them, provided always that the force EXTENSION DIRECTLY AS FORCE AND LENGTH. 213 applied shall not be so great as to destroy the elasticity of the material ; shall not so injure it as to prevent it from recovering its original length upon the removal of the force. 277. Extension Directly a the Length. The above shows the relation between the weight and the extension. The relation will now be shown between the extension and the length of the piece extended. At 5 inches from the upper end of a strip of rubber attach a one-pound weight. This will produce an extension of, say a quarter of an inch. Detach the weight and re-attach it at double the length, or at 10 inches from the upper end. It will now be found that the 10 inches has become ioj inches; the elongation being a half inch, or double what it was before. Again, remove the weight and attach it at 15 inches from the upper end, and the strip will be extended to I5f inches; an elongation of three quarters of an inch, or three times the amount of the first trial. From this we conclude that, under the same amount of pressure, the extensions will vary directly as the lengths of the pieces extended. 278. Amount of Deflection. When the projecting beam ABCD, Fig. 57, is deflected by a . weight, P, suspended from the | free end, it bends the beam, not | only at the point A, at the wall, | but also at every point of its length | from A to B, so that the line | AB becomes a convex curve, as | shown. The exact shape of this elastic curve is defined by writers upon that subject. A full discussion FIG. 57. 214 DEFLECTING ENERGY. CHAP. XIII. of the laws of deflection would include the development of this curve. The purpose of this work, however, will be at- tained without carrying the discussion so far. All that will here be attempted will be to show the amount of deflection ; or, in the present example, the distance, EB, which the point B is depressed from its original position. 279, The First Step. In bending a beam, the fibres at the concave side are shortened and those at the convex side are lengthened. The first step, therefore, in finding the amount of deflection, will be to ascertain the manner of this change in length of fibre, and the method by which the amount of alteration may be measured. 280. Deflection to bo Obtained from the Extension. It is manifest that the elongation of the fibres in the upper edge of the beam AC, Fig. 57, must occur not only at A, but at every point in the length of the line AB. The fibres at every point suffer an exceedingly small elongation, and if we can determine the sum of this large number of small elongations, we shall have the amount of extension of the line AB. This may be done in a simple manner, for we may, without serious error in the result to be obtained, consider them all as though they were collected and concen- trated at one place in the line, instead of considering each one at the point where it occurs. To effect this, let the line AB be drawn straight, as in Fig. 58, and the line FG be drawn at right angles to FK, the neutral line the line which divides between those fibres which are extended and those which are compressed, DEFLECTION DIRECTLY AS EXTENSION. 215 and therefore a line in which the fibres are not altered in length. The line AG maybe taken as the sum of the numerous small extensions which have oc- curred in the fibres at the line AB of Fig. 57. In order to show the relation between the extension and the de- flection, we will investigate the proportion between AG, the mea- sure of the one, and EB, the measure of the other. 281. Deflection Directly a the Extension. Make GJ, Fig. 58, equal to AG, and draw JL parallel with AE. The two triangles AGF and JGL are both right-angled triangles, and if AGF be revolved ninety degrees upon G as a centre, then the line AG will coincide with the line GJ, the line GF with the line GL, and AF with JL ; and we have the triangle JGL, equal in all respects to the triangle A GF. The triangle GJL is homologous with the triangle EBA, for the right line AB cuts the two parallel lines AE and JL, making the angles GLJ and EAB equal; the angles at E and G are by construction right angles, and hence the remaining angles at J and B must be equal, and the two triangles, having all their respective angles equal, must have their respective sides in proportion, or be homologous. Now, since the triangle JGL is iden- tical with the triangle AGF, we have the two triangles AGF and BEA with their corresponding sides in propor- tion, or GF-.AE-.AG'.EB and as AG measures the extension and EB the deflection, 2l6 DEFLECTING ENERGY. CHAP. XIII. it results that the extension is in direct proportion to the deflection. 282. Deflection Directly as the Force, and a the Length. By the experiment of Art. 276, it was shown that the extensions are in proportion to the forces producing them, and since, as just shown, they are also in proportion to the deflection, therefore the deflections are in direct pro- portion to the forces producing them. In the case of a semi-beam pro- jecting from a wall, as AC, Fig. 59, the force producing the deflection EBj is the product of the weight P, into the arm of leverage AE, at the end of which the weight acts ; or, the force producing the deflection is in proportion to the weight and the length. This is shown in Fig. 60. Here let it be required that the weight P remain constant in amount and location, while the length of the semi-beam be increased. We shall then FIG. 59. FIG. 60. have at E, in Fig. 60, the same deflection as at E in Fig. 59, because the force producing the deflection (PxAE) is the same in each figure. But at F, the end of the increased DEFLECTION DIRECTLY AS FORCE AND LENGTH. 2 1/ length, the deflection is greater, owing to an increase in the size of the triangle AEB, from AEB to AFC. The in- crease at F over that at E is in proportion to the increase of AF over AE, because EB and FC, the lines measur- ing the deflections, are similar sides of the two homologous triangles AEB and AFC \ and AE and AF, the lines measuring the lengths, are also similar sides of these tri- angles. For example, if AF equal twice AE, then we will have FC equal to twice EB ; or, in whatever proportion AF is to AE, we shall have the like proportion between FC and EB. In every case, the deflections will be in direct proportion to the lengths. 283. Deflection Directly a the Length. Again: If the weight be moved from E to F, Fig. 61, the end of the above increased length, then the force with which it acts is in- creased, and the deflection FC, caused by the weight when FIG. 61. located at E, now becomes FJ. If AF equals twice AE, then the force producing deflection is doubled, because the leverage at which the weight acts is doubled ; and since the deflections are in proportion to the forces producing them, FJ is double FC\ and in whatever proportion the arm of leverage be increased, it will be found that the deflec- tions at the two locations will be in proportion to the dis- 218 DEFLECTING ENERGY. CHAP. XIII. tances of the weights from the wall AD, or in proportion to the lengths. 284. Deflection Directly as the Length. Once more : When the weight was located at E, the length of fibres suf- fering extension was from A to E, but now this length is increased to AF. This increase in length of fibres will increase the exten- sion (Art. 277), and consequently the deflection (Art. 281). If AF, Fig. 62, be double the length of AE, then, owing to the extension of double the length of fibres, the deflection FJ, Fig- 6l > w iU be doubled, or increased to FK, Fig. 62 ; FIG. 62. and in whatever proportion the beam be lengthened, the deflection will increase in like proportion, or the deflections will be in proportion to the lengths. 285. Total Deflection Directly a the Cube of the Length. Summing up the results as found in the above several steps in the increase of deflection, we find, by a com- parison of Figs. 59 and 62, that, owing to an increase of the beam to twice its original length, we have an increase in deflection to eight times its original amount. If EB I, TOTAL DEFLECTION DIRECTLY AS CUBE OF LENGTH. 2 19 then FC=2, F?=2FC=4, and FK=2F?=SEB. With lengths of beam in proportion as i to 2, the deflections are as i to 8, or as the cubes of the lengths. This is true not only when the length is doubted, but also for any increase of length, for a reference to the discussion will show that the deflection was found to be in proportion to the length on three several considerations: fast (Art. 282), on account of an increase in the size of the triangle contain- ing the line measuring the deflection ; second (Art. 283), on account of the additional energy given to the weight by the increase of the leverage with which it acted ; and, third (Art. 284), on account of the extension of an additional length of fibres. The deflection and the length being neces- sarily of the same denomination, and the deflection being taken in inches, we therefore take the length, N, in inches, and we have the deflection in proportion to NNN or to N s . 286. Deflecting Energy Directly as the Weight and Cube of the Length. From Art. 276 the extensions are in proportion to the weights, and since, from Art. 281, the de- flections are as the extensions, therefore we have the deflec- tions in proportion to the weights. Combining this with the result in the last article, we have, for the sum of the effects, the deflection in proportion to the weight and the cube of the length ; or, 6 ; PN* 220 DEFLECTING ENERGY. CHAP. XIII. QUESTIONS FOR PRACTICE. 287. The rules given in former chapters for beams exposed to cross strains were based upon the power of resistance to rupture. Upon what power of the material may other rules be based ? 288. To what degree may beams be deflected without injury ? 289. What relation exists between extensions and the forces producing them ? 290. What relation exists between extensions and deflections? 291. What relation, in a beam, is there between the deflections, the weights and the lengths? CHAPTER XIV. RESISTANCE TO FLEXURE. V ART. 292. Reistance to Rupture, Directly a the Square of the Depth. Having considered, in the last chap- ter, the power exerted by a weight in bending a beam, atten- tion will now be given to the resistance of the beam. It was shown in the third chapter, that the resistance to rupture is in proportion to the square of the depth of the beam. It will now be shown that the resistance to bending is in proportion to the cube of the depth. 293. Resistance to Extension Graphically Shown. For the greater convenience in measuring the extension of the fibres at the top of a bent lever (Fig. 57), it was proposed in Art. 280 to consider this extension as occurring at one point ; at the wall. In an investigation of the resistance to bending, the whole extension may still be considered as being concentrated at that point. Let the triangle AGF, Fig. 63, represent the triangle AGF of Fig. 58, in which AF is the face ot the wall, and AG, at the top edge of the lever, is the measure of the ex- tension of the fibres there; while at F, the location of the neutral line, the fibres are not extended in any degree. It is evident that the fibres suffer extension in proportion to their distance from F towards G, so that the lines BC y DE, etc., severally measure the extensions at their respec- tive locations. Within the limits of elasticity, the resistance 222 RESISTANCE TO FLEXURE. CHAP. XIV. FIG. 63. of a fibre to extension is measured by its reaction when re- leased from tension. Thus, the line BC measures the extension of the fibres at that location, and when the load is removed from the lever these fibres contract and resume their original length. Hence, BC also measures the resist- ance to extension. The resist- ance of the lever to bending, therefore, is in proportion to the sum of the extensions. The extensions of that por- tion of the lever occurring be- tween the lines A G and BC is measured by the sum of the lengths of all the fibres within the space ABCG. The average length of these fibres will be that of the one at the middle, and the number of fibres is measured by CG, the width they occupy. The sum, there- fore, of the lengths of all the fibres will be equal to the area of the figure ABCG. Again, the sum of the lengths of all the fibres between the lines BC and DE is equal to the area of the figure BDEC; so in each of the other figures into which the triangle AGF is divided a similar result is found. From this we conclude that the sum of the lengths of all the fibres exposed to tension is equal to the area of the whole triangle AGF] and, therefore, that the resistance of the lever is in proportion to the area of this triangle. 294. Reitance to Extension in Proportion to the Number of Fibrc and their Distance from Keutral Line. In the measure of the extensions, we have the reaction or power of resistance ; but there is still another fact connected RESISTANCE OF FIBRES TO EXTENSION. 223 with the act of bending which needs consideration. The power of a fibre to resist deflection will be in proportion to its distance from F, the location of the neutral line ; or, to the leverage with which it acts, as was shown in Figs. 8 and 9. Thus at AG a fibre will resist more than one at DE, while farther down, each fibre resists less until at F, where there is no leverage, the power to resist entirely disappears. It may, therefore, be concluded that the power of each fibre to resist is in proportion to its distance from F ; and adding this power of resistance to that before named, we have, as the total resistance, the sum of the products of the lengths of the several fibres into their respective distances from F. 295. Illustration. As an illustration of the above, we may find an approximate result thus : Let the line FG, Fig. 63, be divided into any number of equal parts, and through these points of division draw the lines BC, DE, etc., parallel with AG. These lines will divide the triangle into the thin slices ABCG, BDEC, etc. Now, the resistance of the top slice, ABCG, will be approx- imately equal to its area into its distance from F\ or, if CG, the thickness of the slice, be represented by t, and the average length of fibres in the slice, \(A G + BC\ by b,, then the area of the slice will equal b t t ; and, if a t be put for FG, the average distance of the slice from F will be a t \t\ and therefore the resistance of the top slice will be R = *X,-iO In like manner, if c t be put for the average length of the fibres of the second slice, we shall have, to represent its resistance, 224 RESISTANCE TO FLEXURE. CHAP. XIV. For the third we shall have R = Thus, obtaining the resistance of #// the slices and adding the results, we have the total resistance. 296. Summing up tlic Reistaiicc of the Fibre. To make a general statement, let x be put for the distance from F to the middle of the thickness of any one of the slices into which the triangle is divided, and let r, a constant, be the length of an ordinate, as DE, located at the distance unity from F. Then we have by similar triangles the proportion I : r : : x : xr and therefore xr will equal the breadth of the slice at any point distant x from F, or putting x equal to the dis- tance from F to the middle of the slice, then xr will, be equal to the average length of the fibres of the slice. The resistance then of one of the slices, say the top slice, will be For the top slice, x=a,%t, therefore (a l J/)V/ = R Again; for the second slice, x a t %t therefore For the third slice we have In like manner we obtain the resistance of each successive slice, each result being the same as the preceding one, ex- cepting the fractional coefficient of /, which differs as shown, the numerator increasing by the constant number 2. When SUMMING THE RESISTANCES OF FIBRES. 225 n represents the total number of slices, then the last result or the resistance of the last slice will be and the sum of all the resistances, or R, + R u -f R ni + etc. +R n = M will equal the total resistance of all the fibres, thus : M = (a t \tjrt + (a^tjrt + etc. + (a %2n itfrt M= rt -i/ + - Now the number of slices multiplied by the thickness of each will equal FG, or nt = a , from which / = ', and, n by substituting this value, X / i \ 2ni a.\t a\- a .( i 1 = a. - ' n ' \ 2nj ' 2n and (a, -\tj = a^^- therefore 4# M=rt -V,^z3)! + etc. + ^=E M = --[(2- 1) 3 + (2n 3) 2 + etc. 4- Now, (2 i) 3 = 4^ i x (2n 3)' = 4* 3 x 4;* + 9 (2 5) a = 4^5 x 4^ + 25 To get the sum of these, we have, first, for the sum of the first terms, n x 4n* = 4^'. 226 RESISTANCE TO FLEXURE. CHAP. XIV. The coefficients of the second terms, namely, i, 3, 5, etc., equal in amount the sum of an arithmetical series com- posed of these odd numbers; or, n 2 (Art. 200), and hence the sum of these several second terms is n 2 x ^n = 4n 3 . The first and second terms summing up alike cancel each other, and we have but the third terms remaining. The sum of these is that of the squares of the odd numbers i, 3, 5, etc., and our last formula becomes M = ^ [i 2 4- 3* + 5 2 + etc. + (2n- 1) 2 ] a. , a*rt afr . Now, / = - and '--? -' , therefore 297. True Value to wliicli these Results Approximate. As an example to test this formula, let n 3, then Again, let n = 4, then and if n = 5, then If n = 10, then EXACT AMOUNT OF RESISTANCE. 22? If n 20, then Reducing these five fractions to their least common denominator, 43,200, we have When n = 3, the numerator = 14,000 n= 4, " " = H,i75 n = 5, " " = J 4,256 n = 10, " " = 14,364 n 20, " " = 14,391 It will be noticed that these numerators increase as n in- creases, but not so rapidly. As n becomes larger, the in- crease in the numerator is more gradual, but still remains an increase, for however large n becomes, the numerator will still increase, until n becomes infinite, when its limit is reached. This limit is equal in this particular case to 14,400, or one third of 43,200, the denominator ; or, in general, the value of the fraction tends towards ^ and M = %afr 298. True Value Defined by the alculu. This definite result is reached more easily and directly by means of the calculus. Taking the notation of Art. 296, we have, for the resist- ance of one of the slices, the expression This gives the resistance for a slice at any distance, x, from F, and if the thickness of the slice be reduced to the smallest conceivable dimension, then /, its thickness, may 228 RESISTANCE TO FLEXURE. CHAP. XIV. be taken for the differential of x, or dx y and we have as the differential of the resistance dR = x*rdx from which, by integration, is obtained (Art. 463) and when the result is made definite by taking the integral between limits, or between x = o and x= a t , we have R = a ?r (108.) 299. Sum of the Two Resistances, to Extension and to Compression. The foregoing discussion has been confined to the resistance offered by that portion of the lever the fibres of which suffer extension. A similar result may be obtained from a consideration of the resistance offered by the remaining fibres to compression. If c be put to represent the depth of that part of the beam in which the fibres are compressed, then it will be found that the resistance to compression will, from (108.), be equal to R = tfr (109.) and the total resistance offered by the lever will be \ar -f tfr = It may be shown also, by a farther investigation, that in levers suffering small deflections, or when not deflected be- yond the limits of elasticity, a t = c y or the neutral line is at the middle of the depth. In the latter case, we have ttj = c = -J</, and therefore RESISTANCES TO EXTENSION AND COMPRESSION, - 22Q 300. Formula for Deflection in Levers. The above is the result in a lever one inch broad. A lever two inches broad would bear twice as much ; one three inches broad would bear three times as much ; or, generally, the resistance will be in proportion to the breadth. We have then for a lever of any breadth (110.) This expression gives the resistance to the deflecting energy, which is (Art. 286) equal to PN 3 . This power, PN 3 , however, not only overcomes the resistance, ^rbd 3 , but in the act also accomplishes the deflection ; moves the lever through a certain distance. Representing this distance by eJ, we have, as the full measure of the work accom- plished, d x -^rbd 3 . When the power and the work are equal, we have PN 3 = -i^rbd 3 from which, PN 3 301. Formula for Deflection in Beams. The expression (111.) is for a semi-beam or lever. When a full beam, sup- ported at each end, is deflected by W, a weight located at the middle, we have to consider that for P we must take %W, and for N take \L (see Art. 35). These alterations will produce WL* = for the deflection of a beam supported at both ends and loaded in the middle. 230 RESISTANCE TO FLEXURE. CHAP. XIV. 302. Value of F, the Symbol for Resistance to Flexure. In formula (112.) the dimensions are all in inches. As it is more convenient that .the length be taken in feet, let / represent the length in feet, then y^=/, L=i2l and L 3 = ~i2l 3 = 172*1* By substitution in formula (112.) we have 1728 J^7 5 _ 1296*07* ~ rbd 3 Wl 1296 M'd The symbol r is a measure of the extension, differing in different materials, but constant, or nearly so, in each. Putting for - ^ the letter F we have The respective values of F for several materials have been obtained by experiment, and may be found in Table XX. Its value in each case is that for a beam supported at each end, and with the load in pounds applied at the middle of /, the distance in feet between the bearings ; while b, d and 6 are in inches ; 6 being the deflection within the limits of elasticity. 303. Coniparion of F with E 9 the Modulus of Elas- ticity. The common expression for flexure of beams when laid on two supports and loaded at the middle is [Tate's Strength of Materials, London, 1850, p. 24, formula (49*)] COEFFICIENT OF ELASTICITY. 231 in which E represents what is termed the Modulus or Co- efficient of Elasticity, which is (same work, p. 3), " that force, which is necessary to elongate a uniform bar, one square inch section, to double its length (supposing such a thing possible) or to compress it to one half its length" ; and / represents the Moment of Inertia (Arts. 457 to 4-63) of the cross-section of the beam. In this expression the dimensions are all in inches. To change L to feet we have = / equals the length in feet, or L 3 = i27*= i;28/ 5 . Substituting this value in (114-} we obtain 1728 Wl 3 $6Wl 3 6 or _ 36 In formula (113.} we have F^- 1 Multiplying this by 12 gives 48^7 El E Wl s Wl and since -fabd* I (see Art. 463) Comparing this with above value of ^ we have jg -- = \2F or E 232 RESISTANCE TO FLEXURE. CHAP. XIV. 304. Relative Value of F and E. In Table XX. the value of F for wrought iron is, from experiments on rolled iron beams, 62,000. Then 432^ = E 432 x 62000 = 26784000, cquab the modulus of elasticity for the wrought-iron of which these beams were made. They were of Ameri- can metal. Tredgold found the value for English iron to be = 24,900,000; and Hodgkinson from 19,000,000 to 28,000,000. An average of the results in seven cases gives 25,300,000 as the modulus of elasticity for English wrought-iron. 305. Comparison of F with E common, and with the E of Barlow. Barlow, in his " Materials and Con- struction/' p. 93, foot-note (Ed. of 1851), uses the expression L 3 W L 3 W -T-TS* = E, instead of TT^ > f r a lever loaded at one end ; UW ' and on p. 94, - f- h iiT = 1 ne dimensions are all in inches. Changing the length to feet, we have bd't E wr 108 Comparing this with (113.), which is ld'6 we have - '$- = F, or ioSF=E. We found before (Art. IOo 303) that E = 432^, and since 4 x 108 = 432, therefore COMPARISON OF CONSTANTS. 233 the E of Barlow equals one quarter of the E in common use, and his values of E are equal to 108 times the values of F as given in this book. For example ; on p. 147, in an experiment on New Eng- land fir, he gives, by an error in computation, E = 54780x5, but which, corrected, equals 373326. Dividing this by 108 as above, gives By reference to Table XX. we find that for spruce, the wood most probably intended for New England fir, F = 3500 Again; taking Barlow's four experiments on oak, p. 146, and correcting the arithmetical errors, we have E = 361758, 482344, 291227 and 242860. This gives an average of 344547, and dividing it by 108 as above, we have F= 3190 By reference to Table XX. we find that by my experi- ments F = 3100 306. Example under the Rule for Flexure. To make a practical application of the rule in formula (113.\ let it be required to find the depth of a white pine beam 10 feet long between bearings and 4 inches broad; and which, with a load of 2000 pounds at the middle of it^length, shall be deflected 0-3 of an inch. 2 34 RESISTANCE TO FLEXURE. CHAP. XIV. We obtain from (113.) or, in this case, Wl 3 d 3 - ~ Fbd d 3 200 x IC)8 2900 x 4 x o 3 2000000 or the depth should be 8-31, say 8 inches. QUESTIONS FOR PRACTICE. 307. How may the resistance of a fibre to extension be measured when the elasticity remains uninjured ? 308. In a beam exposed to transverse strain, what is the resistance to extension in proportion to ? 309. When the bending energy and the resistance of a beam are in equilibrium, what is the expression for this relation ? 310. Given a white pine beam 20 feet long, 6 inches broad arid 12 inches deep, and loaded with 1000 pounds at the middle. What will be the deflection, the value of ' F being 2900 ? CHAPTER XV. RESISTANCE TO FLEXURE LIMIT OF ELASTICITY. ART. 311. Rule for Rupture and for Flexure Com- pared. The rules for determining the strength of materials differ from those denoting their stiffness. The former are more simple ; all their symbols being unaffected except one, and this only to the second power, or square ; in the latter, two of the symbols are involved to the third power or cube. Many, in determining the dimensions of timbers exposed to transverse strains, are induced, by the greater simplicity of the rules for strength, to use them in preference to those for stiffness, even when the latter only should be used. A beam apportioned by the rules for strength will not bend so as to strain the fibres beyond their elastic limit, and will therefore be safe ; but in many cases the beam will bend more than a due regard for appearance will justify. When timbers, therefore, as those in the ceiling or floor of a room, might deflect so much as to be readily percep- tible, and unpleasant to the eye, they should have their dimensions fixed by the rules for stiffness only. 312. The Value of a, the Symbol for Safe Weight. In order that the symbol a in the rules for strength, denoting the number of times the safe weight is contained in the breaking weight, may be of the proper value to preserve the fibres of the timber from being strained beyond the elastic 236 RESISTANCE TO FLEXURE LIMIT OF ELASTICITY. CHAP. XV. limit, a few considerations will now be presented showing the manner in which this value is ascertained. In T^. 64 let ABCD represent a lever with one end, AD, imbedded in a wall, AD being the face of the wall, and car- rying at the other end, BC, a weight P\ the weight de- flecting the lever from the line AE to the extent EB. The line FH is the neutral line, and FG is drawn at right angles to FH. As in Figs. 58 and 63, so here the triangle AFG shows the elongation of the fibres in the upper half of the beam, and AG the elongation to the limits of elasticity of the fibres at the upper edge AB. The triangle AFG is in pro- portion to the triangle ABE, as shown in Art. 281. If AB=N (this being a semi-beam), and e equals the exten- sion per unit of N, then A G = eN. We have by similar triangles AF : AB :: AG : EB Then if AD = d and EB = d \d : N : : eN : 6 = 2eN* FIG. 64. The dimensions here are all in inches. To change N in inches to n in feet, we have N = n, NlIn and </V* = 144;** MEASURE OF EXTENSION OF FIBRES. 237 from which and from this we obtain dd e = 288^ in which d is the deflection when at the limit of elasticity, and in which e, d and <? are in inches, and n in feet. This is for a semi-beam, and it will be perceived that the deflection EB, in Fig. 64, caused by the weight P, is pre- cisely the same as would be produced in a full beam by dou- ble this weight placed at Z>, the beam being in a reversed position. When, therefore, / equals the length of the full beam in feet, n will equal \l . Substituting this value of n in the above expressions, we have d (116 > and for the value of e, dd In Art. 302 we have, for the stiffness^ of materials, formula (113.), F^W* For (5 substitute *-j , its value as just found, and, in order to distinguish the weight used to produce flexure from that used to produce rupture, let us for the moment indicate the former by , and the latter by W. Then, 238 RESISTANCE TO FLEXURE LIMIT OF ELASTICITY. CHAP. XV. from the above, Gl 3 = a Gl j2Fbd 2 e The relation between F, the measure of the elasticity of materials, and >, the resistance to rupture, may be put thus : F B : F : : i : m = -^ ; or, F = Bm > Substituting this value for F in the above, we have Gl = J2Bmbd*e Gl = Bbd* Wl Now the formula lor strength, B ,-j^, ^form. (10.) in Art. 36] gives Wl Bbd s ; a comparison of this value of Bbd s with that above shown gives Gl 7 2 em = Wl Since G is the deflecting weight which bends the lever to the limit of elasticity, it is therefore the ultimate weight which may be trusted safely upon the beam, and as a is a symbol put to denote the number of times G is contained in W y the breaking weight, therefore W G : W : : I : a = ~ and Ga = W (jr Substituting this value for W in the above, we have Gl == Gal ~ MEASURE OF SYMBOL FOR SAFETY. 239 p As above found, m = -5- , therefore Z) From this expression the values of a for various ma- terials have been computed, and the results are to be found in Table XX. 313. Rate of Deflection per Foot Length of Beam. The value of a as just found is based upon the elasticity of the material, and is measured by this elasticity at its limit. This limit is that to which bending is allowable in beams apportioned for strength. In beams required to sustain their loads without bending so much as to be perceptible or offen- sive to the eye, the bending is generally far within the elastic limit. The deflection in these beams is rated in proportion to the length of the beam ; or, when r in inches equals the rate of deflection per foot in length of the beam, then rl= rf. The deflection by formula (116.) is d therefore r i r = This gives r at its greatest possible value, and shows that it should never exceed 72 times the ratio between the length and depth, multiplied by e ; e being the measure of extension as recorded in Table XX. The ratio between 240 RESISTANCE TO FLEXURE LIMIT OF ELASTICITY. CHAP. XV. the length and depth is to be taken with / in feet and d in inches. The value of r as required in beams of the usual pro- portions and deflection, will not be as great as that here shown to be allowable. In cases where the rate of deflection, r, is as great as 0-05 of an inch per foot, and the length of the beam is short in comparison with the depth (say -j- is as small as - V then there will be danger of r exceeding the limit fixed by this rule. When the fraction -7- is less than - then the rate r should be tested to know whether it has exceeded the proper limit. It is seldom, however, that a beam 7 inches high is used shorter than 5 feet, or one 14 , inches high shorter than 10 feet. Generally the num- ber of feet in the length exceeds the number of inches in the depth. 3(4. Rate of Deflection in Floors. The rate of deflec- tion allowable so as not to be unsightly is a matter of judg- ment. Tredgold, in his rules for floor beams, fixed it at ^ of an inch per foot of the length, or 0-025. This is thought by some to be rather small, especially since in floors the limit of the rate is seldom reached ; in fact never, except when the floor is loaded to its fullest capacity, a circumstance which occurs but seldom, and then only for a limited period. For this reason, it is proper to fix the rate at say - s , or 0-03 of an inch per foot. With this as the rate for a full load, the usual rate of deflection under ordinary loads will probably not exceed o-oi or 0-015. In the rules, the symbol r is left undetermined, so that the rate may be fixed as judg- ment or circumstances may dictate in each special case. QUESTIONS FOR PRACTICE. 315. What is the distinction between the rules for strength and those for stiffness! 316. What expression shows <?, the deflection at the elastic limit? 317. What expression gives the measure of extension at the elastic limit ? 318. What expression shows the ultimate value of a, the factor of safety ? 319. What expression gives the ultimate value of r, the rate of deflection ? CHAPTER XVI. RESISTANCE TO FLEXURE RULES. ART. 320. Deflection of a Beam, with Example. The formula (113.) for the deflection of beams supported at each end and loaded at the middle, is E* __ . from which, d = -^7-75 (120.) This is the deflection of any beam placed and loaded as above. For example: \Vhatisthedeflectionofawhitepine beam of 4 x 9 inches, set edgewise upon bearings 16 feet apart, and loaded with 5000 pounds at the middle ; the value of F being 2900, the average of experiments, the results of which are recorded in Table XX. ? The deflection in this case will be 5000 x i6 3 50 x 1024 = 2-4218 3 2900 x 4 x 9 29 x 729 This is a large deflection, much beyond what would be proper in a good floor, for at 0-03 inch per foot of the length of the beam, the rate of deflection adopted (Art. 314), we should have 6 = 16 x 0-03 == 0-48 or, say half an inch, whereas the 5000 pounds upon this PRECAUTIONS IN REGARD TO CONSTANTS. 243 beam produces five times this amount. Although so greatly in excess of what a respect for appearance will allow, it is still, however, within the limits of elasticity, as will be seen by the use of formula (116.), in which we have Obtaining from Table XX. the average value of e, equal 0-0014, we have 72 x 0-0014 x i6 a d = - - = S x 0-0014x256 = 2-8672 as the greatest deflection allowable. 321. Precautions as to Values of Constants F and e. - The above is the ultimate deflection within the limits of elasticity, and is 0-4454 in excess of the 2-4218 produced by the 5000 pounds. In general, it would be undesirable to load a beam so heavily as this, or to deflect it to a point so near the limit of elasticity, and, unless the timber be of fair quality, would hardly be safe. Some pine timber would be deflected by this weight much more than is here shown in fact, beyond the limits of elas- ticity. In the above computation, F was taken at 2900, the average value, and the measure of elasticity, e, was taken at 0-0014, also the average value ; whereas, had these constants been taken at their lowest value, such as pertain to the poorer qualities of white pine, and in which F= 2000 and ^ = 0-001016, the limits of elasticity would have been found at a trifle over 2 inches, while the deflection would have reached 3^ inches. 244 RESISTANCE TO FLEXURE RULES. CHAP. XVI. 322. Values of Constants J^ and c to foe Derived from Aetual Experiment in Certain Cases. For any important work, the capacity of the timber selected for use should be tested by actual experiment. This may be done by submit- ting several pieces to the test of known weights placed at the centre, by increasing the weights by equal increments, and by noting the corresponding deflections. From these deflec- tions, the specific values of F and e for that timber may be ascertained ; and with these values the timber may be loaded with certainty as to the result. In the absence of a knowledge of the elastic power of the particular material to be used, a sufficiently wide margin should be allowed, in order that the timber may not be loaded beyond what the poorer kinds would be able to carry safely. 323. Deflection of a L,ever. The rule for deflection, as discussed in these last articles, is appropriate for a beam sup- ported at both ends and loaded in the middle. A rule will now be developed for a semi-beam or lever; a timber fixed at one end in a wall,, and with a weight suspended from the other. The deflection in this case is precisely the same as that produced by twice the weight, laid at the middle of a whole beam, double the length of the lever, and supported at each end. Let the weight at the end of the lever be represented by P, and the length of the lever by n, then W of formula wr (120), which is d = -pi~r 3 **i equal 2P, and / will equal 2-n, and we have, by substituting these values for W and / Fbd DEFLECTION WITHIN THE ELASTIC LIMIT. 245 324. Example. The deflection above found is that pro- duced in a lever by a weight suspended from its free end. As an example : What would be the deflection caused by a weight of 1500 pounds suspended from the free end of a lever of Georgia pine, of average quality, 3x6 inches square and 5 feet long ? Here we have P= 1500, n= $, F= 5900, b = 3 and d = 6 ; and therefore 16 x 1500 x 5 3 80 x 125 6- 4- = - ^ = 0-7847 5900 x 3 x 6 59 x 216 325. Tet by Rule for Elastic Limit in a Lever. To test the above, to ascertain as to whether the deflection is within the limits of elasticity, take / = 2n = 10, and by formula (116.) we get 72d 2 72 X 0-00109 X I0 2 d = - L r = L - ~~g~~ -=12x0-109=1.308 This is satisfactory, as it shows that the lever has a de- flection (0-7847) of not much more than half that within the elastic limit (i -308), and therefore a safe one. 326. Load Producing a Given Deflection in a Beam. By inversions of formulas (120.) and (121), we may have rules for ascertaining the weight which any beam or level will carry with a given deflection. First ; for a beam, we take formula (120.) Wl 3 ~~Fbd s and have 246 RESISTANCE TO FLEXURE RULES. CHAP. XVI. 327. Example. Foran example: What weight upon the middle of a beam of spruce, of average quality, 5 inches broad, 10 inches high, and 20 feet long between the bear- ings, will produce a deflection of 0-03 inch per foot, or 0-6 inch in all? Here we have F= 3500, = 5, d 10, 6 = 0-6 and / 20; therefore 3500 x 5 x io 3 x 0-6 10500 W.**.r -53- - = -g =1312.5 328. Load at the Limit of Elasticity in a Beam. Again : What weight could be carried upon this beam if the deflec- tion! were permitted to extend to the limit of elasticity ? Formula (116.) gives us and from Table XX. we have the average value of e for spruce equal to 0-00098, and therefore 72 x 0-00098 x 20 2 d= ~ -^ 72 x 0-00098 x 40= 2-8224 Substituting this new deflection in the former statement, we have W= 25 UJIJ^.8224 = 49_39 = ^ This 6174 pounds for good timber would be a safe load, but if there be doubts as to the quality, the load should be made less according to the lower Values of F and e. DEFLECTION AT THE ELASTIC LIMIT. * 247 329. Load Producing a Given Deflection in a Lever- Example. Second ; for a lever, we take formula i6/V 6= and find by inversion . An application of this rule may be shown in the answer to the question : What weight may be sustained at the end of a hemlock lever, 6 inches broad and 9 inches high, firmly imbedded in a wall, and projecting 8 feet from its face ? The hemlock is of good quality, and the deflection is limited to i inch. Here we have ^=2800, b = 6, d = 9, d=i, and n = 8 ; therefore _ 2800 x 6 x 9 3 x i _ ~H6~^V~ that is, 1495 pounds at the end of the lever would deflect it one inch. 330. Deflection in a Lever at the Limit of Elasticity. What deflection in this lever would mark the limit of elas- ticity ? Formula (110.) is Taking / at twice n we have /=: 16, ^==9, and 0-00095 ; and as a result 72x0-00095 x i6 2 os=v~ ^ -==8x0.00095x256=1.9456 248 . RESISTANCE TO FLEXURE RULES. CHAP. XVI. 331. Load on Lever at the Limit of Elasticity. What weight would deflect this lever to the limit of elasticity ? For this we have 2800 x 6 x o 3 x i -9456 p =~ - This is nearly double the weight required to deflect it one inch, as before found ; and the deflection is also nearly double. The weight and the deflection are directly in pro- portion. If 1500 pounds deflect a beam one inch, 3000 pounds will deflect it two inches. 332. Value of W, I, 6, d and 6 in a Beam. By a proper inversion of the formulas for beams, any one of the dimensions may be obtained, provided the other dimensions and the weight are known. Thus we have (form. 3 and from this find the length, the breadth, b = ^ (125) and the depth, d = and, as in formula (120) t Wl 9 the deflection, <J = -757-73 rba DEFLECTION DIMENSIONS OF BEAM. 249 333. Example Value of I in a Beam. Take an exam- ple under formula (124-}- What should be the length of a beam of locust of average quality, 4 inches broad and 8 inches high, to carry 5000 pounds at the middle, with a deflection of one inch ? In formula (124} ^=5050, = 4, d=%, 6 = i and W= 5000; hence ___ x 4 x 8 3 x i =12-74 5000 or the answer is \2\ feet. 334. Example Value of b in a Beam. As an example under formula (125.}, let it be required to know the proper breadth of an oak beam of average quality. The depth is 6 inches and the length 10 feet. The load to be carried is 500 pounds placed at the middle, and the deflection allowed is 0-3 inch. In this case, ^=500, / 10, ^=3100, d= 6 and 6 = 0-3 ; and by substitution 500 x io 3 _ 50000 _ r ~~~6 3 x 0-3" ~ 2^088 ~ or 2\ inches for the breadth. 335. Example Value of d in a Beam. As an example under formula (126.}, find the depth of a beam of maple of average quality, which is 5 inches broad and 20 feet long, and which is to carry 3000 pounds at the middle, with one inch deflection. Here we have F $i$o, W 3000, /=2O, =5 and <J = i ; and hence // 3OOO X 20 3 </=r = 0.768 5i$oxsx i or a depth of Qf inches. 250 RESISTANCE TO FLEXURE RULES. CHAP. XVI. 336. Values of r, n, l>, a and 6 in a Lever. The rules for the quantities in a semi-beam or lever are derived from formula (12 '!), which is i6/V and are as follows : The load, d = Fbd 3 P = Fbd 3 6 i6n 3 The length, n = V . The breadth, b = d _ Vi6/V Fbd (123) (127) . (128) (129) 337. Example Value of n in a Lever. As an example under (127) : What length is required in a semi-beam or lever of ash of average quality, 3x7 inches cross-section, and carrying 200 pounds at the free end, with a deflection of half an inch ? In this example, P= 200, ^=4000, = 3, d= 7 and 6 = o- 5 ; and we have _ 1/4000 x 3 x 7 3 x p. 5 = 16x200 or the length is to be 8 feet J\ inches. _ 338. Example Value of 6 in a Lever. Under formula (128) : What is the proper breadth for a lever of hickory of average quality, 3 inches deep, projecting 4 feet from DEFLECTION DIMENSIONS OF LEVER. 25 I the wall in which it is fixed, carrying a load of 200 pounds at the free end, and having a deflection of one inch ? In the formula, F 3850, d^, 6= i, p 200 and n = 4. Substituting these values, we have > i6x 200 x4 3 _ "7- [ ' 97 The breadth must be 2 inches. 339. Example Value of d in a Lever. What must be the depth of a bar of cherry of average quality, i^ inches broad, projecting 3 feet from the wall in which it is im- bedded, and carrying at its end a load of 100 pounds, with a deflection of f of an inch ? Here P 100, n = 3,.F=2%$o, =1-5 and d = o-75; and formula (129.) becomes d = V' l6xIOQX 3 3 = 2 2850 x i -5 x 0-75 The depth required is 2f inches. 340. Deflection Uniformly Distributed Load on a Beam. The cases hitherto considered in this chapter have all had the load concentrated either at the middle of a beam or at the end of a lever. When the weight is distributed equably over the length of the beam or lever, the deflection is less than when the same weight is so concentrated. In comparing the values of the deflecting energies pro- ducing equal deflections in the two cases, we have [formula (511), p. 477, of " Mechanics of Engineering and Architec- ture," by Prof. Moseley, Am. ed. by Prof. Mahan, 1856, 2$2 RESISTANCE TO FLEXURE RULES. CHAP. XVI. and changing the symbols to agree with ours], for a beam loaded at the middle, WL 3 o and [formula (530.}, p. 484, same work], for a beam uni- formly loaded, UD Comparing these two equal values of eJ, we have WL 3 UU Dr ' or, with equal deflections, the weight at the middle of the beam is equal to f of the uniformly distributed load. Thus, 100 pounds uniformly distributed over the length of a beam will deflect it to the same extent that 62^ pounds would were it concentrated at the middle of the length. Then, since U represents a uniformly distributed load, % U will equal the W of formula (120. \ which formula is Wl 3 Substituting the value of W, as above, and transposing, we have for the relation of the elements in the deflection jof a beam by a uniformly distributed load. DIMENSIONS OF BEAM LOAD DISTRIBUTED. 253 341. Value of U, 1 9 b, d and 6 in a Beam. By in- versions of formula (130.) we have the following rules namely : The weight, U = ^^ (131.) The length, / = The breadth, b = ^ (133.) The depth, d = *^j^ (134.) The deflection, 6 = L^ (135) 34-2. Example Value of U, the Weight; in a Beam. In a spruce beam of average quality, 20 feet long between bearings, 4 inches broad and 12 inches deep : What weight uniformly distributed over the beam will deflect it 2 inches ? In this example, F = 3500, b = 4, d = 12, 6 = 2 and /= 20; and by formula (131.) 20 or the weight required is 9677 pounds. 343. Example Value of , the Length, in a Beam. In a 3 x 10 white pine beam of average quality : What is the proper length to carry 6000 pounds uniformly distributed, with a deflection of 2 inches? 254 RESISTANCE TO FLEXURE RULES. CHAP. XVI. Here F = 2900, b 3, d = io, 6 2 and U = 6000 ; and by the substitution of these in formula (132.} A/2QOO =Y - =16-68 _ x 6000 or the required length is 16 feet 8 inches. 344. Example Value of &, the Breadth, in a Beam. Given a beam of average quality of Georgia pine, 20 feet long and 10 inches deep. If this beam carry a uniformly distributed load of 8000 pounds, with a deflection of if inches, what must be the breadth ? We have, as values of the known elements, U = 8000, 120, F= 5900, d 10 and 6=1.75; an d formula (133.) gives us 5 x 8000 x 2o 3 = 8 x 5900 x io 3 x i~7s =3-874 The breadth must be 3 j- inches. 34-5. Example Value of d, Hie Depth, in a Beam. A girder of average oak, 8 inches broad, and io feet long between bearings, is required to carry 10,000 pounds uni- formly distributed over its length, with a deflection not to exceed -^ of an inch. What must be its depth ? The elements of this case are U =. 10000, / = io, F= 3100, b = 8 and d = 0-3. Applying formula (13 4-) we find or we must make the depth 9^ inches. DEFLECTION LOAD DISTRIBUTED. 255 346. Example Value of (5, the Deflection, in a Beam. We have a 3 x 6 inch beam of hemlock of average quality, 10 feet long. What amount of deflection would be produced by 3000 pounds uniformly distributed over its length ? 7=3000, / 10, F = 2800, b 3 and d 6; and the for- mula applicable, (13u.\ becomes 5 x 3000 x io 3 8 x 2800 x 3 x 6 3 " or a resulting deflection of i inch. 347. Deflection Uniformly Distributed Load on a Lever. For a load at the free end of a lever [Moseley's Me- chanics (cited in Art. 340), formula (509. \ p. 476, changing the symbols] we have 6 = and [page 482, same work, formula (525.)~\ for a lever with a uniformly distributed load, we have 6 = Comparing these equal values of 6 we have PN 3 UN 3 '- TEI Dr ' u 8 or, the deflection by a uniformly distributed load is equal to that which would be produced by f of that load if suspended from the end of the lever. 256 RESISTANCE TO FLEXURE RULES. CHAP. XVI. 348. Values of C7, n, b, d and fi in a Lever. In for- mula (123.), which is P g-y-, we have the relations exist- ing between the elements involved in the case of a lever under strain. If the weight uniformly distributed over the length of the lever be represented by /, then P = -f U and formula (123.) becomes and from this we have the following: The weight, V = ~r (136.) The length, n = \ -^r ( 137 ^ The breadth, b = -^/^ (138.) The depth, d = The deflection, d = -^ s (140.) 349. Example Value of U, the Weight, in a Lever. In a Georgia pine lever of average quality, 6 inches broad and 10 inches deep, and projecting 10 feet from the wall in which it is imbedded : What weight uniformly distributed over the lever will deflect it 2 inches? In this example, F= 5900, b = 6, d=io, 6 = 2 and n = 10 ; and by formula (136. \ TT 5ooox6x io 3 x 2 U = -= 11800 OX IO DIMENSIONS OF LEVER LOAD DISTRIBUTED. 257 or the uniformly distributed weight required is 11,800 pounds. Three eighths of this weight, or 4425 pounds, concen- trated at the free end of the lever, will deflect it the same amount, viz. : 2 inches. 350. Example Value of n 9 the Length, in a Lever. In a lever of the same description as in the last article, except as to length and load : What is the proper length to carry 8000 pounds uniformly distributed, with a deflection of 2 inches ? Here we have ^=5900, # 6, d= 10, 6=2 and U= 8000 ; and by the substitution of these in formula (137.) SQOO x 6 x io 3 x 2 3 . -- = n-383 or the required length is 1 1 feet 4^ inches. 351. Example Value of 6, the Breadth, in a Lever. Given a lever of like description as in Art. 349, except as to breadth and load. If this lever carry a uniformly distrib- uted load of 6000 pounds, what must be the breadth? We have, as values of the known elements, U= 6000, n=io, ^=5900, d= io and d=2\ and formula (138.) gives us 6 x 6000 x io 3 * = i - = 3-051 59OO X IO X2 The breadth must be 3 inches. 352. Example Value of d, the Depth, in a Lever. A lever of like description as in Art. 349, except as to depth and load, is required to carry 10,000 pounds uniformly dis- tributed over its length : What must be its depth ? 258 RESISTANCE TO FLEXURE RULES. CHAP. XVI. The elements of this case are U = 10000, n = io, F = 5900, b = 6 and 6 = 2. We apply formula (139.) and find //6x loooox io 3 d = V - ? 9 403 5900 x 6 x 2 or we must make the depth 9^ inches. 353. Example Value of d, the Deflection, in a L-ever. We have a lever of like description as that in Art. 349, ex- cept as to load and deflection : What amount of deflection would be produced by 5000 pounds uniformly distributed over its length ? /r= 5000, 7210, 7^=5900, b = 6 and d= io; and the formula applicable, (140), becomes 6 x 5000 x io a 6=- = 08475 5900 x6x io or a resulting deflection of J of an inch. QUESTIONS FOR PRACTICE. 354. Given a beam loaded at middle : What are the rules by which to find the weight, length, breadth, depth and de- flection ? 355. Given a lever loaded at the free end: What are the rules by which to find the weight, length, breadth, depth and deflection ? 356. In a beam with the load uniformly distributed: What are the rules by which to obtain the weight, length, breadth, depth and deflection? 357. In a lever with the load uniformly distributed : What are the rules by which to obtain the weight, length, breadth, depth and deflection ? CHAPTER XVII. RESISTANCE TO FLEXURE FLOOR BEAMS. ART. 358. Stiffness a Requisite in Floor Beams. The rules given in Chap. VI. for the dimensions of floor beams are based upon the ascertained resistance of the material to rupture, and are useful in all cases in which the question of absolute strength is alone to be considered. For warehouses and those buildings in which strength is principally required, the rules referred to are safe and proper ; but for buildings of good character, in which the apartments are finished with plastering, the floor timbers are required to possess stiffness as well as strength ; for it is desirable that the deflection of the beams shall not be readily noticed, nor be injurious to the plastering. 359. General Rule for Floor Beams. The relations of the several elements in the question of stiffness, in beams uni- formly loaded throughout their entire length, are found in formula (130.), Fbd't The load upon the floor beam is here represented by U, and its value is U = cfl (see Art. 92) ; in which c is the distance apart between the centres of the floor beams, / is the number of pounds weight upon each square foot of the floor, and / is the length of the beam ; c and / both GENERAL RULE FOR FLOOR BEAMS. 261 being in feet. If for U we substitute this value, and for 6 put rl (see Arts. 313 and 314), we have = Fbd 3 r (141 .) 360. The Rule modified. For the floors of dwellings and assembly rooms, f t the load per foot, may be taken (see Art. 115) at 70 pounds for the loading and 20 pounds for the weight of the materials, or 90 pounds in all; and r, the rate of deflection per foot of the length, at 0-03 (see Art. 314). Formula (141-) thus modified becomes 90 x %cl 3 = = FM . 8x0-03 = Fbd 3 cl 3 = p This coefficient, ~7> taking F at its average value for six of the woods in common use, reduces to firf =3-15 for Georgia pine, = 2-69 " locust, = 1-65 " oak, f|4f =1-87 " spruce, = 1-55 " white pine, = 1-49 " hemlock. 361. Rule for l>Aveliin^ and Assembly Rooms. For p the coefficient in (14&-), ~^ putting the symbol i, we 262 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. have this simple rule for problems involving the dimensions of floor beams in dwellings and assembly rooms, namely, cl 3 = ibd 3 (143.) and we have the value of i for average qualities of six of the more common woods, as taken in Art. 360, as follows : For Georgia pine, i = 3-15 " locust, i = 2-69 " oak, i 1-65 " spruce, i 1-87 " white pine, i= 1-55 " hemlock, i= 1-49 362. Rules giving the Values of c, J, 6 and d. Tak- ing formula (14$ ) we derive by inversions the following rules, namely : The distance from centres, c = -j- r (144-) The length, / = \ - (145.) C The breadth, b = j^ 3 The depth, d = V~ (147.) 363. Example Distance from Centres. At what dis- tance from centres should 3x12 inch Georgia pine beams of average quality, 24 feet long, be placed in a dwelling- house floor? Here we have 2':= 3 -15, = 3, d = 12 and /= 24; and by formula (144-) 3-15 x 3 x 12* _ or the distance c should be about 14^ inches. EXAMPLES OF DIMENSIONS. 263 364. Example Length. Of what length may average quality white pine beams 3 x 10 inches square be used, when placed 16 inches from centres ? In this case 2 = 1.55, = 3, d=io and c= i^; and formula (14&*) gives / - ^3487-5 = I5-I65 or these beams may be used 1 5 feet 2 inches long between bearings. 365. Example Breadth. In floor beams 20 feet long and 12 inches deep, of oak of average quality, placed one foot from centres : What should be the breadth ? Here, c i, I 20, d = 12 and 2=1-65. With for- mula (146. \ therefore, we have 1-65 x I2 3 The breadth should be nearly 2-J , or say 3 inches. 366. Example Depth. What should be the depth of spruce beams of average quality when 3 inches broad and 20 feet long, and placed 20 inches from centres? The sym- bols in this case are <r=if, /=2O, =3, and i = 1-87; and by formula (14? ) we find W^ 1-87x3 or the depth required is 13! inches. Beams 3x13 could be used, provided the distances apart from centres were cor- respondingly decreased. The new distance would be (form. 144-) !&2 instead of 20 inches. 264 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. 367. Floor Beams for Store. The several values of / for dwellings and assembly rooms, as given in Art. 361, will be appropriate also for stores for light goods, because timbers apportioned by the rules having these values of t, will bear a load of 200 pounds per superficial foot before their deflection will reach the limit of elasticity. For first-class stores those intended for wholesale busi- ness, as that of dry-goods the values of i, as above given, are too large. The proper values for this constant may be derived as below. 368. Floor Beams of First-elass Stores. The load upon the floors of first-class stores may be taken at 250 pounds per superficial foot, and the deflection at 0-04 of an inch per foot lineal (see Arts. 313 and 3I4-). Beams proportioned by these requirements will bear a load of about 3 x 250 = 750 pounds per foot before the deflection will reach the limit of elasticity. With 250 as the loading, and, say 25 pounds (Art. 99) for the weight of the materials of construction, we have 7-275. Formula (141 )> modified in accordance herewith, putting r = 0-04, becomes 5 x 2'j^cl 3 = 8 x o-o<\Fbd s 369. Rule for Beams of First-elass Stores. Reducing zp the above constant, , for six of the more common 4296^ woods of average quality, and putting the symbol k for the results, we have for BEAMS FOR FIRST-CLASS STORES. 265 Georgia pine, k = 1-37 Locust, k i- 1 8 Oak, k 0-72 Spruce, = 0-81 White pine, k = 0-67 Hemlock, = 0-65 With this symbol k, the rule for floor beams of first-class stores is reduced to this simple form, d 3 = kbd 5 (149) 370. Values of c, I, b and d. By proper inversions, we obtain from formula (149.), rules for the several values required, thus : The distance from centres, c = ^- (150.) The length, / = j/^! (151.) The breadth, b = |~ (152) The depth, d = V-^ (153) 371. Example Distance from Centres. In a first-class store : How far from centres should floor beams of Georgia pine of an average quality be placed, when said beams are 4 x 12, and 20 feet long between bearings? In this example, we have k = i -37, b = 4, d = 12 and / 20. Then by formula (150.) or the distance from centres is 1-184 f eet > equal to about inches. 266 .RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVIL 372. Example Length. At what length may 4x10 inch beams of average oak be used in the floors of a first- class store, when placed 12 inches from centres? Here we have = 0-72, b = 4, ^=10 and c\\ and by formula (151) j 4/0-72 Xx io /= - - = 14-23 or the length should be 14 feet 3 inches. 373. Example Breadth. The floor beams in a first- class store are to be 20 feet long and 14 inches deep, of white pine of average quality. When placed 12 inches from cen- tres, what should be their breadth ? Taking formula (152.} we have, as values of the symbols, c=i, I 20, k = o-6j and ^=1; and TU = 4-35 0-67 x 14 The breadth should be 4^ inches. 374. Example Depth. What should be the depth, in a first-class store, of spruce beams, of average quality, 4 inches thick and 16 feet long, and placed 14 inches from centres? In this case, we have c i|, / = 16, k = 0-81 and b = 4. Therefore, by formula (153*) d = 0-81x4 or a depth of I if inches. 375. Headers and Trimmers. In Chap. VII., in Arts. 143 to 158, rules for headers and trimmers, based upon the resistance of the material to rupture, are given. These rules STRENGTH AND STIFFNESS COMPARED. 267 contain the symbol a, which represents the number of times the weight to be carried is contained in the breaking weight. The value of this symbol may be assigned at any quantity not less than that which is given for it in Table XX., and, when made so great that the deflection shall not exceed 0-03 of an inch per foot of the length, the rules referred to will be proper for use for headers and trimmers for the floors of dwellings and assembly rooms. 376. Strength and Stiffness Relation of Formulas. The value of a, the symbol for safety, may be determined from the following considerations : Taking formula (113.), which is " bd'd and substituting G for W we have Gl 3 = Fbd 3 S A comparison of the constants for rupture (B) and for elasticity (F) shows that p B : F : : I : m = -^ or Bm = F and putting rl equal to d we have, by substitution, Gl 3 = Bmbd'rl Gl 2 = Bmbd'r dmr 268 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII, We have by formula (10.), Art. 36, Wl *w or Wl = Bbd 2 Comparing this value of Bbd* with that above, we have Gl a Wl = dmr In this formula, G is the weight which may be carried by the beam, with a deflection per foot of the length equal to r; and W is the breaking weight. Putting these symbols in a proportion, we have W G-. W:: i :a=^r G or Ga = W Substitute for W this value of it, and we obtain r j Gr Gal = -j dmr I -j ~ dmr F d- B r 377. Strength and Stiffki ess Value of <i, in Terms of B and F. The values of B and F (form. 154-} are found in Table XX., and r = 0-03. The ratio -7- ( / in feet and d in inches) cannot be exactly determined until the length and depth have been established. An approximation may be MEASURE OF THE SYMBOL FOR SAFETY. 269 assumed, however, for a preliminary calculation, and then, if found to err materially, it may be taken more nearly cor- rect in a final calculation. In all ordinary cases, the ratio -T- will be found nearly equal to | = i - 7. Taking this value in formula (IS 4-) we have 378. Example. Let us apply this in the use of formula .), namely : Wai = Bbd* What weight may be carried at the middle of a Georgia pine beam of average quality, 3 x 10 inches x 17 feet, so as to deflect it no more than would be proper for the floors of a dwelling? Here = 3, d= 10, a - ^ , ^=5900 and I = 17 ; therefore B x 3 x io 2 _ 5900 x 3 x io a yy - - jjr 17 i 770000 379. Tet of the Rule. To test the accuracy of the result just found, the same problem may be solved by formula (113.), W>_ F ''~~bd 3 d from which we have, when <? =rl, and substituting G for W> Gl* = Fbd s r _ Fbd'r and 6- j- 2 2/0 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. In this expression, in the above example, F = 5900, b = 3, d= 10, 1=17 and r = 0-03 ; and hence t 5900 x 3 x 10 s x 0-03 531000 "" ~~ =l837 ' 4 380. Rules for Strength and Stiffne Resolvable. The result in the last article is the same as the one before found, and indeed could not be otherwise, since the one formula is derived directly from the other, and is readily resolvable into it ; for if, in formula (21.), Wai = Bbd 2 we substitute for a its equivalent as in formula (154*), we have Fdr Wl* = Fbd'r so that instead of computing the value of a for use in any particular case by formula (155.), we may introduce into the rule its value as given by (154), and reduce to the lowest terms, as in the next article. 381. Rule for the Breadth of a Header. A rule for a header is given in formula (27 .\ Art. 145. Substituting for a its value as in (154), we have, taking ^U instead of J/, Art. 340, In this expression, / and g are the same, both represent- ing the length of the header, and the (d\) is put for the HEADERS FOR DWELLINGS, 271 effective depth, and is equal to the d of the first member; therefore, reducing, we have which is a rule for the breadth of a header, based upon the resistance to flexure. 382. Example of a Header for a Dwelling. In a dwelling having spruce floor beams of an average quality, 10 inches deep : What would be the required breadth of a header of the same material, 10 feet long, carrying tail beams 12 feet long? The values of the symbols are, f= 90 (Art. 115), n = 12, =10, ^=3500 (Table XX.), r = 0.03 and d=io\ and , 5 xgox 12 x io 5 b = ? 9 = 4-409 16x3500x0-03 XQ or the required breadth is 4| inches full. 383. Example of a Header in a Firt-clas Store. In a first-class store, where the beams are 14 inches deep, what is the required breadth of a header of Georgia pine of aver- age quality, 16 feet long, and carrying tail beams 17 feet long ? Here /= 275, r = 0-04 (Art. 368), n = 17, g 16, F = 5900 (Table XX.) and </= 14; and by formula (156. \ , 5x275x17x16" b = ~~ - - '- - 3= II'54I 16 x 5900 xo-O4x i3 3 The breadth should be 1 1| inches full. 272 RESISTANCE TO FLEXURE -FLOOR BEAMS. CHAP. XVII. 384. Carriage Beam with One Header. (See Art. 389.) In Art. ISO a rule (form. 29.) is given for this case, based upon the resistance of the material to rupture. As with a header (Art. 381), so here, the rule given may be resolved into one depending upon the resistance to deflection. Taking formula (29.), and for a substituting its value as per formula (154-), we have, taking / instead of /, Art. 340, ) = Fbd*r (157.) which is the required rule. 385. Carriage Beam with one Header, for I>welling. In this rule, putting f= 90 and r = 0-03, we obtain 3000 (bcl*+gn*m) = Fbd* which is a rule for carriage beams with one header, in dwellings and assembly rooms. (See Art. 389.) 386. Example. What should be the breadth, in a dwelling, of a carriage beam of average quality white pine, 20 feet long by 12 inches deep, and carrying a header 16 feet long at a point 5 feet from one end ? The floor beams among which this carriage beam is placed are set at 16 inches from centres. Here c i^, / 20, g = 16, n = 15, m 5, F = 2900 and d = 12 ; and by formula (158.) b = ^ 2900 X I2 8 The breadth should be 12} inches. CARRIAGE BEAMS WITH ONE HEADER. 273 387. Carriage Beam with One Header, for First-class Stores. If in formula (157.) we take the value of / equal to 275, and of r equal to 0-04, we shall then have 6875 (ficl*+gn % m) = Fbd* which is the required rule (see Art. 389). 388. ExampDc. Of what breadth, in a first-class store, should be a Georgia pine carriage beam of average quality, 25 feet long, and carrying at 6 feet from one end a header 16 feet long; the floor beams being 15 inches deep, and placed 15 inches from centres? Here c = ij, / = 25, g 16, n 19, m = 6, d = 15 and F = 5900 ; and formula (159.) becomes 6875 [(A x i} x 25 3 ) + (16 x iQ 2 x 6)] / J LMT - 2 - /j _ 5900 X i$ 3 or the breadth required is 14^ inches. 389. Carriage Beam with One Header, for Dwellings- More Precise Rule. The rules above given (157., 158, and 159.) are not strictly correct : they give a slight excess of material (see Art. 241). The rule shown in formula (86.), taking f /", Art. 340, is accurate* and should be the one employed in special cases * Except when h is less than n {Art. 240). In this case the result is slightly in excess, but so slightly that the difference is unimportant. 274 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. in which a costly material is used. Substituting for a in this formula, its value, as in formula (154>), we have Bl mn . A' + f U) = Fbd*r (160.) in which A' is the concentrated load, and U the uniform- ly distributed load. Formula (160.) may be modified, in the case of a carriage beam, by using for these symbols their values, thus: From Arts. 92 and 150, A' = fng, and U=$cfl, and hence fmn (ng + %cl) = Fbd*r (161.) which is a more precise general rule for a carriage beam carrying one header. If, now, we put f equal to 90, and r equal to 0-03, we shall have yxx>mn (ng 4- \cl) Fbd* , yxxmn (ng + frZ) Fd* which is a more precise rule for carriage beams with one header, in floors of dwellings and assembly rooms. 390. Example. Taking the example given in Art. 386, we have m = 5, =I5, = 16, r = ij, / = 20, ^=2900 and d = 12 ; and, in formula (162.) = I2> 2900 X I2 3 showing that by this, the more exact rule, the breadth should be \2\ inches, while by the former rule it was deter- mined to be I2j inches. PRECISE RULE FOR CARRIAGE BEAMS. 275 391. Carriage Beam witli one Header, for First-class Stores More Precise Rule. Modifying formula (161.), by putting 275 for /", and 0-04 for r, we have 6875*** (rig + Id) = Fbd* Fd" (163 1 which is the more precise rule required. 392. Example. Applying this rule to the example given in Art. 388, we find, m 6, n 19, g 16, c ij, I 25, F 5900 and d = 15 ; and hence ^6875 x6x 19(19 x i6 + f x ijx 25) _ 5900 x I5 3 giving the breadth, by this more precise rule, at 13^ inches. This is nearly half an inch less than by the former rule, which gave for the breadth, 14-073, or 14^ inches nearly. 393. Carriage Beam with Two Headers and Two Sets of Tail Beams, for Dwelling*, etc. Formula (32.) in Art. 155 gives the relations of the symbols referring to a case in which a carriage beam has to carry two headers, with two sets of tail beams. From this formula we have, taking f U, Art. 340, b TrWVfi If in this equation the value of a, as in formula (154-), be substituted, there results which is a general rule for these cases. 2/6 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. Putting/ =90 and r o-o^, we have b = ~? \_grn (mn + s 2 ) + T V/ 8 ] (165.) which is a rule for a carriage beam, carrying two headers, with two sets of tail beams, in the floor of a dwelling or assembly room. (See Arts. 402, 405, 415 and 417.) 394. Example. Under rule (165.) take the example given in Art. 156, in which F = 5900, ^=14, g =. 12, c = i-J- and /=25. For m and s there are given 5 and 15, and taking m as the larger, m =15, n = 10, s = 5 and r = 20 ; so that (165.) becomes or the breadth should be j\ inches. 395. Carriage Beam with Two Headers and Two Sets of Tail Beams, for First-elass Stores. If, in formula (164), f be put at 275 and r at 0-04, we shall have 6875 which is a rule for a carriage beam carrying two headers, with two sets of tail beams, in a first-class store (see Arts. 402, 407 and 4(7). 396. Example. Referring to the same example (Art. 156) we have F = 5900, d 14, g=. 12, m 15, n = 10, s = 5, c = i and / = 25 ; and the formula is b = 5QOx 5 i 4 3 [ 12 x I5 (I5 x 1Q + 5 2 ) + TVx H x 2 5 3 ] = 16-486 or the breadth should be i6 inches. CARRIAGE BEAM WITH TWO HEADERS. 277 397. Carriage Beam with Two Headers and One Set of Tail Beams. Formula ($4-), in Art. 157, is a rule for a carriage beam with two headers, carrying but one set of tail beams. Substituting, in this formula, for a its value /?/ (form. 154.) -- , we have, taking U, Art. 340, from which which is a general rule for a carriage beam carrying two headers, with but one set of tail beams, with a given rate of deflection. (See Arts. 402, 409, 411, 419 and 421.) 398. Carriage Beam with .Two Headers and One Set of Tail Beams, for Dwellings. If, in formula (167.), f be put at 90 and r at 0-03, we shall have b = {Jgm (n+s) + fcl*] (168.) a rule for a carriage beam with two headers, carrying only one set of tail beams, in a dwelling or assembly room. (See Arts. 402, 409, 411, 419 and 421.) 399. Example. Let it be required to find, under this rule, the breadth of a carnage beam 20 feet long, of spruce of average quality ; said beam carrying two headers, each 12 feet long, with tail beams 11 feet long between them, leaving an opening 4 feet wide on one side, and another 5 feet wide on the other side. The beams among which this carriage beam is placed are 12 inches deep and 16 inches from centres. 278 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. For the symbols we have, ^=3500, d = 12, jii, g\2, c = 1% and / = 20. Having for m and s the values 4 and 5, we make m equal to the larger one, and therefore m = 5, n 15 and s 4. These values substi- tuted in formula (168.) produce _ 3000 [i i x 12 x 5 (i 5 + 4) + A x ij x 2Q 3 ] 3500 XI2 3 - 7 ' 8;4 The breadth should be, say 7^ inches. 400. Carriage Beam with Two Headers and One Set of Tail Beams, for First-class Stores. If, in formula (167.), we put 275 for / and 0-04 for r, we shall have (169.) which is a rule for carriage beams carrying two headers, with one set of tail beams between them, in a first-class store. (See Arts. 4-02, 409 and 413.) 401. _ Example. What should be the breadth, under this rule, of a carriage beam of average quality Georgia pine, 25 feet long, with two headers each 20 feet long, carrying tail beams 10 feet long between them ? The tail beams are so located that there is an opening 10 feet wide at the left-hand end, and one 5 feet wide at the right-hand end. The tier of beams is 15 inches deep and placed 15 inches from centres. Here F= 5900, d = i$, j = 10, g = 20, ci\ and 12$. For the values of m and s we have 10 and 5; and 10 being the larger it follows that m= 10, n 15 and s = 5 ; and by formula (169.), = Ii; lg 5900 x 1 5 8 or the breadth should be 15! inches. MORE PRECISE RULES FOR- CARRIAGE BEAMS. 279 4-02. Carriage Beam with Two Headers and Two Sets of Tail Beams More Precise Rules. The rules for carriage beams given in Arts. 393 to 401 are drawn from formulas which arc but close approximations to the truth. The resulting dimensions are always in excess slightly of the true amounts, and the rules therefore are safe. The rule embodied in formula (92.), however, is deduced from exact premises, and its results are precise. If for a its value (form. 1>4*) b.e substituted in formula (92.), we shall have, taking f C7, Art. 340, (170.) and, as auxiliary thereto, -~(rs + When h is equal to or exceeds n, then n is to be substituted for h, and the portion of formula (170.) equals a' (see Art. 248), and the formula itself reduces to 280 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. Substituting for a' its value (form. 171.) we have b = - We have here, in formula (170.), a general rule, and in formula (174-), a rule, general when h equals or exceeds n, for a carriage beam carrying two headers, with two sets of tail beams, with a given deflection. 403. Example/*, les than n. Let it be shown, under these rules, what should be the breadth of a carriage beam of spruce of average quality, 20 feet long and 12 inches deep, carrying two headers each 12 feet long, so placed as to leave an opening 41 feet wide ; said opening being 7^ feet distant from one wall and 8 feet from the other. The floor is to carry 100 pounds per superficial foot, with a deflection of 0-03 per foot, and the beams are placed 15 inches from centres. Here we have /= 100, g= 12, m 8, ! = 20, n 12, s=7h r=\2^ y *=ii, d' = l-(m+s) = 20 (8 + ;) = 20 I5i 4j, ^=3500 and d= 12. Preliminary to finding the value of h we have to deter- mine the values of a' and b' . By formulas (171.) and (172.) ICO X 12 X 8 a ' '' 4x20 ( 8xI2 + 7'5 3 ) =18270 ICO X 12 X 7-5 b '-~ 4x20 , -("-S* 7-5 + )= 17746-875 a' b' = 523-125 CARRIAGE BEAM SPECIAL RULES. 28 1 From these and formula (173.} we have So h = 11-49, an< 3 since it is less than n (as n equals 12) is therefore to be retained ; and we have (form. 170.) - fV x i^x 100 x n -49x8- 51 + 17746 -875 + * ^ O * O^ L 3500 X I2 J 523.125 :~\ ^y^X(!I. 49-7. 5)J =9.714 or the required breadth is 9f inches. 404. Example h greater than n. What should be the breadth of a white pine carriage beam 20 feet long, 12 inches deep, and carrying two headers 10 feet long one located at 9 feet from one wall and the other at 6 feet from the other wall ; the floor to carry 100 pounds per foot super- ficial, with a deflection of 0-03 of an inch per foot lineal, and the beams to be placed 15 inches from centres ? Here /= 100, F 2900, d 12, r = 0-03, c i-J, 1=20 and -=10. Comparing m and s we have m = 9, n = 1 1 and s = 6. Proceeding as in the last article, we find that h exceeds n, therefore, according to Art. 402, we have formula (174>) appropriate to this case ; from which * = 2900x^x0.03 [(** i*x it xao) + io( 9 x u+6*)] = 10-140 or the breadth should be loj inches. 282 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. 405. Carriage Beam with Two Headers and Two Sets of Tail Beams, for Dwellings More Precise Rule. If, in formula (17 4>), f = 90 and r = 0-03, we shall have which is a precise rule for carriage beams carrying two headers, with two cets of tail beams, in dwellings and assem- bly rooms. (See Arts. 393 and 402.) 406. Example. An example under this rule may be had in that given in Art. 404 ; in which we have F= 2900, d= 12, c = ij, / 20, ^=10, m = 9, n n and s = 6. Then by formula (175.} 2 a Ki x r x J J x 20)4-10(9x11 +6 2 )] = 9.126 or the breadth should be 9^ inches. 407, Carriage EScam with Two Headers and Two et of Tail Beams, for First-class Stores More Precise Rule. If, in formula (174.), f 275 and r = 0-04, we shall have V J-, J Q Fd* which is a precise rule for carriage beams carrying two headers, with two sets of tail beams, in first-class stores. (See Arts. 395 and 402.) 408. Example. What should be the breadth, under this rule, of a carriage beam of Georgia pine of average quality, 23 feet long, 14 inches deep, carrying two headers each 17 feet long, with tail beams on one side 7 feet long, and on the other 10 feet long ; the beams being placed 14 inches from centres ? CARRIAGE BEAMS FOR FIRST-CLASS STORES. 283 Here ^=5900, d 14, c = i%> / = 23 and g\T. Taking the larger of the two, 10 and 7, for ;, we have m = 10, n = 13, and s = 7 ; and by formula (176.) 14. the breadth should be, say 14! inches. = 14-774 409. Carriage Beam with Two Headers and One Set of Tail Beams More Precise Rule. In a case where there are two openings in the floor, one at each wall, then the two headers carry but one set of tail beams, and these are be- tween the headers. The load at each header is the same ; and when g equals the length of header, j the length of tail beams, and / the load per superficial foot, then the load at each end of each header is W=\fgj and the expression for the load at one point, as in Art. (53, wi IVi'fz -j-(Wn+Vs\ becomes --j--(*'+ f), and therefore (A rt. 243) (177.) and fi' = (r + w) (178.) 4/ In the case under consideration, these two expressions are auxiliary to formula (170.), in the place of those given in for- mulas (171.) and (172.), and with h equal to, or exceeding n, formula (170.) becomes 284 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. Substituting for a' its vajue, as in formula (177.) t we have (179.) which is a precise rule for carriage beams carrying two headers, with one set of tail beams, and with a given rate of deflection. (See Arts. 397, 398 and 402.) 410. Example. What should be the breadth of a car- riage beam of locust of average quality, 16 feet long and 8 inches deep, carrying two headers of 8 feet length, with one set of tail beams 7 feet long between them, so placed as to leave an opening of 6 feet width at one wall, and another of 3 feet at the other? The floor beams are placed 15 inches from centres, and are to carry 90 pounds per superficial foot, with a deflection of 0-04 of an inch per foot lineal. We have from this statement f = 90, m = 6, ;/ = 10, /= 16, r = 13, 5=3, c= ij, F= 5050, d = 8, r 7 = 0-04, g = 8 and j = 7. To test the value of h we have, preliminary thereto, formula (177.), which gives oox 8 x 7x 6 a = 2- x 10 + 3 = 6142 5 and, formula (178.), 90x8x7x3 - - V a'-V = 1653.75 CARRIAGE BEAMS FOR DWELLINGS. 285 Then, by formula (173.), As n = 10, h exceeds n. We must, therefore, substi- tute n for h ; and by formula (179.) we have or the breadth should be 5J, say 5 inches. 4(1. Carriage Beam witli Two Header and One Set of Tail Beams, for Dwellings lHore Precise Rule. If, in formula (179.), f = 90 and r' = 0-03, we shall have (180.) which is a precise rule (in cases where h exceeds n ) for carriage beams carrying two headers, with one set of tail beams, in a dwelling or assembly room. (See Arts. 398, 402 and 409.) 4(2. Example. What should be the breadth, in a dwelling, of a carriage beam of spruce of average quality, 1 8 feet long and 10 inches deep, carrying two headers of 12 feet length, with a set of tail beams between them 7 feet long? The headers are placed so as to leave an opening of 8 feet on one side and 3 feet on the other, and the beams are set 15 inches from centres. Here / = 90, g 12, j = 7, m = 8, n 10, s = 3, r = 15, / = 18, F = 3500, d 10, r' 0-03 and c = \\. 286 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII Preliminary to seeking the value of h we find, by for mulas (177.) and (17 8. \ , 90x12x7x8. " ~- = 7245 a'-b'= 3675 Now, by formula (173.), But n= 10; therefore n is to be used in the place of 7z, and formula (180.) is the proper one to use in this example. This latter formula gives us x8 ; 10x18) + (12x7x10 + 3)] =9- 417 Thus the breadth should be 9! inches ; or the beam be of x 10 inches. 4-13. Carriage Beam wittli Two Headers ancl One Set of Tail Beams, for First-elass Stores More IPrecfse RuBc. If, in formula (17 9.) t f 275 and r 0-04, we shall have which is a precise rule, when h exceeds ?z, for a carriage beam carrying two headers, with one set of tail beams, in a first-class store. (See Arts. 400 and 402.) CARRIAGE BEAM WITH TWO HEADERS. 287 i 4-14. Example. The example given in Art. 412 may be used to exemplify this rule, excepting the depth, which we will put at 14 inches instead of 10. Formulas (180.) and (181.) are alike, with the exception of the numerical constant. The result found in Art. 4-12, b = 9-417, multiplied and divided to correct the constant, will give the result required in this case. The constant 6875 is to take the place of 3000, and the depth 14 is to replace 10. With these changes, we have 6875 TOGO b 9-417 x x - - = 7-865 3000 2744 or the breadth should be 7-86; say 7J inches. 415. Carriage Beam with Two Headers, Equiclituiit from Centre, and Two Sets of Tail B earn Precise Rule. In case the opening in the floor be at the middle, leaving tail beams of equal length on either side, then the moments of the two concentrated loads upon the carriage beam are equal, or a' = b' and, in formula (170.), and the formula itself becomes in which b' represents the combined effect of the two loads, as acting at the location of either of them. This effect is shown (Art. 153) to be 288 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. In the case under consideration, W V and m = s, and therefore '= W~(n + m) = W-~ = Wm Now, W represents the weight concentrated in one end of one of the headers. The load on a header is %fgm, and the load at one end of the header is \fgrn ; therefore b' = and formula (182.) becomes *=Tfi By formula (178.) a'-b' and since in this case a' b' = o ^lt and and therefore which is a precise rule for carriage beams carrying two headers, equidistant from the centre, with two sets of tail beams, and with a given rate of deflection. (See Arts. 393, 396 and 402.) . Example. Under this rule, what should be the breadth of a Georgia pine carriage beam of average quality, 20 feet long and 12 inches deep, to carry two headers each CARRIAGE BEAMS WITH TWO HEADERS. 289 12 feet long ; the headers so placed as to leave an opening 6 feet wide in the middle of the width of the floor ? The floor beams are set 16 inches from centres, and are to carry 200 pounds per foot superficial, with a deflection of 0-04 of an inch per foot lineal. l-d' 20-6 Here /=2O, m = = - = 7; ^=12, ^=12, c = i^-, F== 5900, / 200 and ^ = 0-04; and by formula (183.) , 200 x 20 r/ . N ,.... * ;= 5900 x 12 x 0.04 * x H x 20') + (12 x 7')] = 7-402 or the breadth should be 7 inches. 417. Carriage Beams with Two Headers, Equidistant from Centre, and Two Sets of Tail Beams, for Dwellings and for First-class Stores Precise Rules. If, in formula (183.), f= 90 and r = 0-03, we shall have which is a precise rule for carriage beams carrying two headers, equidistant from the centre, with two sets of tail beams, in a dwelling or assembly room. (For an example, see Art. 418.) But if, instead, /= 275 and ? = 0-04, then we shall have which is a precise rule for carriage beams carrying two headers, equidistant from the centre, with two sets of tail beams, in a first-class store. (See Arts., 393, 395 and 402.) 2QO RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. 4-! 8. Examples. Formulas (184.) and (185.) are alike, except in the numerical coefficient. One example will therefore suffice for an exemplification of the two. Let it be required to show what, in a dwelling, should be the breadth of a carriage beam, 20 feet long and 12 inches deep, of average quality of spruce, carrying two headers 10 feet long ; these headers being so placed as to leave at the middle of the width of the floor an opening 8 feet wide. The beams are to be placed 16 inches from centres. Here we have /= 20, ;;/ = 6, g =. 10, ^/= 12, c= i$ and F= 3500 ; and by formula i )] = 5-225 or the breadth should be, say $J inches. For a first-class store this carnage beam, if of Georgia pine, would be required to be 7- 103, say 7J inches broad. This result is found by eliminating the two con- stants 3000 and 3500 in the above and replacing them by those required by the new conditions, namely, 6875 and 5900. Doing this, we find 6875 3500 = 5-225 x- -x- -=7-103 3 J 3000 5900 ' 419. Carriage fleam with Two Header, Equidistant from Centre, and One Set of Tail Beam Precise Rule. In some cases the wells or openings are at the wall on each side, and the tail beams at the middle of the floor. In this arrangement, if / equals the length of the tail beams, \fgj will equal the load at the end of one header. By Art. 415, b' Wm, from which V = Wm = CARRIAGE BEAM WITH TWO HEADERS. 29! and formula (182.) becomes and since (Art. 415) h = t = I/, therefore fckt = W By substituting this in the above, which is a precise rule for carriage beams, carrying two headers, equidistant from the centre, with one set of tail beams, the rate of deflection being given. (See Arts. 397, 398, 402, 409 and 411.) 420. Example. What should be the breadth of a car- riage beam of hemlock of average quality, 16 feet long and ii inches deep, carrying two headers, each 10 feet long, placed equidistant from the centre of the width of the floor, and having between them one set of tail beams 6 feet long? The floor beams, placed 1 5 inches from centres, are to carry 100 pounds per foot superficial, with a deflection of 0-035 of an inch per foot lineal. Here we have /= 16, m = 5, g-= 10, j = 6, df= 11, c i, F 2800, /= 100 and r= 0-035 ' an d by formula (186.) b = 28oox?i-x 1 o. 035 [(A X '* X I6 ' )+ ( ' x 6 x 5)] = 4-907 or the breadth should be 4| inches. RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. 421. Carriage Beams \viili Two Headers, Equidistant from Centre, and One Set of Tail Beams, for Dwellings and for First-class Stores Precise Rules. If, in formula (186.), f= 90 and r = 0-03, then we shall have which is a precise rule for carriage beams, carrying two headers, with one set of tail beams between them, at the middle of the floor, in a dwelling or assembly room. For an example, see Art, 4-22. But if, instead of these, /= 275 and r 0.04, we shall have which is a precise rule for carriage beams, carrying two headers, with one set of tail beams between them, at the middle of the floor, in a first-class store. 422. Example. Formulas (187.) and (188.) are alike, except in the numerical coefficient. One example will suffice to show the application of both. Take one coming under formula (187.), and in which / = 20, m = 6, g 10, j =8, d 12, =! and JF=.$$oo. Then, by the formula, = 6-415 or the breadth should be 6 inches full. 423. Beam with Uniformly Distributed and Three Con- centrated Loads, the Greatest Strain being Outside. In Art. 256, formula (96.) is a general rule for this case, but BEAM CARRYING THREE CONCENTRATED LOADS. 293 based upon the resistance to rupture. This rule may be modified so that it shall be based upon the resistance to flexure. To this end let a, in formula (96.), be substituted /?/ by its value in formula (154-), r~> an d we have, taking (189.) which is a rule, based upon the resistance to flexure, for a beam uniformly loaded, and also carrying three concentrated loads, the largest of which is not between the other two. 424. Example. What ought to be the breadth of a beam of Georgia pine of average quality, 20 feet long and 12 inches deep, carrying an equally distributed load of 4000 pounds, together with three concentrated loads, namely, 7000 pounds at 7 feet from the right-hand end, 4000 pounds at 7 feet from the left-hand end, and 3000 pounds at 3 feet from the same end. (See Art. 264.) Allotting the symbols to accord with the arrangement required under rule (189.), (the largest strain, as in Fig. 55, not between the other two), we have U 4000, A' = 7000, B' = 4000, O = 3000, /= 20, m = j, n=i$, s=7, z> = 3, d= 12 and /**= 5900, and let r = 0-04 ; and by formula (189.) b = 5900 x 4 i2' 7 x 0-04 [( x 4000 x 13) + (7000 x 13) + (4000 x 7) +. (3000X3)] = 11-020 or the breadth should be 1 1 inches. 294 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. 425. Carriage Beam with Three Headers, the Greatest Strain being at Outside Header. If, in formula (97.), (Art. JR7 258), we substitute for a its value, ~ (form. 154.), we shall have, taking Z7, Art. 340 , _ mf r 5 / / g_ >yi (2QQ \ ZTV/^/** L4 ' 6 \ /_J \ / which is a rule, based upon the resistance to flexure, for car- riage beams carrying three headers, with two sets of tail beams, so located (as in Figs. 54 and 55) that the header at which there is the greatest strain shall not be between the other two headers. 426. Example. What should be the breadth of a car- riage beam of Georgia pine of average quality, 20 feet long and 12 inches deep, carrying three headers 15 feet long, two of them, for a light-well 6 feet wide, located centrally as to the width of the floor, and the third header, at the side of an opening for a stairway 3 feet wide at one of the walls? The floor beams, placed 15 inches from centres, are to carry 200 pounds per superficial foot, with a deflection of 0-04 of an inch per foot lineal. (See Art. 264.) Allotting the symbols as in Fig. 55, we have / = 20, *=7, =i3, * = 7, v = 3i = J 5> </=i2, r=ii, F 5900, f= 200 and r = 0-04 ; and by formula (190.) we have X 2OO or the breadth should be 8i inches. CARRIAGE BEAM WITH THREE HEADERS. 2Q5 427. Carriage Beam with Three Headers, the Greatest Strain being at Outside Header, for Dwellings. If, in for- mula (190), f go and r= 0-03, we shall have which is a rule, based upon the resistance to flexure, for car- riage beams in dwellings and assembly rooms, to carry three headers, with two sets of tail beams, so located that (as in Fig. 55) the header at which there is the greatest strain shall not be between the other two, For an example, see Art. 429. 428. Carriage Beam with Three Headers, the Greatest Strain being at Outside Header, for First-elass Stores. If, in formula (190.), f 275 and r = 0-04, we shall have (192.) which is a rule, based upon the resistance to flexure, for car- riage beams in first-class stores, to carry three headers, with two sets of tail beams, so located that (as in Fig. 55) the header at which there is the greatest strain shall not be located between the other two. 429. Examples. Formulas (191.) and (192) are alike, except in the numerical coefficient, which, in the rule for dwellings and assembly rooms, is 3000, while for first-class stores it is 6875. An example under one rule will serve to illustrate the other, by a simple substitution of the proper coefficient. As an example under rule (191) : What should be the 296 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. breadth, in a dwelling, of a carriage beam of white pine of average quality, 20 feet long and 12 inches deep, carrying three headers 12 feet long, so placed as to provide an open- ing 4 feet wide for a stairs at one wall, and a light-well 6 feet wide at the middle of the width of the floor? The floor beams are placed 16 inches from centres. (See Art. 264.) Allotting the symbols as in Fig. 55 we have / = 20, m 7, n = 13, s= 7, v = 4, -12, d = 12, c = ij and .F= 2900; and by formula (191.) b = -^v*, v Ki x i^ x 13 x 20)+ 12 (7^13 + 7 3 -4')] = 8-052 2QOO X 1 ^ or the breadth should be 8 inches. This is the breadth when of white pine, and in a dwelling. If, instead, it be required of Georgia pine, and for a first- class store, then the breadth just obtained, treated by the proper constant and numerical coefficient, and at the same time relieved from those applying to the previous case, will be 6875 200O b 8-CX2 x x - - = 0-070 3000 5900 or the breadth, when of Georgia pine, and for a first-class store, should be 9^ inches. 4-30. Beam* with Uniformly Distributed and Three Concentrated Loads, the Greatest Strain being at middle Load. In Art. 262 a rule is given for beams uniformly loaded, and also carrying three concentrated loads, the mid- dle one of which produces the greatest strain. This rule is based upon the resistance to rupture. It may be modified to BEAM WITH THREE CONCENTRATED LOADS. 2Q/ depend upon the resistance to flexure by substituting, in for- /?/ mula (99.), for a its value -=j- (form. 1&4-), (taking | U, Art. 340), thus (193.) which is a rule, based upon resistance to flexure, for beams carrying a uniform load (U) and three concentrated loads (A, B' and C), the middle one of which produces the greatest strain of the three, as in Fig. 56. 431. Example. What should be the breadth of a beam of Georgia pine of average quality, 20 feet long and 14 inches deep, and carrying 4000 pounds uniformly distrib- uted, 6000 pounds at 4 feet from one end, 6000 pounds at 9 feet from the same end, and 7000 pounds at 6 feet from the other end ; with a deflection of 0-04 of an inch per lineal foot? (See Art. 264.) Assigning the symbols as per figure, we have, 7 = 4000, A = 6000, B' = 7000, C = 6000, 1=20, m = 9, =n, s = 6, v = 4, */= 14, F = 5900 and r = 0-04 ; and by for- mula (19 3. \ (6000 x 1 1 x 4)] = 9- 163 or the breadth should be 9^ inches. 432. Carriage Beam with Three Headers the Oreatet Strain being at Middle Header. If, in formula (106.), (Art. /?/ 267), there be substituted for a its value - (form. we shall have, taking f 7, Art. 340, b = ~ m 298 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. which is a rule, based upon resistance to flexure, for carriage beams carrying three headers and two sets of tail beams, so placed (as in Fig. 56) that of the strains produced at the headers, the greatest shall be at the header which is between the other two. 4-33. Example. What should be the breadth of a car- riage beam, 20 feet long and 12 inches deep, of Georgia pine of average quality, carrying three headers 14 feet long, so placed as to provide a stair opening 4 feet wide at one wall, and a light-well 5 feet wide 6 feet from the other wall ? The floor beams, placed 15 inches from centres, are to carry 200 pounds per foot superficial, with a deflection of 0-04 of an inch per foot lineal. (See Art. 264.) Assigning the symbols as per Fig. 56 we have, / = 20, m = 9, n = 1 1, s = 6, v = 4, g = 14, d = 12, <?,== if, F = 5900, f= 200 and r = 0-04; and by formula 2OO _ _ = S900xi2'xo.o 4 L9(* XI * x 11 x 20 + 14 x 6') -.- (14 x ii xy' 4 2 )] = 8-651 or the breadth should be, say 8g inches. 434. C?arr5age CScaiai with Three Headers, the Greatest Strain being at Middle Header, for Dwellings. If, in for- mula (194-), f= 90 and r 0-03, we shall have b = {,m(lcnl+gs^+gn (m*-v*)] (195.) which is a rule, based on resistance to flexure, for carriage beams in dwellings and assembly rooms, to carry three headers, with two sets of tail beams relatively placed as in Fig. 56, so that, of the three strains produced at the headers, the greatest shall be at the header which is between the other two. (For an example, see Art. 436.) CARRIAGE BEAM WITH THREE HEADERS. 299 4-35. Carriage Beam with Three Headers, the Greatest Strain being at middle Header, for First-class Stores. If, in formula (194*), f= 275 and r 0-04, then we shall have b = \m (Icnl+gs*} +gn K-tf)] (196.) which is a rule, based on resistance to flexure, for carriage beams in first-class stores, to carry three headers, with two sets of tail beams relatively placed as in Fig. 56, so that, of the three strains produced at the headers, the greatest shall be at the header which is between the other two. 4-36. Example. Formulas (193.) and (106.) being alike, except in the numerical coefficient, a single example will suffice to illustrate them. In a dwelling, what should be the breadth of a carriage beam of oak of average quality, 20 feet long and 12 inches deep, to carry three headers 15 feet long, with two sets of tail beams, so placed as to provide a stair opening 4 feet wide at one wall, and a light-well 7 feet wide, distant 5 feet from the other wall ? The beams are to be placed 1 5 inches from centres. (See Art. 264.) Arranging the symbols in the order in which they appear in Fig. 56, we have, / = 20, m = 8, n = 12, s = 5, v = 4, =15, d = 12, c= i and F 3100; and, by formula or the breadth should be, say 8J inches. 300 RESISTANCE TO FLEXURE FLOOR BEAMS. CHAP. XVII. QUESTIONS FOR PRACTICE. 437. In a dwelling: What should be the depth of white pine beams of average quality; they being 18 feet long and 3 inches broad, placed 18 inches from centres, and allow- ed to deflect 0-03 of an inch per foot? 438. In a first-class store : What should be the breadth of the floor beams of spruce of average quality, 19 feet long, 13 inches deep, placed 13 inches from centres, and with a deflection of 0-04 of an inch per foot ? 439. In a dwelling: What ought to be the breadth of a header of white pine of average quality, 14 feet long and 13 inches deep, carrying one end of a set of tail beams 15 feet long, and with a rate of deflection of 0-03 of an inch per foot ? 440. In the floor of an assembly room, in which the beams are 15 inches from centres: What should be the breadth of a carriage beam of spruce of average quality, 20 feet long and 12 inches deep, carrying one header 13 feet long, located at 5 feet from open end ? The deflection allowable is 0-03 of an inch per foot. 441. In the floor of a first-class store, where the beams are 15 inches deep and set 14 inches from centres : What should be the breadth of a carnage beam 24 feet long, of Georgia pine of average quality, carrying two headers QUESTIONS FOR PRACTICE. 3OI 1 6 feet long, located, one at 9 feet from one end, and the other at 7 feet from the other end, with two sets of tail beams? The deflection is 0-04 of an inch per foot. 4-4-2. In the floor of a first-class store, with beams 16 inches deep placed 15 inches from centres : What should be the breadth of a carriage beam of Georgia pine of average quality, 26 feet long, and carrying three headers 18 feet long, located as in Fig. 54, one at 4 feet from one wall, another at 8 feet from the same wall, and the third at 8 feet from the other wall ? The deflection to be 0-04 of an inch per foot. CHAPTER XVIII. BRIDGING FLOOR BEAMS.* ART. 443. Bridging Defined. Bridging is a system of bracing floor beams. Small struts are cut to fit between each pair of beams, and secured by nails or spikes ; as shown in Fig. 65. The effect of this bracing is decidedly FIG. 65. beneficial in sustaining any concentrated weight upon a floor. The beam immediately beneath the weight is materially as- * The principles upon which this chapter is based the author first made public in an article which appeared in the Scientific American, July a6th, 1873, entitled "On Girders and Floor Beams The Effect of Bridging." EXPERIMENTAL TEST. 303 sisted, through these braces, by the beams on each side of it. It is customary to insert rows of cross-bridging at every five to eight feet in the length of the beams. It is the usual practice, where the ceiling of a room is plastered, to attach the plastering laths to cross-furring, or narrow strips of boards crossing the beams at right angles, and nailed to their bottom edge. These strips are set at, say 12 inches from centres, and when firmly nailed to the beams act as a tie to sustain the lateral thrust of the bridg- ing struts. The floor plank at the top serve a like purpose. . Experimental Test. To test the effect of bridging, about three years since I constructed a model, and sub- jected it to pressure. It was made upon a scale of 1% inches to the foot, or \ of full size, and represented a floor of seven beams placed 16 inches from centres, each beam being 3 x 10 inches and 14! feet long. These beams were con- nected by two rows of cross-bridging, and secured against lateral movement by strips representing floor plank and ceil- ing boards, which were nailed on top and beneath. There were four strips at each row of bridging, two above and two below. Before putting these beams in position in the model, I submitted each beam to a separate test, and ascertained that to deflect it one tenth of an inch required from 37 to 40 pounds, or on the average 38^ pounds. With the model completed, the beams being bridged, it required a pressure of 1554 pounds applied at the centre of the middle beam to deflect it as before, one tenth of an inch. And while this pressure deflected the central beam to this extent, the beam next adjoining on each side was deflected 0-0808 of an inch, the ones next adjoining these were each deflected 0-0617 of an inch, while the two outside beams 34 BRIDGING FLOOR BEAMS. CHAP. XVIII. were each depressed 0-0425 of an inch. Had there been more than seven beams, and all bridged together, the effect would doubtless have been still better. As the result of this test of the effect of bridging, we have one beam sustaining 155! pounds with the same deflection that was produced by 38^ pounds before bridging, or an in- crease of H7 T V pounds; an addition of more than three times the amount borne by the unbridged beam. 445. Bridging Principle of Resistance. The assist- ance contributed by the adjacent beams to a beam under pressure may be computed, but preliminary thereto we have these considerations, namely : First. The deflections of a beam are (within the limits of elasticity) in proportion to the weights producing the de- flections. Thus, if one hundred pounds deflect a beam one tenth of an inch, two hundred pounds will deflect it two tenths of an inch. From which, knowing the deflection of a beam, we can compute the resistance it offers. Second. The resistance thus offered, being at a distance from the beam suffering the direct pressure, is not so effect- ual as it would be were it in direct opposition to the pres- sure. It is diminished in proportion to its distance from that beam. 446. Resistance of a Bridged Beam. Based upon the two preceding considerations, we will construct a rule by which to measure the increase of resistance derived from the adjacent beams through their connection by cross- bridging. Let Fig. 66 represent the cross-section of a tier of floor INCREASED RESISTANCE OF BRIDGING. 305 beams connected by cross-bridging, in which C is the lo- cation of a concentrated weight, AB the distance on one side of the weight to which the deflecting influence acting FIG. 66. through the cross-bridging is extended, BC the deflection at the weight, and DE the deflection of one of the beams E, caused by the weight at C. The triangles ABC and ADE are similar, and tkeir sides are in proportion. Put- ting DE, p for AB, we have AB P m for AD, a' for BC, and V for BC : : AD : DE = ? This is the measure of the deflection at E, or at any one of the beams the distance of which from A is equal to m, and, since the deflections are as the weights producing them, therefore b ', the deflection at E, measures the strain there, when a' measures that at C. It is required, however, to know not only the resistance offered by each beam, but also what weight r, acting at C, would be required to overcome this resistance. The line AB (or /) may be considered to serve as a lever, hav- ing its fulcrum at A. The weight r, at B, acting in the line BC, is opposed at D by b' ' , the resistance of the beam at E, acting in the line ED, with the leverage m. The weight r will act with the moment rp, and I' will resist with the moment b'm. Putting these moments in equilibrium, we have b'm rp, or r = b'. 306 BRIDGING FLOOR BEAMS. CHAP. XVIII. In this, substituting for b' its value as above found, we have ,m m r = a x P P This weight r represents the effect at C of the resist- ance to deflection of any beam whose distance from A is equal to m, and where a' equals the load borne by the beam at B, and / is put for the distance AB. 44-7. Summing tlie Resistances. Let the distance from centres between the floor beams be represented by c, and the number of spaces from A to any beam, as, for example, that at D, by n ; then m = nc t and substituting this value for m in (197.) we have r = n*~ (198.) In this expression, a', c* and p 3 are constants, or quanti- ties which remain constant for the several values of r which are to be obtained from the resistances of the several beam?. For convenience, put / for j and then r = vtt (199.) With this expression, the various values of r may be ob- tained and grouped together. In doing this, we have, for the first beam from A, n = i ; for the second, n = 2; for SUM OF INCREASED RESISTANCES. 307 the third, n 3, and so on to the middle or point of great- est depression. Therefore the whole resistance will be R' = t + 2V + 3V + 4V + etc. R' t (i + 4 + 9 + 16 + etc.) This gives the resistance on one side of the point C. The beams on the other side afford a like resistance ; and the sum of the two resistances will be R = 2t (i + 4 + 9 + 16 + etc.) R = 2 T- (i + 4 + 9 + 16 + etc.) 448. Example. When a concentrated weight deflects six beams on each side of it, they being placed 16 inches from centres : What will be the amount of resistance to de- flection offered by the twelve beams, the beam upon which the weight rests being capable of sustaining alone, unaided by the adjoining beams, 1000 pounds ? Here a' = 1000, c = i% and p = ?xi$ = 9^. There- fore, by formula (200.}, 2 X IOOO X = 37H-3 This 3714 pounds is the resistance offered by the twelve beams, through the means of bridging, and is nearly four times the amount that the centre beam, unaided by the bridging, would carry with a like deflection. The combined resistance of all the beams would be 3714+1000 = 4714 pounds. 308 BRIDGING FLOOR BEAMS. CHAP. XVIII. 44-9- Assistance Derived from Cross-bridging. Just how many beams on each side will be affected, and by their resistance contribute in aiding the beam at 7, will depend upon circumstances. The bridging will be effective in re- sisting deflection in proportion to the elevation of the angle at which the bridging pieces are placed, which will be directly as the depth of the beams and inversely as their distance apart. It will also be in proportion to the faithful- ness with which the work of bridging is executed. From these considerations, and from the experiment of Art. 44-4, we conclude that, in well-executed work, we shall have d An equally distributed load upon a floor beam is represented (Art. 92) by cfl. A load at the centre of the beam produc- ing an equal effect will be f of this, or \cfl. The symbol a' (form. 200.) represents the load at the middle of a floor beam, and therefore a' = fc/7 These values of / and a' may be substituted for these symbols in formula (00.\ and the result will be R= 2 -TT-j (i + 4 + 9 + etc.) or, (901.. In this rule R equals the additional resistance to a concen- trated weight on a beam, obtained from adjacent beams through the cross-bridging. USEFUL FOR CONCENTRATED WEIGHTS. 309 450. Xumfoer of Beams Affording Assistance. The value of /, as above, is -. The symbol -n being put for the number of spaces on each side of the beam sustaining the concentrated weight, over which this weight exerts an in- fluence ; or /, the distance AB of Fig. 66; and c for the distance apart from centres at which the beams are placed ; then, / = = nc ; from which we have n = % (02.) To apply this rule : How many beams on each side of a concentrated weight would contribute towards sustaining it, when they are 12 inches deep, and 16 inches from centres? Here we have d = 12 and c = i-J-, and therefore n = -p = 6f say 7 spaces. 3 In seven spaces, six beams will be affected. 451. Bridging Useful in Sustaining Concentrated Weights. The results shown in Art. 448 illustrate the ad- vantage of cross-bridging in resisting concentrated weights, and show the importance of always having floor beams bridged, and the work faithfully executed. The advantage, however, of cross-bridging inheres only in the case of concen- trated weights. For, although in the example of Art. 44-8, the 13 beams sustained by their united resistance a concentrated weight of 4714 pounds, yet it will be observed that this is not the limit of their power, for they are each capable of sustaining 1000 pounds placed at the middle, or together, 13,000 pounds; nearly three times the previous amount. 310 BRIDGING FLOOR BEAMS. CHAP. XVIII. 4-52. I5icreaed Resistance Due to Bridging. A useful application of the results of this investigation is found in determining the amount of concentrated weight which may be borne upon a floor beam. As an example : In a dwelling with well-bridged floor beams of an average quality of white pine, 3 x 10 inches, and 16 feet long, what concen- trated weight may be safely sustained at the middle of one of them ? The distance from centres at which these beams should be placed is had by formula (144-), Art. 362, ibd 3 i -55 the value of i being taken as found in Art. 361. With the above value, c 1-135, we may, by formula (202.), find the distance to which the effect of the weight extends on each side, thus : io io say 8 spaces, or 7 beams. The symbols of formula (20 l.\ applied in this case, will be as follows : c = i 135, / = 90, /= 16 and d = io, and the squares in the parenthesis extend to 7 places. Therefore n 5xi- i35xgox 16 , R 4 x IQ^" -(1+4+9+16 + 254-36 = 18 x i-"i35~ ft x 140 EXAMPLE, BY LOGARITHMS. 311 The product of these factors, one of them being raised to a high power, will best be obtained by logarithms, thus : Log. 1-135 = 0-0549959 5 0-2749795 Log. 1 8 = 1-2552725 Log. 140 = 2-1461280 4746-6 = 3-6763800 The product of the factors, or the value of R, is there- fore equal to 4746-6 pounds. This is the increased resist- ance. The resistance offered by the beam upon which the weight is laid equals (Art. 449) To this, adding the increase = 4746-6 we have 5768- 1 as the total resistance to a concentrated load at the middle of the beam, when assisted by 7 beams on each side by cross-bridging. CHAPTER XIX. ROLLED-IRON BEAMS. ART. 453. Iron a Substitute for Wood. When the beams composing a floor are of wood, they are of rectangular form in cross-section. Investigations into the philosophy of the transverse strain, by which the importance of depth was developed, led to the use of beams of which the rectangle of cross-section was narrow and high. Owing to the liabil- ity, in wooden beams as generally used, of destruction by conflagration and by other causes, iron was introduced as a substitute. The greater cost of this material over that of wood, made it important, now more than ever, to give to the beam that shape which should prove the strongest. 454. iron Beam Its Progressive Development. In the use of iron as a floor beam, economical considerations reduced the breadth until the beam became weak laterally. To remedy this defect, metal was added at the top and bottom in the form of horizontal plates, and these were connected to the thin vertical beam by angle irons as in Fig. 67 ; the whole forming what is known as the plate beam or girder. This expedient served FlG - 6 7- not only to stiffen the thin vertical beam laterally, but added very greatly to its ab- PROPORTIONS BETWEEN FLANGES AND WEB. 313 solute strength. The added material had been placed just where it would do the greatest possible good ; at a point far removed from the neutral axis of the beam. 455. Rolled-Iron Beam Its Introduction. The in- crease of strength obtained in the plate beam (Fig. 67) was so great that it became popular. To supply the demand, iron manufacturers, at great expense, made rolls similar to those for making railroad iron, by which they were enabled to fur- nish beams (Fig. 68) rolled out in one piece, with all the best features of the plate beam, and which could be much more readily and cheaply made. Owing to the large cost of the rolls, only a very few sizes were at first made, but these few only increased the demand. The man- ufacturer, thus encouraged, made rolls for other sizes, and thus the number of beams was increased, until now we have them in great variety, from 4 to 15 inches high.* FlG - 68 - 4-56. Proportions between Flanges and Web. These beams, as usually made, have the top and bottom plates, or flanges, of the same form and size. In wrought-iron the resistances to rupture, by compression and by tension, are not equal. When the load upon the beam, however, is not so large as to strain the metal beyond the limits of elasticity, * There were exhibited at the Centennial Exposition at Philadelphia, by the Union Iron Co., of Buffalo, a 15 inch beam 52 feet in length, and a 9 inch beam 80 feet long. This is believed to be the limit reached in American manufacture at the present time. The English and Germans, however, are roll- ing them larger. A German exhibit in Machinery Hall contained beams from Burbach half a metre (19-69 inches) high by 15 metres (49-21 feet) long. 3H ROLLED-IRON BEAMS. CHAP. XIX. then it resists both compression and tension equally well, and hence the propriety of having- the top and bottom flanges equally large. The manifest advantage of having the material accumu- lated at a distance from the neutral axis, has led to putting as much as possible of the area of the whole section into the flanges, and thereby reducing the web or vertical part to the smallest practical thickness. The web is required to main- tain the connection between the top and bottom flanges, and to resist the shearing effects of the load. In rolled-iron beams, as usually made, the thickness of the web is more than sufficient to resist these strains. 457. The Moment of Inertia Arithmetically Considered. For the intelligent use of the rolled-iron beam as a substi- tute for the wooden beam in floors, as well as for other uses, the rules already given need modification. The resistance of a beam to flexure or bending is termed its moment of inertia. This is represented in symbolic for- mulas by the letter 7. In formula (111.), (Art. 300), the coefficient -% and the symbols bd 3 represent the moment of inertia, and /, its symbol, may be substituted for them, thus: PN S PN 3 6 = rl The moment of inertia for any cross-section is equal to the sum of the products of each particle of the area o the cross-section, into the square of its distance from the neutral axis.** For example : in a beam with a cross-section of the I form, a horizontal line drawn through the centre of area of the cross-section will be the neutral line for strains within * Rankine's Applied Mechanics, Art. 573. MOMENT OF INERTIA. 315 the limits of elasticity. Let the area be divided into a large number of small areas. Then, for the portion of the figure above the neutral line, multiply each of these small areas by the square of its vertical distance above the neutral line, and the sum of these products will equal the moment of inertia for the upper half of the section. A like process will give the moment of inertia for the lower half. The two in this case will be equal, and their sum is the moment of inertia for the whole section. The result thus obtained will not be exact, but will approach accuracy in proportion to the small- ness of the parts into which the area of the cross-section is divided. 458. Example A. As an illustration, let A BCD, in Fig. 69, represent the cross-section of a beam ; MN, drawn through the middle of the height AD, being - _ . the neutral axis ; and let the lines EF, GH, IJ, KL, OP, QR, and ST divide the area ABMN into twenty equal* parts. The four squares in each horizontal row are equally distant from the neutral line MN, and may therefore M be taken together. Suppose each of these squares to measure 2x2 inches, then the area of each will be 4 inches, and of the four in each horizontal row will be 4x4=16 inches D area. The distances from the neutral line to the centre of each square will be as follows : In the first row, D = i " " second " D = 3 " " third " D 5 " " fourth " D = 7 " " fifth " D = 9 P R T FIG. 69. 3*6 ROLLED-IRON BEAMS. CHAP. XIX. Their moment of inertia will be as follows : In the first row, 7, = 16 x i 2 = 16 x i " " second " I 2 = 16 x 3' = 16 x 9 " " third " I 3 16 x 5 2 = 16 x 25 " " fourth " 7 4 = 16 x f = 16 x 49 " "fifth " 7 5 = 16 x 9 2 = 16 x 81 and their sum 7= 16(1 + 9 + 25 + 49 + 81)= 16 x 165 = 2640. 459. Example B. If we subdivide each of the squares in Fig. 69, and take the sum of the products as before, the result will be larger and nearer the truth. For example : divide each of the squares into four equal parts, each one inch square. There will be eight of these parts in a row, and ten rows. The area of each row will be 8x 1=8, and their distances from the neutral line will be -J, f, f , J, f , V- ~/> V-> an d V- respectively. The moments of inertia will be as follows : In the first row, /, = 8 x () a = 8 x l 2x i " second " 7, = 8 x (f) 2 :=8x f = 2 x 9 " third " /, = 8 x (f) 2 = 8 x *. = 2 x 25 " fourth " 7 4 = 8 x Q-) 2 = 8 x **- = 2 x 49 " fifth " 7, = 8x (|)* = 8x ^L=2x 81 " sixth " I 6 = 8 x (^ = 8 x -LfL = 2 x 121 " seventh " 7 7 = 8 x (-V 3 -)' = 8 x .IJA = 2 x 169 " eighth u /, = 8 x (iff = 8 x AjA = 2 x 225 " ninth " /. = 8 x (- 1 /) 2 = 8 x i fi = 2 x 289 " tenth N " 7 /0 = 8 x (J/) 2 = 8 x AJJ. 2 x 361 which is equal to twice the sum of the series of 1+9+25+ etc. or, 7=2x1330=2660 This result exceeds in amount the previous one (2640). MOMENT OF INERTIA COMPUTED. 317 460. Example C. If the eighty squares of this last trial be each subdivided into four equal parts, the whole cross-section will contain 4x80=320 parts; there will be twenty rows, with sixteen in each row ; the area of each part will be -J-x| = J; and the perpendicular distance from the neutral line to the centres of these 320 parts will be : In the first row, J " second " f " third " J " fourth " I and so on, each distance being a fraction having 4 for a de- nominator, and for a numerator one of the arithmetical series of the odd numbers i, 3, 5, 7, 9, 11, etc., to 39. The moment of inertia will be the sum of the products, as follows : In the first row, /, = i6xx() 2 " second " /,= i6xix() 2 " third " /, = 1 6 x i x () 2 etc. These are equal to : In the first row, 7, = 16 x ix^ x i 2 = x i a " second " I 2 = 16 x \ x -^ x f = x 3 2 " third u I 3 = i6xix-^x 5' = x 5' etc. Thus the sum of all the products will be equal to a quarter of the sum of the squares of the arithmetical series of the odd numbers i, 3, 5, 7, 9, 11, etc., to 39. Selecting the squares of these numbers from a table of squares, we find their sum to equal 10,660, and then, as above, / = x 10660 = 2665 318 ROLLED-IRON BEAMS. CHAP. XIX. 4-61. Comparison of ResuBt. We have now the three results, 2640, 2660, and 2665, gradually increasing as the number of parts into which the sectional area is divided in- creases, and tending towards the true amount, to which it can only arrive when the parts become infinitely small and in- finite in number. To compute these by the arithmetical method would be impossible, but by the calculus it is exceed- ingly simple and direct. The formula for the moment of inertia, as generally used, is complicated, but for a rectangu- lar section in a horizontal beam, subject to limited vertical pressure, is simple. 462. Moment of Inertia, toy the Calculu Preliminary Statement. Let A BCD in Fig. 70 represent the rectangu- lar cross-section of a beam ; let MN be the neutral line, and the two lines at EF be drawn parallel to MN. Let the breadth of the section EF equal 7, JM and the perpendicular distance from the neutral line to the lower line EF equal x. The two parallel lines at EF may be taken at any distance, x, from the neu- tral line. This distance is variable ; x is a variable representing any and every dis- tance possible on the line GH, from zero to its full length. It is always the distance from the line MN to 7, the lower line at EF, wherever 7 be taken. The ver- tical distance between the two lines at EF is termed dx, which means the differential of x, or the difference in the length of x when slightly increased by the movement of 7 farther from MN. This augmen- tation, dx, is taken infinitesimally small. Now the area of the space between the two lines at EF will be the product of its length by its height, or 7 x dx. M- MOMENT OF INERTIA, 43Y THE CALCULUS. 319 4-63. Moment of Inertia, by the Calculus. The mo- ment of inertia is equal (Art. 457) to the sum of the products of each particle of the area of the cross-section, into the square of its distance from the neutral axis. In the last article, the expression ydx represents the area of the in- finitesimally small space at the lines EF, Fig. 70. The dis- tance from this small area to the neutral axis is x, and the square of the distance is x* ; therefore x*ydx equals the area into the square of its distance, equals the moment of inertia of the small area ydx\ or, the differential of the moment of the area of the whole figure ABMN. This differential is expressed thus, dl=x*ydx (203.) This expression represents the moment of only one of the infinitesimal parts into which the area ABMN is supposed to be divided. To obtain the moment of the whole area, it is requisite to add together the moments of all the infinitesi- mal parts ; or, to obtain from the differential (form. 203.) its integral. The rule for this is,* " Add one to the index of the variable, and divide by the index thus increased and by the differential of the variable." Applying this rule to formula (203.) it becomes This is in its general form. To make it definite, we have y = b, the breadth ; and -r, at its maximum, equals %d, half the depth. These values substituted for y and x> we have (204) * Ritchie, Dif. and Integ. Calculus, p. 21. 320 ROLLED-IRON BEAMS. CHAP. XIX. This result is the moment of inertia for the upper half of the section of the beam, and represents the resistance to com- pression. The resistance to tension in the lower half of the beam is (under the circumstances of the case we are consider- ing) an equal amount ; hence for the two we have* (205.) 464-. Application and Comparison. This formula gives the value of the moment of inertia for the whole section; for the two parts, one above and the other below the neutral line. To obtain the value of the part above the line, for com- parison with the results obtained in Arts. 4-58 to 460, we take formula (204.) in which b is the breadth and d the depth of the beam. The section of beam given in Art. 4-58, Fig. 69, was proposed to be 8 inches broad and 20 inches high, or AB = b = 8 and AD d 20. With these figures in the formula, we have / = ^ x 8 x 20 3 = 2666f This is the exact amount. In the three trials of Arts. 4-58 to 4-60, we had the approximate values 2640, 2660 and 2665. In the last trial, in which the parts were small and numerous, the result was a close approximation. * Moseley, Am. Ed. by Mahan, Art. 362. MOMENT OF INERTIA GRAPHICAL REPRESENTATION. 321 465. Moment of Inertia Graphically Represented. The two processes, arithmetical and by the calculus, are graphically represented in Y Fig. 7 1 , in which the area of the figure contained within the straight lines OB and AB and the curved line OA, is the correct area by the calculus, to which the sum of the squares of the arithmetical progression I, 3, 5, 7, 9 and u closely approximates. Here x and y, indicating the distances along the axes OX and OY, are co-ordinates to points in the curve, as Aj C, D, E, etc., these points being midway in the difference be- tween the sides of each two con- tiguous squares. The values of y for these points are 2, 4, 6, 8, 10 and 12 ; a difference between FIG. 71. each two consecutive values equal to 2. The consecutive ordinates x are i, 4, 9, 16, 25 and 36; or i\ 2\ 3", 4 2 , 5 2 and 6 2 . Comparing these values ol y and x in each pair, we have In the first pair, y " " second " " " third " " " fourth "' " " fifth , " " " sixth " y = 2 and x I = i" y - 4 X 4 = 2 2 y = 6 " X ~~~ 9 = 3' y = 8 X 16 = 4' y ~~~ 10 " X = 25 = 5 y 12 x 36 = 6' 322 ROLLED-IRON BEAMS. CHAP. XIX. From this, the relation between y and x is readily seen to be represented by the following expressions : (D- - = "v f = 4* (206.) 4-66. Parabolic Curve Area of Figure. The ex- pression just obtained is the equation to the curve, and this curve is a parabola, with / = 2, or y* 2px.* By formula (206.) any number of points in the curve may be found, and the curve itself drawn through them. Also, by it and by the rules of the calculus, the area of the figure inclosed between the curved line and the two lines AB and BO may be found. To this end, let the narrow space in- cluded between the two lines GH, drawn perpendicular to OB from H to G (a point in the curve), be a small por- tion of the area of the whole figure ; dx, the distance be- tween the two lines, being exceedingly small. The area of this narrow space will be the product of its length by its breadth, or y x dx. The differential of formula (206.), the equation to the curve,f is 2ydy = Afdx \ydy dx Multiplying both sides by y gives = ydx * Robinson's Conic Sections and Analytical Geometry, 1863, p. 50. f Ritchie's Dif. and Integ. Calculus, p. 20. MOMENT OF INERTIA PARABOLIC CURVE. 323 which equals the differential of the area as above shown. The integral of this value of ydx is, by the rule (Art. 4-63), \ffdy = or the area This is the area of the figure bounded by the curved line OA and the straight lines AB and BO. 467. Example. The example given in Art. 460 may be taken to show an application of the last formula. The number of horizontal rows of parts into which the area is there divided is 20, and the last number of the arithmetical series is 39. By an examination of Fig. 71, it will be seen that AB, the base of the figure, is equal to the side ol the last square plus unity. Therefore, 39+1 =40 is the base of the area proposed in Art. 460, or. y = 40. From the dis- cussion in that article, it appears that the small squares con- sidered are each J of unity in area, from which the area of the figure in that case is found to be one quarter of the sum of the squares of the arithmetical series ; or, by formula (207.}, A = i x i/ = To apply this result to the present case, where y = 40, we have A = -fa x 4O 3 = 2 the same result as in Art. 464. 324 ROLLED-IRON BEAMS. CHAP. XIX. 468. Moment of Inertia General Rule. That formula (207 '.) may be general in its application, we need to find a proper coefficient. Let the beam, instead of being 8 inches wide, as in Fig. 69, be only one inch wide, and let the portion above the neutral line be divided by horizontal lines into any number of equal parts. Put n for the number of parts, and / for the thickness of each part. The area of each part will be I x t = t inches, and the several distances from the neutral line to the centre of each part will be, respectively, -J/, f/, ^ 2^ _ T |/, |^, etc., to the last, which will be "- - 1. Now, the moment of inertia of each part being its area into the square of the distance to its centre of gravity, there- fore the several moments will be as follows : In the first piece, t ( i x -J == i 2 x \f second " / (3 x ^) = 3* x i' 3 ( t \ 2 third " /^5 x -J = 5 2 x^t 3 f t \ 2 fourth " / ^7 x -J = f x J/ 3 last " t ((2n i)^J = (2n i) 2 x \f The sum of these will be 5= i/ 3 [i 2 + 3 2 + 5 2 +7 2 + ...... (2n i) 2 ] (208.) But the sum of the series I 2 + 3" + 5 2 + etc., is the area of the parabolic figure (Fig. 71), ancl has been found to be equal to \f (form. 207.) " " " " MOMENT OF INERTIA RULE. 325 Now y, when at its maximum, coincides with the base AB of Fig. 71, and is equal to the side of the last square plus unity. As above, the side of the last square is 2n I, from which y 2n, and and therefore formula (208.) becomes 5 = \fvt (209.) which is a rule for ascertaining correctly the moment of inertia for a beam one inch broad. 469. Application. To show the application of the above, take the example of Art. 458, where the number of slices is 5 and the thickness is 2, and we have, by the use of formula (209. \ The formula gives the result for a beam one inch broad. The beam in Art. 458 is 8 inches broad. Therefore, for the full amount we have 8 x 3334 = 2666f Again, take the example of Art. 460, where t = % and n = 20, and we find as the result and 8 x 333^- = 2666f Thus in both cases we have the same result as that obtained directly by the calculus. If b, for the breadth, be added to formula (209.) we shall have the complete rule, thus : / = fif 326 ROLLED-IRON BEAMS. CHAP. XIX. and since /// equals the height above the neutral line, equals , the half of the depth of the beam, t 3 n 3 = (tn) 3 & (%dj m \d s and this value of t s n 3 substituted for it in the above equa- tion, gives This is for one half the beam. For the whole beam we have twice this amount, or the same as found directly by the calculus in formula (205.} 470. Rolled-Iron Beam Moment of Inertia Top Flange. An expression for the moment of inertia appropri- ate to rolled-iron beams of the I form of section (Fig. 68) A B may be obtained directly from the for- mula (205.) for the rectangular section. In Fig. 72, showing the cross-section required, b equals the breadth of the beam, or the width of the top and bot- tom flanges, and t equals the width or thickness of the web ; b minus t equals b /y d equals the entire height of the section, and d f the height be- tween the flanges. MN is the neu- FlG - 72. tral line drawn at half height. By formulas (203.) and (204.) the moment of inertia for the part above the neutral axis is M T MOMENT OF INERTIA FOR FLANGE AND WEB. 327 If this be applied so that x = \d, the result (%bd 3 ) y as in (204*), is the moment for the rectangle ABMN. Again, if it be applied with x = \d t , the result (^bdf) will be the moment for the rectangle EFMN. Now, if the latter re- sult be subtracted from the former, the remainder will be the moment for the area ABEF, the upper flange, or 4-71. Rolled-Iron Beam Moment of Inertia Web. Formula (210.) is the moment of inertia for the top flange. The moment of inertia for the upper half of the web is that due to a rectangle having for its breadth y = t, and for its height x = \d t , and by Art. 463, and since / = b b t , therefore 4-72. Rolled-Iron Beam moment of Inertia Flange and Web. Formula (211.) is the moment of inertia for the upper half of the web. Added to formula (210.), the sum, representing the moment of inertia for all of the beam above the neutral line, will be 328 ROLLED-IRON BEAMS. CHAP. XIX. 4-73. Rolled-Iron Beam Moment of Inertia Whole Section. Formula (212.) is the moment of inertia for that half of the rolled-iron beam which is located above the neu- tral line. The moment for the portion below the line will be equal in amount ; and therefore, for the moment of the entire section, we have twice the amount of formula (212.) or / = T V(WW,//) (213.) 474. Rolled-Iron Beam Moment of Inertia Compari- son with other Formulas. Formula (213.) is the same as that given by Professor Rankine* and others, and is in gen- eral use. Canon Moseleyf gives an expression which is complicated. Mr. Edwin Clark, in his valuable work on the Britannia and Conway Tubular Bridges, Vol. I., p. 247, gives the formula in which d 2 is the distance between the centres of gravity of the top and bottom flanges, a is the area of the top or bottom flange, and a / is the area of the web. This is more simple than the common formula (213.), but is not exact. It is only an approximation. Its relation to the true formula will now be shown. From formula (213.) we have, multiplying by 12, Of these symbols we have (Fig. 72, putting h = AE ), d = d 2 + h, b t = bt and d t = d 2 h By substitution, we now have 1 2/ = b(d 2 + //)' - (b-t}df * Rankine's Applied Mechanics, pp. 316 and 317. f Moseley's Mech. of Eng., Am. Ed. by Mahan, Art. 504. MOMENT OF INERTIA FORMULAS COMPARED. 329 and since (bt)df bd?td? = bd?a t d* (putting a t for the area of the web); and since dj d 2 h, therefore we have = b(d 2 4- h) 3 - \b(d 2 -h) 3 -a t d? 1 2/ = b(d, 4- //)* - b(d s h) 3 + a t df Then we have (d, + /i) 3 = d/ + (d 2 + h) 3 -(d 2 -/i} 3 = o + &//// + o 4- 2k 3 Substituting these in the above, we have I2/ = b(6dfh 4- 2k 3 ) + a t df The area of the top flange equals bh = a, therefore 1 2/ = 6a(d;+ \h 2 } + a t d? or, / = &\6a(d;+ i//) + *,<//] (215.) In Mr. Clark's formula, (2 14-), we have + y ) or, + a.df) Comparing this with the reduction of the common formula as just found [form. (215.)'], the difference is readily seen to be, that while in the one the quantities a and a t are each multiplied by the factor df, in the other the factor for a is (X/+i//) and that for a t is df. 330 ROLLED-IRON BEAMS. CHAP. XIX. 475. Rolled-Iron Beam moment of Inertia Compari- son of Itctult*. To show, by an application, the difference in the results obtained by the two formulas (214-} and (215.), let it be required to find the moment of inertia for a rolled- iron beam 12 inches high and 4 inches broad, and in which the top and bottom flanges are one inch thick, and the web one half inch thick. Here we have d 12, d t 10, <4=n, / = J, =4, ^ = 3i, # = 4x1=4, a / and h = i ; and by formula (214) we have /= .^x 1 1 2 x (6x4 + 5) = 292^5- The value by formula (215.) is The value by the common formula, (213.), is 1 = AK4 x I23 )~(3 5 x io 3 )] = 284!- Thus we have by either of the two formulas (213..) or (215.) the exact value, /= 284^, while by formula (214) the value obtained is /= 292^. 476 Rolled-Iron Beam moment of Inertia Remarks. When, in a rolled-iron beam, the top and bottom flanges are comparatively thin, the difference between d t and d a will be small, and in consequence the value of 7 as derived by formula (214.) will differ but little from the truth. This formula, therefore, for such cases, is a near approximation, and for some purposes may be useful ; but formula (215.), and that from which it is derived, (213.), are exact in their results, and should be used in preference to formula (2 14-) in all important cases. LOAD AT MIDDLE RULES. 331 477. Reduction of Formula Load at Middle. The expression (213.), then, is that which is proper for the moment of inertia for rolled-iron beams namely : In Art. 303, formula (115.), we have wr " 16 This is for a beam supported at each end, with the load in pounds at the middle, the length in feet and the other dimensions in inches. F is a constant, which, from an average of experiments (Art. 701) upon rolled-iron beams, has been ascertained to be 62,000. The value of /, the moment of inertia, has been computed, for many of the sizes of beams in use, by formula (213.), and will be found in Table XVII. We have, therefore, from (115.) Wl s 12 X 62000 =-vr- 16 744000 = ~ (216) 478. Rules-Values of JF, I, 6 and JT. Rule (216) is for a load at the middle of a rolled-iron beam. The values of the several symbols in (216) may be had by transpositions, as follows : The weight, W = (217.) length, 7 = ^44^. Wl 8 t deflection, 6 = -, (219.) 744000/ Wl a moment of inertia, /= - 744000<5 332 ROLLED-IRON BEAMS. CHAP. XIX. 479. Example Weight. Formula (217.) is a rule by which to find the weight in pounds which may be carried at the middle of a rolled-iron beam, with a given deflection. As an example : What weight may be carried at the middle of a 9 inch 90 pound beam, 20 feet long between bear- ings, with a deflection of one inch ? Here we have 6 = i, / = 20 and (from Table XVII.) /= 109-117 ; and, by the formula, TT _ 744000 x log- H7 x i W= / - -^- -=10147-881 or the weight to be carried equals 10,148 pounds, or say 5 net tons. 480. Example Length. Formula (218.) is a rule by which to find the length at which a beam may be used when required to carry at the middle a given load, with a given deflection. For example : To what length may a Buffalo 6 inch 50 pound rolled-iron beam be used, when required to carry 5000 pounds at the middle, with a deflection of -^ of an inch ? Here 7=29-074 (from Table XVII. ), 6 = 0-3 and W 5000 ; and, by the iormula, /= //7440QQX 29-074x0.3 = IQ 5OOO or the length may be 10 feet 1 1 inches. 481. Example Deflection. Formula (219.) is a rule for finding the deflection in a rolled-iron beam, when carry- ing at the middle a given load. As an example : What de- flection will be caused in a Phoenix 9 inch 70 pound beam 20 feet long, by a load of 7500 pounds at the middle ? LOAD AT ANY POINT GENERAL RULE. 333 Here ^=7500, / = 20 and (from Table XVII.) / = 92-207 ; and, by the formula, 75OO X 2O 8 =087461 744000 x 92 207 or the deflection will be -J of an inch. 482. Example moment of Inertia. In formula we have a rule by which to ascertain the moment of inertia of a rolled-iron beam, laid on two supports, and carrying a load at the middle. To exemplify the rule : Which of the beams in Table XVII. would be proper to carry 10.000 pounds at the middle, with a deflection of one inch ; the length between the bearings being twenty feet ? Here W 10000, / = 20 and 6 = i, and by the for- mula, i oooo x 2o 3 / - - = 107-527 744000 x i or the required moment of inertia is 107-527. The nearest amount to this in Table XVII. is 107-793, pertaining to the Phoenix 9 inch 84 pound beam. This beam, therefore, would be the one required. 483. Load at Any Point General Rule. The rules just given are for cases where the loads are at the middle. Rules for loads at any other place in the length will now be developed. Formula (%3*\ is 334 ROLLED-IRON BEAMS. CHAP. XIX. If bd* be multiplied by -^ its value will not be changed, and there will result \2bd 3 i2 and formula &f. becomes Bv formula (154.\ in ^4r/. 376, Bl and as rl = 6, or r = -j, therefore a ~ d - Fdd Fdj For a in the above, substituting this value, we have mn _ - a = 12/73 484. Load at Any Point on Rolled-Iron Ream*. The moment of inertia, /, in formula (221.) is [form. (205.)'], I -?bd 3 for a rectangular beam. For a tube, or for a beam of the I form, it is, by formula (213.), If in (221) we substitute for / this value of it, we have iWlmn = Fd(bd 3 -b { d?) (222.) LOAD AT ANY POINT. 335 This is a rule for rolled-iron beams supported at each end and carrying a load at any point in the length, with a given deflection ; and in which W is the weight in pounds, m and n the distances from the load to the two supports, and m plus n equals / equals the length ; m, n and / all being in feet ; 6 is the deflection, b and d are the breadth and depth of the beam, b t and d t the breadth and depth of the part which is wanting of the solid bd (Art. 470) ; <?, b, d, b t and d t all being in inches ; and F is the constant for rolled iron (Table XX.). 485. Load at Any Point on Rolled-Iron Beams of Table XVII. The value of F is 62,000. If it be substituted for F in (221) we shall have 4 Wlmn 1 2 x 62OOO/d ^Wlmn = 744000! d Wlmn = i86ooo/<? i W = . Imn which is a rule for ascertaining the weight which may be carried, with a given deflection, at any point in the length of any of the rolled-iron beams of Table XVII. 486. Example. What weight may be carried on a Paterson 12^ inch 125 pound rolled-iron beam, 25 feet long between bearings, at 10 feet from one of the bearings, with a deflection of 1-5 inches ? Here we have 6 = 1-5, m 10, n = 15, / = 25 and / = 292-05 (from Table XVII.) ; and hence 186000x292.05x11 25 x lox 15 or the weight allowable is, say 21,730 pounds. 336 ROLLED-IRON BEAMS. CHAP. XIX. 487. Load at End of Rolled-Iron Lever. In formula (113.) we have Wl 3 12F =~I6- or > Wl' = I2FI6 This expression is for a beam supported at each end and loaded at the middle. In a lever the strains will be the same when the weight and length are each just one half those in a beam supported at each end. Hence if for W we take 2P, and for / take 2n, P being the weight at the end of a lever and n the length of the lever, we shall have, by substitution in the above, 2Px 2H 3 = \2FId i6Pn*= \2FId i6Pn 3 = Fd (bd 3 -b t df) (224.) and Pn 3 = %Fl6 (225.) and further, since F= 62000 (Table XX.), therefore Pn s = 465001$ 46 5 oo IS P = n~ which is a rule for ascertaining the weight which may be supported at the free end of a lever, with a given deflection, the lever being made of any one of the rolled-iron beams of Table XVII. 488. Example. Let it be required to show the weight which may be sustained at the free end of a Trenton 15^- inch 150 pound rolled-iron beam, firmly imbedded in a wall, and projecting therefrom 20 feet; the deflection not to exceed 2 inches. LOAD UNIFORMLY DISTRIBUTED. 337 Here 7=528-223 (Table XVIL), <S = 2 and n = 2O', and by formula (226.) or the weight which may be carried is 6140 pounds. 489. Uniformly Distributed Load on Rolled-Iron Beam. By formula (115.) we have This is for a load at the middle of a beam. Let U rep- resent an equally distributed load; then %U will have an effect upon the beam equal to the concentrated load W, (Art. 340), and hence, substituting this value, f// 3 = 12FI6 \Ul 3 = F6 (bd 3 -b t d^j (227.) By Table XX. F = 62000, and the formula reduces to Ul 3 1190400/6 u= ^ol ^ which is a rule for ascertaining the amount of weight, equally distributed, which, with a given deflection, may be borne upon any of the rolled-iron beams of Table XVIL 490. Example. What weight, uniformly distributed, may be sustained upon a Buffalo 10^ inch 105 pound rolled-iron beam; 25 feet long between bearings, with a deflection of f of an inch ? 338 ROLLED-IRON BEAMS. CHAP. XIX. Here 7=175-645 (Table XVII.), rf = o-75 and 7=25; and therefore, by (228.), u= 1190400 xj 7^645 ^0.75 25 or the weight uniformly distributed is 10,036 pounds. 491. Uniformly Distributed Load on Rolled-Iron Lever. A rule for a lever loaded at the free end is given in formula (225.), Pn 3 = When a load concentrated at the free end of a lever is equal to f of a load uniformly distributed over the length of the lever, the effects are equal. (Art. 347.)* If U equals the load equally distributed, and P the load concentrated at the free end, then f U P, and sub- stituting this value for P in formula (225.) gives = \2FI6 6 Un 3 = F6 (bd s -b t df) (229.) Putting for F its value 62,000, and reducing, we have Un s 1 24.000/6 124.000/6 ~~" which is a rule for ascertaining the load, uniformly distrib- uted, which may be sustained upon any of the rolled-iron beams of Table XVII. , with a given deflection, when used as a lever. * Rankine, Applied Mechanics, p. 329. LOAD ON A FLOOR ITS COMPONENTS. 339 492. Example. What weight, uniformly distributed, may be sustained upon a Trenton 6 inch 40 pound rolled-iron beam, used as a lever, and projecting 10 feet from a wall in which it is firmly imbedded; the deflection not exceeding f of an inch ? Here / = 23 761 , 6 = f and n = 10 ; and by (230.) io j or the weight will be 1965 pounds. 493. Component* of Load on Floor. When rolled- iron beams are used as floor beams, they have to sustain a compouncj load. This load may be considered as composed of three parts, namely : First : The superincumbent load, or load proper; Second : The weights of the materials within the spaces between the beams, and of the covering ; and, Third : The weight of the beams themselves. 494. The Superincumbent Load. This will be in pro- portion to the use to which the floor is to be subjected. If for the storage of merchandise, the weight will vary accord- ing to the weight of the particular merchandise intended to be stored. Warehouses are sometimes loaded heavily, and for these each case needs special computation. For general purposes, such as our first-class stores are intended for, the load may be taken at 250 pounds per superficial foot (Art. 368). A portion of the floor may in some cases be loaded heavier than this, but as there is always a considerable part kept free for passage ways, 250 pounds per foot will in general be found ample to cover the heavier loads on floors of this class. 340 ROLLED-IRON BEAMS. CHAP. XIX. On the floors of assembly rooms, banks, insurance offices, dwellings, and of all buildings in which the floors are likely to be covered with people, the weight may be taken at 66, or say 70 pounds per foot ; 66 pounds being the weight of a crowd of people (Art. 114). 495. The Materials of Construction Their Weight. These (not including the iron beam) will differ in accordance with the plan of construction. As usually made, with brick arches, concrete filling, and wooden floor laid on strips bed- ded in the concrete, this weight will not differ much from 70 pounds per superficial foot, and, in general, it may be taken at this amount. 496. The Rolled-Iron Beam Its Weight. The differ- ence in the weight of rolled-iron beams is too great to per- mit the use in the rule of a definite amount, taken as an average. To represent this weight, therefore, we shall have to make use of a symbolic expression. Let y equal the weight of the beam in pounds per lineal yard, and c equal the distance in feet between the centres of two adjacent beams. Then \y will equal the weight of the beam per lineal foot ; and this divided by c will give, as a quotient, equals the weight of beam per superficial foot of the floor. 497. Total Load on Floors. Putting together the three weights, as above, we have the total weight per super- ficial foot as follows : LOAD PER SUPERFICIAL FOOT. 341 For the floors of dwellings, assembly rooms, banks, etc., the superincumbent load is 70 pounds ; the brick arches, concrete, etc., equal 70 " y and the rolled-iron beams equal These amount in all to /= HO + For the floors of first-class stores, the superincumbent load is 250 pounds; the brick arches, concrete, etc., equal . 70 " y and the rolled-iron beams equal or, in all, 498. Floor Beam Distance from Centres. In formula >.) U stands for the weight uniformly distributed over the length of the beam. When / is taken to represent the total load in pounds per superficial foot of the floor, c the distance apart in feet between the centres of two adjacent beams, and / the length of the beam in feet, then Substituting for U in formula (228.} its value as here shown, we have 342 ROLLED-IRON BEAMS. CHAP. XIX. When r represents the rate of deflection per foot lineal of the beam, we have rl d, equals the whole deflection. Substituting for 6 in formula (234-) this equivalent value we have J * I 3 Again ; for f substituting its value as in (232^) t we have / ( ?- 140 + , = --- which is a rule for ascertaining the distance apart from cen- tres between rolled-iron beams, in the floors of assembly rooms, banks, etc., with a given rate of deflection. 4-99. Example. It is required to show at what dis- tance from centres, Paterson io| inch 105 pound rolled- iron beams, 25 feet long, should be placed in the floors of a bank, in which the rate of deflection is fixed at 0-035 f an inch. Here we have 7=191.04 (Table XVII.), r ~ 0-035, /= 25 and y = 105 ; and by (236.) 85024x191.04x0-035 105 ' = - -- -- = ' DISTANCE BETWEEN CENTRES, IN DWELLINGS. 343 or the distance from centres should be, say 3 feet 4! inches. 5 CO. Floor Beams Distance from Centre* Dwellings etc. If the rate of deflection be fixed, and at 0-03 (Art. 314), then formula (236.), so modified, becomes _ 420 which is a rule for ascertaining the distance apart from cen- tres of rolled-iron beams, in the floors of assembly rooms, banks, etc., with a rate of deflection fixed at 0-03 of an inch per foot lineal of the beam. 501. Example. What distance apart from centres should Buffalo \2\ inch 125 pound rolled-iron beams 25 feet long be placed, in the floor of an assembly room ? Here 7=286.019 (Table XVII.), 7=25 and 7=125; and by formula (237.) 25_l;0frx 286.019 _ 125 or the distance from centres should be 4! feet, or 4 feet 4J inches. The distances from centres of various sizes of beams have been computed by formula ($7*\ and the results are re- corded in Table XVIII. 344 ROLLED-IRON BEAMS. CHAP. XIX. 502. Floor Beams Distance from Centres. If in for- mula (235.) we substitute for / its value in (233.) we shall have ngoAOoIr i 90400 Ir *-i- i i goAooIr y 320^ == - Zj s - --- - i i 90400 Ir y y 320/ 5 320 x 3 c = I 3 960 This is a rule for ascertaining the distance apart from cen- tres between rolled-iron beams, in floors of first-class stores, with a given rate of deflection. 503. Example. At what distance apart should Phoenix 15 inch 150 pound beams 25 feet long be placed, with a rate of deflection of r = 0-045 ? Here we have 7=514-87 (Table XVII.), r = 0-045, /= 25 and y = 150; and in formula (238.) 3720x514-87x0-045 150 _ ~~ or the distance required is 5-36 feet, or 5 feet 4^ inches. 504. Floor Beams Distance from Centres First-class Stores. If the rate of deflection be fixed, and at 0-04 of an inch (Arts. 3l3 f 314 and 368), then formula (238.) becomes DISTANCE BETWEEN. CENTRES, "iX STORES FLOOR ARCHES. 34$ which is a rule for ascertaining the distance apart from cen- tres of rolled-iron beams, in floors of first-class stores, with a rate of deflection fixed at 0-04 of an inch per foot lineal of the beam. 505. Example. At what distance apart should Buffalo I2j inch 1 80 pound rolled-iron beams 20 feet long be placed, in a first-class store? Here 7 = 418.945 (Table XVII.), I = 20 and 7=180; and, by the above formula, 148-8 x 418-945 180 or the distance from centres should be 7-6 feet, or 7 feet 7J inches nearly. The distances from centres, as per formula (239!), have been computed for rolled-iron beams of various sizes, and the results are recorded in Table XIX. 506, Floor Arches General oniderafion. If the spaces between the iron floor beams be filled with brick arches and concrete, as in Art. 495, care is necessary that these arches be constructed with very hard whole brick of good shape, be laid without mortar, in contact with each other, and that the joints be all well filled with best cement grout and be keyed with slate. As to dimensions, the arch when well built need not be over four inches thick for spans of seven or eight feet, except for about a foot at each spring- ing, where it should be eight inches thick, and where care should be taken to form the skew-back quite solid and at right angles to the line of pressure. In order to economize the height devoted to the floor, it 346 ROLLED-IRON BEAMS. CHAP. XIX. is desirable to make the versed sine or rise of the arch small. But there is a limit, beyond which a reduction of the rise will cause so great a strain that the material of which the bricks are made will be rendered liable to crushing. Experi- ments have shown that this limit of rise is not much less than ij inches per foot width of the span, and in practice it is found to be safe to make the rise i inches per foot. 507. Floor Arclic Tie-Rods. The lateral thrust ex- erted by the brick arches may be counteracted by tie-rods of iron. The arches, if made with a small rise, will differ but little in form from the parabolic curve. Let Fig. 73 repre- sent one half of the arch and tie-rod. Draw the lines AD and DC tangent to the points A and C. Then AE = EB* FIG. 73. equals i of the span, or i^, and DE BC equals the versed sine, or height of the arch. If DE, by scale, be equal to the load upon the half arch AC, then AE equals the horizontal strain ; or DE : AE :: v : j : : \ H : H (240.) in which U is the load, in pounds, and s is the span and v the versed sine, both in feet. To resist this strain * Tredgold's Elementary Principles of Carpentry, Art. 57 and Fig. 28. FLOOR ARCHES TIE-RODS. 347 the rod must contain the requisite amount of metal. The ultimate tensile strength of wrought-iron may be taken at an average of 55,000 pounds per inch. Owing, however, to de- fects in material and in workmanship (such, for instance, as an oblique bearing, which, by throwing the strain out of the axis and along one side of the rod, would materially increase the destructive effect of the load), the metal should be trusted with not over 9000 pounds per inch. If a rep- resent the area of the tie-rod in inches, then 90000 = H Substituting this value of H in formula (240.) we have 9000* = g (*4-Z.) For U we may put its equivalent, which is the load per foot multiplied by the superficial area of the floor sustained by the rod, or U=cfs c being the distance from centres between the rods, and s the span of the arch, both in feet, and f the weight of the brick-work and the superimposed load, in pounds, or 70+^. If the arch be made to rise \\ inches per foot of width, or \ of the span, then 8v = s, and formula (2Jf.l.) becomes 7Q + ? a = - -- cs 9000 Putting q, the superimposed load, at seventy pounds, we have 140 a = -- cs 9000 which is a rule for the area, in inches, of a tie-rod in a bank, office building, or assembly room floor. 34 8 ROLLED-IRON BEAMS. CHAP. XIX. If q be put equal to 250 pounds, then 320 a = - cs 9000 a = o 03-! x cs (^ 44-) which is a rule for the area, in inches, of a tie-rod in the floor of a first-class store. For general use, the diameter, rather than the area, of the tie-rod is desirable. We have as the area of any rod, *,= .7854^ and therefore -7854^* = o-oi^ x cs and d Vo-oi^cs (45.) which is a rule for banks, etc. ; and d = t/o -04527^ which is a rule for first-class stores. 508. Example. In a first-class store, with beams 20 feet long, and arches 6 feet span : What is the required diameter of tie-rods ? Here s = 6, and if there are to be, say two rods in the length of each arch, then c 6|, and therefore d= Vo- 04527 x 6fx 6= 1-35 or the required rods are to be if inches diameter. Tie-rods should be placed at or near the bottom flange, and so close together that the horizontal strain between them from the thrust of the arch shall not be greater than the bottom flange of the beam is capable of resisting. HEADERS FOR DWELLINGS AND ASSEMBLY ROOMS. 349 509. Headers. In Art. 381 we have the expression a rule for a header of rectangular section. We have also in formula (205.) or 1 2/ = bd 3 Substituting this I2/ for b(di)* in the above equa- tion gives which is a rule for rolled-iron headers ; and in which f is the load in pounds per superficial foot, n is the length of the tail beams having one end resting on the header, and g is the length of the header ; n and g both being in feet. 510. Hcader for Dwelling*, etc. If in (2 4? ) we sub- stitute for f its value as per formula (232.), and for F its value 62,000 (Table XX.), and make r 0-03 (Art. 314), we shall have / 38-4x62000x0-03 140 + 71424 which is a rule for ascertaining the moment of inertia of a rolled-iron header, in a floor of an assembly room, bank, etc. ; from which an inspection of Table XVII. will show the required header. 350 ROLLED-IRON BEAMS. CHAP. XIX. 511. Example. In the floors of a bank, constructed of Buffalo ioj- inch 105 pound beams, placed 4 feet from centres : What ought a header to be which is 20 feet long, and which carries tail beams 16 feet long? Here y = 105, c 4, ;/ = 16 and = 20; and by 71424 or the beam should be of such size that its moment of inertia be not less than 266-577. By reference to Table XVII. we find the beam, the moment of inertia of which is next greater than this, to be the Pottsville 12 inch 125 pound beam, for which 7=276-162. This may be taken for the header, although it is stronger than needed. Should the depth be objectionable, we may use two of the Pottsville io| inch 90 pound beams, bolted together; for of this latter beam I = 150-763, and 2x 150-763 = 301-526 considerably more than 266-577, tne result of the computa- tion by formula (248.). But these two beams, although nearer the required depth, yet, when taken together, weigh 1 80 pounds per yard ; while the 12 inch beam weighs but 125 pounds. On the score of economy, therefore, it is preferable to use the 12 inch beam. 512. Header for Firt-cla Stores. If, in formula .), for /, F and r, there be substituted their proper values, namely, /= 320 + (form. 233 ), F= 62000 and r 0-04, as in Arts. 367 and 368, we shall have 320 + (249.) 95232 which is a rule for rolled-iron headers in the floors of first- class stores. CARRIAGE BEAM WITH ONE HEADER. 351 As this expression is the same as (248.), excepting the numerical coefficients, the example of the last article will suffice to illustrate it, by simply substituting the coefficient y y 320 + 140 + - - in place of 95232 71424 5(3, Carriage Beam with One Header. Formula is appropriate for a case of this kind, but it is for a beam of rectangular section. To modify it for use in this case, we have (205.) I = ^bd* ; or I2/ = bd\ Substituting for bd*, in (161.), this value, we have fmn (ng+^cl') \2lFr which is a general rule for this case. 514. Carriage Beam with One Header, for Dwellings V etc. In formula (250.), putting for f its value 140 + - (form. 232^ for F its value 62,000 (Table XX.), and for r its value 0-03 (Art. 314), we have - 1 mn /= 12 x 62000 xo-03 22320 which is a rule for the moment of inertia of a rolled-iron car- riage beam, with one header, in floors of assembly rooms, banks, etc. With the moment of inertia found by this rule, the required beam may be selected from Table XVII. 352 ROLLED-IRON BEAMS. CHAP. XIX. 515. Example. In a dwelling floor of Paterson 9 inch 70 pound beams, 20 feet long and 2 T 8 feet from centres : Of what size should be a carriage beam which at 5 feet from one end carries a header 17 feet long, with tail beams 1 5 feet long ? Here 7=70, <r=2-8, m = 5, n = 15, =17 and / 20; and by (251.) we have 140 '22320' 05 x 17 + f x 2-8 x 20) = 161 -990 or the moment of inertia required is 161-990. By reference to Table XVII. we find /= 159-597 as the moment of inertia of the 9 inch 135 pound Pittsburgh beam, being nearly the amount called for. If the construc- tion of the floor permit the use of a beam i-J inches higher, then it would be preferable to use for this carriage beam one of the Trenton zoj inch 90 pound beams; as these beams, although stronger than we require, are yet (being 45 pounds lighter) more economical. 516. Carriage Beam with One Header, for First -class Stores. If, in formula (250.), f be substituted by its value 520+ (form. 233. \ F by its value 62,000 (Table XX.), and r by 0-04 (Arts. 367 and 368), we shall have I = (320 + \mn V W 1 2 X 62OOO X O 04 320 +]*** which is a rule for the moment of inertia for rolled-iron car- riage beams, carrying one header, in first-class stores. CARRIAGE BEAM WITH TWO HEADERS. 353 517. Example. Of what size, in a first-class store, should be a rolled-iron carriage beam 25 feet long, which carries at 5 feet from one end a header 20 feet long, with tail beams 25 feet in length ; the tail beams being Trenton 12^ inch 125 pound beams, placed 2f feet from centres? Here 7125, c 2f, m 5, n = 20, g = 20 and I 25 ; and by formula (252.) we have i x 5 x 2O f== ~ 2 9 ?60 - X (20X20 + 1 X2|X 25)^545.090 or the moment of inertia required is 545-090. To supply the strength needed in this case, we may take a Trenton io| inch 135 pound beam, with one of their I2j inch 125 pound beams; as these two bolted together will give a moment of inertia a little less than the com- puted amount. It will be more economical, however, to take two of the 12 inch 125 pound beams, since the weight of metal will be less, although the strength will be greater than required. 518. Carriage Beam with Two Headers and Two Sets of Tail Beams. Formula (170 ) contains the elements appro- priate to this case, but is for beams of rectangular section. It is quite general in its application, although somewhat complicated. A more simple rule is found in formula (174.). This is not quite so general in application, but still suffi- ciently so to use in ordinary cases (see Art. 402), In any event, the result derived from its use, if not accurate, devi- ates so slightly from accuracy that it may be safely taken. We will take, then, formula (17 4-) and modify it as required 354 ROLLED-IRON BEAMS. CHAP. XIX. for the present purpose. For bd* putting I2/, its value (form. 205.), we have 1 2lFr fm [%cnl+g (mn + r )] )] (253.) which is a general rule for the case above stated (see Arts. I53 and 243). 5 1 9. Carriage Beam with Two Header and Two Set* of Tail Beam*, for Dwellings, etc. If, in formula (253. \ 140 + be substituted for / (form. 232.), 62,000 for F (Table XX.), and 0-03 for r (Art. 314), then we have as a result 140 + -- which is a rule for the moment of inertia for this case as above stated (see Arts. 153 and 243). 520. Example. In a dwelling having a floor of Pater- son loj inch 105 pound rolled-iron beams, 20 feet long, and placed 5 84 feet from centres : Which of the beams of Table XVII. would be appropriate for a carriage beam to carry two headers 16 feet long, one located 9 feet, and the other 1 $ feet, both from the same end of the carriage beam? (See Arts. 153 and 243.) Here the two headers are respectively 9 feet and 5 feet from the walls. The one 9 feet from its wall, being farther away than the other, will create the greater strain, CARRIAGE BEAM WITH TWO HEADERS, FOR STORES. 355 and therefore m = 9, n = n, r 15, J = 5, / 20, r = 5-84 and y 105 ; and by formula (254-} we have 105 te?+jjm or the required moment of inertia is 211-337. By reference to Table XVII., we find, as the nearest above this amount, the Trenton or Paterson 10^ inch 135 pound beam, of which / = 241-478, and which will be the proper beam for this case. 521. Carriage Beam with Two Header and Two Sets of Tail Beams, for First-elass Stores. If, in formula (253.), there be substituted for F its value 62,000 (Table XX.), v for f its value 320 H (form. 233.), and for r its value 0-04 (Arts. 367 and 368), we shall have y 320 + - 7 = which is a rule for the moment of inertia required in this case, as above stated (see Arts. 153 and '243). 522. Example. In a store having a floor of Trenton inch 150 pound rolled-iron beams, 25 feet long and 4-87 feet from centres : What ought a carriage beam to be which carries two headers 20 feet long, one located 10 feet from one wall, and the other at 7 feet from the other wall ? Here the distances to the header more remote from its wall are to be called (see Arts. 153 and 243) m and n. 356 ROLLED-IRON BEAMS. CHAP. XIX. Then m = 10, = 15, r = 18, 5 = 7, ^=20, /=25, r = 4-87 and 7 150; and by formula (255.) 1 50 320 + -^- 4-87 x 15 x 25) + 20(10 x 15+7')] = 695027 or the moment required is 695027. By an examination of Table XVII. , we find that the moment of the Trenton 15^ inch 200 pound beam is more than enough for this case, and its use more economical than any combination of other beams affording the requisite strength. 523. Carriage Beam with Two Headers, Equidistant from Centre, and Two Sets of Tail Beams, for Dwellings, etc. If for /, F, bd 3 and r, in formula (183.), their re- I/ spective values be substituted, namely, /= 140+ (form. >.), F= 62000 (Table XX.), bd 3 = I2/ (form. 205. \ and r 0-03 (Art. 314); then formula (183.) becomes y 140 + - 22320 (256.) which is a rule for a rolled-iron carnage beam, carrying two headers equidistant from the centre, with two sets of tail beams, in assembly rooms, banks, etc. 524. Example. In an assembly room, having a floor of Buffalo io inch 105 pound rolled-iron beams, 20 feet long and 5-35 feet from centres: What ought a carriage beam to be which carries two headers 16 feet long, located equidistant from the centre of the width of the floor, with an opening between them 6 feet wide ? CARRIAGE BEAM WITH TWO EQUIDISTANT HEADERS. 357 Here ^=5.35, jj/ = 105, / = 20, g = 16 and m = 7 ; therefore by formula (256.) we have 105 140 ' 2232Q 5 ' 35 x 20 (Ax 5-35x20'+ 16x7*)= 190761 By reference to Table XVII. we find that either the Paterson or Trenton ioj inch 105 pound beam is sufficiently strong to serve for the required carriage beam. 525. Carriage Beam with Two Headers, Equidistant from Centre, and Two Sets of Tail Beams, for First-elass Stores. In formula (183^ if we substitute for /, F, bd 3 v and r their respective values, as follows, f= 320 + -- (form. 233.), F = 62000 (Table XX.), bd 9 = 12! (form. 205.) and r = 0-04 (Arts. 367 and 368), we shall have 320+ which is a rule for a rolled-iron carriage beam carrying two headers equidistant from the centre, with two sets of tail beams, in first-class stores. 526. Example. In a first-class store, having a floor of Phoenix 15 inch 150 pound beams 25 feet long and 4-75 feet from centres : What ought a carnage beam to be which carries two headers 20 feet long, located equidistant from the centre of the width of the floor, with an opening between them 8 feet wide ? 358 ROLLED-IRON BEAMS. CHAP. XIX. Here we have ^=150, = 4-75, /=2$, g 20 and m = 8^; therefore formula (257.) becomes 320 3 x 475 29760 or the moment required is 658-813. Table XVII. shows that either of the 15 inch 200 pound beams is of sufficient strength to satisfy the requirements of this case. 527. Carriage Beam with Two Headers and One Set of Tail Beams, for Dwellings, etc. If, in formula (179.), we substitute for the symbols bd 3 , /, F and r, their respec- y tive values, as follows, bd 3 = 12! (form. 205.), f 140 + -; 6 C (form. 232.), F = 62000 (Table XX.) and ^' = 0-03 (Art. 314), we shall have 140 + I= 22320* m H /+ > ( + *M (258-) which is a rule for the moment of inertia of a rolled-iron carriage beam, carrying two headers with one set of tail beams, for floors of assembly rooms, banks, etc. (See Arts. 153 and 409.) 528, Example. In a bank having a floor of Paterson loj- inch 105 pound rolled-iron beams, 20 feet long and 5 84 feet from centres : What ought a carriage beam to be which carries two headers 16 feet long, located one at 5 feet from one wall and the other at 6 feet from the other wall, the tail beams being between them? CARRIAGE BEAM WITH TWO HEADERS, FOR STORES. 359 Here (Art. 157) m is to be put at the wider opening, hence m 6, 71=14, s = $, I 20, ^=5-84, = 16, y = /(/#+<$) = 20 ii ="9 and jy = 105 ; and by formula (258.) 105 /= -- 22320 * ' 84 x6[fx5>84xi4x20+i6x 9 x(i 4 + 5)] = 187-593 or, the moment required is 187-593. Referring to Table XVII. we find that either the Paterson or Trenton loj inch 105 pound beam will be suitable for this case. 529, Carriage Beam with Two Headers and One Set of Tail Beams, for First-class Stores. If, in formula (258), y y 320 + -- 140 + -^- we substitute (as in Art. 525) ^ for ~ we 29760 22320 shall have 320 + -^- 7 = ' m which is a rule for the moment of inertia for a rolled-iron carriage beam, carrying two headers and one set of tail beams, in a first-class store. 530. Example. In a first-class store having a floor of Buffalo 15 inch 150 pound beams 25 feet long and 4^ feet from centres : What ought a carriage beam to be which carries two headers 20 feet long, located, one at 5 feet from one wall, and the other at 8 feet from the other wall, with tail beams between them ? 360 ROLLED-IRON BEAMS. CHAP. XIX. Here (Art. 157) m = 8, n = 17, s = 5, /= 25, = 20, /= 25 -(5 +8)= 12, ^ = 4i and j> = 150; and by (269.) 320 + /= " 29760 = 682-750 which is the moment required. Either of the 15 inch 200 pound beams of Table XVII. will serve the present purpose. 531. Carriage Beam with Three Headers, the Oreatet Strain being at Outside Header, for Dwellings, etc. As in Fig. 54, floor beams are sometimes framed with two openings, one for a stairway at the wall, and another for light at or near the middle of the floor. In this arrangement the carriage beams are required to sustain three headers. Formula (190.) in Art. 425 is appropriate to this case, but is adapted to a beam of rectangular section. Substituting for bd 3 its value I2/ (form. 205), for / its value 140 + (form. 232), for F its value 62,000 (Table XX.), and for r its value 0*03 (Art. 314), we have 140 + 7 = ~ 22320* m \il+g(n+s*-v>}-\ (260) which is a rule for the moment of inertia for a rolled-iron carriage beam carrying three headers, in an assembly room, bank, etc. ; the headers placed, as in Fig. 54, so that the one causing the greatest strain shall not be between the other two. (See Arts. 252 to 254.) CARRIAGE BEAM WITH THREE HEADERS. 361 532. Example. In an assembly room having a floor of Trenton 9 inch 70 pound beams 20 feet long and 2-80 feet from centres : Of what size should be a carriage beam carrying, as in Fig. 54, three headers 1 5 feet long ; two of them located at the sides of an opening 6 feet wide, which is placed at the middle of the width of the floor, and the other header located at 3 feet from one of the side walls ? As two of these headers are equidistant from the centre of the floor, the one carrying the longer tail beams will pro- duce the greater strain upon the carriage beam (Art. 253). The distances from this header, therefore, are to be desig- nated by m and ;/ (Art. 244-), while r and s are to represent the distances from the other, and v and u are to be the distances from the third header ; the one at the stairway. Here m 7, w = 13, s = 7, t>=3, /=2O, ^=15, c 2 - 8 and y = 70 ; and by formula (260.) we have 70 140 + ' 7 = 22/20 2 ~ x 7 [(* x 2 ' 8 x 13 x 20) + 15 = 133746 which is the required moment. An examination of Table XVII. shows that the Trenton 9 inch 125 pound beam will be more than sufficient for this case. 533. Carriage Beam with Three Headers, the Greatest Strain being at Outside Header, for First-class Stores. Here, with the headers located, as in Fig. 54, so that the one causing the greatest strain in the carriage beam shall not be between the other two, the rule is the same, with the excep- 362 ROLLED-IRON BEAMS. CHAP. XIX. tion of the coefficient, as in the case last presented (form. 60.). Substituting therefore, in formula (260.), 320+-- (see form. 259.) in place of --- ~L we shall have 22320 320 + -- 1= n \%cnl+g(mn + J-v*)-\ (261.) which is a rule for the moment of inertia required for a rolled-iron carriage beam carrying three headers, in a first- class store * the headers being placed, as in Fig. 54, so that the one carrying the greatest strain shall not be between the other two. (See Arts. 252 to 254.) The example given in Art. 532 will serve to illustrate this rule, for the two rules are alike except in the coefficient, as above explained. 534. Carriage Ream with Three Headers, the Greatest Strain being at Middle Header, for Dwellings, etc. If the headers be located as in Fig. .56, so that the header causing the greatest strain in the carriage beam shall be between the other two (Arts. 260 and 264), then we have formula (194-) (in Art. 432) appropriate to this case, except that it is for a beam of rectangular section. To modify it to suit our pres- ent purpose, we have only to substitute for bd 8 , /, F and r, their respective values as in Art. 531, and we have 1= -^\?*(b*l+g?)+gn(m*-it)-] (262.) as a rule for the moment of inertia required for rolled-iron CARRIAGE BEAM WITH THREE HEADERS. 363 carriage beams carrying three headers, in an assembly room, etc. ; the headers so located that the one causing the greatest strain shall be between the other two. (See Art. 264.) 535. Example. In a bank, having a floor of Phcenix lo^ inch 105 pound rolled-iron beams, 20 feet long and placed 5 - 59 feet from centres : Of what size ought a car- riage beam to be which carries three headers, 16 feet long, placed, as in Fig. 54, so that the opening in the floor at the wall shall be 4 feet wide, and the other opening 5 feet wide, and distant 6 feet from the other wall ? The middle header in this case being the one which causes the greatest strain in the carriage beam, the distances from it to the two walls are to be called m and n. (See Arts. 244 and 253.) The header carrying the tail beams, one end of which rest upon the wall causing the next great- est strain, the distances from it to the walls are to be called r and s. The distances from the third header are v and u. We have, therefore, m = 9, n = 11, s = 6, v 4, 120, g 16, y 105 and ^=5-59; and by formula (262.} have 105 140 + - 5-59 16(11 x c/- 4 *)]= 199-597 or the required moment is 199-597. From the moments in Table XVII. we find that the Phcenix and Pittsburgh loj inch 135 pound beams are a trifle stronger than the required amount. The Trenton and Paterson ioj inch 135 pound beams are still stronger than this. Being of the same weight, either of the four named beams will serve the purpose. 364 ROLLED-IRON BEAMS. CHAP. XTX. 536. Carriage Beam with Three Headers, the Greatest Strain being at middle Header,, for First-clas Stores. Take a case where the header causing the greatest strain in the carriage beam occurs between the other two, as in Fig. 56. Formula (262.) is suitable for this case, except in its coeffi- 320 + J cient. To modify it to suit our purpose, let - g~- in 140 + formula (261.) be substituted for - - in formula (262.)] 22320 and we have y 320 + -^ which is a rule for the moment of inertia for rolled-iron car- riage beams carrying three headers, in first-class stores ; the header causing the greatest strain being between the other two. (See Art. 264.) The example given in Art. 535 will be sufficient to illus- trate this rule, as the two formulas are alike, except in their coefficients. QUESTIONS FOR PRACTICE. 537. What is the moment of inertia for a beam having a rectangular section ? 538. What is the moment of inertia for a beam of I section, or of the form of rolled-iron beams ? 539._Which of the beams of Table XVII. would be ap- propriate, when laid upon two supports 25 feet apart, to sustain 15,000 pounds at the middle, with a deflection of f of an inch ? 54-0. What weight could be sustained at 10 feet from one end of a Trenton 10^ inch 105 pound beam, 25 feet long between bearings, with a deflection of one inch ? 541. What weight uniformly distributed could be sus- tained upon a Buffalo 9 inch 90 pound beam, projecting* as a lever 1 5 feet from a wall (in which one end is firmly imbedded), with a deflection of \ an inch ? 542. In the floors of a first-class store, constructed with Phoenix 12 inch 125 pound beams, 3^ feet from centres : Which of the beams of Table XVII. ought to be used for a header 15 feet long, carrying one end of a set of tail beams 12 feet long ? 366 ROLLED-IRON BEAMS. CHAP. XIX. 543. In the floor of a first-class store, constructed with 12 inch 125 pound beams 2\ feet from centres : Which of the beams of Table XVII. ought to be used for a carriage beam 25 feet long between bearings, carrying, with 0-04 of an inch per foot deflection, a header 20 feet long, located at 7 feet from one end of the carriage beam, and carrying one end of a set of tail beams 18 feet long? 54-4. In the floor of a first-class store, constructed of 15 inch 150 pound beams 4^ feet from centres : What size should be a carriage beam 25 feet long, which carries two headers 19 feet long, one located at 9 feet from one wall, and the other at 8 feet from the other wall ; the two head- ers having an opening between them ? 545. In the floor of a bank, constructed of loj inch 105 pound beams 22 feet long, and placed 4 feet 4 inches from centres : Of what size should be a carriage beam which carries three headers, 16 feet long, and located, as in Fig. 56, so that one opening at the wall shall be 3 feet wide, and the other opening 6 feet wide, with a width of floor of 6 feet between the two openings ? CHAPTER XX. TUBULAR IRON GIRDERS. ART. 546. Introduction of the Tubular Girder. Dur- ing the construction of the great tubular bridges over the Con way River and the Menai Straits, Wales (1846 to 1850), engineers and architects were moved with new interest in discussions and investigations as to the possibilities of con- structions involving transverse strains. Since the complete success of those justly celebrated feats of engineering skill, the tubular girder (Fig. 74), as also the plate girder (Fig. 67), and the rolled-iron beam (Fig. 68), all of which owe their utility to the same principle as that involv- ed in the construction of the tu- bular girder, have become deserv- edly popular. They are now extensively used, not only by the engineer in spanning rivers for the passage of railway trains, but also by the architect in the lesser, but by no means unimportant, work of constructing floors over FlG - 74- halls of the largest dimensions, without the use of columns as intermediate supports. 547. Load at Middle Rule Essentially the Same as that for Rolled-Iron Beams. The capacity of tubular gir- TUBULAR IRON GIRDERS. CHAP. XX. ders may be computed by the rules already given. For example : Formula (216.) affords a rule for a load at the middle of a rolled-iron beam, in which (form. 213.), whereof b is the width of top or bottom flange, and b t equals b, less the thickness of the two upright parts, or webs ; d is the entire depth, and d, is the depth, or height, in the clear between the top 'and bottom flanges, bd then is the area of the whole cross-section, measured over all, while b t d t represents the area of the vacuity, or of so much of the cross- section as is wanting to make it a solid. The numerical coef- ficient in formula (216) is based upon a value of F equal to 62,000, which is the amount derived from experiments on solid rolled-iron beams. For built beams, such as the tubular girder, F by experiment would prove to be less, but the formula (216) may be used as given, provided that proper allowance be made in the flanges on account of the rivet holes ; that is, taking instead of the actual breadths of the flanges only so much of them as remains uncut for rivets. 548. Load at Any Point Load Uniformly Distributed. For a load at any point in the length of a beam, formula (222.) will serve, while for a load uniformly distributed, for- mula (228) affords a rule. In general, any rule adapted to rolled-iron beams will serve for the tubular or plate girder, by taking as the areas of metal the uncut portion only. 54-9. Load at Middle Common Rule. The rules just quoted are not those which are generally used for tubular beams. Preliminary to planning the Conway and Britannia LOAD AT MIDDLE. 369 tubular bridges, the engineers tested several model tubes, and from them deduced the formula in which C is a constant, found to be equal to 80 when W represents gross tons. Changing W to pounds, we have 2240 x 80 x a'd a'd W- j -=i792oo-y- This is for the breaking weight. Taking the safe weight as 9000 pounds per inch, or of the breaking weight, we have - = 35840 and, as an expression for the safe weight, the area of the bottom flange equals Wl a = 35840^ or, if instead of the above constant, 80, we put 80-357, we shall have our constant in round numbers, thus, Wl a ' = which is a rule for the area of the bottom flange of a tubular girder, with the load at the middle; a' being in inches, / and d in feet, and W in pounds. This rule is identical with formula (265.), deduced in another manner. 550. Capacity by the Principle of Moments. Gene- rally, the strength of tubular beams is ascertained by the principle of moments or leverage. Sufficient material must be 37 TUBULAR IRON GIRDERS. CHAP. XX. provided in the top flange to resist crushing, and in the bot- tom flange to resist tearing asunder, while the material in the web or upright part should be adequate to resist shearing. 551. Load at Middle Momeiat. We will first con- sider the requirements in the flanges. The leverage, or action of the power tending to break the beam, as also that of the resistance of the materials, is repre- sented in Figs. 8 and 9. When the load upon a beam is con- centrated at the middle, it acts with a power of half the weight into half the length of the beam (Art. 35), and the tension thereby produced in the bottom flange is resisted by a leverage equal to the height of the beam ; or, if d equals the height of the beam between the centres of gravity of the cross-sections of the top and bottom flanges, and T equals the amount of tension produced in the lower flange by the action of a weight W upon the middle of the beam, then Again, if k equals the pounds per square inch of section with which the metal in the lower flange may be safely trusted, and a' equals the area in inches in the bottom flange, then a'k = T, and = a'kd Wl which is a rule for the area of the bottom flange of a tubular girder, loaded at the middle, and in which W and k are in pounds, a is in inches, and d and / are in feet. (The COMPUTATION BY MOMENTS. 371 area of the top flange is to be made equal to that of the bottom flange. See Art. 456.) If k be taken at 9000, as in Art. 549, then 4/ = 36000, and formula (265) becomes identical with formula (264). 552. Example. What area of metal would be required in the bottom flange of a tubular girder 40 feet long and 3 feet high, to sustain at the middle 75,000 pounds; 9000 pounds being the weight allowed upon one inch of the wrought-iron of which the flanges are to be made ? Here W =. 75000, / = 40, d 3 and k 9000 ; and we have, by formula (265), 75000 x 40 "-27.77 4 x 3 x 9000 or the area equals 27^ inches. This is the amount of metal in addition to that required for rivet holes. 553. Load at Any Point. A load concentrated at any point in the length of the beam acts with a leverage equal to W y- (see Art. 56), and the resistance is Td=a'kd', therefore which is a rule for this case, as above stated, in which a f is in inches, W is in pounds, and m, n, d and / are in feet. 554. Example. What amount of metal would be re- quired in the bottom flange of a tubular girder 50 feet long 37 2 TUBULAR IRON GIRDERS. CHAP. XX. and 3^ feet high, to sustain a load of 50,000 pounds at 20 feet from one end, when k = 9000 ? Here W 50000, ;;/ = 20, n = 30, d = 3^, k 9000 and / = 50 ; and, by formula (266.), 50000 x 20 x 30 *'= *- = 19-05 3^ x 9000 x 50 or the area should have 19 inches of solid metal, uncut by rivet holes. The top flange should contain an equal amount. (See Art. 456.) 555. Load Uniformly Distributed. For this load the effect at any point in the beam is equal to that of half the load, if concentrated at that point (see Art. 214); or, from formula which is a rule for the area of the bottom flange at any point in its length, and in which a' is in inches, U is in pounds, and m, n, d and / are in feet. 556. Example. In a tubular girder 50 feet long, 3^ feet high, and loaded with an equably distributed load of 120,000 pounds : What should be the area of the bottom flange at the middle, and at each 5 feet of the length thence to each support, k being taken at 9000 ? Here U = 1 20000, d 3 , k = 9000 and / = 50 ; and by formula (267.) we have 12OOOOMH a > . 2 x 3J * 9000 x 50 UNIFORMLY DISTRIBUTED LOAD. 3/3 When m = n 25, then a' = 0-038095 x 25 x 25 = 23-81 or the area required in the bottom flange at mid-length is 23-81 inches. When m = 20, then n = 30, and a' = o 038095 x 20 x 30 = 22 86 or the required area at 5 feet from the middle, either way, equals 22-J inches. When m=i$, then n 35, and a' = 0-038095 x 15 x 35 = 20-00 or, at 10 feet each side of the middle, the area should be 20 inches. When m = 10, then n 40, and a' = 0-038095 x 10 x 40 = 15-24 or, at 15 feet each side of the middle, the area should be I5J- inches. When m = 5, then n = 45, and a' = 0-038095 x5 x45 = 8-57 or, at 20 feet each side of the middle, the area should be 84 inches. 557. Thickness of Flanges. In the results of the ex- ample just given, it will be observed that the area of metal required in the flanges increases gradually from the points of support each way to the middle of the beam (see Art. 178). In practice, this requirement is met by building up the flanges with laminas or plates of metal, lapping on according 3.74 TUBULAR IRON GIRDERS. CHAP. XX. to the computed necessary amount. In this process, the plates used are generally not less than J of an inch thick. For an example, take the results just found. Adding, say | for rivet holes, and dividing the sum by the width of the girder, which we will call 12 inches, there results as the thickness of rnetal required, at the middle, 2-31, say 2^ inches ; " 5 feet from middle, 2-22, " 2j " " 10 " " 1-95, " 2 " 15 " " 1-48, " i " " 20 " " 0-83, " i inch. 558. ontruetion of Flanges. The girder of the last article might be built with the two flanges in plates 12 inches wide, thus : Lay down first a plate one inch thick the whole length of the girder. (With an addition for supports on the walls, say -fa of the length, or 2^ feet at each end, this plate would be 55 feet long.) Upon this place a plate inch thick and 40 feet long ; on this a plate \ inch thick and 30 feet long; on this a plate J inch thick and 20 feet long; and on this a plate \ inch thick and 10 feet long. The plates are all to extend to equal length each side of the middle of the girder, and to be well secured together by riveting. The longer plates, probably, will have to be in more than one piece in length. Where heading joints occur, a covering plate should be provided for the joint and riveted. 559. Shearing Strain. A sufficient area having been provided in the top and bottom flanges to resist the com- pressive and tensile strains, there will be needed in the web metal sufficient to resist only the shearing strain. This strain SHEARING STRAIN. 375 is, theoretically, nothing at the middle of a beam uniformly loaded, but from thence increases by equal increments to each support, at which place it is equal to one half of the whole load (Arts. !72 and 174). For example : In the case considered in Art. 556, the beam, 50 feet, long, carries 120,000 pounds uniformly distributed over its whole length ; half of the load over half of the beam. At the centre, the shearing strain is nothing ; at 5 feet from the centre, it is equal to -- of half the load, or is equal to 12,000 pounds ; at 10 feet it is 24,000; at 15 feet it is 36,000; at 20 feet it is 48,000; and at 25 feet, or at the supports, it is 60,000 pounds, or half the whole load. 560. Thickiies of Wefo. If G be put for the shear- ing stress, then G = a'k' in which a is the area in inches of the web at the point of the stress, and k' is the effective resistance of wrought- iron to shearing, per inch area of cross-section. If t equals the thickness, and d the height of the web, then a' = td, and the above equation becomes G = k'td * = w which is a rule for the thickness of the web, at any point in the length of the beam, and in which t and d are in inches. 561. Example. What should be the thickness of the web of the tubular girder considered in Art. 556, computed 376 TUBULAR IRON GIRDERS. CHAP. XX. at every 5 feet in length of the girder ? If k' be taken at 7000 pounds, it will be but little more than three quarters of 9000, the amount taken in tensile strain (Art. 173),* and taking d at, say 38 inches, we have, by formula (268.\ 38 x 7ooo 266000 Therefore, when G equals 60,000 (Art. 559), then 60000 t = -z =0-225 266000 When G equals 48,000, then t = ^, - = o- 180. When G equals 36,000, then = -ig = 0-135. As these are the greater of the strains, and are all below the practical thickness in girders, it is not worth while to compute those at the remainder of the stations. 562. Conitruction of Web. From the results in the last article, it appears that in this case the web is required, of necessity, to be only a quarter of an inch thick in its thickest part, at the supports. With an increase of load, the thick- ness of the web would increase, for by the formula it is directly as the load. The thickness of web just computed is the whole amount required in the two sides of the girder. In practice, it is found unwise to use plates less than a quarter of an inch thick. Following this custom, the two sides of the girder * The resistance to shearing is generally taken at three quarters of the ten- sile strength (see Haswell's Engineers' and Mechanics' Pocket Book, p. 485 Weisbach's Mechanics and Engineering, vol. 2, p. 77). CONSTRUCTION OF WEB. 377 taken together would be half an inch thick, more than twice the amount of metal actually required. Hence it may justly be inferred that in similar cases the plate beam (Fig. 67) would be preferable to the tubular girder, as its web, being single, would require only half the metal that would be re- quired in the two sides of the tubular girder. It is also pre- ferable for the reason that it is more easily painted, and thus kept from corrosion. On the other hand, a tubular beam is stiffer laterally. In the construction of the web, as a precau- tion to prevent buckling, or contortion, it is requisite to pro- vide uprights of T iron, at intervals of, say 3 feet on each side, to which the web is to be riveted. 563. Floor Girder Area of Flange. If for U in for- mula (267.), there be substituted its value in a floor, c'fl, of which c' is the distance from centres between girders, or the width of floor sustained by the girder, / is the length of the girder between supports (both in feet), and / is the load per foot superficial upon the floor, including the weight of the materials of construction, then which is a rule for the area of the bottom flange of a tubular girder, sustaining a floor, and in which a' is in inches and c' t m, n and d are in feet. 664. Weight of the Girder. In estimating the load to be carried by a girder, the estimate must include the weight of the girder itself, It is desirable therefore to be able to 378 TUBULAR IRON GIRDERS. CHAP. XX. measure its weight approximately before its dimensions have been definitely fixed. The weight of a tubular girder will be in proportion to its area of cross-section (which will be approximately as the load it has to carry), and to its length (form. 265^) ; 'or, when U is the gross load to be car- ried, and / the length between bearings, then the weight of the girder between the bearings is in which n is a constant, and U is the whole load, includ- ing that of all the materials of construction. The value of n, when derived from so large a structure as that of the tu- bular bridge over Menai Straits, is about 600, but from sev- eral examples of girders from 35 to 50 feet long, in floors of buildings, its value is found to be about 700. For our purpose, then, we have n 700. If for U we put its equivalent c'fl, as in Art. 563, then (270.} 7oo This is the weight of so much of the girder as occurs within the clear span between the supports. 565. Weight of Girder per Foot Superficial of Floor. The area of the floor supported by a girder is c'l. Dividing K by this, the quotient will be /', the weight of the girder per foot superficial of the floor, thus : j^X. -72 - fL J ~ c'l ~ c'l ~~ 700 Now /, the total load per foot superficial of the floor, com- WEIGHT OF GIRDER. 379 prises the superimposed load, the weight of the brick arches, etc., and the weight of the girder /'; and, putting m for the weight of all else save that of the girder, we have f m +/' and, from the above, 1L~ (? 700 Im+fl = = * 700 700 * 700 ;oo/' a= lm 700/-/7 = lm f = 700 / which is a rule for ascertaining the weight per foot superfi- cial of the floor due to the tubular girder. 566. Example. A floor, the weight of which, including that of the superimposed load, is 140 pounds per superficial foot, is carried upon a girder 50 feet in length between its bearings. What additional amount per foot superficial should be added for the weight of the girder? Here / 50 and m 140, and by (27 1.\ _ 700 50 or the weight to be added for the girder is lof pounds. Then f= m+f 140+ lof = 150} pounds. 567. Total Weight of Floor per Foot Superficial, in- eluding Girder. In the last article m represents the weight of one foot superficial of a floor, including the load to be car- 380 TUBULAR IRON GIRDERS. CHAP. XX. ried ; also, f represents the weight due to the girder ; or, for the total load, f=m+f. Using for /' its value as in formula (27 1.) we have / m+f = 7oo / /= m(i -f 7 ) J \ 700 // f=m 700- 700 700 / and for m, taking its value as given in formula (232.\ it being there represented by f, which is the value of f, the total load per superficial foot of the floors of assembly rooms, banks, etc., to be used in the calculation of tubular girders ; and taking the value of m, as expressed in formula (233.) we have /=( which is the corresponding value of f for the floors of first- class stores. 568. Girders for Floors of Dwellings, etc. If in formula (269.), we substitute for / its value as in formula (272.), we shall have / y\ 700 c'mn a' = (ijp + f-Jife-r? x 3<V700- which is a rule for the area of the bottom flange of a tubular girder, supporting the floor of an assembly room or bank, and in which a' is in inches, and c, /, c', m, n and d are in feet. TO SUPPORT FLOORS OF DWELLINGS. 381 569. Example. In a floor ot 9 inch 70 pound beams, 4 feet from centres: What ought to be the area of the bottom flange of a tubular girder 40 feet long between bearings, 2| feet deep, and placed 17 feet from the walls, or from other girders ; the area of the flange to be ascer- tained at every five feet in length of the girder ? Here y = 70, c = 4, /= 40, c' = 17 and d 2f. Putting k at 9000 we have, by (2 74-), , ( 70 \ 700 17 a = ( 140 + x - x mn 3x4/700-40 2X2f The values of m and n are as follows: At the middle, m = 20 and n = 20 " 5 feet from middle, m 15 " n = 25 " 10 " " " m = 10 " n = 30 " 1-5 " " " m = 5 " n = 35 from which the values of a' are as follows: At the middle, a' = 0-05478 x 20 x 20 = 21 -91 " 5 feet from middle, a' = 0-05478 x 15 x 25 = 20-54 " 10 " " " a' 0-05478 x 10 x 30 = 16-43 " 15 " " " a' = 0-05478 x 5X35= 9-59 These are the areas of cross-section of the lower flange, at the respective points named. The top flange is to be of the same size. (See Art. 456.) 570. Girder for Floors of Firt-cla Storci. If, in for- mula (2 74,), 320 be substituted for 140 (see form. 233.), we shall have 7oo ( TUBULAR IRON GIRDERS. CHAP. XX. which is a rule for the area of the bottom flange of a tubular girder in a first-class store. [The area of the upper flange should be made equal to that of the bottom flange (Art. 456).] As this rule is similar to (274-}, the example given to illustrate that rule will suffice also for this. 571. Ratio of Depth to Length, in Iron CJirder. In order that the requisite strength in tubular girders may be attained with a minimum of metal, the depth of a girder should bear a certain relation to the length. To deduce a rule for this ratio from mathematical considerations purely, is not an easy problem. Baker in his work on the Strength of Beams, p. 288, discusses the subject at some length. No more will be attempted here than to obtain a rule based upon some general considerations, and upon results tested and corrected by experience. 572. Economical Depth. In the construction of tubular girders for the floors of large buildings, it is found in practice* to be unadvisable to use plates of a less thickness than one quarter of an inch. If each side of the girder be a quarter of an inch thick, then the least thickness for the web (using this term technically) is a half inch. This is more than is usually found necessary, in this class of girders, to resist shearing (Art. 562). As the thickness is thus fixed, there- fore the area of the web will be in proportion to its height, and consequently it is advisable, in so far as the web is con- cerned, to have the depth of the girder small ; but, on the other hand, as the area of the flanges is inversely propor- tional to the 'depth (see form. 65.), a reduction of the flanges will require that the depth be increased. The cost of the girder is in proportion to its weight, which is in proportion ECONOMICAL DEPTH. 383 to its area of cross-section, and hence the desirability of making both as small as possible. The area of the flanges is, by formula (265.), in propor- Wl tion to -T7, and, as before shown, the area of the web will be in proportion to its height ; or the whole area will Wl be in proportion to -rr + d ; and the problem is to find such dk a value of d as will make this expression a minimum. Putting the differential of this equation equal to zero, we find that the area of the cross-section of the beam will be a minimum when Wl in which x is a constant, to be derived from experience, and which, by an application of the formula to girders of this class, is, when the weight is equally distributed, found to be equal to 30. This reduces the formula to and when for U its value c'fl is substituted which is a rule for ascertaining the economical depth of a tubular girder ; a rule useful in cases where the depth is not fixed by other considerations. 573. Example. In a floor where the girders are 50 feet long and placed 15 feet from centres, and where the total load per foot superficial is 155 pounds: What would be the most economical depth for the girders ? TUBULAR IRON GIRDERS. CHAP. XX. Here /= 50, c =15, /= 155 and k 9000, equals the safe tensile power of wrought-iron ; and by (277.) 30 x 9000 or the depth should be 4 feet ;| inches. The depth may be found by this formula, and then the area of flanges by formula (2 74-) for assembly rooms, banks, etc. ; or, by for- mula (275.} for first-class stores. QUESTIONS FOR PRACTICE 574. In a tubular girder 50 feet long, 3 feet 4 inches high, and loaded with 100,000 pounds at the middle : What ought to be the area of each of the top and bottom flanges, when the metal of which they are made may be safely trusted with 9000 pounds per inch ? 575. In the same girder: What should be the area of the top or bottom flange, if the load of 100,000 pounds be placed at 15 feet from one end, instead of at the middle of the beam? 576. In a tubular girder 50 feet long, 40 inches high, and uniformly loaded with 200,000 pounds : What should be the area of the top and bottom flanges, at every five feet of the length of the girder? 577. In the same girder: What ought to be the thick- ness of the web, at every five feet of the length of the girder, to effectually resist the shearing strain ? QUESTIONS FOR PRACTICE. 385 578. In a tubular girder 40 feet long, 32 inches high, sustaining, with other girders and the walls, the floor of an assembly room, composed of 9 inch 70 pound beams 5 feet from centres, the girders being placed 16 feet from centres : What should be the area of each of the top and bottom flanges, at every five feet of the length, the metal in the flanges being such as may be safely trusted with 9000 pounds per inch ? 579. In a floor, where the depth of the tubular girders is not arbitrarily fixed, where the girders are 42 feet long and placed 17 feet from centres, and where the total load to be carried is 160 pounds per superficial foot : What would be the proper depth of the girders, putting the safe tensile strain upon the metal at 9000 pounds ? CHAPTER XXL CAST-IRON GIRDERS. ART. 580. Cast-Iron Superseded by Wrouglit-Iron. The means for the manufacture, of rolled-iron beams (Chapter XIX.) have so multiplied within the last ten years that the cost of their production has been much reduced, and as a consequence this beam is now so extensively used as to have almost entirely superseded the formerly much used cast-iron beam or girder. Beams and girders of cast-iron, however, are still used in some cases, and it is well to know the proper rules by which to determine their dimensions. A few pages, therefore, will here be devoted to this purpose. 581. Flanges Their Relative Proportion. In Fig. 75 we have the usual form of cross-section of cast-iron beams, in which the bottom flange AB contains four times as much metal as the top flange CD. It was customary, fifty years since, to make the top and bottom flanges equal. (See Tredgold on Cast-Iron, Vol. I., Art. 37, Plate I.) Mr. Eaton Hodgkinson (who in 1842 edited a fourth edition of Tredgold's first volume, and in 1846 added a second volume to that valuable work) made many important experiments on cast-iron. Among the valuable deduc- FIG. 75. tions resulting from these experiments was this : that cast- PROPORTIONS BETWEEN FLANGES AND WEB. 387 iron resists compression with about seven times the force that it resists tension (Vol. II., Art. 34) ; and that the form of section of a beam which will resist the greatest transverse strain, is that in which the bottom flange contains six times as much metal as the top flange (Vol. II., Art. 138, page 440). If beams of cast-iron for buildings were required to serve to the full extent of the power of the metal to resist rupture, the proportion between the areas of top and bottom flanges should be as i to 6. If, on the other hand, they be subjected only to very light strains, the areas of the two flanges ought to be nearly if not quite equal. In view of the fact that in practice it is usual to submit them to strains greater than the latter, and less than the former, therefore an average of the proportions required in these two cases is that which will give the best form for use. Guided by these considerations, it is found that when the flanges are as i to 4, we have a proportion which approximates very nearly the requirements of the case. 582. Flaiige ami Web Relative Proportion. The web, or vertical part which unites the top and bottom flanges, needs only to be thick enough to resist the shearing strain upon the metal ; a comparatively small requirement. Owing, however, to a tendency in castings to fracture in cooling, the thickness of the web should not be much less than that of the flanges, and the points of junction between the web and flanges should be graduated by a small bracket or easement in each angle. (Tredgold's Cast-Iron, Vol. II., Art. 124.) The thickness of the three parts web, top flange and bottom flange may with advantage be made in propor- tion as 5, 6 and 8. Made in these proportions, the width ot the top flange will be equal to one third of that of the bottom flange ; for if w t equal the width of the bottom flange and w tl that of the top flange, t t equal the thick- 388 CAST-IRON GIRDERS. CHAP. XXI. ness of bottom flange and t u that of the top, a, equal the area of the bottom flange and a a the area of the top flange, then a = wf t and w t ~~ ; also, a a = w a t a and w u = - ; and from these, remembering that a t = 4^, and that /, : t, : : 6 : 8, we have or the width of the top flange equals one third of that of the bottom flange. 583. Load at middle. Mr. Hodgkinson found, in his experiments, that the strength was nearly as the depth and as the area of the bottom flange. For the breaking weight, IV, he gives an expression for the relative values of the dimensions and weight ; in which W is the breaking weight at the middle, / the length of the beam, d its depth, a t the area of the bottom flange, and c is a constant, to be derived from ex- periment. This constant, when the weight was in tons and the dimensions all in inches, he found to be 26. Taking the weight in pounds and the length in feet, we have 4853^ for the constant, or say 4850, and therefore jr=4 85ogX When a is the factor of safety, _ al LOAD AT MIDDLE. 389 which is a rule for the area of the bottom flange of a cast- iron beam, required to sustain safely a load at the middle. The area of the top flange is to be made equal to , and 4 the thicknesses of the web and top and bottom flanges are to be in proportion as 5, 6 and 8. 584. Example. What should be the dimensions of the cross-section of a cast-iron beam 20 feet long between supports and 24 inches high at the middle, where it is to carry 30,000 pounds ; with the factor of safety equal to 5 ? Here W ' 30000, 0=5, / = 20 and d= 24 ; and, by formula (279.), 30000 x 5 x 20 ' =25 ' 773 or the area of the bottom flange should be 25! inches. Now the thickness will depend upon the width, and this is usually fixed by some requirement of construction. If the width be 12 inches, then the thickness of the bottom flange will be ^?=2 -IS, or 2\ inches full. The width of the top flange will equal - = 4 (see Art, 582), and its thickness will be |/ / = |-x2-i5=i.6i, or if inches ; while the thickness of the web will be -/, = |- x 2 1 5 = I 34, or i- inches. 585. Load Uniformly Distributed. A load uniformly distributed will have an effect at any point in a beam equal to that which would be produced by half of the load if it were concentrated at that point (Art. 214). Therefore, if 390 CAST-IRON GIRDERS. CHAP. XXL U equals the load uniformly distributed, %U W in for- mula (879.), or \Ual a, = 4850^2? Ual a. = - (280.) which is a rule for cast-iron beams to carry a uniformly distributed load. This is precisely the same as the previous rule, except in the coefficient. The example given in Art. 584 will therefore serve to illustrate this rule, as well as the previous one. 586. Load at Any Point Rupture. From formula (78.) we have Wl = ca,d and, by a comparison of formulas (&1.) and therefore, in the above, substituting this value of Wl, we obtain (*) which is a rule for the area of the bottom flange at any point in the length of the beam. The weight given by this rule is just sufficient to rupture the bottom flange. RULES FOR SAFE LOADS. 39! 587. Safe Load at Any Point. The value of c, for a concentrated load, is (Art. 583) 4850, hence 4 = 4 i C ~~ 4850 ~~ 1212^ In formula (281.), substituting for : this value, and in- serting a, the factor of safety, then Wamn which is a rule for the area of the bottom flange at any point ; W, the safe load, being concentrated at that point. 588. Example. In a cast-iron beam 20 feet long be- tween bearings : What should be the area of the bottom flange at eight feet from one end, at which point the beam is 20 inches high and carries 25,000 pounds; the factor of safety being equal to 5 ? Here ^=25000, 0=5, m = S, n = 12, d2Q and l2Q\ and by formula (282.) 25000 x 5 x 8 x i2 I2I2|X 20 X 2C or the area should be 24f inches. 1212-j-X 20 X 20 2 4'74 589. Safe Load Uniformly I>itributed Effe<* at Any Point. This effect at any point is equal to that produced by half the load were it all concentrated at that point (Art. 214); therefore, if U represent the uniformly distributed load, then by formula (282.) 392 CAST-IRON GIRDERS. CHAP. XXI. which is a rule for the area of the bottom flange of a cast- iron beam at any point, to carry safely a uniformly dis- tributed load. If the depth of the beam remain constant throughout the length* then a t will vary as the rectan- gle mn. From formula (283.) we have which is a rule for the depth of a beam at any point, to carry safely a uniformly distributed load. If the area of the bottom flange remain constant throughout the length, then d will vary as the rectangle mn. 590. Form of Web. By the last formula, (284.), it will be seen that when a t , ' the area of the bottom flange, re- mains constant throughout the length of the beam, then the depths will vary in proportion to the rectangle of the two segments, m and ;/, of the length. The corresponding curve which may be drawn through the tops of the ordinates denoting the various depths, is that of a parabola (Art. 212). Instead of computing the depths at frequent intervals, there- fore, it will be sufficient to compute the depth at the centre only, and then give to the web the form of a parabola. 591. Two Concentrated Weight Safe Load. Formula is appropriate for a concentrated load at any point in the length of a beam, and formula (30.) is for two concen- trated loads at any given points. A comparison of these formulas shows that ~ LOADED WITH TWO WEIGHTS. 393 In Art. 586 we have which is an expression for the breaking load. Inserting a, the symbol of safety, in this expression, we have mn ca t d 4 Wa , an expression for the safe load for cast-iron beams. If for the second member of this equation there be substituted its value as above, we shall have ca / d = 4a-(Wn+Vs) an expression for two concentrated safe loads. From this we have A 1 1 J Vs) In Art. 587 we have = =, therefore c 1212%' m or t d = a -,- ( Wn + Vs) m a-(Wn+Vs) which, in a beam carrying two concentrated loads, is a rule 394 CAST-IRON GIRDERS. CHAP. XXL for the area of the bottom flange at the location of of the loads, as in Fig. 76; and (see Art. 153) one Wm) (286.) which, in a beam carrying two concentrated loads, is a rule for the area of the bottom flange at the location of F, one of the loads, as in Fig. 76. 592. Examples. As an application of rules (285.) and (286), let it be required to ascertain the dimensions of a cast- iron girder to sustain a brick wall in which there are three windows, as in Fig. 76, so disposed as to concentrate the weight of the wall into two loads, as at W and V. Let /, the length in the clear of the supports, = 20, m 7, n = 13, s = 6 and r = 14 feet, and the height of the girder at W and V equal 25 inches. Also, let the wall be 16 inches thick, and so much of it as is sustained at W measure 250 cubic feet, at no pounds per foot, or 27,500 pounds. Likewise, suppose the weight upon V to equal 27,000 pounds. Taking the factor of safety at 5 we now have, by formula (285. \ FIG. 76. a 5 x A K 2 75QQ x 13) + (27000 x 6)] _ X 25 29-99 SUSTAINING TWO BRICK PIERS. 395 or the area of flange is required to be 30 inches at W\ and, by formula (286.), 5 x A [(27000 x 14) + (27500 x 7)] a, =. -- 1 -- - -- = 2o-23 ' 1212^x25 or the area of flange is required to be 28^ inches at V. As the wall is 16 inches thick, the width of the bottom flange should be 16 inches, and its thickness therefore should be ^ = 1-875 inches at W ? 8 ? i ft . i 764 inches at V From W to V the thickness is to be graded regularly from 1-875 to 1-764; while from W to the end 'next W it is to be equal to that at W, i-J- inches thick, and from V to the end next V it is to be if inches thick. The width of the top flange is to be (Art. 582) one third of the width of the bottom flange, or ig- = 5-J- inches. Pro- portioning the three parts as 5, 6 and 8 (Art. 582), the thickness of the top flange will be -| x if = i^ inches at V I x i-J= i-J-f inches at W The thickness is to be graded regularly between W and V, and thence to each end of the beam the thickness is to be that of W and V respectively. The web is to be of the shape shown in Fig. 76, and is to be (Art. 582) m \ x if = i^\- at V and |xi|=iii at W or, say i inches, averaging it throughout. 39 6 CAST-IRON GIRDERS. CHAP. XXI. 593. Arclied Girder. A beam such as shown in Fig. 77 is known as the " bow-string girder," in which the curved part is a cast-iron beam of the T form of cross-section, and the feet of the arch are held horizontally by a wrought-iron tie-rod. This beam, although very pop- FIG. 77. u lar with builders, is by no means worthy of the confidence which is placed in it. With an appearance of strength, it is in reality one of the weakest beams used. Without the tie-rod its strength is very small, much smaller than if the T section were reversed so as to have the flange at the bottom, thus, J. (Tredgold, Vol. II. , pp. 414 and 415). 594. Tie-Rod of Arched Girder. The action of a con- centrated weight at the middle of a tubular girder, in producing tension in the bottom flange, is explained in Art. 551. The tension in the tie-rod of an arched girder is produced in precisely the same manner, and therefore the rule {form. 265.) there given will be applicable to this case, when modified as required for a uniformly distributed load ; or, for W substituting its value, \U (Art.SQS). Then, upon the presumption that there is sufficient material in the cast arch to resist the thrust, we have in which d is in feet. If d be taken in inches, then Ul (287.) TIE-ROD OF ARCHED GIRDER. 397 which is a rule for the area of the cross-section of the tie- rod in an arched girder ; in which a t is the area of the cross-section of the rod, U is the weight in pounds equally distributed over the beam, / is the length in feet between the supports, d, in inches, is the depth or versed sine of the arc, or the vertical distance at the middle of the beam from the axis of the tie-rod to the centre of gravity of the cross- section of the cast-iron arch, and k v is the weight in pounds which may safely be trusted when suspended from the end of a vertical rod of wrought-iron of one square inch section. If this latter be put at 9000 pounds, then Ul Now a t is the area of the tie-rod. The area of any circle is equal to the square of its diameter multiplied by -7854, or ,== -7854^ and since, by formula (287. \ therefore and If k, the safe resistance to tension per inch, be taken at 9000- pounds, the rule becomes which is a rule for the .diameter of the tie-rod of an arched girder. 398 CAST-IRON GIRDERS. CHAP. XXI. 595. Example. What should be the diameter of the tie-rod of an arched girder, 20 feet long in the clear between supports, and 24 inches high from the axis of the tie-rod to the centre of gravity of the cross-section at the middle of the arched beam ; the load being 40,000 pounds equally dis- tributed over the length of the beam ? Here we have U = 40000, / 20 and ^=24; and therefore, by formula (289.\ 4712 x 24 or the diameter of the rod, with the safe resistance to tension taken at 9000 pounds, should be 2f inches. 596. Substitute for Arched Girder. The cast-iron arch of an arched girder serves to resist compression. Its place can as well be filled by an arch of brick, footed on a pair of cast-iron skew-backs, and these held in position by a pair of tie-rods, as in Fig. 78. FIG. 78. To obtain a rule for the diameter of each rod, we have as above, in Art. 594, *,= -7854^ BRICK SUBSTITUTE FOR IRON ARCH. 399 This is for one rod. When a t is put for the joint area of two rods, we will have Comparing this with formula (287. \ we have til or f X2 X 78540$ and when k is taken at 9000 (Art. 594) Ul (290.) 9425^ This is a rule in which D f represents the diameter of each of the two required rods. For example, see Art. 595. An arch of brick, well laid and secured in this manner, will serve quite as well as the cast-iron arch, and may be had at less cost. The best supports, however, to carry brick walls are those made of rolled-iron beams, putting two or more of them side by side and bolting them together. (See Art. 489, form. 400 CAST-IRON GIRDERS. CHAP. XXI. QUESTIONS FOR PRACTICE. 597. What should be the dimensions of cross-section of a cast-iron girder, 23 feet long between supports, and 27 inches high at the middle, at which point it is to carry 40,000 pounds ; with 5 as the factor of safety ? The width of bottom flange is 16 inches. 598. In a girder of the same length, height and width : What should be the cross-section if the weight be 60,000 pounds and be uniformly distributed ; the factor of safety being 5 ? 599. In a girder of the same length, and of the same height and width at 8 feet from one end, where it is required to carry 50,000 pounds, with a factor of safety of 5 : What should be the dimensions of cross-section ? 600. In a girder 25 feet long between bearings, carry- ing a load of 40,000 pounds at 10 feet from one end, with 5 as the factor of safety, and having 30 inches area of cross- section in the bottom flange : What should be the depth of the girder ? 60 1. A girder, 25 feet long and 30 inches high, is re- quired to carry, with 5 as a factor of safety, two weights, one of 25,000 pounds at 8 feet from one end, and the other of 30,000 pounds at 6 feet from the other end : What should QUESTIONS FOR PRACTICE. 4OI be the dimensions of cross-section at each weight, the bottom flange being 16 inches wide? 602. In an arched girder, 24 feet long between bear- ings, with a versed sine or height of 30 inches from the axis of the rod to the centre of gravity of the arched beam at the middle, and with the load on the girder taken at 80,000 pounds uniformly distributed : What ought the diameter of the tie-rod to be ? CHAPTER XXII. FRAMED GIRDERS. ART. 603. Transverse Strains in Framed Girders. This work, a treatise elucidating the Transverse Strain, would seem to have reached completion with the end of the discus- sion on simple beams ; but when it is recognized that the formation of a deep girder, by a combination of various pieces of material, is but a continuation of the effort to gain strength in a beam, by concentrating its material far above and below the neutral axis, as is done in the tubular girder and rolled-iron beam, it is clear that the subject of framed girders is properly included within a treatise upon the transverse strain. The subject of framed girders, however, will here be discussed so far as to develop only the more im- portant principles involved. For examples in greater vari- ety, the reader is referred to other works (Merrill's Iron Truss Bridges, and Bow's Economics of Construction). 604. Device for Increasing the Strength of a Beam. The use of simple beams is limited to comparatively short spans ; for beams cut from even the largest trees can have but comparatively small depth. The po\ver of a beam to resist cross-strain can be considerably increased by a very simple device. Let Fig. 79 represent the side view of a long beam of wood, from which let ACDB, the upper part of the beam, be cut. With the pieces thus removed, and the addi- tion of another small piece of timber, there may be con- INCREASING STRENGTH OF BEAM. 403 structed the frame shown in Fig. 80, which is capable of sus- taining a greatly increased load. This increase will be in FIG. 79. FIG. 80, proportion to the depth of the frame (Art. 583), and is ob- tained here by increasing the distance between the fibres which resist compression and those which resist tension. It is upon this principle that roof trusses and bridge girders, alike with common beams, all depend for their stability. 605. Horizontal Thrust. In a frame such as Fig. Bo, the horizontal strains produced by the weight W are bal- anced ; or, the tension caused in the tie CD is equal to the compression caused in the short timber on which the weight rests. If the tie CD were removed, it is obvious that the weight W, acting through the two struts AW and BW, would push the two abutments AC and BD from each other, and, descending, fall through between them ; unless the abutments were held in place by resistance other than that contained in the frame such, for instance, as outside buttresses. 404 FRAMED GIRDERS. CHAP. XXII. From this we learn the importance of a tie-beam ; or, in its absence, of sufficient buttresses. From this we may also learn why it is that roof trusses framed without a horizontal tie at foot so invariably push out the walls, when constructed without exterior buttresses. 606. Parallelogram of Forces Triangle of Forces. A discussion of the subject of framed girders can only be in- telligently understood by those who are familiar with some of the more simple and fundamental principles of statics. One of these principles is known as the parallelogram of forces, or the triangle of forces, and is useful to the archi- tect in measuring oblique strains due to vertical and hori- zontal pressures. Proof of the truth of this principle may be found in most mathematical works. (See Cape's Math., Vol. II., p. 118 ; chap, on Mechs., Art. 20.) In this chapter its application to construction will be shown. In Fig. 81, let the lines A W and BW represent the axes of two timber struts, which, meeting at the point W t sus- FIG. 81. tain a weight, or vertical pressure, as indicated by the arrow at W. Then, let the vertical line WE, drawn by any PARALLELOGRAM OF FORCES. 405 convenient scale, represent the number of pounds, or tons, contained in the vertical weight at W. From E, draw ED parallel with A W, and EC parallel with BW. CWDE is the parallelogram of forces, and possesses this important property namely, that the three lines WE, EC and CW, forming a triangle, are in proportion to three forces ; the weight at W, the strain in WB, and the strain in WA. The same is true of the other triangle WED ; or, to des- ignate more particularly, we have : as the line WE is to the weight at W, so is the line CE, or WD, to the strain in WB; and also : as the line WE is to the weight at W, so is the line DE, or WC, to the strain in A W. In- dicating the lines by the letters a, b and c t as in the fig ure, we have c : a : : W / : A, in which A t equals the strain caused by the weight W t through the line WA ; and c : b : : W, : B t B, =- W~ in which B t equals the strain caused by the weight W f through the line WB. 60 7 B Line and Force in Proportion. The above pro- portions hold good when the two lines A W and BW are inclined at any angle, and whether they are of equal or of un- equal lengths ; indeed, the principle is general in its applica- 406 FRAMED GIRDERS. CHAP. XXII. tion, for in all cases where the three sides of a triangle are respectively drawn parallel to the direction of three several forces which are in equilibrium, then the lengths of the three lines will be respectively in proportion to the three forces. 608. Horizontal Strain Measured Graphically. In Fig. 81, and in the triangle WCE, draw, from C, the horizon- tal line CF, or h ; then we have the line , in proportion FIG. 81. to the line //, as B n the strain in WB, is to H t , the horizontal strain ; or, b : h :: B, : ff,-B^ and by substituting the value of B t in formula (292.) have H, =: Bj- = W~ = W ' <b 'cb 'c or the horizontal strain is measured by the quotient arising from a division by the line c, of the product of the weight FORCES IN EQUILIBRIUM, W t into the line h ; or, c : h : : W, : H, 407 (293.) This measures the horizontal strain at AB, or at W, for it is the same at all points of such a frame, whatever the angle of inclination of the struts, or whether they are inclined at equal or unequal angles. 609. Measure of Any Number of Forces in Equilibrium. In Fig. 82, let AB be the axis of a horizontal timber, sup- ported at A and B, and let AC, CD and DB be three iron rods, with two weights R and P suspended from the FIG. 82. points C and D. The iron rods being jointed at A, C, D and B, so as to permit the weights to move freely, and thus to adjust themselves to an equilibrium, the whole frame ABDC will be equilibrated. From D erect a vertical, DH, and by any convenient scale make DG equal to the weight P, and GH equal to the weight R. From , draw GE parallel with CD, and from H draw HE parallel with AC. The sides of the 408 FRAMED GIRDERS. CHAP. XXII, triangle GED are parallel with the three lines CD, DB and DP, and consequently are in proportion as the strains in the three lines CD, DB and DP. Again ; the sides of the triangle HEG are parallel with the lines AC, CD and CR, and consequently are in proportion as the strains in the lines AC, CD and CR. From E draw EF horizontal. Then the sides of the triangle FED, being parallel with the lines BA, BD and BK, are in proportion to the strains in these lines. Also, the sides of the triangle HEF, being parallel to the lines AB, AC and AL, are in proportion FIG. 82. to the strains in these lines. Thus, in the triangles within HDE, we have the measures of all the strains of the funicu- lar or string polygon ABDCA ; FE being the horizontal strain, FD the vertical strain or load on BK, and HF the vertical strain or load on AL. 610. Strain* in ait Equilibrated Truss. In Fig. 82 the strains in the lines AC, CD and DB are tensile, while that in AB is compressive. If the lines AC, CD and DB were above the line AB, instead of below it, then these strains would all be reversed ; those which are tensile in the figure would then be compressive, while that which is com- EQUILIBRATED FRAME. 409 pressive would then be tensile ; but the amount of strain in each would be the same and be measured as in Fig. 82. For example : Let Fig. 83 represent an equilibrated frame ; the pieces A C, CD, DE, EF and FB suffering compression FIG. 83. from the vertical pressures indicated by the arrows at C, D, E and F, while AB, a tie, prevents the frame from spread- ing. Draw the vertical line LQ, and from B draw radiating lines, parallel respectively with the several lines AC, CD, DE and EF, and cutting the line LQ at the points Q, P, O and M. Then the several lines BQ, BP, BO, BM and BL will be in proportion, respectively, to the strains in AC, CD, DE, EF and FB ; and the lines LM, MO, OP and PQ will be in proportion, respectively, to the -vertical pressures at F, E, D and C ; while the line LN will represent the vertical pressure on B, and NQ that on A, and the line NB the horizontal thrust in AB. 611. From Given Weigh!* to Construct a Scale of Strains. The construction of the scale of strains LBQ, as here given, is proper in a case where the points C, D, E 4io FRAMED GIRDERS. CHAP. XXII. and F are fixed, and the weights and strains are required. When the weights at C, D, E and F, with their horizontal distances apart, and the two heights RC and UF, are given ; then, to find the scale of strains, and incidentally the heights of the points D and E, proceed thus : From B FIG. 83. draw BQ parallel with AC, make the vertical QL equal, by any convenient scale, to the sum of the weights at C, D, E and F, and upon this vertical lay off in succession the distances LM, MO, OP and PQ, equal respectively to the weights at F, E, D and C. Then, the several lines BL, BM, BO, BP and BQ will, by the same scale, measure the several strains in BF, FE, ED, DC and CA, and BN will measure the horizontal strain. 612. Example. In constructing Fig. 83, the weights given are 11,899 pounds at F, 4253 pounds at E, 4464 at D and 11,384 at C\ being a total of 32,000 pounds. The distances AR, RS, etc., are successively 8, 13, 20, 24 and 13; in all 78 feet. The height RC = 15, and UF = CONSTRUCTING A SCALE OF STRAINS. 411 With these dimensions all laid down as in Fig. 83, draw BQ parallel with AC. Draw the line LQ vertical, and at such a distance from B as that its length shall, by a scale of equal parts, be equal to the total load on the four points C, D, E and F ; or to a multiple of the total load. For example : a scale of 100 parts to the inch will be convenient ki this case, by appropriating 4 parts to the thousand pounds. The 32,000 pounds require 32x4=128 parts for the length ol the line LQ, and the several other weights and distances require as follows : LM 4 x 1 1 899 = 47 - 596 MO 4 x 4-253 17-012 OP 4 x 4-464= 17-856 PQ = 4x 11-384 ^ 45.536 The sum of these, LQ = 47-596 4- 17-012 +'17-856 + 45-S3 6 = 128 as before. Therefore, draw LQ at such a distance from B that it will, by the scale named, equal 128 parts. On this line lay off the distances LM = 47-596, MO = 17-012, etc., as above given. Join B with each of the points P, O and M. These lines give the directions of the lines CD, DE and EF ; therefore, draw FE parallel with BM, ED parallel with BO, and DC parallel with BP. By applying the scale to the lines radiating from B, the strains in the several lines AC, CD, etc., will be shown. BQ, by the scale, measures 80 parts, therefore \- = 20 ; or the strain in A C is 20,000 pounds. BP measures 45 parts, and -\ 5 = n; or the strain in CD is 11,250 pounds. BO measures 38, and $- = 9^ ; or the strain in DE is 9500. BM measures 39, and - 3 T 9 - = 9! ; or the strain in EF is 9750. 412 FRAMED GIRDERS. CHAP. XXII. (5o ^ BL measures 69.5, and =17-375; or the strain in FB is 17,375. BN measures 37-5. and -- = 9.375; or the horizon- tal strain is 9375 pounds. Also, as LN measures 58, therefore ^- = 14,500, equals the load on B \ and as NQ measures 70, therefore, -\- = 17,500, equals the load on A ; and the two loads A t and B, together equal 17500 + 14500 = 32000, equals the total load. In practice the diagram should be large, for the accuracy of the results will be in proportion to the size of the scale, as well as to the care with which it is drawn and measured. The size above taken is large enough for the purposes of illustration merely, but in practice the diagram should be drawn at a scale of 12 feet to the inch ; or, still better, at 8 feet. (See Art. 615.) 613. Horizontal Strain Measured Arithmetically. In the last article, directions were given for locating the line LQ, Fig. 83. This line may be located more precisely by arithmetical computation, and the horizontal thrust be thus denned more accurately than is there done. In Fig. 84, showing parts of Fig. 83 enlarged, we have the triangles ACR and BFU, the same as in Fig. 83. The triangle ACR, as stated (Art. 612), has a base of 8 and a height of 15. Make A Y equal 10, and draw YZ vertical. We now have this proportion, AR : RC : : A Y : YZ or 8 : 15 :: 10 : YZ = --^ = 18-75 o Again ; triangle BFU, as stated (Art. 612), has a base of 13 and a height of 2oJ. Make BX equal 10, and draw XV MEASURING HORIZONTAL STRAIN. 413 vertical. We have now this proportion, BU\ UF \\BX\XV or 10 x 2oJ 13 :2oi:: 10 : XV = = 15-577 Thus we have the two angles at A and B comparable, for, with a common base of 10, the one at A has a height of 18-75, while the one at B is 15-577. The two triangles may now be put together at the line BU. Extend the verti- cal line VX to W, make XW equal to YZ = 18-75, and join W and B. Then the triangle BXW equals the triangle A YZ, and BW is parallel with AZ. FIG. 84. The problem now is to locate the point N, so that the vertical LQ drawn through N shall be equal, by any given scale, to the total of the loads at C, D, E and F. To do this we have FRAMED GIRDERS. CHAP. XXII, BX x NL BN:NL::BX:XV = BN:NQ::BX:XW = BN BXx NQ BN By addition we have _ BXxNL BXxNQ _ BX(NL + NQ) By the diagram, XV+XW= VW, and NL + NQ = LQ, and therefore . BN VW: BX-.: LQ: BN = or BX^LQ VW ~ The total load is 32,000, for which we may, putting one for a thousand, make LQ equal to 32 ; and, since VW equals YZ+ VX = 18-75 + 15-577 = 34-327. and BX = 10, we have PRESSURE ON THE TWO POINTS OF SUPPORT. 415 defining accurately the horizontal thrust BN as 9-322, or 9322 pounds. Vertical Pressure upon the Two Points of Support. This pressure was shown in Art. 612 by the diagram, but may be more accurately determined by arith- "metical computation, as follows : In the last article the horizontal thrust BN was shown to be 9322 pounds. We have the proportion BX\ XV :: BN : NL 15-577 x 9-322 10 : 15- 577 1:9-322 :NL = - ^--^ - = 14-521 or the vertical strain upon the support B is 14,521 pounds. To find that upon the support A we have BX : XW :: BN : NQ 18-75x9.322 10 : 18-75 :: 9-322 : NQ = - - = 17.479 10 or the vertical strain upon the support A is 17,479 pounds and the two, 17479 + 14521 = 32000, equals the total load. 615. Strains measured Arithmetically. The resulting strains in Fig. 83, as obtained by scale in Art. 6(2, are close approximations, and are near enough for general purposes. The exact results can be had arithmetically, as in Arts. 613 and 614. For example : In Art. 612 the horizontal thrust was found by scale to be 9375, while in Art. 613 it was more accurately defined by arithmetical process to be 9322. So, also, the portions of the total load borne by the two 41 5 FRAMED GIRDERS. CHAP. XXII. supports, A and B, were found by scale to be 17,500 and 14,500, respectively, while in Art. 614 they were accu- rately fixed at 17,479 on A and 14,521 on B. A carefully drawn diagram at a large scale will generally be sufficient for use, but it is well, in important cases, to compute the dimensions also. When both are done, the scale measure- ments serve as a check against any gross errors in the com- putations. In Art. 612 the strains in the several timbers are given, as ascertained by scale. By the rules for computing the sides of a right-angled triangle (the 47th of first book of Eu- clid), the several strains, as represented by BL, BM, BO, etc. (Fig. 83), may be found arithmetically. The following list shows the results by this method, side by side with those by scale : By Scale. By Computation. BL = 17,375 and 17,256 BM 9,750 " 9,684 BO 9,500 " 9,464 BP 11,250 ' 11,138 BQ -- 20,000 " 19,810 616. Curve of Equilibrium Stable and Unstable. When the positions of the supporting timbers AC, CD, DE, etc. (Fig. 83), are regulated in accordance with the weights upon the points C, D, E, etc., and, as shown in Art. 611, the frame is in a state of equilibrium ; and a curve drawn through the points A, B, C, D y etc., is called the curve of equilibrium. When the several weights are numerous, are equal, and are located at equal distances apar.t ; or, when the load is uniformly distributed over the length of the frame, this curve is a parabola. In these cases, if the rise CURVE OF EQUILIBRIUM. 417 be small in comparison with the base, the curve is nearly the same as a segment of a circle, and the latter may be used without serious error. (Tredgold's Carpentry, Arts. 57 and 171.) The pressures in an equilibrated frame act only in the axes of the timbers composing the frame, and these carry the effects of the several weights on, from point to point, until they successively arrive at A and B, the points of final support. A frame thus balanced is not, however, stable, for if subjected to additional pressure, however small, at any one of the points of support, it is liable to derange- ment ; and if so deranged it has no inherent tendency to recover itself, but the distortion will increase until total failure ensues. A frame thus conditioned is therefore said to be in a state of unstable equilibrium ; while a frame of suspended pieces, as in Fig. 82, is said to be in a state of stable equilibrium, since, if disturbed by temporary pressures, it will recover its original position when they are removed. 617. Trussing- a Frame. The tendency to derangement and consequent failure, .in a frame such as Fig. 83, can be counteracted by additional pieces termed braces, located in any manner so as to divide the inclosed spaces into triangles. For example: it may be divided into the triangles ACS, CSD, DST, DTE, RTF, TFU and UFB. If these additional pieces be adequate to resist such pressures as they may be subjected to, and be firmly connected at the joints, the frame will thereby be rendered completely stable. Treated in this way the frame becomes a truss, from the fact that it has been trussed or braced. 618. Forces in a Truss Graphically Measured. When a frame is divided into triangles, as proposed in the last article, sometimes three or more pieces meet at the same 418 FRAMED GIRDERS. CHAP. XXII. point. In such a case, owing to the complexity of the forces, it becomes difficult to trace, and, by the parallelogram of forces, to measure them all. Professor Rankine, in his " Applied Mechanics," somewhat extended the use of the triangle of forces in its application to such cases. It was afterward more fully developed and generalized by Professor J. Clerk Maxwell in a paper read before the British Asso- ciation in 1867, and by him termed ''Reciprocal Figures, Frames and Diagrams of Forces ;" and Mr. R. H. Bow, C.E., F.R.S.E., in his " Economics of Construction," has simplified the method in its use, by a system of reciprocal lettering of lines and angles. By Professor Maxwell's method, the forces in any number of pieces converging at one point are readily determined. The principle involved is very simple, and is this : Construct a closed polygon, with lines parallel to the direction of, and equal in length to, the amount of the forces which in the framed truss meet at any point. A system of such polygons, one for each point of meeting of the forces, so constructed that in it no line representing any one force shall be repeated, is termed a diagram of forces. 619. Example. Let Fig. 85 represent a point of con- vergence of parts of a framed truss, and Fig. 86 be its corresponding diagram of forces, in which latter the lines are drawn parallel to the lines in Fig. 85. Designate a line in the diagram in the usual manner, by two letters, one at each end of the line, and indicate the corresponding line in Fig. 85 by placing the same two letters one on each side of the line. For instance, the line AB of 86 is parallel with that line of 85 which lies between the spaces A and B ; and so of each of the other corresponding lines. In Fig. 86 the lines are in proportion to each other, respectively, as the forces in the corresponding lines of Fig. 85. Thus if AD (86), by any scale, RECIPROCAL FIGURES. 419 represents the force in the line between A and D (85), then will the line AB equal the force in the line between A and B; and in like manner for the other lines and strains. 620. Another Example. Let Fig. 87 represent the axial lines of the timbers of a roof truss, and its two sustaining piers, and let Fig. 88 be its corresponding diagram of forces. FIG. 85. The truss being loaded uniformly, the three arrows (87) one at the ridge and one at the apex of each brace repre- sent equal loads. Let these three loads be laid down to a suitable scale on the line FJ (88), one extending from F to G, another from * G to ff, and the third from H to y. The half of these, or FE, is the load sustained on one of the supports of Fig. 87, and the other half, EJ, is the load upon the other support. In Fig. 87 a letter is placed in each triangle, and one in each partly-enclosed space outside of the truss. Each line of the figure is to be designated by the two letters which it separates; thus, the line between A and E is called line AE, the line between A and B is called line AB, etc. In Fig. 88 the corresponding lines are designated by the same letters ; the letters here being, as usual, at the ends of the lines. Also, it will be observed that while in the diagram any point is designated in the usual way by the letter standing at it, it is the practice in 420 FRAMED GIRDERS. CHAP. XXII. the frame to designate a point by the several letters which cluster around it ; for example, the point of support FAE (Fig. 87). The diagram, Fig. 88, is constructed by drawing a line parallel with each of the lines in Fig. 87. Thus the three FIG. 87. FIG. 88. forces converging in Fig. 87 at FAE, the left-hand point of support, are EF, FA and AE\ and in Fig. 88 the lines EF, FA and AE, drawn parallel with the corresponding lines of Fig. 87, form the triangle EFA. Taking the forces at DIAGRAM OF FORCES. 421 point GBAF, 87, we find them to be AF, FG, GB and BA, and drawing, in 88, lines parallel with these, we obtain the quadrangle AFGBA. Again, in 87 the forces at point GBCH are BG, GH, HC and CB, and drawing, in 88, lines parallel with these, we obtain the closed polygon BGHCB. At point HCDJ (87) the forces are CH, HJ, JD and DC, and, in 88, drawing the corresponding lines pro- duces the quadrangle CHJDC. In 87 the forces at point JED, the right-hand support, are DJ, JE and ED, and in 88, the corresponding lines produce the triangle DJE. In 87, at point ABODE, we have the five forces EA, AB, BC, CD and DE, and, in 88, the corresponding lines give the closed polygon EABCDE. This completes the dia- gram of forces, Fig. 88, in which the several lines, measured by the same scale with which the three loads were laid off on FJ, are equal to the corresponding forces in the similar parts of Fig. 87. 621. Diagram of Force. In this manner the diagram of forces may be drawn to represent the strains in a framed truss, by carefully following the directions of Art. 618 ; com- mencing by first laying down the forces which are known; from which the ones to be found will gradually be determined until the whole are ascertained. 62 2 Diagram of Forces Order of Development. When more than two of the forces converging at any one point are undefined in amount, the diagram can not be completed. Thus, where three forces converge it is requisite to know one of them. Of four forces, two must be known. Of five forces, three must be known. In constructing a diagram, the first thing necessary is to establish the line of loads, as FJ in Fig. 88, then to ascertain the portion of the total load which bears upon each of the 422 FRAMED GIRDERS. CHAP. XXII. points of support, AEF and JED (Art. 56) (one half on each when the load is disposed symmetrically), and, with this, to obtain the first triangle FEA. From this proceed up the rafters, or to where the points of convergence have the fewest strains, leaving the more complex points to be treated later. In this way the most of the forces which affect the crowded points will be developed before reaching those points. 623. Reciprocal Figures. By comparing Figs. 87 and 88, we see that the lines enclosing any one of the lettered spaces in the former are, in the latter, found to radiate from the same letter. The space A (87) has for its boundaries the lines AF, AB and AE, and these same lines in 88 radiate from the letter A. The space E (87) has for its enclosing lines EF, EA, ED and EJ, and these same lines are found to radiate from the point E (88). Thus the diagram (88) is a reciprocal of the frame (87). 624. Proportions in a Framed Girder. In order to treat of the method of measuring strains in trusses, we have digressed from the main subject. Returning now, and refer- ring to the relations existing between a girder and a simple beam, as in Arts. 603 to 605, we proceed to develop the proportion in a girder, between the length and depth. A girder, as generally used, serves to support a tier of floor beams at a line intermediate between the walls of the building, and when sustained by posts at points not over 12 to 15 feet apart, may be made of timber in one single piece. But when a girder is required to span greater dis- tances than these, it becomes requisite, by some contrivance, to increase its depth, in order to obtain the requisite strength. An increase of depth, however, may interfere with the demand tor clear, unobstructed space in rooms so PROPORTION BETWEEN DEPTH AND LENGTH. 423 large as those in which girders are required. To prevent this interference, the depth of the girder should be the least possible ; although diminishing the depth will increase the cost ; for the cost will be in proportion to the amount of material in the girder, and this will be in proportion to the strains in its several parts, and the strains will be inversely as the height. For economy's sake, therefore, as well as for strength, the girder should have a fair depth; modified, how- ever, by the demand for unobstructed space. Where other considerations do not interfere to prevent it, the depth of a framed girder should be from y 1 ^ to -J- of the length ; the former proportion being for girders 25 feet long, and the latter for those 125 feet long. If these two rates be taken as the standard rates, respectively, of two girders thus differing in length 100 feet, and all other girders be required to have their depths proportioned to their lengths in harmony with these standards, their rates will be regularly graduated. In order to develop a rule lor this, let the two standards be reduced to a common denominator, or to ^ and / . If their difference, -%, be divided into 100 parts, each part will equal i i 24 x 100 2400 and will equal the difference in rate for every foot increase in length of girder; for the two standards are 100 feet apart. The scale of rates thus established is for lengths of girder from 25 to 125 feet, but it is desirable to extend the scale back over the 25 feet to the origin of lengths. To do this, we have for the difference in rates for this 25 feet, 25 x ^Vo = *Hhr = ?V Deducting this from T L (= -&), the rate at 25 feet, we have -fa -fa = -fa, the rate be- tween depth and length at the origin of lengths (if such a thing were there possible). Now if to this base of rates we 424 FRAMED GIRDERS. CHAP. XXII. add the increase (y^V^ of the length) the sum will be the rate at any given length. As an example: What should be the rate, by this rule, for a girder 125 feet long ? For this the difference in rate is 125 x -^fa = ^WV = -fa. Adding this to the base of rates, or to -fa, as above, and the sum ^ + ^_ II _ } the required rate. This is one of the standard rates. The other standard may be found by the rule thus, 25 x ^Vir = ^j-jfo = -fa. Adding this to the base -A gives -ft- = T^ the standard rate. We have therefore for the rate at any length r = . -L + _!_/= -!Zi_ 1 96 24OO 24OO 24OO 24OO r = 2400 This gives the rate of depth to length, and since the depth is equal to the rate multiplied by the length, therefore 2400 d = (994.) 24OO in which d is the depth between the axes of the top and bottom chords, and / is the length (between supports), both being in feet. This rule will give the proper depth of a girder, and may be used when the depth is not fixed arbitrarily by the cir- cumstances of the case. (See Art. 572.) 625. Example. What should be the depth of a girder which is 40 feet long between supports? By formula (294.), + 40)x 4 x) = 2400 ECONOMICAL DEPTH. 425 or the economical depth is 3 feet and 7 inches, measured from the middle of the depths of the top and bottom chords. Again : What should be the depth of a girder which is 100 feet long in the clear between supports ? By (294>\ (175 + 100) x IPO a = -- ---- 1 1 -450^ 2400 or the depth between the axes of the chords should be n feet 5f inches. A girder 125 feet long would by this rule be 15 feet 7-J- inches, or -J- of its length, in depth ; while a girder 25 feet long would be 2 feet and i inch deep between the axes, or T ^ of the span. 626. Trussing, in a Framed Girder. One object to be obtained by the trussing pieces the braces and rods is to transmit the load from the girder to the abutments. The braces and rods forming the trussing may be arranged in a great variety of ways (see Bow's Economics of Construction), but that system is to be preferred which will take up the load of the girder at proper intervals, and transmit it to its two supports in the most direct and economical manner. Just which of the great number of systems proposed will the more nearly perform these requirements it will perhaps be somewhat difficult to determine, but the one in which the struts and ties are arranged in a chain of isosceles triangles is quite simple, and offers advantages over many others. It is therefore one which may be adopted with good results. 627. Plaaniiis a Framed Girder. After fixing upon the height (Art. 624), the next point is as to the number of panels or bays. These should be of such length as to afford points of support at suitable intervals along the girder, and the rods and struts should be placed at such an angle as will 4^6 FRAMED GIRDERS. CHAP. XXII. secure a minimum for the strains in the truss. To set the braces and ties always at the same angle, would result in fur- nishing points of support at intervals too short in the girders of short span, and too long in those of long span. So also, if the width of a bay be a constant quantity, there would be too great a difference in the angles at which the rods and struts would be placed. To determine the number of bays, so as to avoid as far as practicable these two objections first, we have the number of the bays directly as the length of the truss and inversely as the depth, arid, second (to vary this pro- portion as above suggested), we may deduct from this result a quantity inversely proportioned to the length of the girder. Combining these, we have, n being the number of bays, / 120 / n ~j -- d c and by substituting for d its value as in formula I 120-1 2400 2400 1 20 / " 175 + ' ~ ~~c~ in which / is the length of the girder in feet, and c is a constant, to be developed by an application to a given case. To this end we have, from the last formula, 1 20 /_ 2400 =l7sT7- I2O / 2400 With n =4-5 and /= 20, we have DETERMINING THE NUMBER OF BAYS. 427 12020 100 C = -- - = - ; - = 12 SO7 2400 _ 12-3084.5 175 + 20 ~~ or, say c = 12-8 ; and with this value I75 + / 12-8 which is a rule for determining the number of bays in a truss, when not determined arbitrarily by the circumstances of the case, and when the height of the girder is obtained as in formula (294-}. In the resulting value of ;/, the fraction over a whole number is to be disregarded, unless greater than , in which latter case unity should be added to the whole number. 628. Example. What should be the number of bays in a truss 80 feet long ? Here / = 80, and, by the formula, 2400 1 20 80 _ 2400 40 n= i75 + 8o~~T^~8~ : "255" ~~~i2T8 - or the required number of bays is six ; disregarding the decimal 287 because it is less than . 629. Example. How many bays are required in a girder no feet long? Here /= no, and, by formula 2400 1 20 no = ~ or, adding unity for the decimal -64, the number required is 8. FRAMED GIRDERS. CHAP. XXII. 630. Number of Bays in a Framed Oirder. According to the above rule, the number of bays or panels required in framed girders of different lengths is as follows : Girders from 20 to 59 feet long should have 5 bays. " 59 85 " " " " 6 " 85 " 107 " " " " 7 " " " 85 " 107 " " " " IQ y tt I2 y u u 8 " " 127 " 146 " " " " 9 In cases where the length exceeds 120 feet, the quantity of the formula to be deducted f - - j becomes a nega- V 12-5 i tive quantity, and, since deducting a negative quantity is equivalent to adding a positive one, the result may be added, thus: 120144 24 631. Forces in a Framed Oirder. Let Fig. 89 represent the axial lines of a framed girder, or the imaginary lines passing through the axes of the several pieces composing the frame. Let the load, equally distributed, be divided into six parts, one of which acts at the apex of each lower triangle. We may notice here that in a truss with an even number of lower triangles, as in Fig. 89, there is an even number of loads, one half of which are carried by the struts and rods to one point of support, and the other half to the other support. Thus the load PQ, at point PEFGQ, is sustained by the top chord, and by the strut EF. The portion passing down this strut is carried by the rod DE up to the top chord, and thence, together with the load OP, at point OCDEP, down by the strut CD to the bottom chord. This accu- mulated load is carried by the rod BC up to the top chord, DETERMINING THE PRESSURES. 429 and thence, with the addition of the last load AO, at ABCO, finally reaches, through the strut AB, the point of support for that end of the truss. The three weights on the other side are in like manner conveyed to MNT, the other point of support. We here see the manner in which, in a framed girder upon which the load is uniformly dis- tributed, one half is carried by the trussing pieces to each point of support. 632. Diagram for the above Framed Girder. Fig. 90 is a diagram constructed as per Arts. 619 and 620, and repre- sgpfo-.iK^v^ , trie sides ot which measure the torc&> 5. 89. s converging at the point BCDT of 89. The next ir er is the point OCDEP. Of the five forces concentrat here, we already have, in Fig. 90, three, PO, OC an '. To find the other two, draw from D the line D -allel with the line DE of 89, and from P draw PL rallel with line PE of 89. These two lines will meet a and complete the polygon POCDEP, which measures 3 forces in the lines concentrating at point OCDEP oceeding now to the point DEFT of Fig. 89, we find foui ces, two of which, TD and DE., are already deter ned. For the other two, draw from E the line El rallel with EF in 89, and from T, TF parallel with th e TF of 89. These two lines meet in F and complete 3 polygon TDEFT, which measures the forces in the es converging at the point DEFT of Fig. 89. The nex order is the >oint PEFGQ in 89, where five IWe FIG. 90. To construct this diagram, we proceed as follows: Upon the vertical line AN lay off the several distances AO, OP, PQ, QR> RS and SN\ each equal by any convenient 430 FRAMED GIRDERS. CHAP. XXII. scale to one of the six equal loads resting upon the top of 89. The load at the apex of the triangle ./?, or point ABCO (89), is placed from A to O in 90. The load OP, at point OCDEP (89), is placed from O to P in 90 ; and so on with the other loads. The other lines of Fig. 90 are obtained by drawing them parallel with the corresponding lines of Fig. 89, as per directions in Art. 618. Commencing at the point ABT (89), we draw (in 90) three lines parallel with the direc- tion of the forces at that point. The first of these is the ver- tical pressure upon the point of support ABT, which in this case equals one half the total load, or AQ, or AT of eu .. uvj UC UCLlUCLCci i Q / ~ " ~ re quantity, and, since deducting a negative quantity [uivalent to adding a positive one, the result may be adde us: 120 631. Force in a Framed Girder. Let Fig. 89 represen ^ axial lines of a framed girder, or the imaginary line ssing through the axes of the several pieces composing th ime. Let the load, equally distributed, be divided into si: rts, one of which acts at the apex of each lower triangle e may notice here that in a truss with an even number o wer triangles, as in Fig. 89, there is an even number c ids, one half of which are carried by the struts and rods t e point of support, and the other half to the other suppon FIG. 90. Fig. 90. Next, from T, draw the horizontal line Tfi y and from A, the inclined line AB, parallel with the brace AB of 89. These two lines meet at B, and we have the tri- angle ABT, representing the three forces converging at. the point of support ABT. For the four forces at the point CONSTRUCTING DIAGRAM OF FORCES. 43! ABCO in 89, we proceed as follows : We already have the forces AO and AB. From B in 90, draw BC parallel with the rod BC of 89 ; and from O in 90, draw OC par- allel with OC of 89. These two lines intersect at C, com- pleting the polygon ABCOA, the sides of which are in pro- portion as the forces in the several lines converging at the point ABCO of Fig. 89. Proceeding to the point BCDT (Fig. 89), we find, of the four forces converging there, that two are already drawn, 777 and BC. From C draw CD parallel with the brace CD of Fig. 89 ; and from T draw TD. These two lines will meet at D and complete the poly- gon TBCDT, the sides of which measure the forces in the lines converging at the point BCDT of 89. The next in order is the point OCDEP. Of the five forces concentrat- ing here, we already have, in Fig. 90, three, PO, OC and CD. To find the other two, draw from D the line DE parallel with the line DE of 89, and from P draw PE parallel with line PE of 89. These two lines will meet at E and complete the polygon POCDEP, which measures the forces in the lines concentrating at point OCDEP. Proceeding now to the point DEFT- of Fig. 89, we find four forces, two of which, TD and DE., are already deter- mined. For the other two, draw from E the line EF parallel with EF in 89, and from T, TF parallel with the line TF of 89. These two lines meet in F and complete the polygon TDEFT, which measures the forces in the lines converging at the point DEFT of Fig. 89. The next in order is the point PEFGQ in 89, where five lines con- verge. The forces in three of these we have already namely, QP y PE and EF. Draw from F a. line parallel with the line FG of 89, and from Q a line parallel with QG -of 89. These two intersect at G and complete the polygon QPEFGQ, which gives the forces in the lines around the point PEFGQ of Fig. 89. 432 FRAMED GIRDERS. CHAP. XXII. In this last proceeding we meet with a peculiarity. The line FG has no length in Fig. 90. It commences and ends at the same point, since G is identical with F. This seems to be an error, but it is not. It is correct, for an ex- amination of Fig. 89 will show that the two inclined lines meeting at the foot of the triangle G do not assist in carry- ing the weights upon the top chord, and may therefore, in so far as those weights are concerned, be dispensed with, so that the space occupied by the three triangles F, G and H may be left free, and be designated by one letter only in- stead of three. In place of five, there are in fact only four forces meeting at the point PEFGQ, and these four are rep- resented in Fig. 90 by the polygon QPEFQ. The above analysis is in theory strictly correct, and yet in practice it is not so, for in such cases as this there is al- ways more or less weight on the lower chord at the middle point. If nothing more, there is the weight of the timber chord itself, and this should be considered. In Art. 634 a truss with weights at the points of each chord will be found discussed, and the facts as found in prac- tice there developed. The construction of one half of the diagram (Fig. 90) has now been completed. The other half is but a repetition of it in reversed order, and need not here be shown in detail. In drawing the lines for the latter half, it will be seen that the point H is identical with the point F, and that K and M coincide respectively with D and B. 633. Gradation of Strains in Chords and Diagonals. In considering the forces shown in Ftg. 90, we find that those in the chords increase towards the middle of the girder, while the forces in the diagonals decrease towards the middle. Thus, in Fig. 90, of. the lines representing the upper chord, LOADS ON EACH CHORD. 433 PE is longer than OC, and QG is longer than PE, in- dicating a corresponding increase in the lines OC, PE and QG of 89. So in the lower chord, we have a successive in- crease of forces, as seen in a comparison of the lengths of the lines TB, TD and TF of Fig. 90, representing the chord at the several bays B, D and F of Fig. 89. The diagonal lines AB, CD and EF in 90 show decreasing forces in the diagonals AB, CD and EF in 89 decreasing towards the middJa/to a point beyond the middle of tnYi/P remem- ber wVphe remainder of Fig. 92 may be traced for thevledge of thjf 9I ^ by a continuance of the process used in tracing 6 di- a g" ol half. Since, in this instance, the loading and plan of the dialer are symmetrical, and hence the several forces in 2S of one half of the girder respectively equal to those 3 other half, the lines of the diagram as laid down for e may be used for the other half. Let Fi rre- sp< -am, 635. Gradation of Strains in Chord and Diagonal . we the e gradation of the forces in Fig. 92 may (as was remai . Art. 633) be observed in the diagonals representing .28 KA, AB, BC, CD and DE, which diagonals decrt , m the end towards the middle of the girder ; and also \. . ^ lines representing the chords A U, BL, CT, DM an( which gradually increase from the end towards the pai J idle. two J. In tl ^~d being symmetrically ui^^ , 1- cwo parts are equal, or KU = UV. From U and K draw lines parallel to the corresponding lines UA and KA (91). These will meet at A and complete the triangle of forces for the point A UK of Fig. 91. From A in 92 draw the Une AB, and from L the line LB. These meet at B and com- plete the polygon KLBAK for the forces at the point KABL of 91. Starting from U, set off upon the vertical 434 FRAMED GIRDERS. CHAP. XXIT. line KV the several distances UT, TS, SR and RQ, respectively equal to the several loads UT, TS, SR and RQ as found in 91. For the forces at the point ABCTU, draw the line BC from B, and the line TC from T, each parallel with its corresponding line in 91. These lines meet at C and complete the polygon ABCTUA, which gives the forces converging in the point UABCT. occupied by the three triangles . , y be left free, and be designated by one letter OL of three. In place of five, there are in fact only 3S meeting at the point PEFGQ, and these four are i nted in Fig. 90 by the polygon QPEFQ. The above analysis is in theory strictly correct, and } >ractice it is not so, for in such cases as this there is i s more or less weight on the lower chord at the midd t. If nothing more, there is the weight of the timb' id itself, and this should be considered. i Art. 634 a truss with weights at the points of eac 3 will be found discussed, and the facts as found in pra ythere developed. /The construction of one half of the diagram (Fig. 90) h jw been completed. The other half is but a repetition / in reversed order, and need not here be shown in deta (n drawing the lines for the latter half, it will be seen that t oint H is identical with the point F y and that K a Coincide re - FIG. 92, For the point LBCDM, draw from C the line CD, and from M the line MD^ each parallel with its corre- sponding line in 91. These lines, meeting in D, complete the polygon MLBCDM^ which gives the forces surround- ing the point LBCDM. For the point TCDES, draw from D the line D, and from 5 the line SE, respectively GRADATIONS OF STRAINS. 435 parallel with the corresponding lines of Fig. 91. They will meet at E and complete the polygon TCDEST, which measures the forces around the point TCDES. For the point MDEFN y draw from E the line EF, and from N the line NF, parallel with EF and NF of 91 ; and they, meeting at F, will complete the polygon MDEFNM, thus giving the forces converging at the point MDEFN. The correspondence of lines in the two figures has now been traced to a point beyond the middle of the framed gir- der. The remainder of Fig. 92 may be traced for the other half of 91, by a continuance of the process used in tracing the first half. Since, in this instance, the loading and plan of the girder are symmetrical, and hence the several forces in the lines of one half of the girder respectively equal to those in the other half, the lines of the diagram as laid down for the one may be used for the other half. 635 a Gradation of Strains in Chords and Diagonals. The gradation of the forces in Fig. 92 may (as was remarked in Art. 633) be observed in the diagonals representing the lines KA, AB, BC, CD and DE, which diagonals decrease from the end towards the middle of the girder ; and also in the lines representing the chords A U, BL, CT, DM and ES, which gradually increase from the end towards the middle. 636. Strains Measured Arithmetically. Let Fig. 93 represent a framed girder, in which the loads are symmetri- cally placed, and where L is put for the load on each point of bearing of the upper chord, and N for that suspended at each bearing point of the lower chord. Let a represent the vertical height of the girder, c the length of a diagonal, and b the base of the triangle formed with c and a. 436 FRAMED GIRDERS. CHAP. XXII. 637. Strains in the Diagonals. To analyze these, we commence at the middle of the girder. There being an odd number of loads upon the upper chord, one half of the Fin. 93- central one, Z, is carried at Q, one of the points of sup- port, and the other half at F, the other point. The effect of this upon .the brace MC may be had from the relation of the sides of the triangle abc, for a : c : : \L : L 2a 2a : c : : L : L 2a D equals the strain in the diagonal ; or, when W equals the vertical load, equals JZ, D = W- a (296.) The vertical effect of this at M is JZ-, the same as it is at C. This amount, added to the suspended load N at M, equals \L + A 7 ", equals the total vertical force acting at M. This is sustained by the lines MK and BM, the latter standing at the same angle with MK as did MC. Hence the effect upon the diagonal is STRAINS IN THE DIAGONALS. 437 equals the strain on the diagonal BM ; and the vertical effect at M is equal to \L-\- N. Adding this to the load on the top chord at B, the sum, fZ+TV, is the total load at B, and it is supported by the forces in the lines PB and BC, constituting, with the weight, three forces, acting in the directions of the three sides of the triangle abc. The effect in the diagonal BP is therefore, as before, the load into the ratio , or (IL + N) . The vertical effect of this a V2 } a at P is equal to the vertical effect at B, or %L + N. Add- ing to it the load N at P, their sum, J-Z 4- 2N, is the total vertical effect at P ; and, as before, the effect of this on the diagonal AP which carries it is (fZ + 2N), with a vertical effect at A of f Z 4- 2N, the same as at P. Adding the load Z, at A, the sum, L+2N, equals the total vertical pressure at A. This is sustained by the forces in the lines QA and AB, which, with the weight, act in the direction of the sides of the triangle abc, and therefore the effect in the diagonal, as before, is (-4-2^), while the vertical a effect of this at Q is equal to the same effect as at A, or Thus, the loads on half the girder have, one by one, been picked up and brought along, step by step, until they are finally received upon Q, their point of support at one end of the girder. It will be observed that this accumulated load, %L+2N, coincides with the sum of the loads as seen upon one half of the figure, that is, to the 2-J- loads on the top chord and the two loads suspended from the bottom chord. 638. Example. Let it be required to show the strains in the diagonals of a framed girder 50 feet long, of five bays and 4^ feet high. 438 FRAMED GIRDERS. CHAP. XXII. Here b, the base of the measuring triangle, is equal to |~o = 5, and a, its height, equals the height of the girder, equals 4-5 ; and <:, the hypothenuse of the right-angled triangle, is therefore c 1/20-25 + 25 =6-7268 The load L upon each point of the upper chord is 10,000 pounds, while N, the suspended load at each point of the bottom chord, is 2500 pounds. FIG. 93. The strain upon the diagonals is, by formula (296. \ The load on CM is %L, and therefore the strain in the diagonal CM is c 6-7268 D, = L~ = loooo x r = 74741 pounds. The strain in the diagonal MB is c ,6-7268 D a ($L + M) - = (5000 + 2500) - = 1 121 it pounds. STRAINS IN LOWER CHORD. 43C The strain in the diagonal BP is c ,6-7268 D 3 (f-Z, + N ) - = (i 5000 + 2500) - - = 261 59! pounds. 'a 4-5 The strain in the diagonal PA is D & = (4Z, + 2N) - = (i 5000 + 5000) - = 29896! pounds ; a 4*5 and the strain in the diagonal A Q is D 5 = (%L + 2N) - (25000+ 5000) - - = 448451 pounds. 639. Strains in tlie L,ower Chord. From the measuring triangle abc of Fig. 93 we have a : b :: W : H H=W- (MT-) a in which H is the strain in the horizontal lines due to W the weight ; and with this formula we may ascertain the horizontal forces in the chords of the girder. First. In the lower chord. At the point Q we have, for W in the formula, one half the total load, or (JZ, + 2N\ and therefore equals the horizontal strain in QP. For the next bay, PM, we have, for W, the same amount, plus that caused by the thrust in the strut BP, plus that due to the tension in the rod AP. These three amounts 440 FRAMED GIRDERS. CHAP. XXII. are respectively f Z, + 2N, f + N and f L + 2N, and their sum is (f -f 27V) + (f 4- JV) f (f -f 2^0 = JT = -y- + 5^ equals the total weight causing- horizontal strain in PM. From this, the horizontal strain in PM is .. For the third, or middle bay, MK, we have the weight the same as for PM, together with that coming from the 1 . FIG. 93. thrust of the strut CM, and from the tension of the rod BM. These three weights are ty-L -f $N> \L and %L+ N, or together, (V + $N) + \L + (f + N)~ W=L + 6A? and for the horizontal strain in MK we have This completes the strains in the lower chord, for those of the other end are the same as these. 640. Strains in the Upper Chord. For the first bay, AB, there are two compressions, namely: that due to the reaction from the strut AQ, and that from the tension in STRAINS IN UPPER CHORD. 441 the rod AP. The weight causing thrust in the strut is equal to half the total load, or fZ 4- 27V, and the weight causing tension in the rod is f Z + 2N ; or, together, we have for the weight 4Z 4- 47V; and for the compression in AB, H' = (AL 4- 47V) d For the second bay, BC, we have this same thrust, plus that due to the reaction of the strut PB, plus that due to the tension in the rod BM. The three weights are 4Z + 4/V, fZ + N and -JZ 4- N, and their sum is (4Z 4- 47V) 4 (f Z 4- TV) 4- (JZ 4- TV) = 6Z 4- 67V and the horizontal compression in BC is 77" = (6Z + 6yV)- i? 641. Example. What are the horizontal strains in a girder of five bays, it being 50 feet long and 4^ feet high, and having 10,000 pounds resting upon each bearing point of the upper chord, and 2500 pounds at each point of sus- pension in the lower chord? Here, in the measuring triangle abc, b -f-^- = 5 and a = 4- 5 ; from which - = - = i4. Hence, for each hor- a 4-5 izontal strain, we have Now, in the lower chord, we have, as in Art. 639, for the bay QP, the weight W= L-\-2N and, therefore, H f = - 1 / [(2^- x 10000) + (2 x 2500)] = 33333-} pounds ; equals the horizontal tension in QP. 44 2 FRAMED GIRDERS. CHAP. XXII. For the next bay, PM, we have for the weight, as per Art. 639, W = V 1 7- f 5^ and, therefore, -ft = -V-KSi x 10000) + (5 x 2500)] = 75000 pounds ; equals the horizontal tension in PM. For the third, or middle bay, MK, for the weight, as per Art. 639, we have W= ifL -f 6N and, therefore, H s = -V-[( 6 i x i oooo) -h (6 x 2500)] = 88888| pounds ; equals the horizontal tension in MK. This completes the work for the lower chord, as the ten- sions in the other half are the same as those here found for this. In the upper chord the weight causing compression in the first bay, A, is, as per the last article, W 4L i 4N and, therefore, H' = -[(4>< 10000) + (4x2500)] = 555551 pounds; equals the horizontal compression in AB. For the next bay, BC, for the weight causing compres- sion we have, as per last article, W 6L -f 6N and, therefore, H" = W 6 x 10000) -f- (6 x 2500)] = 83333^ pounds ; equals the horizontal compression in BC. HORIZONTAL STRAINS TENSION. 443 This completes the strains for the upper chord. Tabu- lated, these several horizontal strains stand thus : For the lower chord : In QP and JF the strains are 33,333^ pounds. " PM " KJ " " tk 75,ooo " MK " strain is 88,888| " For the upper chord : In AB and DE the strains are 55,555| pounds. " BC " CD " " " 83,333^ To test the correspondence of these results with those shown by the graphic method in Figs. 91 and 92, the student may make diagrams with the given figures at a scale as large as convenient, giving to L and N the proportions above assigned them, namely, L 4.N, and making the bays with a base of 10 and a height equal to 4-5. The results ob- tained should approximate those above given, in proportion to the accuracy with which the diagrams are made. 642. Resistance to Tension. Only in so far as tension is incidental to the transverse strain would it be proper to speak of the former in a work on the latter. In a framed girder, the lower chord and those diagonals which tend downwards towards the middle of the girder are subject to tension. The better material to resist this strain is wrought- iron, and this, in the diagonals at least, is usually employed. The weight with which this material may be safely trusted per square inch of sectional area varies according to the quality of the metal, from 7000 to 15,000 pounds. Ordi- narily, it may be taken at 9000 pounds, but when the metal 444 FRAMED GIRDERS. CHAP. XXII. and the work upon it are of superior quality, it is taken at as much as 12,000, or even higher in some special cases This is the safe power of the metal per square inch of the sectional area. Let k equal this power, W equal the load to be carried, and A the sectional area of the bar, then Ak= W W A = - (298.) K As an application of this formula, take the case of the di- agonal AP, Fig. 93 ; the strain in which is 29,896^ pounds. Putting k = 9000, we have 29896! A = - -^ = 3-3218 9000 or the rod should contain 3^- inches in its sectional area. Referring to a table of areas of circles, we find that the rod, if round, should be a trifle over 2 inches in diameter, or, if a flat bar 4 inches wide, it would need to be -J- of an inch thick, since 4x1 = 3-333. ^ The above is for the diagonals. The chords are usually of wood. When so made, the value per square inch sec- tional area may be taken at.one tenth of the ultimate tensional power of the materials as given in Table XX. Since a chord is usually compounded of three or more pieces in width, and of lengths less than the length of the chord, it is necessary to see that the area of material determined by the use of for- mula (298.) is that of the uncut material, or of the uncut sec- tional area at all points in the length. Thus, were the pieces so assembled as to have no two heading joints occur at the same point in the length, or so near each other that the re- quisite bolts for binding the pieces together could not be in- troduced between the two joints, then the uncut sectional TENSION IN LOWER CHORD. 445 area would be equal to that of all the pieces in the width ex- cept one. Should two joints occur at or near one point in the length, then the sectional area of all but two pieces in width must be taken ; and so on for other cases. Where care is exercised in locating the joints, the allow- ance for joints, bolt holes, and other damaging contingencies may be taken as amounting to as much as the net size ; or, ordinarily, the net size should be doubled. Then for the total sectional area we have . IO X 2 2O ~^r\ 20 (299.) in which T is the ultimate resistance to tension, as found in Table XX. As an illustration, take the case of the lower chord of Fig. 93, which, at the middle bay, has a horizontal strain of 88,889 pounds. From Table XX. we have the resistance to tension of Georgia pine equal to 16,000 pounds. By formula (299.) 2oW 20x88880 A = ~- - -~ =in inches T 16000 or the area should be not less than 1 1 1 inches. The chord may be 10 x 12 120 inches, and may be compounded of three pieces in width a centre one of 4x 12 and two out- side pieces of 3x12 each. 643. Resistance to Compression. The top chord of a framed girder, and the struts or diagonals directed down- 44-6 FRAMED GIRDERS. CHAP. XXII. ward towards the points of support, are in a state of com- pression. The rules for determining- the resistance to compression in posts or struts are numerous, and their discussion has oc- cupied many minds. The theory of the subject will not be rehearsed here. For this the reader is referred to authors who have made it a special point, such as Tredgold, Hodg- kinson, Rankine, Baker, Francis and others. For short columns, the resistance is, approximately, in proportion to the area of cross-section of the post. As the post increases in length, the resistance per square inch of cross-section gradually diminishes. In framed girders, the struts, and also the chords, when properly braced against lateral motion, are in lengths com- paratively short, and hence the resistance which the material in them offers is not much less than when in short blocks. Baker* gives as the strain upon posts =<:(< L* Reducing this expression and changing the symbols to agree with those of this work, we have in which / equals the ultimate resistance of the post per inch of sectional area, C equals the ultimate resistance to compression of the material when in a short block, e the extension of the material per foot due to flexure, within * Strength of Beams, Columns and Arches, by B. Baker, London, 1870, p. 182. RESISTANCE TO COMPRESSION. 447 the limits of elasticity, as found in Table XX., and d is the dimension in the direction of the bending. This in a post will be the smaller of the two, or the thickness. Let h rep- resent this thickness and be substituted in the above for d\ then r = j is the ratio of the length to the thickness or It smallest dimension of the cross-section ; / and h both being taken of the same denomination, either inches or feet. The safe limit of load for posts is variously estimated at from 6 to 10. Putting a to represent this, and taking C for the ultimate resistance, as in Table XX., we have for the safe resistance f __ _ (300.) - and when W equals the load to be carried, and A equals the sectional area, we have W Af= W or A = -j and, by substituting for /) its value, as in (300.), W A = C (S01} As an application, let it be required to find the area of the Georgia pine strut AQ in Fig. 93, the strain in which is (Art. 638), say 45,000, and the length of which is 6-73. The ratio r can not be assigned definitely in the formula, as h is unknown. From experience, however, a value may be assigned it approximating its true value, and after computation, if the result shows that the assigned value deviates materially from the true value, then a nearer ap- 443 FRAMED GIRDERS. CHAP. XXII. proximation may be made for a second computation. The ratio in the case now considered is probably about equal to 12. We will take it at this amount for a trial. Take, from Table XX., the values of 6^=9500 and e = 0-00109, for Georgia pine. Make a, the factor of safety, equal to 10. The value of W is 45,000. Then, by formula (301.), 10 [i +(f x 0-00109 x I2 2 )] x 45000 A = - - = 58-521 9500 or the area should be 58^ inches. Having taken the ratio at 12 we should have the thick- ness in inches equal to the length in feet, or 6-73. Divid- ing the area 58-521 by this gives a quotient of 8-696 as the breadth. The dimensions of the piece are 6f x 8f . If it be desirable to have the thickness greater than here given, then a second trial may be had with a less ratio. 644. Top Chord and Diagonals Dimensions. By transformation of formula (301.) a rule may be arrived at which shall define the breadth of a diagonal or post exactly. Let A = hb, and let //, the thickness, bear a certain relation to b, the breadth ; or nh = b, n being a con- stant assumed, at will (for example, if n = 1-2, then i-2/i b). Then A = nh 2 . Putting also for r its value TO/ (/ being taken in feet) we have h 2 n = C Ctfn = Wa + Ctfn = Wak 3 + (| x I2 2 Wael*} Ctfn - Wah 3 = | x 12' Wael* _ Cn 2 C TOP CHORDS AND DIAGONALS. 449 Completing the square and reducing gives Cn \ 2Cn) 2Cn Let W~- be called G ; then we have 2Cn 2Cn and by substitution the above formula becomes h = V432Gel'+G' + G which is a rule to ascertain the thickness or smallest diame- ter of a strut or post, and in which / is in feet and the other dimensions are in inches. This rule, owing to its complication, will be found to be tedious in practice. For this reason, formula (301.) ordi- narily, for its greater simplicity, is to be preferred ; although, from the necessity of assuming the value of r, a second computation may be required. 645. Example. What is the value of h, the thickness of the strut at AQ, *Fig. 93 ; the length being 6-73, and the force pressing in the line of its axis being 45,000 pounds. Putting 10 for a, the factor of safety, putting 1-2 for n, the factor defining the relation of the breadth to the thickness, and taking from Table XX. the values of the con- stants C and e for Georgia pine, we have F= 45000, a = 10, *? 0-00109, /=6-/3, C =9500 and #=i-2. 450 FRAMED GIRDERS. CHAP. XXII. By formula (302.) we have r a 45000 x 10 G = W^- = 19-737 2Cn 2 x 9500 x i -2 Then, by formula (303. \ x 19-737 x 0-00109 x ) 4- 19-737 + 19-737 = 6 '943 or the thickness of the strut is required to be, say 7 inches. As nh = b, therefore /; = I -2 X 6-943 = 8-332 equals the breadth of the strut ; and since hb = A, there- fore A = 6-943x8-332 = 57-849 equals the area of the strut ; a fraction less than was before found by formula (301.). That value would have been the same as this had the value of r been correctly assumed. Its exact value is 11-632 instead of 12, the amount there taken. 64-6. Derangement from Shrinkage of Timber*. Ow- ing to the natural shrinkage of timber in seasoning, the most carefully framed girder will settle or sag more or less, pro- vided adequate measures are not taken to prevent it. The ends of the struts press upon the inside of the chords, while the iron rods have their bearing at the outside. The conse- quent diminution in height of the girder will be equal to the shrinkage of both the top and bottom chords, and the rods which at first were of the proper length will be found cor- respondingly long. By screwing up the nuts upon the rods as the shrinkage progresses, the sagging may be prevented ; but this would be inconvenient in most cases. It is better, in constructing the girder, to provide bearings of metal extending through UNEQUAL LOADS, IRREGULARLY PLACED. 451 the depth of each chord, and so shaped that the strut and rod shall each have its bearing upon it. The shrinkage will then have no effect upon the integrity of the frame. 64-7. Framed Girder with Unequal Loads Irregularly Placed. Let Fig. 94 represent such a case, wherein A, B and C are the loads upon the top chord, and D and E the loads on the bottom chord, all located as shown. As FIG. 94. in other cases, the first requirement is to know the reactions at the two supports R and P. In a girder symmetrically loaded this involves but little trouble, as the half of the total load equals the reaction at each support. In our present case, we can not thus divide the load, since the reactions are not equal. To obtain the required division of the total load, we must consider each of the several weights separately, dividing it between the two supports according to its dis- tances from them. Thus, putting m and n for the dis- tances of the load A from the two supports, the portion of A bearing upon R is shown by formula (#.) (placing A for W), in which A, the weight, is multiplied by n, its distance 45 2 FRAMED GIRDERS. CHAP. XXII. from the opposite support, and divided by /, the length or span. In like manner, each of the other weights may be divided, and the portion bearing upon each support found. Putting the letters o, p, q, r, s, t, n and v to repre- sent the distances shown in the figure, we have, as the total effect upon one of the supports, An Bp Cr Dv Et * = + 7- + T + T + T R = and for the total effect upon the other support, _ A m + Bo + Cq + Du + Es (SO 5 ^ Adding these two formulas, we have as the total effect upon both supports, A(m + ri) + B(o+p)+C(q + r) + D(u + v) + E(s + f) ./v. ~T~ Jr Here the sum of the two quantities within each parenthesis is equal to / the length, and consequently _ J\.-\- r ~ / R+P = A+B+C+Di E or the sum of the reactions of the two supports is equal to the sum of all the weights. In this we have proof of the accuracy of the two formulas (304.) and (305.). 648. Load upon Each Support Graphical Represen- tation. The value of R in formula (304*) may be readily DIVIDING THE LOADS BETWEEN SUPPORTS. 453 found, either arithmetically or graphically. The formula for one weight, R< = Aj (d), gives R,lAn, or two equal rectangles. Having three of these quantities, /, A and n, the fourth quantity, R may be graphically found thus : In Fig. 96 let AB, by any convenient scale, equal n. Draw AC at any angle with AB, and equal in length to FIG. 96. /. Lay off AD equal to A. Join B with C, and from D draw DE parallel with CB. AE will equal R t the required quantity, for, from similar triangles, we have AC : AB :: AD : AE I : n : : A : R, = Aj To obtain the value of R for all of the weights, proceed as in Fig. 97, in which the parallel lines FL, GM, HN, JO H J S l< R U TCL FIG. 97. w and KP are each equal to /, the span RP of Fig. 94. 454 FRAMED GIRDERS. CHAP. XXII. From F lay off upon FL, the first of these lines, the distance FV equal by scale to the weight A of Fig. 94, and from F on line FW place FQ equal to n. Con- nect Q with L. From V draw VG parallel with LQ. FG will represent R,. From G draw GM parallel with FL. Make GV equal to the weight B (94), and GR equal to /. Con- nect R with M, and from V draw VH parallel with MR. GH will represent R s . H J S K R U TO. FIG. 97. From H draw //N parallel with FL. Make equal to the weight C (Fig. 94), and HS equal to r. Connect 5 with N, and parallel with NS draw VJ. HJ will represent R 3 . In like manner, with the weight D and distance v of Fig. 94, obtain ^A" equal to R t ; and with the weight E and distance t obtain KU equal to R 5 . We now have the line FU equal to the sum of equals that portion of the total load on the girder which presses upon the support R. IRREGULAR FORCE DIAGRAM. 455 Similarly, the amount of pressure upon the support P may be obtained. The two, R and P, should together equal the sum of the weights A, B, C, D and E. 649. Girder Irregularly Loaded Force Diagram. Having accomplished the division of the total weight, we FIG. 94. H E FIG. 95. may now construct -upon the same scale with that of Fig. 97, the force diagram, Fig. 95, for the girder represented in Fig. 94 and described in Art, 647. On a vertical, RP, make 456 FRAMED GIRDERS. CHAP. XXII. RM equal to FU (97); and RS equal to the weight A, 57' equal to the weight B, and TP equal to the weight C, all as in Fig. 94. Now, since RM equals FU (97), equals the reaction of the support R. therefore, from R draw RE parallel with RE (94), and from M draw ME parallel with ME (94). From M make ML equal to the weight E (94), and LK equal to the weight D (94). Draw the other lines all parallel with the corresponding lines of Fig. 94, as per Arts. 618 and 619, and the force dia- gram will be complete. 650. Load upon Each Support, Arithmetically Obtained. The reaction of the two supports may be found arithmet- ically, as before stated, by the use of formulas (304) and (305.). Thus, let the several weights A, B, C, D and E of Fig. 94 be rated, by the scale of the diagram, at 15, 23, 17, 22 and 19 parts respectively. These parts may represent hundreds or thousands of pounds, or any other denomination at will. Let /, the span, equal 64, and the several distances n, /, r, v and t measure respectively 54, 34, 8, 26 and 44 by the same scale. Formula (304) now gives (15. x 54) + (23 x 34) + (17 * 8) + (22 x 26) + (19 x 44) . ~6^~ = 49 Formula (305) gives p = 05 x io) + (23x3o) + (i;x 56) + (22x 38) + (19x20) _ 64 R -r P 49 + 47 = 96 The sum of the weights is W 15 + 23 + 17 + 22+19 = 96 the same amount, thus proving the above computation cor- rect. QUESTIONS FOR PRACTICE. 651. Given a frame similar to Fig. 87, with a span of 40 feet, a height of 23 feet, with the length of the vertical BC equal to 15 feet, and with AF and BG equal. Draw a diagram of forces, and show what the strains are in each line of the frame; the three loads FG, GH and HJ being each equal to 5000 pounds. 652. According to the rule given in Art. 624, show what should be the height of a framed girder which is 75 feet between bearings. 653. According to Art. 627, show how many bays the girder of the last article should have. 654. Show, by the diagram of forces, what are the strains in the several lines of a girder 55 feet long between centres of bearings and 5-27 feet high between axes of chords ; the girder to be divided into five equal bays, each being an isosceles triangle as in Fig. 93. The load upon the apex of each triangle is 5000 pounds, and that suspended from the lower chord at each point of intersection with the diagonals is 1250 pounds. Letter the girder as in Fig. 91. 655. To test the accuracy of the results obtained in the last article compute the strains arithmetically. FRAMED GIRDERS. CHAP. XXII. 656. What should be the areas of cross-section of the bottom chord of the girder of Art. 654, at the several bays? What should be the sizes of the upper chord and of the diagonal struts? The timber is to be of spruce ; a, the fac- tor of safety, to be taken at 10, and n at 1-2. What should be the areas of cross-section of the diagonal rods, taking the safe strength of the metal at 9000 pounds ? In the questions of this Art. take the strains given by the diagram of forces. CHAPTER XXIII. ROOF TRUSSES. ART. 657. Roof Trusses considered as Framed Girders. It is proposed, in this chapter, to discuss the subject of roof trusses in so far only as they may be considered to be framed girders, placed in position to carry the roofing mate-' rial. A full treatise on roofs would include matter extending beyond the limits of a work on the transverse strain. Those desirous of pursuing the subject farther are referred to Tred- gold, Bow and others* who have written more fully on roofs. 658. Comparison of Roof Trusses. Designs for roof trusses, illustrating various principles of roof construction, are herewith presented. The designs at Figs. 98 to 102 are distinguished from those at Figs. 103 to 106, by having a horizontal tie-beam. In the lat- ter group, and in all designs similarly destitute of the horizon- tal tie at the foot of the rafters, the strains are much greater than in those having the tie, unless the truss be protected by exterior resistance, such as may be afforded by competent buttresses. To the uninitiated it may appear preferable, in Fig; 103, to extend the inclined ties to the rafters, as shown by the dotted lines. But this would not be beneficial : on the con- * Tredgold's Carpentry. Bow's Economics of Construction. 460 ROOF TRUSSES. CHAP. XXIII. trary, it would be injurious. The point of the rafter where the tie would be attached is near the middle of its length, and consequently is a point the least capable of resisting transverse strains. The weight of the roofing itself tends to bend the rafter ; and the inclined tie, were it attached to the rafter, would, by its tension, have a tendency to increase this bending. As a necessary consequence, the feet of the rafters would separate, and the ridge descend. 98. 99- 100. K 101. 102. 103. 104. 105. 106. In Fig. 104 the inclined ties are extended to the rafters; but here the horizontal strut or straining beam, located at the points of contact between the ties and rafters, counteracts the bending tendency of the rafters and renders these points stable. In this design, therefore, and only in such designs, is it permissible to extend the ties through to the rafters. Even here it is not advisable to do so, because of the in- creased strain produced. (See Figs. 118 and 120.) The design in Fig. 103, 105 or 106 is to be preferred to that in Fig. 104. LOAD UPON EACH SUPPORT. 461 659- Force Diagram Load upon Each Support.- By a comparison of the force diagrams hereinafter given, of each of the foregoing designs, we may see that the strains in the trusses without horizontal tie-beams at the feet of the rafters are greatly in excess of those having the tie. In constructing these diagrams, the first step is to ascertain the reaction of, or load carried by, each of the supports at the ends of the truss. In symmetrically loaded trusses, the weight upon each support is always just one half of the whole load. 660. Force Diagram for Trus in Fig. 98. To obtain the force diagram appropriate to the design in Fig. 98, first letter the figure as directed in Art. 619, and as in Fig. 107. Then G- FIG. 108. draw a vertical line, EF (Fig. 108), equal to the weight W at the apex of the roof; or (which is the same thing in effect) equal to the sum of the two loads of the roof, one extending on each side of W half-way to the foot of the rafter. Di- vide EF into two equal parts at G. Make GC and GD each equal to one half of the weight N. Now, since EG is equal to one half of the upper load, and GD to one half of the lower load, therefore their sum, EG + GD = ED, is equal to one half of the total load, or to the reaction of each support, E or F. From D draw DA parallel with DA of Fig. 107, and from E draw EA parallel with EA of Fig. 107. The three lines of the triangle AED rep- 462 ROOF TRUSSES. CHAP. XXIII. resent the strains, respectively, in the three lines converging at the point ADE of Fig. i7. Draw the other lines of the diagram parallel with the lines of Fig. 107, and as directed in Arts. 619 and 620. The various lines of Fig. 108 will repre- sent the forces in the corresponding lines of Fig. 107; bearing in mind (Art. 619) that while a line in the forge diagram is designated in the usual manner by the letters at the two ends of it, a line of the frame diagram is designated by the two letters between which it passes. Thus, the horizontal lines AD, the vertical lines AB, and the inclined lines AE have these letters at their ends in Fig. 108, while they pass between these letters in Fig. 107. 661. Force Diagram for Tru in Fig. 99. For this truss we have, in Fig. 109, a like design, repeated and lettered as FIG. 109. FIG. no. required. We here have one load on the tie-beam and three loads above the truss ; one on each rafter and one at the ridge. In the force diagram, Fig. no, make GH, HJ and JK, by any convenient scale, equal, respectively, to the weights Gff, HJ and JK of Fig. 109. Divide GK into two equal parts at L. Make LE and LF each equal to one half the weight EF (Fig. 109). Then GF is equal to one half the total load, or to the load upon the support G (Art. 660). Complete the diagram by drawing its several lines parallel with the lines of Fig. 109, as indicated by the letters (see Art. 660), commencing with GF, the load on FORCE DIAGRAMS. 463 the support G (Fig. 109). Draw from F and G the two lines FA and GA, parallel with these lines in Fig. 109. Their point of intersection defines the point A. From this' the several points B, C and D are developed, and the figure completed. Then the lines in Fig. no will represent the forces in the corresponding lines of Fig. 109, as indicated by the lettering. (See Art. 619.) 662. Force Diagram for Truss in Fig. 100. For this truss we have, in Fig. in, a similar design, properly prepared B D FIG. 112. by weights and lettering ; and in Fig. 112 the force diagram appropriate to it. 464 ROOF TRUSSES. CHAP. XXIII. In the construction of this diagram, proceed as directed in the previous example, by first constructing NS, the ver- tical line of weights ; in which line NO, OP, PQ, QR and RS are made respectively equal to the several weights above the truss in Fig. in. Then divide NS into two FIG. ii2. equal parts at T. Make TK and TL each equal to the half of the weight KL. Make JK and LM equal to the weights JK and LM of Fig. in. Now, since MN is equal to one half of the weights above the truss, plus one half of the weights below the truss, or half of the whole weight, it is therefore the weight upon the support N (Fig. FORCE DIAGRAMS. 465 in), and represents the reaction of that support. A horizon- tal line drawn from M will meet the inclined line drawn from Nj parallel with the rafter AN (Fig. in), in the point A t and the three sides of the triangle AMN (Fig. 112) will give the strains in the three corresponding lines meeting at the point AMN (Fig. in). The sides of the tri- angle HJS (Fig. ii2) give likewise the strains in the three corresponding lines meeting at the point HJS (Fig. in). Continuing the construction, draw all the other lines of the force diagram parallel with the corresponding lines of Fig. in, and as directed in Art. 619. The completed diagram will measure the strains in all the lines of Fig. in. 663. Force Diagram for Truss in Fig. 101. For the roof truss at Fig. 101 we have, in Fig. 113, a repetition of it, and in Fig. 114 its force diagram. FIG. 113. FIG. 114. 466 ROOF TRUSSES. CHAP. XXIII. The dimensions on the vertical line HL (Fig. 114) are made respectively equal to the weights in Fig. 113, as indi- cated by the lettering. With GH equal to half the whole weight on the truss (Art. 660), the triangle AGH is con- structed, giving the strains in the three lines concentrating at the point AGH (Fig. 113). Then, drawing the other lines parallel with the corresponding lines of Fig. 113, the com- pleted diagram gives the strains in the several lines of that figure, as indicated by the lettering. (See Art, 619.) 664-. Force Diagram for Trus in Fig. 102. The roof truss indicated at Fig. 102 is repeated in Fig. 115, with the addition of the lettering required for the construction of the force diagram, Fig. 116. In this case, there are seven weights, or loads, above the truss, and three below. Divide the vertical line OV at W y into two equal parts, and place the lower loads in two equal parts on each side of W. Owing to the middle one of these loads not being on the tie-beam with the other two, but on the upper tie-beam, the line GH, its representative in the force diagram, has to be removed to the vertical BJ, and the letter M is duplicated. The line NO equals half the whole weight of the truss, or 3^ of the upper loads, plus one of the lower loads, plus half of the load at .the upper tie- beam. It is therefore the true reaction of the support NO, and AN is the horizontal strain in the beam there. It will be observed also, that while HM and GM (Fig. 116), which are equal lines, show the strain in the lower tie-beam at the middle of the truss, the lines CH and FG, also equal but considerably shorter lines, show the strains in the upper tie-beam. Ordinarily in a truss of this design, the strain in the upper beam would be equal to that in the lower one, which becomes true when the rafters and braces above FORCE DIAGRAMS. 467 the upper beam are omitted. In the present case, the thrusts of the upper rafters produce tension in the upper beam FIG. 115. FIG. 1 1 6. equal to CM or FM of Fig. 116, and thus, by counteract- ing the compression in the beam, reduce it to CH or FG of the force diagram, as shown. 468 ROOF TRUSSES, CHAP. XXIII. 665. Force Diagram for Truss In Fig. 103. The force diagram for the roof truss at Fig. 103 is given in Fig. 118, while Fig. 117 is the truss reproduced, with the lettering requisite for the construction of Fig. 118. FIG. 117. FIG. 118. The vertical EF (Fig. 118) represents the load at the ridge. Divide this equally at W, and place half the lower weight each side of W, so that CD equals the lower weight. Then ED is equal to half the whole load, and equal to the reaction of the support E (Fig. 117). The lines in the triangle ADE give the strains in the corresponding lines converging at the point ADE of Fig. 117. The other lines, according to the lettering, give the strains in the cor- responding lines of the truss. (See Art. 619.) 666. Force Diagram for Truss in Fig. 104. This truss is reproduced in Fig. 119, with the letters proper for use in the force diagram, Fig. 120. Here the vertical GK, containing the three upper loads GH, HJ and JK, is divided equally at W, and the lower load EF is placed half on each side of W, and extends from E to F. Then FG represents one half of the whole load of the truss, and therefoie the reaction of the sup- port G (Fig. 119). Drawing the several lines of Fig. 120 parallel with the corresponding lines of Fig. 119, the force FORCE DIAGRAMS. 469 diagram is complete, and the strains in the several lines of 119 are measured by the corresponding lines of 120, (See Art. 619.) FIG, 120. A comparison of the force diagram of the truss in Fig. 117 with that of the truss in Fig. 119 shows much greater strains in the latter, and we thus see that Fig. 117, or 103, is the more economical form. 667. Force Diagram for Truss in Fig. 105. This truss is reproduced and prepared by proper lettering in Fig. 121, and its force diagram is given in Fig. 122. Here the vertical JM contains the three upper loads JK, KL and LM. Divide JM into two equal parts at 47 ROOF TRUSSES. CHAP. XXIII. G, and make FG and GH respectively equal to the two loads FG and GH of Fig. 121. Then HJ represents one half of the whole weight of the truss, and therefore the reac- tion of the support J. From H and J draw lines par- allel with AH and AJ of Fig. 121, and the sides of the tri- FlG. 121. M FIG. 122. angle AHJ will give the strains in the three lines concen- trating in the point AHJ (Fig. 121). The other lines of Fig. 122 are all drawn parallel with their corresponding lines in Fig. 121, as indicated by the lettering. (See Art. 619.) FORCE DIAGRAMS. 471 FIG. 123. FIG. 124. 472 ROOF TRUSSES. CHAP. XXIII. 668. Force Diagram for Truss in Fig. 106. This truss is reproduced in Fig. 123 with the lettering proper for its force diagram, as given in Fig. 124. The five external weights of Fig. 123 make up the line LQ y and the two internal weights are set, one on each side of y, the middle point of LQj extending to H and K. KL equals one half the weight of the whole truss, and equals the reaction of the point of support L (Fig. 123). The sides of the triangle AKL, therefore, give the respective strains in the three lines converging at the point AKL of Fig. 123. The other lines of Fig. 124 are found in the usual manner. (See Art. 619.) 669. Strains in Horizontal and Inclined Ties Compared. A comparison between a truss with a horizontal tie at the FIG. 125. feet of the rafters, and one without such tie will now be given. The truss without a horizontal tie shown in Fig. 103 is one of the simplest in construction, and is suitable for the comparison. Repeating it in Fig. 125, and adding the dotted lines, we have likewise the form of a truss with a horizontal tie. From Art. 608 we have, in formula (293.\ for the hori- zontal strain, H, = W in which W t .equals the total weight of the truss and its load (Fig. 125), // equals half the span, equals AD, and c HORIZONTAL AND INCLINED TIES. 473 W equals twice the height, equals 2DE. By putting P equals the reaction of one of the supports A or B, and putting d for DE, we have or, from Fig. 125, DE : AD : : P : H d : h : : P : H = P-, a that is to say, when the vertical DE represents half the weight of the truss, then AD may be put to represent the horizontal strain. Draw CF horizontal, and by similar tri- angles we have DE : AD : : CE : CF or CE : CF :: P : H = P^ LJtL or, with CE put to represent one half the weight of the whole truss, then CF, by the same scale, will measure the horizontal strain. Under these conditions, CF measures the horizontal strain in either truss, whether with or without a tie-beam. If the truss have a horizontal tie AB, then CF measures the tension in this tie. If it be without the tie AB, having instead thereof the raised tie ACB, then still CF mea- sures the horizontal strain at A or B J but not the strain in the raised tie AC. The strain in this inclined tie is measured by the line AC, for the three sides of the triangle ACE are in propor- tion as the strains in these lines respectively (see Art. 619), therefore the strains in the ties of the two trusses are compa- rable by the two lines CF and AC. 474 ROOF TRUSSES. CHAP. XXIII. The compressive strain in the rafter is also correspond- ingly increased; for just in proportion as AE exceeds F, so does the compressive strain in the rafter of a truss with an inclined tie exceed that of one with a horizontal tie. 670. Vertical Strain in Tru with Inclined Tie. In Fig. 125, if the inclined tie were lowered, so that the point C, FIG. 125. descending, should reach the point D ; or, if the inclined tie become the horizontal tie AB ; then the vertical rod DE would be subject to no strain from the weight of the rafters and the load upon them. In the absence of the horizontal tie, or when the inclined tie is depended upon to resist the spreading of the rafters, the vertical rod CE is strained directly in proportion to CD, the elevation of the tie, and inversely as the height CE. This relation may be shown as follows : Let P be put for DE (Fig. 125) and represent one half the weight of the truss. Then AD will represent the horizontal strain at A ; or, representing the span AB by then equals AD equals the horizon- the symbol tal strain. Putting a for CD and d for DE we have the proportion ENHANCED STRAINS FROM INCLINED TIES. 475 DE : AD : : P : H or d : - :: P' H=P-^ 2 2d and also, AD : CD : : H : V S - : a :VJ5T: V = H^ = P 2 s d by substitution, or This gives the vertical strain in CE, due to the raising of the tie from D to C, but it is not the whole of the strain ; it is only so much of the vertical strain as is due to the weight of the roof. The tension thus found in CE is sustained at E by the two rafters, and, passing through them to A and /?, creates horizontal and vertical thrusts precisely as did the original weight. The vertical tension thus brought to CE again acts as a weight at , and, passing down the rafters and through the tie back to C, again adds a load at C. This in turn passes around and re- turns to C, adding to the load ; and so on in an endless round to infinity. But the successive strains thus generated are in a decreasing series, and they may therefore be summed up and defined. Thus, as has just been shown, the vertical effect from the weight of the roof is The vertical effect of this latter is d: a :: V : V = V--^p d 47 6 ROOF TRUSSES. CHAP. XXIII. The vertical effect of this is d : a :: V : V" = V'r = P\- d \d The next term in the series will be and the sum of all the terms will be a / a showing that the several values of the fraction by which the weight P is multiplied constitute a geometrical series, with j- for the first term and -j- for the ratio. Since j- is a da less than unity, we have a geometrically decreasing infinite series, the sum of which is equal to the first term divided by one minus the ratio,* or a c- ^ a a d a and, since da~b ot Fig. 125, We have, therefore, as the total vertical effect due to the elevation of the middle of the tie from D to C, Ray's Algebra, Part Second, Art. 299, INCLINED TIES ILLUSTRATIONS. 477 or the vertical effect is directly in proportion to CD, the elevation of the tie, and inversely in proportion to CE, the length of the vertical tie-rod. 671. Illustrations. To illustrate the effect of the eleva- tion of the tie-rod, upon the vertical strain in the suspension- rod, let the point C, Fig. 125, be elevated -J- of the verti- Fio. 125. cal height of the truss above the horizontal line AB. Here a = i and b == 4, and -y- = \ ; or When the elevation equals of the entire height, then When the elevation equals \ of the entire height, then When the elevation equals \ of the whole height, then Thus it is seen, in this last case, that the effect due to the 4/8 ROOF TRUSSES. CHAP. XXIII. elevation of the tie-beam is equal to that of doubling- the whole weight of the roof, and this increase affects not only the vertical suspension rod at the middle, but also the rafters and inclined ties, as was shown at Art. 669. When, therefore, in order to gain a small additional height to the interior of a building, it is proposed to raise the middle point of the tie-rod, it would seem advisable to con- siaer whether this small additional height be an adequate compensation for the increased strains thereby induced, and the consequent enhanced cost for material necessary to re- sist these strains ; and also, whether it be not more advisable to raise the walls of the building, rather than the ties of the trusses. 672. Planning a Roof. In designing a roof for a build- ing, the first point requiring attention is the location of the trusses.' These should be so placed as to secure solid bear- ings upon the walls ; care being taken not to place either of the trusses over an opening, such as those for windows or doors, in the wall below. Ordinarily, trusses are placed so as to be centrally over the piers between the windows ; the number of windows consequently ruling in determining the number of trusses and their distances from centres. This distance should be from ten to twenty feet ; fifteen feet apart being a suitable medium distance. The farther apart the trusses are placed, the more they will have to carry ; not only in having a larger surface to support, but also in that the roof timbers will be heavier ; for the size and weight of the roof beams will increase with the span over which they have to reach. In the roof-covering, itself, the roof-planking may be laid upon jack-rafters, carried by purlins supported by the trusses ; or upon roof beams laid directly upon the back of PLANNING A ROOF LOAD ON TRUSS. 479 the principal rafters in the trusses. In either case, proper struts should be provided, and set at proper intervals to re- sist the bending of the rafter. In case purlins are used, one of these struts should be placed at the location of each purlin. The number of these points of support rules largely in determining th'e design for the truss, thus : For a short span, where the rafter will not require sup- port at an intermediate point, Fig. 98 or 103 will be proper. For a span in which the rafter requires supporting at one intermediate point, take Fig. 99, 104 or 105. For a span with two intermediate points of support for the rafter, take Fig. 100 or 106. For a span with three intermediate points, take Fig. 102. Generally, it is found convenient to locate these points of support at nine to twelve feet apart. They should be suf- ficiently close to make it certain that the rafter will not be subject to the possibility of bending. 673 Load upon Roof Truss. In constructing the force diagram for any truss, it is requisite to determine the points of the truss which are to serve as points of support (see Figs. 109, in, etc.), and to ascertain the amount of strain, or loading, which will occur at every such point. The points of support along the rafters will be required to sustain the roofing timbers, the planking, the slating, the snow, and the force of the Avind. The points along the tie- beam will have to sustain the weight of the ceiling and the flooring of a loft within the roof, if there be one, together with the loading upon this floor. The weight of the truss itself must be added to the weight of roof and ceiling. ROOF TRUSSES. CHAP. XXIII. 674. Load on Roof per Foot Horizontal. In any im- portant work, each of the items in Art. 673 should be care- fully estimated, in making up the load to be carried. For ordinary roofs, the weights may be taken per foot superficial, as follows : Slate, about 7-0 pounds. Roof plank, " 2-7 " Roof beams, or jack-rafters, " 2-3 " In all, 12 pounds. This is for the superficial foot of the inclined roof. For the foot horizontal, the augmentation of load due to the angle of the roof will be in proportion to its steepness. In ordinary cases, the twelve pounds of the inclined surface will not be far from fifteen pounds upon the horizontal foot. For the roof load we may take as follows : Roofing, about 15 pounds. Roof truss, " 5 Snow, " 20 " Wind, " 10 Total on roof, 50 pounds per square foot horizontal. This estimate is for a roof of moderate inclination, say one in which the height does not exceed i of the span. Upon a steeper roof, the snow would not gather so heavily, but the wind, on the contrary, would exert a greater force. Again, the wind acting on one side of a roof may drift the snow from that side, and perhaps add it to that already lodged upon the opposite side. These two, the wind and the snow, are compensating forces. The action of the snow is vertical : that of the wind is horizontal, or nearly so. The power of the wind in this latitude is not more than thirty LOADING SELECTION OF DESIGN. 481 pounds upon a superficial foot of a vertical surface ; except, perhaps, on elevated places, as mountain tops for example, where it should be taken as high as fifty pounds per foot of vertical surface. 675. Load upon Tie-Beam. The load upon the tie- beam must of course be estimated according to the require- ments of each case. If the timber is to be exposed to view, the load to be carried will be that only of the tie-beam and the timber struts resting upon it. If there is to be a ceiling attached to the tie-beam, the weight to be added will be in accordance with the material composing the ceiling. If of wood, it need not weigh more than two or three pounds per foot. If of lath and plaster, it will weigh about nine pounds ; and if of iron, from ten to fifteen pounds, according to the thickness of the metal. Again, if there is to be a loft in the roof, the requisite flooring may be taken at five pounds, and the load upon the floor at from twenty-five to seventy pounds, according to the purpose for which it is to be used. 676. Selection of Design for Roof Trus. As an ex- ample in designing a roof truss : Let it be required to provide trusses for a building measuring 60 x 90 feet to the centre of thickness of the walls, with seven windows upon each side, and with a roof having its height equal to one third of the span. The roofing is to be of plank and slate, the ceil- ing is to be finished with plastering, and the space within the roof is to be used for the storage of light articles, not to exceed twenty-five pounds to the square foot. Here, in the first place, we have to determine the number of trusses. As there are seven windows on a side, there should be six trusses, one upon each pier between the win- dows. The six trusses and the two end walls will afford 4^2 ROOF TRUSSES. CHAP. XXIII. eight lines of support for the roofing. There will thus be seven bays of roofing of Sf- = \2-\ feet each, and this is the width of roofing to be carried by each truss. In the next place, the points of support in the truss are to be ascertained. If these are provided at every ten feet horizontally, they will divide the half truss into three spaces, and there will be two intermediate points of support. For this arrangement, such a roof truss as is shown in Fig. 100 will be appropriate, but if the space in the roof is required to be quite unobstructed with timber at the middle, then a modifi- cation of this design may be used, as in the form shown in Fig. 126 ; each rafter being still divided into three equal parts. 677. Load on Each Supported Point in Tru. The horizontal measurement, then, of the roofing to be carried by each supported point in the truss, will be 10 feet along the line of the truss and 12-f- feet across the truss (this latter being the width of each bay as above found); or lox 12-f- = 128^- feet. With a weight per foot of 50 pounds, as estimated in Art. 674, we have, for the load upon each supported point of the truss, 1284- X 50 = 64284 or, say 6500 pounds. 678. Load on Each Supported Point in Tie-Beam. The tie-beam having two points of support, we have *- = 20 feet for the length of the surface to be carried. This, multiplied by the width between trusses, gives 20 x \2\ = 257^ feet area of surface to be carried by each point of support. We will estimate the weight per foot in this present case as follows: LOADING CONSTRUCTING FORCE DIAGRAM. 483 Load upon the floor, 25 pounds. Flooring, with timber, 5 Plastering, 9 Tie-beam, etc., i pound. Total at tie-beam, 40 pounds. This gives 2571- x 40 = 102856- or, say 10,300 pounds upon each supported point. Therefore, the two balls GH and HJ, suspended from the tie-beam of Fig. 126, are to be taken as weighing 10,300 pounds each, while the five balls located above the rafters are to be understood as weighing 6500 pounds each (Art. 677). 679. Contracting the Force Diagram. We may now proceed to construct the force diagram, Fig. 127, as follows: Upon the vertical line KP lay off in equal parts KL, LM, MN, NO and OP, according to any convenient scale, each equal to 6500 pounds the weight of the balls above the rafters (Art. 677). If a scale of 100 parts to the inch be selected for the force diagram, and each part be understood as representing 100 pounds, then *- = 65, equals the number of parts to assign to each of the distances KLj LM, etc., and each will be T 6 ^ 5 y of an inch in length. Dividing KP at H into two equal parts, lay off on each side of H the distances GH and HJ^ each equal, by the scale, to 10,300 pounds. This distance is found by dividing 10,300 by IOD ; the quotient 103 is the number of parts, and the distances will each be fjj-J, or one inch and of an inch in length.* * The scale here selected, although sufficient for the purposes of illustration, 484 ROOF TRUSSES. CHAP. XXIII. HK now represents one half the weight upon the rafters, and HJ one half the load upon the tie-beam, and their sum, JK, equals one half the total load of the truss, equals the load upon the point of support K. FIG. 126. FIG. 127. From y, H and G draw the horizontal lines JA, HD and GF. From K, L, M, N, O and P draw would be too small for a working drawing. For the latter, a scale should be selected as large as can be conveniently used, such as 10 parts to the inch, and too pounds to each part. This would give 1000 pounds to the inch, and each of the distances KL, LM, etc., would measure 6^ inches. It must also be remembered that the accuracy of the force diagram depends upon the care with which the distances upon the vertical line are laid off and the lines drawn. The drawing implements should be examined to know that they are true, and each line should be drawn carefully parallel with the corre- sponding line of the truss. Unless this care is exercised, the results may differ considerably from the truth. MEASURING THE STRAINS. 485 lines, as shown, carefully parallel with the rafters. From F and A draw the lines FE and AB, parallel with the two braces. Connect B and E by the vertical line BE, and then the force diagram is complete. 680. Uleasanrtiig tlie Force Diagram. After drawing the lines of the diagram as above directed, they should all be carefully traced to know that the required conditions are fulfilled, or that each set of lines, drawn parallel, in the dia- gram of forces, to the lines converging to a point in the truss, forms a closed polygon. (See Arts. 618, 619 and 620.) The diagram, by this test, having been found correct, the force in each line of the truss may be measured by applying the scale to the corresponding line of the diagram. For example, take the strains in one of the rafters. At its lower end, or the part A K, its corresponding line AK of Fig. 127 measures 478 parts, by the same scale with which the weights on the vertical line KP were laid off. This, at 100 pounds to the part, gives 47,800 pounds as the strain in the foot of the rafter. The next section of the rafter is designated by the letters BL, and the line BL (Fig. 127) measures 420, and indicates a strain in this part of the rafter of 42,000 pounds. The third or upper portion of the rafter is designated by the letters CM, and the cor- responding line in Fig. 127 measures 58 parts, indicating 5800 pounds as the strain in the upper end of the rafter. For the brace AB we have the line AB (Fig. 127), measuring 58 parts of the scale, and indicating 5800 pounds as the strain in the brace. For the vertical BD we have the line BD (Fig. 127) measuring 135 parts of the scale, and indicating 13,503 pounds as the strain in the vertical. For the horizontal strains, we have for CD, the corre- sponding line in Fig. 127, which measures 301 parts, and 486 ROOF TRUSSES. CHAP. XXIII. gives 30,100 pounds as the strain. For DH, the middle portion of the tie-beam, DH (Fig. 127) measures 350, showing the strain to be 35,000 pounds; and lor AJ, or one end of the tie-beam, AJ (Fig. 127) measures 398 parts, and gives 39,800 pounds as the strain. The strains in the other and corresponding parts of the truss are the same as these, so that we now have all the strains required. 681. Strains Computed Arithmetically. Instead of de- pending solely upon the scale, the lengths of the lines in the force diagram may be computed arithmetically. The sizes measured by the scale, when the diagram is carefully drawn, are sufficiently accurate for all practical purposes ; but in some cases, such, for instance, as when the implements for making a correct diagram are not at hand, and in all cases as a check upon the accuracy of the results obtained by the graphic method, to be able to arrive at the correct results arithmetically would be useful. Preparatorv to computing r the lengths of the lines, it will be observed that the triangle KAJ, Fig. 127, is precisely proportionate to the triangles formed by the inclination of the rafters of Fig. 126 with the vertical and horizontal lines ; that all the inclined lines of Fig. 127 are drawn at equal angles of elevation ; and that the triangles formed by these inclined lines with the vertical and horizontal lines are all homologous. Since the height of the roof is given at 20 feet, and half the span is 30 feet, therefore the perpendicular and base lines of each triangle are in like proportion namely, as 20 to 30, or as i to ij. The perpendicular being the weight in each case, which is known, we may, therefore, by this proportion obtain the base. Having both base and perpendicular, the length of the hypothenuse may be found by Euclid's 47th of ist book COMPUTING THE STRAINS. 487 the length of the hypothenuse equals the square root of the sum of the squares of the base and perpendicular. If the hypothenuse of one triangle be computed by this method, that of the others (since the triangles are homologous) may be found by the more simple method of proportion. Taking a triangle having the perpendicular and base equal to i and i|, we find, by the above rule, that its hypothenuse equals 1-802776 nearly. The hypothenuses of the other triangles, therefore, may be found by the proportion : I : 1-802776 : : / : h h = i- 802 776^ and for the base we have I : i - 5 : : / : b b= i - 5/ With these formulas, the lines in Fig. 127 have been computed. The strains in the proposed truss (Fig. 126), by both methods, have been found to be as follows : HY SCALE. BY COMPUTATION. AK 47,800 pounds ; 47,864 pounds. BL = 42,000 42,005 CM = 5,800 " 5,859 AB = 5,800 " 5,859 CD 30,100 " 30W5 DH = 35,000 " 34,950 AJ = 39,800 " ' 39,825 BD = 13,500 " 13,550 682. I>imeiision of Part Subject to Tension. With these forces, and the appropriate rules hereinbefore given, the dimensions of the several parts of the truss may now be determined. ROOF TRUSSES. CHAP. XXIII. Commencing with the tie-beam, KP, it may be observed, preparatory to computing- its dimensions, that while this piece, in resisting the thrust of the rafters, is subjected to a tensile strain, it is also subject to a transverse strain from the weight of the ceiling and floor which it has to carry. These two strains, however, are of such a nature that in their effect upon the beam they do not conflict ; for the tensile strain from the thrust of the rafters, acting, as it will usually, in the upper half of the beam, serves to counteract the compression produced by the transverse strain in this part of the beam, and the fibres near the middle of the beam, owing to their proximity to the neutral line, being strained very little by the transverse strain, have a large reserve of strength available to assist in resisting the tensile strain. It will be sufficient, therefore, to provide a piece of timber for the tic-beam of sufficient size to resist only one of the two strains ; not necessarily that strain, however, which is the greater, but that one which requires the larger piece of timber to resist it. The computations of dimensions required to resist the two strains will now claim attention. For the tensile strain we have, 'by formula (299.)^ 20 x 39800 _ 16000" or say 50 inches area of cross-section, for Georgia pine. For white pine the area should be 65 inches, The load producing transverse strain is (Art. 678) 10,300 pounds. The rule for determining the proper area of cross- section is to be found in formula (130.\ which may be modified for this case by substituting rl for <5, the symbol for deflection, and by putting for r the rate 0-04 of an inch. With these substitutions, we have STRAINS IN TIE-BEAM. 489 Fixing upon a proportion for b in terms of d, say, for ex- ample, b f^> and substituting this value for b, we have IU1 2 0-04 x IFd* MfWV ,< F ~ If the timber is to be of white pine, then F equals 2900 (Table XX.), and we have 4/2O| =r y - X 10300 X 20 . - = 13- 116 2900 or the depth will need to be I3-J- inches. Three quarters of this, or 9!, will be the breadth. The tie-beam, of white pine, will need to be, therefore, say lox 13 inches. If of Georgia pine, instead of white pine, then 5900, the value of F for Georgia pine, must be substituted for 2900 in the formula, and the results, 8-237 and 10-982, will show, say 8J- x 1 1 inches as the size of timber required. The dimensions thus found, to resist the transverse strain, being in excess of those required to resist the tensile strain, are to be adopted as the dimensions of the required tie- beam. The length of the tie-beam, 60 feet, being greater than can readily be obtained in one piece, it will have to be built up. In doing this, it is necessary that each piece be of the full height of the beam, or that the joints of the make-up be vertical and not horizontal. These vertical laminas should be in pieces of such lengths that no two heading joints occur within five feet of each other, and that these joints shall be as near as practicable to the two vertical suspending rods. The laminas need to be well secured together with proper iron bolts. The feet of the rafters should be provided with iron clamps of sufficient area to resist the horizontal strain there, and should be secured to the tie-beam with bolts of corresponding resistance. 490 ROOF TRUSSES. CHAP. XXIII. If the iron in the bolts and clamps of the truss be of aver- age good quality, it may be calculated on as resisting effec- tually 9000 pounds per square inch (see Art. 642). The vertical suspension rods BD and DE, Fig. 126, may also be calculated for a like strain. 683. Dimenion of Part Subject to ompreion. The rafters, straining beam and braces are all subject to compression, and their dimensions may now be obtained. The areas of these pieces may be had by the use of formula (301.) ; or, as this in some cases is objectionable, for the reason that the ratio between the length and thickness has to be assumed in advance, we may find in formula (303.) a rule free from this objection, but encumbered with more intricate computations. Formula (301.), when used by those having experience in such work, is far preferable^ on account of its greater simplicity. Taking first the rafter, and the portion of it at the foot, where the strain is greatest, 47,800 pounds, we have for its length about 12 feet. If of Georgia pine, its thinnest dimension of cross-section will probably be about 8 inches. Then r= - = 18 (see Art. 643). The value of C is 9,500 and the value of e is 0-00109, both by Table XX. Making the symbol for safety, a, equal 10 we have io[i +(f x 0-00109 x 1 8 2 )] 47800 A m 70 9500 / or the area of the rafter should be 77, say 8 x 9f inches. If computed by formula (303.), putting n = 1-2, the exact size will be found at 8-006 x 9-607 = 76-92 inches area. PIECES SUBJECT TO COMPRESSION. 684. _ I>imeiiion of Mid-Rafter. In the rafter at BL the strain is 42,000 pounds. The length and ratio here will be the same as at AK, and the dimensions of AK and BL are therefore in proportion to the weights (form. 301.\ or 47800 : 42000 : : 76-92 : A so that 68 inches of sectional area, or 8 x 8J inches, is the size required. 685. Dimension of Upper Rafter. The upper end of the rafter has only the weight at the ridge, 5,800 pounds, to bear. The thickness of the rafter here will probably be but 4 inches. This gives a ratio of i|^ = 36. With this ratio, with 5,800 for the weight, and with the other quantities as before, a computation by formula (301^ will result in show- ing the required area to be 19-04, or, say 4x5 inches; but, in order to resist effectually the distributed load of the roofing, this part of the rafter should not be less than 4x8 inches. 686. l>imenion of Braee. The brace, AB, being of equal length and carrying an equal load with the upper end of the rafter, may be made of the size there found necessary, or, say 4x6 inches. 687. Dimeiiiions of Straining-Beam. The straining- beam CD is compressed with a strain of 30,100 pounds, 492 ROOF TRUSSES. CHAP. XXIII. and its length is 20 feet. Assuming its thickness to be that of the rafter, we have r = = 30, and in formula (301.) )1 30100 -=78-29 9500 or its area should be ;8i, or, say 8 x 10= 80 inches. With this result, the computation of the dimensions of all the pieces of the truss is completed ; for the other rafter and brace are in like condition with those computed, and should therefore be of the same dimensions. QUESTIONS FOR PRACTICE. 688. In a roof truss similar to that shown in Fig. 109, of 42 feet span and 14 feet height, measuring from the axial lines : What will be the strains in the various pieces of the truss, with a load of 5,000 pounds at each of the three points above the rafters, and a load of 10,000 pounds suspended from the centre of the tie-beam ? Draw the appropriate force diagram, and give the strains from measurement. 689. Draw a force diagram for a roof truss similar to the design in Fig. in, with a span of 54 feet and a height QUESTIONS FOR PRACTICE. 493 of 1 8 feet; the upper weights being taken at 6,000 pounds each, the central weight under the tie-beam at 5,000 pounds, and each of the two other weights at 7,000 pounds. Show, from the diagram, the strain in each line of the truss. 690. In a truss similar to that in Fig. 121, show, by a force diagram, what would be the strains in each line, when the span is 40 feet and the height 20 feet. The weights FG and GH are so located as to divide the span into three equal parts, the three loads above the rafters are each 7,000 pounds, and the two loads below each 4,000 pounds. The point JABK is to be taken at the middle of the rafter, and the line AB is to be drawn at right angles with the rafter. 691. In a roof with an elevated tie-beam, such as in Fig. 125, with a span of 40 feet and height of 20 feet, and with the tie elevated at the middle 8 feet above the level of the feet of the rafters, compute the strain in the suspen- sion-rod at the middle, due to the elevation of the tie ; the weight upon one half of the truss being 24,000 pounds. 692. In a building 119 feet long, and 80 feet wide to the centres of bearings, and having the side walls pierced for seven windows each, state how many roof trusses there should be. Which of the designs given, having a tie horizontal from the feet of the rafters, would be appropriate for the case ? The roof is to be 25 feet high at middle, and to have the interior space along the middle free from timber. The 494 ROOF TRUSSES. CHAP. XXIII. load upon the roof is to be taken at 50 pounds per foot horizontal, upon the tie-beam at 40 pounds to the foot, and upon the straining beam at 5 pounds per foot. Make a force diagram, and from it show the strains in each piece. .Compute the dimensions of the several timbers, which are all to be of Georgia pine ; the rafter being 9 inches thick below the straining-beam and 6 inches above, and the iron work being subjected to a tensile strain of 9000 pounds per inch. CHAPTER XXIV. TABLES. ART. 693. Tables I. to XXI. Their Utility. Rules for determining the required dimensions of the various timbers in floors are included in previous chapters. These rules are carefully reduced to the forms required in practice. In using them, it is only needed to substitute for the various alge- braic symbols their proper numerical values, and to perform the arithmetical processes indicated, in order to arrive at the result desired. To do even this simple work, however, requires care and patience, and these the architect, owing to the multiplicity of detail demanding time and attention in his professional practice, frequently finds it difficult to exercise. To relieve him of this work, the first twenty-one of the following tables have been carefully computed. Tables I. to XXI. afford the data for ascertaining readily the dimensions of the beams and principal timbers required in floors of dwellings and first- class stores. Tables XVII., XVIII. and XIX. refer to beams of rolled-iron ; the others to those of wood. 694. Floor Beams of Wood and Iron (I. to XIX. and XXI.) In these tables will be found the dimensions of Floor Beams and Headers, of Hemlock, White pine, Spruce and Georgia pine ; for Dwellings and for First-class stores. Tables XVIII. and XIX. exhibit the distances from cen- tres at which Rolled-iron Beams are required to be placed 496 TABLES. CHAP. XXIV. in Banks, Office Buildings and Assembly-Rooms, and in First-class Stores. 695. Floor Beams of Wood (I. to Till.). In these tables the recorded distance from centres is in inches, and is for a beam one inch thick, or broad. The required distance from centres is to be obtained by multiplying the tabular dis- tance by the breadth of the given beam. For example : Let it be required to ascertain the dis- tance from centres at which white pine 3 x 10 inch beams, 1 6 feet long in the clear of the bearings, should be placed in a dwelling. By reference to Table II., " White Pine Floor Beams One Inch Thick, for Dwellings, Office Buildings, and Halls of Assembly," we find, vertically under 10, the depth, and opposite to 16, the length, the dimension 4-5. This is the distance from centres for a beam one inch broad. Then, since the given beam has a breadth of 3 inches, 3><4-5= 13-5 equals the required distance from centres for beams 3 inches broad. Therefore, 3 x 10 inch white pine beams with 16 feet clear bearing, should, in a dwelling, etc., be placed 13^ inches from centres. Tables I. to IV. were computed from formula (14$-), cl* = ibd 3 which, with b= I, and putting c in inches, becomes ' (308.) FLOOR BEAMS AND HEADERS. 497 Tables V. to VIII. were computed from formula (149.), cl 3 = kbd s which, with =i, and with c in inches, becomes \2kd 3 c = - (307.) 696. Hcader of Wood (IX. to XVI.). (See Art. 142.) The results recorded in these tables show the breadth of headers which carry tail beams one foot long. The tabular breadth, if multiplied by the length in feet of the given tail beam, will give the breadth of the required header. For example : Let it be required to ascertain the breadth of a Georgia pine header 20 feet long, 15 inches deep, and carrying tail beams 12 feet long, in the floor of a first- class store. By referring to Table XVI., 4 ' Georgia Pine Headers for First-class Stores," at the intersection of the vertical column for 15 inches depth and the horizontal line for 20 feet length, we find the dimension io6. This is the breadth of the header for each foot in length of the tail beams. As the tail beams in this case are 12 feet long, therefore 12x1-06=12.72, equals the required breadth of the header in inches. The first four (IX. to XII.) of these tables were computed from formula (156.), i6Fr(d-i? which, when reduced (putting r = 0-03, / = 90 and n = i) becomes 498 TABLES. CHAP. XXIV. The second four (XI H. to XVI.) of these tables were com- puted from the same formula, (156-}, by putting r = 0-04, f= 275 and n = i ; which reduction gives (309 , 697. Elements of Rolled-Iron Beams (XVII.). Table XVII. contains the dimensions of cross-section and the values of /, the moment of inertia, for 1 19 of the rolled- iron beams of American manufacture in use. These values are required in using the rules in Chapter XIX., by which the capacities of the beams are ascertained. (See Arts. 479 to 482, 485 to 492, 501, 511, 512, 514, 517, 519, 521, 523, etc.) The values of / were computed by formula (213.) 698. Rolled-Iron Beams for Office Buildings, etc. (XVIII.). Table XVI II. contains the distances from centres, in feet, at which rolled-iron beams should be placed, in the floors of Dwellings, Banks, Office Buildings and Assembly Halls. (See Arts. 500 and 501.) These distances were computed by formula (237.), __ y_ ' I 3 420 699. Rolled-Iron Beams for First-class Stores (XIX.). Table XIX. contains the distances from centres, in feet, at which rolled-iron beams should be placed, in the floors of First-class Stores. (See Arts. 504 and 505.) ROLLED-IRON BEAMS. 499 These distances were computed by formula (239.\ 148-87 y I 3 960 700. Example. As an example to show the uses of Tables XVIII. and XIX. : Let it be required to know the dis- tances from centres at which 9 inch 84 pound Phcenix rolled-iron beams should be placed, on walls with a span or clear bearing of 18 feet, to form a floor to be used in an Office Building or Assembly Room. In Table XVIII., the one suitable for this case, at the intersection of the vertical column for 18 feet, with the horizontal line for the given beam named above, we find 4-51, or 4^ feet, the required distance from centres. For a First-class Store (see Table XIX.), these beams, if of the length stated, should be placed 2-66, or 2 feet and 8 inches from centres. 701. Constants for Use in the Rules (XX.). Constants for use in the rules in previous chapters are to be found in Table XX. These constants, for the 13 American woods named and for mahogany, have been computed from experiments made by the author in 1874 and 1876 expressly for this work (Arts. 704 to 707). For the values of B and F, the lowest and highest of the two series of experiments are taken, and the average given for use in the rules. The constants for the other woods named in the table have been computed for this work from experiments made by Barlow, and recorded in his work on the Strength of Materials. The constant F, for American wrought-iron, was com- 5OO TABLES. CHAP. XXIV. puted by the author from six tests made by Major Anderson on rolled-iron beams at the Trenton Iron Works, and from two tests made at the works of the Phoenix Iron Co. of Philadelphia. The beams upon which these tests were made were from 6 to 15 inches deep and from 12 to 27 feet long. The values of F for the other metals, ajid of B for all the metals, have been computed from tests made by trust- worthy experimenters, such as Hodgkinson, Fairbairn, Kir- kaldy, Major Wade and others. The average of these values may be used in the rules, for good ordinary metal. For any important work, however, constants should be derived from tests expressly made for the work, upon fair specimens of the particular kind of metal proposed to be used. 702. Solid Timber Floors (XXI.). The depths re- quired for beams when placed close to each other, side by side, without spaces between them, may be found in Table XXI. This is not an economical method of construction. More timber is required than in the ordinary plan of narrow, deep beams, set apart. But a solid floor has the important characteristic of resisting the action of fire nearly as long, if not quite, as a floor made with rolled-iron beams and brick arches. A floor of timber as usually made, with spaces between the beams, resists a conflagration but a very short time. The beams laid up like kindling-wood, with spaces between, afford little resistance to the flames ; but, when laid close, they, by the solidity obtained, prevent the passage of the air. The fire, thus retarded and confined to the room in which it originated, may be there extinguished before doing SOLID TIMBER P^LOORS. 5<DI serious damage. Floors built solid should be plastered upon the underside. The plastering lath should be nailed to narrow furring strips, half an inch thick, and the plastering pressed between the lath so as to till the half inch space with mortar. The mortar used should contain a large portion of plaster of Paris, and be finished smooth with it. Owing to the fire-proof quality of this material, it will pro- tect the lath a long time. Thus constructed, a solid floor will possess great endurance in resisting a conflagration. The timbers should be attached to each other by dowels. These will serve, like cross-bridging, to distribute the pressure from a concentrated weight to the contiguous beams. The depths given in Table XXI. were computed by formulas (311.) and (312.). These were reduced from form- ula (130.). which is In this formula U cfl, c and / being taken in feet. If c be taken in inches, then for c we have , and U=-fL Putting rl for 8 (Art. 313) we have In a solid floor the breadth of the beams will equal the distances from centres, or b c (c now being in inches). In the formula these cancel each other ; or ? = Fd'r and 8x 12 <* 3 =~l; (*) 502 TABLES. CHAP. XXIV. For dwellings and halls of assembly, we have taken (Art. 115) f at 90, or 70 for the superincumbent load and 20 for the materials of construction. In a solid floor, however, the weight of the timbers differs too much to permit an average of it to be used as a constant in the formula. The weight of the plastering, furring and floor- plank is constant, and may be taken at 12 pounds. To this add 70 for the superincumbent load, and the sum, 82, .plus the weight of the beam, will equal /, the total load. The weight of the beam will equal the weight of a foot superficial, inch thick, of the timber, multiplied by the depth of the beam ; or, putting y equal to the weight of one foot, inch thick, of the timber, we have its total weight equal to yd ; or, f = %2 + yd. Substituting this value for f in formula (310.), and putting r = 0-03, then we have 19-2 x o-o^F This formula is general for floors of dwellings, office buildings, and halls of assembly. As the symbol for the depth is found on both sides of this equation, the depth for any given length can not be directly obtained by it ; a modi- fication is needed to make the formula practicable. An inspection of the formula shows that the depth will be very nearly in direct proportion to the length. By a simple transformation of the symbols, a formula is obtained which will give the length for any given depth. By an ap- plication of this formula to the two extremes of depth and length for each kind of material, the relative values of d and / may be found. The results for the two extremes in each case will differ but little. An average may be used as a constant for all practical lengths, without appreciable THICKNESS OF SOLID TIMBER FLOORS. 503 error. The values of d have been computed for the four woods named below, and the average value found to be for Georgia pine, d = O-3I4/ Spruce, d =. o-$6$l White pine, d = 0-3897 Hemlock, d o-^gl An average value of y, the weight per foot superficial, inch thick, may be taken as follows : for Georgia pine, y 4 Spruce, y = 2% White pine. y = 2-J Hemlock, y 2 With these values of y and d, formula (311.} becomes practicable, and will give the required depth for any given length of floor beams, of the four woods named, for the solid floors of dwellings, office buildings, and halls of assembly. For the floors of first-class stores, taking 250 pounds as the superincumbent load and 13 pounds as the weight of the plastering, flooring, etc., and putting r = 0-04 we have, in formula (310.), This formula is general for floors of first-class stores. The values of d have been computed for the extremes of lengths, and an average found to be as follows : for Georgia pine, d = -4/ Spruce, d = White pine, d = Hemlock, d = -so6/ 504 TABLES. CHAP. XXIV. With these values of d, and the above values of y, for- mula (312.) will give the depths of solid floors for first-class stores. The depths of solid floors in Table XXL, for dwellings, office-buildings and halls of assembly, were computed by formula (311.), and those for first-class stores by formula (312.) 703. Weights of Building Materials (XXII.). Table XXII. contains the weight per cubic foot of various build- ing materials. 704. Experiments on American Woods (XXIII. to XE, VI.). Tables XXIII. to XLVL, inclusive, contain the results of experiments upon six of our American woods such as are more commonly used as building material. These experiments, as well as those of 1874 (Art. 701), were made upon a testing machine constructed for the author, and after his plan, by the Fairbanks Scale Co. It is a modification of the Fairbanks scale, a system of levers working on knife edges, and arranged with gearing and frame by which a very gradual pressure is brought to bear upon the piece tested, which pressure is sustained by the platform of the scale and thus measured. By an application of clock-work, devised by Mr. R. F. Hatfield, son of the author, the poise upon the scale beam is kept in motion by the pressure upon the platform, and is arrested at the instant of rupture of the piece tested. For the moderate pressures (under 2000 pounds) required, this machine is found to work satisfactorily. 705. Experiments by Transverse Strain (XXIII. to XXXV., Xi.il. and XLIII.). Tables XXIII. to XXXV. EXPERIMENTS ON WOODS. 505 contain tests by Transverse Strain, upon six of the thirteen woods tested by the author for this work. At intervals, as shown, the pressure was removed and the set, if any, measured. It was found that in many instances a decided set had occurred before the increments of deflec- tion had ceased being equal for equal additions of weight. It was thus made plain that some modification of this rule for determining the limit of elasticity must be made. To fix this limit clearly inside of any doubtful line, 25 per cent of the deflection obtained, while the increments of de- flection remained equal for equal additions of weight, was deducted, and the remainder taken as the deflection at the limit of elasticity. With this deflection, the values of the constants e and a in Table XX. were computed (Art. 701). The load upon a beam, determined by the rules with the constants restricted within this limit, will not, it is confident- ly believed, be subject to set ; or if, as is claimed by Professor Hodgkinson, any deflection, however small, will produce a set, that this set will be so slight and of such a nature as not to be injurious, or worthy of consideration. A resume of the results of Tables XXIII. to XXXV. is given in Tables XLII. and XLIII. The values of F and B, given in Table XX., were derived, not alone from the results given in these tables, but also from results of the other experiments made in 1874. (Art. 701.) 706. Experiments by Tensile and Sliding Strains (XXXVI. to XXXIX., XL.IV. and XLV.). Tables XXXVI. and XXXVII. contain tests of the resistance to tensile strain of six of the more common American woods. A rhiimt of the results is given in Table XLI V. 506 TABLES. CHAP. XXIV. Tables XXXVIII. and XXXIX. give tests made to show the resistance to sliding of the fibres in six of the more common American woods. These experiments were made to ascertain the power of the several woods to resist a force tending to separate the fibres by sliding, in the longitudinal direction of the fibres. The rafter of a roof, when stepped into an indent in the tie-beam, exerts a thrust tending to split off the upper part of the end of the tie-beam. A pin through a tenon, when subjected to strain, tends to split out the part of the tenon in front of it. These are instances in which rupture may occur by the sliding of the fibres longi- tudinally, and a knowledge of the power of the various woods to resist it, as shown in these tables, and as condensed in Table XLV., will be useful in apportioning parts subject to this strain. The symbol G, in Table XX., represents in pounds the sliding resistance to rupture per square inch superficial, and is equal to the average of the results of the experiments in Table XLV. A discussion to show the application of these results is omitted as being uncalled for in a work on the Transverse Strain. For its treatment, see " American House Carpenter/' Arts. 301 to 303, where H, the value of each wood, is taken at \ of the resistance to rupture. 707. Experiments by Crushing Strain (XL., XLI. and XL.VI.). Tables XL. and XLI. contain tests of resistance to crushing, in the direction of the fibres, of six of the more common of our American woods. The pieces submitted to this test were from one to two diameters high. A rtsumt of the results is given in Table XLVL TAB L E S. TABLE I. HEMLOCK FLOOR BEAMS ONE INCH THICK, FOR DWELLINGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. DISTANCE FROM CENTRES (in inches). For Beams Thicker than One Inch, see Arts. 693 and 695. LENGTH DEPTH OF BEAM (in inches). BETWEEN BEARINGS (in feet). 6 7 8 9 1O 11 12 13 14 7 n-3 8 7-6 12-0 9 5-3 8-4 12-6 10 3-9 6-1 9-2 I3-I 11 2-9 4-6 6-9 9-8 12 .. 3-6 5-3 7-6 10-4 13 .. 2-8 4-2 5-9 8-2 10-9 14 . 3-3 4-8 6-5 8-7 n-3 15 , 2-7 3-9 5-3 7-i 9-2 u-7 16 3-2 4-4 5-8 7-6 9-6 17 . , .. 2-7 3-6 4-9 6-3 8-0 IO-O 18 .. .. 3-r 4-1 5-3 6-7 8-4 19 .. .. 2-6 3-5 4-5 5-7 7-2 20 3-o 3-9 4.9 6-1 21 3. <j 4. 2 5. o 22 o 2-Q 5.7 J 4-6 23 V O I 3. o *T 4-O 24 * 2-8 ** 3-6 508 TABLE II. WHITE PINE FLOOR BEAMS ONE INCH THICK, FOR DWELLINGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. DISTANCE FROM CENTRES (in inches}. For Beams Thicker than One Inch, see Arts. 693 and 695. LENGTH DEPTH OF BEAM (in inches}. BETWEEN BEARINGS (in feet}. 6 7 8 9 10 11 12 13 14 7 n-7 8 7-8 12-4 9 5-5 8-7 13-0 10 4-0 6-4 9'5 11 3-o 4-8 7-i 10-2 12 43-7 5-5 7-8 10-7 13 2-9 4-3 6-2 8-4 II-2 14 .. 3-5 4-9 6-8 9-0 H. 7 15 2-8 4-0 5-5 7-3 9'5 16 . . 3-3 4-5 6-0 7-8 IO-O 17 2-8 3-8 5-o 6-5 8-3 10-4 IS .. 3 . 2 4-2 5-5 7-0 8-7 19 .. 2-7 3-6 4-7 5-9 7.4 2O 3-1 4-0 5-i 6-4 21 2-7 3-5 4.4 5-5 22 3-0 3-8 4-8 23 a. 4 4-2 24 2-9 3-7 509 TABLE III. SPRUCE FLOOR BEAMS ONE INCH THICK, FOR DWELLINGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. DISTANCE FROM CENTRES (in inches). For Beams Thicker than One Inch, see Arts. 693 and 695. LENGTH BETWEEN BEARINGS (in feet}. DEPTH OF BEAM (in inches}. 6 7 8 9 10 11 12 13 14 7 14-1 8 9'4 9 6-6 10-5 1O 4-8 7-7 n-5 11 3-6 5-8 8-6 12-3 12 2-8 4-4 6-6 9-4 13 3-5 5-2 7'4 TO -2 14 2-8 4-2 6-0 8-2 10-9 15 3-4 4-8 6-6 8-8 n-5 16 2-8 4-0 5-5 7-3 9'4 17 .. .. 3-3 4-6 6-1 7-9 IO-O 18 .. .. .. 2-8 3-8 5-i 6-6 8-4 10-5 19 .. .. .. .. 3-3 4-3 5-6 7-2 9-0 2O ' 2-8 3-7 : 4-8 6-2 7-7 21 3-2 \ 4-2 5-3 6-6 22 .. .. 2-8 3-6 4-6 5-8 23 a.'?. 4.0 c . i 24 ' 2-8 3-6 4-4 510 TABLE IV. GEORGIA PINE FLOOR BEAMS ONE INCH THICK, FOR DWELL- INGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. DISTANCE FROM CENTRES (in inches). For Beams Thicker than One Inch, see Arts. 693 and 695. LENGTH BETWEEN BEARINGS (in feet). DEPTH OF BEAM (in inches). 6 7 8 9 10 11 12 13 14 9 II-2 10 8-2 13-0 11 6-1 9-7 12 4'7 7'5 II-2 13 3-7 5-9 8-8 14 3-o 4-7 7-0 1O-O 15 3-8 5-7 8-2 II-2 16 3-2 4-7 6-7 9-2 17 18 2-6 3'9 3-3 5-6 4-7 7-7 6-5 10-2 8-6 II-2 19 20 2-8 4-0 3-4 5-5 4'7 7-3 6-3 9-5 8-2 10-4 21 3-o 4-1 5-4 7-0 9-0 11-2 22 .. .. .. .. 3-5 4'7 6-1 7-8 9-7 23 .. 3-i 4-1 5-4 6-8 8-5 24 2-7 3-6 4-7 6-0 7-5 511 TABLE V. HEMLOCK FLOOR BEAMS ONE INCH THICK, FOR FIRST-CLASS STORES. DISTANCE FROM CENTRES (in inches). For Beams Thicker than One Inch, see Arts. 693 and 695. LENGTH DEPTH OF BEAM (in inches). BETWEEN BEARINGS (in feet}. 8 9 10 11 12 13 14 15 16 17 18 8 7-8 9 5-5 7-8 1O 4-0 5-7 7-8 11 3-o 4-3 5-9 12 2-3 3-3 4-5 6-0 13 1-8 2-6 3-6 4-7 6-2 14 2-1 2-8 3-8 4-9 6-3 15 i-7 2-3 3-1 4-0 5-i 6-4 16 1-9 2-5 3-3 4-2 5-2 6-4 17 .. 2-1 2-8 3-5 4-4 5-4 6-5 18 1-8 2-3 2-9 3 7 4'5 5-5 6-6 19 2-0 2-5 3-i 3-8 4-7 5-6 6-6 20 2-1 2-7. 3-3 4-0 4-8 5-7 21 1-9 2-3 2-8 3'5 4-1 4*9 22 2-O 2-5 3-o 3'6 4-3 2-2 1 2-6 V2 ! V7 44. I O 2 a 2-8 V3 OK 1 . o 2-5 2-0 ' 26 2-2 2-6 512 TABLE VI. WHITE PINE FLOOR BEAMS ONE INCH THICK, FOR FIRST- CLASS STORES. DISTANCE FROM CENTRES (in inches). For Beams Thicker than One Inch, see Arts. 693 and 695. LENGTH DEPTH OF BEAM (in inches}. BETWEEN BEARINGS (in feet}. 8 9 1O 11 12 13 14 15 16 17 18 8 8-1 9 5-7 8-1 10 4-1 5-9 11 3'i 4-4 6-1 12 2-4 3-4 4-7 6-2 13 1-9 2-7 3-7 4-9 6-4 14 2-2 3-o 3'9 5'i 6-5 15 i-7 2-4 3-2 4-1 5-3 6-6 16 2-0 2-6 3-4 4-3 5-4 6-7 17 2-2 2-8 3-6 4'5 5-6 6-8 18 .. 1-8 2-4 3-i 3-8 4-7 5-7 6-8 < 19 2-0 2-6 3-2 4-0 4-8 5-8 6-9 20 2-2 2-8 3'4 4-1 5-o 5-9 21 1-9 2-4 3-o 3-6 4-3 5-1 22 2- I 2-6 <2 - I -i .7 4. A 23 1-8 2-2 2-7 3-3 3-9 *>4. 2-O 2-d 2-Q 3. /i Q K 2- I 2- ^ vo 26 1-9 2-3 2-7 513 TABLE VII. SPRUCE FLOOR BEAMS ONE INCH THICK, FOR FIRST-CLASS STORES. DISTANCE FROM CENTRES (in inches). For Beams Thicker than One Inch, see Arts. 693 and 695. LENGTH BETWEEN BEARINGS (in feet). DEPTH OF BEAM (in inches). 8 9 10 11 12 13 14 15 16 17 18 9 6-9 1O 5-o 7'i 11 3-8 5-4 7-3 12 2-9 4-1 5-7 7-5 13 2-3 3-2 4-4 5'9 14 1-8 2-6 3-6 4'7 6-2 15 2- I 2-9 3-9 5-0 6-4 16 i-7 2-4 3-2 4-1 5-2 6-5 17 2-0 2-6 3-4 4-4 5-5 6-7 18 .. 2-2 2-9 3-7 4-6 5-7 6-9 19 .. .. 1-9 2-5 3'i 3-9 4-8 5-8 20 2-1 2-7 3-4 4-1 5-0 6-0 21 1-8 2-3 2-9 3-6 4-3 5-2 6-2 22 .. .. .. 2-0 2-5 3-1 3-8 4'5 5'4 23 2-2 2-7 3-3 3-9 4'7 24 I -Q 2-4 2-9 3-5 4" z je 2- I 2-6 3' * 3-6 26 1 1-9 2-3 2-7 3-2, . . | TABLE VIII. GEORGIA PINE FLOOR BEAMS ONE INCH THICK, FOR FIRST- CLASS STORES. DISTANCE FROM CENTRES (in inches). For Beams. Thicker than One Inch, see Arts, 693 and 695. LENGTH BETWEEN BEARINGS (in feet}. DEPTH OF BEAM (in inches) 8 9 10 11 12 13 14 15 16 17 18 11 6-3 12 4'9 7-0 13 3-8 5-5 14 3'i 4.4 6-0 15 2-5 3-6 4'9 6-5 16 2-1 2-9 4-0 5'4 7-0 It i-7 2-4 3-4 4-5 5-8 18 2-1 2-8 3-8 4'9 6-2 19 1-8 2-4 3'2 4-2 5'3 6-6 20 , 2-1 2-7 3'6 4'5 5*7 21 1-8 2'4 3'i 3'9 4'9 6-0 22 2-1 2-7 3'4 4-2 5'2 6.3 23 1-8 2-3 3-0 3'7 4-6 5'5 6-7 24 2-1 2-6 3'3 4-0 4.9 5'9 25 1*8 2-3 2-9 3-6 4'3 5-2 6-2 26 - - 2-1 2-6 3-2 3-8 4-6 5'5 515 TABLE IX. HEMLOCK HEADERS FOR DWELLINGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. THICKNESS OF HEADER (in inches) FOR TAIL BEAMS ONE FOOT LONG. For Tail Beams Longer than One Foot, see Arts. 693 and 696 a LENGTH BETWEEN BEARINGS (in feet). DEPTH OF HEADER (in inches). 6 7 8 9 10 11 12 13 14 5 33 19 12 .08 6 58 33 21 .14 10 07 7 92 53 33 22 16 ii 09 8 1-37 79 50 33 23 17 13 IO 08 9 i-95 1-13 71 .48 33 24 18 14 II 10 2-68 i-55 98 65 46 33 25 19 15 11 ^ . ''. 2-06 1-30 87 61 45 33 26 20 12 2-68 1-69 1-13 79 58 43 33 26 13 .. 2-14 i-44 I-OI 73 55 43 33 14 ' * . $ 2.68 1-79 1-26 92 .69 53 .42 15 ... 2-21 i-55 I-I3 85 .65 5i 16 . . . . . . 2-68 1-88 i-37 1-03 79 62 17 .. . 3.21 2-26 1-64 1-24 95 75 18 . 2-68 i-95 1-47 I-I3 -89 19 .. 3-15 2-80 1-72 i-33 1-04 2O 3-67 2-68 2-OI 1-55 1-22 516 TABLE X. WHITE PINE HEADERS FOR DWELLINGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. THICKNESS OF HEADER (in inches) FOR TAIL BEAMS ONE FOOT LONG. For Tail Beams Longer than One Foot, see Arts. 633 and 696. LENGTH BETWEEN BEARINGS (in feet}. DEPTH OF HEADER (in inches). 6 7 8 9 1O 11 12 13 14 5 32 .19 12 .08 6 56 32 20 14 IO .07 7 .89 51 32 22 15 IT 08 8 1.32 77 48 32 23 16 12 IO 07 9 1-88 1-09 .69 .46 32 24 18 14 ii 1O 2-59 1.50 94 63 44 32 .24 19 .16 11 1.99 1-25 -8 4 59 43 32 25 20 12 2-59 1-63 I.Og 77 .56 .42 32 25 13 .. 2-07 1-39 97 7i 53 41 32 14 .. 2-59 i-73 1-22 .89 67 51 40 15 3-i8 2-13 1-50 1-09 82 63 50 16 . . 2-59 1.82 1-32 99 77 .60 17 3-io 2-18 i-59 1-19 92 72 18 2-59 1-88 1.42 1-09 86 19 3-04 2-22 1-67 1-28 I-OI 20 3-55 2-59 1.94 1.50 1.18 517 TABLE XL SPRUCE HEADERS FOR DWELLINGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. THICKNESS 'OF HEADER (in inches) FOR TAIL BEAMS ONE FOOT LONG. For Tail Beams Longer than One Foot, see Arts. 693 and 696. LENGTH BETWEEN BEARINGS (in feet). DEPTH OF HEADER (in inches). 6 7 8 9 10 11 12 13 14 5 .27 15 .10 06 6 .46 27 17 ii .08 7 73 42 27 .18 13 .09 8 1. 10 .63 .40 .27 .19 .14 IO 08 9 1-56 .90 57 33 27 .19 15 II 09 1O 2-14 1.24 .78 52 37 27 20 15 12 11 1-65 1-04 7 ' -49 36 27 21 l6 12 2-14 i-35 .90 .63 .46 35 27 21 13 2-72 1.72 i-i5 81 59 44 34 27 14 2-14 i-43 I-OI 73 55 .42 33 15 2.63 i-77 1.24 .90 68 52 .41 16 . . 3-20 2-14 1.50 I-IO .82 63 50 17 2-57 i. 80 1.32 99 .76 60 18 .. 3-05 2-14 1-56 r.zjf .90 71 19 2.52 1.84 1.38 i. 06 .84 20 2-94 2.14 1.61 1-24 97 518 TABLE XII. GEORGIA PINE HEADERS FOR DWELLINGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. THICKNESS OF HEADER (in inches) FOR TAIL BEAMS ONE FOOT LONG. For Tail Beams Longer than One Foot, see Arts. 693 and 696. LENGTH BETWEEN BEARINGS (in feet). DEPTH OF HEADER (in inches). 6 7 8 9 1O 11 12 13 14 5 16 .09 6 27 .16 10 07 7 44 25 .16 II .07 8 .65 38 24 16 II 08 9 93 54 34 23 .16 12 09 1O 1-27 73 .46 3i .22 16 12 -09 11 1-69 .98 62 .41 .29 21 16 12 10 12 2- 2O 1-27 .80 54 38 27 21 16 T2 13 1.62 ! 02 .68 .48 35 26 .20 16 14 2-02 1-27 85 .60 44 33 25 20 15 2-48 1-56 1-05 73 54 40 31 24 16 . . 1.90 1-27 .89 65 49 3 8 30 17 2-28 1-52 1.07 .78 59 45 35 18 2-70 i. 81 1.27 93 70 54 42 19 2-13 1.49 1-09 82 63 .50 2O 2.48 1.74 1-27 95 73 .58 519 TABLE XIII. HEMLOCK HEADERS FOR FIRST-CLASS STORES. THICKNESS OF HEADER (in inches) FOR TAIL BEAMS ONE FOOT LONG. For Tail Beams Longer than One foot, see Arts. 693 and 696. LENGTH BETWEEN BEARINGS (in feet). DEPTH OF HEADER (in inches]. 8 9 10 11 12 13 14 15 16 17 18 5 .28 19 13 IO .07 6 .48 32 23 17 .12 .10 07 7 77 51 .36 26 2O .15 12 10 08 8 1.14 77 54 39 .29 23 .18 .14 12 .10 08 9 1.63 1-09 77 .56 .42 32 25 20 17 .14 II 1O 2-24 1.50 1.05 77 58 44 35 .28 ( 23 .19 .16 11 2.98 1.99 1-40 i -02 77 59 .46 37 30 25 21 12 2-59 1-82 i-33 LOO 77 .60 .48 39 32 -27 13 3-29 2.31 1-69 1.27 97 77 .61 .50 4* 34 14 ... , .. . 2-89 2-10 1-58 1-22 .96 .77 ' -62 5i 43 15 3-55 2-59 i-95 1.50 LI8 94 -77 63 53 16 . . . . . . 3-14 2.36 L82 i-43 i-i4 -93 77 64 17 3-77 2.83 2.1-8 1-72 i-37 1-12 .92 77 18 .. ... 3-36 2-59 2-04 1-63 i-33 1.09 .91 19 20 3-95 4-61 3-5 3-55 2-39 2-79 1-92 1-56 2-24 1.82 1.28 1-50 1.07 1-25 520 TABLE XIV. WHITE PINE HEADERS FOR FIRST-CLASS STORES. THICKNESS OF HEADER (in inches) FOR TAIL BEAMS ONE FOOT LONG. For Tail Beams Longer than One Foot, see Arts. 693 and 696. LENGTH BETWEEN BEARINGS (in feet}. DEPTH OF HEADER (in inches'). 8 9 10 11 12 13 14 15 16 17 18 5 .27 .18 13 .09 .07 6 47 31 .22 .16 .12 .09 07 7 74 50 35 25 19 15 12 .09 .07 8 i.ii 74 .52 .38 28 .22 17 14 .11 09 .08 9 i-57 1-05 -74 54 .41 31 25 .20 .16 13 ii 10 2.16 1-46 1.02 74 .56 43 34 .27 22 .18 15 11 2-87 i-93 1-35 99 74 57 45 .36 .29 2*4 20 12 2.50 1.76 1.28 .96 74 .58 47 - 3 8 31 26 13 3-i8 2-23 1.63 1.22 94 74 59 . 4 8 40 33 14 2.79 2-03 i-53 1.18 .92 74 .60 50 .41 15 3-43 2.50 1.88 i-45 1-14 .91 . -74 .61 5i 16 . . 3-03 2-28 1.76 1-38 i-ii 90 74 62 17 3-64 2-73 2. II i-66 i-33 I- 08 89 74 18 4-32 3-25 2-50 i-97 1-57 1-28 1-05 88 19 3-82 2-94 2.31 1-85 1.50 1.24 1-03 20 - 4-45 3-43 2-70 2-16 1.76 1-45 1. 21 1 521 TABLE XV. SPRUCE HEADERS FOR FIRST-CLASS STORES. THICKNESS OF HEADER (in inches) FOR TAIL BEAMS ONE FOOT LONG. for Tail Beams Longer than One Foot, see Arts. 693 and 696. LENGTH BETWEEN BEARINGS (in feet). DEPTH OF HEADER (in inches). 8 10 11 13 13 14 15 16 17 18 5 .22 15 IO .08 6 39 .26 18 13 10 .08 7 61 .41 29 21 .16 .12 IO .08 8 92 .61 43 31 .24 .18 14 II 09 08 9 1-30 .87 .61 45 34- 26 20 .16 13 II 09 10 1.79 1-20 .84 .61 .46 35. .28 22 .18 15 12 11 2-38 I- 60 I-I2 82 61 47 37 30 24 2O 17 19 3-09 2-07 i-45 1-06 80 61 .48 39 31 .26 .22 13 2.6 3 1-85 i-35 1. 01 .78 .61 49 .40 33 27 14 3-29 2.31 i-68 1-26 97 77 61 50 .41 34 15 2-84 2.07 1.56 1-20 94 75 61 5i 42 16 3-45 2-51 1.89 i-45 1-14 92 74 61 5i 17 3-02 2-27 1.74 i-37 I-IO .89 74 61 18 .. .. .. 3-58 2-69 2-07 1-63 1.30 I- 06 87 -73 10 4-21 3-i6 2.44 1-92 i-53 1-25 1.03 86 20 5.60 2-84 2-23 i-79 i-45 1-20 I- 00 522 TABLE XVI. GEORGIA PINE HEADERS FOR FIRST-CLASS STORES. THICKNESS OF HEADER (in inches) FOR TAIL BEAMS ONE FOOT LONG. For Tail Beams Longer than One Foot, see Arts. 693 and 696. LENGTH BETWEEN BEARINGS (in feet}. DEPTH OF HEADER (in inches). 9 10 n 13 13 14 15 16 17 18 5 .13 -eg .06 6 23 : -15 II .08 7 36 ! -24 .46 .12 09 .07 8 54 36 25 I 9 14 II .08 9 77 ! -52 36 26 .20 15 .12 .IO 08 10 i. 06 71 50 36 27 21 17 13 II 09 11 1-41 95 .67 . 4 8 36 .28 22 .18 .T 4 . 12 10 12 1.83 1-23 86 63 47 36 29 23 .I 9 15 13 13 2-33 1-56 I-IO .80 60 .46 .36 .29 24 .19 .16 14: 2-91 i-95 1-37 I-OO 75 58 45 .36 30 .24 20 15 2-40 1.69 1-23 .92 71 56 45 36 30 25 16 . . 2.91 2.05 1-49 I- 12 86 68 54 44 36 30 17 3-49 2-45 1.79 i-34 1-03 81 .65 53 44 36 18 .. 2-91 2-12 i-59 1-23 97 77 63 52 43 19 .. 3-43 2-50 1.88 1.44 1.14 .91 74 61 5i 30 2-gi 2-19 1-69 1-33 i. 06 .86 7i 59 1 523 TABLE XVII. ELEMENTS OF ROLLED-IRON BEAMS. See Art. 697. NAME. i Q 1! ** WEIGHT PER YARD. * = BREADTH. AVERAGE THICKNESS OF FLANGE. i THICKNESS OF WEB. 6, d t X ~oT ii i s | Pittsburgh . 3 21 2.316 359 .191 2.125 2.281 3.109 Pittsburgh . 3 27 2.516 -359 39 1 2. 125 2.281 3-559 Phoenix . . . 4 18 2. .278 .2 T.8 3.444 ; 4.539 Trenton . . 4 18 2. .29 .187 I.8I3 3.42 i 4.623 Pottsville. . 4 18 2.125 .271 .187 1.938 3.458 ; 4.655 Paterson . . 4 18 2.25 .281 .156 2.094 3.438 4-909 Pittsburgh . 4 24 2.481 .328 .231 2.25 3.344 6.221 Pittsburgh . 4 30 2-631 .328 .381 2.25 3.344 i 7-021 Pottsville. . 4 30 2.25 5 25 2. 3- 7-5 Buffalo 4 30 2,75 4 25 2-5 3-2 7.840 Paterson . . 4 30 2-75 4 2 5 2-5 3-2 7.840 Phoenix . . . 4 30 2-75 4 25 2-5 3 2 7.840 Trenton. . . 4 30 2-75 -25 2-5 3-2 7.840 Paterson . . 4 37 3- -456 .312 2.688 3.088 9.404 Trenton. . . 4 37 3- 456 .312 2.688 3 .o8S 9.404 Buffalo 5 3^ 2-75 35 25 2-5 4-3 12.082 Paterson . . 5 30 2-75 -35 25 2-5 4-3 12.082 Phoenix . . . 5 30 2-75 35 -25 2-5 4-3 12.082 Trenton.. . 5 30 2-75 35 2-5 4-3 12.082 Pottsville. . 5 30 3.062 3ii 25 2.812 4-378 12.232 Pittsburgh. 5 30 2.725 375 225 25 4-25 12.393 Pittsburgh . 5 39 2.905 375 4^5 2-5 4-25 14.268 Phoenix . . . 5 36 3- 389 -3 2.7 4.222 I4-3I7 Paterson . . 5 40 3- .438 333 2. 667 4.125 15-650 Trenton.. . 5 40 3- 454 .312 2.688 4.092 15-902 Pottsville. . 5 40 3-125 434 .312 2.813 4.132 16.015 Phoenix . . . 6 40 2 75 5 25 2-5 5- 23-458 Buffalo 6 40 3- 454 25 2.75 5.091 23-761 Paterson . . 6 40 3- 454 -25 2-75 5-9i ' 23.761 Trenton. . . 6 40 3- 454 25 2-75 5-9* 23.761 524 TABLE XVII. (Continued^ ELEMENTS OF ROLLED-IRON BEAMS. See Art. 697. i s . E 11 t/2 g X HI NAME. S 1 * 11 ^ rj W ^ rv' w fa b t d, 1 s ! w PQ < ^o H Pottsville.. 6 40 3-375 4 25 3.125 5-2 24-T33 Pittsburgh. 6 4i 3-237 437 237 3- 5-125 24.613 Pittsburgh. 6 54" 3.462 437 .462 3- 5-125 28.663 Buffalo 6 50 3-25 532 312 2.938 4-935 29.074 Pottsville. . 6 50 3-437 .500 .312 3.125 5.080 29.314 Phoenix . . 6 50 3-5 .492 31 3-19 5.016 29-451 Paterson . . 6 SO' 3-5 5 3 3-2 5- 29.667 Trenton. . . 6 50 3-5 5 3 3-2 5- 29.667 Phoenix . . . 7 55 3-5 484 35 3.15 6.032 42.430 Pottsville. . 7 55 -510 .312 3-25 5.980 43-897 Trenton . . . 7 55 3-75 493 3 3-45 6.014 44.652 Pittsburgh. 7 54 3.604 562 .229 3-375 5-875 45-983 Buffalo . 7 60 3-5 54 375 3-125 5-92 46.012 Paterson . . 7 60 3-5 54 375 3-125 5.92 46.012 Phoenix . . . 7 69 3-687 .476 56 3.127 6.048 47-739 Pottsville.. 7 65 3.625 596 375 3-25 5-8oS 50.553 Paterson . . 6 90 5- .667 5 4-5 4-667 51.881 Trenton. . . 6 90 5- .667 4-5 4.667 51.881 Pittsburgh. 7 75 3-94 .562 .529 3-375 5-875 54.558 Buffalo 8 65 3-5 56 375 3-125 6.880 64.526 Paterson . . 6 1 20 5-5 789 75 4-75 4.422 64.773 Phoenix . . . 8 65 4- .478 38 3-62 7.044 65.232 Trenton. . . 6 120 5-25 .892 .625 4-625 4.216 65.618 Pottsville. 8 65 4- 543 .312 3.688 6.914 69 . 089 Paterson . . 8 6 5 4- -554 3 3-7 6.892 69.729 Trenton. . . 8 65 4- = 554 3 3-7 6.892 69.729 Pittsburgh. 8 66 3.806 593 306 3-5 6.813 70.153 Pottsville.. 8 80 4.187 542 5 3-687 6.916 77.007 Phoenix . . . 8 81 4.125 556 3-6i5 6.888 77-552 Buffalo 9 70 3-5 .500 437 3-063 8.000 81-937 525 TABLE XVII. (Continued.) ELEMENTS OF ROLLED-IRON BEAMS. See Art, 697. NAME. d = DEPTH. g el s> M *s= BREADTH. AVERAGE THICKNESS OF FLANGE. THICKNESS OF WEB. *, 4 V <f ii i s l Trenton . . . 8 80 4-5 .606 -375 4.125 6.788 84.485 Paterson . . 8 80 4-5 .610 -37 4-13 6.780 84-735 Pottsville.. o. 70 4.125 483 375 3-75 8.034 88-545 Pittsburgh. 8 105 4-293 -593 -793 3-5 6.813 90.932 Phoenix . . . 9 70 3-5 .660 3i 3-19 7.680 92.207 Paterson . . 9 70 3-5 .672 3 3-2 7.656 92.958 Trenton. . . 9 70 3-5 .672 3 3-2 7-656 92.958 Pittsburgh. 9 70-V 4.012 -625 .262 3-75 7-75 98.265 Pottsville. . 9 90 4-5 .501 562 3-938 7.998 105.480 Phoenix . . . 9 84 4- .667 4 3-6 7.667 107.793 Buffalo 9 90 4- .643 5 3-5 7-714 109.117 Paterson . . 9 85 4- .697 384 3.616 7.605 110.461 Trenton. . . 9 85 4- .707 375 3-625 7-586 in . 124 Pittsburgh. 9 99 4-329 .625 579 3-75 7-75 117-523 Pottsville.. ioi 90 4-25 .578 437 3-8i3 9-344 150.763 Pittsburgh. 10 90 4-325 .718 325 4- 8-563 151-123 Buffalo ioj 90 4-437 55i 437 9-397 151-436 Trenton.. . 9 125 4-5 937 -57 3-93 7-125 i54-9 T 7 Pittsburgh. 9 135 4-931 .812 744 4.187 7-375 159-597 Pittsburgh. 10$ 94* 4-534 .625 .409 4.125 9-25 165-327 Pottsville.. 10} 105 4-562 575 .562 4- 9-350 167.624 Trenton.. . 10} 90 4-5 .683 .312 4.188 9.434 168.154 Pittsburgh . 9 150 5-098 .812 .911 4.187 7-375 169.742 Buffalo. . . . 10* 105 4-5 656 -5 4- 9.187 175-645 Phoenix . . . io| 105 4-5 .724 44 4.06 9.052 183.164 Pittsburgh. 10 135 4-775 .718 775 4- 8-563 188.623 Phoenix . . . 9 150 5-375 1.005 .6 4-775 6.99 190.630 Paterson . . io.V 105 4-5 795 375 4.125 8.909 191.040 Trenton. . . roi 105 4-5 795 375 4-125 8.909 191.040 Phoenix . . . H>* 135 4-875 .614 .81 4-065 9.272 200.263 525 a TABLE 1KM\\. (Continued.) ELEMENTS OF ROLLED-IRON BEAMS. See Art. 697. NAME. | W Q 1! > WEIGHT PER YARD. b = . BREADTH. AVERAGE THICKNESS OF FLANGE. i THICKNESS OF WEB. ft < N V nf s l <w Pittsburgh. 10 V i35 4.920 .625 795 4.125 9 25 202.564 Paterson . . 10:} 135 5- 945 47 4-53 8.609 241.478 Trenton. . . io| 135 5- 945 47 4-53 8.609 241.478 Pottsville. . 12 -25 4.562 .800 5 4.062 10.400 276.162 Pittsburgh. 12 126 4-638 .781 5i3 4.125 10.438 276.946 Phoenix . . . 12 125 4-75 777 49 4.26 10.446 279.351 Buffalo. . . . I2 4 L 125 4-5 797 -5 4- 10.656 286.019 Paterson . . 12\ 125 4-79 .768 .48 4-3i 10.714 292.050 Trenton. . . TO 1 I2 4 - 125 4-79 .780 47 4-32 10.690 293.994 Pittsburgh. 12 1 80 5.088 .781 9 6 3 4 125 10.438 341.746 Pottsville.. 12 170 5- i.oS6 .625 4-375 9.828 373-907 Phoenix . . 12 170 5-5 I.OTO 59 4.9: 9.980 385.284 Paterson . . I2i 170 5-5 .985 .6 49 10.28 398.936 Trenton. . . 12 fs 170 5 5 .981 .6 49 10.351 402.538 Buffalo I2f 1 80 5-375 1.089 .625 4-75 10.072 418.945 Buffalo 15 150 4-875 .761 562 4-3*3 13-477 491.307 Paterson . . i5-, 3 6- 150 5- 731 56 4.440 13-725 502.883 Phoenix . . . 15 ' 150 4-75 .882 -5 4-25 13-235 514.870 Pottsville. . 15 150 5.062 .822 5 4.562 13-356 517.948 Trenton.. . ISA 150 5- .822 -5 4-5 I3-542 528.223 Pittsburgh . 15 150 5 030 .875 .468 4.562 13-25 530.343 Pittsburgh. 15 195 5-330 .875 .768 4.562 13-25 614.718 Pittsburgh. 15 2OI 5-545 .031 .670 4-875 12 938 679.710 Phoenix . . . 15 2OO 5-375 .085 -65 4-725 12.830 680. 146 Buffalo.... 15 20O 5-375 .118 .625 4-75 12.763 688.775 ! Paterson . . 15* 200 5-5 .048 -65 4.85 13.028 692.166 Pottsville. . 15 2OO 5.687 .04x5 .625 5.062 1 2 . 902 693.503 Trenton. . . i5i 2OO 5-75 .060 .6 5-15 I3-004 714.205 Pittsburgh. 15 240 5-805 .031 930 4-875 12.938 752.835 ! 525 b TABLE XVIII. ROLLED-IRON BEAMS IN DWELLINGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. DISTANCES FROM CENTRES (in feet}. See Arts. 694, 698 and 7OO. NAME. DEPTH. WEIGHT PER YARD. LENGTH (in feet) BETWEEN BEARINGS. 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Pittsburgh.. 3 21 3.62 2.26 L50 Pittsburgh.. 3 27 -.14 2.58 1.71 .18 Phoenix 4 18 5.32 3.33 2.22 .54 .11 Trenton.. . . 4 18 i-4 2 1.39 2.26 57 T 4 Pottsville... 4 18 5.45 2.28 .59 .14 Paterson . . . 4 18 3.61 2.4O .67 .21 Pittsburgh.. 4 2 4 * [57 3.4 .12 .53 1.13 Pittsburgh.. Pottsville... 4 4 30 3 ' 39 1.27 Buffalo 4 30 5^6 3.83 .67 .93 L43 Paterson . . . 4 3 5.76 3.83 .67 93 !-43 Phoenix.... 4 3 5.76 383 .67 93 L43 Trenton... . 4 3 5.76 3.83 .67 93 Paterson . . . 4 37 4.60 3.20 3 1 1.71 .30 Trenton.... 4 37 4.60 3.20 3 Buffalo 5 3 5.95 4.16 3.01 2.24 7 1 33 Paterson . . . 5 3 5-95 4.16 3.01 2.24 7 1 33 Phoenix.... 5 3 5-95 4.16 3.01 2.24 7 l .33 Trenton 5 3 5-95 4.16 3.01 2.2^ 71 33 Pottsville.. . 5 3 .. .. 6. 02 4.21 3.05 2.27 73 35 Pittsburgh. Pittsburgh. 5 5 30 39 6.10 7.02 4.26 4.90 3.9 3.55 i-.i: .76 .01 37 56 1.23 Phoenix . . 5 36 7-5 4.92 3.57 2.66 .03 .58 1.24 Paterson . . 5 40 \\ 5.38 3.9 2.90 .21 .721.36 Trenton . . . 5 40 . 5.47 3.96 2.95 .25 75 1.38 Pottsville. . 5 40 5.51 3-99 2.97 2.27 .76 L39 Phoenix... Buffalo . . . 6 6 40 40 '* '* 8. ii 8.22 5.89 5-97 4.40 4.46" 3-37 .63 2.09 .662.11 1.68 1.70 ;-j Paterson . . 6 40 8.22 5-97 4.46 3.4' .662.11 i.7 c r^s Trenton . . . 6 40 8.22 5-97 4.46 3.41 .662.11 1.70 !.} 1 526 TABLE XVIIL (Continued.} ROLLED-IRON BEAMS IN DWELLINGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. DISTANCES FROM CENTRES (in feet). See Arts. 694, 698 and 7OO. NAME. DEPTH. WEIGHT PER | YARD. LENGTH (in feet) BETWEEN BEARINGS. 6 7 8 9 1O 11 12 13 14 15 16 17 18 19 20 21 22 23 Pottsville... Pittsburgh.. 6 6 40 5.ob 6.18 4.53 -1.62 3.47 3.54 2.71 2.76 .15 .IQ 73 .76 .44 i Pittsburgh.. 6 S4 7 18 4.10 3.20 54 .04 66 1 .36 Buffalo .... 6 50 .. 7.30 5.45 4.17 3.26 S8 .08 .69 1.39 Pottsville... 6 5 i -6 5 5 4.21 7 > .61 .10 .71 1 .40 Phoenix.. . . 6 5 o (r, Paterson . . . 6 so 7-45 S.S7 4 26 3.33 2.64 2.12 73 1.42 Trenton 6 50 7.45 "57 4.26 2 .64 2.12 .73 1.42 Phoenix 7 55 8.00 6.n t-79 3 .8i! 3 .c8 Si 2.07 .72 L45 Pottsville .. 7 55 8.28 6.35 4-97 3-95 3-19 .60 2.15 79 i-5C Trenton Pittsburgh.. Buffalo 7 7 7 55 54 60 .. .. 8.43 8.68 6.46 6.6f 6.6=; 5.C-5 5.21 3.20 4.0213.24 4- I 53.35 4.1313.33 .65 73 .72 2.19 2. 2O 2.25 .82 .88 87 1.53 1.58 .29 .34 .32 Paterson. . . 7 60 6.6= 5.20 4.!3 3-33 .72 2.25 .87 1 .57 .32 Phoenix.... 7 69 8.98 6.88 5.38 4.27 3-44 .81 2.31 92 i .61 36 Pottsville... 7 65 7-3 1 5.7* 4.54 3.67 2.09 2.47 .06 1.72 >4 f i Paterson . . . Trenton Pittsburgh.. Buffalo 6 6 7 8 90 90 65 ' 7-44 7.87 9-37 -.8l 5.8i 5.i6 7-34 4'.6\ 4.8 9 5.84 3.71 3-7 1 3-94 4.72 3-02 3.02 3.22 3.86 2.48 2.48 1% .05 .05 .21 .67 1.71 L7 1 1.85 2.24 44 44 .56 .90 .32 .62 1.39 Paterson . . . Phoenix Trenton Pottsville... Paterson. . . 6 8 6 8 8 120 65 120 ;; - 9.28 9-47 9-4C 10.04 10.14 7.23 7.42 7-33 7.87 7-94 5-73 5.9 1 S.oi 6.27 6.33 4.6i 4.77 4.67 5.07 3.75 3^0 4.15 4.19 3.08 3.23 3.12 3-43 3.46 .55 .'87 2.12 2.27 2.15 2.41 2.44 .7* .92 .05 .07 501.27 .64 .41 .52 .29 .75, -So 77[ .5* 1.31 Trenton Pittsburgh.. Pottsville... Phoenix Buffalo 8 8 8 8 9 65 66 80 81 70 .. 10.14 IO.2O 7-94 7-99 3 75 6.33 6.36 6.97 7.02 7.45 5." 5-'4 5.63 5.67 6.03 4.19 4-21 4.61 4.64 4.94 3.46 3.48 3.8i 3.83 4-09 .89 9 1 3-i8 3.20 3.4 2 2.44 2:67 2.69 .07 .08 .26 .28 45 77 .77 .09 'S !so 1.31 1.42 1.43 i.55 . 8. 8 1 9-35 527 TABLE XVIII. (Continued^ ROLLED-IRON BEAMS IN DWELLINGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. DISTANCES FROM CENTRES (in feet). See Arts. 694, 698 and 7OO. 9* ^ ^R!?!5 00 e* O O * M C\oo is.0 t^ t^ !>. t^. t^vO t**.co OO OO ON ON O sir .0-^ ^" a - a " & ;- g - B " ft : ma e* "5 ' S ro 1 ? X& Kvg vS S c? c?^ S -< S 5? 5 5 5-vo 1 JH ^ Sco cS" S vM 8 2 $ ? IS Evo K R oSoctwON^ K>^vSc? ooco^^S^ ^^cT-cT-f^ z " I w ^ S^vo^vo ^ C ri. C ^. V 8 "o O M w N C\ ON ON ON O IN S CO 8 5 S vo vo r- t-xOO W HHMHM MHHP.CN d 0) IN (N <N c. N <Nroro rororororo ro ro ro ro ro i g 00 OO ON ON O O o M\O iO -^TJ-'< J -U-)'^- ^'i-^-mr^ t^co t^ONw conroro-'^- ^^NC?^ r^r^Svo^^ ON ^vo O "^ invo t^r^ro t^NCsONON r^ONwi-i^ t>.oo OOOON ONONONOro ro-^roinfN. ooooOO" o * ? ir, irivo vD t>. OO OO O\ w (N "grJ-.rotnS ^vSv?^^ g^Ho'SS & fLorTorTI) . * a C& oTN S 5 J?^ 8* c? ? r.cS oo"oNONtH > ro ^.WVOMCVI co g-vo.covo ON ro in in ro s - N N rororo ro ro ro ro ro rororon-^ in m vn in *> . inxo >nvo vo vo vo vo vo ^ a . 2>Jn^Kco SNci^S-tn \rt\O O ON ro ^ ** TJ-VO O OHO^t^ O^ON^*-"-^- i-< fc. O H ro t^ cvi vo vo rOv.o O N N * -^-VO VO vo ON N ** 5S5?oo'0 vOvoS:^^ ^fn^-OT^ ^^vS-0 C H CO g.OMni-irx CMNiniOM ino c^ O M cvi <N rovo O M co w vo w ^ mvo vo *"* HI 2" N Jnvo oo oo oo "8 ^. ON ro in ^100 M ro _ -i- PH vovovoovo vovo r^^^ OOCOOOCO H M M M M ^ NO vo"^ 5- ^ ^ o^ Jnoo C> 0" ^vo" 1 ' - 1 t^ t^OO 00 00 00 00 00 ON ON ON O O N Win^w-3- roro-^f ONONOOO O O ' QHVA H3d J.HDIHAY cooot>.ot^ t-.t^S. o-oo are gg, 8.&SBS ?ao'o' gijss HI.d3Q 00 OO O CO ON ON ON ON ON ON ON ON ON ON*O O 1) ON ON'O "o "o ON"O "o O ON'O O O H jj^tfj J^Jj iliii lg|ii urtO.-_d rt^-.-Oj^ pH QH CU PH PH Q-t ^ P-i Q-* Pi |ll|l I1H1 is||| l|||| 527 a TABLE lVl\\. (Continued.) ROLLED-IRON BEAMS IN DWELLINGS, OFFICE BUILDINGS, AND HALLS OF ASSEMBLY. DISTANCES FROM CENTRES (in feet). See Arts. 694, 698 and 7OO. LENGTH (infect) BETWEEN BEARINGS. O M O O ON O CN ONNO CO -* HCNt^OlO COO CONO f^ ONNO CO Tf CO COCO ON O CO CJ IN LDNO NO NO t~- t^ M LOND t^ CO ON C^ ON O O M w O-.NO NO r^ t-. l-^ ON CO M 0) 0) CN] N <N) CN] N <N co rororororo <*- -*t- 10 10 LO o LONO NO NO NO NO NO t^. 30 (M ff^^&Sx ss'ss?; s^ffsr? ^^^^ S5-5K ffi" < s OMMCOO>. CN1MONHO * ONNO W~0 - NO IN "1 ON M O CO * to V) M CO ON CNOOOOMtNl VOiO-*-t^CO (M ONCO 1- LO r^ Tf H NO M TJ-NO ON ON M CN O- OMO NO M M r^ i^, t^.co ON OMO O >i co TT-NO I^ONHMCO co-^-coroio in loco co .Nrorocoro rororoco^ ^.01010,0 NONO^^^ t-OOOONON ON.ONONO e5 ONNO NOCNCN NCOV-iot-, t-,CO w n * NO CO O w N CO VONO NO t^ t^OO H . 3 CO " Tl- Tl- TC T.-^J-LOLOIO VONONO^^ COCOOXONON ONO M 0^?:^?" "Lo^corJrT* ^^^0^ ^2>" ^ s ro O NO <N| ro ON LO O ^NO LO CN LO ^*- >-< H i '*-* co co roior-i>.t^ 1000 MMNO <#>'..< ^-LOIONONO NONONONOt- OOOOONONON M FH NOrorOMCO OOOlOOCO ONMCOOOM <N ro ro co co Tt- LO t^oo O-OOINIONOM .... ^"o^o "^vo^o'o^So NO'^^OO'OO- oo'oo*a,ovO H M NO NO g. ON ^^^ t^OO 00 ON C O O O GO FH ^t* rh -^-oo co O O M FH o o FH FH H M FH aav A H3J -LHDI3M. co co co o?lN c? c? c? -c?co S..R.S.oo toS^OLOLO LoSoSS 8885- fff . . B S g ss ?fS -j-fsp-f jo.,.,.,.,, S-jf., H "^cj^-ij^ WtniJ 2 -^ 4j5ajCtk2 if^QjW^lc* ^ r ^*^cutlH QjiQCc/) 527 b TABLE XIX. ROLLED-IRON BEAMS IN FIRST-CLASS STORES. DISTANCES FROM CENTRES (in feet). See Arts. 694, 697, 699 and 7OO. NAME. X K n WEIGHT PER YARD. LENGTH (infect) BETWEEN BEARINGS. 5 6 7 8 9 10 11 1| 13 14 15 16 17 18 19 2O Pittsburgh 3 21 3.68 2.12 1.33 Phoenix SB e -38 Trenton Pottsville 4 18 18 5.48 3.17 [.99 1.32 18 e 82 Pittsburgh 4 24 4,26 2.67 1.78 1.24 Pittsburgh Pottsville 4 4 3 3 4.81 3.01 3.22 2.OI 2.15 1.40 1.50 Buffalo 4 30 5-37 3-37 2.25 1-57 1.14 Paterson 4 30 5.37 3.37 2.2S 1-57 .14 Phoenix 4 30 5.37 3.37 2.2.S 1-57 .14 4 4 30 37 " 5-37 3-37 4.04 2.2 5 2.6q i.57 1.88 .14 .36 Paterson Trenton " fin i 88 36 Buffalo 5 S 30 30 5.21 5.21 3.48 3.48 2-43 2.43 .77 77 1.32 1.32 Paterson Phoenix T 4.8 Trenton Pottsville 5 5 30 30 ;; 5.21 5.27 3.48 3.52 2.43 2-47 77 79 1.32 1-34 Pittsburgh Pittsburgh 5 S 30 39 5-34 3.57 4.11 2.50 .81 .08 !-35 .IQ Phoenix n fi OQ rv\ z: Paterson 6* 17 * 6*86 A Cfi Pottsville 40 6.QI .61 3.23 2 .34 1.75 Phoenix o 58 Q8 1.55 i 23 Buffalo 6 86 4 81 2 6l Paterson Trenton 6 6 4 40 .86 .86 4.81 4.8i 3-49 3-49 2.61 2.6l .OO .OO i.57 1-57 1.22 1.22 528 TABLE XIX. (Continued^ ROLLED-IRON BEAMS IN FIRST-CLASS STORES. DISTANCES FROM CENTRES (in feet}. See Arts. 694, 697, 699 and 7OO. NAME. DEPTH. WEIGHT PER 1 YARD. LENGTH (in feet} BETWEEN BEARINGS. 5 6 7 8 9 10 11 12 13 11 15 16 17 18 19 *o Pottsville Pittsburgh 6 6 6 6 40 54 ! 8R 3 55 -> 66 .27 .23 2 4 .25 .26 .26 .81 .88 .91 97 97 97 3 .16 .19 .19 [78 .73 .81 .77 .98 3.01 3.01 3.02 3.3 1 3-33 3-54 1.48 1.54 i!6i i!6i 1.66 1-77 i.79 1.79 2.23 2.30 .26 .44 .46 .46 .48 7 1 .73 .9 1.27 29 34 .33 .33 .37 .46 .48 .48 57 89 .84 .02 .04 .04 .06 .25 .26 4 1 : 7 1 71 72 .88 89 .02 1.33 i.28 35 30 43 .44 44 45 59 .00 .70 :1 7- 11 8.27 R 10 4-98 3.62 4.21 4 2 7 2.71 3.20 2.41 ~ .45 '.88 .92 93 94 !?6 2.82 2.92 2.97 3.06 3-5 3.05 3.16 3.36 3.42 3-42 3.62 4-30 4.26 4-35 4.32 4.61 4.65 4.65 4.68 5.13 5.17 5.48 .29 .50 .52 54 !s6 24 32 36 44 43 43 52 .67 .72 .72 2.88 3-43 3.39 3.47 3.43 3.68 3^73 4.09 4.12 4-37 Buffalo Pottsville 6 6 6 6 7 7 7 7 7 7 7 7 6 6 6 8 6 8 8 8 8 8 8 9 50 50 50 So 55 55 55 & 60 69 65 9 90 75 65 120 65 120 65 65 65 66 80 81 7 8.47 1:11 8.57 5-93 5.96 6.00 6.00 8.60 8.90 9.06 9-33 9-33 9-33 4-3 1 4-33 4-36 4.36 6.26 6.47 6.59 6.79 6.78 6.78 7-03 7-45 7.63 7-63 8.04 9-53 9-51 9.64 9.64 10.21 10.31 10.31 10.37 3.22 3-24 3.26 3.26 4-69 4-85 4-93 5.o8 5.o8 s!?6 5.58 5.7 1 5-7 1 6.02 7. T 5 7.12 7.22 7!66 7-73 7.73 7.77 8.53 8.59 9.09 2.47 2.48 2.50 2.50 3 .6o 3-72 3.79 3-9 3-9 3-9 4.04 4.28 4-37 4-37 4.62 5.49 5.45 5-55 I'M 5.94 5.94 5.97 6.55 6.59 6.98 Paterson Trenton Phoenix Pottsville Trenton Pittsburgh Buffalo Pottsville Trenton Pittsburgh Buffalo Paterson Trenton Pottsville Trenton Pittsburgh Pottsville Phoenix Buffalo V 529 TABLE XIX. ( ROLLED-IRON BEAMS IN FIRST-CLASS STORES. DISTANCES FROM CENTRES (in fcef). See Arts. 694, 697, 6O9 and 7OO. o eo 00 Ok 1 (M J$ voS P.S.E: s *$- Tj-OO OO ON t>sOO \O 10 in u-i\o vo c^o t^oo CO^ ON^ 00 1 ?5 f?B? !??!? JJJJ* I 3 H ^^5- 5.3.^^ gvg^^-S ,3-^ 'invS^^o? 0? SN'ON^O I <N c O ? ? ^^nvo" 58 !<& 5 O> C ON C O S. S. K. Kw % S o 8 "S g, r^ r^ $ * ! 63 M f O in in invo m -i- -<j-vo o>'m t^ H M moo . coONrowo-. roinroovo ^ t^oo OOON ONONO^M Ntncn-^-i-i MHCNCO-^- ininm r^oo S'ONO o 8 S lH s X ^ ? S 'S ^> ^ ^ ^ K K EN R ^ J^oo a. M E^ 8 M ro^ ^g. ^^^ g; a ^ i s If "S-^vo vS"?L ^^SvS'S c,^^^-^ 5- S^hviT SN ^ 8 'ON S ? invo vo VO ON ,1 ** o CNONWMN COfOiO t^.CO OO ON C\ M CO ^ "* u^v^ Q\ Q\ Q O CJ IO t^ t^OO 00 M * .NiNroroco MmrorOM mmro^in ^m^^u. mvovovovo VO VO VO VO t^ U5 NO"VO > OO ONN ggvOvO^O NOOMOOm r^OO O ON ON OO (M ro CO (^ C? M rovg" *"* 00 00 00 CO 00 <* & ^> Reg S Ss Ss^ vo'^ do'SNO^Jr'o 225- Res' 'oxOo'^-cS ONOO in in CNI O w IN N t^ -Tj-Tr-^-rj- Tj-^inuim mininvcoo* oo'oo'oo'oo'oo* OOO\ONO>O^ 00000 M vo"vo ovo'S" WN^SpN rom ^oo w ^"S "rovo" 2 " " m S ro rt in in invo" vo* vo NO vo" t^. t^. f*. t^. t^ t^ o' 6666" H M w H CM e* ^N KS-o-r ^^N CT ^o;2 N ro^-o ** ^ t^* t^ t^- ^ t^ t^-OO OO O\ Os ON O\ O ^ ^i ^xxP'S c? fn co ON R. tH ON t> a\ o* o" o* o o" -* -QHV Had iHi A 3M NM N^4 gftjiii a OO OO ^OO O^ O>OONOOv ^^O>OsO OOONONO OOO^OC ON'O'O'O 1 lll-J ll^ll ill| li^ll |sll| w j c " J- j lili *- O ii aj o .ti J= rt >- J2 529 a TABLE XIX. (Continued.} ROLLED-IRON BEAMS IN FIRST-CLASS STORES. DISTANCES FROM CENTRES (in feet}. .See Arts, 694, 697, 699 and 7OO. LENGTH (in feet) BETWEEN BEARINGS. M S. g oiS S F 2 fev5S8.K fcsssa vSvSsa " H MWONO NN(NNC1 NCOfOCOm m CT f*> CO ITlVU VD M_> UN MM(NOrO OCONOUO ooiATj-^-ox MCJIT>-^- QiOONONON OOwm 00 SK ^3ffa? ^?KS^ ^,?KK^ ^5-^ ^3tf# 1 as^^s. Nb-ds^ ^Sof.sr'S, ^'S^oS- u-iTrromO (Nroo^ OO^-ONOO O O H --^ ^ in ON CN Ci 01 oojrorot^ ON o ci a ro o >- w d ro RSSffi^ ^ys M grg^8& SR^S- %^5SJ8 MR^ ^^srs ^^^SN s s-^^oS-^ 3b$ ^s^a^s, '^^g-s s^sss ^^^ CO ^.S^ 8^ a ^K^^ ? ^R^Sft (M \O N <N l' t^ I- 00 O O m O (N -t- Tf VO t^-OO O O (N Nrorororo ro ro ro ro ^ in in in in in vo o r. t- ^ t^OO ON ON ON O^ ON O\ O - MtT^ror^, V>^n ' c S? fT Jo Os 2 ~ 8 ^^^. ^?^ : O ' fpTi s-^assr fffp ^g^^ .> ON" Ci '*& ^oSS'' SN^^^SN AlCS'S^ in CO UNO 000 t.^r.^w O-ONOOO 5 i-O^^OOOO 000000000 ^^_ 2 u ro ro o ro N o co in tvoo oo ON ON O O O O 13 H t?&>&0 >< OO O O W CM 2 O w IH 1 2 H H H3J XHOI37V\. inininino Jf JJ> jo Jp^ O O O O^ O^OOOO^ SS?8 8cT | H H M H 1- Hid3Q 'O'O'ONN (N'N^CNI'CNIN N N W inininm'u-) ^^ ti X |1|1| 3i|| lj||| ll's| fti fti --i ft j- z: cu DM ft< ft< H CQ P9 0U'ft H f ? ? y - c = c ' u '|X 3^ SH P-i i5 PUP^HPn 529 b TABLE XX. See Arts. 7OI, 7O5 and 7O6. The larger figures five the average^ for use in the rules. See Table XLIL For- mula (10.) It See Table XLllI. For- mula IIS II fe. Formula (117.) 11 <a For- mula "ft 11 St* Table XLIY. D OJJ^ 2 hi) See Table XLV. i 0~(i D '. W HC/3 . . See Table XLVI. Z H *& gg*n 2o'i8< 11>1 e^ ' & Georgia Pine < 850 1176 952 I2OO 1406 460 650 875 417 550 722 420 500 643 280 450 707 580 600 700 442 480 520 860 9OO 9 60 1067 IIOO 1167 1040 1050 IIOO 616 650 746 725 750 824 507 650 790 813 850 920 394 557 490 460 475 589 4807 5900 6990 4470 5050 5650 1704 3100 4444 2209 3500 4819 2026 2900 3766 1660 2800 4000 3450 3556 2300 2550 2824 3800 4000 4248 4962 5150 5333 374 3850 4000 2800 2850 2933 3619 3900 4211 3273 3600 3894 4545 4750 5000 2022 3350 2697 2249 2580 4466 001069 OOIO9 OOII12 001239 OOI5 001764 000791 00086 00093 0008646 OOOgS 0010987 OOIOI56 OOI4 001791 000937 00095 000971 0009375 OOO96 0009896 0008854 OOIO3 OOII77 OOIO42 OOIII 001177 0013854 0014 0014063 001198 0013 001406 001563 001563 001563 00099 OOIO4 001094 001146 OOII6 001177 001042 OOIOg 001146 0009014 0007256 0009014 OOO8IO4 000834 OOO6I2 i-35i4 1-8357 2-1013 2-3874 2 2002 1-9593 4-7400 2-9405 3'3 2 4 2-227I i 8940 2-8350 I -7105 i 3240 2.5002 2 3496 2 5282 2 5512 2'5l6l 2-7628 3 0146 2--5382 2-I728 3-OI66 2-8153 2 6667 2-I558 2-IIgO 2-l6l2 3-2552 2-9I38 2 7165 1-9549 2-0266 2 2601 2-8I05 2-5682 2 4842 1-8773 2-1618 2-3940 2-3843 2 2SO2 2-2300 3-0024 3-I826 2-7994 3-5054 3-0660 2-9930 11671 16000 21742 11487 24800 33882 "453 19500 I57I9 19500 22069 I200O 9871 7^3 840 934 970 Il6o 1389 1076 1250 1474 463 540 647 433 4 80 530 322 370 410 8i 7 o 9500 ; JI 5<>3 i 1 1009 11700 12582 6531 8000 9775 7166 7850 ! 8408 58-9 i 6650 7502 ! 5213 5-00 6281 White Oak . . . -! Spruce . . -< White Pine \ Hemlock \ Whitewood -) Chestnut \ Ash . -j Maple \ Hickory .... . \ Cherry . J Black Walnut -J Mahogany. St. Dom . . J Bay Wood. J Oak, English " Dantzic. . . . Adriatic 1 Oaks, Average of I Oak, Canadian TABLE XX. (Continued.} See Arts. 7O 1 , 7O3 and 7O6. The larger figures gh>e the aver -age \ for use in the rules. See Table XL1I. For- mula (10.) li CQ See Table XLIII. For- nmla (113.) ff M* ii s Formula (117.) n 1! * For- mula (118.) iR t>. II i See Table XLIV. 0, Z U t>< D O Id tt *> II 2 W I T h H H u j t l ^ M 6-j f M ""' < ^" sssip See Table XLV. :-di I! ^C/J HC/3 .,c 8^ S ^ r?-- fe < - ! of n agS c RESISTANCE TO ^ CRUSHING, PER SQ. IN. ^^^ SEC. AREA, SHORT \^ | BLOCKS. = C. ^ e< Ash 675 519 338 544 447 367 369 350 360 38i 421 408 284 277 376 330 491 2OOO 2500 3OOO l6oO 2100 2600 2400 26OO 2800 1600 19OO 22OO 3200 6OOO 72OO 3810 3134 1590 2836 4259 3454 3080 2293 2686 1858 2013 1961 1422 2078 2437 2093 3375 41500 50000 58500 27700 40000 53 2co 55500 62OOO 69000 53000 6OOOO 67000 60000 65000 71500 67000 7OOOO 74000 0007177 000582 0009552 0006428 0004279 0005277 0004932 0006813 0005868 cooS i 74 0007762 0007903 0010686 0006265 0006412 000744 0006173 3-4285 3.9520 3-0910 4-1446 3-4066 2 7966 3-3738 3-III7 3-I723 3-4843 3-7423 3-6564 2-5958 2-9552 3.3420 2-9433 3-2732 2OOOO 27000 45000 T3000 17OOO 26000 40000 6OOOO 80000 3OOOO 50000 65000 1I57 80 155500 190262 80000 I2OOOO 170000 Socoo IOOOOO 140000 40000 70000 looooo 40000 5OOOO 65000 IOOO 3000 6000 IOOO 1 2000 3000 4000 j 20000 Beach Elm Pitch Pine Red Pine . . Fir New England . . . 41 Ricra " Average of Riga . . *' Mar Forest . . Av'ge of Mar Forest Larch Average of. Norway Spar Cast-Iron, American.. J English.. ...j "Wrought-I ron, Amer . 3 English J "- Swedish 3 Steel Bars J .0002 " Chrome < Blue Stone Flagging J Sandstone -j Brick, Common. 3 *' Pressed 122 2OO 251 33 59 94 20 33 43 37 147 Marble, Eastchester. . . . ' 531 TABLE XXI. SOLID TIMBER FLOORS. DEPTH OF BEAM (in inches). See Art. 7O2. LENGTH BETWEEN 1 BEARINGS (in feet). || DWELLINGS AND ASSEMBLY ROOMS. FIRST-CLASS STORES. i2 . td O W U ei Z > H W 5 2 HEMLOCK. GEORGIA PINE. SPRUCE. M t w HEMLOCK. 8 9 1O 2-4O 2-72 3-03 2-83 3-20 3-56 3-oi 3-40 3-79 3-04 3-43 3-82 3-i5 3-55 3-95 3-73 3-97 4-20 4-47 4-68 4.'. 8 4-01 4-52 5-03 11 3-35 3-93 4-18 4-21 4-35 5-15 5-48 5-54 12 3-67 4-30 4-58 4-61 4-76 5-63 5-99 ! 6 '5 13 3-99 4-68 4-98 5-01 5-16 6- 10 6-50 6-56 14 4-32 5-05 1 5-38 5-4i' 5-57 6-58 7-01 -07 15 4-64 5-43 5-78 5-Si 5-98 7 -06 7-52 7-59 16 4-97 5-8i 6-19 6-21 ^39 7-55 8-03 8-io 17 18 5-^4 6-19 6-58 6 '59 7-00 6-62 7-03 6 -So 7-21 8-03 8-51 8-55 8-62 9-06 9-14 19 20 ,5-98 6-32 6-97 7-35 7-83 7-44 7-85 7-^3 8-05 9-00 9-48 9-5S 10- 10 9-66 10-18 21 6-66 7-75 8-24 8-27 8-46 9-97 10-61 10-70 22 7-00 8-14 S-66 8-68 S-S8 10-46 11-13 11-22 23 7-35 8-53 9-08 9-10 9-30 10-95 n-66 11-74 24 7-70 8-93 9-5i 9-52 9-72 11-44 12-18 12-27 25 8-05 9-33 9'93 9'94 10-15 11-94 12-71 12-79 26 8-40 9-73 10-36 10-37 10-57 12-43 13-23 13-32 27 8-76 10-14 10-79 10-79 II -OO 12-92 13-76 I3-85 23 9- II 10-54 11-22 11-22 II -42 13-42 14-29 14-38 29 9-47 10-95 II -65 II-65 11-85 13-92 14-82 14-91 3O 9-83 1 1 - 36 I2-Og 12-08 12-28 14-42 15-35 15-44 532 TABLE XXII. MATERIALS USED IN THE CONSTRUCTION OR LOADING OF BUILDINGS. WEIGHTS PER CUBIC FOOT. As per Barlow, Gallier, Ilaswell, Hursf, Rankine, Tredgold, Wood and the Author. MATERIAL. I o K fe o H AVERAGE. MATERIAL. ~ ta AVERAGE. WOODS. 41 35 49 4i 39 35 59 27 47 3^ 3 2 29 27 27 21 6 9 51 5i 51 57 53 49 65 35 57 38 46 4i 55 4i 33 83 46 38 50 49 46 42 62 83 64 31 52 34 40 39 35 41 15 34 4O 44 1? IS 43 46 27 37 47 53 62 37 26 49 52 33 43 23 62 46 57 38 Mahogany, St. Domingo. . . . Maple 45 33 35 60 38 57 47 43 4 38 63 49 55 66 79 54 57 44 S^ 55 41 45 62 63 54 ! 47 ! 54 68 i 51 5O 58 44 42 48 ! 43 32 i 37 39 i 28 : 33 45 30 44 23 45 3O n 38 51 30 8O 33 49 27 50 614 5O6 544 531 516 Mulberry ... !Oak, Adriatic " Black Bog " Canadian Dantzic English " Red."i ' White Olive Orange Pear-tree. Pine, Georgia (pitch) Mar Forest . . Alder Apple-tree Ash Beech Birch. .. . Box " French Brazil-wood Cedar " Canadian " Palestine. " Virginia Red Cherry Chestnut, Horse 11 Memel and Riga " Red .:. ... '' Scotch " White '* Yellow... Plum Poplar. . .... 29 27 21 2 7 41 2 3 55 24 36 4i g 40 25 487 528 508 35 Si 35 39 49 37 59 36 8 3 4 o 58 29 525 S34 524 Cork Cypress Spanish Deal, Christiania '' English u (Norway Spruce). ... Dogwood Ebony 1 Quince ..... Redwood Rosewo >d Elder Sassafras Satin wood Spruce Sycamore. Elm Fir (Norway 3p r u<-e) . 33 21 30 59 33 44 " (Ued Pine) " Ri g;l Teak Tulip-tree Vine. Walnut, Black " White ' Water 3* 58 63 35 54 83 Si 40 ' Hackmatack 21 40 41 31 31 41 41 35 Hemlock Hickory Whitewood . Yew Larch METALS. Bismuth, Cast ,. brass, Cast '' (Gun-metal) . . , Plate " Red. " White Lignum-vitae Logwood Mahogany, Honduras i Bronze . . 533 TABLE XX II. (Continued^ MATERIALS USED IN THE CONSTRUCTION OR LOADING OF BUILDINGS. WEIGHTS PER CUBIC FOOT. As per Barlo-w, Gallier, Haswell, Hurst, Rankine, Tredgold, Wood and the Author. MATERIAL. I h O H AVERAGE. MATERIAL. K t* o C-i ui a < M H < 1O4 100 112 no 130 81 10O 113 145 122 16O 96 83 79 85 54 13O 106 110 126 126 250 160 185 163 16O 183 165 163 174 165 164 166 185 166 105 134 140 52 169 146 162 170 166 170 Copper, Cast 537 549 543 556 544 1206 1108 509 481 454 475 480 7O9 717 713 851 849 837 488 453 975 1345 1379 142 636 655 658 644 489 462 439 173 156 8O 277 171 139 129 160 10-' 138 1O7 100 1O5 Brick-work ... 9 6 112 " Plate ... 12O Gold " in Mortar 100 475 434 487 474 Iron, Bar " Cast " Malleable Roman, Cast and Sand, equal parts.. Chalk iie 119 'is 77 46 125 J 74 125 102 90 81 6J '35 125 ^65 195 " Wrought 474 486 Lead Cast Clay '' with Gravel Coal, Anthracite English Cast " Milled ... Bituminous " Cannel " 60 Nickel, Cast .. Coke Pewter.. Concrete, Cement Platina, Crude k * Pure F t h C ' ' 95 " Rolled Plumbago Silver, Parisian Standard... Pure Cast " " " Hammered " Standard Steel 486 456 429 165 492 468 449 180 " Loamy Emery Feldspar Flagging, Silver Gray . ... Flint 155 171 Tin, Cast " Flint Zinc, Cast STONES, EARTHS, ETC. Alabaster. Asphalt, Gritted " Plate 153 167 158 173 181 172 " White Granite " Egyptian Red Asphaltum Barytes, Sulphate of Basalt Bath Stone 57 250 155 122 124 '85 103 304 18 156 '34 119 " Quincy Gravel 90 120 Gypsum 135 '45 199 Beton Coignet Blue Stone, Common Brick. . . . Limestone 39 Aubigne ' Fire- | N. R. common hard... ... Marble 161 178 ' Philadelphia Front... ... 534 TABLE XX 1 1. (Continued.) MATERIALS USED IN THE CONSTRUCTION OR LOADING OF BUILDINGS. WEIGHTS PER CUBIC FOOT. As per Barlow, Gallier, Hasivell, Hurst, Rankine, Tredgoid, Wood and the Author. MATERIAL. i H AVERAGE. MATERIAL. o AVERAGE. Marble, Eastchester 167 178 173 167 166 167 I4O 125 175 155 98 103 107 9 86 105 1OO 118 83 146 72 80 18O 175 147 56 165 165 124 112 1O5 105 123 105 97 172 126 144 133 142 134 141 134 162 15O Serpentine. 165 144 152 95 159 167 157 18O 135 151 14O 160 14O 124 115 170 76 58 49 59 62 26 20 58 57 61 17 61 69 114 73 131 559 68 171 133 131 14 8O 62* 64 81 Chester, Pa '' Green ... ... " Italian Marl 165 100 no 169 179 140 Slate 137 181 150 io Mica 120 1 2O '87 88 109 118 " Welsh Mortar Stone, Artificial Stone-work Hewn " Rubble " Hair, incl. Lath and Nails, per foot sup. Hair, dry 44 new... " Sand 3 and Lime paste 2 well beat together 7 ir Tiles, Common plain . Trap Rock ... MISCELLANEOUS. Ashes. Wood .... Peat, Hard Petrified Wood Pitch ... Plaster. Cast Porphyry, Green Red Portland Stone Pumice-stone . . Puzzoiana Quartz, Crystallized Rotten-stone 132 161 Butter Camphor Charcoal 17 14 52 IO 56 34 25 '62 24 66 Cotton, baled Fat Gutta-percha Hay baled Sand, Coarse Common Dry Moist 4 Mortar.. 92 90 118 118 120 128 101 158 ... Pit 92 Quartz Red Lead with Gravel Resin Sandstone Amherst, O. . Belleville, N.J... Berea, O Dorchester, N. S. Little Falls, N. J. 44 Marietta, O. . . ** Middletown, Ct.. 130 Rock Crystal... Salt 20 100 Saltpetre. . .. 8 60 Water, Rain ... . Sea Whalebone 535 TABLE XXIII. TRANSVERSE STRAINS IN GEORGIA PINE. LENGTH i 6 FEET. BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF / EXPERIMENT. ( 1 2 3 4 5 6 7 8 9 i DEPTH I 1 (/ inches). \ 1-04 i 04 1-03 i 04 I 1-03 j i 03 1-03 1-03 I 02 BREADTH ( (in -inches), j i 05 1-04 i 03 1-03 I 1-04 1-04 1-04 1-03 1-04 PRESSURE (in founds). DEFLECTION (in inches). 000 ooo ooo ooo ooo ooo coo oco 000 25 020 015 020 01 5 015 OiD 010 020 015 50 040 030 040 030 025 035 025 040 030 75 OSS 040 060 04S 045 050 035 OSS 045 100 070 050 oSo 060 065 06 5 -050 070 -060 ooo ooo 000 ooo 000 000 000 oco oco 100 070 050 080 060 c6 5 065 050 070 ceo 125 085 065 095 75 080 080 070 090 075 i5O TOO 080 1 10 090 100 090 085 no 085 175 "5 095 -125 105 115 1 05 ICO 130 100 200 130 no 140 120 130 1 20 115 15,0 115 000 ooo ooo 000 ooo ooo ooo 000 coo 200 130 no 140 1 20 130 120 i '5 150 115 225 145 120 160 '35 145 '35 125 170 130 250 160 135 175 ISO 160 150 140 i$p -J45 275 -I 75 150 IQO <6s 180 ico 160 2IO 160 300 190 160 -210 180 IC;5 7i *T5 225 175 | O ooo ooo OOO ooo ooo ooo ooo ooo coo 1 3OO IQO 160 210 180 !95 175 175 22<y 175 325 2IO '75 230 200 215 190 190 ? 45 j IQO 350 225 IQO 2^O 215 230 2IO 2IO 26^ 205 375 240 205 265 230 245 225 225 285 220 Eoo 255 220 280 245 260 240 245 310 235 o oco CCO 005 coo ooo 005 005 -CO5 co^ 4OO 255 22O 2 o 245 260 240 245 -310 240 425 275 -235 300 265 280 255 255 330 ; -255 45O 290 2 5 320 280 295 270 270 355 -275 475 310 265 340 295 '3 '5 285 28; 385 -2co 5OO 330 280 3 80 3 J 5 335 3 00 3CO 4'5 "SOS ooo OIO 030 000 005 005 005 030 cos 50O 330 280 3 80 3'S 340 300 300 430 305 525 35 295 330 360 '3'5 32O -470 325 550 370 -3'0 350 380 330 33* 345 575 390 330 375 405 -350 355 370 GOO 4i5 35 395 435 "375 375 400 020 02O 015 025 020 025 025 600 420 350 .... 45 .440 380 380 410 625 460 370 435 475 405 405 455 65O 560 400 470 535 430 43 675 440 460 475 700 480 BREAKING WEIGHT (In 674- 719- 5 l8- 6C2- 652- 6 99 - 685- 536- 642- pounds). 536 TABLE XXIV. TRANSVERSE STRAINS IN LOCUST. LENGTH 1-6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ) EXPERI- V 1O 11 12 13 14 15 16 17 18 19 20 MENT. \ DEPTH | (in inches), f 1-03 1-08 1-05 I -08 1-07 1-65 1-07 I -08 I -07 1-09 I -08 BREADTH j (in inches). \ I -02 1-05 I -08 I -02 1-09 I -08 1-68 1-04 i-oS I -08 1-03 PRESSURE (in pounds). DEFLECTION (in inches'). OOO OOO OOO OOO OOO OOO 660 OOO OOO OOO 660 25 015 015 015 015 010 -015 -015 615 015 -615 -015 5O 035 030 030 030 625 0301 -030 , -035 030 -030 -030 75 055 050 045 045 635 '656: -045 656 645 645 -645 100 075 065 060 060 O5O -065 ; -O6O 676 660 666 -060 000 000 000 OOO OOO OOO OOO 66O 000 OOO -OOO 100 075 065 060 j 060 656 065 : -O6O 670 060 060 e66 125 095 080 075 1 -075 -065 080 1 -075 QgO 075 075 -080 15O no 100 090 090 080 095 -090 no 090 090 -095 175 125 -115 105 110 696 115 -105 125 165 IOO -IIO 200 145 130 115 125 165 130 -120 145 120 115 125 O OOO OOO OOO 000 OOO OOO OOO OOO 000 000 666 200 145 130 115 125 165 130; -120 145 120 115 125 225 165 150 136 140 126 145 135 165 136 130 1401 25O 180 170 145 155 135 165 156 190 145 145 155 275 200 185 160 175 145 180 I6 5 2IO 160 160 170 3OO 22O 200 175 -19 166 195 -180 230 175 175 190 O OOO 005 OOO -OOO 600 OOO OOO OIO ooe 000 005 300 226 205 175 -19 160 195 -180 230 175 175 190 325 240 220 190 -2IO 175 210 -195 250 190 185 205 350 26O 240 -205 -225 190 230 -2IO 270 205 200 22O 375 275 265 22O -240 205 245 -22O 290 22O 215 240 40O 295 290 235 -255 22O 265 -235 315 235 230 255 1 005 02O 005 -005 OOO 005 ! -005 OOO 665 OIO | 4OO 295 295 235 -255 22O 265 235 : .... 235 236 255: 425 335 250 275 235' 280 250; .... 250 245 275 45O .... 265 295 250 295 265 .... 265 266 290 ; j 475 280 , o rr\ * f) " 280 5OO 290 325 280 335 295 .... 295 275 290 3 IQ ; 336 ; 537 TABLE XXIV. (Continued.} TRANSVERSE STRAINS IN LOCUST LENGTH 1-6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ) i EXPERI- V MENT. ) 1O 11 12 13 14 15 16 1 17 18 19 20 DEPTH \ ' (in inches), f BREADTH ( (in inches). \ 1-03 I -O2 I -08 1-05 1-05 i -08 i -08 i -02 1-07 1-09 1-05 i -08 1-07 i -08 i -08 j 1-07 \ 1-09 ; I 04 I I 08 I O8 I -08 I -03 PRESSURE (in pounds). DEFLECTION (in inches). 500 525 550 575 600 O 600 625 65O 675 7OO 700 725 750 775 800 SOO 825 850 875 90O O 900 925 950 975 1000 005 290 305 320 335 350 OIO 350 365 385 400 4i5 015 4i5 435 455 475 495 035 505 530 555 585 615 065 635 .66e 005 325 345 365 385 405 020 405 430 455 485 510 035 515 540 575, 610 640 075 650 -690 725 005 280 295 310 325 345 OIO 345 365 385 405 425 030 430 455 475 505 535 060 545 580 615 645 675 015 335 355 375 395 415 020 420 455 49 52< 560 040 565 595 635 005 '-95 310 325 340 -360 015 .360 375 390 405 420 020 420 440 460 4 80 500 035 505 530 550 575 605 050 610 640 670 700 735 .... 005 295 310 325 340 355 OIO 355 375 39 405 425 025 425 445 465 490 515 045 520 545 -570 600 630 075 640 675 005 290 305 320 335 350 -015 355 370 390 410 430 025 435 455 480 500 520 045 530 560 585 615 650 015 330 350 3/0 39 4i5 025 4i5 440 460 485 510 050 520 .... .... .... .... .... .... 695 7^ i BREAKING j I WEIGHT (in > pounds). ) | 449- 425- 1046- 956- looi- 860- 1037- 402- 937' ;I027- 7i5- TABLE XXV. TRANSVERSE STRAINS IN WHITE OAK. LENGTH 1-6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ) EXPERI- y 21 22 23 24 25 26 27 28 29 3O MENT. ) DEPTH j (in inches), j I -06 I -06 I -08 1-07 i -08 I -06 ,. i -08 1-07 1-06 BREADTH (in inches). I -08 i -08 I -06 i 06 i 06 i -08 1-07 1-07 I -06 i -08 PRESSURE (in pounds). DEFLECTION (in inches'). ooo ooo ooo ooo ooo ooo 000 ooo ooo ooo 25 030 -040 035 040 045 045 035 040 035 045 5O 060 -075 065 080 090 085 075 080 070 085 75 090 -115 095 115 135 125 115 125 105 125 10O 120 -155 130 150 180 170 155 165 140 -165 ooo ooo ooo 000 ooo -GOO 000 000 ooo ooo 1OO 1 20 155 130 i -150 185 -170 155 165 140 170 125 150 195 165 190 235 215 200 215 175 215 150 180 | -235 195 -230 295 260 245 260 2IO 260 175 215 280 225 -275 355 310 285 310 240 310! 200 245 325 265 315 410 355 330 1-360 275 365 ooo 020 OIO O2O 025 020 015 O2O OIO 025 200 250 1 -330 265 325 410 365 345 360 280 365 225 285 -380 310 380 480 420 395 420 320 420 250 320 -440 350 -430 -555 480 450 485 360 480 1 275 360 -500 390 -480 <>35 545 SIS 560 405 545 30O 400 560 440 -540 715 615 580 640 450 615 030 060 040 065 100 075 065 080 045 080 300 415 530 440 560 735 635 '595 660 455 640 325 465 650 .490 -620 705 665 510 725 350 515 7 J 5 545 -680 *6s 375 570 605 760 .... 62S *4OO 630 670 600 105 ion 400 690 . 710 425 760 BREAKING ) WEIGHT (in > pounds). ) 520- 404- 510- 475' 368- 430- 426- 391- 504- 40!' 539 TABLE XXVI. TRANSVERSE STRAINS IN SPRUCE. i LENGTH 1-6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ) EXPERI- v MENT. J 31 32 33 34 35 36 37 38 39 40 DEPTH \ (in inches). ) 1-09 1-05 I -08 1-04 I -07 1-04 1-03 1-07 I -08 I -04 BREADTH / (in inches), f 1-04 I -08 1-03 1-07 1-04 I -08 1-07 1-04 1-04 I -08 PRESSURE (in pounds). DEFLECTION (in inches). O ooo 000 ooo 000 ooo ooo ooo ooo 000 ooo 25 020 025 040 025 025 030 025 025 025 02O 50 040 045 075 050 050 060 045 045 045 040 75 060 070 no 070 075 090 065 070 065 060 1OO 080 090 145 095 ICO I2O 085 090 085 080 OOO ooo 000 ooo 000 OOO ooo ooo ooo OOO 100 080 090 145 095 IOO 120 085 090 085 080 125 IOO "5 180 115 125 145 no 115 105 IOO 150 125 135 215 135 150 175 135 140 125 125 175 145 155 250 155 170 200 155 160 145 145 200 165 175 285 175 190 230 175 .180 165 165 ooo 000 010 ooo OIO OO5 005 ooo ooo 005 200 165 175 285 180 190 230 175 I 80 170 165 225 185 2OO 325 200 215 265 195 2OO 190 190 250 2IO 225 3/o 220 240 295 220 225 210 2IO 275 230 245 415 245 260 330 240 250 235 235 30O 250 265 465 27O 285 370 260 2 7 255 255 O 005 005 045 005 OIO 025 OO5 005 005 005 30O 250 270 475 275 285 375 260 275 255 255 325 275 295 530 300 310 410 285 300 275 280 1 35.O 300 320 -600 330 335 450 310 325 300 305 375 330 350 680 355 360 495 335 355 325 330 4OO 380 390 760 385 390 540 365 395 350 360 1 O 035 030 .... 020 020 070 020 025 OIO O2O 4OO 390 4OO .... 395 395 570 370 410 355 370 425 460 440 .... 425 430 620 400 455 385 400 45O .... 495 ... 455 460 680 440 445 440 475 .... 505 .... 485 .... 500 5OO .rfir ^7O O :> u :> j / u O7O 500 59O J? BREAKING WEIGHT (in 445' 487- 400- 502- 470- 465- 498- 441- 475' 527- Pounds). 540 TABLE XXVII. TRANSVERSE STRAINS IN SPRUCE. LENGTH i 6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ) EXPERI- > 41 42 43 44 45 46 47 48 49 50 MENT. ) DEPTH I (in inches). \ 1-52 i-55 1-56 1-56 1-56 i-55 i-55 1-56 i-55 i'57 BREADTH I (in inches). ] 1-09 I IO i -06 I IO I IO i 09 1-08 I-IO I- IO 1-09 PRESSURE (in pounds). DEFLECTION (in inches). O OOO OOO OOO 000 1 OOO OOO 000 OOO OOO OOO 25 OIO OIO OIO OIO OIO OIO OIO OIO 005 005 50 O2O O2O O2O 015 O2O O2O O2O 015 OIO OIO 75 025 030 025 025 030 025 030 025 O2O O2O 100 035 040 035 035 035 035 035 030 025 025 OOO OOO OOO OOO OOO OOO OOO OOO OOO OOO 100 035 040 035 035 035 035 035 030 025 025 125 045 050 045 040 045 045 040 035 030 035 ISO 050 060 055 050 055 050 050 040 040 040 175 060 070 065 060 065 060 055 050 045 045 200 065 075 070 065 070 065 065 055 055 055 000 OOO OOO OOO OOO OOO OOO 000 000 OOO 200 065 075 070 065 070 065 065 -055 055 055 225 070 -085 080 075 080 070 070 -065 060 060 250 080 -095 090 085 090 080 080 070 065 065 275 090 IOO IOO 090 095 085 -090 -075 075 075 300 IOO no no 095 105 090 095 085 080 080 OOO OOO 000 OOO OOO OOO OOO OOO OOO OOO . 300 IOO no no 095 105 090 095 085 080 080 325 105 US 120 105 115 IOO IOO 090 085 085 35O 5 I2O 130 115 125 105 IIO IOO 090 095 375 120 130 140 125 130 110 120 105 IOO IOO 400 125 140 150 135 140 I 20 125 IIO 105 IIO OOO OOO 005 000 000 OOO 000 OOO OOO 000 400 125 140 150 135 140 I2O J 25 IIO 105 IIO 425 135 150 160 140 145 125 130 120 IIO I2O 450 145 160 170 145 155 135 140 125 120 -125 475 150 165 180 155 165 140 150 130 125 130 500 160 175 190 165 175 150 155 140 I3 o 135 O 005 OOO OIO OOO 000 OOO 000 OOO OOO OOO 500 160 175 190 165 175 150 155 140 130 135 525 170 180 205 170 185 155 160 145 140 145 550 175 190 215 180 195 160 170 150 145 155 575 185 200 225 185 205 170 180 160 150 160 600 190 2IO ;>40 195 215 180 185 170 : i6o 170 541 TABLE XXVIL (Continued) TRANSVERSE STRAINS IN SPRUCE LENGTH 1-6 FEET BETWEEN BEARINGS. 'See Arts. 7O4 and 7O5. NUMBER OF ) i EXPERI- > 41 42 43 44 45 46 47 48 49 50 MENT. ) DEPTH | (in inches), j 1-52 i-55 1-56 1-56 1-56 i'55 i-55 1-56 i-55 i-57 BREADTH ) (in inches). \ 1-09 1 1O I -06 ] -10 1-10 i -09 i -08 1 -10 I -10 i -09 PRESSURE (in pounds). DEFLECTION (in inches). I 005 005 015 005 005 005 005 005 ooo ooo 600 IQO 2IO 240 195 215 180 185 170 160 165 625 195 215 250 205 225 185 195 175 170 175 650 205 225 265 215 235 195 200 185 175 I 80 675 215 230 275 220 245 200 2IO 190 180 190 700 225 240 290 230 255 2IO 220 200 190 195 005 005 025 OIO OIO 005 005 005 005 ooo 700 225 240 290 230 255 210 22O 200 190 195 725 235 250 305 240 265 22O 230 2IO 200 205 750 245 260 320 245 275 230 -240 220 2IO 215 775 255 265 335 255 290 240 25O 230 22O 225 8OO 265 275 350 265 305 250 255 240 230 235 OIO OIO 040 OIO 025 OIO 010 -015 OIO 005 80O 265 275 350 275' 310 250 255 -240 23O 235 825 275 285 370 285 330 260 265 250 240 245 850 285 295 335 295 345 270 275 260 255 255 875 .300 305 405 310 365 280 285 275 27O 265 900 310 315 420 .320 405 290 295 2 9 2 9 275 O2O O2O 060 025 025 02O 030 O25 015 900 315 320 430 325 .... 295 295 295 295. 275 925 365 330 460 340 310 310 310 315 290 950 345 .... 355 .... -325 320 325 340 305 975 .... 360 .... 370 340 335 345 470 320 1000 370 335 .... 365 350 365 335 030 045 050 030 060 035 1OOO .... 380 395 .... 375 355 400 345 1025 .... 390 .... 400 375 450 365 1O50 .... 405 .... .... 430 400 385 1075 415 BREAKING ! WEIGHT (in pounds). 950- 1074- 926- ! ioo i 900- 1067- 1071- 1028- 977' 1078- 542 TABLE XXVIII. TRANSVERSE STRAINS IN SPRUCE. LENGTH 1-6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ) EXPERI- > 51 52 53 54 55 56 57 58 59 60 MENT. ) DEPTH 1 (in inches), f 2-OI 2-00 1-99 1-99 | 2-02 1-99 1-98 2-01 2-01 2-02 BREADTH 1 (in inches), f I -08 i i -08 I- 08 1-08 i i. 08 i i -06 I -08 1-09 1-07 1-09 PRESSURE (in pounds). DEFLECTION (in inches). o OOO OOO OOO OOO 000 OOO OOO OOO OOO 000 5O 015 015 OIO OIO OIO OIO OIO OIO 015 005 100 025 025 O2O 015 O2O 020 O2O 020 025 015 150 035 035 030 025 025 O25 025 025 035 020 200 040 045 '035 030 035 030 035 035 C45 030 O OOO OOO OOO OOO OOO 000 OOO 000 000 000 200 040 045 035 -030 035 030 035 035 045 030 250* 050 055 045 040 045 035 040 040 055 040 300 060 065 050 050 055 045 050 045 070 050 350 070 075 060 055 065 050 055 055 080 055 400 075 085 065 065 -070 060 065 060 090 -065 O OOO OOO OOO OOO OOO 000 000 OOO 000 OOO 400 075 085 065 065 070 060 065 060 090 065 450 080 095 070 070 080 065 070 070 IOO -070 500 090 105 -080 080 090 070 080 075 IIO 080 55O IOO 115 090 085 IOO 080 085 080 120 090 60O no 125 095 095 IIO 085 095 090 130 IOO O OOO 005 OOO OOO -000 OOO 000 OOO 005 OOO 60O no 125 095 095 IIO 085 C95 090 135 IOO 65O I2O 135 105 IOO "5 090 IOO 095 140 IIO TOO 130 145 no I IO 125 IOO 105 105 150 -115 750 135 155 I2O 115 135 105 115 IIO 160 125 8OO 145 165 125 125 145 115 125 120 175 135 005 OIO 005 000 005 000 005 005 OIO 005 80O 145 .165 125 125 145 115 125 120 175 135 85O 155 175 135 130 155 I2O 130 125 185 140 90O 165 185 140 140 165 130 140 130 195 150 950 175 195 150 145 175 135 145 140 2IO 160 1OOO 185 210 160 155 .[85 145 155 145 225 170 543 TABLE XX VI 1 1. (Continued) TRANSVERSE STRAINS IN SPRUCE LENGTH i 6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF i 1 EXPERI- V 51 52 53 54 55 56 57 58 59 60 MENT. ) DEPTH I (in inches), f 2-01 2-00 1-99 1-99 2 -O2 1-99 1-98 2-01 2-OI 2-02 , BREADTH 1 (in inches). ) I -08 I -08 i -08 i -08 i i -08 i -06 I -08 ,.0 9 1-07 1-09 PRESSURE (in founds). DEFLECTION (in inches). O OIO 015 005 005 OIO 000 OIO OIO 015 OIO 1000 IQO 210 160 155 190 MS 155 145 22S i/o 1O5O 2OO 225 165 160 2OO 150 160 150 235 180 | 110O 210 240 175 170 210 160 170 1 60 250 195 1 115O 230 255 185 1 80 22O 165 175 170 270 205 1200 245 27O 195 190 235 "175 185 175 28 5 215 O 030 02 5 OIO OIO 025 OIO OIO OIO 030 O2O 1200 250 275 195 190 240 1 80 185 175 -28 5 22O 125O 270 290 205 2OO 255 190 195 185 3OO 230 13OO 295 305 215 2IO 270 200 205 195 320 245 1350 325 325 -230 220 28 S 210 215 205 345 260 140O 360 355 240 235 305 22O 225 215 365 275 070 060 O2O O2O 040 020 O2O O2O 055 030 1400 375 370 245 235 310 22O 230 215 370 280 145O 415 400 255 250 330 235 240 230 390 295 15OO 430 270 265 355 250 255 245 415 320 1550 . , . . 290 280 265 275 260 350 160O ' *?IS 3O^ 28s 2Q^ 280 380 "OSS 040 .... 040 040 O4O O7O 16OO 330 310 290 300 2 9 165O 375 335 .... 310 325 3 IO 1 TOO ^7^ 335 175O 370 BREAKING WEIGHT (in pounds). 1472- I536- 1675- I7I7' 1519- 1653- 1686- 1800- 1545' I6OO- 544 TABLE XXIX. TRANSVERSE STRAINS IN WHITE PINE. LENGTH i 6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ) EXPERT- > 61 62 63 64 65 66 67 68 69 MENT. ) , DEPTH ) (in inches). ( 1-02 99 99 1-03 1-02 99 99 I-OO 99 BREADTH | (in inches), j I -OO 1-02 1-02 I-OO I-OI I -01 I-OI 99 i -02 PRESSURE (in pounds). DEFLECTION (in inches). O OOO OOO OOO 000 000 000 000 000 000 25 035 030 040 030 035 040 030 035 030 50 070 060 075 060 070 080 065 065 065 15 IOO 095 'US 090 IOO 115 095 095 095 100 130 125 150 120 130 150 130 125 125 O OOO OOO 000 OOO OOO OOO OOO OOO OOO 10O 130 125 150 120 130 150 130 125 -125 125 160 155 185 150 165 185 160 160 155 15O 190 185 220 180 195 22O 190 190 185 175 220 215 260 210 230 250 220 225 215 2OO 250 245 295 240 260 285 250 255 245 O 005 OOO OO5 -000 OOO OIO OOO 005 OOO 200 2 3 250 295 240 260 290 250 255 245 225 280 280 335 270 295 325 285 285 280 250 315 310 380 300 .330 365 320 320 310 275 345 345 425 335 365 410 350 355 345 30O 380 375 470 365 405 460 385 385 375 O OIO 005 .... 005 OIO 030 005 OIO OIO 300 385 380 .... 370 .410 465 385 390 380 325 430 415 .... 405 460 520 430 425 415 350 485 450 440 e ^ 5 .465 4cc 375 500 ?*?** 540 DJO *r v -'D ^3^ TO J 4QC Jjj T^V J BREAKING WEIGHT (in pounds). 376- 385- 300- 382 356- ; 350- 349' 383- 399- i 545 TABLE XXX. TRANSVERSE STRAINS IN WHITE PINE. LENGTH 1-6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ) EXPERI- v 70 71 72 73 74 75 76 77 78 MENT. J DEPTH | (in inches). } i-53 1-50 1-52 i-53 I-5I 1-49 i-53 1-52 1-49 BREADTH I (in inches). ) 1-02 1-03 1-03 1-02 1-03 I -OI I -02 i -02 I -OI PRESSURE (in poundsy. DEFLECTION (in inches). O ooo ooo OOO ooo ooo ooo 000 000 000 25 015 OIO OIO 015 OIO 015 OIO OIO OIO 50 025 O2O 020 - -025 020 025 020 015 O2O 75 035 030 030 035 030 035 030 025 030 100 045 035 040 045 040 045 040 035 O4O O ooo 000 000 ooo OOO 000 000 000 000 100 045 035 040 045 040 045 040 035 040 i 125 055 045 050 055 050 055 045 040 050 ij 15O 065 050 060 065 055 065 055 050 1 -060 175 075 060 070 075 065 075 065 060 070 2OO 085 070 080 090 075 085 075 070 080 O ooo ooo ooo 000 000 ooo ooo 000 ooo 20O 085 070 -080 090 075 085 075 070 080 225 095 080 090 IOO 085 095 080 075 090 25O 105 090 IOO no 095 105 090 085 IOO 275 115 100 no I2O 105 115 IOO 095 no 30O 125 105 I2O 130 no 125 105 105 -125 O 005 oco 000 ooo ooo ooo ooo ooo ooo 3OO 125 105 I2O 130 no 125 105 105 125 325 135 115 130 140 120 135 115 115 135 35O 145 125 140 150 125 145 125 I2O 145 375 155 135 145 160 135 155 135 130 160 4OO 165 145 155 170 145 165 145 140 I/O 005 005 005 005 ooo 000 000 000 ooo 40O 165 145 155 170 145 165 145 140 170 425 175 150 I6 5 180 150 175 150 145 180 45O 185 160 175 190 160 185 lt)O 155 190 475 195 170 I8 5 200 170 195 170 165 200 50O 205 180 195 2IO 180 205 175 175 210 546 TABLE XXX. (Continued^) TRANSVERSE STRAINS IN WHITE PINE. LENGTH 1-6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. 1 NUMBER OF ) EXPERI- > 70 71 72 73 74 75 76 77 78 MENT. ) (in inches), f i-53 1-50 1-52 i 53 i-5i 1-49 i-53 1-52 i-49 BREADTH | (in inches), f I 02 I 03 1-03 i 02 1-03 I OI 1-02 1-02 I 01 PRESSURE (in pounds). DEFLECTION (in inches). O 005 005 005 005 000 005 000 005 ooo 5OO 205 180 195 215 180 205 175 175 21O 525 215 190 205 225 190 220 I8 5 I8 5 225 550 225 200 215 235 200 230 195 195 235 575 235 2IO 225 250 210 240 205 205 250 60O 245 220 235 2^0 22O 25O 215 215 260 O 005 005 OIO OIO 005 OIO OO5 005 005 GOO 245 22O 235 260 22 5 255 215 215 260 625 -260 230 245 -275 235 265 -225 -225 27O 650 275 240 255 290 245 280 235 235 285 675 2 9 2<0 26$ 305 255 295 245 245 300 700 305 260 275 325 26 5 310 255 255 32O O O2O 010 OIO 025 OIO 020 OO5 005 015 700 3*5 265 275 335 265 315 260 255 320 725 345 275 290 275 335 275 265 340 75O 445 290 .300 .... 285 355 285 280 365 775 310 315 .... 305 375 3OO 295 SOO 330 335 325 395 320 315 O 030 020 020 040 025 025 800 .... 340 340 335 405 325 32O 825 365 .... 355 430 355 .340 85O SQO 380 jet 7QO ^^ 875 465 37^ 90O . 400 O O45 9OO 4IO I 925 450 H i BREAKING ) I WEIGHT (in V Pounds). ) 753- | 824 877- : 720 874- 854- 869- 947' 773- i 547 TABLE XXXI. TRANSVERSE STRAINS IN WHITE PINE. LENGTH i 6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ) EXPERI- V 79 SO 81 82 83 84 85 86 87 MENT. J DEPTH \ (in inches). ) 2-11 2-IO 2-05 2-09 2-09 2-08 2-06 2-09 2-II BREADTH \ (in inches). J I 04 1-04 i 03 1-03 1-03 1-03 1-04 1-03 1-03 PRESSURE (in pounds). DEFLECTION (In inches). 1 000 000 000 000 ooo ooo ooo 000 ooo 5O 010 OIO 0!5 015 OIO 015 OIO OIO 015 too O2O 020 025 025 O2O 030 O2O 020 025 15O 030 030 040 035 030 040 030 030 035 200 035 040 050 040 035 050 040 040 040 O 00 5 000 005 005 005 005 ooo ooo 005 200 035 040 050 040 035 050 040 040 040 250 045 045 O6o 050 045 060 045 050 050 30O 350 055 O6O 055 060 O7O O8o 055 065 055 060 070 080 055 065 060 065 055 065 400 070 070 Ogo 075 070 090 070 075 070 O 400 010 005 OIO OIO OIO OIO 005 005 OIO 450 5OO 55O 070 075 -085 070 080 -085 090 IOO no 075 080 090 070 075 085 090 IOO IIO 070 080 090 075 080 090 070 080 085 600 095 095 I2O 095 090 120 IOO IOO 095 105 IOO 130 105 IOO 130 IIO IIO I 'JO O 015 OIO 015 015 015 020 OIO OIO 015 60O 105 IOO -130 105 TOO 130 IIO IIO IOO 650 no no 140 115 no I4O -I2O 120 IIO 700 120 US 155 125 120 ISO -130 130 115 75O 125 125 165 135 130 160 140 140 '1*5 8OO 135 -130 i/5 145 135 170 -150 150 135 548 TABLE XXXL (Continued) TRANSVERSE STRAINS IN WHITE PINE. LENGTH 1.6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ) i EXPERI- V MENT. ) 79 SO 81 82 83 84 I 85 86 87 DEPTH \ (in inches), f 2-II 2'10 2-05 2-og 2-09 2-08 2-06 2-09 2-It BREADTH I (in inches), f I 04 I 04 1-03 1-03 1-03 1-03 1-04 1-03 1-03 PRESSURE (in pounds). DEFLECTION (in inches}. o 020 015 020 O2O 020 030 015 015 O20 8OO 135 130 175 145 135 170 150 150 135 85O 140 140 190 'ISO 'MS 180 160 160 140 90O 150 145 200 160 150 195 170 170. 150 950 155 155 -215 170 160 -2 5 180 180 155 1OOO I6 5 160 -230 185 170 220 190 . -I 9O I6 5 025 020 -035 025 025 035 -O2O 025 O2i; 1OOO I6.S 160 -235 -185 170 220 190 195 I6 5 105O 175 170 -250 195 180 235 200 205 175 11OO I 80 180 275 205 190 250 215 215 I8 5 115O I8 5 190 22O 200 27O 235 230 195 12OO 195 2OJ 240 2IO 295 255 245 205 030 O25 045 030 -060 035 040 035 1200 205 2OO 245 2IO 310 265 255 2IO 125O 210 210 .... 270 22O 340 285 275 22O i 1300 22O 22O ... 305 23O 310 335 230 135O 230 230 .... 390 245 .... .... 245 1/1 on 240 240 260 26o 040 o^o 'OJC O^'s 140O 245 245 .... 270 265 | 145O 26O 260 ^oo 2QO i 150O 28 5 280 .... .... .... 315 155O ^1^ 160 16OO 355 BREAKING ) WEIGHT (in V pounds). ) 1629- 1536. 1150- I383' 1500- I280' 1349- 1303- 1553- 549 TABLE XXXII. TRANSVERSE STRAINS IN WHITE TINE. LENGTH i FOOT BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ] _ . EXPERI- 278 279 28O 281 282 283 284 285J 286 MENT. ) DEPTH 1 (in inches), j 251 253 258 498 502 S3 - 74 8 74 6 747 BREADTH | (in inches). J I -OOO I-OOO I-OOO i -ooo I -OOO I-OOO r -ooo I-OOO I'OOO 1 PRESSURE (in pounds). DEFLECTION (in inches). PRESSURE (/* pounds). DKFLECTION (in inches). PRESSURE (in pounds). DEFLECTION (in inches). ooo ooo 000 ODO 000 000 ooo ooo ooo 1 019 022 016 4 cog on 009 10 007 009 007 1 038 057 044 066 032 048 8 12 0,8 026 O2I 032 018 027 20 30 014 O2I 018 027 014 022 4 076 o38 064 16 034 042 35 40 029 036 029 5 095 no 080 20 043 052 044 50 036 045 37 6 114 132 096 24 o 5 r 062 053 6O 43 054 045 7 '33 '54 . 112 28 059 O72 -062 7O 050 .063 052 8 152 176 -128 32 068 082 071 80 057 072 059 9 17' , "9'3 , 144 36 076 092 079 90 .064 08 1 066 1O 190 -220 160 4O 085 103 088 100 072 090 074 11 209 -242 176 44 094 114 096 110 079 .099 082 12 228 -264 -192 48 102 125 105 120 086 108 08 9 13 247 -286 -208 52 1IO 136 114 13O 093 118 097 14 15 267 I -308 286 -330 225 241 60 iiS 127 I 47 I 57 123 132 140 15O 100 107 127 136 I0 4 112 16 305 | -352 257 64 136 l6 7 141 160 114 -146 119 17 324 '374 273 68 J 45 I ?8 150 170 - 121 127 18 343 -396 290 72 '54 ,89 159 18O 128 165 *35 19 20 362 381 419 442 306 - 3 22 76 80 -.63 172 '99 210 168 176 19O 20O 135 I 4 2 176 143 152 21 400 466 339 84 181 221 -185 210 '49 200 161 22 23 419 438 490 515 356 373 88 92 -191 200 2 3 2 2 44 '94 203 220 230 156 164 2I 3 227 171 181 24 25 '457 477 : 567 390 407 96 100 209 -219 2 5 6 268 213 225 240 250 171 179 247 '95 210 26 497 594 425 104 229 280 237 260 -186 233 27 28 518 443 462 108 112 239 249 294 308 250 262 270 280 194 201 29 30 583 482 505 116 120 258 268 323 338 277 296 29O 3OO 20-) 2l3 31 32 33 606 629 654 .... 11 124 128 132 2 7 8 288 299 354 373 317 310 320 228 239 .251 136 310 340 26 S 140 321 144 332 148 344 152 '357 156 371 160 3*7 550 TABLE XXXIII. TRANSVERSE STRAINS IN HEMLOCK. LENGTH i 6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF J EXPERI- V 88 89 90 91 92 93 94 95 96 MENT. \ DEPTH I (/# inches), f 1-07 r -08 I 07 1-07 I 09 I -10 1-09 I 08 I II BREADTH 1 (in inches). \ 1-07 I 06 I 09 i -06 I 05 i -06 1-07 1-08 1-09 PRESSURE DEFLECTION (in inches). (in pounds'). O ooo OOO ooo 000 000 000 000 ooo 000 25 050 050 030 025 035 025 040 035 040 50 ogo 085 060 050 070 050 080 065 075 75 i 20 125 090 075 100 070 12O 095 110 100 150 165 115 105 135 095 160 125 145 O OIO 005 005 ooo ooo 005 ooo coo ooo 100 160 165 1 20 105 140 095 160 125 145 125 190 210 150 130 -170 1 20 200 155 185 150 225 25O .185 160 2OO 140 240 .185 22O 175 265 295 215 190 230 165 285 220 260 200 300 340 245 220 260 190 330 250 500 OIO 015 005 OOO 005 005 OIO O05 OIO 200 305 345 250 220 265 185 330 250 300 225 340 400 285 2 5 305 210 .380 -28 5 340 250 380 455 315 2SO 235 430 385 275 -420 510 350 310 260 480 .... 430 300 395 345 285 540 475 O 025 005 .OIO 045 040 30O . . .... 400 350 285 570 490 325 .... 390 .... '315 .... 545 35O 350 375 *8* 40O J J 445 O 045 400 46=; 425 *T W D 530 BREAKING WEIGHT (in 292- 277 324 350 234 433 3i3- ,33 348- pounds). TABLE XXXIV. TRANSVERSE STRAINS IN HEMLOCK. LENGTH 1-6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF) EXPERI- > 97 98 100 101 102 103 104 105 106 MENT. ) DEPTH ) (in inches), f 1-56 I -60 1-60 i-59 1.56 I -60 i-54 i'54 1.58 1-58 BREADTH 1 (in inches). ) 1-04 I -06 1-07 1-03 I-OI I -08 1-09 i- IT I -08 1-09 PRESSURE (in pounds). DEFLECTION (in inches). ooo ooo ooo 000 ooo ooo ooo ooo 000 ooo 25 OIO OIO OIO 015 OIO 015 015 015 OIO 015 50 O2O 025 025 030 015 030 030 030 025 030 75 030 035 035 040 025 045 045 045 040 045 100 040 050 050 055 035 060 060 055 055 055 OOO ooo 000 000 000 000 ooo ooo ooo ooo 100 040 050 050 055 035 060 060 055 055 055 125 045 060 060 065 045 075 075 070 070 070 150 055 075 070 080 055 090 090 085 085 080 175 065 085 080 090 065 105 105 IOO IOO 090 200 070 095 090 105 075 115 125 115 115 105 ooo 000 ooo 000 ooo ooo ooo ooo 000 ooo 200 070 095 090 -105 075 115 125 115 115 105 225 080 105 105 115 080 130 140 130 130 120 250 085 115 120 130 090 145 155 145 140 130 275 095 130 130 145 095 160 -170 160 155 145 300 100 140 140 160 105 175 185 175 yo 155 005 000 005 ooo 000 005 005 coo 000 ooo 3OO IOO 140 145 160 105 175 190 175 170 155 325 no 155 155 175 115 190 205 190 185 165 350 I2O 170 170 190 125 2IO 225 205 2CO 175 375 125 180 I 80 205 135 225 240 22O 215 190 400 135 190 190 22O 145 240 260 235 230 200 552 TABLE XXXIV. (Continued) TRANSVERSE STRAINS IN HEMLOCK LENGTH i 6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. ! NUMBER OF ) EXPERI- > 97 98 99 1OO 1O1 1O2 1O3 1O4 105 1O6 MENT. ) DEPTH 1 (in inches). ) 1-56 I -60 I -60 1-59 1-56 I -60 i-54 i-54 1-58 1-58 BREADTH 1 (in inches), ) 1-04 I -06 1-07 1-03 I-OI I -08 1-09 i -ii I -08 1-09 PRESSURE (in pounds). DEFLECTION (in inches). 005 000 OIO 005 005 015 015 OIO OIO coo 40O 135 190 190 220 145 240 265 235 230 200 425 145 205 205 240 150 255 285 255 250 215 45O 150 220 '220 255 160 275 305 270 265 23O 475 160 230 23O 270 170 290 325 285 285 240 500 170 240 245 295 175 310 345 305 305 255 O 005 OIO 015 020 OIO 025 030 020 020 OIO 500 170 245 245 300 180 315 355 310 305 255 525 -180 260 260 340 190 335 380 330 325 270 55O iSS 275 275 200 355 405 350 345 285 575 195 300 290 .... 210 380 .430 370 365 300 600 205 305 22O -400 460 385 315 -OIO O25 015 045 060 .... 030 O2O 600 205 310 .... 225 405 470 .... 390 320 625 215 .... 325 235 435 500 .... 415 335 . 22^ "3AO 24. ^ 465 435 3^0 o5U 675 .240 360 .... 265 .... .... 460 370 70O 26O 380 290 .... .... .... .... 385 O2O O^O 035 TOO 260 ^o^ 3QO 280 4(X 1 Ah T "iO 2Q^ 425 . 360 450 BRFAKING ) WKIGHT (in r 777- 575- 700- 548- 727- 651- 650- 600- 687- 800 pounds). ) 1 1 553 TABLE XXXV. T R A N S VE R S E STRAINS IN HEMLOCK LENGTH 1-6 FEET BETWEEN BEARINGS. See Arts. 7O4 and 7O5. NUMBER OF ) EXPERI- V 107 108 109 no 111 112 113 114 115 MENT. J DEPTH | (in inches), j 2 -O2 2 02 2-00 2-03 2-01 2-01 1-99 2-00 2-03 BREADTH | (in inches), j 1-03 1-05 I-0 3 1-04 1-05 1-04 1-02 1-03 1-05 PRESSURE (in founds). DEFLECTION (in inches). O ooo 000 oco ooo 000 000 ooo ooo ooo 5O 015 OIO OIO OIO 015 015 O2O OIO OIO 100 025 020 O2O 020 025 030 030 020 025 15O 035 035 030 035 035 045 045 035 35 2OO 045 045 045 050 045 055 055 045 045 O ooo 000 000 ooo ooo ooo 000 000 ooo 200 045 045 045 050 045 055 055 45 045 250 '055 060 55 065 055 -.065 065 055 55 300 35O 065 075 070 085 065 075 075 090 070 085 075 085 075 090 065 75 065 075 40O 085 095 083 IOO 095 095 IOO 090 085 40O 005 085 ooo 095 CDS 085 005 IOO 005 095 005 9S ooo IOO 000 090 000 085 45O 095 105 095 115 105 105 no IOO IOO 500 105 I2O 105 130 120 "5 . I2O no no 55O "5 130 "5 MS 135 125 130 I2O 1 20 6OO 125 145 125 155 'MS 135 140 130 130 O 005 005 005 005 005 005 005 005 005 60O 125 >45 125 '55 140 130 130 650 135 160 *3S 170 160 'MS 150 140 140 70O 145 170 150 185 170 165 *5S 150 75O 160 185 165 200 . 185 165 175 165 165 800 170 200 175 220 205 175 190 180 180 O OIO 005 005 OIO 005 OIO OIO 005 OIO 80O 85O 170 185 200 2I 5 III! 220 2 4 205 225 175 185 190 200 180 190 180 '95 9OO 200 230 195 260 245 215 205 2,0 95O 215 2 S 2IO 28 5 265 2,0 235 215 225 1OOO 230 275 220 35 220 2OO j .... 2 5 O 1000 025 240 025 280 OIO 22 S .... 025 310 015 220 02 5 270 020 255 !().,() 255 '3 I S 240 235 280 1100 280 370 260 115O 320 290 BREAKING ) WEIGHT (in > founds). ) "54- ,1,1. i, Si- 991- 1049- 1099- 1036- 98S- 1075- 554 TABLE XXXVI. TENSILE STRAINS IN GEORGIA PINE. See Arts. 7O4 and 7O6. NUMBER OF ) i | EXPKRI- v 116 117 118 119 12O 121 122 123 124 MENT. ) DIAMETER { (in inches). \ 355 355 350 355 355 345 345 355 365 BREAKING ) Less than Less than WEIGHT (in > founds). ) 2005- I6OO- I300- 2152- 1400- 1924- 1091- 1306- 1268- TENSILE STRAINS IN LOCUST. NUMBER OF ) EXPERI- V 125 ! 126 127 128 129 13O 131 132 133 MENT. ) DIAMETER I * (in inches'), f 355 345 305 305 .300 300 300 300 300 BREAKING 1 More than WEIGHT (in V pounds). ) "37- 2265- 24OO- I592- 2074- 1561- 2131- 1799- 2395- TENSILE STRAINS IN WHITE OAK. NUMBER OF ) EXPERI- > 134 135 136 137 138 139 140 141 142 MENT. ) DIAMETER I (in inches). | 355 3^5 345 355 305 300 305 .300 300 BREAKING ) WEIGHT (in v founds). ) 1908- 1303- 1182- 2375' 1700- 1442- 1003- 1319- 2205- 555 TABLE XXXVII. TENSILE STRAINS IN SPRUCE. See Arts. 7O4 and 7O6. NUMBER OF ) EXPERI- \\ 143 144 145 146 147 148 149 15O 151 MENT. ) | ! DIAMETER | (in inches). \ . 305 305 300 305 305 305 355 365 360 BREAKING J WEIGHT (in V pounds). ) 1573- I402- 1560- 1368- I385- 1533- 1882- 2078- 1600 TENSILE STRAINS IN WHITE PINE. NUMBER OF ) 1 EXPERI- V 152 153 154 155 156 157 158 159 160 MENT. ) 1 (in inches), f 395 365 365 360 -365 355 350 365 360 BREAKING ) WEIGHT (in V pounds). ) 1363- H57 1127- 1316' '431 1487- 1192- 1024- 1400 TENSILE STRAINS IN HEMLOCK. NUMBER OF ) , EXPERI- >- 161 162 163 164 165 MENT. ) DIAMETER } (in inches), f 365 355 345 36o j -355 BREAKING ) WEIGHT (in V fiounds). \ 645- 897- 864- 999. 863- 166 167 168 169 355 350 335 355 726- 895- 809- 977- 556 TABLE XXXVIII. SLIDING STRAINS IN GEORGIA PINE. See Arts. 7Q4- and 7O6. ^ ' NUMBER OF ) \ EXPERI- > 170 171 172 ! 173 | 174 175 176 177 178 MENT. ) ! DIAMETER ) (in inches), f 525 520 530 530 520 520 525 520 530 LENGTH I (in inches), j 1-065 Broke in two. I-O2O I -OIO Broke in two. 1-040 1-020 I-OI5 1-050 BREAKING ) WEIGHT (in > 1546- ,1295- I4II- 1571- I28l- 1347- I520- 1401 1247- pounds). ) SLIDING STRAINS IN LOCUST. NUMBER OF ) EXPERI- V MENT. ) 179 180 181 182 183 184 185 186 187 DIAMETER I (in inches). \ 530 530 535 525 525 530 535 525 525 LENGTH I (in inches), f "735 730 715 745 Broke in two. 760 730 715 760 BREAKING ) WEIGHT (in V 1490- 1236- 1533- 1192- I492- I758- ! I403' I33I- 1483- pounds). ) SLIDING STRAINS IN WHITE OAK. NUMBER OK 1 | EXPERI- > 188 i 189 19O 191 192 193 194 195 196 MEXT. ) DIAMETER ) (in inches), f 530 525 535 540 535 530 530 535 530 LENGTH i (in inches). \ 730 755 740 750 750 740 725 725 730 BREAKING ) WEIGHT (in V 1308- 1801- 1834- 1502- 1701- 1359' 1667- 1321- 1399' pounds). ) 557 TABLE XXXIX. SLIDING STRAINS IN SPRUCE. See Arts. 7O4 and 7O6. NUMBER OF ) EXPERI- > MENT. ) 197 19S 199 200 201 202 2O3 204 205 DIAMETER j (in inches), j 565 535 550 525 550 550 545 550 550 LENGTH \ (in inches). ) I-OIO 990 I-OIO I-OIO 1-030 1-020 1-005 I-OIO -990 BREAKING j WEIGHT (in > founds). ) 988- 770- 1130- 882- 927- 976- 1043- 838- 902- SLIDING STRAINS IN WHITE PINE. NUMBER OF ) 1 i EXPERI- v 206 207 208 209 : 210 211 212 213 214 MENT. ) I ! DIAMETER ) (in inches), f 540 545 555 545 545 545 550 545 545 LENGTH 1 (in inches), f 995 I-OOO 990 I-OIO i -025 1*005 i-oio i -040 '995 BREAKING ) WEIGHT (in V pounds). ) 730- 907- 792- 803- 842- 8co- 881- 852- 885- SLIDING STRAINS IN HEMLOCK. NUMBER OF ) I EXPERI- H 215 216 217 218 219 220 221 222 223 MENT. ) DIAMETER 1 (in inches), f 540 540 545 '540 530 540 540 535 530 LENGTH ) (in inches'), f 995 I-OIO 995 Broke in two. 1-025 1-015 '995 I -010 99 BREAKING ) WEIGHT (in V 607- 702- 620- 796- 700- 674- ss&. 627- 530- founds). ] ! 558 TABLE XL. CRUSHING STRAINS IN GEORGIA PINE. See Arts. 7O4 and 7O7. NUMBER OF ) EXPERI- V MENT. ) 224 225 226 227 228 229 230 231 232 DIAMETER ) (in inches), f 515 515 520 52O 505 515 510 500 515 LENGTH ) (in inches), f 1-035 1-025 I -040 1-035 T-035 505 515 505 510 BREAKING ) WEIGHT (in V pounds}. \ 2010- 1878- 2061 1735- 2304- 2002- 1845- I705- 2141- CRUSHING STRAINS IN LOCUST. NUMBER OF ) EXPERI- > MENT. ) 233 234 235 236 237 238 239 240 241 DIAMETER j (in inches), j 520 520 520 525 530 520 525 515 520 LENGTH ) (in inches), f 1-055 1-020 1-035 1-045 -490 515 500 490 495 BREAKING ) WEIGHT (in \ pounds}, ) 2338- 2 39 I- 2547- 2539- 2695' 2500- 2495' 2374- 2672- CRUSHING STRAINS IN WHITE OAK. NUMBER OF ) EXPERI- V MENT. ) 242 243 244 245 246 247 248 249 25O DIAMETER | (in inches), j 525 530 520 530 525 520 525 520 515 LENGTH | (in inches}. \ 1-035 1-035 1-035 1-030 1-035 505 500 485 515 BREAKING ) WEIGHT (in > pounds), j 1546- 1978- 1992- 1455- 1989- 1650- 2116- I387- 1376- 559 TABLE XLI. CRUSHING STRAINS IN SPRUCE. See Arts. 7O4 and 7O7. NUMBER OF ) EXPERI- V WENT. ) 251 252 253 254 255 256 257 258 259 DIAMETER \ (in inches), f 535 535 535 535 535 530 540 530 530 LENGTH 1 (in inches), f 1-035 1-025 1-040 i -030 490 480 485 495 490 BREAKING ) WEIGHT (in V Jouneis). } 1692- I7I5- 1611 1633- 1871- 1818- 1812- 1855- 1832- CRUSHING STRAINS IN WHITE PINE. NUMBER OF ) I i 1 1 EXPERI- V 26O 261 262 263 264 265 i 266 267 268 MENT. ) 1 1 DIAMETER 1 (in inches}, f 540 525 535 515 530 535 525 -510 540 (in inches), f 1-035 1-035 i -040 1-040 1-030 495 49 50 495 BREAKING ) I WEIGHT (in > pounds). ) 1454' I536- 1473- I322- 1297- 1503- 1624- 1353- 1540- CRUSHING STRAINS IN HEMLOCK. NUMBER OF ) 1 EXPERI- U 269 MENT. ) 1 270 271 272 273 274 275 276 277 DIAMETER [ (in inches) ) 520 520 525 530 530 520 520 525 -520 LENGTH 1 (in inches), f 1-035 1-030 1-030 1-030 480 525 520 495 490 BREAKING ) WEIGHT (in \ founds). \ II37' 1178- 1130- 1150- 1150- 1334' 1290- I3I7- 1320- 560 TABLE XLII. TRANSVERSE STRAINS. BREAKING WEIGHTS (in pounds) PER UNIT OF MATERIAL = B. See Arts. 7O4 and 7O5. i w w w M ^ 2 2 2 ^ 2; ^ jj s/ PU: ' w ~ t&~r+ Q. - M C/5 M **' M U M Cj ^ I-JH ** M M o 3 x U X W x 5 x S X 5 x w x W x W X J X O-. P4 - J" ^i FW - CL, ^ r" ^ H &; w ~ i w ~ w^ W O ^ J ^ * ffl i-U M s 950- 664- 686- 576- 604- 540- 578- 504- 563. 381. 491- 439' 1023- | 555- 533' 6 54 - 650- 569- 616- 569- 536- 358- 339" 415- 758- 1406- 660- 533- 574' 627- 480- 590- 425' 415- 409- 459' 95i- 1286- 626- 694- 598- 642- 576- 482- 492- 461- 337- 370- 945" 1283- 476- 632- 538- 552- 542- 595- 533- 300- 473- 39 6 ' 1014- 1156- 567- 637- 652- 630- 566- 609. 460- 540- 377- 418- 993- 1342- S3 6 ' 702- 660- 637- 564- 582- 489- 394- 402- \ 410- 785- 530- 501- 593- 614- 6 54 - 619- 643- 463- 302- 365- 383- 949' 1212- 664- 627. 592- 572- 639- 552- 542- 4i5- 408- 398- I28l- 529- 722- 642- 576- 470- 952- AVERAGE BREAKING WEIGHTS, = B. 930- 1061- 578- 637- 612- 600- 576- 570. 500- 396- 407- 4IO' 561 TABLE XLIII. DEFLECTION. VALUES OF CONSTANT,'^. See Arts. 7O4 and 7O5. w - 1 M I'M O " H X 'RUCE. 'x i". & w 2 w x H - H K M w x i? c? 1 \ M K-M w M Cfl%H x M M S M ffi M * ^ * 5155- 4983- 2599" 3649- 3329- 2577' 3088- 2746. 2484. 2083- 3112- 2316- 6302- 4645- 2033- 3640- 2909. 22c; 9 - 3378- 3191- 2658- 1859. 1986- 1958. 5199- 5616- 2360- 2215. 2679 ' 2962- 2806- 2891- 2130- 2667. 2077- 2386- 5555- 5000- 2057- 3867. 2965 ' 3105- 3124- 2624- 2489. 2868- 2940- 1822- 5498- 5442- 1704- 3384- 2766- 2539- 2940- 3176- 2581- 2259- 3052- 1988- 6007- 4998- I837- 2932- 329I- 3382- 2850- 2932- 2040- 3056- 1606- 2190- 6007- 5239- 1907- 4004. 3302- 3I27- 3257- 3101- 2371- 1847. 1660- 2184- 4807- 4312- 1842- 3572- 3344' 3I9 1 ' 3267- 3225- 2323. 2441- 1725- 2294- 6336- 5239- 2253- 3678-: 3714- 2176- 3338- 2919. 2509- 1895. 1683. 2152- 5033- I930- 39 6 7- 3387- 2682- 1864. 4952- AVERAGE VALUES OF CONSTANT, F. 5652- 5042- 2052- 349 1 ' 3I 6, 2804- 3Il6- 2978- 2398- 2331- 2170- 2143- 1 562 TABLE XLIV. TENSILE STRAINS. BREAKING WEIGHTS (in pounds) PER SQUARE INCH OF SECTIONAL AREA, == See Arts. 7O4 and 7O6. GEORGIA PINE. LOCUST. WHITE OAK. SPRUCE. WHITE PINE. HEMLOCK. 20257- 11487- 19277- 21530- 11123- 6164- / 24229 21790- 12453- 12644- 23995. 19189- 22069- 18724- 11057- 10771- 12929- 9062- 9242- 9815- 21742- 14144- 20582- 29341- 22084- 23268 20400- 18957- 20982- 13676- 15023- 8719- 7335- 11671- 30147- 13728- 19014- 12389- 9302- I3I95- 25451- 18660- 19860- 9786 9178- 12118- 33882- 31194- 15719- T3754- 9871- AVERAGE WEIGHTS PRODUCING RUPTURE, = T. 16244- 24801 19513- 19560- 12279- 8743- 563 TABLE XLV. .SLIDING STRAINS. BREAKING WEIGHTS (in pounds) PER SQUARE INCH OF SLIDING SURFACE, = G. See Arts. 7O4 and 706. GEORGIA PINE. LOCUST. WHITE OAK. SPRUCE. WHITE PINE. HEMLOCK. 880- 1218- 1076- 55i- 433' 360- 1017- 1447- 463- 530- 410- 831- 1275- 1474. 647- 459' 364- 934" 970- 1181- 530- 464- 1349- 521- 480- 410- 793" 1389- 1103- 554' 465- 392- 904- ii43- 1381- 606- 505- 329- 845- 1129- 1084- 480- 479' 369- 7i3- 1183- 1151- 527- 520- 322- AVERAGE RESISTANCE TO RUPTURE PER SQUARE INCH, = G. 843- 1165- 1250- 542- 482- 369- 56 4 TABLE XLVI. CRUSHING STRAINS. CRUSHING WEIGHTS (in founds) PER SQUARE INCH OF SECTIONAL AREA, = C. See Arts. 7O4 and 7O7. GEORGIA PlNE: LOCUST. WHITE OAK. SPRUCE. WHITE PINE. HEMLOCK. 9649- 11009- 7142- 7527- 6349- 5354- 9015- 9705- 8170- 11259- H993- 11729- 8966- 9380- 6595- 7629- 7166- 7264. 7095- 6552- 6346- 5547- 5220 5240- II503- 12216 9188- 8323- 5879- 5213- 9611- 11772- 7769- 8240 6686- 6281- 9032- 11525- 9775- 7912- 7502- 6074- 8683- 11396- 6531- 8408 6623- 6084- I0278- 12582- 6606- I 8304- 6724- 6216 AVERAGE RESISTANCE TO CRUSHING PER SQUARE INCH, = C. 9516- 11720- 7995- 7864- 6640- 5692- 565 DIRECTORY, DIGEST OF THE PRINCIPAL RULES. BELOW may be found the numbers of such formulas, arti- cles, figures and tables as are particularly applicable in any given problem. By reference to these, the rules needed in any certain case, occurring in practice, may be more readily found than by either the index or table of contents. LEVERS WOOD. ' Strain at wall, " any point, . Figs. 27, 28, 33, (4.), Size when at the point of rupture, " to resist rupture safely, f Weight, . .* . a5 I Length, | -I Breadth, E Depth, Deflection, . . . . . .(6.) (19. \ (36.} . (123.) (127.) . (128) (129) (121.) ~o >% I I "5 3 DIRECTORY. 567 f Strain at wall, . . . (75.) " " any point, . . Fig. 46, (76.) Size when at the point of rupture, . . (18.) 1 ' to resist rupture safely, . . (20.), (77.) " at any point to resist rupture safely, (77.) [ Shape of lever, Fig. 47 Weight, . (186.) Length, (187.) | \ Breadth, (188.) Depth, (189.) _ Deflection, . . (140) Strain at wall, Figs. 45, 51 " any point, . .Figs. 45, 48, 50, 51 Size " " " Art. 227 Shape of lever, Figs. 31, 49 Depth at any point, (80.) LEVERS ROLLED-IRON. Load at end. Flexure. Weight, . . . . Size, . . . . (224), Load uniformly distributed. Flexure. Weight, ... (230) Size,. ;68 DIRECTORY. 1 SINGLE BEAMS WOOD. ' Strain at middle, . . . " " any point, .... Size when at the point of rupture, (9.), (11.) 44 to resist rupture safely, " at any point to resist rupture safely, | -I Weight, Length, '.".*.". Breadth, Depth, *. Y '. Constant B, . . . . i Pressure on each support, ['Weight, .... . . g j Length, . . . . J Breadth, . \ . . Depth, Deflection, . (9.) Fig. 29 (12.),(14.) . .(21.) . (37.) . .(13.) .(11.) ..(10.) (<?), (4-) (122.) (125.) (126.) (120.) E J |J b X " Strain at middle, . ... (44-) " " the load, Art. 190 - any point, . Figs. 34, 35. (44-), (45.) Size when at the point of rupture, . . . (16.) " to resist rupture safely, (23.), (46.), (47.), (48.), (49.), (50.) El 3 I f f Strain at middle, 44 " any point, . | i Size when at the point of rupture, to resist rupture safely, I-EH | Shape of beam, .... ( Pressure on each support, . ['Weight, g Length, . . .... I -j Breadth, Depth, Deflection, .... Art. 59, (72.) Fig. 42, (71.) . . (17.) - Figs. 43, 44 . (3.) t (4) . (131.) ? .) . (135.) DIRECTORY. 569 f r Strain at any point, Figs. 36, 37, (53.), (54.), (55.) " locations of weights, Art. (53, (51.), (52.) Size to resist rupture safely, (30.), (31.), (56.), (57.) . f f I Strain at any point, Fig. 38, (61.), (62.), -i J 2 1 W o rt -< I o " locations of weights, (58.), (59.), (60.) I Size to resist rupture safely, (65. \ (66.), (67.) ~ f Rupture. Strain at any point, Figs. 52, 53, (<$-*) (95-), 1 1 31 I Rupture. Size to resist safely, | I Flexure. " " <4 "... (189. \ (193.) Loaded ( Rupture. Strains, Figs. 39, 4, 41, Arts. 196 to 203, 210, 211 SINGLE BEAMS ROLLED-IRON. f General rule, ....... (216.) t | Weight, (217.) | -! Length, (218.) E Deflection, . .... (219.) ( Moment of inertia > , (220.) Load at any point. Flexure. Weight, .... " '! " - Size, . . . (221.), (222.) Load uniformly distributed. Flexure. Weight, '. . . (228.) Size, 5/0 DIRECTORY. I FLOOR TIMBERS WOOD. General rule, ....... Dwellings, assembly rooms, etc., . . . General rule, ....... f General rule, .... (142.), I Distance from centres, . I. to IV., (144), \ Length, ....... \ Breadth, ..... . . ( Depth, . . ...... f General rule, ..... (U8.), \ Distance from centres, V. to VIII., (150.), Length, ....... Breadth, ....... I Depth, . ..... Solid floors of wood, XXL, (310.), (311.), (25.) (141-) (143.) (306.) (145.) (146.) (307.) (151-) (152.) (153.) (312) f Rupture. General rule, ! f General rule, I j | J Dwellings, etc., j UH j First-class stores, . . . (27.) . . . . (156.) IX. to Xn., 382, (308.) XIII. to XVI., 383, (309.) With one header (29.) " two headers, (32.), (33.), (34.), (35.), (92.). (93.) " three " .... (97.),(106.) g - [ General rule, .... (157.), (161.) || j Dwellings, etc., . . (158.), (162.) I First-class stores, . . . (159), (163.) ( General rule, . (164.). (107.), (170.), (174), (179.), (183.), (186.) rgl , Dwellings, etc., (165.), (168.), (175.), (180.), g | (184\(187.) | First-class stores, (166),(169.\ (176), (181.), (185.), (188.) S!j [ General rule, . Figs. 55, S^(190), (194) 1 1 i Dwellings, etc., . . (191.), (195.) [ First-class stores, . . . (192.), (196.) Girders. Rupture. General rule, . . . Art. (37 X * o DIRECTORY. 571 FLOOR REAMS ROLLED-IRON. | f ,j f General rule (234.) Dwellings, etc., . . . XVIII., (236.\ J i E | First-class stores, . XIX., fa I 2 f d f General rule, ..... | Dwellings, assembly rooms, etc., . . . (248.) E [ fe I First-class stores, . . . . . (249.) f f ,.: f General rule, (250.) gl -\ Dwellings, assembly rooms, etc., . (251.) ~ L First-class stores, . . . . (252.) f General rule, . .... (253.) 2% j Dwellings, assembly rooms, etc., ' 1 First-class stores, . (255. \ (257.), (259.) f General rule, .... Art. 531 (A I . J Dwellings, assembly rooms, etc., . (260. )\ First-class stores, . FRAMED GIRDERS. Proportionate depth, (294-) Number of bays, ........ (295.) Strains in a framed girder, .... Pigs. 93, 94 " diagonals, (296.) Tensions in lower chord, ...... (297) Areas of cross-section in lower chord, . . . (299) " " " " upper " . . (301), (303) Unsymmetrical load, divided between the two supports, Figs. 96, 97, (304.), (305.) 572 DIRECTORY. TUBULAR IRON GIRDERS. Load at middle. Area of flange, . (264), (65) " "any point. " " " (266.) >s% f General rule, " " " . . . . (267 .) o o 1 \ Banks and assembly rooms. Area of flange, (274-) J 'l| ! First-class stores, " " t4 (275.) Thickness of web to resist shearing, . . . (268.) Weight of girder, (270.), (271.) Economical depth of girder, .... (276.), (277.) CAST-IRON GIRDERS. Load at middle. Breaking weight, (278) - Safe area of flange, . (279.) " " any point. Breaking weight, .... (281.) Safe area of flange, . . . (282.) Two concentrated loads. Safe area of flange, . (285), (286) Safe area of flange at middle, . . . (280.) " " " " " any point, . . . (283.) " depth at any point, . . . . (284.) Arch girder, safe area of tie-rod, . . . (287.) safe diameter of tie-rod, . (288.), (289.) . Brick arch, " " < t (290.) ROOF TRUSSES. Comparison of designs, Art. 658 Strains derived graphically, .. . Arts. 660 to 668, 679 Horizontal and inclined ties, . Fig. 125, Arts. 669 to 671 Designing a roof, Arts. 672, 676 Load upon a roof, . . . . Arts. 673, 674, 675 Load upon each supported point in a truss, . Art. 677 " " the tie-beam, Art. 678 Measuring the strains, as in force diagram, . Art. 680 Arithmetical computation of strains, . . Art. 681 Dimensions of parts suffering tensile strains, Art. 682 " " " " compressive strains, Arts. 683 to 687 DIRECTORY. 573 FLOOR-ARCHES TIE-RODS. Horizontal strain, . (240.) Uniformly distributed load, area of rod, . . . (241) Load per superficial foot, " " " , . . (242.) Banks, assembly rooms, etc., " " " . . . (243.) " " Diameter of rod, . (245.) First-class stores, " " . . (246.) " " " area of rod, ; (244) SHEARING. With compound load on lever, . ... (38.) " load at end of lever, (39.) " on beam, (40.) Nature of the strain, . . Fig. 30, Arts. 172, 173, 174 Web of tubular girder, . . . . . . (268.) PROMISCUOUS. Bridle irons, for headers, (28.) Bearing surface of beam on wall, .... (41-) Shape of beam and lever, . . Figs. 31, 43, 44, Art. 178 " depth at any point, . . . (?4) Cross-bridging, .... Chap. XVIII., (201.) Deflection illustrated, * Figs. 57 to 64 Moment of inertia illustrated, . . . Figs. 69 to 72 Forces in equilibrium illustrated, . . Figs. 81 to 84 Diagrams of forces illustrated, . . . Figs. 85 to 88 Force diagrams, . . . Chapters XXII. and XXIII. Building materials, weights of ... Table XXII. INDEX. PAGE American House Carpenter, sliding strains 506 " manufacture of rolled iron -beams, . . 313 " woods, constants (or, 499 " experiments on 504 " wrought-iron, constant for, 499 elasticity of, 232 Anderson, experiments made by Major 500 Angle irons in plate beam, 312 Approximate formulas discussed, 183 " value of resistances, 226, 227 Arch, area of cross-section of tie rod of floor 347 Arches and concrete floors, weights of, 340 " for floors, general considerations, 345 tie -rods for brick 346 " where to place tie-rods in brick 348 Arched girder of cast-iron, and tie-rod 396 " substitute for iron, 398 Architect, his liability to err 28 tables save time of the, 495 too busy to compute by rules, 495 Architect's knowledge of construction 27 Area of cross-section, resistance, 31 " of tie-rod of floor arch, 347 Arithmetical computation of strains in truss, 486 progression, the sum, 147, 148 series , 151 " coefficients form an, 226 Arithmetically computed strains 168 Ash, resistance of, ... 120 INDEX. 575 PAGE Assembly halls, formula for solid floors of, 502 rolled-iron beams for, 498 rooms, strains in, the same as in dwellings, 88 " and banks, load on floors of, 340, 341 " " load on floors of, 88 " " " " tubular girders for, 380 " " rolled-iron beams for, 495 " " " " Table XVI1L, 526, 527 " " carriage beams with two headers and one set of tail beams, for 358 " rolled iron carriage beams with two headers and two sets of tail beams, for, 354, 356 " rolled-iron carriage beams with three headers, for, 360, 362 " rolled-iron headers for floors of, 349 " rule for floors in, 261 " " " tubular girders for, 380 tie-rods for floor arches of, 347 Auxiliary formula for carriage beams, 193 Baker, Strength of Beams, Columns and Arches, 446 " " " " " " " ratio by, ...... 382 '* formula for posts, 446 " on compression of materials, 446 Banks, formula for solid floors of, 502 " load on tubular girder for 380 " rolled-iron beams for, 495, 498 " Table XVIII. , 526, 527 " " carriage beams with two headers and one set of tail beams, for, 358 " " carriage beams with two headers and two sets of tail beams, for, 354, 356 " carriage beams with three headers, for 360, 362 " headers for floors of, 349 " tie-rods for floor arches of, 347 " rule for tubular girders for, 380 load on floors of, 340, 341 Barlow's constants for use in the rules, 499 " experiments on woods, 233 " expression for elasticity, 232 Bays in a framed truss, number of, 426, 428 Beam and lever compared, 244 576 INDEX. PAGE Beam and lever compared, deflection in, 237 " " " their symbols compared, 49 " device for increasing the strength of, ... 402 " distributed load on rolled-iron 337 " ends shaped to fit bearings, 122 " load for a given deflection in a, 245 " of economic form, 163 " " equal strength, 163 " rules for dimensions of deflected 248 " shaped as a parabola, 124 " values of U, /, b, d and & in a, 253 " " W, /, b, d " 6 " " . 248 " with load distributed, rules for size of, 253 Beams, formula for deflection of, . 229 " general rule for strength of, 92 " of dwellings, general rule for strength of, 89 " " wood, their weight 79 " . " warehouses to resist rupture, 260 " comparison of rolled-iron, plate and tubular, 367 " should not only be, but also appear safe, 211 strains in, graphically expressed, 177 Bearing surface, 122 " of beams on walls, 121 Bearings, beams shaped to fit, 122 Bending, a beam is to resist 211 and appearing dangerous, beam safe, yet 235 " in good floors far within the elastic limit, 239, 243 its effects on the fibres, 35 " moment of inertia, resistance to, 314 rafter to be protected from, 479 " resistance to 221 Bent lever, equilibrium in, 42 Bow-string iron girder, 396 " " " " substitute for, 398 " " " " unworthy of confidence, 396 Bow, Economics of Construction, 402,418,425 " has written on roofs, 459 Braces in truss, dimensions of, 490, 491 Breaking and safe loads compared, 68 " load of unit of material, 69 INDEX. 577 PAGE Breaking load, the portion to be trusted, 69 '* weight, 267. " " compared with safe weight, 235 " " index of, 51 " " per inch sectional area, tensile, Table XLIV., . . . 563 " " " " surface, sliding, Table XLV., ...... 564 " " " unit of material, transverse, Table XLII., . . . 561 Breadth from given depth and distance from centres 92 " in first-class stores, 265, 266 " its relation to depth, 33 " of beam, rule for 248, 249 " " " in dwellings, 262, 263 " " " with distributed load, rule for, 254 " " header, rule for, 271 " lever " " . . . 250, 256, 257 " proportioned to depth, rule for 73 Brick arch a substitute for iron arch girder 398 " " for floor, rate of rise, 346 " " less costly than cast-iron arch, 399 " arches and concrete filling, 345 " for floors, general considerations 345 " " tie-rods for, 346 " where to place tie rods in, 348 Bridge, greatest load on, 80 Bridges, Conway and Menai Straits tubular, 367, 368, 378 Bridged beam, resistance of a, 304 Bridging causes lateral thrust, 303 " for concentrated loads, 88 " floor beams 302 " in floors tested, ... 303 " increased resistance due to, . - .... 310 " measure of resistance of, 304 " number of beams resisting by, ... . 309 principles of resistance by, . 304 " useful to sustain concentrated loads, . . 309 Bridle iron and carriage beam 98, 195 " " load upon a, * 98 " " rule for a, 98, 99 " " to be broad, 99 Britannia and Conway tubular bridges 328, 368 5/8 INDEX. * PAGE Buckling or contortion of a tubular girder 377 Building materials, weights of, Table XXII., 533,534,535 Buildings require stability, 27 " requisites for stability in, 28 Burbach, large rolled-iron beams from 313 Buttresses to support roofs without ties 459 Calculus and arithmetic compared, 318, 320, 321 " scale of strains 161 " applied, result by the , 323, 325 " coefficient defined by the 227 " strain by distributed load, 157 " " defined by differential 180, 18.;. " strains in lever by differential 168 Cape's Mathematics, forces shown in, 404 references to, 160, 164, 169 Carriage beam and bridle irons 98 " " " headers 94 " auxiliary formula, 193 " definition, 95 " for dwellings, precise rule, 273, 285 " " first-class sto.res, precise rule, 275, 282, 286 " ' formula not accurate, 183 " load on a, 98, 107 " of equal cross-section, 103 " precise rule, h greater than n, 281 " " " " h less " n 280 " special ruleSj 281 " with one header, rule, 99 " " ." . " " rolled-iron 351 " " " for assembly rooms, rolled-iron . . . 351 " banks, rolled-iron 351 " " " " " " dwellings 272 " . " " " rolled-iron 351 " first-class stores, 273 " rolled-iron . . . 352 " two headers, 101 " " - " for dwellings, precise rule, 292 " 4< " first-class stores, precise rule, . . 292 " " equidistant headers, precise rule, 287 " for dwellings, precise rule, 289 INDEX. 579 PAGE Carriage beam, with two equidistant headers, for first-class stores, precise rule 289 " headers and one set of tail beams, 106 * " " 'I " " " " " " precise rule, . 283 " " " " equidistant headers and one set of tail beams, precise rule, 290 * " " " headers and one set of tail beams, for dwellings, 277 rolled-iron 358 " " " " headers and one set of tail beams, for first-class stores 278 " " " " headers and one set of tail beams, for first-class stores, rolled-iron 359 " " " headers and two sets of tail beams, . 104, 192, 194 c it . . " ... precise rule, 279 " " " " ' " rolled-iron . 353 ) 275 precise rule, 282 " " " " headers and two sets of tail beams, for dwellings, rolled-iron 354, 356 " " " " headers and two sets of tail beams, for first-class stores . 276 " " " " headers and two sets of tail beams, for first-class stores, rolled-iron 355, 357 " " " three headers 195, 196, 197 " " for dwellings, rolled iron . , . 360, 362 " " " " " " first class stores, rolled-iron 361, 364 " " " " the greatest strain at middle header, . 297 " " " " outside " . 294 " " ' " " " middle >; for dwellings 298 " " " " headers, the greatest strain at outside header, for dwellings, 295 " " ' " headers, the greatest strain at middle header, for first-class stores, 299 " " " " headers, the greatest strain at outside header, for first-class stores, 295 " " " headers and two sets of tail beams, . . . 200,207 Cast-iron, compression and tension in, 387 S8O INDEX. PAGE Cast iron resists compression more than tension, 45 44 " superseded by wrought-iron, 386 beam, load at middle, Hodgkinson, 383 44 " arched girder with tie-rod, 396 " " ' 4< tie-rod for, 396 " " 4 ' 4t form of web of, 392 " " ' " for brick wall with three windows, ..... 394 * 4 4< 4< 4 ' load at any point of, rupture, 390 - " " middle of, 389 " " proportion of flanges of, 386 44 " 44 " safe distributed load, effect at any point on, . . 391 " " " " safe load at any point on 391 44 44 ' 4 " two concentrated weights on 392 44 " girders, chapter on, 386 Ceiling of room plastered, 303 44 to be carried by roof truss, weight of, 481, 483 44 weight of, .......'-. 78 Cement grout for brick arches of floors, 345 Centennial Exposition, rolled-iron beams at, 313 Centre of gravity, load concentrated at the 60 Centres, distance from, 91 Cherry, resistance of, . . . . 120 Chestnut, 4f " 120 Chord, framed girder with loads on each 433 44 of framed girder, allowance for joints, etc., in, 445 area of uncut part of, 444 44 strains in lower 439 44 " upper 440 41 and struts of framed girder, upper 448 " compression in upper 445 Chords and diagonals, gradation of strains in 432, 435 of framed girders usually of wood 444 Civil Engineer and Architects' Journal, 82 Clark, moment of inertia, by Edwin 328 Clark's formula only an approximation 328 useful in certain cases, 330 Clay has but little elasticity, 211 Coefficient of strength for tubular girder, 368 Coefficients in rule for floors of dwellings, 261, 262 44 4t 4< " 4t " first-class stores 264, 265 INDEX. 581 PAGE Components of load on floor, 339 Compound load, assigning the symbols, 187 44 " dimensions of beam, 187 " " general rule, 199 44 " greatest strain from, T82 44 " maximum moment, 188 " strain analyzed, 178 44 " strains and sizes, 171 44 on floors, , . 339 " " " lever, the effect of, 171, 174 strains graphically expressed, 177 Compressibility of fibres 37 Compression balances extension, 45 dimensions of parts subject to 490 graphically shown, 115 resistance to, 45 " and extension of fibres, strength, 35 " " " summed up, 228 44 4 ' tension, fibres resisting, 403 " 44 " of cast-iron, 387 " " 44 rupture by, 313 of fibres at top of beam, 42 44 44 struts, rule for, . 447, 449 44 " application of rule, 447 in struts and chord of framed girder, 445 Rankine, Baker and Francis on, . 446 44 Tredgold and Hodgkinson on, 446 Compressive and tensile strains, 408 strain in rafter increased, 474 Computation by logarithms, example of, 311 4 ' of moment of inertia, 315 strains in framed truss by, 416 44 to check graphic strains, 132 Concave side of beam, fibres compressed at, 37, 45 Concentrated and half of distributed load equal in effect at any point, . . 162 " load, bridging useful in sustaining, 302, 309 " " resistance of bridging to a, 308 " " location of greatest strain 181 " loads, a series of, 155 " " approximates a distributed load, . . . 155 5cS2 INDEX. PAGE Concentrated and distributed loads, 155,179,181,182,183,274 " " " compared, 61, 62, 63, 161 " " " graphically expressed, 177 " " " " on beam, 252 " " " size of beam, 182,191 " on lever, 171, 174, 255 loads, a distributed and two 184,187,191 " " " " " three . . . 195, 197, 198, 199, 203, 292 " " middle load of distributed and three 296 Concrete and brick arches, weight of, 340 " filling over floor arches, 345 Conflagrations resisted by solid floors, 500 Constant F for deflection, values of, Table XLIII., 562 Constants for tubular girders. 369 " " use in the rules, . . * 499 Table XX 530, 53i " from experiments in certain cases, 244 " how derived 505 precautions in regard to, 243 Converging forces readily determined 418 Convex side of beam, fibres extended at 37, 45 Conway and Britannia tubular bridges, 328, 367, 368 Construction defined, 27 Tredgold an authority on, 81 " weight of the materials of, 78,261,264 weights of materials of, Table XXII., 533, 534, 535 " in a roof, weight of the materials of, 480, 483 Cross-bridging, 303 " " assistance derived from, 308 " " dowels act as, 501 Cross-furring, . 303 Cross-section, moment of inertia proportioned to, 314 Crush, bricks liable to 346 Crushing strains, tests of woods by, 506 " " in Georgia pine, locust and white oak, Table XL., . . 559 " " " spruce, white pine and hemlock, Table XLL, . . . 560 " weights per inch sectional area, Table XLVI 565 Curve and tangent, point of contact defined, 179, 184 " of equilibrium is a parabola, 416 " " " stable and unstable 416 INDEX. 583 PAGE Dangerous, beam though safe may bend and appear 235 Deflected lever, rules for size of, . , 256 Deflecting energy, 211 " of weight on lever, 229 energies in beam and lever, .... * 251 power of concentrated and distributed loads, 252 Deflection and rupture compared, 211 excessive under rules for strength, 77 resistance to, 221 by distributed load, rule for, 255 directly as the weight, 304 " " " extension, 214,215 " " . " " length, . . 217,218 " " " " force and length, 216 total, directly as the cube of the length, 218 " " " weight and cube of length, 219 values of constant F, Table XLIII 562 of beam, effect at bearing 121 " " formula for, 229 " " load forgiven, 245 with load at middle, 242 in floors, rate of, 240 not to be excessive, 211 to the limit of elasticity, 237,243,246,247 within elastic limit, 245 of beams not to be perceptible, 260 per lineal foot, rate of, 239, 261, 264, 267, 342 " injurious to plastering, perceptible 260 in good floors far within the elastic limit, 239, 243 ' of beam with distributed load, 251 rule for dimensions of beam 248 of bridged beams tested, 303 " rolled-iron beams, load at middle, 331, 332 " " lever and beam compared 237 " amount of, ... t 213 " test of, 245 " load for a given, 247 " " " by distributed load, . 255 " " to limit of elasticity, 247 " " " rule for, 229, 244, 256, 258 584 INDEX. PAGE Peflection, dimensions of lever, 251 " 4t rules for, 250 is as the leverage, the power to resist 223 Demonstration of scale of strains 134 Depth, its value, test by experience, 33 " and length, ratio between, 240 " relation to weight and fibres 36 " in proportion to weight, 44 " denned for compound load, 178 " relation to breadth, 33 " proportional to breadth, rule, 73 " from given breadth and distance from centres, 92 " of simple Jaeams necessarily small, 402 in a beam, the importance of, 312 of beam proportioned to load, square of, 123 " " " rule for, 248, 249 " with load distributed, rule for, 254 " " " in dwellings, 262, 263 " " " first-class stores, 265, 266 " lever, uniform load, 169 " " " promiscuous load 175 " " " rule for, 251, 256, 257 " " framed girder, rule for, 424 " in " " objectionable, 422 " and length of framed girder 422 " to length in tubular girders, ratio of, 382 Depths analytically defined, varying 137 " expressions for varying 141 " demonstrated, rule for varying, 135 " compound load on lever, scale of, 172, 173 Design for a roof truss, selecting a 481 Destructive energy, , . 47, 48, 116, 121, 151 " its measure, 53 " symbol of safety, 71 " and resistance, 53 " load at any point, 57, 129 " from two weights, 133 " several weights, 62, 66, 67 " on lever, in power of weight and resistance of material, 68 INDEX. 585 PAGE Diagonals, gradation of strains in chords and, 432, 435 " of framed girder, strains in the, 436 top chord and, 448 Diagram of forces described 418,419,421 " " " order of development of, 421 " " " gradation of strains in, 433 " " " in fra'med girder 429 Diagrams and frames, reciprocal, 418 " correspondence of lines in frames and, ......... 462 Differential calculus, 158 " computation by, 228 " " strain denned by . 180,184 " " strains in lever 168 of variable, moment of inertia, 318,319, 322 Digest or directory of this work, 566 Dimensions of beam for compound load, 182,187,189 " 4t at given point, for compound load, . ....... 189 " " load at any point, 129 " " when h equals , 190 " " h exceeds n, 191 Directory or digest of this work, , . 566 Distance from centres of beams 80, 91 " girders, ... 94 " " rolled-iron beams, 341, 342, 498 in dwellings, 262 rolled-iron beams, 343 " first-class stores, 265 " rolled-iron beams, .... 344 Distributed load, strain by the calculus, 157 " effect at any point, . 161 " " equal in effect at any point, concentrated and half of, . 162 " " on floors '77 " " " beam, 75 " " deflection of beam under, 251 " " shape of side of beam, . . . . . . 162 " " on lever, 74 " deflection of lever by, , 255 " " shape of side of lever, 170 on rolled-iron beam, 337 " " " cast iron girder, . . . 389 586 INDEX. PAGE Distributed load on tubular girder, 368 " " " size at any point, 372 " . and concentrated loads compared 61,62,63,155,161 " " " '' on beam compared, 252 " " " " " lever " 255 ' " one concentrated load, . 179, 182, 183, 274 " graphic representation, .... 177 " size of beam 182 " " on lever, 171, 174 " two loads, . 184, 187, 1 91, 195 size, 191 " three .... 195, 197, 198, 199, 203, 292 " middle load, 296 Drury, testimony on loading, 82 Dwellings, load on floors of, - .,88, 340, 341 " floor beams and headers for 495 " rule for floors in 261 " " " headers in 271 " values of c, /. b and d in floors of, 262 " rule for solid floors of, 502 . " hemlock beams, Table 1 508 headers, Table IX 516 " Georgia pine beams, Table IV., 511 " " " headers, Table XII., 519 " spruce beams, Table III 510 headers. Table XL, 518 " white pine beams, Table II 509 " " " headers, Table X., 517 carriage beam, precise rule, 282 " with one header 272 " " " " two headers, precise rule, ...... 292 " " " " equidistant headers, precise rule, . . 289 " " headers and one set of tail beams. . 277 " " " " two sets " " . 275 three " the greatest strain at middle header, . 298 " " " the greatest strain at outside header, 295 rolled-iron beams for, . 498 " Table XVIII., 526, 527 INDEX. 587 PAGE Dwellings, rolled-iron beams for, distance from centres, 343 tie-rods for floor arches of 347 " rolled-iron headers for, 349 " " " carriage beams with two headers and one set of tail beams, 358 V " " " " " " headers and two sets of tail beams, . . . 354, 356 " <4 " " " three headers, .... 360, 362 " load on tubular girders for 380 rule for " " 380 Economical depth of framed girder, 425 " tubular '.' 382 " form of beam, . . 163 " " " rolled-iron floor beam 312 " " " roof truss, more, 469 Elastic curve defined by writers 213 " limit, bending in good floors far within the, 239, 243 " " fibres strained beyond the, 235 " " important to know the, 212 44 " in elongation of fibres, 236 " " symbol for safety at the, 239 " power of material, knowledge of, 244 " substance in soles of feet, 84 Elasticity are exceeded, rupture when limits of 212 defined, limits of, 212 " for wrought-iron, modulus of, 232 " of floor, moving bodies, 84 " possessed by all materials, 211 Elements of rolled-iron beams, Table XVII., 524, 525 Elliptic curve for side of beam, 164 Elongation of fibres 214 " " " graphically shown, 236 English rolled-iron beams, large, 313 44 wrought iron, elasticity of, , 232 Equal weights equally disposed, 141. 143, 144, 146 4< " general results, 146 " " strain at first weight, 147 44 " 4< " second weight 148 44 44 4t " any weight, 150 Equally distributed safe load, rule for, 70 588 INDEX. PAGE Equation, management of an, 71 to a straight line, 171 Equilibrated truss, strains in an 408 Equilibrium at point of rupture, . 68 " measure of forces in 407 Equilibrium of pressures , 38 " " resistances of fibres 45 " stable and unstable 416 " three forces in 406 Error in rules on safe side, ... 183 Euclid's proposition in a triangle, 486 Excess of material by rule for carriage beam, 183 Experiment as to action on fibres 36 " on India-rubber, ; 212, 213 " " New England fir. 233 " " white pine units, 32 Experiments by transverse strain, 504 on American woods, 504 " cast-iron, Hodgkinson, 386 " model iron tubes 369 " side pressure . 120 ." " tensile and sliding strains 505 " timber, 30 " units, conditions 32 ." " weights of men, . 85 " " woods, by crushing 506 " " wrought-iron, 232 " rules useful in, . . 68 Experimental test of cross-bridging, 303 Extension and compression of fibres, strength, 35 " " summed up, 228 as the number of fibres, resistance to 222 balances compression, . . 45 directly as the area and depth, 222 . " force 212 " length 213 graphically shown, resistance to 221 measured by reaction of fibres, 222 of fibres 37 " " at bottom of beam, 42 INDEX. 589 PAGE Extension, resistance to, 45 Factory floors, load on, 78, 79 Fairbairn's experiments, 500 Fairbanks Scale Co., testing machine by 504 Falling body, the force exerted by a, 84 Feet, elastic substance in soles of, . . , 84 Females, weights of, 83 Fibres, crushed on wall, 121 " crushing in direction -of, 506 " elongated to elastic limit, 236 " end and side pressure on, 120 " extended or compressed 212, 214 " extension of, graphically shown, * ... 222 " in a tie-beam, consideration of, . i . 488 " load should not injure the, 69 " measuring extension of the, , 221 " power of resistance as the depth, , 46 resistance as the depth of beam, 43 " " " " leverage, 223 " " directly as the depth, 36 " " to change of length, 46 " " " extension, 222 " " horizontal strain, 43 " " side pressure, 120 " resisting compression and tension, , . 403 " strained beyond elastic limit, 235 " strength due to their coherence, 35 Fire, resisted by solid timber floors 500 " wooden beams liable to destruction by, 3 12 Fireplaces, framing for 95 First-class stores, carriage beams with one header, 273 " . " " floor beams, 264 " and headers, 495 " " " rule for headers, 2 7 l " " " formula for solid floors, 5C-3 " * load on floors, 339> 34 1 " " " " " tubular girders, 380 " " " rule for tubular girders, 381 " * ' rolled iron beams, 495 " " " " " " distance from centres 344 590 INDEX. PAGE First-class stores, rolled iron headers, 350 " " " " carriage beams with one header, .... 352 " <v " " " two headers and one set of tail beams, . . . 359 " " " two headers and two sets of tail beams, . . . 357 " " " " " " " " three headers, . . 361, 364 " " " tie-rods in floor arches, 348 " " " values of c, /, b and d y . : 265 Five equal weights, graphic strains 144 Flanges an element of strength, 313 and web, proportions between, . . .- 313 " " " in cast-iron girders, relation of, 387 " " " moment of inertia for, . 327 in cast-iron girders, proportion of, 387 ' equal, top and bottom 314 " of cast-iron girders, proportion of, 386 " " tubular girders, construction of 374 " equal, top and bottom 37 1 tension in lower, 370 " for floors, area of. k . 377 minimum area of, 3 8 3 " " " thickness of, . 373 to predominate over the web, , 314 Flexure and rupture compared, 267 " rules compared 235, 293 weights producing compared, 237 floor beams by rules based on, . . . , 77 formula for denned, 230 " moment of inertia, resistance to 3 r 4 of floor beams, resistance to. 260 " resistance to, : 221 " rules for, 2 4 2 value of F % the symbol of resistance to, 230 Floors, application of rules for strength of, 77 load on rolled iron beam 34 not always strong 29 of solid timber, Table XXI 532 1 warehouses, factories and mills, , 7& per superficial foot, load on, 2 ^ r INDEX. 59 1 PAGB Floors, safe, 28 " severest tests on, 85 " strength of, 29 " beams in, rule for, . . , . 77 " " " test by specimens, 29 " tubular girders for, rule for, 377 " weights of, in dwellings, 80 Floor arches, general considerations, 345 " of parabolic curve, 346 " " tie-rods for, . . . ^ . . 346 " " " " area of cross-section of, 347 " " " " where to place 348 " beams, general rule for, , 89, 92 " " bridged 302 " " nature of load on, 78 " load on, rule for, 78 " " of dwellings, modified rule for, 2bi " resistance to flexure of, 260 " " stiffened by bridging, 310 " " stiffness of, rule for 260 " of wood, Tables of, 496 " " " and iron, Tables of, 495 " " " iron, distance from centres, 341 " " Georgia pine, for dwellings, Table IV 511 " " " ' first-class stores, Table VIII., 515 " " hemlock, for dwellings, Table I., 508 " ' " first-class stores, Table V., 512 " " spruce, " dwellings, Table III 510 " " " first-class stores, Table VII., 514 " " white pine, for dwellings, Table II., 509 " " " " " first-class stores, Table VI., . 513 bridging tested, 303 " " openings, carriage beams, 195, 196 " " planks, their weight, 79 Force exerted by a falling body, 84 and frame diagrams correspond, lines of, 462 Forces and lines in proportion, 405 described, diagram of, 418, 419, 421 " in a framed girder, 428 " " " truss, graphically shown, 417 592 INDEX. PAGE Forces shown by a closed polygon, 418 Force- diagram, example of constructing a, ?.;..., 483 " = for a roof truss, . . . . . 461, 462, 463, 465, 466, 468, 469, 472 " . ** form a closed polygon, lines in a, 485 " line of weights for a, 464, 466 " " of a roof, measuring the, 485 " " of an unsymmetrically loaded girder, 455 " " scale of weights in a, 483 " diagrams, strains in trusses compared by, 469 Form of beam for distributed load, 162 " *" lever " " 171 '" " " " compound " 173 " " iron beam, economical 312 Formula, comparison of F with E of common, . , 232 " for resistance to flexure, , 232 " solid floors, , 502 ' " "- reduction, 501 " management of a, 89,90 " practical application, , 71 Four equal weights, graphic strains, 143 Frames and diagrams, reciprocal, 418 Framed girder, allowance for joints, etc., in chord, 445 " " area of imcut part of chord, 444 " bearings of metal for struts, 450 " compression in chord and struts, 445 " compromise of objections, 423 cost inversely as the depth, ... , 423 " diagram offerees in, 429 " economical depth, 425 '* forces in, 428 " horizontal thrust in, 403 '* irregularly loaded, 451 '* its relation to a beam 402 liable to sag from shrinkage, 450 " minimum of strains in, , 426 " number of bays or panels, 425,428 " peculiarity in strains of, 432 " proportions of, .- 422 " resistance to tension in, 443 " rule for depth, 424 " " series of triangles in, . 425 . INDEX. 593 PAGE Framed girder, strains in diagonals of, 436 44 " " " lower chord of, 439 " " upper " " 440 " " system of trussing in, . . t 425 41 " top chord arid diagonals of, 448 tracing the strains in, 437 trussing in, . . . . . . ' 417,425 unequal reactions of supports of, 451 with loads on each chord, . 433 wrought-iron ties, in 443 " girders, chapter on 402 compression in, rule for, 447 " usually of wood, chords of 444 truss, reaction of supports of, 415 France, testing bridges in 82 Francis on compression of materials 446 Funicular or string polygon, 408 Furniture reduces standing room, 88 Galileo's theory of the transverse strain, 36 Geometrical approximation to moment of inertia, 315 series of values of strains, 476 Georgia pine, resistance of, 120, 121 " beams, their weight, 79 41 floor beams and headers, 495 " coefficient of in rule, 261,265 German rolled-iron beams, large, 313 Girder defined, rule, 94 " history of tubular iron, 367 " plate and jolled-iron, compared with tubular, 367 Girders, distance between, , 94 " headers and carriage beams 94 Graphic representation of strains, 127 " strains checked by computations, 132 " " from two weights 133 " three " 138 " " " ' 4 equal weights, 141 " four " 143 Graphical representations, ' in of compound loads, . 177 " " of moment of inertia, 321 594 INDEX. PAGE Graphical strain at any point, . '. 127 " strains in a beam 114 " " " double lever, 113 Graphically shown, horizontal strains 406 " resistance of fibres 223 Gravity, its prevalence, 27 load concentrated at centre of, 60 Greatest load on floor, 80 Hatfield's, R. F., clock-work motion, 504 Headers, definition, 95 " load upon, . 96, 196 " allowance for damage to. 97 " formulas for 96, 97 11 " " tables of, 497 " ' " breadth of 270 " and trimmers 266 " wooden floor . 495 " rolled-iron floor 349 *' for dwellings and assembly rooms, 271 " " " " rolled-iron 349 " " first' class stores, 271 " " " " " rolled-iron 350 '* carriage beams and girders, 94 ". in carriage beam, two 104 " one set of tail beams and two 106 " carriage beam with three 200 " of wood, Tables of, . . 497 " Georgia pine, for dwellings, Table XII 519 " ' " " first class stores, Table XVI 523 " hemlock, for dwellings, Table IX 516 " first-class stores. Table XIII 520 " spruce, for dwellings. Table XI 518 " first class stores, Table XV 522 white pine, for dwellings, Table X 517 " first-class stores, Table XIV., 521 Hemlock, coefficient in rule for, 261, 265 Hemlock, resistance of, 120, 121 beams, their weight, 79 floor beams and headers 495 Hickory, resistance of, 120 INDEX. 595 PAGE History of the rolled-iron beam 3 T 3 V " tubular iron girder, 367 Hodgkinson on compression of materials 44 6 Hodgkinson's edition of Tredgold on Cast-iron, - . 386 experiments, . .'.... 50 " rule for cast-iron, load at middle 388 " "set" in testing, , , . . 55 value of elasticity of iron, 232 Hoes' foundry, weight of men at 85 Homologous triangles, proportions by 487 Hooke's contribution to the science 37 Horizontal and inclined ties compared, strains in 472 strain in roof truss, . 466,473,474 " " resisted by iron clamps 489 " strains in framed girders, 439, 440, 441, 443 " " measured arithmetically, , 412 " " shown by bent lever, 42 " " " graphically, 406 " thrust in a framed girder 403 " tie, raise wall of building to get 478 Hypothenuse of right-angled triangle 486 Important work should be tested, materials in 244 Inclined tie-rod of truss, enhanced strain, 477 Increased strains in roof truss from inclined tie, 474, 478 Index of strength for unit of material. 48 India-rubber, experiment on, 212 " " largely elastic, 211 Infantry, space required for, .,.......' 83 Infinite series, sum of an, 476 " " value of coefficient, 227 Infinitesimally small, differential is , . 318, 319 Insurance offices, load on tloor of, 340, 341 Integral of moment of inertia, . . . , , 319 " calculus, maximum ordinate, 180, 184 Integration, computation by, 228 " rule for strain in lever by 169 " strain by, 159 Iron a substitute for wood 312 " bolts and clamps for tie-beam, 489 " load upon wrought, 99 'C)6 INDEX. Iron beam, load at middle upon, 331, 332, 333 " " progressive development of, 312 Jackson's foundry, weight of men at, 85 Kirkaldy's experiments, 500 Laminated and solid beams compared 34 Lateral thrust by cross-bridging, 303 Lead has but little elasticity, 211 Leibnitz's theory of transverse strains 37 Length and weight, relation between, 65 " " depth, ratio 240 " of beam, rule for 248, 249 " '.' " with load distributed, rule for, 253 " " " in dwellings, . 262, 263 " " " " first-class stores, . 265,266 " " rolled-iron beam, load at middle, 331, 332 " " lever, rule for, 250, 256, 257 " and depth of framed girder, 422 " to depth in tubular girder, ratio of, 382 Lever and beam compared, 244 deflection in 237 " " " " i strength " 55 symbols " 49 " arms in. inverse proportion as the weights, 39 " at limit of elasticit)', load on 248 " by distributed load, deflection of 255 deflection in a, 213 destructive energy in a. , 55,111 " dimensions of a deflected, 251 " distributed load on rolled-iron, 338 " effect of weight at end of, 47 formula for deflection in a, 229 " modified to apply to a, 54 " graphical strains in a double, 113 load at end of rolled-iron, 336 principle in transverse strains, 38 demonstration 39 effect of several weights, 62 " unequal weights, 39 " promiscuously loaded, 175 depth of, . ... 175 INDEX. PAGE Lever, rule for deflection of, 244 " " resistance of, 48 " " " strength " 55 " rules for dimensions of deflected, 250,256 " safe load, rule, 70 " distributed load, rule, 70 " shape of side of 123 " shaped as a parabola, 124 " showing elongation of fibres, , 236 " strains like two weights at ends, 45 " measured by scale, in symbol showing strength of, 47 " test of deflection in a, 245 " to compression, resistance of fibres of . 228 44 " extension . . 228 ." " limit of elasticity, deflection of, 247 " uniformly loaded, strains in, 168 4 ' values of P, n, b, d and d in a, 250 " " U, n, b, d " 6 " ' 256 41 effect of weight at end of, , . . 58 ." with compound load, strain in and size of, . 171 " " distributed load, the form a triangle, 170 " 4 ' unequal arms, strain in, 127 " " uniformly distributed load, 74 Leverage, arm of, 47 capacity of tubular girder by, 369, 370 graphic representation, in resistance of fibre is as the, 223 Light-well in tier of floor-beams, 201 Light-wells, carriage beams, 195, 196, 198 " " framing for, . 95 Lignum-vitae, resistance of, 120 Limit of elasticity, . . 212,213,235 " " . " deflection to the, 237, 243 " " " in floor beam, . 264 load on beam at, 246 " 4 ' 4t " ". lever " 248 strain beyond the 313, 315 " " " testsofthe, 505 Limited application of formula for value of h in carriage beam, .... 181 598 INDEX. PAGE Lines and forces in proportion 405 Live load, measurement of a, .*.>.. . , . 80 " " weight of people, . . 84 Load and strain, various conditions ....:.... in ' at limit of elasticity in a beam ...,,,, 246 " " any point, effect on beam, . , , . 56 " " " " test of rule ..*.= . , 57 " " " " rule for strength, -; 58 " " " " safe rule. . _'. . , 70 " " " " strain at any point, . . , , 128 " to rupture a cast-iron girder. ..... T ... 390 Load at any point on tubular girder, ............. 371 ' " end of rolled-iron lever, , 336 " " middle, pressures, 39 " " " of beam . 75 " " " deflection, 242 " i4 ** - M safe rule, 70 " " " " cast-iron girder 388 " " '* " rolled-iron beam, . . , 331, 332, 333 " analyzed, compound ..'.,-. 178 44 strains and sizes, compound , 171 " deflection of beam with distributed 251 " lever " 255 " distributed, rules for size of beam with 253 safe rule, .- .- 70 " equally distributed, effect, 58 " "at middle 60 " for given deflection in a lever, 247 " not at middle, effect at middle, 59 " ** " pressure on supports e 40, 41 " on beam, at middle and distributed, . . , 252 rule for distributed , , 253 ** " lever, distributed and concentrated, 255 " " bridge, Tredgold's 80 " " floor, components of 339 " Tredgold's remarks 80 estimate, 81 " " " the greatest . . .- 80 " per superficial foot 261 " " " beam, its nature . 78 INDEX. 599 PAGE Load on floor-beam, rule, 78 " " rolled-iron floor beam, 340 " " header. . . 196 " " carriage beam, . .-..-.,...,.. 105, 107 " " " " with one header, , 99 * roof per foot horizontal, . . .......... 480 " inclined foot superficial, ........... 480 " supports arithmetically computed, ...... 456 " proportion of, .......... . . 119 " " from weight not at middle, ....;.,.... 56 " each support from unsyrnmetrical loading, ....... 451 ' per foot on floor, for people ,..,.. 83 * " 66 pounds, .... 87 4 superficial of floor, 70 pounds 88, 264 on lever, promiscuous .... .... ... . 175 " proportioned to square of depth of beam, ...,.,.... 123 " upon a header, ....,.,. 96 " " " bridle iron, . 98 ' carriage beam, .......... ^ . 98 " " roof truss 479, 483 " " each supported point in a truss, . , . 482 " tie-beam of a roof. . , 481, 483 Loads between the supports, dividing unsyrnmetrical 453 compared, concentrated and distributed ..,.,,.,. 61 Loaded, framed girder irregularly 451 loo heavily, a beam 243 Locust, coefficient in rule for. 261, 265 resistance of 120 Logarithms, example of computation by, 311 Mahan's edition of Moseley's work, 251 Mahogany, resistance of. . . 120 Males, weight of , , . t 83 Maple, resistance o(. . . 120 Mariotte's theory of transverse strains, 37 Material, knowledge ol elastic power of any , 244 defined, unit of, , . 29 Materials for important work to be tested, 244 " weights of building . . . 504 Table XXII.. ....... 533,534,535 of construction, weight of, ....... 78, So, 339, 340, 379 600 INDEX. PAGE Materials of construction of floors, 502 " in a roof, weight of, 480, 483 Maximum moment defined, 188 " ordinate by the calculus, 180, 184 " strain analytically denned. ..,..,,',..,. 181. 184 " -" graphically shown, , . . , 178 compound load , 186, 189 " " location analytically defined, , . '. . 179, 184 " " three concentrated loads, . . 202 " " on middle one of three headers, 201, 204 " " " outside " " " " 196, 204 " "of three loads on carriage beam, 197, 198 Maxwell, reciprocal frames and diagrams by Prof 418 Measure of extension of fibres, 237 " " forces in equilibrium 407 " " symbol for safety tested, 269 " resistance of cross-bridging, 304 " " strains in truss with inclined tie, ......... 475 " symbol for safety, 239, 269 Measured arithmetically, strains 415 horizontal strains . 412 Measuring strains in roof truss, . , 485 Men, actual weight of, , 85 effect of when marching, 87 space required for standing room 82 Menai Straits and Conway tubular bridges 367, 368 " " weight of bridge over, 378 Merrill's Iron Truss Bridges, 402 Methods of solving a problem, various 72 Military, estimate of space required by, . 83 " weight of, . . ..,,..,. 85 step, the effect of, ....,,... c 86 Mill floor, load on. . . . , 78, 79 Minimum of strains in framed girder, 426 area of tubular girder, 383 Model of a floor of seven beams, 303 " iron tubes experimented on, 369 Modulus of elasticity for wrought-iron, 232 " rupture by Prof. Rankine, 50 Moment of inertia defined, 314, 319 INDEX. 60 T PAGE Moment of inertia, value of 320 arithmetically considered, 314 geometrical approximation, , . , 315 " " " by the calculus, 318, 319 " " " area of parabola .... 322 " " " shown graphically, 321 " computed, 315, 316, 317 general rule for 324 " " " . comparison of formulas, 328, 330 " proportioned to cross-section, . ... . . 314 " " resistance to flexure, .... 314 " " for rolled- iron beams. . 326, 328 " " " " " " load at middle, .... 331, 333 " " " " " " Table of, 498 41 " " " flange and web 327 " " " " rolled-iron header 349 " " weight defined 47 " " on lever, . in " " ". arm of lever, 56 " at middle of beam, 48 Moments of compound load, 188 capacity of tubular girder by 369 load at middle, tubular girder by, 370 Momentary extra strains, . 87 Mortising, the weakening effect of, . . 195 damaging to a header. t 97 carriage beams to be avoided - . . 98 Moseley. moment of inertia, by Canon 328 " modulus of rupture by Prof. 50 " symbol for strength ' 49 Moseley's work on Mechanics of Engineering and Architecture, , 251, 255 Movement of men, effect of . 86 Negative equals adding a positive, deducting a . 428 Neutral axis, distance from, 315, 318, 319, 324 " " in a framed girder 402 " flange to be distant from, 313, 314 line * 45 " '' denned 37 ' " distance of fibres from 222 " " at middle of depth, 45 602 INDEX. PAGE Neutral axis at any depth, effect 46 ' " in a deflected lever, 236 " " " tie-beam, fibres near 488 New England fir, experiment on, 233 Oak, coefficient in rule for, . 261, 265 " live, resistance of, .... 120, 121 Office buildings, formula for solid floors of. 502 rolled-iron beams for, 495, 498 " Table XVIII., 526, 527 Openings in floors, framing for, . , 95 Ordinate. location of longest, compound load, 182, 184. 185 Ordinates measure strains, 128, 130, 134, 136, 139, 140, 167, 168, 172, 173, 177, 179, l8l, 183, 184, 197, 198. 201, 202 Ordinates measure strains in lever 175 Panels in a framed truss, number of, 426, 428 Parabola, a polygonal figure 161 " the curve of equilibrium is a 416 expression for the curve, 160 form of scale of strains, 177 side of beam from a 124, 163 " " lever " . 124. 172, 173 defines strains in lever, , 169 form of web of cast-iron girder is a 392 Parabolic curve, moment of inertia, . 322 " " limits the strains, , 184, 197, 201 " form of floor arches, 346 Parallelogram of forces in framed girders, , : 404 People as a live load, weight of 84 to weigh them, 81 floors covered with 340 required for a crowd of, ' . 77 on floor, crowd of, 81 their weight, . . . 81 per foot, ....... 83 their weight, authorities, 82 Philadelphia, iron beams at Exposition at, . . 313 Phoenix Iron Co., beams tested by 500 Planning a roof, general considerations, ..... 478 an example in 481 Plaster of Paris, fire-proof quality of, f . . 501 INDEX. 603 PAGE Plastered ceiling of a room, 303 Plastering, weight of, 79 perceptible deflection injurious to, . 260 Plate beam formed with angle irons, 312 girder or beam, . . . 367 and tubular girders compared. 377 " over a tubular girder, advantages of a 377 Polygon forces shown by a closed . 418 funicular or string 408 " lines in force diagram form a closed 485 Polygonal figure, parabola, 161 Pores of wood, size of, 120 Position of weight on a beam, 54 Positive quantity, deducting a negative equals adding 428 Post, rule for thickness of a, 447, 449 Posts, Baker's formula for, 446 Precautions in regard to constants, 243 Precise rule for carriage beams, for dwellings 273, 282, 285 " first class stores 275, 282, 286 with two equidistant headers, .... 287 " headers, for dwellings, . . . 292 " first-class stores, .* 292 and one tail beam, . 283 equidistant headers and one tail beam 290 " headers and two tail beams, . . 279 Pressure, conditions in loaded beam, 39 on support from load not at middle 40, 41 Problem, various methods of solving a, 72 Promiscuous load, scale of strains, 175 " on lever 167 Proportion between flanges and web, 313 Proportions of a framed girder, 422 Quetelet on weight of people, 83 Rafters, dimensions of, 490, 491 increased, compressive strain in ... . . 474 " to be avoided, transverse strains in . 460 " " " protected from bending, 479 Rankine on compression of materials, 446 " converging forces 418 604 INDEX. PAGE Rankine on modulus of rupture 50 " moment of inertia 3 28 Rate of deflection per foot lineal 239,261,264,267,342 11 " " in floors 240 " " rise in brick arch in floor 34 6 Ratio of depth to length in tubular girders 382 Ray's Algebra referred to 476 Reaction of fibres on removal of force 213, 222 " from points of support . . 58, 465, 466, 470, 473 " of supports equal to load 39 " " " " " shearing strain 119 " " " from unsymmetrical loading, 451 " " " of framed truss 415 Reciprocal figures explained, 4 22 frames and diagrams, 4*8 " lettering of lines and angles, . 4 J 8 Resistance of materials 53 " ** " to destructive energy, 68 " its measure 53 " directly as the breadth 33 " increases more rapidly than the depth, 34 " as the area of cross-section, 31, 32 " not as the area of cross-section, 31, 33 " to compression, . . . 45 " extension 45 " " and compression. 229 " " .* v equal, 45 " " " or to deflection 222 summed up, . , 225 " " flexure 235 " " rules for 242 " " " value of F, the symbol of, . 230 " " of floor beams 260 of a lever, rule for, .... * 48 to rupture 266 elements of, . 46 " cross-strain shown 47 of fibres to change of length, 46 " " as the depth of beam. 43 " " directly as the depth, . . 36 INDEX. 605 PACK Resistance of fibres to extension and compression 35 " " " " " expression for, 224 " to extension as the number of fibres, 222 " as the distance of fibres from neutral line, 222 " of cross bridging, principles of, 304 " increased by cross bridging, 305, 310 of a bridged beam 304 " in cross-bridging, number of beams giving 309 Rise of brick arch in floor, rate of, 346 Rivet holes in iron girders, allowance for 368 Rolled-iron beams, chapter on, 312 " beam, history of the 313 " " beams preferable to cast-iron, 399 " " 4< means of manufacture, 386 " " " have superseded c^st-iron, .... 386 " " " to be had in great variety, 313 " . " " distance from centres, ...... 342 " " beam, moment of inertia for, 326, 328 " " " weight of, . . 340 " " " plate beam and tubular girder, 367 " " " load at any point, 333, 334 " ' Table XVII., 335 " " " " " middle 331 " " " " distributed, - 337 " " beams for dwellings, etc 498 distance from centres, 343 " " " " first class stores 498 distance from centres, .... 344 Table of elements of, 498 " " " elements of, Table XVII., 524, 525 Tables of, 495 " for dwellings, Table XVIII 526,527 " first-class stores, Table XIX... 528, 529 " " lever, load at end. 336 " " " " distributed 338 " " headers for dwellings 349 " " " " first-class stores, 350 " " carriage beam with one header, for dwellings 351 " " " " " " " " first-class stores, . . . 352 606 INDEX. PAOB Rolled-iron carriage beam with two headers and one set of tail beams, for dwellings, etc.. . 358 " two headers and one set of tail beams, for first class stores, . . 359 " two headers and two sets of tail beams, for dwellings, etc 354^ 356 two headers and two sets of tail beams, for first class stores, 355 " three headers, for dwellings, . . . 360, 362 " first-class stores, . 361, 364 Roof, general considerations in planning a 478 an example in planning a .gj beams, increase in weight of 478 trusses, chapter on, . . .v , 459 comparison of designs for 450 selecting a design for 479? 4 8 X considered as girders 4 c ( , with and without tie-beams 4 59 truss, force diagram for a 4 6i horizontal strain in 473 supports, .... '.,. . : II9 Rule for floor beams, using the g~ Rules for rupture, various conditions Gg Rupture the base of rules for strength, 77 resistance to, ... . . , 221,266 " theory of, , ^ 7 " elements of, 46 modulus of, by Prof. Rankme 5 o equilibrium at point of, . . kg by compression and tension, o 3x3 " cross strain, ZII its resistance, tension, 4Q and flexure compared, 211 267 rules compared .... 235 203 compared, weights producing, 237 ensues from defective elasticity 2I2 beams of warehouses to resist 260 resistance of carriage beam to, rule for 100 of cast iron girder, load at any point 300 relation of flanges, . . 387 INDEX. 607 PAGE Safe load at any point on cast iron girder . 39 1 " distributed load, effect of at any point on cast-iron girder 391 " and breaking loads compared 68 " load, value of a, the symbol for a, . 235 " loads, rules for strength, , 7 " by rules for strength yet too small, beam 77 " beam should appear cs well as be, 235 " load on tubular girder 39 Safety in floors 28 precautions to ensure 244 measure of symbol for, , . . . . 239 a, in terms of B and F, symbol for, 268 cautions in regard to symbol for 71 Sagging of framed girder from shrinkage, 450 Scale, strains measured by, m of depths, compound load on lever, 172 " strains and the calculus, 161 " " " demonstrated, 134 ," " " applied practically, . 411, 414 " " " to be carefully drawn 412 " " " for depths, 132 " " " made from given weights, 409 " " load at any point, 128 " " promiscuous load, . . 167, 175 " for two weights 133 " " distributed and one concentrated load, 179 " " two loads 184, 187 " " compound load on beam, 177 " " lever 172, 174 " " carriage beam with three headers, . . . 197, 198, 201, 202, 208 " weights for a force diagram 483 Scientific American quoted, 302 Set produced by strain on materials 505 Shape of beam elliptical, 164 " side of beam under a distributed load 162 " " " from parabola 163 " " " " graphically shown, 122 " lever a triangle under a distributed load, 170 Shearing and transverse strains 116 strain equals reaction of support, 119 6o8 INDEX. PAGE Shearing strain graphically shown, ... . 115 " " provided for 123 ' " at end of beam, 118 ' . " in tubular girder 374, 375 Shrinkage of timbers, derangement from, . 450 Side of beam graphically shown, shape of 122 " pressure, resistance to, 119 Size and strength, relation of, 31 Skew-back of brick arch in floor, 345 brick arch footed on, 398 41 ' " tie rod to hold arch on 398 Slate, brick arches keyed with 345 " on roof, weight of, ... 480 Sliding strains, experiments on, 505, 506 " " in Georgia pine, locust and white oak, Table XXXVIII., 557 " '* " spruce, white pine and hemlock, " XXXIX., . 558 " surface, breaking weight per square inch, " XLV., . . . 564 Snow on roof, weight of, 480 Soldiers on a floor, weight of, 83 Solid and laminated beams compared, 34 " timber floors not so liable to burn 500 " " " reduction of formula for, 501, 502 " " " should be plastered, 501 " Table for, .... 504 " " " thickness of 500, 501 . " Table XXI. 532 Space on a floor occupied by men . , 85 " " " " required by people 83 " men when moving, , \ 86 " " " reduced by furniture 88 Spruce, coefficient in rule for 261, 265 resistance of, , . . . 120, 121 " beams, weight of, 79 " floor beams and headers 495 Square timber, rule for strength of . . . 72 Squares of base and perpendicular of triangle, 487 Stability to be secured in buildings 27, 28 Stable and unstable equilibrium, . 416 Stair header, strain on carriage beam, 197 Stairs, framing for, 95 INDEX. 609 PAGE Stairway opening in floor, 201 " openings, carriage beams, 195, 196 Step, effect of military 86 Stiffness and strength compared, 267, 268 " " resolvable, rules for, 270 " differ from rules for strength, rules for 235 " requisite in tloor beams, 260 Stores for light goods same as dwellings, 264 " floor beams for first-class 264 " headers for first class 271 " carriage beams for first-class, precise rule for, , 275, 282. 286 " " il with one header, for first-class . . ... 273 " 4i " " two headers, " " " precise rule, , . 292 equidistant headers, for first-class, pre- cise rule, 289 " ' " " " headers and one tail beam, for first-class. 278 " " headers and two tail beams ; for first-class, 276 " " " three headers, the greatest strain being at middle header 299 " " three headers, the greatest strain being at outside header, 295 " load on tubular girders for first-class 380 " tie-rods of floor arches of first-class 348 Georgia pine beams for first class, Table VIII., 515 " " " headers " " " " XVI., 523 " hemlock beams " " " " V., ....".. 512 " headers " " ' * XIII 520 " spruce beams " '' l ' " VII., 514 headers " " " ' XV., 522 " white pine beams " VI., 513 - headers " XIV., . . .. ' , . . 521 " rolled-iron beams " " " " XIX., 528, 529 Straight line, equation to a, . r 171 Strains useful, knowledge of gradation of, . . . . 433 " by movement, increase of 84 " analytically defined , 137 " arithmetically computed, 168, 415 " graphically represented, 127 " checked by computations, graphical 132 " graphically shown, shearing 115 6 10 INDEX. PAGE Strains measured by ordinates, 128, 130, 167, 168, 177, 201, 202 " " " lines 405 proportioned by triangles 486 Strain analytically defined, maximum . 181, 184 graphically " . ,178 " analytically " location of, . . , 179, 184 at any given point graphically shown 127 " ' point from a distributed load, 161 Strains demonstrated, scale of. 134 " from given weights, to construct scale of, , 40-) promiscuous load, 167 distributed load, by the calculus 157 two weights, . . . 101 " " graphic 133 of compound load analyzed, 178 -<' " " " maximum. 186, 189 " greatest at concentrated load 181 and dimensions, compound load, . 171 Strain at first weight, with equal weights, 147 .*" " second weight, with equal weights, 148 " any 150 Strains in a beam, graphical 114 Strain " " " loaded at any point. , 128 Strains " beam and lever compared, . . 336 *' lever measured by scale , m " " " computed " calculus 169 " " levers, graphical 167 " " double lever, graphical 113 " " lever with unequal arms, . . 127 promiscuously loaded. . 175 " " uniformly 168 like two weights at ends of lever 45 " from three, headers. 197 " in framed girder arithmetically computed 435 " peculiarity in 432 tracing the 437 " diagonals of framed girder 436 " lower chord of framed girder 439 11 upper " " " " 440 " chords and diagonals, gradation of, 432, 435 INDEX. 6ll PAGE Strains in roof truss compared, 469 " " truss, arithmetical computation of, 486 " " equilibrated truss 408 " " tie-beam of truss, two 488 " " horizontal and inclined ties compared, 472 " " truss with inclined tie may be measured, 475 " " " an infinite series, 476 " " " without tie increased, , 461 " from raising the tie of truss increased, . . 477 " in rafters increased, . 460 " Georgia pine, transverse, Table XXIII., 536 " " hemlock. Tables XXXIII. to XXXV., . 551 to 554 " " locust, Table XXIV 537, 538 " " spruce, Tables XXVI. to XXVIII., . . 540 to 544 " " white pine, " " XXIX. to XXXII., . 545, 546, 547, 548, 5-19, 550 11 " white oak, " Table XXV 539 " Georgia pine, locust and white oak, tensile, Table XXXVI.. . . 555 " " spruce, white pine " hemlock, " " XXXVII., . 556 " " Georgia pine, locust " white oak, sliding, " XXXVIII, . 557 " " spruce, white pine " hemlock, " " XXXIX., . . 558 " " Georgia pine, locust " white oak. crushing, " XL., . . . 559 " " spruce, white pine " hemlock, " " XLI , . . .. 560 Straining beam in a roof truss. . . 460 " dimensions of a, 490, 491 Strength, test of specimens as to, 29 as the area of cross-section 46 not as the area of cross-section, . . . . 33 " directly as the breadth 33 increases more rapidly than the depth . 34 " in more common use rules for 235 and stiffness compared. .... ........ 267, 268 " differ from those for stiffness, rules for . , . 235 " more simple than those for stiffness, rules for ....... 235 " and stiffness resolvable, rules for 270 " size, relation of, . .... 31 " of beams, rule for 49 " " general rule for, 92 rule for load at any point, 58 " beam increased by a device 402 6l2 INDEX. PAGB Strength of floor, by experiment, , 29 44 " " beams, rule for. , 77 " " beam and lever compared, 55 14 " lever, rule for, . . , . -. 55 44 " square timber, rule for 72 44 " wood, unit of material , 30 String polygon, funicular or 408 Strongest form for a floor beam , ^ . 312 Struts of timber under pressure ...... V .". 404 formula for compression of 447 " " " thickness of. ..:.-." * - 447. 449 " of framed girder, compression in . . . . . 445 " or straining beams in trusses, ... 460 " and ties form triangles in a girder 425 " in trusses prevent bending of rafters, . 479 Superficial foot load per . ..'"' . 88 44 " " on floors per 261 " " " 200 pounds per, 264 " " " 250 " " . 264 41 " on roofs per inclined, .,..., 480 " " weight of people, . , 82 44 " " tubular girder per 378. 379 Superimposed load on floor,. . : '. '-; ; '' 78, 80 44 " tubular girder 379 Superincumbent load on floor 339, 502 Support in a roof truss, points of, 479 " " " framed girder, points of, 425 " of a truss, weight upon . . 464. 466,470 Supports" " '' division of load upon 461 unyielding, 119 reaction from, 58 equal to load, reaction of, 39 of framed truss, ;t i: 415 portions of load on 119 shearing strain equals reaction of, . . . . .... 119 Surfaces of contact, resistance of, 119 Suspension bridge at Vienna, 82 44 rod of truss, strains in.. , 477 44 " 4< " iron for 490 Symbol of safety, a, the. 267 41 '* a, in terms of B and F, 268 INDEX. 613 PAGE Symbol of safety, value of a, the 69, 235, 239, 269 ' cautions in regard to, 71 " " unit of materials, the, 49 Symbols, assigning the 101 compound load, assigning the 187 for two weights, " 105 " headers, " 108 " three " "....,.... 201, 204, 206 " beam and lever compared, . . . 49 Symbolic expression, moment of inertia, . . 314 System of trussing a framed girder, 425 Tables, chapter on the, . 495 of beams for dwellings, etc 495 " " of wood. , 495 " " " for first-class stores 495 " " rolled-iron beams 495 save time of architect, . 495 Tables. . ....... 507 Table I., hemlock beams for dwellings ,, 508 " II., white pine beams for dwellings 509 " III., spruce 510 " IV.. Georgia pine " " ........ 511 " V. hemlock " " first-class stores 512 " VI , white pine " " " " 513 " VII., spruce " " " " " 514 " VIII., Georgia pine beams for first-class stores, 515 " IX.. hemlock headers for dwellings. 516 " X., white pine " 517 " XL, spruce 518 " XII,, Georgia pine headers for dwellings, ....,..,. 519 " XIII. hemlock headers for first-class stores 520 " XIV.. white pine " " " " '* , . 521 XV , spruce " 522 " XVI , Georgia pine headers for first-class stores, 523 " XVII., elements of rolled-iron beams, .524,525 " XVIII., rolled-iron beams for dwellings, 526, 527 *' XIX., " " " " first class stores 528, 529 " XX., constants for use in the rules 530, 531 " XXL, solid timber floors, 532 " XXIL, weights of building materials, 533.534,535 614 INDEX. Table XXIII., transverse strains in Georgia pine, " XXIV.. -.;; XXV., ; .-' XXVI., *. r -* XX VII., <_ XXVIIL, " c v". XXIX, > XXX., - " XXXI., .." XXXII., .;:" XXXIIL, " -.-V XXXIV, " ,-V XXXV., . " -! XXXVI., tensile ;" XXXVII., " " XXXVIII.. sliding '# " XXXIX., " . " XL., crushing " PAGE 536 locust 537, 538 white oak 539 spruce 540 . 541, 542 543, 544 white pine, 545 546, 547 . 548. 549 " " 550 hemlock, . 551 552. 553 554 Georgia pine, locust and white oak, . 555 spruce, white pine and hemlock. . . 556 Georgia pine, locust and white oak, . 557 spruce, white pine and hemlock, . , 558 Georgia pine, locust and white oak, . 559 spruce, white pine and hemlock, . 560 XLI., \ '* XLIL, transverse breaking weight per unit of material. . . . . . 561 XLIIL, deflection, values of constant, F. 562 XLIV., tensile breaking weight per square inch area, 563 XLV.. sliding " " ' surface, .... 564 XLVL. crushing weight per square inch sectional area 565 Tail beams, definition of, . 95 two headers and one set of, 106 " sets of, 104 three headers and two sets of, 200 Tangent defined, point of contact with 179, 184, 198 Tensile and compressive strains 408 strains, experiments on, -505 in Georgia pine, locust and white oak, Table XXXVI., . 555 " " spruce, white pine and hemlock, Table XXXVII., . . 556 '-" - breaking weight per square inch area, Table XLIV., 563 strength of wrought-iron 347 Tension measures rupture 49 " graphically shown, .... 115 in a framed girder, resistance to 443, 445 dimensions of parts subject to 487 " and compression, rupture by, 313 INDEX. 6l 5 PAGE Tension and compression in cast-iron, 387 in wrought-iron, 117 " tubular iron girder 37 " bottom flange or tie-rod 39. 397 rule for shearing based on 117 Testing machine 54 Test of deflection in a lever 245 Tests of the materials used desirable 69 " should be made for any special work, 5 "of value of symbol of safety, .... 69, 269 Three headers, the greatest strain at middle one 201, 204, 207 " " " outside " 196, 204 " equal weights on beam 141 weights, graphic strains from 138 Ties compared, strains in horizontal and inclined 472 " in trusses extended through to rafters 460 Tie-beam, importance of a, 404 " " tensile and transverse strains in a, 488 '' ' of roof, load upon the, 48-1. 483 " <l " " truss, strains in 488 " " " truss, built up . . 489 " " " " manner of building, 489 Tie-rod, effect of elevating the 477 " " of brick arch, area of cross section, 347 ' " " where to place, 348 " " " " " in floor 346 11 " " " !< on skew-back 398 " cast-iron arched girder with 396 " " of iron arched girder 39. 397 Timber, experiments on. ...... 30 fl9ors, thickness of solid, . 500 Transverse breaking weights per unit of material, Table XLII., .... 561 force, test, 29 strain, the philosophy of, 312 " experiments by, 504 " " resistance to, 47 " by rupture or deflection, 211 " " lever principle, 38 " in a tie-beam 488 " cast-iron, 387 6l6 INDEX. PAGE Transverse strains, the object of this work 28 " in framed girders, 402 " " shearing and 116, 118 in Georgia pine, Table XXIII 536 " locust, Table XXIV 537, 538 " *' " white oak. Table XXV 539 spruce, Tables XXVI. to XXVIII 54010544 " white pine.. Tables XXIX. to XXXII., 545, 546, 547, 548, 549. 550 " hemlock, Tables XXXIII. to XXXV., . . 551 to 554 " strength of beams, rule for, 48 Tredgold an authority on construction, 81 " on compression of materials, 446 " " cast-iron. 386, 387. 396 " has written on roofs . 459 Tredgold's "Carpentry." ... 417 " estimate of load on floor 81 " rate of deflection. . . 240 " remarks on load on floor. 80 " value of elasticity of iron. , . 232 Trenton Iron Works, beams tested by. . , 500 Triangle, measure of extension, 221 " showing elongation of fibres 236 " resistance of fibres as area of, ....... . . 222, 223 " distributed load, side of lever a, 170 " of forces in framed girders, 404 Triangles " " Professor Rankine, 418 " " strains, ... 128, 168, 413 " in proportion to strains 486 " " framed girder, series of, .... 425 Trimmers, definition of, . . , 96 " and headers, 266 " bridle irons, 98 Truss, arithmetical computation of strains in a 486 " graphically shown, forces in a, 417 " should have solid bearings for support, 478 " with inclined tie, vertical strain in 474 " without tie, increased strains in, . .' . 461 " load upon each supported point in a, 482 " " " " support of a, 464,466,470 INDEX. 617 PAGE Trusses for roofs, 459 " load upon roof 479. 4 8 3 " points of support in roof 479 " division of load upon supports of roof 461 " distance apart for placing roof 478 " required, number of roof ..,,,. 481 dimensions of rafters, braces, etc., in 490, 491 Trussing a framed girder, , . . . 417, 425 Tubular iron girder, history of, , . . . . 367 44 " " useful for floors of large halls, . 367 " " ** coefficient of strength, .... . 368 " " " constants J ........ 369 " " '* by leverage, . . .... 369 " " " '* principle of moments, 369 " *' " " moments, load at middle, . . , 370 " " " allowance for rivet holes 368 * " " shearing strain in 374, 375 '* buckling of sides of, 377 " " uprights of T iron in sides of, 377 " " economical depth , 382 " ratio of depth to length 382 " ' '* approximated, weight of. . 378 " " per superficial foot, weight of, 378, 379 " " minimum area of . . . . 382, 383 " " tension in lower Mange of. 370 " " top and bottom flanges equal, 371 " " construction of flanges of, 374 " " thickness of flanges of, 373 ** " strain in web of. 374 " " construction of web of 376 " " thickness of web of, 375 " " " " plates of, , . 382 " " load at middle, 367 44 " " common rule, 368 " " any point 368, 371 " uniformly distributed, 368, 372 " " size at any point, . . . 372 ** " " weight of load on floor, 377 ** " *' for floors, rule for, 377 *' " " " dwellings, etc., load on, 380 6l8 INDEX. PAGK Tubular iron girder, for dwellings, etc., rule for, 380 " " " compared with plate girder, . 377 advantages of plate girder over, 377 ' " girders, chapter on 367 Two loads, graphic strains, . 133 Two weights, their effect at location of one of them, 103 Union Iron Co. of Buffalo, large beams, , ' . ' , 313 Unit of material, symbol of, ... 49 " " " Moseley's symbol of . 49 " " v ' index of strength, . . 50 41 " " measure of strength, 30 strength and size of, 46 " " " size arbitrary, 32 dimensions adopted 53 and weight 53 harmony necessary between piece taken and .... 54 breaking load of 69 resistance of, 53 Unknown quantities, eliminating, 90 Unstable and stable equilibrium, 416 Unsymmetrical loading, reaction of supports from, 451 Unsymmetrically loaded girder, force diagram of, . . 455 Unyielding supports for load 119 Value of h limited to certain cases 181 Versed sine of brick arch 346 Vertical effect of an infinite series, 476 strain in truss with inclined tie 474 Vienna suspension bridge, , 82 Von Mitis, testimony as to load on bridges, 82 Wade's experiments, Major 500 Wales, tubular bridges of, 367, 368 Walker, James, testimony on loading 82 Walls, bearing surface of beams on 121 of building raised to permit horizontal tie, 478 pushed out for want of tie, 404 Walnut, resistance of, 120 Warehouse beams to resist rupture, 260 load on floors of 78, 79, 339 Web, moment of inertia for the 327 " and flanges, proportion between 313 INDEX. 619 PAGE Web to connect flanges and resist shearing, 314 " usually larger than needed 314 and flanges in cast-iron, relation of, 387 of cast-iron girder, form of, 392 " " tubular girder, strain in 374 construction of, 376 thickness of, 375 " " " area of, 382, 383 Weight and depth of beam, relation, 36 " in proportion to depth 44 " its effect and position 53 " " at end of lever 58 " on a lever, rule for 256 " " " beam, its position, 54 not at middle, effect on supports, 56 " at middle of rolled-iron beam, 331, 332 on each support of a truss, 464, 466, 470 " of military 85 " " people, live load, 84 " beams per superficial foot, 79 " " floors in dwellings, 80 " " timber in solid floors, . 502 " " materials of construction in a roof, 480, 483 " " rolled iron beams, 339, 340 Weights of building materials, 504 " assigning the symbols for, 101, 105 " and deflections in proportion, 304 " lengths, relation between, 65, 66, 67 pressures, equilibrium, 38 " strains put in proportion, . . . 409 in proportion as arms of levers, 39 at location of one, effect of two 101 " for force diagram, line of, 464, 466 producing rupture and flexure compared 237 of building materials. Table XXII., 533, 534, 535 Whalebone largely elastic 211 White pine, coefficient in rule for 261, 265 resistance of, 121 " experiment on units of, 32 beams, weight of, 79 6 20 INDEX. PAGE White pine floor beams and headers. . 495 Whitewood, resistance of, 120, 121 Wind, its effect on a roof. 480 stronger on elevated places 481 Wood, iron a substitute for, 312 Woods, experiments on American 504 Wooden beams liable to destruction by fire, 312 44 weight of. 79 " floors, when solid not so liable to burn, 500 Work accomplished in deflecting a lever, 229 44 to be tested, materials in important 244 Wrought-iron, modulus of elasticity of, 232 " tension, 116 44 tensile strength of, 347 44 for ties in framed girders, 443 Yarmouth, fall of bridge at, 82 ANSWERS TO QUESTIONS. 37. Transverse. 38. In proportion directly as the breadth. 39. In proportion directly as the square of the depth. 4-0. The elements are the strength of the unit of mate- rial, the area of cross-section and the depth. The expression is R = Bbd'. 41. The amount is equal to the total load. 42 One half. 43. The sum is equal to the total load. 44. The portion of the weight borne at either point is equal to the product of the weight into its distance from the other support, divided by the length between the two sup- ports. 45. R = W~ *e.-p = w^ 47. 10000 pounds. 5000 pounds. 48. The moment, or the product of half the weight into half the length of the beam. 622 ANSWERS TO QUESTIONS. 49 Wl = Bbd*. 50. 22666! pounds. 51. As many times as the breadth is contained in the depth, 62. 22781^ pounds. 63. 40500 pounds. 64. 20250 pounds. 65. 5062^ pounds. 84. Depth, 6-6 inches; breadth, 3-3 inches. 85. 5-24 inches square. 86. 6-93 inches. 87. 4 inches. 126. 2 feet lof inches. 127. 2 feet 4^ inches. 128. 2 feet ;| inches. 129. 2 feet if inches. 130. i foot 8f inches. 131. i foot u|- inches. 132. 2 feet o inches. 133. i foot 7f inches. 134. i foot 9! inches. 135. 3 feet i inch. 160. Breadth, 6-78 inches; depth. 12-34 inches. 161. 2-94 inches. ANSWERS TO QUESTIONS. 623 162. 2-86 inches. 163. 0-291! inches. 164. 1-763 inches. 165. 1-244 inches. 166. 3- 1 ii inches. 179. 36720 foot-pounds. 180. 36-72 inches. (81. Ordinates. Strains. For 5 ft., 18-36 18360 foot-pounds. " 6 " 22-032 22032 " " " 7 " 25-704 25704 " " 8 " 29-376 29376 " 9 '" 33-048 33048 " 182. 10-733, 10-182 and 9-6 inches respectively. 183. 6-245, 8-062, 9-539 and 10-198 inches respec- tively. 184. Depth, 8-1565 inches. Weight, 652-218 pounds. Shearing- strain at wall, 733 783 pounds. " " " 5 ft. from wall, 699-798 pounds. 185. 302-222 pounds. 186. Shearing strain, 4973! pounds. Height, o93i inches. 187. 1-46 inches. 204. io666| pounds. 205. 2666f, 5333^ and 8000 pounds. 206. Strain at A, 17142!; at /?, 22285^. 624 ANSWERS TO QUESTIONS. 207. 8571!, 7428f and 21000 pounds respectively. 208. 12750, 22750, 19000, 8500, 9500, 20500 and 21500 pounds respectively. 209. 3920 pounds. 217. A parabolic curve. 218. 800, 1050 and 1200 pounds. 219. 5 1087 inches. 220. Elliptical 228. o, 300, 600, 900 and 1700 pounds. At the wall 2500 pounds. 229. A parabolic curve. 230. 1250 pounds. 231. Triangular. 232. 3450, 6250 and 9450 pounds. 233. 200, 1600, 6150 and 11050 pounds. 269. 19200 pounds; located at the concentrated weight. 270. 7-01 inches. 287. Resistance to flexure. 288. To any amount within the limits of elasticity. 289. The extensions are directly as the forces. 290. The deflections are directly as the extensions. 291. The deflections are as the weights into the cube of the lengths. 307. By the power of reaction. ANSWERS TO QUESTIONS. 625 308. To the number of fibres, to the distance they are extended, and to the leverage with which they act. 309 Wl s = Fbd'd. 3(0. 0-266 of an inch. 315. The rules for strength are the more simple. 3(9. r-- a 354.-Formulas (122.), (124.), (125.), (126.) and (120.). 355. Formulas (123.), (127.), (128.), (129.) and (121.). 356. Formulas (131.), (132.), (133.), (134.) and (135.). 357. Formulas (136.), (137.), (138.), (139.) and (140.). 437. 12-345 inches. 438. 4-176 inches. 439. 7-700 inches. 440. 8-417 inches. 441. 10-779 inches. 442. 9-530 inches. 537. -fabd* (form. 205.). 538. &(bd'-b l d! r ) (form. 213.). 539. The Buffalo I2j inch 180 pound beam. 626 ANSWERS TO QUESTIONS. 540. 9475-58 pounds. 541. 2004-52 pounds. 542. The Pottsville io| inch 90 pound beam. 543. Two 12 inch 170 pound beams. 544. It should be a 15 inch 200 pound beam. 545. A Paterson or Trenton io inch 135 pound beam. 574. 41^ inches. 575. 35 inches. 576,, At 5 feet from the end of girder, 15 inches each ; " 10 " " ' " " " 26f " " .. I5 M U .* M 35 " 20 " " " " " " 40 <; " " 25 " or at middle, 4if <; " 577. At end of girder, 0-38 inch; 5 feet from end of girder, 0-30 " - 15 - 0-15 " " 20 " " " " " 0-08 " 25 " or at middle, o-o " 578. At 5 feet from end of girder, 8-95 inches; " 10 " " " " 15-34 " 15 " " " " " 19.18 " 20 " or at middle, 20-46 579. 4-2155 feet. 597. Bottom flange, 16 X2-I95 inches; Top 5^x1-646 Web, 1-372 u thick. ANSWERS TO QUESTIONS. 627 598. Bottom flange, 16 x 1-646 inches;. Top Six 1-234 Web, 1-029 thick. 599. Bottom flange, 16 x249 inches; Top " Six 1-867 Web, I-556 " thick. 600. 32-99 inches. 601. -^-At the location of the 25000 pounds ; The bottom flange, 16 x 1-663 inches; " top si x i - 247 " web, J-039 " thick. At the location of the 30000 pounds ; The bottom flange, 16 x 1-588 inches; " top " Six 1-191 " web, 0-992 " thick. 602. 3-68 inches. 651. The strain in AB is 3550 pounds; " " BC " 10280 " " AF " 15240 " u 10130 652. 7-8125 feet. 653. Six. 654-. The strain in DE is 3600 pounds ; " CD " 5425 " " 14500 2I72 , At/ " 15700 628 ANSWERS TO QUESTIONS. The strain in S is 41800 pounds; " " BL " 26125 " DM " 39200 655. The strain in DE is 3614 pounds; " CD " 5420 " BC " 12647 " AB " 14454 " AT-4 " 21681 " AU " 15655 " " " CT 35223 " ES " 41746 26091 " 39 r 37 656. The area at ^ f/ should be 16-103 inches; 36-180 42-872 The size of BL 6 - 626 x 7 - 95 1 inches ; " " " DM " 7-715x9.258 3-004x3-605 " 4-612x5.534 <{ 5.651x6-781 " The area of CZ> " 0-603 inches; " " " AB " " 1-611 688. The computed strain in ^(^ is 22535 pounds; " BH " 18028 " ^4^ " 4507 " AF 18750 " " " " ^r u 15000 " " measured " " AG " 22500 ' " 18000 * " 4500 " 18750 " 15000 ANSWERS TO QUESTIONS. 629 689. The strain in AN is 44200 pounds ; " BO " 38720 < t 29160 u " 5400 CD " 13300 ^Jfcf " 36760 6Z " 32240 BC " loooo DE " 26320 690. The strain in AJ is 41000 pounds; " BK " 36150 " u " AB " 4950 " AH " 32400 " BC 4< 24700 " 14500 " 691. The strain is 16000 pounds. 692 Six. That shown in Fig. 115. The strain in AO is 96200 pounds: B p 88200 " " C/7 " 53000 " CQ " 25700 44 DR " 17600 " " " ^^? " 8000 " " CD " 8000 " 81600 " " 74800 DE " 10200 " " 24700 " 630 ANSWERS TO QUESTIONS. AO should be g x 15-80) BP " 9 f x 14-49) 9x16 tapered to 9 x 14 CH 44 9 13-41, or 9x14 CQ " 6 x 8-61) DR " 6 y 5-8 9 ) 6x9 tapered to 6 x 6 AB " " 4 x 6-41, or 4x 7 CD " 4 6-41 " 4x 7 AN "' 8.K . O O '3 x ii' 77 9 x 12 HM " U-8 H X IQ.SO " \ / ">r\ 1-133 area, or ij diameter. 2-744 4< " 2 RETURN CIRCULATION DEPARTMENT TO*- 202 Main Library LOAN PERIOD 1 HOME USE 2 3 4 5 6 ALL BOOKS MAY BE RECALLED AFTER 7 DAYS 1-month loans may be renewed by calling 642-3405 JtaHS^JTS S"-*"* 1 *" 1 * * in fltafl the books to the Circulation Desk Renewals and recharges may be made 4 days prior to due date DUE AS STAMPED BELOW INTERLIBRAF Y LOAN OCT 1 9 QP4 I IF k LlI\/ -r*Lr UNIV. OI- UAJL. F., BERK. UNIVERSITY OF CALIFORNIA, BERKELEY FORM NO. DD6, 60m, 1/83 BERKELEY, CA 94720