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Full text of "Theory of transverse strains and its application in the construction of buildings : including a full discussion of the theory and construction of floor beams, girders, headers, carriage beams, bridging, rolled-iron beams ... and roof trusses; with tables ..."

REESE LIBRARY 



UNIVERSITY OF CALIFORNIA. 

Received.. ^-/tfai*^ ,i8?/ 



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HATFIELD. And its Application to the Construction of Buildings, includ- 
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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 




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