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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 Fran§ais.
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.
260 — 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 wrould 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. Wve 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> and 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 Pre§sures. — 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 Fre§§ure 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 Wt 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 — = 4Q2T4T 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
+ 3°7A = 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 subtracting1 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=Rt
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. — Re§i§taiicc 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. — Re§istance to Comprc§§ion— 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 = Bbd3 (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 Tran§ver§e 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= 4Sbda (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 Bbd2 ; therefore,
4$bd* = Bbd*
hence Wl = Bbd2 (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
Bbd2 — ^Sbd2 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 be3
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.
bda
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
i§B = 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?
43B — 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-\ and that P, the portion resting
upon the more remote end, is (form. 4-) equal to W—,- ;
where W7 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
pn=w™-n = wmf
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 = W1^ = 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, Py 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 — Rc§ult al§o 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 = Sbd2
" 19, $Un - Sbd2
" 20, \Wl= Sbd2
21, \Ul = Sbds
and, since (Art. 35) ^S = By S = J/?. If in the above we
substitute this value for 5, we shall have the following rules :
For case i, ^Pn — Bbd2 (15.)
" 2, 2 Un = Bbd9 (18.)
" 3, Wl = Bbd9 (9.)
" 4, $Ul= Bbd2 (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 = \Bbd2 ; 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 = Bbd2 (21.)
" 4, \Ual- Bbd2
„ 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 = Bbd9
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 — Bbd3}, 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 = Bbd2 ; 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,
bd2 = 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 = bd2 = o-6dx d* = o-6ds
= d3 = 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 = bd2 — 3<af*
; -ii±2i=,/. = 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 bd2 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 = d3 = ^S43
Second. If the breadth and depth are to be as 6 to 10,
56-47 = 0-6 d3
and £ = 4.55x0.6 =
VARIOUS LOADS COMPARED. 75
Third, If the breadth be fixed at 2 inches, then
56.47 = bd3 = 2da
^=j* = 28.24
2
d= 5-31
Fourth. If the depth be fixed at 5 inches, then
56-47 = bd2 — b x 59
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 bd2 in the formula
Wai = Bbds, 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 application1 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
= Bbd2
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.
92B — 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 PIa§terSng. — 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\ = S2^ 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 Ca§es. — 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- acfl2 = Bbd* becomes
•J- ac 90 /* = Bbd3 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,
_L8oT=^ (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.)
i8ox2o2 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
bd2
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 82 102
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 l3O-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. I360 — 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* = Bbd2
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 ; or> 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— Bbd2
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*
lafng2 = Bbd3
I44B — 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 (d—ij, 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
\acfl2 = Bbd2
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
\acfl2 + afg "^- = Bbd9 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 — l—n = 20— 16 = 4, B — 500 and
d= 10. Therefore,
2 + .ioxi6ax-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) = Sbd2 = -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 d2 — 348000
348000
ds—-^^ -=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* =
Bbd2 = 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' = 4XI5of I5(I5XIo+52)^7 + iXIix252
*= '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 Wy
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 Wy 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><iVx202 —4-34
500XI22 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 = Sbd2
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 = Bbd2
and if P be taken as £ W, W being the load at the centre
of a whole beam, we have
2 *%Wal=Bbd*
Wai = Bbd2
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 Cy
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- 29- fore (see Art. 26) the same
formula (36.) is applicable, namely :
tfax = Bbda
P = ^W, and by substitution
(37.)
2 Wax = Bbd3
a rule applicable to this case, conditioned that x shall not
exceed £/.
When x = / then we have
Wai = Bbd2
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+ exy 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
-?-—bd—iQ
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 bda
4 x 10000 x 4x 10 . „
= W = = 3200
and, making d = 16, have
b x i62 = 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 Expres§ed.—
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^Sbd2 and
Px = Sbdf from which
c\y\\ Sbd* : Sbdf and when
S and b are constant
c : y : : d3 : 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, dt, din dilit 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 dn d:t, dllt,
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 d2) d2 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<=WUjx> (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 " d2 = 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
d2 = -- — ^x'—iA.-^x1
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, gt
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 Br.
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 A1 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 A1 and Bf,
**'"\'+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' : yr
h • p-g ::.*": y" -g
h(y"-g) = x»(p-g)
hy"-hg = x"(p-g)
hy" = x"(p—g) + 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 Bf, 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 Gt we have A'^ for A', B'-- for Bf, and
C-r for C ; which together give
n^ A'ms + B'rs + C'rv
±>(JT = —
To find H, we have Af~ for A', B'~ for Bf, and C~
for C ; which added, will equal
CH — -(A'm + Bfr -
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 = tJLx"> + 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 \
d2 = : (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---— ..
0
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,
Thxt T ht
from load at A, L—j- — %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 gL—r, 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 + 2n2—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,
Ay 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 = 22 = 4
3 " « = 3, " " = 32 - 9
4 » = 4» " " = 42 = 16
5 " «=5. " " = 52 = 25
6 0 = 6, " " = 62 = 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 20—1.
na
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 ns we put (n— i)2.
The denominator is found by taking ;/— i for «, or
2(0—1)—!, equal to 20—3, for 20—1. 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
in 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 i2 equals i ; at C there are
two weights preceding, hence the numerator equals 22 equals
4 ; at D there are three weights, hence the numerator equals
32 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> etc. ; 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-fTV
" «= 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 edxt 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 + txtdxt ) = 4 (\hx* + \tx?} — y
By integrating between x — o and x = /, also between
xt •=. o and xt — /i, or making the integral definite, we
have
but h = I - t
therefore ht = (I — t) t
and h2 = (l-t)a
therefore
= (l-t)t+(l-t}s = lt—t*+l*—2lt+t* = Is -It
l6o STRAINS FROM UNIFORMLY DISTRIBUTED LOADS. CHAP. X.
and the formula
et
y = — j (ht + ks) 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™- wht
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 lip§e. — 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
U2 = - (2(IX — X2)
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
ua 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.
I65
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 Fy 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 (n—y) •= du or,
endy—eydy = du
The integral of this expression is (Arts. 462 and 4-63)
/ (endy—eydy) = 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 y2 = -x becomes y2 = 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 = %--n2
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 = Bbd2
[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* = Bbd2 (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>imen§ioiis 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 Pronii§cuoiis 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 = Sbd2 = ±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(l—f), 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. — Greate§t 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 / Aimn 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 +££#) = Sbd2 = \Bbd2 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) = Bbd2
= 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 Oreate§t 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 J1 and is defined thus :
.L
Let HJ = d' EJ ±± x, EG = v = b'+fr EF = «,
AH=af, BJ=V and a'—bf = 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'
af-V
and v — b -\ 77— <
Formula (70.),
gives [putting /// for t(l—t) 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—^ehs + 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.— A§§ignlng 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 = l—h = 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 bf = j (Br r+A' m) = (as above) 2880
and #'— £' = 3680—2880 = 800.
We therefore have, as a resulting1 value of y in formula
1200 + 2880 + -—(4—3) = 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 = Bbd9
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 nt then
t = m, h—s = df , 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'-br
dr
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* — io2, /= 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 + 62) = 15187-5
4 x 20
b' = ico x io x ^— 2- (14 x 6 + 92) == 12375
a'—b1 = 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 ^ JJ
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 Fir§t 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.
I97
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 — (h—s)
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>i§tributecl and Three
Concentrated Loads. — Putting the strain y of last article
equal to the resistance (Art. 35) gives us
% Uni~ + ™(A 'n + B's + Cv) = Sbd*
and with B — ^S and a as the coefficient of safety,
'n + B's + C'v) = Bbd2 (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 Af, 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 J3 + 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(s—v) ; and for the load at C the
same, C' = :kfg(s—v). 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.
1—6
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.
-32)] = 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=cf, 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 Cy 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, af = ™Afn
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, n—ii, 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-.— A§§igning 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 At 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'Hj (103)
" B, " b' = j(Arm+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 == ^(7X7 + 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, Ji—12, 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(in—ii). 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 (m—v)
U— \cfl
n = \cfln
b = jntfcfln + tfS* ^^ + i/^) + \fgnv(m-v)\
b =
t> =
b =
(cnl+gn m—v +g/) + \fgnv (m—v]\
—v) +gms* +gnv(m—v)\
+gn(m—v) (
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'-43)] = 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> wiU 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 Ns.
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. — Re§istance 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 BCy
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. — Re§i§tance 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 btt ; and, if at be put for
FG, the average distance of the slice from F will be
at— \t\ and therefore the resistance of the top slice will be
R = *X«,-iO
In like manner, if ct 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 Re§istaiicc§ 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
(al— J/)V/ = R
Again; for the second slice, x — at—%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, + Ru -f Rni + etc. +Rn = M
will equal the total resistance of all the fibres, thus :
M = (at— \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 \ 2n—i
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, n2 (Art. 200), and hence
the sum of these several second terms is n2 x ^n = 4n3. 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 = ^ [i2 4- 3* + 52 + 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, " " = J4,256
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 dxy 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= at, 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, at = cy 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 PN3. This power,
PN3, however, not only overcomes the resistance, ^rbd3,
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 -^rbd3. When the power and the work are
equal, we have
PN3 = -i^rbd3 from which,
PN3
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 L3 = ~i2l3 = 172*1*
By substitution in formula (112.) we have
1728 J^75 _ 1296*07*
~ rbd3
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.— Conipari§on of F with E9 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 L3= i27*= i;28/5.
Substituting this value in (114-} we obtain
1728 Wl3 $6Wl3
6 — —
or _
36
In formula (113.} we have
F^-1
Multiplying this by 12 gives
48^7 El
E Wls
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
L3 W L3 W
-T-TS* = E, instead of — TT^ > f°r a lever loaded at one end ;
UW '
and on p. 94, - f-hiiT = £• 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.
234 RESISTANCE TO FLEXURE. CHAP. XIV.
We obtain from (113.)
or, in this case,
Wl3
d3 — — -
~ Fbd
d3 — 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, N—lIn 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,
Gl3 =
a
Gl — j2Fbd2e
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 — Bbds ; a comparison of this value of
Bbds 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
Wy 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 ri —
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 i63 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 i6a
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~r3 » **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 53 80 x 125
6—- —4- = - ^ = 0-7847
5900 x 3 x 6 59 x 216
325. — Te§t 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
72d2 72 X 0-00109 X I02
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.)
Wl3
~~Fbds
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 io3 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 202
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 93 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 i62
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 o3 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
Wl9
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 83 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 io3 _ 50000 _
r ~~~63x 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 203
</=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 =
Fbd3
P =
Fbd36
i6n3
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 73 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 x43 _
"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' l6xIOQX33 = 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,
WL3
o —
and [formula (530.}, p. 484, same work], for a beam uni-
formly loaded,
UD
Comparing these two equal values of eJ, we have
WL3 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
Wl3
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, 19 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,
1—20, F= 5900, d— 10 and 6=1.75; and formula (133.)
gives us
5 x 8000 x 2o3
= 8 x 5900 x io3 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 io3
8 x 2800 x 3 x 63 "
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
PN3 UN3
'- 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 io3x 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 n9 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 io3 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 io3
* = — — 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 io3
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, 72—10, 7^=5900, b = 6 and d= io; and the
formula applicable, (140), becomes
6 x 5000 x ioa
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
= Fbd3r (141 .)
360. — The Rule modified. — For the floors of dwellings
and assembly rooms, ft 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 %cl3 =
= FM.
8x0-03
= Fbd3
cl3 =
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,
cl3 = ibd3 (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 I23
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^cl3 = 8 x o-o<\Fbds
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,
d3 = kbd5 (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 feet> 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
Gl3 = Fbd3S
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,
Gl3 = Bmbd'rl
Gl2 = Bmbd'r
dmr
268 RESISTANCE TO FLEXURE— FLOOR BEAMS. CHAP. XVII,
We have by formula (10.), Art. 36,
Wl
*w
or Wl = Bbd2
Comparing this value of Bbd* with that above, we have
Gla
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-Br
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 io2 _ 5900 x 3 x ioa
yy -— -
jjr 17
i 770000
379. — Te§t of the Rule. — To test the accuracy of the
result just found, the same problem may be solved by
formula (113.),
W>_
F''~~bd3d
from which we have, when <? =rl, and substituting G for
W>
Gl* = Fbdsr
_ 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 10s 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 = Bbd2
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 io5
b = —?— —9 = 4-409
16x3500x0-03 XQ
or the required breadth is 4| inches full.
383. — Example of a Header in a Fir§t-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 i33
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 I28
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 253) + (16 x iQ2 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 I23
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" (1631
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 I53
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 + s2) + TV/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 = 5QOx5i43[12 x I5 (I5 x 1Q + 52) + TVx H x 253] = 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, j—ii,
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 2Q3]
3500 XI23 - 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, c—i\ and
1—2$. 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 58
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 + 7'53) =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.)
- fVx i^x 100 x n -49x8- 51 + 17746 -875 +
* ^ O * O^ L—
3500 X I2J
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(9x 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.}
2a Ki x r± x J J x 20)4-10(9x11 +62)] = 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 =0i%> / = 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, r7 = 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, Equicli§tuiit
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
— ^l—t 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 '» and by formula
(186.)
b = 28oox°?i-x1o.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~> and 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 x4i2'7x 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 r5 / / 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 £= J5> </=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 + 73-4')] = 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 Oreate§t
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.o4L9(*XI* x 11 x 20 + 14 x 6') -.-
(14 x ii xy'— 42)] = 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
3°4 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 H7TV 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 = nct and substituting this value
for m in (197.) we have
r = n*~ (198.)
In this expression, a', c* and p3 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. — I5icrea§ed 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,
ibd3 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~ftx 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- 67- 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 bd3 represent the moment
of inertia, and /, its symbol, may be substituted for them,
thus:
PNS PN3
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 i2 = 16 x i
" " second " I2 = 16 x 3' = 16 x 9
" " third " I3 — 16 x 52 = 16 x 25
" " fourth " 74 = 16 x f = 16 x 49
" "fifth " 75 = 16 x 92 = 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-> ¥ and 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 " 74 = 8 x Q-)2 = 8 x **- = 2 x 49
" fifth " 7, = 8x (|)* = 8x ^L=2x 81
" sixth " I6 = 8 x (^ = 8 x -LfL = 2 x 121
" seventh " 77 = 8 x (-V3-)' = 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, /, = i6x£x(£)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 i2 = £x ia
" second " I2 = 16 x \ x -^ x f = £ x 32
" third u I3= 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 203 = 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. 71, 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",
42, 52 and 62.
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
= 22
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 4O3 = 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 == i2 x \f
second " / (3 x ^) = 3* x i'3
( t \2
third " /^5 x -J = 52 x^t3
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[i2+ 32+ 52+72+ ...... (2n— i)2] (208.)
But the sum of the series I2+ 3" + 52 + 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,
t3n3 = (tn)3 & (%dj m \ds
and this value of tsn3 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 df 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 (£%bd3)y as in
(204*), is the moment for the rectangle ABMN. Again, if
it be applied with x = \dt, 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 = \dt, and by Art. 463,
and since / = b — bt, 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
/ = TV(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 d2 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 = d2 + h, bt = b—t and dt = d2—h
By substitution, we now have
1 2/ = b(d2 + //)' - (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 (b—t)df — bd?—td? = bd?—atd* (putting at for
the area of the web); and since dj — d2—h, therefore we
have
= b(d2 4- h)3 - \b(d2-h)3-atd?
1 2/ = b(d, 4- //)* - b(ds—h)3 + atdf
Then we have
(d, + /i)3 = d/ +
(d2 + h)3-(d2-/i}3 = o + &//// + o 4- 2k3
Substituting these in the above, we have
I2/ = b(6dfh 4- 2k3) + atdf
The area of the top flange equals bh = a, therefore
1 2/ = 6a(d;+ \h2} + atd? 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 at are
each multiplied by the factor df, in the other the factor for
a is (X/+i//) and that for at 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, dt — 10,
<4=n, / = J, £=4, ^ = 3i, # = 4x1=4, a/
and h = i ; and by formula (214) we have
/= .^x 1 12 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 io3)] = 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 dt and da
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.)
Wls
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^.
Wl8 t
deflection, 6 = -, (219.)
744000/
Wla
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 2O8
=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 2o3
/ — - - = 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
\2bd3 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 — -£?bd3 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(bd3-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, bt and dt the breadth and
depth of the part which is wanting of the solid bd (Art.
470) ; <?, b, d, bt and dt 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
Wl3
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 2H3 = \2FId
i6Pn*= \2FId
i6Pn 3 = Fd (bd3-btdf) (224.)
and Pn3 = %Fl6 (225.)
and further, since F= 62000 (Table XX.), therefore
Pns = 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
\Ul3 = F6 (bd3-btd^j (227.)
By Table XX. F = 62000, and the formula reduces to
Ul3 — 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.),
Pn3 =
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 Un3 = F6 (bds-btdf) (229.)
Putting for F its value 62,000, and reducing, we have
Uns — 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.)
ioj
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€* — I3
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 (®$7*\ and the results are re-
corded in Table XVIII.
344 ROLLED-IRON BEAMS. CHAP. XIX.
502.— Floor Beams— Distance from Centres. — If in for-
mula (235.) we substitute for / its value in (233.) we shall
have
ngoAOoIr
i 90400 Ir
*-i-
i i goAooIr y
320^ == - Zjs- --- -
i i 90400 Ir y
y
320/5 320 x 3
c =
I3 960
This is a rule for ascertaining the distance apart from cen-
tres between rolled-iron beams, in floors of first-class stores,
with a given rate of deflection.
503. — Example. — At what distance apart should Phoenix
15 inch 150 pound beams 25 feet long be placed, with a
rate of deflection of r = 0-045 ?
Here we have 7=514-87 (Table XVII.), r = 0-045,
/= 25 and y = 150; and in formula (238.)
3720x514-87x0-045 150 _
•~~
or the distance required is 5-36 feet, or 5 feet 4^ inches.
504.— Floor Beams— Distance from Centres— First-class
Stores. — If the rate of deflection be fixed, and at 0-04 of an
inch (Arts. 3l3f 314 and 368), then formula (238.) becomes
DISTANCE BETWEEN. CENTRES, "iX STORES — FLOOR ARCHES. 34$
»
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 €on§iderafion§. — 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.
348 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(d—i)* 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 Fir§t-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 2T8¥ 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 7—125, c — 2f, m — 5, n = 20, g = 20 and
I — 25 ; and by formula (252.) we have
i x 5 x 2O
f==~ 29?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, bd3 and r, in formula (183.), their re-
I/
spective values be substituted, namely, /= 140+ — (form.
>.), F= 62000 (Table XX.), bd3 = 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 '
2232Q5'35 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, bd3
v
and r their respective values, as follows, f= 320 + —--
(form. 233.), F = 62000 (Table XX.), bd9 = 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 bd3, /, F and r, their respec-
y
tive values, as follows, bd3 = 12! (form. 205.), f — 140 + -;
6C
(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 * '84x6[fx5>84xi4x20+i6x9x(i4 + 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 Oreate§t
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 bd3 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 \i™l+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 bd8, /, 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,
1—20, 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 bt
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 btdt
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 af 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
372 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. — €on§truetion 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 Fir§t-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 wt equal the width of the bottom
flange and wtl that of the top flange, tt equal the thick-
388 CAST-IRON GIRDERS. CHAP. XXI.
ness of bottom flange and tu that of the top, a, equal the
area of the bottom flange and aa the area of the top flange,
then a = wft and wt — ~~ ; also, aa = wata and wu = — - ;
and from these, remembering that at = 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, at 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=485ogX
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, d—2Q and
l—2Q\ 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 24'74
589. — Safe Load Uniformly I>i§tributed— 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 at 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 at, ' 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
catd — 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
td = 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 K275QQ 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.
396
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. ular 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 at 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 at 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 at 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 Df 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 Wt 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 ct as in the fig
ure, we have
c : a : : W/ : A,
in which At equals the strain caused by the weight Wt
through the line WA ; and
c : b : : W, : Bt
B, =- W~
in which Bt equals the strain caused by the weight Wf
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 Bn the strain in WB, is to Ht, the
horizontal strain ; or,
b : h :: B, : ff,-B^
and by substituting the value of Bt 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,
Wt 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-S36 = 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 -3T9- = 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 At
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, Dy 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 20 — 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 y1^ 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 TL (= -&),
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)x4x) =
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
120—20 100
C = -- - = - ; - = 12 • SO7
2400 _ 12-308—4.5
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 " " " "
« « IQy tt I2y « 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:
120—144 —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 Tfiy 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,
QPy 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 tracing6 di-
ag"olhalf. 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 Fy 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 MDEFNy 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.
635a — 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 + A7", 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
Da — ($L + M) - = (5000 + 2500) - = 1 121 it pounds.
STRAINS IN LOWER CHORD. 43C
The strain in the diagonal BP is
c ,6-7268
D3 — (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
D5 = (%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,
Hf = -1/ [(2^- x 10000) + (2 x 2500)] = 33333-} pounds ;
equals the horizontal tension in QP.
442 FRAMED GIRDERS. CHAP. XXII.
For the next bay, PM, we have for the weight, as per
Art. 639,
W = V1 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,
Hs = -V-[(6i 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" = W6 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 I22)] 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 = nh2. Putting also for r its value
TO/
(/ being taken in feet) we have
h2n =
C
Ctfn ±= Wa +
Ctfn = Wak3 + (| x I22 Wael*}
Ctfn - Wah3 = | 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
452 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,l—An, 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 Rt 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 Rs.
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 R3.
In like manner, with the weight D and distance v of
Fig. 94, obtain ^A" equal to Rt ; and with the weight E
and distance t obtain KU equal to R5.
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 Tru§s 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. i°7. 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 At 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 Tru§s 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
Wy 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 LQy 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 Wt .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 BJ 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
476 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 d—a~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 Tru§s. — 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 T6^5y 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
IU12 — 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. — Dimen§ion§ of Part§ Subject to €ompre§§ion. —
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 82)] 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>imeii§ion§ 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>imen§ion§ 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* = ibd3
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.),
cl3 = kbds
which, with £=i, and with c in inches, becomes
\2kd3
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 i«o6. 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_
' I3 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
I3 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
0«
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 696a
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
-84
•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
•38
•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
-38
•31
• 26
13
• •
3-i8
2-23
1.63
1.22
•94
•74
•59
.48
•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
0
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
2.63
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
•I9
• 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
.48
•36
.28
• 22
.18
.T4.
• 12
•10
12
1.83
1-23
• 86
•63
•47
•36
•29
•23
.I9
•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,
dt
X
~oT
ii i
s|
Pittsburgh .
3
21
2.316
•359
.191
2.125
2.281
3.109
Pittsburgh .
3
27
2.516
-359
•391
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
• 25
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
bt
d,
1
s!
w
PQ
<^o
H °
Pottsville..
6
40
3-375
•4
•25
3.125
5-2
24-T33
Pittsburgh.
6
4°i
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
3-9°4
.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
65
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
sl
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
i54-9T7
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
sl
<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. . . .
I24L
125
4-5
•797
-5
4-
10.656
286.019
Paterson . .
12\
125
4-79
.768
.48
4-3i
10.714
292.050
Trenton. . .
TO1
I24-
125
4-79
.780
•47
4-32
10.690
293.994
Pittsburgh.
12
1 80
5.088
.781
•963
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-,36-
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-42
1.39
2.26
• 57
• T4
Pottsville...
4
18
5.45
2.28
.59
.14
Paterson . . .
4
18
3.61
2.4O
.67
.21
Pittsburgh..
4
24
*
[•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
• 31
1.71
.30
Trenton....
4
37
4.60
3.20
• 3°
Buffalo
5
3°
5.95
4.16
3.01
2.24
• 71
• 33
Paterson . . .
5
3°
5-95
4.16
3.01
2.24
• 71
• 33
Phoenix....
5
3°
5-95
4.16
3.01
2.24
«7l
.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.7c
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-I53.35
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
7-31
5.7*
4.54
3.67
2.09
2.47
.06
1.72
>4f
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\
4.89
5.84
3.71
3-71
3-94
4.72
3-02
3.02
3.22
3.86
2.48
2.48
1%
.05
.05
.21
.67
1.71
L71
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
5.91
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
• 91
3-i8
3.20
3.42
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.
0
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 £ ro1 ? X&
Kvg vS S c? c?^ S-<S 5? 5 5 5-vo1 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 ^ Cri.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"
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ir, irivo vD t>. OO OO O\ w (N
"grJ-.rotnS ^vSv?^^ g^Ho'SS & fLorTorTI)
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oo"oNONtH>ro ^.WVOMCVI co g-vo.covo ON ro in in ro
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rororon-^ in m vn in *> . inxo >nvo vo vo vo vo vo ^
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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 0 ^100 M ro «_ -i-
PH
vovovoovo vovo r^^^
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^
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C> 0" ^vo"1 • • '
-1
t^ t^OO 00 00 00 00 00 ON ON
ON 0 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 ?a§o'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
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pH QH CU PH PH Q-t ^ P-i Q-* Pi
|ll|l I1H1 is||| l||||
527 a
TABLE l£Vl\\.— (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 0 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
0
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 ....
0
^"o^o "^vo^o'o^So
NO'^^OO'OO- oo'oo*a,ovO H •
0
M NO NO g. 0 ON ^^^
t^OO 00 ON C O 0 O O
GO
FH
•^t* rh -^-oo
co O O M
FH
0
o
o
FH
FH
H
M
FH
•aavA
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^ WtniJ2-^ 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
0
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
2.25
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
nfi
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
•24
.25
.26
.26
.81
.88
.91
•97
•97
•97
• °3
.16
.19
.19
[78
.73
.81
.77
.98
3.01
3.01
3.02
3.31
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
• 71
.73
.9°
1.27
•29
• 34
.33
.33
.37
.46
.48
.48
• 57
•89
.84
£
.02
.04
.04
.06
.25
.26
• 41
:£
• 71
• 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 27
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
4-31
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
5.71
5-71
6.02
7.T5
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>CONCO 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
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If
"S-^vo vS"?L ^^SvS'S c,^^^-^ °5- S^hviT SN ^ 8 'ON S ?
invo vo VO ON
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**
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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° ON§N 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 ^^NCT^o;2N 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
0 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.
0
M
S. g oiS SF2 fev5S8.K
fc«sssa vSvSsa "
H MWONO NN(NNC1 NCOfOCOm m CT f*> CO
0
ITlVU VD M_> UN MM(NO»rO OCONOUO
ooiATj-^-ox MCJIT>-^-
QiOONONON OOwm
00
SK ^3«ffa£? ^?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><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§§ 8§§cT
| 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
960
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
37°4
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'°324
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
480
530
322
370
410
8i7o
9500 ;
JI5<>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
0 W I «T
hH H u
jt«l
^M6-j
f M ""' <
^"Ǥ
sssip
See
Table
XLV.
:-di
«I!
0 ^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
532co
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
1I5780
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
0
AVERAGE.
WOODS.
41
35
49
4i
39
35
59
27
47
3^
32
29
27
27
21
69
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
27
41
23
55
24
36
4i
g
40
25
487
528
508
35
Si
35
39
49
37
59
36
£
83
4o
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 3pru<-e) .
33
21
30
59
33
44
" (Ued Pine)
" Rig;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.
0
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 ...
96
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
J74
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
0
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
i€o
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).
0
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
•065
-050
•070
• -060
0
•ooo
ooo
000
•ooo
000
•000
•000
oco
•oco
100
•070
050
•080
060
c65
•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
0
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
-I75
•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
•250
•320
•280
295
•270
•270
•355 -275
475
•310
•265
•340
•295
•'3 '5
285
•28;
385 -2co
5OO
•330
•280
•380
•3J5
•335
•300
•3CO
•4'5 "SOS
0
•ooo
•OIO
•030
•000
•005
•005
•005
•030
•cos
50O
•330
•280
•380
•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
0
•020
•02O
•015
•025
•020
•025
•025
600
•420
•350
....
•4°5
.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-
5l8-
6C2-
652-
699-
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').
0
•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
0
•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
•I65
•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
0
•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
•3IQ ;
•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).
0
500
525
550
575
600
O
600
625
65O
675
7OO
0
700
725
750
775
800
0
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
•480
•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').
0
• 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
0
• 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
0
•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
0
• 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
•7J5
•545 -680
•*6s
375
•570
•605
• 760
....
•62S
*4OO
• 630
•670
•600
0
• 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
0
•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
0
•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
•270
•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
0
•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
0
•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
0
•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
0
•OOO
•OOO
•005
•000
•000
•OOO
•000
•OOO
•OOO
•000
400
• 125
• 140
•150
•135
•140
•I2O
•J25
•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
•I3o
•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
0
•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
0
•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
0
•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
•290
•290
•275
0
•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
0
•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
0
•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
,.09
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
•285
215
O
•030
•025
•OIO
•OIO
•025
•OIO
OIO
•OIO
•030
•O2O
1200
•250
•275
• 195
190
•240
1 80
•185
175
-285
•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
•28S
•210
•215
•205
•345
•260
140O
•360
•355
•240
•235
•305
•22O
•225
•215
•365
•275
0
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
0
"OSS
•040
....
•040
040
•O4O
O7O
16OO
•330
•310
• 290
•300
•290
165O
•375
•335
....
• 310
•325
•3IO
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
•230
•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
*rv-'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
0
•005
•005
•005
•005
•ooo
•000
•000
•000
•ooo
40O
•165
•145
•155
•170
145
•165
•145
• 140
• 170
425
•175
•150
I65
• 180
150
•175
• 150
• 145
•180
45O
•185
160
•175
•190
160
•185
• lt)O
•155
•190
475
•195
• 170
•I85
•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
•I85
•I85
•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
•225
•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
•290
•2<0
•26$
•305
•255
•295
•245
•245
•300
700
•305
•260
•275
•325
•265
•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
0
•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
•005
•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
•I65
•160 -230
• 185
•170
• 220
•190
. -I9O
•I65
0
•025
•020 -035
•025
•025
•035 -O2O
•025
•O2i;
1OOO
•I6.S
•160 -235 -185
•170
•220
•190
•195
•I65
105O
•175
•170 -250
•195
•180
•235
•200
•205
•175
11OO
•I 80
•180
•275
•205
•190
•250
•215
•215
•I85
115O
•I85
• 190
•22O
•200
•27O
•235
•230
•195
12OO
•195
•2OJ
•240
•2IO
•295
•255
•245
•205
0
•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
0
•040
• o^o
'OJC
•O^'s
140O
•245
•245
....
•270
•265 |
145O
•26O
• 260
• ^oo
• 2QO i
150O
•285
• 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
•S°3
-748
•746
•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).
0
•ooo
ooo
•000
0
•ODO
•000
000
0
•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
•037
6
•114
•132
•096
24
•o5r
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
•089
13
•247 -286 -208
52
• 1IO
•136
•114
13O
• 093
•118
•097
14
15
•267 I -308
•286 -330
•225
•241
60
•iiS
•127
•I47
•I57
•123
132
140
15O
• 100
•107
•127
•136
I04
• 112
16
•305 | -352
•257
64
•136
•l67
•141
160
•114
-146
•119
17
•324
'374
•273
68
•J45
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
-322
76
80
-.63
•172
•'99
•210
•168
•176
19O
20O
•135
I42
•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
•232
244
'94
203
220
230
•156
164
•2I3
•227
171
•181
24
25
'457
•477
:567
•390
•407
96
100
•209
-219
256
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
•278
288
299
•354
•373
317
310
320
•228
•239
.251
136
310
340
•26S
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
0
OIO
015
•005
•OOO
•005
•005
•OIO
•O05
•OIO
200
•305
345
•250
•220
265
•185
•330
•250
300
225
•340
•400
•285
250
•305
•210
.380
-285
•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
*TWD
• 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).
0
•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
0
•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
0
•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
0
•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).
0
•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
0
-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-03
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
0
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
•2I5
•III!
•220
•240
•205
•225
•175
•185
•190
•200
•180
•190
•180
•'95
9OO
•200
•230
•195
•260
•245
•215
•205
•2,0
95O
•215
•2S0
•2IO
•285
•265
•2,0
•235
•215
•225
1OOO
•230
•275
•220
•3°5
•220
•2OO j ....
•250
O
1000
•025
•240
•025
•280
•OIO
•22S
....
•025
•310
•015
•220
•025
•270
•020
•255
!().,()
•255
'3IS
•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-
239I-
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 *™*
0 M 0 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'
654-
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-
396'
1014-
1156-
567-
637-
652-
630-
566-
609.
460-
540-
377-
418-
993-
1342-
S36'
702-
660-
637-
564-
582-
489-
394-
402- \ 410-
785-
530-
501-
593-
614-
654-
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
0
*
^
*
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'
3I91'
3267-
3225-
2323.
2441-
1725-
2294-
6336-
5239-
2253-
3678-: 3714-
2176-
3338-
2919.
2509-
1895.
1683.
2152-
5033-
I930-
3967-
3387-
2682-
1864.
4952-
AVERAGE VALUES OF CONSTANT, F.
5652-
5042-
2052-
3491'
3I6,
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-
564
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 -<
0 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, 312
Fireplaces, framing for 95
First-class stores, carriage beams with one header, 273
• " . " " floor beams, 264
" and headers, 495
" " " rule for headers, 27l
" " " 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 dy . : 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 371
tension in lower, 370
" for floors, area of. k . 377
minimum area of, 383
" " " 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 3r4
of floor beams, resistance to. 260
" resistance to, : 221
" rules for, 242
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.
591
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 3T3
V " tubular iron girder, 367
Hodgkinson on compression of materials 446
Hodgkinson's edition of Tredgold on Cast-iron, - . 386
experiments, . .'.... 50°
" rule for cast-iron, load at middle 388
" "set" in testing, , , . . 5°5
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 328
Rate of deflection per foot lineal 239,261,264,267,342
11 " " in floors 240
" " rise in brick arch in floor 346
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, 422
frames and diagrams, 4*8
" lettering of lines and angles, . 4J8
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? 48X
considered as girders 4c(,
with and without tie-beams 459
truss, force diagram for a „ 46i
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 5o
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 . 391
" 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 3°9
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. 6l5
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 5°4
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. 483
" 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
620 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, o«93i 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 Wls = 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'-bld!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 x2«49 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 "
39r37
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:
« Bp « 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 5-89)
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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