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Full text of "Mechanics of engineering"

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WORKS OF PROF. I. P. CHURCH 

PUBLISHED BY 

JOHN WILEY & SONS. 



Mechanics of Engineering. 

Comprising Statics and Dynamics of Solids, the 
Mechanics of the Materials of Construction or 
Strength and Elasticity of Beams, Columns, Shafts, 
Arches, etc., and the Principles of Hydraulics and 
Pneumatics with Applications. For the Use of 
Technical Schools. 8vo, xxvi + 854 pages,4 half 
tones, 656 figures, cloth, $6.00. 

Notes and Examples in Mechanics. 

With an Appendix on the Graphical Statics of 
Mechanism. 8vo, 167 pages, 178 figures, cloth, $2.00. 

Diagrams of Mean Velocity of Water in Open 
Channels. 

Based on the Formula of Ganguillet and Kutter. 
gX 12 inches, paper, $1.50. 
Hydraulic flotors ; with related subjects ; includ- 
ing Centrifugal Pumps, Pipes, and Open 
Channels. 8vo, 279 pages, 125 figures, cloth, 
$2.00. 




One of the two " clinometers " in use in the Testing Laboratory of the College of Civil En^- 
neering at Cornell University (see p. 241). The main barrel or sleeve of the instrument encircles 
the horizontal shaft or rod (in testing machine) whose angle of torsion is to be obtained, near 
one extremity of the same. At each end of the barrel are four brass screws having smooth 
rounded ends where they bear on the shaft. These are used for centering the barrel on the 
shaft, but do not grip it. The four steel " gripping screws," at the middle of the barrel, are 
thumb-screws with flat heads and hardened sharp points. They serve to grip the shaft after 
the centering is completed. After the shaft has thus been gripped at a certain transverse 
section, the collar carrying the graduated arc is clamped upon the barrel, the plane of the arc 
and its vernier arm being that of the points of the gripping screws. By taking a reading of the 
vernier on the arc at any stage of the test (the vernier-arm being adjusted each time so that the 
bubble of the spirit level carried by this arm is brought to the center ot its scale) and subtract- 
ing its initial reading, the angle through which the transverse section has turned from its initial 
position becomes known. The second clinometer is placed at another transverse section, near 
the other end of the shaft, and serves to measure its turning movement. The difference of 
these movements is the angle of torsion. The verniers read to single minutes. (The shaft in 
above figure is H in. in diameter). 

Frontispiece. 



Mechanics of Engineering 



COMPRISING 

Statics and Kinetics of Solids ; the Mechanics ov the 

Materials of Construction, or Strength and Elasticity 

OF Beams, Columns, Shafts, Arches, etc. ; and thk 

Principles of Hydraulics and Pneumatics, 

WITH Applications. 



FOE USB IN TECHNICAL SCHOOLS. 



IRVING P. OHUROH, O.E, 



Professor of Applied Mechanics and Hydraulics, College of Civil Enginkkring, 
Cornell University. 



REVISED EDITION, PARTLY REWRITTEN 
total issue, fifteen thousand. 



NEW YORK: 
JOHN WILEY & SON& 
T^KDON CHAPMAN & HALL. Limitid. 
1908 



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\ 



O)- 



USHARV of CONGRESS 
IwoCooies Kecetvb« 

JUL 1 2908 

ouvj^iiKiii collar 

COPY d; / 



Up. 



C!opyrisht, 1890. 1908 

BT 

iHviNG P. Church. 



SUibert BrumnwnJi ani> ©ompaMg 
N«n Inrk 




PREFACE. 



In presenting a revised edition of this work for the use of technical 
schools the writer would call attention to the principal changes that have 
been made; omissions as well as additions. 

The chapter on "Continuous Girders by Graphics" has been omitted 
in its entirety, while the graphic treatment of the horizontal straight girder, 
formerly a part of the chapter on "Arch Ribs," has been removed to the 
appendix, in which will also be found varioiis paragraphs involving special 
problems in flexure, once located in the body of the book. Former chap- 
ters V and VI in Part III, on beams under oblique forces and on columns, 
respectively, have been merged in one (Chapter VI), -^e matter having been 
largely rewritten and more fully illustrated, with introduction of the more 
modern formiiisB for columns and some treatment of the problem of eccen- 
tric loading. 

Chapter V in Part III of the revised book, on "Flexure of Reinforced 
Concrete Beams," is entirely new and presents both theory and numerical ' 
illustration; as also diagrams aiding in practical design. New matter will 
also be found in an analytical treatment of "Circular Ribs and Hoops," 
placed at the end of the chapter on "Arch Ribs." Two other new chapters 
in Part III, are XII, on the flexure of beams treated by a geometrical method 
(which, however, does not call for the use of drafting instruments) leading 
to a very simple and available form of the Theorem of Three Moments ; and 
XIII, which gives the analysis of stresses in thick hollow cylinders and 
spheres. A few pages on the strength of plates have also been added in 
Chapter III. 

In Part IV additional matter is presented relating to the differential 
manometer, gas- and oil-engines, the Cippolletti weir, losses of heads in 
pipes and bends, the hydraulic grade-line, the Venturi meter, current-meters, 
Pitot's tube, use of Kutter's formula, etc. In Parts I and II numerous 
additional examples and illustrations are introduced while many pages have 
been rewritten throughout the book, aside from the new chapters already 
referred to. Tables of logarithms, trigonometric functions, and hyperbolic 
sines and cosines, will be found in the appendix. 

Grateful acknowledgment is again due to Dr. H. T. Eddy for the use of 
his methods* in treating arch ribs; to Prof. C. L. Crandall for the chapter 
on retaining- walls ; and to Col. J. T. Fanning for the table of coefficients 
of friction of water in pipes. The writer would also extend his thanks to 
Messrs. Buff and Buff of Boston, for the half-tone cut of their current-meter; 
and to Builders Iron Foundry of Providence, R. I., for the engravings 
illustrating the Venturi meter. 
Cornell University, Ithaca, N. Y, 
June, 1908. 

Note .^Additional matter involving many examples and forming an 
appendix to the present work, but too bulky to be incorporated with it, 
was issued in a separate volume in 1892 and entitled "Notes and Examples in 
Mechanics." A second edition, revised and enlarged, was published in 1897. 

* See pp. 14 and 25 of " Researches in Ghaphical Statics." by Prof. H. T. Eddy, C.E., 
Ph.D., publi.shed by D. Van Nostrand, New York, 1878, reprinted from Van Nostrand's 
Magazine for 1877; or the German translation of the same, " Neue Constructionen aus der 
Graphischen Statik," published by Teubner u. Cie., I.eipsic, 1880. 



INTRODUCTORY NOTES. 



Preparation. — Prior to tlie use of this book the student is supposed to have 
had the usual training given in technical schools in analjrtical geometry and 
in the differential and integral calculus ; and also a year of college physics. 

Gravitation Measure of a Force. Mass and Weight. — Since the gravitation 
measure of a force is the one almost exclusively used by engineers, a brief resume 
of its nature is here given, aside from the paragraph of p. 835, Appendix. 

The amount of matter in a certain piece of platinum, kept by the British 
government, is called by the physicist a pound of mass, but the engineer 
understands by the word "pound" the force of gravitation, or weight, exerted 
by the earth on this piece of metal at London; and if this piece of metal 
be supported, at London, by a spring balance, the scale of which is so grad- 
uated that the pointer now stands at unity, such a balance constitutes a 
standard instrument with which to measure forces for the purposes of the 
engineer. According to the indications of such an instrument the same 
piece of metal, if suspended on the same balance at the equator, at sea-level, 
would be found to weigh only 0.997 lbs. (force) on account of the diminished 
intensity of gravitation; the difference, however, being only about three 
parts in a thousand, or one-third of one per cent. For ordinary engineering 
problems involving the strength of structures, this difference is of no prac- 
tical importance. 

A unit of force based on this gravitation method is called a gravitation 
measure of force. The mass of the piece of platinum, has, of course, suffered 
no change in the transit from London to the equator, and since the fraction 
obtained by dividing the weight (obtained from the spring balance) by the 
acceleration of gravity, g, is constant, regardless of the place where the two 
quantities are measured, it is convenient (though not essential) for the 
engineer to give the name "mass" to this fraction when it occurs in the 
equations of kinetics. For instance, since g (for foot and second) =32.18 at 

1.000 0.997 
London, and 32.09 at the equator (at sea-level), we note that „^ „ = „^ ^„ 

=0.03108. 

Arithmetic. — In arithmetical operations the student should remember 
that the degree of refinement attained or employed does not depend on the 
number of decimal places used, but upon the number of significant figures. 
Thus, each of the quantities 0.0003674 and 510.4 contains four significant 
figures. For instance, let us suppose that the value of x is to be obtained 
from the relation x = a-b, where a = 0.0000568 and b = 0.0000421. Should 
the student conclude that five decimal places would be accurate enough and^ 
thus write 0.00005 for a, and 0.00004 for b, he would obtain a: = 0.00001, con- 
taining only one significant figure; whereas the true result is a; = 0.0000147. 
Hence the former result is seen to be in error to the extent of 47 parts in 
147, or 32 parts in 100, i.e., 32 per cent. ; which is a very gross and totally 
unnecessary error. Values obtained from the ordinary 10-inch slide rule 
usually contain only three significant figures (four if near left of scale). 

Logarithms. — The following facts and operations are not usually fresh in 
the student's mind. The logarithm of a number less than unity is a negative 
quantity but is usually expressed as the algebraic sum of a positive mantissa 
(or decimal part) and a negative characteristic which is a whole number; 
thus, the common logarithm of 0.20 is f.301030 . . . , that is, log. 0.20 = 
-1-0.301030-1.000000 (or, -1-9.301030-10). This should be borne in mind 
in raising such a number to any power. For example: required the value of 

Solution.—^^OMQl and log. 0.8461 = 1.9274, i.e., =0.9274-1.0000. 

Hence 0.71 X log. 0.846_1 = 0.71(0.9274 -1.0000), =0.6584 -0.7100, 
= -0.0516 = 1.9484 = log. 0.8880; therefore a;=0.8880. 
Note that, according to the definition of a logarithm, the statement 
en=m is equivalent to the statement n = loge m. 

iv 



MATHEMATICAL DATA. 



Trigonometry. cos^A +sin^A = 1. 

cos^ A — sin^ A ^ cos 2 A 
sin 2A = 2 sin A cos A 
cos 2A =cos^A — sin^A 
sin A 
1 — cos A 
Solution of Oblique Triangles, etc. 

B 



2 sinM = l — cos 2A. 
2cos2A = l + cos2A. 

tan A 1 



cot iA = . 



sin A = 
cos A = 



Vl+tan^A cosecA 
1 1 




Given a, b, B; to find A : 



a, b, C; 
a,b,C; 
a, b, c: 



A: 

c: 

C: 



sin A = 
tan A = 



Vl+tan^A sec A 

sin A _ sin B _ sin C 
a b c ' 

d = a sin C d = c sin A 

m = c cos A n = o cos C 

d = mtanA d=n tan C 

a sin B 



a sin C 



cos C 



b — a cos C 
c' = a^ + &^ — 2a& cos C 

a^ + b'^ — c^ 



2ab 




^—■{a- 



r 



(the 



Mensuration. Area of a circle =;rr'; 

^ circumference = 2n,r. Area of sector 

^ / a° 

A5C0A=(3gQ, 

latter a ia radians). Vol. of sphere = 

4 

■^Ttr^. Area of the segment, ABC DA, of 

a circle, 






W 



= (area of sector ABCOA)- (area of triangle ACO). 

Area of rightsegment of a para6oto= two- 
thirds that of circumscribing rectangle, 
= f(2/ia). Equation to curve OA is 

v^ X — 

p = -. Distance OC, of center of gravity, 

3 

is X = ?a, from vertex 0. 
ti o 




VI 



MATHEMATICAL DATA. 



Integral Forms. — (Each integral to be taken between limits, or to have 
a constant added and determined). (See also p. 480.) 



ixndx 



xn+1 



J n + 1 

cos X da; = sin x; 
j* dx 



J X 



dx 



\/.T^ ± a' 
dx 



= sin— ix; 

= loge(a; + v'x^ 
1 



isin x dx= — cos x ; 
' dx 



1 + x^ 



'-'y' \^a 



tan— ix; 
dx 



+ hx — cx^ 




, y/ab + bx 
loge -7=- 



l'=-2^'^oge{a-bx^). 



\a-bx^ SVab'"* Vab-bx' ja-bx^ 

Numerical Constants. — The acceleration of gravity, g, (for the English 
foot and second) is 32.16 for the latitude of Philadelphia at sea-level, and 
for any latitude ji, and elevation h above sea-level, is 

32.1723-0.0833 cos 2/?-0.000003/i. 
For ordinary problems in mechanics, however, in the northern United 
States g may be taken as 32.2, for which value we have 



\/23 = 8.025; - = 0.03105; 
9 



and j^ =0.01553. 
22 



The ratio 7r = 3. 141592, or approx. 3J-, i.e., -=-; 

-=0.31831; ;r2 = 9.86960; ^=0.10132; V'7r= 1 .77245. 

1° = 0.01745 radians. One radian = 57° 17' 44.8". 
If n denote, any number, then 

Login (n) = 0.43429 X logs (r>) ; and loge (n) =2.30258 Xlogio (n). 
Base of nat. logs. = e, =2.71828; base of Briggs system = 10. 



GREEK ALPHABET. 



A a 
B p 

ry 

E € 

z c 

H V 
Odd 
I I 
K K 

A X 
Mix 



Names, 

Alpha 

Beta 

Gamma 

Delta 

Epsilon 

Zeta 

Eta 

Theta 

Iota 

Kappa 

Lambda 

Mu 



Letters. 


Names. 


N V 


Nu 


E a 


Xi 


o 


Omicroa 


Htt 


Pi 


P P 


Rho 


2 (T s 


Sigma 


T t 


Tau 


Tv 


UpsiloQ 





Phi 


Xx 


Chi 


W^ ip 


Psi 


n CO 


Omega 



TABLE OF CONTENTS. 
[Mjechanics of solids.} 



PRELIMINARY CHAPTER 

PACS 

§§ l-'lS'i. Definitions. Kinds of Quanlily. Homogeneous Equa- 
tions. Parallelogram of Forces ,... 1 

PAET I. STATICS. 

Chapter I. Statics of a Material Point. 
IP 16-19. Composition and Equilibrium of Concurrent Forces. ... 8 

Chapter II. Parallel Forces and the Centre op Gravity. 

§§ 20-22. Parallel Forces 1^ 

§§23-276. Centre of Gravity. Problems. Centrobaric Method. . . 18= 

Chapter III. Statics of a Rigid Body. 

g§ 28-34. Couples 27 

^§35-39. Composition and Equilibrium of Non-concurrent Forces. 31 

Chapter IV. Statics of Flexible Cords. 

§§ 40-48. Postulates. Suspended Weights. Parabolic Cord. Cat- 
enary 4r 

PART II. KINETICS. 

Chapter I. Rectiijnear AIotion of a Material Poini 

^§ 49-55. Uniform Motion. Falling Bodies. Newton's Laws, 

Mass .... 49 

g§ 56=60= Uniformly Accelerated Motion. Graphic Representa- 

tions. Variably Accelerated Motions. Impact. . . 5^- 

V Chapter II. Virtual Velocities. 

SS Q1-6&, Definitions and Propositions. Connecting-rod. Prob 

lems 67 



yiil CONTENTS., 

•^ChAPTBR III. CUKVILINEAR MOTION OP A MATERIAL POINT. 

PAOS 

§§ 70-74. Composition of Motions, of Velocities, etc. General 

Equations 72 

§§ 75-84 Normal Acceleration. Centripetal and Centrifugal 
Forces. Simple Pendulums. Projectiles. Rela- 
tive Motion 77 

Chapter IV. Moment of Inertia. 

%% 85-94. Plane Figures. Rigid Bodies. Reduction Formulae. 

The Rectangle, Triangle, etc. Compound Plane 

Figures. Polar Moment of Inertia 9t 

§S 95-104. Rod. Tliin Plates. Geometric Solids 98 

■§§105-107. Numerical Substitution. Ellipsoid of Inertia ICo 

Chapter V. Kinetics of a Rigid Body. 

§§108-115. Translation. Rotation about a Fixed Axis. Centre of 

Percussion 105 

|§ 117-121. Compound Pendulum. The Fly-wheel 118 

§§ 122-123. Uniform Rotation. " Centrifugal Action," Free Axes. 125 
|§ 124-136. Rolling Motions. Parallel Rod of Locomotive 130 

^ Chapter VI. Work, Energy, and Power. 

§§ 127-134. Work. Power, Horse-power. Kit etic Energy...... 133 

§§ 135-138. Steam-hammer. Pile-driving. Inelastic Impact -. 188 

§§ 139-141. Rotary Motion. Equivalent Systems of Forces. Any 

Motion of a Rigid Body .„ 143 

§§ 142-146. Work and Kinetic Energy in a Moving Machine of 

Rigid Parts 147 

§§ 147-155. ^ Power of Motors. Potential Energy. Heat, etc. Dy- 
namometers. Boat-rowing. Examples 153 

Chapter VII. Friction. 

§§ 156-164. Sliding Friction. Its Laws. Bent Lever 164 

§§ 165-171. Axle-friction. Friction Wheels. Pivots. Belting. 

Transmission of Power by Belting 175 

|§ 172-177. Rolling Friction. Brakes. Friction of Car Journals; 
and of Well-lubricated Journals. Rigidity of 
Cordage. Examples 186 



CONTENTS. 



IX 



PART III. STRENGTH OF MATERIALS. 

(or mechanics of matekials.) 

OHAPTER I. ELEMENTARY STRESSES AND STRAINa 

^§ 178-183. Stress and Strain i of Two Bands. Oblique Section 

of Rod in Tension . 195 

§§ 183a-190o Hooke's Law. Elasticity. Safe Limit. Elastic 

Limit. Rupture. Modulus of Elasticity. 

Isotropes. Resilience. Ellipse of Stress. 

Classification of Cases 301 

TENSION. 

Hooke's Law by Experiment. Strain Diagrams. 

Lateral Contraction. Modulus of Tenacity. . . . 307 

Besilience of Stretched Prism. Load Applied Sud- 
denly. Prism Under Its Own Weight. Solid 
of Uniform Strength. Temperature Stresses . . 313 



§§ 191-195. 
i§ 196-199b 

§§ 300-30Sp 

§§ 303-306. Tables 



3§ so^-aia 



COMPRESSION OF SHORT BLOCKS. 

Short and Long Columns. Remarks on Grusliing, . . 

EXAMPLES IN TENSION AND COMPRESSION. 

Factor of Safety. Practical 



318 



Examples. 
Notes. .... 



SHEARING. 



Eivets. Shearing Distortion. 
Examples. ............... 



230 



Punching. 

cooo..^ .. 335 



^ CHAPTER II. TORSION, 

§§ 314-330. Angle and Moment of Torsion. Torsional Strengtila, 

Stiffness, and Resilience. Non-Circular Shafts 383 

§§ 331-333. Transmission of Power. Autographic Testing Ma- 
chine. Torsion Clinometers. Examples 238 

OHAPTER in. FLEXURE OF HOMOGENEOUS PRISMS UNDER 
PERPENDICULAR FORCES IN ONE PLANE. 

§§ 334r-333ao The Common Theory. Elastic Forces. Neutral 
Axis. The "Shear" and "Moment." Flex- 
lu-al Strength and Stiffness. Radius of Curva- 
Uire. Resilience c ... c . ^ .-..., ^ - .. . 344 

ELASTIC CURVES. 

^§ 333-338. Single Central Load ; at Rest, and Applied Suddenly. 

Eccentric Load. Uniform Load. Cantilever. . 353 



X co:n'tents. 

SAFE LOADS. 

§§ 239-246. Maximum Mom^ent. Shear = x-Derivative of the 
Moment. Simple Beams With Various Loads. 
Comparative Strength and Stiffness of Rectan- 
gular Beams 262 

§§ 247-252. Moments of Inertia. Rolled Steel I-Beams, etc. 

Cantilevers. Tables. Numerical Examples . . . 273^ 

SHEARING STRESSES IN FLEXURE. 

§g 253-257. Shearing Stress Parallel to Neutral Surface ; and in 
Cross Section. Web of I-Beam. Riveting of 
Built Beams . , 284 

SPECIAL PROBLEMS IN FLEXURE. 

§§ 258-265. Designing Sections of Built Beams. Moving Loads. 
Special Cases of Quiescent Loads. Hydrostatic 
Load 395 

§§ 266-270. Strength of Flat Plates. Weight Falling on Beam. 

Crank Shaft. Other Shafts. Web of I-Beam . 310^ 



§§ 271-276. 



CHAPTER IV. FLEXURE; CONTINUED. 

CONTINUOUS GIRDERS. 

Analytical Treatment of Symmetrical Cases of Beams 

on Three Supports ; also, Built in ; (see p. 499.) . 320' 



THE DANGEROUS SECTION IN NON-PRISMATIC BEAMS. 

§§ 277-279. Double Truncated Wedge, Pyramid, and Cone 332 

NON-PRISMATIC BEAMS OF UNIFORM STRENGTH. 

§§ 280-289. Parabolic and Wedge-Shaped Beams. Elliptical 

Beams. (See Appendix, p. 841, for Cantilevers). 335 

CHAPTER V. FLEXURE OF REINFORCED CONCRETE 
BEAMS. 

§§ 284-287. Concrete, and "Concrete-Steel" Beams. Beams of 
Rectanaiular Section. Horizontal Shearing 

Stresses in Latter. Examples 33& 

§§ 288-294. Concrete-Steel Beams of T-Form Section. Deflection 
of Concrete-Steel Beams. Practical Fotmulse 
and Diagrams 346 



CONTENTS. XI 

CHAPTER VI. FLEXURE. COLUMNS AND HOOKS. OBLIQUE 

LOADS. 

§§ 294-300. Oblique Cantilever. Moment, Thrust and Shear. 
Experimental Proof. Common Theory of 
Crane-Hooks. Winkler-Bach Theory 352 

|§ 301-314. Long Columns. End-Conditions. Euler's and Ran- 
kine's Formulae. Examples. Radii of Gyra- 
tion. Built Columns. Other Formul:e: Merri- 
man-Ritter ; ' ' Straight-Line ; ' ' Parabolic . 
Wooden Posts. Eccentric Loading of Columns. 
Eccentric Loading Combined with Uniform 
Transverse Pressures. . Buckling of Web of 
Plate Girder 360 

CHAPTER VII. LINEAR ARCHES (OF BLOCKWORK.) 

§§ 315-823. Inverted Catenary. Parabolic Arch. Circular Arch. 

Transformed Catenary as Arch 886 

CHAPTER VIII. ELEMENTS OF GRAPHICAL STATICS 

§§ 831-326. Force Polygons. Concurrent and Non-Concurrent 
Forces in a Plane. Force Diagrams. Equili- 
brium Polygons 897 

§§ 327-832. Constructions for Resultant, Pier-Reactions, and 
Stresses in Roof Truss. Bow's Notation. The 
Special Equilibrium Polygon 403 

CHAPTER IX. GRAPHICAL STATICS OF VERTICAL FORCES. 

§g 333-336. Jointed Rods. Centre of Gravity 413 

§§ 337-848. Useful Relations Between Force Diagrams and Their- 

Equilibrium Polygons 415 

CHAPTER X. RIGHT ARCHES OF MASONRY. 

§§ 344-353. Definitions. Mortar and Friction. Pressure in Joints. 
Conditions of Safe Equilibrium. True Linear 

Arch 421 

§§ 353-357. Arrangement of Data for Graphic Treatment 428 

§§ 358-363. Graphical Treatment of Arch. Symmetrical and 

Unsymmetrical Cases 431 

CHAPTER XI. ARCH-RIBS. 

§§ 334-374, Mode of Support. Special Equilibrium Polygon and 
its Force Diagram. Change in Angle Between 
Rib Tangents. Displacement of Any Point on 
Rib 438 

§§ 374a-378a. Graphical Arithmetic. Summation of Products. . 
Moment of Inertia by Graphics. Classification 
of Arch-Ribs 450 



XU CONTENTS. 

§§ 379-388. Prof. Eddy's Graphical Method for Arch-Ribs of 
Hinged Ends ; and of Fixed Ends. Stress 
Diagrams. Temperature Stresses. Braced 
Arches 461 

§§ 389-391. Cu-cular Ribs and Hoops 479 

CHAPTER XII. FLEXURE OF BEAMS, SIMPLE AND CONTIN- 
UOUS, GEOMETRICAL TREATMENT. 

§§392-398. Geometrical Treatment Defined. Angle between End- 
tangents. Relative Displacement of Any Point 
of Elastic Curve. Deflections and Slopes by 
Calculus. Examples. Properties of Moment 
Diagrams. Deflections and Slopes by Geomet- 
ric Method. Examples 485 

§§ 399-405. The "Normal Moment Diagram." The Theorem of 
Three Moments. Values of Products of Moment- 
Areas by "Gravity x's." Continuous Girders 
Treated by the Theorem of Three Moments. 
Continuous Beam with "Built-in" Ends. 
Deflections Found by Theorem of Three Mo- 
ments 494 

CHAPTER XIII. THICK HOLLOW CYLINDERS AND SPHERES. 

§§ 405a-405sr. General Relations between Stress and Strain. The 
" Elongation Theory" of Safety. Thick Hollow 
Cylinder under Internal Fluid Pressure; also 
under External Fluid Pressure. Equalization 
of Hoop Stress in Compound Cylinder. Equa- 
tion of Continuity for Cylinder and Sphere. 
Thick Hollow Sphere under Internal Fluid 
Pressure 507 



[CONTENTS OF ''MECHANICS OF FLUIVS,"] 
PAET lY.— HYDKAULICS. 

CHAPTER I.— DEFINITIONS. FLUID PRESSURE. HYDRO= 
STATICS BEGUN. 

PAGE 

§§ 40&-417. Perfect fluids. Liquids and Gases. Principle of ' ' Equal 

Transmission of Pressure." Non-planar Pistons .. . 515 

§§ 418-427. Hydraulic Press. Free Surface of Liquid. Barometers 
and Manometers. The Differential Manometer. 
Safety-valves. Strength of Thin Hollow Cylinders 
against Bursting and Collapse 526 

CHAPTER II.— HYDROSTATICS CONTINUED. PRESSURE OF 
LIQUIDS IN TANKS AND RESERVOIRS. 

§§428-434. Liquid in Motion, but in "Relative Equilibrium." 
Pressure on Bottom and Sides of Vessels. Centre 
of Pressure of Rectangles, Triangles, etc 540' 

§§ 435-444. Stability of Rectangular and Trapezoidal Walls against 
Water Pressure. High Masonry Darns. Proposed 
Quaker Bridge Dam. Earthwork Dam. Water 
Pressure on both Sides of a Gate 554 

CHAPTER IIL -EARTH PRESSURE AND RETAINING WALLS. 

§§ 445-455. Angle of Repose. Wedge of Maximum Thrust. Geo- 
metrical Constructions. Resistance of Retaining 
Walls. Results of Experience 573 

CHAPTER IV.— HYDROSTATICS CONTINUED. IMMERSION 
AND FLOTATION. 

§§456-460. Buoyant Effort. Examples of Immersion. Specific 

Gravity. Equilibrium of Flotation. Hydrometer.. 586- 

§§ 461-465. Depth of Flotation. Draught and Angular Stability of 

Ships. The Metacentre 593. 

xiii 



XIV CONTENTS. 

CHAPTER v.— HYDROSTATICS CONTINUED. GASEOUS 

FLUIDS. 

§§466-478. Thermometers. Absolute Temperature. Gases and Va- 
pors. Critical Temperature. Law of Charles. 
Closed Air-manometer. Mariotte's Law. Mixture of 
Gases. Barometric Levelling. Adiabatic Change.. G04 

§§479-489. Work Done in Steam-engine Cylinders. Expanding 
Steam. Graphic Representation of Change of Stat? 
of Gas. Compressed-air Engine. Air-compressor. 
Hot-air Engines. Gas-engines. Heat- efficiency. 
Duty of Pumping-engines. Buoyant Effort of the 
Atmosphere 624 

CHAPTER VL— HYDROKINETICS BEGUN. STEADY FLOW OF 
LIQUIDS THROUGH PIPES AND ORIFICES. 

§§ 489a-495. Phenomena of a " Steady Flow." Bernoulli's Theorem 
for Steady Flow without Friction, and Applications, 
Orifice in " Thin Plate" 646 

§§ 496-500. Rounded Orifice. Various Problems involving Flow 
through Orifices. Jet from Force-pump. Velocity 
and Density; Relation. Efflux under Water. Efflux 
from Vessel in Motion. Barker's Mill 663 

§§ 501-508. Efflux from Rectangular and Triangular Orifices. Pon- 
celet's Experiments. Perfect and Complete Con- 
traction, etc. Overfall Weirs. Experiments of 
Francis, Fteley and Stearns, and Bazin. The Cip- 
poUetti Weir. Short Pipes or Tubes 672 

§§ 509-513. Conical Tubes. Venturi's Tube Fluid Friction. 
Fronde's Experiments. Bernoulli's Theorem with 
Friction. Hydraulic Radius. Loss of Head. Prob- 
lems involving Friction Heads in Pipes. Accumu- 
lator 693 

§§ 513«-518. Loss of Head in Orifices and Short Pipes. Coefficient 
of Friction of Water in Pipes. Fanning's Tabic. 
Petroleum Pumping. Flow through Long Pipes . . 703 

§§519-526. Chezy's Formula. The "Hydraulic Grade-Line." 
Pressure-energy. Losses of Head due to Sudden 
Enlargement of Section; Borda's Formula. Dia- 
phragm in Pipe. Venturi Water-meter 71'! 

.?§ 537-536. Sudden Diminution of Section. Losses of Head due to 
Elbows, Bends, Valve-gates, and Throttle-valves. 
Examples, Capt. Bellinger's Experiments on Elbows. 
Siphons. Branching Pipes. Time of Emptying 
Vessels of Various Forms; Prisms, Wedges Pyra- 
mids, Cones, Paraboloids, Spheres, Obelisks, and 
Volumes of Irregular Form using Simpson's Rule. . 727 



CONTEXTS. XV 

CHAPTER VII.— HYDROKINETICS, CONTINUED; STEADY 
FLOW OF WATER IN OPEN CHANNELS. 

PAGE 

§§ 538-542a. Nomenclature. Velocity Measurements and Instru- 
ments for the same. Ritchie-Haskell Direction 
Current-meter. Change of Velocity with Depth. 
Pitot's Tube and W. M. White's Experiments. 
Current-meters. Gauging Streams. Chezy's For- 
mula for Uniform Motion in Open Channel. 
Experiments 749 

§§ 54Ji6-647. Gutter's Formula. Sections of Least Resistance. Trape- 
zoidal Section of Given Side Slope and Minimum 
Friction. Variable Motion in Open C'hauuel. 
Bends. Formula introducing Depths at End Sec- 
tions. Backwater 758 



CHAPTER VIII— KINETICS OF GASEOUS FLUIDS. 

§§ 548-556. Theorem for Steady Flow of Gases without Friction. 
Flow through Orifices by Water-formula; with 
Isothermal Expansion; with Adiabatic Expansion. 
Maximum Flow of Weight. Experimental Co- 
efficients for Orifices and Short Pipes. Flow con- 
sidering Velocity of Approach - W3 

§^ 567-561a. Transmission of Compressed Air through Long Pipes. 
Experiments in St. Gothard Tunnel. Pipes of Vari- 
able Diameter. Tiie Piping of Natural Gas. ...... 786 

J CHAPTER IX.— IMPULSE AND RESISTANCE OF FLUIDS 

§§ 562-569. Reaction of a Jet of Liquid. Impulse of Jet on Curved 
Vanes, Fixed and in Motion. Pitot's Tube. The 
California "Hurdy-gurdy." Impulse on Plates. 
Plates Moving in a Fluid. Plates in Currents of 
Fluid .... 798 

B§ 57C'-575. Wind-pressure. Smitlisoniau Scale. Mechanic-;- of the 
Sail-boat. Resistance of Still Water to ImmersL'-'l 
Solids in Motion. Spinning Ball, Deviation fiom 
Vertical Plane. Robinson's Cup anemometer. Re- 
sistance of Ships. Transporting Power of a Cur- 
rent. Fire-streams, Hose- friction, etc - 818 

Appendix. Miscellaneous Addenda and Tables 835-854 



MEOHAmCS OF ENGINEERING. 



PEELIMINARY CHAPTER. 

1. Mechanics treats of the nnitual actions and relative mo- 
tions of material bodies, solid, liquid, and gaseous ; and by 
Mechanics of Engineering is meant a presentment of those 

principles of pnre raeclianics, and their applications, which are 
of special service in engineering problems. 

2. Kinds of Quantity. — Mechanics involves the following 
fundamental kinds of quantit}' : Space, of one, t\vo, or three 
dimensions, i.e., length, surface, or volume, respectively ; time, 
which needs no definition here; force and mass, as defined be- 
low; and abstract numbers, whose values are independent of 
arbitrary units, as, for example, a ratio. 

3. Force. — A force is one of a pair of equal, opposite, and 
simultaneous actions between two bodies, by which the state* 
of their motions is altered or a change of form in the bodies 
themselves is effected. Pressui-e, attraction, repulsion, and 
traction are instances in point. Muscular sensation conveys 
the idea of force, while a spring-balance gives an absolute 
measure of it, a beam-balance only a relative measure. In 
accordance with Newton's third law of motion, that action and 
reaction are equal, opposite, and simultaneous, foi'ces always 
occur in pairs; thus, if a pressure of 4:0 11)S. exists between 
bodies A and B, if A is considered by itself (i.e., " free"), 
apart from all other bodies whose actions upon it are called 
forces, among these forces will be one of 40 lbs. directed from 
B toward A. Similarly, if B is under consideration, a force 

* The state of motion of a small body under the action of no force, or of 
balanced forces, is cither absolute rest, or uniform motion in a right line. 
If the motion is different from this, the fact is due to the action of an un- 
balanced force (§ 54), 



2 MECHANICS OF ENGINEBRINO. 

of 40 lbs. di]-ected from A toward £ takes its place ;imoiig the 
forces acting on £. This is the interpretation of Newton's 
third law, 

[Note.— In some common phrases, such as " The tremendous force '' o^ a heavy Dody in 
rapid motion, the word force is not used in a technical sense, but signifies energy (as ex- 
plained in Chap. VI.). The mere fact that a body is in motion, whatever its mass and 
velocity, does not imply that it is under the action of any force, necessarily. For instance, 
at any point in the path of a cannon ball through the air, the only forces acting on it ara 
the resistance of the air and the attraction of the earth, the latter having a vertica in(J 
downward direction.] 

4. Mass is the quantity of matter in a body. The masses of 
several bodies being proportional to their weights at the same 
locality on the earth's surface, in physics the weight is taken 
as the mass, but in practical engineering another mode is used 
for measuring it (as explained in a subsequent chapter), viz.'. 
the mass of a body is equal to its weight divided by the ac- 
celeration of gravity in the locality where the weight is taken, 
or, symbolically, M= G -r- g. This quotient is a constant 
quantity, as it should be, since the mass of a body is invariable 
wherever the body be carried. 

6. Derived Quantities. — All kinds of quantity besides the 
fundamental ones just mentioned are compounds of the latter, 
formed by multiplication or division, such as velocity, accele- 
ration, momentum, work, energy, moment, power, and force- 
disti'ibution. Some of these are mej-ely names given for 
convenience to certain combinations of factors which come 
together not in dealing with first principles, but as a result of 
common algebraic transformations. 

6. Homogeneous Equations are those of such a form that they 

are true for any arbitrary system of units, and in which all 

terms combined by algebraic addition are of the same kind. 

of 
Thus, the equation s = ~ (in which g = the acceleration of 

gravity and t the time of vertical fall of a body in vacuo, 
from rest) will give the distance fallen through, «, whatevei 
units be adopted for measuring time and distance. But if foi 



PRELIMIlSrARY CHAPTER. S 

g we write the niimerlcal value 32.2, which it assumes when 
time is measured in seconds and distance in feet, the equation 
s = IQ.lf is true for those units alone, and the equation is not 
of liomogeneous form. Algebraic combination of homogeneous 
equations should always produce homogeneous equations ; if 
not, some error has been made in the algebraic woi'k. If any 
equation derived or proposed for practical use is not homogene 
ous, an explicit statement should be made in the context as to 
the proper units to be employed. 

7. Heaviness. — By heaviness of a substance is meant tlie 
weight of a cubic unit of the substance. E.g. the heaviness of 
fresh water is 62.5, in case the unit of force is the pound, 
and the foot the unit of space; i.e., a cubic foot of fresh 
water weighs 62. 5 lbs.* In case the substance is not uniform 
in composition, the heaviness varies from point to point. If 
the weight of a homogeneous body be denoted by G, its volume 
by F", and the heaviness of its substance by y, then G = Yy, 



Weight in Pounds of a Cubic Foot (i.e., the heaviness) of vakious 

MATEIIIAL& 



Anthracite, solid 100 

" broken 57 

Brick, common hard 125 

" soft 100 

Brick-work, common 112 

Concrete 125 

Earth, loose 72 

" as mud 102 

Granite 164 to 172 

Ice 58 

Iron, cast 450 

" wrought 480 



Masonr}^ dry rubble 138 

" dressed granite or 

limestone 165 

Mortar 100 

Petroleum _ 55 

Snow 7 

" wet 15 to 50 

Steel 490 

Timber 25 to 60 

Water, fresh 62. 5 

sea 64.0 



8. Specific Gravity is the ratio of the heaviness of a material 
to that of water, and is therefore an abstract number. 

9. A Material Point is a solid body, or small particle, whose 
dimensions are practically nothing, compared with its range of 
motion. 

- Or, we may write 62.5 lbs. /cub. ft.; or 62.5 Ibs./ft.^ 



4 MECHANICS OF ENGINEERING. 

10. A Eigid Body is a solid, M-liose distortion or change of 
form under anj system of forces to be brought upon it in 
practice is, for certain purposes, insensible. 

11. Equilibrium. — When a system of forces applied to a 
body produces the same effect as if no force acted, so far as 
the state of motion of the body is concerned, they are said to 
be balanced, or to be in equilibrium. [If no force acts on a 
material point it remains at rest if already at rest ; but if 
already in motion it continues in motion, and uniformly 
(equal spaces in equal times), in a right line in direction 
of its original motion. See § 54.] 

12. Division of the Subject. — ^to^^'c* will treat of bodies at 
rest, i.e., of balanced forces or equilibrium; kinetics, of 
bodies in motion ; strength of materials will treat of the effect 
of forces in distorting bodies ; hydraulics, of the mechanics 
of liquids and gases (thus mcXxx&mg j)7ieumatics). 

13. Parallelogram of Forces. — Ducliayla's Proof. To fully 
determine a force we must have given its amount, its direc- 
tion, and its point of application in the body. It is generally 
denoted in diagrams by an arrow. It is a matter of experience 
that besides the point of application already spoken of any 
other may be chosen in the line of action of the force. This 
is called the transmissibility of force; i.e., so far as the state of 
motion of the body is concerned, a force may be applied any- 
where in its line of action. 

The Resultant of two forces (called its components) applied 
at a point of a body is a single force applied at the same point, 
which will replace them. To prove that this resultant is given 
in amount and position by the diagonal of the parallelogram 
formed on the two given forces (conceived as laid off to some 
scale, so many pounds to the inch, say), Duchayla's method 
requires four postulates, viz. : (1) the resultant of two forces 
must lie in the same plane with them ; (2) the resultant of two 
equal forces must bisect the angle between them ; (3) if one of 
the two forces be increased, the angle between the other force 
and the resultant will be greater than before; and (4) the trans- 
missibility of force, already mentioned. Granting these, we 
proceed ns follows (Fig. 1) : Given the two foi'ces P and Q - 



PRELIMINARY CHAPTER. 5 

P' + P" {P' and P" being each equal to P, so that Q = 2P), 
applied at 0. Transmit P'[ to A. Draw the parallelograms 
OP and AP ; OP will also be a parallelogram. By postulate 
(2), since OP is a rhombus, P and P' at may be replaced by 
a single force P acting through P. Transmit P' to P and 
replace it by P and P\ Transmit P from P to A, P' from 
P to i?. Similarly P and i-*", at A, may be replaced by a 
single force P" passing through P ; transmit it there and re- 
solve it into P and P" . P' is already at P, Hence P and 
P' -\- P'\ acting at J?, are equivalent to P and P' -f- P" act- 
ing at {?, in their I'espective directions. Therefore the result- 
ant of P and P' -\- P" must lie in the line OP^ the diagonal 
of the parallelogram formed on P and Q = 2P at O. Similarly 



SLg-a 





C/ FV /B 



--:^...N^.;::JD 
H\E 

Fig. 2. 



this may be proved (that the diagonal gives the direction of 
the resultant) for any two forces P and mP ; and for any two 
forces nP and mP, m and i^ being any two whole numbei-s, 
i.e., for any two commensurable forces. When the forces are 
incommensurable (Fig. 2), P and Q being the given forces, 
we may use a reductio ad ahsurdum^ thus : Form the parallelo- 
gram OP on P and Q applied at 0. Snppose for an instant 
that P the resultant of P and Q does not follow the diagonal 
OP, but some other direction, as OP'. Note the intersection 
H, and draw HG parallel to PP. Divide P Into equal parts, 
each less than HP ; then in laying off parts equal to these from 
O along OP, a point of division will come at some point F 
between C and P. Complete the parallelogram OFEG. The 
force Q" = OF is commensurable with P, and hence their 



6 MECHANICS OF ENGINEERING. 

resultant acts along OE. Now Q is greater than Q'\ while R 
makes a less angle with P than OE^ which is contrary to pos- 
tulate (3); therefore R cannot lie outside of the line OD. 
Q. E. D. 

It still remains to prove that the resultant is represented in 

amount, as well as position, by the diagonal. OD (Fig. 3) is 

••. /p' the direction of M the resultant of P and 

/F ''\^ Q ; required its amount. If P' be a force 

^"~— "^^r:;^ y equal and opposite to P it will balance P 

■••• i^ "''nD/. ^^^ Q 5 ^'^j tl^^ resultant of P' and P 
P l^'-< must lie in the line QO prolonged (besides 

^^**' ^' being equal to Q). We can therefore de- 

termine P' by drawing PA parallel to DO to intersect QO 
prolonged in A ; and then complete the parallelogram BF on BO 
and BA as sides. Since OFAB and AODB are paraUelograms, 
OF must=5A and BA must = OL'. Hence OF and OD are 
equal and lie on the same right line. Evidently if R^ were 
any shorter or any longer than OF the resultant of it and 

OB(=P) would not take the direction QOA. Hence R^ must 

= 0F, i.e., =0D', and hence R=zOD in amount. Q. E. D. 

Corollary. — The resultant of three forces applied at the same 
point is the diagonal of the parallelopiped formed on the three 
forces. 

14. Concurrent forces are those whose lines of action intersect 
in a common point, while non-concurrent forces are those which 
do not so intersect ; results obtained for a system of concurrent 
forces are really derivable, as particular cases, from those per- 
taining to a system of non-concurrent forces. 

15. Resultant. — A single force, the action of which, as re- 
gards the state of motion of the body acted on, is equivalent to 
that of a number of forces forming a system, is said to be the 
Resultant of that system, and may replace the system ; and con- 
versely a force which is equal and opposite to the resultant of 
a system will balance that system, oi', in other words, when it 
is combined with that system there will result a new system in 
equilibrium ; this (ideal) force is called the Anti-resultant. 

In general, as will be seen, a given system of forces can al- 



PRELIMINARY CHAPTER. 7 

ways De I'eplaced by two single forces, but tliese two can be 
combined into a single resultant only in particular cases. 

15a. Equivalent Systems are those which may be replaced by 
the same set of two single forces — or, in other words, those 
which have the same effect, as to state of motion, upon the 
given body. 

15b. Formulae. — If in Fig. 3 the forces P and $ and the angle or = 
PO Q are given, we have, for the resultant. 



JS = OD = V-f" + §' + 2 Pg cos tx. 

(If a is > 90° its cosine is negative.) In general, given any three parts 
of either plane triangle D Q, or D B, the other three may be obtained 
by ordinary trigonometry. Evidently if a = 0, R = P + Q; ifa = 
180°, i? = P - ^ ; and if a =t 90°, R = V -?" + Q"- 

15c. Varieties of Forces. — Great care should be used in deciding 
what may properly be called forces. The latter may be divided into ac- 
tions by contact, and actions at a distance. If pressure exists between two 
bodies and they are perfectly smooth at the surface of contact, the pressure 
(or thrust, or compressive action), of one against the other constitutes a force, 
whose direction is normal to the tangent plane at any point of contact (a 
matter of experience) ; while if those surfaces are not smooth there may also 
exist mutual tangential actions or friction. (If the bodies really form a 
continuous substance at the surface considered, these tangential actions are 
called shearing forces.) Again, when a rod or wire is subjected to tension, 
any portion of it is said to exert a pull or tensile force upon the remainder ; 
the ability to do this depends on the property of cohesion. The foregoing 
are examples of actions by contact. 

Actions at a distance are exemplified in the mysterious attractions, or re- 
pulsions, observable in the phenomena of gravitation electricity, and mag- 
netism, where the bodies concerned are not necessarily in contact. By the 
term weight we shaU always mean the force ot the earth's attraction on the 
body in question, and not the amount of matter in it. 

lad. Example 1. — If OD, = R, is given, =40 lbs., while the angle BOD 
is 110° and QOD = 40° (also = ODB), find the components P and Q. 
Solution. — From the triangle BOD, OB.OD: :sin 40°: sin 30°; whence 

P, or OB, = (40X0.6428) -^ 0.5000 = 51.42 lbs. 
Similarly, from triangle BOD, we have BD:OD: :sin 110°: sin 30°, 

.-. Q, or 5Z), = (40X0.9397) ^0.05000 = 75.17 lbs. 
Example 2.— Given P = 20 lbs., Q = 30 lbs., and angle a{ = POQ), =115°, 
find the resultant R in amount and direction. As to amount 

R^s/{20)'+{30) +2X 20X30 X( -0.4226) = V792:88 = 28.16 lbs. 
As to direction, let /? denote the angle ODB,==QOD; we then have, 
from triangle OBD, 20:28.16: :sin /?:sin 65°; whence, solving, 

sin/? =(20X0.9063)^-28.16 = 0.6437; i.e., angle ^ = 40° 4'. 



PART I.-STATICS. 



CHAPTER I. 

STATICS OF A MATERIAL POINT. 

16. Composition of Concurrent Forces. — A system of forces 
acting on a material point is necessarily composed of concurrem: 
forces. 

Case I. — All the forces in One Plane. Let be the 
material point, the common point of apph'catiou of all the 
forces ; Pj, P^, etc., the given forces, making 
""j?z angles tVj, a^^ etc., with the axis X. By the 

-■/A -p^p, parallelogram of forces P, may be resolved 

^/J2^4^i i i^to and replaced by its components, P^ cos or, 
-^-* — '- — *^— ^ acting along JT, and P^ sin a^ along Y. 
Fig. 4. Similarly all the remaining forces may be re< 

placed by their X and Y components. We have now a new 
system, the equivalent of that first given, consisting of a set of 
^forces, having the same line of application (axis X^^ and a 
set of I^ forces, all acting in the line Y. The resultant of the 
X forces being their algebraic snm (denoted by "^X^ (since 
they have the same line of application) we have 

^X=^ P^ cos a, -\- P^ COS fl'j + etc. = '2{^P cos «), 

and similarly 

^Y =^ P^ sin ar, + P^ sin a^ -|- etc. = 2{P sin a). 

These two forces, 2X and ^Y^ may be combined by the 
parallelogram of forces, giving P = VCSXY -\- i^Y^ ^^ ^^^^ 
single resultant of the whole system, and its direction is deter- 

:sY 

mined by the angle or; thus, tan a = ^^r^-; see Fig. 5. For 
eQuilibrium to exist, R must = 0, which requires, sejparately^ 



STATICS OF A MATERIAL POINT. 



9 



'2X^=0, and ^1^= (for tlie two squares {2X^y and 
{2 Yy can neither of them be negative quantities). 

Case II. — The forces having any directions in space, 
but all applied at 0, the material point. Let ^j, P^, 
etc., be the given forces, jP^ making the angles a^, ^j, and y^, 
respectively, with tliree arbitrary axes, X^ T^, and Z (Fig. 6), 
at right angles to each other and intersecting at 0, the origin. 
Siniilai-ly let a^, /3^, y^, be the angles made by jP^ with these 
axes, and so on for all the forces. By the parallelepiped of 
forces, 7^1 may be replaced by its components. 

Xi = Pi cos ofj, Yi = Pi cos /3i, and Z^ = JP^ cos ;/, ; and 





Y 


2Y' 


R 





^^ : X 


sx 




Fig. 5. 





Fig. 6. 



Fig. r. 



similai-ly for all the forces, so that the entire system is now- 
replaced by the tliree forces, 

2X = F, cos a^ + J\ cos a^ -\- etc ; 
2 T = P, cos ^, + P, cos 13, + etc ; 
^Z = P, cos y, + P^ cos /^ + etc ; 

and finally by the single resultant 

R = V{2Xf + [2 ry + {2zy. 

Therefore, for eqnilibrinm we mnst have separately, 
:SX= 0, :SY = 0, and 2Z= 0. 
^s position may be determined l)y its direction cosines, viz., 



cos 



2x , ^r ^z 

a — —^ ; cos // == -jy- ; cos ;k = -^. 



17. Conditions of Equilibrium. — Evidently, in dealing with 
a system of concurrent forces, it would be a simple matter to 



10 . MECHANICS OF ENGINEERING. 

replace any two of the forces by their resultant (diagonal 
formed on them), then to combine this resultant with a third 
force, and so on until all the forces had been combined, the 
last resultant being the resultant of the whole system. The 
foregoing treatment, however, is useful in showing that for 
equilibrium of concurrent forces in a plane onlj' two conditions 
are necessary, viz., ^ JT = and 2 JT = 0; while in space 
there are three, 2^= 0, 2 Y = 0, and 2Z = 0. In Case I., 
then, we have conditions enough for determining two unknown 
quantities ; in Case II., three. 

18. Problems involving equilibrium of concurrent forces. 
(A rigid body in equilibrium under no more than three forces 
may be treated as a material point, since the (two or) three 
forces are necessarily concurrent.)* ■ 

Problem 1. — A body weighing G lbs. rests on a horizontal 

table: required the pressure between it and the table. Fig. 8. 

Consider the body free, i.e., conceive all other bodies removed 

, (the table in this instance), being replaced by the 

forces which they exert on the first body. Taking 



the axis J" vertical and positive upM^ard, and not 
+X assuming in advance either the amount or drrec- 
|IM tion of JV, the pressure of the table against the 

I body, but knowing that G, the action of the earth 

Fig. 8. ^^ ^j^^ body, is vertical and downward, we have 
here a system of concurrent forces in equilibrium, in which 
the ^ and Y components of G are known (being and — 
G respectively), while those, iVx ^"^ -^^ of JV are unknown. 
Putting 2^ = 0, we have JV^ -|- = ; i.e., iVhas no hori- 
zontal component, .'. iV is vertical. Putting 2 Y = 0, we 
have iVy — G = 0, .". JV^ =^ -\- G; or the vertical component 
of JV, i.e., JV itself, is positive (upward in this case), and is 
numerically equal to G. 

Peoblem 2. — Fig. 9. A body of weight G (lbs.) is moving 
in a straight line over a rough horizontal table with a uniform 
velocity v (feet per second) to the right. The tension in an 
oblique cord by which it is pulled is given, and = P (lbs.), 

* Three parallel forces form an exception ; see §§ 20, 21, etc. 




STATICS OF A MATERIAL POINT. 11 

•which remains constant, the cord making a given angle of 
elevation, a^ with the patli of the body. Required the vertical 
pressure iV (lbs.) of the table, and also its ^y' 

horizontal action F (friction) (lbs.) against 
the body 

Referring by anticipation to Newton's fii'st 
law of motion, viz., a material point acted 
•on by no force or by balanced forces is either fig. 9. 

.at rest or moving uniformly in a straight line, we see that this 
problem is a case of balanced forces, i.e., of equilibrium. Since 
there are only two unknown quantities, iV and F, we may 
•detei-mine tliem by the two equations of Case I., taking tlie 
axes Xand Y as before. Here let us leave the direction of 
iVas well as its amount to be determined by the analysis. As 
^must evidently point toward the left, treat it as negative in 
summing the X components ; the analysis, therefore, can be 
•expected to give only its numerical value. 
2X = gives P 0,0^ a — F = 0. .-. F = P cos a. 

^^I^= gives iV+P sin «- G = 0. .-. I^= G - Psin a. 
.". iV is upward or downward according as 6^ is > or < P 
sin a. For i\^ to be a downward pressure upon tlie body would 
require the surface of the table to be above it. The ratio of the 
friction F to the pressure iV" which produces it can now be 
•obtained, and is called the " coefficient of friction." It may 
Tary somewhat with the velocity. (See p. 168.) 

This problem may be looked npon ns arising fi'om an experi- 
ment made to determine tlie coefiicieiit of friction between the 
given surfaces at the given uniform velocity. 

19. The Eree-Body Method. — The foregoing rather labored so- 
lutions of very simple problems have been made such to illus- 
trate what may be called the "free-body method" of treating any 
problem involving a body acted on by a system of forces. It 
consists ill conceiving the body isolated from all others which 
act * on it in any way, those actions being introduced as so many 
forces known or unknown, in amount and position. The sys- 
tem of forces thus formed may be made to yield certain equa- 
tions, whose character and number depend on circumstances, such 
as the behavior of the body, whether the forces are confined to 

* That is, in any "force-ahle" way. 



12 



MECHANICS OF ENGINEERING. 



a plane or not, etc. , and which are therefore theoretically avail- 
able for determining an equal number of unknown quantities. 

li'a. Examples. — 1. A cast-iron cylinder, with axis horizontal, rests 
against two smooth inclined surfaces, as shown in Fig. 9a. Its length, 
I, is 4 ft., diameter, d, is 10 in., and "heaviness" (p. 3) 480 lbs. /cub. ft. 
Required the pressures (or "reactions," or " supportivg forces"), P and 
Q at the two points of contact A and B. (Points, in the end view.) 
These pressures on the cylinder are shown pointing normal to the sur- 
faces {smooth surfaces) and hence pass through the center of the body, 



p; 


i \ 40° , 


/ 


,'^ "■^^ 


-^20° 


J^ 


;i§p|| 






G 




\'?0'' 












^] X 4X480= 1047.6 lbs. 



Fig. 9a. Fig. 9h. 

C, where we may consider the resultant weight, G, of the body to act. 
These' three forces, then, form a concurrent system, and the body is 
in equilibrium under their action. 

4~''^4 . 

i:X = gives: +P cos 40°-Q cos 20° + = 0; (1) 

IY = Q " +Psin40° + Q sin 20°-G = 0; . (2) 

that is, numerically, 0.7660P-0.9396Q = 0; (3) 

and 0.6428P + 0.3420Q = 0.1047.6 lbs (4) 

From (3) we have P=1.227Q, which in (4) gives 

(0.7887 + 0.3420)Q = 1047.6 lbs. ; and hence Q = 926.4 lbs. \ . 
Therefore P, =1.227Q, =1127.6 lbs. / 

Example 2. — Fig. 96. The 4-ton weight is suspended on the bolt C, 
which passes through the ends of boom OC^ and tie-rod DC. Bolt C 
is also subjected to a horizontal pull tov/ard the left, due to the 2-ton 
weight, suspended as shown. . Find the pull P in the tie and the thrust 
Q in the boom. Note that the boom is pivoted at both ends and hence 
(if we neglect its weight) is under only two pressures; both of which, 
therefore (for the equilibrium of the boom), m,ust point along its length.. 
Hence the thrust Q on bolt C makes an angle of 41° with the horizontal. 
Similarly, P, the action of tie-rod on C, is at 15°. 

Solution. — At (?>) we see the bolt as a "free body"; in equilibrium 
under the four concurrent forces. 

2X = Qcos41°-Pcosl5°-(?2-0 = 0; (5) 

iF = Q sin41°-Psin 15°-Gi-0 = 0; (6) 

or, numerically, 0.7547Q-0.9659P-2 = 0, (7) 

and 0.6560Q-0.2588P-4 = (8) 

From (7), Q = 2.6514- 1.279P, which in (8) gives 

0.6560(2.651 + 1.279P) - 0.2588P = 4 ; 
that is, 1.740-H0.8390P-0.2588P = 4; and hence, finally, 

P= 2.260 ^0.5802 = 3.896 tons, and .-. Q = 7.633 tons. Ans. 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 13 



CHAPTER 11. 

PARALLEL FORCES AND THE CENTRE OF GRAVITY. 

20. Preliminary Remarks. — Althongli by its title tliis section 
sliould be restricted to a treatment of tiie equilibrium of forces, 
certain propositions involving the composition and resolution 
of forces, without reference to the behavior of the body under 
their action, will be found necessary as preliminary to the prin- 
cipal object in view. 

As a rigid body possesses extension in three dimensions, to 
deal with a system of forces acting on it we require three co- 
ordinate axes : in other words, tlie system consists of " forces 
in space," and in general the forces are non-concurrent. In 
most problems in statics, however, the forces acting are in one 
plane: we accordingly begin by considering non-concurrent 
forces in a plane, of which the simplest case is that of two 
parallel forces. For the present the body on which the forces 
act will not be shown in the figure, but must be understood to 
be there (since we have no conception of forces independently 
of material bodies). The device will frequently be adopted of 
introducing into the given system two opposite and equal forces 
acting in the same line : evidently this will not alter the e£fect 
of the given system, as regards the rest or motion of the body. 

21. Resultant of two Parallel 
Forces. 

Case I. — The two forces have 
the same d'u'ection. Fig. 10. 
Let P and Q be the given forces, 
and AJB a line perpendicular to 
them {P and Q are supposed to have sL---3/---Js 

been transferred to the intersections ^^<*- ^^^ 

A and B). Put in at A and B two equal and opposite 
forces 8 and S^ combining them with P and Q to form P' 



6 "S A 



TP t 

^ i Q Q 

< VOr -"^ "f; 

■X--H 



D /B S 



14 



MECHANICS OF ENGINEERING. 



shaded by dots, .•, 

iave* -7^ == 

Q X 



and Q'. Transfer P' and Q' to tlieir intersection at C, and thera 
resolve them again into S and P, /S'and Q. 8 and /iS^ annul each 
other at C', therefore P and Q^ acting along a common line CD, 
replace the P and Q first given ; i.e., the resultant of the origi- 
nal two forces is a force R =^P -\- Q, acting parallel to them 
through the point P, whose position must now be determined. 
The triangle CAP is similar to the triangle shaded by lines, 
.'. P : S :: GP : a?; and CPB being similar to the triangle 
8 '. Q :: a — a? : CP. Combining these, we 

■'• "^ ^ '^TO "^ %• ^^^ write this 
Px = Qa, and add Pc, i.e., i^c-j- Qg, to each member, c being 
the distance of (Fig. 10), any point in AP produced, from 
A. This wull give P{x -\- c) = Pc -\-Q{a -{- c), in which c, 
■a-\~ c, and x -\- c are respectively the lengths of perpendiculars 
let fall from upon P, Q, and their resultant P. Any one of 
these products, such asPc, is for convenience (since products of 
this form occur so frequently in Mechanics as a result of alge- 
braic transformation) called the Moment of the force about the 
arbitrary point 0. Hence the resultant of two parallel forces of 
the same direction is equal to their sum, acts in their plane, in 
a line parallel to them, and at such a distance from any arbi- 
trary point in their plane as may be determined by writing 
its moment about equal to the sum of the moments of the 
two forces about 0. O is called a centre of moments, 'dud each 
of the perpendiculars a lever-arm. 

Case II. — Two parallel forces i^ and Q of opposite direc- 
11. By a process similar to the foregoing, we 
obtain P =P- Q and {P — Q)x 
= Qa, i.e., Px = Qa. Subtract 
each member of the last equation 
from Pc (i.e., Pc—Qc), in wliicli c 
is the distance, from A, of any arbi- 
trary point in A£ produced. This 
gives P{c — x) = Pc — Q{a -j- c). 
But ((? — «), G, and {a-{-G) are re- 
FiG. 11. spectively the perpendiculars, from 

* That is, the resultant of two parallel forces pointing in the same direc- 
tion divides the distance between them, in the inverse ratio of those foi'ces. 



tions. 




PARALLEL FORCES AND THE CENTRE OF GRAVITY. 15 

O., upon i?, P^ and Q. That is, i?(c — x) is the moment of R 
about 0\ Pc, that of P aboiit 0; and ^(«+c), that of Q 
about 0. But the moment of Q is subtracted from that of P, 
which corresponds with the fact that Q in tliis figure would 
produce a rotation about opposite in direction to that of P. 
Havi.jg in view, tlien, this imaginary rotation, we may define 
the moment of a force && positive when tlie indicated direction 
about the given point is against the hands of a watch; as nega- 
tive when with the hands of a watch.* 

Hence, in general, the resultant of any two parallel forces is, 
in amount, equal to their algebraic sum, acts in a parallel direc- 
tion in the same plane, while its moment, about any arbitrary 
point in the plane, is equal to the algebraic sum of the mo- 
ments of the two forces about the same point. 

Corollary. — If each term in the preceding moment equations 
be multiplied by the secant of an angle {a, Fig, 12) thus; 



p.. 



%-'^"" 



^^^ 



..^'i' 



Of^- — -a-i- 

jt 1 0, y j 

k — - fta 'A 

Fig. 13. 




Fig. 13. 



(using tlie notation of Fig. 12), we have 

Pa sec a = Pxai sec a+P^jii sec a, i.e., P6 = Pi6i -I-P2&25 
in which, h, h\ and 62 are tHe oblique distances of the three 
lines of action from any point in tlieir plane, and lie on the 
same straight line ; P is the resultant of the parallel forces P^ 
and P2' 

22. Resultant of any System of Parallel Forces in Space. — 
LetP*i, ^2? Pii 6tC'5 t>e the forces of the system, and a?,, y„ 
s„ a?,, ^j, Sj, etc., the co-ordinates of their points of application 
as referred to an arbitrary set of three co-ordinate axes X, Y^ 
and Z, perpendicular to each other. Each force is here re 

* These two directions of rotation are often called counter clockwise, and 
clockwise, rescectively. 



16 MECHANICS OF ENGINEERING. 

stricted to a definite point of application in its line of action 
(with reference to establishing more directly the fundamental 
equations for the co-ordinates of the centre of gravity of a 
body). The resultant P' of any two of the forces, as 
Pj and /*„ is = P, + P^ ^^^ ™^y be applied at C, the in- 
tersection of its own line of action with a line BD joining 
the points of application of P^ and P^^ its components. 
Produce the latter line to^, where it pierces the plane ^Y^ 
and let 5„ &', and 5^, respectively, be the distances of B^ (7, 
D^ from A. \ then from the corollary of the last article we have 

p'y^Ph^p^K', 

but from similar triangles 

V \\\\\\z' : z, : 0„ .-. P'z' = P,z, + P,3,. 

Now combine P., applied at C^ with P^^ applied at E^ calling 
their resultant P" and its vertical co-ordinate z'\ and we obtain 

P"z" = P'z' + P3S3, i-e., P"z" = Pa + P.\ + ^3^3, 
also 

P-=P' + P3 = P,+P, + P,. 

Proceeding thus until all the forces have been considered, we 
shall have finally, for the resultant of the whole system, 

P-P. + i'.+ i^s + etc.; 
and for the vertical co-ordinate of its point of application, 
which we may "write 3, 

Rz — P,z, + P,s, + P3S3 + etc ; 

- P,z, + P,z, + P,z,... _^{Pz)^ 

..e.,2 _ p^_^p^_^p^:^ - ^p , 

and similarly for the other co-ordinates. 

In these equations, in the general case, such products as P,j!!i» 
etc., cannot strictly be called moments. The point whose co» 



PARALLEL FORCES ANB THE CENTRE OF GRAVITY. 11 

ordinates are the x, y, and b, just obtained, is called the Centre 
of Parallel Forces, and its position is independent of the {com- 
mon) direction of the forces concerned. 

ExaTYijple. — If the parallel forces are contained in one plane, 
and the axis I^be assumed parallel to the direction of the 
forces, then each product like P^x^ will be a moment, as de- 
fined in § 21 ; and it will be noticed in the accompanying nu- 
merical example, Fig. 14, that a detailed substitution in the 

equation R □ iY R ra 
_ t f^ L-i-l 1 
i?a?=:P,a?, + P,a?,+ etc., . . . (1) U -i..i^ |_ 

having regard to the proper sign of each ^, 0| +X 

force and of each abscissa, gives the same fig. i4. 

result as if each product Px were first obtained numerically, 
and a sign affixed to the product considered as a moment 
about the point 0. Let P^ = — 1 lb.; P, = + 2 lbs.; P^ = 
+ 3 lbs.; P^^~-& lbs.; a?^ = + 1 ft.; a;^ = -f 3 ft.; a?, = — 2 
ft.; and a?^ = — 1 ft. Required the amount and position of the 
resultant R. In amount R = -SP =— 1 + 2 + 3 — 6 = — 2 
lbs.; i.e., it is a downward force of 2 lbs. As to its position, 
Rx= 2{Px) gives ( — 2)« = ( - 1) X (+ 1) + 2 X 3 + 
3 X (- 2) + (-6) X(-l) = -l + 6-6 + 6. Now from 
the figure, by inspection, it is evident that the moment of P, 
about is negative {with the hands of a watch), and is numer- 
ically = 1, i.e., its moment = — 1 ; similarly, by inspection, 
that of Pj is seen to be positive, that of P^ negative, that of 
P, positive; which agree with the results just found, that 
(- 2)^ = - 1 + 6 - 6 + 6 = + 5 ft. lbs. (Since a moment 
is a product of a force (lbs.) by a length (ft.), it may be called 
so many foot-pounds.) Next, solving for a?, we obtain 
X = (+ 5) -f- ( — 2) = — 2.5 ft.; i.e., the resultant of the given 
forces is a downward force of 2 lbs, acting in a vertical line 
2.5 ft. to the left of the origin. Hence, if the body in question 
be a horizontal rod whose weight has been already included in 
the statement of forces, a support placed 2.5 ft. to the left of 
and capable of resisting at least 2 lbs. downward pressure 
will preserve equilibrium ; and the pressure which it exerts 



18 MECHANICS OF ENGINEERING. 

against the rod must be an upward force, P^, of 2 lbs., i e. tiie 
equal and opposite of the resultant of P^, P^? P^^ Pa- 

Fig. 15 shows the rod as a fi-eo body in equilibrium under 
the live forces. P^ = -|- 2 lbs. — the reaction of the support. 
Of course P^ is one of a pair of equal 
and opposite forces ; the other one 
J is the pi'essure of the rod against the 



I' 



"—I \ ' ^ ^ 

?^\ — -gi-s- iO support, and would take its place among 

Fig. 15. ■ the forces acting on the support. 

23. Centre of Gravity. — Among the forces acting on any 
rigid body at the surface of the earth is the so-called attraction 
of the latter (i.e., gravitation), as shown by a spring-balance, 
which indicates the weight of the body hung upon it. The 
weights of the different particles of any rigid body constitute a 
system of parallel forces (practically so, though actually slightly 
convergent). The point of application of the resultant of these 
forces is called the centre of gravity of the body, and may also 
be considered the centre of onass, the body being of very small 
dimensions compared with the earth's radius. 

If a?, y, and z denote the co-ordinates of the centre of gravity 
of a body referred to three co-ordinate axes, the equations 
derived for them in § 22 are directlj' applicable, with slight 
changes in notation. 

Denote tlie weight of any particle * of the body by dG, its 
volume by d F, by ;^its heaviness (rate of weight, see § 7) and 
its co-ordinates by a?, y, and z ; then, using the integral sign as 
indicating a summation of like terms for all the particles of the 
body, vs^e have, \v: heterogeneous bodies (see also p. 119, Notes). 

r~_fy^dy^ -_frydV_, - _fr^. ,-,x 

^- fydV' y ~ fydV' ^ - fydV^ * ' ^^^ 
while, if the body is homogeneous, y is the same for all its ele- 
ments, and being therefore placed outside the sign of sumnu\- 
tion, is cancelled out, leaving for homogeneous bodies {Y de- 
noting the total volume) 

-„ _-«I. ^ -Ml., and I --^I f2) 

a? — p:— , y — y , ana z — y . . . \^z) 

* Any subdivision of the body may be adopted for use of equations (1) 
and (2), etc.; but it must be remembered that the ai (or y, or s) in each term 
of the summations, or integralSj is the co-ordinate of the center of gravity of 
the subdivision employed. 



PARALLEL FOECES AND THE CEN"TRE OF GRAVITY. 19 

Corollary. — It is also evident that if a homogeneous body is 
for convenience considered as made up of several finite parts, 
whose volumes are y^, F^? etc., and whose gravity co-ordinates 
are a?„ y„ z^ ; «„ y,, z^ ; etc., we may write 

. = -^-j-^-^-— .... (3) 

If the body is heterogeneous, put G^ (weights), etc., instead 
Df T^i, etc., in equation (3). 

If the body is an infinitely thin homogeneous shell of uni- 
form thickness = h, then dV =^ hdF {dF dawoimg an element, 
and J^the whole area of one surface) and equations (2) become, 
after cancellation, 

^-f^il. z-Ml. -,-fi^ u) 

For a thin homogeneous plate, or shell, of uniform thick- 
ness, and composed of several finite parts, of area Fi, F2, etc., 
wdth gravity co-ordinates Xi, X2, etc., we may write 

_ FiXi+F2X2+ . . . , . - 

^= F,+F2+... • • • • (^«) 
Similarly, for a homogeneous wi?'e of constant small cross- 
section (i.e.. a geometrical line, having weight), its length 
being s, and an element of length ds, we obtain 

3=^;^=>^»;i=-^. ... ^5) 

24. Symmetry. — Considerations of symmetry of form often 
determine the centre of gravity of homogeneous solids without 
analysis, or limit it to a certain line or plane. Tlius the centre 
of gravity of a sphere, or any regular polyedron, is at its centre 
of figure;, of a right cylinder, in the middle of its axis; of a 
thin plate of the form of a circle or regular polygon, in the 
centre of figure ; of a straight wire of uniform cross-section, in 
the middle of its length. 

Again, if a homogeneous body is symmetrical about a plane, 
the centre of gravity must lie in that plane, called a plane of 



20 



MECHANICS OF ENGINEEUING. 



gravity; if about a line, in that line called a line of gravity; 
if about a point, in that point. 

25. By considering certain modes of subdivision of a homo- 
geneous body, lines or planes of gravity are often made appar- 
ent. E.g., a line joining the middle of the bases of a trape- 
zoidal plate is a line of gravity, since it bisects all the strips 
of uniform width determined by drawing parallels to the 
bases; similarly, a line joining the apex of a triangular plate to 
the middle of the opposite side is a line of gravity. Other 
cases can easily be suggested by the student. 

26. Problems.— (1) Required the position of the centre of 
A, gravity of %fine homogeneous wire of the 

,,. r-g^ form of a circular arc, A£, Fig. 16. Take 
the origin at the centre of the circle, and 
the axis ^ bisecting the wire. Let the 
length of the wire, s, = 2Si ; ds = ele- 
ment of arc. We need determine only the 
X, since evidently y ^ 0. Equations (5), 

fxds 




Fig. 16. 



23, are applicable here, i.e., x 



From similar triangles we liave 



7 ^^V 

ds : dy :: r : x; .-. ds = — -; 



,?/ = + a ^ra 



:^ I dy — -^r—, i.e., = chord X radius -r- length of 



2s 



wire. For a semicircular M'ire, this reduces to x == 2r -~ 7t. 

Problem 2. Centre of gravity of trapezoidal {and trian- 
giilar) thin plates, homogeneous, etc. — Prolong the non-parallel 
sides of the trapezoid to intersect at 0, which take as an origin, 
making the axis X perpendicular to the bases h and &,. We 
may here use equations (4), § 23. and may take a vertical strip 
for our element of area, dF, in determining x; for each point 
of such a strip has the same x. Now dF ^ {y -f- y')dx. and 

* The two triangles meant {m being any point of the wire) are the 
finite triangle Omc, and the infinitely small one at m formed by the 
infinitesimal lengths dy, dx, and ds. 



PARALLEL POKCES AND THE CENTKE OF GRAVITY. 21 



from similar triangles y-{-y' = jx. 'NowF, = -{hh — bji,),"^ 
can be written ^ , {^^ ~ K'), and x = --^ — becomes 



= 


Ji UK 


-2a(^-^0 = 3;,. 


-K 


for the trapezoid. 




For a ti 


•iangle h^ = 


— 2 
0, and we liave x = h ; 


that 



centre of gravity of a triangle is one tliird the altitude from the 
base. The centre of gravity is finally determined by knowing 





Fig. 17. 



Fig. 18. 



that a line joining the middles of h and h^ is a line of gravity; 
or joining O and the middle of h in the case of a triangle. 

Problem 3. Sector of a circle. Thin plate, etc. — Let the 
notation, axes, etc., be as in Fig. 18. Angle of sector = 2<ar; 
a? = ? Using polar co-ordinates, the element of area dF (a 
small rectangle) = pdqi . dp, and its a? = p cos qj ; hence the 
total area = 



9 



F= J'^'^X fpdiP\dcp = y^+"^ r'd^ = 
i.e., F:= T^a. From equations (4), § 23, we have 

- \ n 

X = -jp i xdF 

♦Note that h\•.'h^'.•.'h'.h, so that ?)ifei=(&-^^)i'ii'. 



22 MECHANICS OF ENGINEERING. 

{Note on double integration. — The quantity 
cos (p J p' dp \dq), 

is that portion of the summation / / cos cpp'dpdq) which 

belongs to a single elementary sector (triangle), since all its 
elements (rectangles), from centre to circumference, have the 
same q) and dcp.) 
That is, 

— 1 r^ n + a. 'p^ r+a 2 y sin or 
a? = ^^-o / cos ^c?ffi> = 5-^ sinffl = -5-. : 

o c^ T 1 ^ — 4 7* sin -I /? 
or, putting p =z 'za z= total angle 01 sector, a? = -^ -z • 

— 4:7' 

For a semicircular plate this reduces to a? = 7;—. 

\_Mote. — In numerical substitution the arcs a and /? used 
above (unless sin or cos is prefixed) are understood to be ex- 
pressed in circular measure (;r-measure) ; e.g., for a quad- 
rant, yS = I = 1.5707*+ ; for 30°, /? = ^ ; or, in general, if fi 

m degrees := , then p in ;r-measure = — . 

° n n J 

Problem 4. Sector of a flat ring ; thin 

_ -^^ plate, etc. — Treatment similar to that of 

,^y \.---^...\\- Problem 3, the difference being that the 

#^^P^ If - _ ... . P' 

limits of the interior integrations are 

instead of | . Result, 

FiQ, 19. 1-0 

- _ 4 T^ — r^ sin ^/? 
^""l- r/ - r: ' ~~W~° 

* "Radians." 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 23 



Pkoblem 5. — Segment of a circle ; thin plate, etc. — Fig. 20. 



Since each rectangular element of any ver- 
tical strip has the same x, we may take the 
strip as dF \w finding x, and use y as the 
half-height of the strip. dF = 2ydx, and 
from similar triangles x : y :: {— dy):dx,^ 
i.e., xdx = — ydy. Hence from eq. (4), 



- ^/(vdF ^ /x^ydx - 2XVVZy _2__ 



F 



F 



F 



SF 




but a = the half-chord, hence, finally, x = 



12F. 




Fio. 21. 



Problem 6. — Trajpezoid ', thin plate, etc., 
by the method in the corollary of § 23 ; equa- 
tion (4a). Kequired the distance x from the 
base AB. Join BB^ thus dividing the trape- 
zoid ABCB into two triangles ABB = F^ 
and BBC = F^, whose gravity a?'s are, re- 
spectively, x^ = ^h and x^ = |A. Also, F^ 
= iMj, F^ = ^hh^, and F (area of trape- 
zoid) = ih{h, + h,). Eq. (4a) of § 23 gives 
Fx = F^x^ -\- F^x^ ; hence, substituting, 
(6i +62)^ = k^ih^lhji. 

_^h (61 + 2&2) 
•'• ^-3 * 61+62 • 

The line joining the middles of 5, and h^ is a line of gravity, and 
is divided in such a ratio by the centre of gravity that the fol- 
lowing construction for finding the latter holds good : Prolong 
each base, in opposite directions, an amount equal to the other 
base; join the two points thus found: the intersection with 
the other, line of gravity is the centre of gravity of the trape- 
zoid. Thus, Fig. '21, with BF= h &ndBF= \, join FF, 
etc. 

* The minus sign is used for dy since, as we progress from left to right 
in bringing into account all the various strips, x increases while y diminishes; 
i.e., dx is an increment and dy a decrement. At the point of beginning of 
the summation, on left, y= +a; while at the extreme right, y = 0. 



24 



MECHANICS OF ElSTGnsrEERING. 



Peoblem 1. Homogeneous ohlique cone or pyramid. — ■ 
Take the origin at the vertex, and the axis X perpendicular to 
the base (or bases, if a frustum). In finding x we may put 
dY^ =^ vohirae of any lamina parallel to YZ, ^ being the base 
•of such a lamina, each point of the lamina having the same x. 
Hence, (equations (2), § 23), (see also Fig. 22). 

x= ^fxdV, V=/dV=/Fdxi 
but, from the geometry of similar plane figures, 



F:F, :: x' : h,% .-.F. 



F 



and 



^=i-^»'*'^^=§'' 






-',fxdY=^Jx'dx=-!^ 



F 

K LI* 



Q Z 4 Z, 4 

-For a frustum, x = 7 • , '3 ~ y\ I while for a pyramid, Aj, be- 

— 3 

ing = 0, a? = jA. Hence the centre of gravity of a pyramid 

is one fourth the altitude from the base. It also lies in the line 
joining the vertex to the centre of gravity 
of the base. 

Pkoblem 8. — If the heaviness of the ma- 
terial of the above cone or pyramid varied 
directly as x, y^ being its heaviness at the 
base F^, we should use equations (1)5 § 23, 

putting y = j^ x\ and finally, for the frustum, 

- 4 h:-h: 




Fig. 23. 



r« 



h:-hr 



and for a complete cone m = — A,. 



27. The Centrobaric Method. — ^If an elementary area dF he 
revolved about an axis in its plane, through an angle a < 'Itt. 



PAEALLEL FORCES AND THE CENTRE OF GKAVITY. 25 




the distance from the axis being = x, tlie volume generated is 

^Y =z axdF^ and the total volume generated by all the dF''% 

of a finite plane figure whose plane con- ^^,g 

tains the axis and which lies entirely on one 

;side of the axis, will be T^ = fd V = 

afxdF. But from §23, afxdF^aFx\ 

ax being the length of path described by 

the centre of gravity of the ])lane iigure, Ym. 23. 

we may write : The vohime of a solid of revolution generated 

hi/ a plane figure, lying on one side of the axis, equals the 

area of the figure multiplied hy the length of curve descrihed 

hy the centre of gravity of the figure. 

A corresponding statement may be made for the surface 
generated by the revolution of a line. The arc a must be ex- 
pressed in It measure in numerical work. 

27a. Centre of Gravity of any Cluadrilateral. — Fig. 23a. 
Construction', ABOD being any quad- 
rilateral. Draw the diagonals. On the 
long segment DK of DB lay off BE = 
BK, the shorter, to determine E\ simi- 
larly, determine iV^on the other diagonal, 
by making GN = AK. Bisect FK in H 
and KN in M. The intersection of FM 
and NH\& the centre of gravity, C. 
p.poof—R being the middle of BB, and AH and HG 
Slaving been joined, I the centre of gravity of the triangle 
ABB is found on AH, by making ^/= i-^iZ; similarlj^, by 
makino- HB = ^HG, B is the centre of gravity of triangle 
BBG. . ' . IB is parallel to AG and is a gravity-line of the 
whole figure; and the centre of gravity Cmay be found on it 
if we can make CB : CI :: area ABB : area BBG (§ 21). 
But since these triangles have a common base BB, their areas 
are proportional to the slant heights (equally inclined to BB) 
AK and KG, i.e., to GN and NA. Hence HN, which di- 
vides IB in the required ratio, contains C, and is .'. a gravity- 
line. By similar reasoning, using tlie other diagonal, AG, and 




Fig. 23a. 



26 MECHANICS OF ENGINEERING. 

the two triangles into wliicb it divides the whole figure, we 
may prove E2i to be a gravity-line also. Hence the construc- 
tion is proved. 

27b. Examples. — 1. Required the volume of a sphere bj 
the centrobaric method. 

A sphere may be generated by a semicircle revolving about 
its diameter through an arc a = 27r. The length of the path 

descj'ibed by its centre of gravity is = Stt^— ("see Prob. 3, § 
26), while the area of the semicircle is |-7^r^ Hence by § 27, 

4r 4 

Yolume venerated = 27r . 7-— . ^nr^ = — nr'. 

2. Tiequired the position of the centre of gravity of the sectoi* 

of a flat ring in which '}\ = 21 feet, r^ = 20 feet, and /3 = 80° 

(see Fig. 19', and § 26, Prob. 4). 

/3 . 
sin — = sin 40° = 0.64279, and y5 in circular measure =• 

80 4 
T^7^=-q7r = 1.3962 radians. By using ri and r2 in feet, 

X will be obtained in feet. 

• I 
_ 4 ri^-r2^ ^^^- 2 4 1261 0.64279 

.'.T=— — — = - -, • - = 18 87 feet 

"^ S'n^-r^^- /? 3*41 '1.3962 ^^-^'leei. 

3. Find tlie height (z, = OC) of the center of gravity of 
05" uo.e'tj the plane figure in Fig. 23& 

"I above its base OX. 
L This figure is bounded 

i by straight lines and is an 
j_ approximation to the shape: 

'^ o -15- — -»«5-i of the cross-section of a steel 

^'^- 2^^- , "channel" (see p. 275). 

Dividing it into three rectangles and two triangles (see 
dotted lines in figure) and applying eq. (4a) of p. 19, we have 

i.r 



J. 



15X.6X.3-f2[3.4x.6x2.3]+2 
z = — 



71 
1.7X.5X^ 

— = 0.882 in. 



15X0.6 + 2[3.4X0.6] + 2[1.7X0.5] 
(The student should carefully verify these numerical details.) 



STATICS OF A RIGID BODY. 27 

CHAPTER III. 

STATICS OF A RIGID BODY. 

28. Couples. — On account of the peculiar properties and 
utility of a system of two equal forces acting in parallel lines 
and in opposite directions, it is specially ^^ 
considered, and called a Couple. The z^::::^^^^^^^^ t 
ar7n of a couple is the perpendicular « fr^^V^^ ""'^'^^^^ 
distance between the forces ; its TwomeTi^, P S'* lQJ<i J^ 
the product of this arm, by one of the |<^ ^^'"'^^^ .^^^^^^^^^ 
forces. The axis of a couple is an ^"^''^^:>^^,^^<^^^^ 
imaginary line drawn perpendicular to y\q. 24. 

its plane on tliat side from which the rotation appears positive 
(against the hands of a watch). (An ideal rotation is meant, 
suggested by the position of the arrows ; any actual rotation 
of the rigid body is a subject for future consideration.) In 
dealing with two or more couples the lengths of their axes are 
made proportional to their moments; in fact, by selecting a 
proper scale, numerically equal to these moments. E.g., in Fig. 
24, the moments of the two couples there shown are Pa and 
Qh\ their axes p and q so laid off that Pa : Qh '.: p : q, and 
that the ideal rotation may appear positive, viewed from the 
outer end of the axis. 

For example, if each force P of a couple is 60 lbs., and the 
arm is a=6 ft., its moment is 360 jbot-pounds; or 0.180 foot- 
tons; or 4320 inch-pounds; or 2.16 inch-tons. 

29. No single force can halance a couple. — For suppose the 
couple P^ P, could be balanced by a force P', then this, acting 

?f at some point C, ought to hold the couple 

ni /..:■-. -P- -Q in equilibrium. Draw CO throuo-h 0, the 

tT /p p^f centre of symmetry of the couple, and 

Fig. 25. make OD = OC. At D put in two op- 

posite and equal forces, S and T, equal and parallel to P', 
The supposed equilibrium is undisturbed. But if P\ P, and 



"^S MECHANICS OF ENGINEERING. 

P are in equilibrium, so ought (by symmetry about 0) S, jP, 
and P to be iu equilibrium, and they may be removed without 
disturbing equilibrium. But we have left Tand P', which are 
evidently not in equilibrium ; .•. the proposition is proved by 
this reductio ad absurdum. Conversely a couple has no singlo 
resultant. 

30, A couple may he transferred anywhere in its own plane. 

— First, it may be turned through any angle «', about any 

p* point of its arm, or of its arm produced. 

Gt- -; j^- -|- Let {P^ /*')be a couple, G any point of its 

\-.-'-j^-4 yp' arm (produced), and a. any angle. Make 
^^^■.,_ i 0G= GA, CD — AB, and put in at G, 
'^ \ ^ \; I P^ and P^ equal to P {or P'), opposite to 
--* "^ each other and perpendicular to GC; and 



'® "" \ R*' P^ and P^ similarly at P. IS^ow apply and 

Fig. 26. combine P and P, at 0, P' and P, at 0'\ 

then evidently P and R' neutralize each other, leaving P^ and 
P^ equivalent to the original couple {P^ P'). The arm 
CD = AB. Secondly, if G be at infinity, and or = 0, the 
same proof applies, i.e., a couple may be moved parallel to 
itself in its own plane. Therefore, by a combination of the 
two traiisferrals, the proposition is established for any trans- 
ferral in the plane. 

31. A coujple. may he replaced hy another of equal moment 
in- a parallel plane. — Let {P, P') be a couple. - Let CD, in a 
parallel plane, be parallel to AB. At D put in a pair of equal 

and opposite forces, ^3 and S^., parallel to P and each = ^=:iP. 

ED 

Similarly at (7, 8^ and 8^ parallel to P and each = ==-P. 

sLkj 

But, from similar triangles, 

^ — ^. . o _ c _ o _ e 
pjjy — PC'' ' ' "^ — 5 — ' — *' 

* See Fig. 27, which is a perspective view. The arm of the couple (P, P') 
is AB, in the background. The length of CD, which is in the foreground, 
may be anything whatever. 



STATICS OF A KIGID BODY. 



29- 



[Note. — The above values are so chosen that the intersection point E 
may be the point of application of (P' -|- JS2), the resultant of F and /6a;. 
and also of {P-\- Sa), the resultant of Pand S3, as follows from § 21; thus 
(Fig. 28), Ji, the resultant of the two parallel forces Pand iSs, is = P-f-xSg, 
and its moment about any centre of moments, as E, its own point of ap- 
plication, should equal the (algebraic) sum of the moments of its com- 

AhJ 

ponents about E; i.e., B X zero = P . AE — Sz . DE; .-.83 = == . P.] 



UE 



lA 



S,| I 



hk-ff" I 






R! 



Fig. 27. 



D E 

Fig. 28. 



I 



Replacing P' and S, by {P' + S,\ and P and S, hj 
{P -f- ^,), the latter resultants cancel each other at E^ leaving 
the couple {S^, 8^ with an arm CD^ equivalent to the original 

couple P, P' vi^ith an arm AB. But, since 8^ = ===. P = 

MjL/ 

-=r. . P, we have S.xOP = PxAB ; that is, their moments 
ai'e equal. 



32. Transferral and Transformation of Couples. — In view of 
the foregoing, we may state, in general, that a couple acting on 
a rigid body may be transferred to any position in any parallel 
plane, and may have the values of its forces and arm changed 
in any way so long as its moment is kept unchanged, and still 
have the same eifect on the rigid body (as to rest or motion, 
not in distorting it). 

Corollaries. — A couple may be replaced by another in any 
position so long as their axes are equal and parallel and simi- 
larly situated with respect to their planes. 

A couple can be balanced only by another couple whose axis 
is equal and parallel to that of the jfirst, and dissimilarly situ- 
ated. For example. Fig. 29, Pa being = Qb^ the rigid body 
AB (here supposed without weight) is in equilibrium in each 



:30 



MECHANICS OF ENGINEERING. 



case shown. By " reduction of a couple to a certain arm «" 
is meant that for the original couple whose arm is a' ^ with 
forces each = P\ a new couple is substituted whose arm shall 
be = «, and the value of whose forces P and P must be com- 
puted from the condition 

Pa = P'a\ i.e., P = P'a' -^ a. 




Fig. 29. 



Fig. 30. 



33. Composition of Couples. — Let (P, P') and {Q, Q') be two 
-couples in different planes reduced to the same arm AB = a, 
which is a portion of the line of intersection of theii' planes. 
That is, whatever the original values of the individual forces 
and arms of the two couples were, they have been transferred 
and replaced in accordance with § 32, so that P . AP, the 
moment of the first couple, and the direction of its axis, p, 
have remained unchanged ; similarly for the other couple. 
Combining P with Q and P' with Q', we have a resultant 
couple {P, i?')M^hose arm is also AP. The axes p ^.nd q of 
the component couples are proportional to P . AP and Q . AB, 
i.e., to P and Q, and contain the same angle as P and Q. 
"Therefore the parallelogram p . . . q\& similar to the parallelo- 
gram P . . . Q\ whence p '. q : r'.'.P '. Q : P, or p : q : ri: 
Pa : Qa : Pa. Also r is evidently perpendicular to the plane 
of the resultant couple (P, i?'), whose moment is Pa. Hence 
r, the diagonal of the parallelogram on p and q, is the axis of 
the resultant couple. To combine two couples, therefore, we 
have only to combine their axes, as if they were forces, by v> 
parallelogram, the diagonal being the axis of the resultant 
couple ; the plane of this couple will be perpendicular to tlie 



STATICS OF A RIGID BODY. 31 

axis just found, and its moment bears the same relation to the 
moments of the component couples as the diagonal axis to the 
two component axes. Thus, if two couples, of moments Pa 
and Q}}^ lie in planes perpendicular to each other, their result- 
ant couple has a moment Re = ^{Paf + {Qbf' 

It three couples in different planes are to be combined, the 
axis of their resultant couple is the diagonal of the parallelo- 
piped formed on the axes, laid off to tliesame scale 2a\d point- 
ing in the proper directions, the proper direction of an axis 
being away from the plane of its couple, on the side from 
which the couple appears of positive rotation. 

34. If several couples lie in the same plane their axes are 
parallel and the axis of tlie resultant couple is their algebraic 
sum ; and a similar relation holds for the moments : thus, in 
Fig. 24, the resultant of the two couples has a moment = Qh 
— Pa, which shows us that a convenient way of combining 
couples, when all in one plane, is to call the moments positive 
or negative, according as the ideal rotations are against, or with, 
the hands of a watch, as seen from the same side of the plane ; 
the sign of the algebraic sum will then show the ideal rotation 
of the resultant couple. 

35. Composition of Non-concurrent Forces in a Plane. — Let 

Pj, Pa, etc., be the forces of the system ; x^, y^, x^, y^, etc., the 






y: 



■-x^, -/-^■y^^ 



/ 

J' y;'' 

Fig. 31. 



co-ordinates of their points of application ; and a^, a^, . , . etc., 
their angles with the axis X. Replace P^ by its components 
Xj and ]rj, parallel to the arbitrary axes of reference. At the 
origin put in two forces, opposite to each other and equal and 
parallel to X^ ; si'milarly iovY ^. (Of course X^ = P^ cos oc and 
Y^ = P^ sin a.) We now have P^ replaced by two forces X. 



32 MECHANICS OF ENGINEERING. 

and y, at the origin^ and two couples, in the same plane, whose 
moments are respectively — X{y^ and + ^ x^x-: ^"^ ^^^ there- 
fore (§34) equivalent to a single couple, in the same plane with 
a moment = {Y ^x^—X^y^. 

Treating all the remaining forces in the same way, the whole 
system of forces is replaced by 

the force :2{X) =X, +X, + . . . attlie origin, along the axisX; 
the force ^{Y) = Y,^ Z, + . . . at the origin, along the axis Y- 

and the couple whose mom. G= ^ { Yx — Xy), which may be 
called the couple C (see Fig. 32), and may be placed anywhere 
in the plane. Now -5'(X) and 2( Y) may be combined into a 
force jR i i.e., 

, ... ^X 

R = V[^Xf -\- 2 Yy and its direction-cosine is cos a = —p-. 

Since, then, the whole system reduces to C and i?, we must 

have for equilibrium B = 0, and G = ; i.e., for equilibrium 

2X= 0, ^r= 0, and ^{Yx-Xy) = 0. . eq. (1) 

If i? alone = 0, the system reduces to a couple whose mo- 
ment is 6^ = ^( Yx—Xy) ; and if G alone = the system re- 
duces to a single force i?, applied at the origin. If, in general, 
neither I^ nor G = 0, the system is still equivalent to a single 
force, but not applied at the origin (as could hardly be ex- 
pected, since the origin is arbitrary) ; as follows (see Fig. 33) : 

Replace the couple by one of equal moment, G, with each 

G 
force = jR. Its arm will therefore be -^. Move this couple 

in the plane so that one of its forces i? may cancel the i? al- 
ready at the origin, thus leaving a single resultant i? for the 
whole system, applied in a line at a perpendicular distance, 

G 
c = -p , from the origin, and making an angle or whose cosine = 

2X 

■^ , wdtb the axis X. It is easily preyed that the " moment" 

Re, of tbe single resultant, about tbe origin 0, is equal to the 
algebraic sum of those of its " components " (i.e., the forces of 
the system. 

36. More convenient form for the equations of equilibrium 
of non-concurrent forces in a plane. — In (I.), Fig. 34, being 



STATICS OF A EIGID BODY. 



33 



any point and a its perpendicular distance from a force P\ 
put in at two equal and opposite forces P and P' = and || 
to P, and we have P replaced by an equal single force P' at 
0, and a couple whose moment is -|- Pa. (II.) shows a simi- 
lar cunstrtcrion, dealing- with the JTand 1^ components of P, 
so that in (II.) P is replaced by single forces ^' and Y' at 




^.....^......Liix 

^_ 1! „ 


X X' 
Y^ (TI.) 



(I.) 

Fig. 34. 

(and they are equivalent to a resultant P', at 0, as in (I.), and 
two couples whose moments are -|- Yx and — ^y. 

Hence, being the same point in both cases, the couple Pa 
is equivalent to the two last mentioned, and, their axes being 
parallel, we must have Pa = Yx — Xy. Equations (1), 
§ 35, for equilibrium, may now be written* 

:SX- 0, 2Y = 0, and :S{Pa) = 0. . . (2) 

In problems involving the equilibrium of non-concurrent 
forces in a plane, we have three independent conditions^ or 
equations.^ and can determine at most tliree unknown quantities. 
For practical solution, then, the rigid body having been made 
free (by conceiving the actions of all other bodies as repre- 
sented by forces), and being in equilibrium (which it must be 
if at rest), we apply equations (2) literally ; i.e., assuming an 
origin and two axes, equate the sum of the JT components of 
all the forces to zero; similarly for the 1^ components ; and 
then for the "moment-equation," having dropped a perpen- 
dicular from the origin upon each force, write the algebraic 
sum of the products {moments) obtained by multiplying each 
force by its perpendicular, or " lever-armj^'' equal to zero, call- 
ing each product + or — according as the ideal rotation ap- 
pears against, or with, the hands of a watch, as seen from the 
same side of the plane. (The converse convention would do as 
■well.) 

* Another proof is given on p. 15 of th« " Notes and Examples in Mechanics," 



34 MECHANICS OF ENGHNEERING. 

Sometimes it is convenient to use three moment equations, 
takkig a new origin each time, and then the 2X =i and 2Y 
= are superfluous, as they would not be independent equa- 
tions. 

37. Problems involving Non-ooncurrent Forces in a Plane. — 

Remarks. The weight of a rigid body is a vertical force 
through its centre of gravity, downwards. 

If the surface of contact of two bodies is smooth the action 
(pressure, or force) of one on the other is perpendicular to the 
surface at the point of contact. If a cord must be imagined 
cut, to make a body free, its tension must be inserted in the 
line of the cord, and in such a direction as to keep taut the 
small portion still fastened to the body. In case tiie pin of 
a hinge must be removed, to make the body free, its pressure 
against the ring being unknown in direction and amount, it is 
most convenient to represent it by its unknown components X 
and J^, in known directions. In the following problems there 
is supposed to be no friction. If the line of action of an un- 
known force is known, but not its direction (forward or back 
ward), assume a direction for it and adhere to it in all the three 
equations, and if the assumption is correct the value of the 
force, after elimination, will be positive ; if incorrect, negative. * 
ProhleTTh 1. — Fig. 35. Given an oblique rigid rod, with two 
loads (xj (its own weight) and G^ ; required the reaction of the 
smooth vertical wall at A, and the direction and amount of the 
A^^^e-pressure at 0. The reaction at A 
must be horizontal ; call it X'. The pres- 
^1 sure at 0, being unknown in direction, will 
have both its X and ]P" components un- 
known. The three unknowns, then, are 
^^^^^^'^^ ^M ^' 1 and J^o, while G^, G^, «„ a„ and 

■^''^ h are known. The figure shows the rod 

Fis. 35. ^g ^ y^^^ hody, all the forces acting on it 

have been put in, and, since the rod is at rest, constitute a sys- 
tem of non-concurrent forces in a plane, ready for the condi' 
tions of equilibrium. Taking origin and axes as in the figure, 

* That is, the force must point in a direction opposite to that first 
assumed for it. 



6 



STATICS OF A RIGID BODY. 



35 



2X= gives +X„ - X' = ; :S Y = gives -^ T, - G, 
— G, = 0; while 2{Fa) = 0, about 0, gives + XA — 
(?,«j — G^a^ = 0. (Tiie moments of JC^ and Y^ about 
are, each, = zero.) By elimination we obtain Y^ ^ ^i 4" 
6^2 ; Xj = X' = [6^iaj -f- ^2*^2] -i- '^^j while the pressure at 
= VX^ -[" ^o^ ^"d makes with the horizontal an angler, 
whose tan = I^o -r- X,,. 

[N.B. A special solution for this problem consists in this, that the result- 
ant of the two known forces Oi and O2 intersects llie line of X' in a point 
which is easily found by § 21. The hinge-pressure must puss through this 
point, since three forces in equilibrium must be concurrent.] 

IN'ote that the line of action of the pressure at 0, i.e. , of the 
resultant of Xq and Yq, does not coincide with the axis of the 
rod; the rod being subjected to more than the two forces at 
its extremities. The case therefore differs from that presented 
by the boom in Ex. 2 of p. 12. 

Problem, 2. — Given two rods with loads, three hinges (or 
" pin -joints"), and all dimensions: required the three hinge- 





Fig. 36. 



FiGf. sr. 



pressures; i.e., there are six unknowns, viz., three Xand three 
Y components. We obtain three equations from each of the 
two free bodies in Fig. 37. The student may fill out the de- 
tails. Notice the application of the principle of action and 
reaction at B (see § 3). 

ProMem 3. — A Warren bridge-truss rests on the horizontal 
smooth abutment-surfaces in Fig. 38. It is composed of equal 
isosceles triangles ; no piece is 
continuous beyond a joint, each 
of which is a, pin connection. All 
loads are considered as acting at 
the joints, so that each piece will lj j j i^i 
be subjected to a simple tension pjo ^g. ' 

or compression. " Two-force ipieces ; Bee -p. 18, Notes.) 




MECHANICS OF ENGINEERING. 



First, required the reactions of the supports "Fj and T^' 
these and the loads are called the external forces. '^[Pd) 
about ■=■ ^ gives (the whole truss is the free body) 



F,3« - P, 



\a 



I\.%a- P3.f« = 0; 



while ^{Pa) about K =^ gives 



and 



- F, . 3« + P3 . i« + P^^a + P,|a = 0; 
V. = iCA + 3P. + 5^3]. 



Secondly, required the stress (thrust or pull, compression or 
tension) in each of the pieces A, P, and Cent by the imaginary 
line PP. The stresses in the pieces are called internal forces. 
These appear in a system of forces acting on a free body only 
when a portion of the truss or frame is conceived separated 
from the remainder in such a way as to expose an internal 
plane of one or more pieces. Consider as a free body the por- 
tion on the left of PE (that on the right would serve as well, 
I p p but the pulls or thrusts in A, P, and 

6^ would be found to act in directions 
opposite to those they have on the 
other portion ; see § 3). Fig. 39. The 
^iIq arrows (forces) A, B, and C, are as- 
sumed to 23oint, respectively, in the 
directions shown in the figure. 
They, with Vi, P\, and P2, form a system holding the body 
in. equilibrim. 

For this system, I (Pa) about (9=0 gives 

0+Ah-Vi2a+Pi ■ fa+P2 • ia = 0; 
and hence A = {ia^h)[Wi-3Pi^P2l 

which is positive; since, (see above), 4Fi is >3Pi+P2. 

Therefore the assumption that A points to the left is con- 
firmed and A is a thrust, or compression ; (its value as above.) 
Again, taking moments about Oi (intersection of A and B), 
we have an equation in which the only unknown is (7, viz. , 
C/i-7i|a+Pia=0; /. C=(ia-J-/i)[37i-2PJ, 




Fig. 39. 



STATICS OF A RIGID BODY. 



37 



a positive value since 37i is >2Pi; :.C must point to the 

right as assumed; i.e., is a tension, and=— [STi— 2Pi]. 

Finally, to obtain 5, j)ut 2'(vert. comp8.)=0; i.e. 
5 cos <j6 + 7i-Pi-P2 = 0. 
.*. B cos 9S =Pi +P2— Vi ; but, (see foregoing 
value of V\) we may write 

K = (P. + PJ - (iP, + IP^ + ^P3, 

/. P cos cp will be + (upward) or — (downward), and P will 
be compression or tension^ as ^P^ is < or > [^P, -j- iPj. 



P = [P,+ P,-FJ--cos9> 



^^-V.+P.-rj. 



Problem 4. — Given the weight G^ of rod, the weight G^, 
and all the geometrical elements (the student will assume a 

w\|Pi m 





Gi AG, 
Fig. 40. ■ Fig. 41. 

convenient notation); required the tension in the cord, and the 
amount and direction of pressure on hinge-pin. Fig. 41. 

Problem 5. — Roof-truss; pin-connection; all loads at joints ; 
wind-pressures W and TF, normal to OA ; required the three 
reactions or supporting forces (of the two horizontal surfaces 
and one vertical surface), and the 
stress in each piece. All geomet- 
rical elements are given ; also P, 
P,P„Tr(Fig.40). 



38. Composition of Non-concur- 
rent Forces in Space. — Let P„ P„ 

etc., be the given forces, and a?j, y^^ 
2!„ 35,, y,, s,, etc., their points of ap- 
plication referred to an arbitrary 
origin and axes; a^^ /?j, y^^ etc., 




Fig. 42. 



the angles made by their lines of application with Xs. Y^ and Z, 



38 



MECHANICS OF ENGINEERING. 



1 

Considering tlie first force* i^j, replace it by its three com- 
ponents parallel to the axes, J^^ = P^ cos a^^ Y^ = P^ cos /?,; 
and Z, = P^ cos y^ {P^ itself is not shown in the figure). At 
(?, and also at A^ put a pair of equal and opposite forces, 
each equal and parallel to Z^ ; Z, is now replaced by a single 
force Z^ acting upward at the origin, and two couples, one 
in a plane parallel to YZ and having a moment = — Z^y^ (as 
we see it looking toward from a remote point on the axis 
-\- JT), the other in a plane parallel to XZ and having a mo- 
ment := -f~ ^\^\ (seen from a remote point on the axis -^ Y^. 
Similarly at and G put in pairs of forces equal and parallel 
to -Z^, and we have X^^ at B^ replaced by the single force X^ 
at the origin, and the couples, one in a plane parallel to XY^ 
and having a moment -|- X{y^^ seen from a remote point on 
the axis -|- Z, the other in a plane parallel to XZ. and of a 
moment ■= — ^i^i, seen from a remote point on the axis -\-Y\ 
and finally, by a similar device, Y^ at B is replaced by a force 
Y^ at the origin and two couples, parallel to the planes XY 
and YZ^ and having moments — Y^x^ and -f^ i^2j, respective* 
ly. (In Fig. 42 the single forces at the origin are broken 
lines, while the two forces constituting any one of the six 

couples may be recognized as being 
equal and parallel, of opposite di- 
rections, and both continuous, or 
both dotted.) We have, therefore, 
replaced the force P^ by three 
forces Xj, y"j, Z^, at 0, and six 
couples (shown more clearly in 
Fig. 43; the couples have been 
transferred to symmetrical posi- 
tions). Combining each two couples 
whose axes are parallel to X^ Y^ 
or Z, they can be reduced to three, viz., 

one with an X axis and a moment = Y^z^ — Z^y^ ; 
one with a T^axis and a moment = Z-^o^ — X^z^\ 
one with a Z axis and a moment =: X^y^ — Y^x^. 

* This "first force," Pj, is applied at the point B, whose co-ordinates 
are Xi, y^, and 2i, and is typical of all the other forces of the system. 




Fig. 43. 



STATICS OF A RIGID BODY. 39 

Dealing with each of the other forces P^, P^, etc., in the same 
manner, the whole system maj finally be replaced by three 
forces 2X, ^Y, and 2Z, at the origin and three couples 
whose moments are, respectively, (ft-lbs., for example) 

Z, = 2( Yz — Zy) with its axis parallel to X\ 
M = 2{Zx — Xs) with its axis parallel to JT; 
JV = -2'(.Zy — Yx) with its axis parallel to Z. 

The " axes" of tliese couples, being parallel to the respective 
co-ordinate axes JT, Y, and Z, and proportional to tl]^ mo- 
ments Z, 2f, and JV, respectively, tlie axis of their resultant 
C, whose moment is G, must be the diagonal of a parallelo- 
pipedon constructed on the three component axes (propor- 
tional to) Z, M, and iT. Therefore, G = VZ' ~{- M' -^ ]V% 
while the resultant of ^X, -2 Y, and 2Z is 



p = Vi^xy + (^ Yy + {:szy 

acting at the origin. If a, y5, and y are the direction-angles 

^X , 2Y 2Z 

of P, we have cos oc = — ^, cos p = ~ti-, and cos ;^ i= -^ ; 

while if A, ju, and r are those of the axis of the couple C, we- 

Z J^ . ^ 

have cos A, = -p, cos >u = --^, and cos r = --^. 

For equilibrium we have both G = and ^ = ; i.e., 

separately, six conditions, viz., 

:^X= 0, 2 r = 0, 2Z=:0 ; and Z=0, Jf=0, JV=0 . (1) 

Now, noting that :SX = 0,:2Y= 0, and ^(Xy - Yx) = 
are the conditions for equilibrium of the system of non-concur- 
rent forces which would be formed by projecting each force of 
our actual system upon the plane XY, and similar relations 
for the planes YZ and XZ, we may restate equations (1) in 
another form, more serviceable in practical problems, viz. : 
Note. — I]f a system of non-concurrent forces in space is in 
equilibrium, the plane systems formed hy projecting the given 
systein upon each of three arbitrary co-ordinate planes will eaah 
be in equilibrium. But we car obtain only six independent 



40 



MECHANICS OF ENGINEERING. 



equations in any case, available for six unkno'wns. If H alone 
^ 0, we have the system equivalent to a couple C^ whose 
moment = ^ ; if 6^^ alone = 0, the system has a single re- 
sultant R applied at the origin. In general^ neither i? nor G 
being = 0, we cannot further combine i? and G (as was done 
with non-concurrent forces in a plane) to produce a single re- 
sultant unless B, and C happen to be in parallel planes ; in which 
case the system may be reduced to a single resultant by use 
•of the device explained near foot of p. 32. 

Remark. — In general, R and C not being in parallel planes, the system 
may be reduced to two single forces not in the same plane, b^ assigning 
any value we please to P, one of the forces of the couple C, computing 
the corresponding arm a = G-i-P, transferring C until one of the P's has 
the same point of application as R, and then combining these two forces 
into a single resultant. This last force and the second P are, then, the 
equivalent of the original system, but are not in the same plane. (See 
§§ 15 and 15a.) 

Again, if a reference plane be chosen at right angles to R, and the 
couple C be decomposed into two couples, one in the reference plane and 
the other in a plane at right angles to it, this second couple and R may 
be replaced by a single force (as on p. 32) and we then have the whole 
system replaced by a single force and a couple situated in a plane perpen- 
dicular to that force; (and this may be called a "screwdriver action.") 

Example. — A shaft, with crank and drum attached and supported 
horizontally on two smooth cylindrical bearings, constitutes a hoisting 
device. See Fig. 43a. | „ ,,^ 

A force P is to be 
applied to the crank 
handle at 30° with the 
horizontal (and T to 
the crank), and acting 
in a plane at right 
angles to the shaft; 
and is to be of such 
value as to preserve 
equilibrium when the 
weight of 800 lbs. is 
sustained, as shown. 
The weight of the 
shaft, etc., is 200 lbs., 
and its center of grav- 
ity is at C in the axis of the shaft. (Counterpoise for crank not shown.) 
The reactions at the two bearings will lie in planes T to the axis of 
the shaft {smooth cylindrical surfaces), making unknown angles with 
the vertical; and will be represented by their X and Z componentsi 




8001 



Fig. 43o. 



STATICS OF A EIGID BODY. 



41 



as shown. It is required to find the proper value for P and the amount 
and position of the two reactions. 

Solution. — The seven forces shown in the figure (of which five are un- 
known) constitute a non-concurrent system of forces in space; in equi- 
librium. Since there are no Y-components the condition -- F = is already 
satisfied. Let us now apply the statement of the "note" on p. 39, 
first projecting the forces on the plane ZX (vertical plane T to the shaft). 
(That is, we take an "end-view" of the system.) Each of the seven forces 
projects in full length, or value, since they are all parallel to that plane. 
Treating the plane system so formed as in equilibrium and taking mo- 
ments about the point 0, we find (feet and lbs.) 

+ PX1.5-800XH0 = 0; .-. P= 177.77 lbs. . . . (1) 
Next projecting on the vertical plane ZY, containing axis of shaft 
(i.e., taking a "side-view" of the system) we note that#the projection 
of P is P sin 30° and those of Xq and X^ each zero, while Z^, Z^, and 
the 200 and 800 lbs., project in full length; hence taking moments about 
we have 

-|-200Xlf+800X2-ZiX3-PX0.50X4 + 0=0 ... (2) 

while moms, about Oigives+ZoX 3 -Px0.5Xl-200Xli- 800- 1 = (3) 

Finally, projecting on the horizontal plane XZ ("top-view"), the 

forces in this projection are P cos 30°, X^, and X^; so taking moms. 

about point Oj, 

+ZoX3-Px0.8660Xl = 0; .-. X„= +51.34 lbs.; . (4) 

whne from ^X = 0, X^-X^ = 0, or X^=Xq; i.e., Xi= +51.34 lbs. . (5) 
From (2) and (3) we have Zi= +525.93 lbs., and Zo= +385.18 lbs. 
All these + signs show that the arrows for X^, X^, Zg, and Z^ have been 
correctly assumed (Fig. 43a) as to direction. Combining results, we 
find that the pressure or reaction at O is Rq, =VXo^ + Zo^ = 388.6 lbs- 
and makes an angle whose tang, is Xq-hZq, (i.e., 0.1333), Viz., 7° 36', 
on the left of the vertical; also that the pressure or reaction at 0^ is 
Pi, =V'Xi^ + Zi^ = 528.4 lbs., at an angle on the right of the vertical 
whose tang., =X^^Z^, ( = 0.09763); i.e., 5° 34'. 

39. Problem (Somewhat similar to the foregoing.) — Given all geo- 
metrical elements (including a, 
/?, ;-, angles of P) , also the weight 
of Q, and weight of apparatus 
G; A being a hinge whose pin 
is in the axis F, a ball-and- 
socket joint -.vequiredthe amount 
of P (lbs.) to preserve equi- 
librium, also the pressures 
(amount and direction) at A 
and O; no friction. Replace 
P by its X, Y, and Z com- 
ponents. The pressure at A 
will have Z and X components; 
that at 0, X, Y, and Z com- 
ponents. [Evidently there are six unknowns; Yq will come out negative. 




Fig. 44. 



42 



MECHANICS OF ENGINEERING. 



CHAPTER lY. 



STATICS OF FLEXIBLE CORDS. 



40. Postulate and Principles. — The cords are perfectly flexi- 
ble and inexteiisible. All problems will be restricted to one 
plane. Solutions of problems are based on two principles, 
viz.: 

Pein. I. — The strain or tension, in a cord at any point can 
act only along the cord, or along the tangent if it be curved. 

Pein. II. — We may apply to flexible cords in equilibrium 
all the conditions for the equilibrium of rigid bodies ; since, 
if the system of cords became rigid, it would stiU, with 
greater reason, be in equilibrium. 

41. The Simple Pulley. — A "simple pulley" is one that is 
acted on by only one cord (or belt) and the reaction of the 
bearing supporting its axle (or "journal"). 

A cord in equilibrium over a simple pulley whose axle is 
smooth is under equal tensions on both sides; for, Fig. 46, 





Fig. 46. 



Fig. 47. 



considering the pulley and its portion of cord free 2(Pa) — 
about the centre of axle gives I^'r =^ Pr, i.e., J*' = I^ = ten- 
sion in the cord. Hence the pressure i? at the axle bisects 
the angle ex, and therefore if a weighted pulley rides upon a 
cord ABC, Fig. 47, its position of equilibrium, B, may be 
found by cutting the vertical through A by an arc of radius 
CD = length of cord, and centre at C, and drawing a horizon- 
tal through the middle of AD to cut CD in B. A smooth 
ring would serve as well as the pulley ; this would be a slip- 
knot. From Fig. 46, R = 2P cos Ja. 



STATICS OF FLEXIBLE COKDS. 



43 



42. If tliree cords meet at 2i fixed Tcnot, and are in equilib- 
rium, the tension in any one is the equal and ymy/»y^///.. 
^opposite of the resultant of those in the other y _ 
two. 

43. Tackle. — If a cord is continuous over a 

number of sheaves in blocks forming a tackle, 

neglecting the weight of the cord and blocks and 

friction of any sort, we may easily find the ratio 

between the cord-tension P and the weight to be 

sustained. E.g., Fig. 48, regarding all the straight 

cords as vertical and considering the block B 

free, we have, Fig. 49 (from •2Y=%^P- G 
ri 

= 0, .*. P = -T-. The stress on the support G will = 5P. 
4 





Fig. 49^ 



Fig. 49c. 



G 

Fig. 49c?. 



Other designs of tackle are presented in Figs. 49a, 496, 49c, and 49<^, 
and should be worked out as exercises by the student. In each case 
the weight G is supposed to be given and [a value of the smaller 
weight (or pull) P must be determined for the equilibrium of the tackle. 
Friction, and the weights of the pulleys and cords, are neglected and 
all straight parts of cords (or chains) are considered vertical. 
. All of the pulleys shown are "simple pulleys," except the one at A in 
Fig. 49d, which represents a "differential pulley" tackle. Pulley A 
consists of two ordinary pulleys fastened together, the groove in each 
being so rough, or furnished with "sprocket-teeth" in case a chain is 
used, that slipping of the cord or chain is prevented. The chain or cord 
is endless, the loop C being slack. B is a simple pulley. In this case, 
for equilibrium the pull P must =W(r^~r)-r-ri. The other results 
.are p = iG for Fig. 49a; iG for 496; and IG for 49c. 



44 



MECHANICS OF ENGINEERING. 



44. Weights Suspended at Fixed Knots.— Given all the geo. 

metrical elements in Fig. 50, iind 
one weight, G^\ required the re- 
maining weights and the forces 
iZoi 1^05 Hn ^i'^ y^ni at the points 
of support, that equilibrium may 
obtain. H^^ and Fq are the hori- 
zontal and vertical components of 
the tension in the cord at 0\ 
simihvrlj 11^ and Y^ those at n. There are ^ -f 2 unknowns.. 
(The solution of this problem is deferred. See p. 420.) 





Fig. 50a. 

45. Example. — The boom OD, tie-rod RT), with four simple pulleys 
and a cable, form a crane as shown in Fig. 50a. Find the necessary 
vertical force P to be exerted on the piston at H, that the load of 
800 lbs. may be sustained. Also find the pressure of pulley B on its 
bearing, the pull T in the tie-rod and the pressure Pq (amount and 
position) at pin O; neglecting all friction and rigidity (p. 192) and the 
weights of the members. Dimensions as in figure. Since all puUeye 
are "simple" the tension in cable is the same at all points; and is. 
= 400 lbs. since the straight parts of cable adjoining pulley A are parallel. 
For a similar reason P = 800 lbs. 

Pressure at B bisects the angle (50°) between adjoining straight parts 
of cable; i.e., is 25° with vertical, and =2X400Xcos 25° = 725 lbs., (§ 41). 

Next take the free body in Fig. 506 (boom and pulley '5 together 
with a part of cable.) Three unknowns and three equations. 
^(moms.)o = 0; .-. -F TXQXtan 40° + 400X 1-400X9-400X7 = (1) 
i.e., TX 9 X 0.8391 = 6000 ft.-lbs.; .-. 2" = 794.3 lbs., (tension in tie-rod.) 
IX = Xi, .•.Xo-400cos40°-r = 0; .-. Xo = 794.3-H400X .766= 1100.7 lbs. 
^F=0, .-. Yo-^OO sin40°-400-400 = 0; 

.-. y„ = 400 X. 6428 + 800 =1057. 12 lbs. 

Hence P„, = VZo^ -F Fo^ = 1 526 lbs. at tan-* Fo/Zq, or 43° 50', with horiz.. 



STATICS OF FLEXIBLE CORDS. 



45 



Note. — If the weight 800 lbs. were attached directly to cable on right. 
of pulley B, the value of P would need to be 1600 lbs. 

46. Loaded Cord as Parabola. — If the weights are equal and 
inlinitelj small, and are intended to be uniformly spaced 
along the horizontal, when equilib- 
rium obtains, the cord having no 
weight, it will form a parabola. Let 
q = weight of loads per horizontal 
linear unit, O be the vertex of the 
curve in which the cord hangs, and 
Tn any point. We may consider 
the portion Om as a free body, if 
the reactions of the contiguous portions of the cord are put in, 
J3q and T, and these (from Prin. I.) must act along the tangents 
to the curve at O and m, respectively ; i.e., Sq is horizontal, 

and T makes some angle cp (whose tangent = —, etc.) with 
the axis X! Applying Prin. II., 

2X = gives Tcos, cp — Hq = ; i.e., T^~^ = jSq ; 




ds 



2 Y= gives T sin (p — qx = ', i.e., T-j- = 



. (1) 
qx. . . (2) 

Dividing (2) by (1), member by member, we have — = -^ ; 

q 
dy = -^xdx, the differential equation of the curve ;, 



B, 



(/ -_-._ / tJuCLvO — -r~r 



or X 



y, the equation oi a 



parabola whose vertex is at (9 and axis vertical. 

Note. — The same result, ~ = %- , mav be obtained by considering that 

we have here (Prin. II.) ?ifree rigid hody acted on 
by three forces, T, Hq, and R = qx, acting verti- 
cally through the middle of the abscissa x; the 
resultant of Hq and R must be equal and oppo- 




site to T, Fig. 53. 



R dy qx 
tan o) = — -, or -f =-- 
Ho dx 



H^ 



Evidently also the tangent-line bisects the ab- 
scissa X. (Try moments about m.) 
Example. — Let g = 800 lbs. "per foot run" and .'c = 100 ft., with 2/ = 20 ft. 
Then we have, for the value of the tension at the vertex of the parabola, 
F = ?x= ^ 2j/ = 800 X (100) => -^ 40 = 200,000 lbs. 



46 



MECHANICS OF ENGINEERING. 



47. Problem under § 46. [Case of a suspension-bridge m 
which the suspension-rods are vertical, the weight of roadway 
is uniform per horizontal foot, and large compared with that 
of the cable and rods. Here the roadway is the only load : it 
is generally furnished with a stiffening truss to avoid deforma- 
tion under passing loads.] — Given the span = 2J, Fig. 53, 
Y; vf 71 ^^1® deflection = a, and the rate of loading 

j. ^_ —j^y^^ = q lbs. per horizontal foot ; required the 

tension in the cable at 0, also at m ; and 
^ the length of cable needed. From the 
equation of the parabola qx^ = 'iH^y, put- 
ting a? = 5 and y = a, we have Mq = qjf -r- 2« — the tension 
at 0. From -S'T' = we have Y^ = qb, while ^^= gives 




qr--x- 

Fis. 53. 



M, = U,\ .-. the tension at m = \^ H^ + Y^= ^-{_qb V4:a'-\- b']. 

Act 

The semi-length, Om , of cable (from p. 88, Todhunter's In- 
tegral Calculus) is (letting n denote Hq -t- 2^', = 5^ -f- 4a) 

Otn = Vna -\- a" -{- n . log^ [( Va -j- Vn -j- a) -^- Vti]. 

48. The Catenary.* — A flexible, inextensible cord or chain, of 
uniform weight per unit of length, hung at two points, and 
supporting its own weight alone^ forms a curve called the 
catenary. Let the tension Hq at the lowest point or vertex be 
represented (for algebraic convenience) by the weight of an 
imaginary' length, c, of similar cord weighing q lbs. per unit 
of length, i.e., 11^=. qG\ an actual portion of the cord, of 
length 5, weighs qs lbs. Fig. 54 shows -as, free and in equilib- 
-f-^^ ^ rium a portion of the curve of any 
length s, reckoning from the 
vertex. Requii-ed the equation of 
the curve. The load is uniformly 
spaced along the curve, and not 
horizontally, as in §§ 46 and 47. 




Fig. 54. 



2T 



gives 2-J^ = ^s; while 



rdx 



SX - gives T-^- = qG. 
squaring, c^«?2/'' = ^do^. . 



Hence, by division, cdy = sdx, and 
• ■ (1) 



* For the " transformed catenary," see p. 395. 



STATICS OF FLEXIBLE COEDS. 47 

Put dy^ = ds^ — dx^, and we have, after solving for dx 
ods /*s (is rs 

and X =0 . log, [(s + V? + c') -^ c], . . . (2) 

a relation between the horizontal abscissa and length of curve 
Again, in eq. (1) put dx^ = ds^ — dy"^, and solve for dy. 

This gives dy = , = ^r . \ , o cr. Therefore . 

^ ^ i/c^ -I- 5« 2 (c^+5')* • 

2/ = iX'{o' + 5')~*c?(c' + «') = i '2(c^ + sy, and finally 



y = |/*^ 4- c" _ c (3) 

Clearing of radicals and solving for c, we have 

c = («'-y')-% (4) 



Kow T, the tension at any point, = V(qs)~ + (56)2, and 
from (3) we obtain 

T=q(y + c) (4a) 

Example. — A 40-foot chain weighs 240 lbs., and is so hung from 
two points at the same level that the deflection is 10 feet. Here, for 
s=20ft., ?/ = 10; hence eq. (4) gives the parameter, c = (400 — 100) -^- 20 = 15 
feet. 5 = 240-^-40 = 6 lbs. per foot. .•. the tension at the middle is if (; = gc 
=6X15 = 90 lbs.; while the greatest tension is at either support and 
= \/90M^120'=150 lbs. 

Knowing c=15 feet, and putting s = 20 feet = half length of chain, 
we may compute the corresponding value of x from eq. (2) ; this will 
be the half-span. That is, 

.x = 15 . loge 3 = 15 X2.303 X 0.4771 = 16.48 ft. 

To derive s in terms of x, transform eq. (2) in the way that 

?i = logg m may be transformed into e^ = m, clear of radicals 

and solve for s, obtaining "^ 

or, s = c.sinh(— j. . . (5) 

Again, eliminate s from (2) by substitution from (3), trans- 
form as above, clear of radials, and solve for y+c, whence 

y +c = lc[ee -{-e « J; or, ?/-l-c = c. cosh|— j. (6) 

which is the equation of a catenary with axes as in Fig. 54. 
If the horizontal axis be taken a distance = c below the vertex, 
* sinh and cosh denote "hyperbolic" sine and cosine; see table, appendix. 






48 



MECHANICS OF ENGINEERING. 



the new ordinate z = y-\-c, while x remains the same; the last 

equation is simplified. See figure below. 

If the span and length of chain are given, or if the span 

and deflection are given, c can be determined from (5) or (6) 

only by successive assumptions and approximations. 
48a. Catenary (Chain or Cable) with Supports at Different Levels. — Given 

the span k, the difference of 
elevation d of the two supports, 
and the whole length of chain, Z; 
it is required to find x' and y' 
(see Fig. 54a) and thus deter- 
mine the position of the vertex, 
or lowest point, 0, of the cate- 
nary. By applying the equa- 
tions of p. 47 to parts A'O and 
B'O, in turn, and combining, 
we may finally deduce (see 
p. 179 of Rankine's Applied 
Mechanics) 




Fig. 54a. 



-d^ = 2c.sinh( k" 



(7) 



(8) 



-l/P-d^ = ch2c-e 2cJ; i.e., ^/P- 

and also the relation x' — x" = c-\oge\ y—i • 

From (7) we find the "parameter," c, by trial; then the value of x' — x" 
from (8) ; whence, finally, we obtain x' and x" separately (since x'-f a;" = fc). 
With x' known y' is found from (6) ; i.e., 

[x' x'-i 

^ . - ^- ec +e c\ 



or, y' + c-- 



=c-coshl — 



(9) 



Thus the position of the vertex is located. The greatest tension 
will be at the highest point A', viz., TA' = q{y' + c) (10) 

[The expression ^-[e^— g-w] is called the "hyperbolic sine" of the number 
u, or sinh (u); and ^[e^+e-u] the "hyperbolic cosine" of w, or cosh (u); 
e being the Naperian base 2.71828 . . . Tables of sinh u and cosh u 
will be found in the appendix.] 

Example. — A chain 100 ft. long is supported at two points 80 ft. apart 
horizontally and 30 ft. vertically; find the position of its lowest point. 
That is, (Fig. 5 4a) giv en Z=100 ft., k = 80 ft., d = 30 ft. 

Solution.— ■\/P-d^ = \/9100 = 95A ft., the left-hand member of (7). 
Assuming c = 20 ft. as a first trial, we find fc-r-2c = 2.00 and sinh (2.00) 
= 3.6269, so that 2c sinh (2.00) is 145.176, which is much larger than 
95.4. Next, with c assumed as 40, 39, and 38.3 ft., we find 2c sinh (A;-r-2c) 
to be 2X40X1.1745 = 93.96; 2X39X1.2153 = 94.8; and 2X38.3 X 1.2444 
= 95.3, respectively; and hence conclude that c=38.2 ft. will satisfy 
eq. (7) with sufficient accuracy. Eq. (8) now becomes 
a;'-x" = 38.2X2.303X0.2689 = 23.66 ft.; 
and finally we obtain x' = 51.83 and x" = 28.17 ft. From eq. (9) we now 
have 2/' + 38.2 = 38.2X2.07 13 = 79. 10 ft. and .-. j/' = 40.9 ft. With 
?=1.5 lbs. per foot, the tension at A'=l. 5X79. 1 = 118.6 lbs. 



PART II.-KINETICS. 



CHAPTEE I. 

EECTILINEAR MOTION OF A MATERIAL POINT, 

49. Uniform Motion implies that the moving point passes 
over equal distances in equal times ; variable motion, that un- 
equal distances are passed over in equal times. In uniform 
motion the distance passed over in a unit of time, as one sec- 
ond, is called tlie velocity (= v), which may also be obtained 
by dividing the length of any portion (= s) of the path by 
the time (= t) taken to describe that portion, however small or 
great ; in variable motion, however, the velocity varies from 
point to point, its value at any point being expressed as the 
quotient of ds (an infinitely small distance containing the 
given point) by dt (the infinitely small portion of time in 
which ds is described). 

49«. By acceleration is meant the rate at which the velocity 
of a variable motion is changing at any point, and may be a 
uniform acceleration, in which case it equals the total change 
of velocity between any two points, however far apart, divided 
by the time of passage ; or a variable acceleration, having a 
different value at every point, this value then being obtained 
by dividing the velocity -increment, dv, or gain of velocity 
in passing from the given point to one infinitely near to it, by 
dt, the time occupied in acquiring the gain.* (Acceleration 
must be understood in an algebraic sense, a negative accelera- 
tion implying a decreasing velocity, or else that the velocity in 
a negative direction is increasing.) The foregoing applies to 
motion in a path or line of any form whatever, the distances 
mentioned being portions of the path, and therefore measured 
along tlie path. (Sue p. 43 in the ' ' Notes, ' ' etc. 

* See addendum on p. 836. 

49 



50 MECHANICS OF EJSGINEERING. 

50. Eectilinear Motion, or motion in a straight line. — The 
general relations of the quantities involved may be thus stated 
(see Fig. 55) : Let v = velocity of the body at any instant ; 

-s g .as as, t^ ^^^^^ ^^ ~ ^^^^ ^^ velocity 

• ? \ \ I in an instant of time dt. Let 

idtidt'k i i^ = time elapsed since the 
body left a given fixed point, 
which will be taken as an origin, 0. Let s = distance (-f- or 
— ) of the body, at any instant, from the origin ; then ds = 
distance traversed in a time dt. Let^ = acceleration = rate 
at which v is increasing at any instant. All these may be 
variable ; and t is taken as the independent variable, i.e., time 
is conceived to elapse by eqical small increments, each = dt ; 
lience two consecutive dsh will not in general be equal, their 
difference being called d^s. Evidently d^t is = zero, i.e., dt is 
constant. 

Since -,- = number of instants in one second, the velotity at 
any instant (i.e., the distance which would be described at that 

•IN . 7 1 ^^ /TV 

rate m one second) \q v =■ ds . -n-', .' . v ■= -^- (L) 

Similarly, J> =^ dv . -t> and I since dv = di^^J =~^ J' 

dv d^s 

Eliminating dt, we have also vdv = pd^. (HI.) 

Tliese are the fundamental differential formulae of rectilinear 
motion (for curvilinear motion we have these and some in ad- 
dition) as far as kinematics, i.e., as far as space and time, is 
concerned. The consideration of the mass of the material 
point and the forces acting upon it will give still another rela- 
tion (see § 55). 

Example. — If we have given s = [6<^ + 3<^ + 2<] ft., for a certain motion, 
then the velocity, v, at any time, =ds^^dt, =[18i'*+6< + 2] ft. per sec; 
and the acceleration, p, =dv-^dt, =[36t + 6]ft. per sec. per sec. 

61. Rectilinear Motion due to Gravity. — If a material point 
fall freely in vacuo, no initial direction other than vertical 
having been given to its motion, many experiments have 



RECTILINEAR MOTION OF A MATERIAL POINT. 51 

sho^*•n that this is a uniformly accelerated rectilinear motion 
in a vertical line having an acceleration (called the accelera- 
tion of gravity) equal to 32.2 feet per square second,* or 9.81 
metres per square second ; i.e., the velocity increases at this 
constant rate in a downward direction, or decreases in an up- 
ward direction. 

[Note. — By " square second " it is meant to fay stress on tlie fact that an 
acceleration (being = d^s -h df^) is in quality equal to one dimension of 
length divided by two dimensions of time. E.g., if instead of using the 
foot and second as units of space and time we use the foot and the minute, 
g will = 33.3 X 3600; whereas a velocity of say six feet per second would 
= 6 X 60 feet per minute. The value of g = 33.3 implies the units foot 
and second, and is sufficiently exact for practical purposes.] 

52. Free Fall in Vacuo.— Fig. 56. Let the body start at 

v,dth an initial downward velocitj^ = c. The accelera- _s 
tion is constant and = -\- g. Reckoning both time and I 

distance (-|- downwards) from O, required the values of „i^\ 

the variables s and v after any time t. From eq. (II.), \ \ c 

§ 50, we have -{- g = dv -^ dt; .'. dv = gdt, in which the " k 

two variables are separated. I 1 

dv = gj dtt\ i.e., v ^=^ g\ t\ ox v — c =^ ""j 

gt — ^\ and finally, -y = c + (/^ (1) fig. 56. 

(ISTotice the correspondence of the limits in the foregoing 
operation ; when ^ = 0, -y = -f- <^') 

From eq. (I.), § ^^^v ^= ds -^ dt\ .'. substituting from (1), 
ds ^ {c -{- gt)dt, in which the two variables s and t are sepa- 
rated. 

ds = cj^ dt + gj^ tdt ; i.e., [_^5 = e\j + ^[^ ^ ' 

or 5 = c? + ^gf (2) 

Again, eq. (Ill-), § 50, vdv = gds, in which the variables v 
and ,9 are already separated. 

/v ps r"u r~s 

vdv = gj^ ds; or \ iv' = g s; i.e., ^{v' — c') = gs, 



■_.c 



L-0 



* Or, 82.3 "feet per second per second." 



52 MECHANICS OF ENGINEERING. 

If the initial velocity = zero, i.e., if the body falls from rest, 

t 



eq. (3) gives s=^-audv= V^gs. [From the frequent re- 



currence of these forms, especially in hydraulics, ^is called the 

"height due to the velocity t;," i.e., the vertical height through 
which the body must fall from rest to acquire the velocity v ; 
while, conversely, V'2gs is called the velocity due to the height 
or "head" s.] 

By eliminating g between (1) and (3), we may derive another 
formula between three variables, s, v, and t, viz., 

S = i{G-i-V)t . (4:) 

Example. — A leaden ball occupies 4.6 seconds in falling from the eaves 
of a tall building to the sidewalk; initial velocity zero. Find the height 
fallen through, =s'. We have from eq. (2) 

s' = + i(32.2)(4.6) = = 341 ft. 

53. Upward Throw. — If the initial velocity were in an up- 
ward direction in Fig. 56 we might call it — c, and introduce it 
with a negative sign in equations (1) to (4), just derived ; but 
for variety let us call the upward direction -|-, in which case 
an upward initial velocity would = -|- c, while the acceleration 
= — g, constant, as before. (The motion is supposed confined 
within such a small range that g does not sensibly vary.) Fig. 

i 67. From p = dv -^ dt we have dii ■= — gdt and 

J^ dv = — gj^ dt; r.v — G = — gt\orv = G — gt. {l)a 

From V ^=ds -^ dt, ds = cdt — gtdt, 

ns r*t r>t 

i.e., J^ ds = cj^ dt — gj^ tdt ; or s=Gt — ^gt\ (2)a 

O i ' p' p^ 

— S vdv = pds gives / vdv = — a I ds, whence 

Fig. 57. .^ & «/c if Jo 

^(^' — c') = — gs, or finally, s = . . (3)a 

And by eliminating g from {l)a and (3)a, 

s = i{G-{-v)f (4)a 

The following is now easily verified from these equations ; 
the body passes the origin again (6' = 0) with a velocity = — c, 
after a lapse of time =2g -r- g. The body comes to rest (for 



if 



EECTILINEAE MOTION OF A MATERIAL POINT. 53 

an instant) (put v = 0) after a time = c -^ ^, and at a distance 
s = c^ -^ 2g (" height due to velocity c") from 0. For t > 
G -^ g, V is negative, sliowing a downward motion ; for t > 
2g -^ g, s is negative, i.e., the body is below the starting-point 
while the rate of change of v is constant and = — ^ at all 
points. 

Example.— Let c be 40 ft./sec. Then in a time = 40 -r- 32.2, =1.24 

sec, the body will reach its maximum height, (40)2^-2X32.2 = 24.84 
ft. above the start. After 3 sec. the body will be found a distance 
S3=40x3-i(32.2)(3)2=-24.9 ft. from the origin, i.e., below it. 

54. Newton's Laws. — As showing the relations existing in 
general between the motion of a material point and the actions 
(forces) of other bodies upon it, experience furnishes the fol- 
lowing three laws or statements as a basis for kinetics : 

(1) A material point under no forces, or under balanced 
forces, remains in a state of rest or of uniform motion in a 
right line. (This property is often called Inertia^ 

(2) If the forces acting on a material point are unbalanced, 
' an acceleration of motion is produced, proportional to the re- 
sultant force and in its direction. 

(3) Every action (force) of one body on another is always 
accompanied by an equal, opposite, and simultaneous reaction. 
(This was interpreted in § 3.) 

As all bodies are made up of material points, the results ob- 
tained in Kinetics of a Material Point serve as a basis for the 
Kinetics of a Rigid Body, of Liquids, and of Gases. 

55. Mass.- — If a body is to continue moving in a right line, 
the resultant force P at all instants must be directed along that 
line (otherwise it would have a component deflecting the body 
from its straight course). (See addendum on p.' 835.) 

In accordance with Newton's second law, denoting by j? the 
acceleration produced by the resultant force {O- being the 
body's weight), we must have the proportion P \ G : : jp \ g \ 
i.e., 

P = — .p , orP=i^. . . (lY.) 

Eq. lY. and (I.), (II.), (III.) of § 50 are the fundamental 
equations of Dynamics. Since the quotient G ■— g \s, invaria- 



54 MECHANICS OF ENGINEERING. 

ble, wherever the body be moved on the earth's surface {O and 
g changing in the same ratio), it will be used as the measure 
of the mass M ov quantity of matter in the body. In this way 
it will frequently happen that the quantities G and g will ap- 
pear in problems where the weight of the body, i.e., the force 
of the earth's attraction upon it, and the acceleration of gravity 
have no direct connection with the circumstances. No name 
will be given to the unit of mass, it being always understood 
that the fraction G -^ g will be put for M before any numeri- 
cal substitution is made. From (lY.) w'e have, in words, 
accelerating force = mass X acceleration^ 
also, acceleration — accelerating force -=- Quass. 

56. Uniformly Accelerated Motion. — If the resultant force is 
constant as time elapses, the acceleration must be constant (from 
eq. (IV.), since of course J/" is constant) and = P -^ M. The 
motion therefore will be uniformly accelerated, and we have 
only to substitute + pi, (constant) , ior g in eqs. (1) to (4) of 
§ 52 for the equations of this motion, the initial velocity being 
= c (in the line of the force). 

v = G-]-pit . . . (1); s = ct-i-ipit^; , . . (2) 

('y*-_c'') 

If the force is in a negative direction, the acceleration will 
be negative, and may be called a retardation; the initial veloc- 
ity should be made negative if its direction requires it. 

57. Examples of Unif. Ace. Motion. — Exomh;ple 1. Fig. 58. 
A small block whose weight is \ lb. has already described a 
— S g P M ^ g distance Ao = 4-8 inches over a 

A SMOOTH — 7-> ^" -0^~ 1 smooth portion of a horizontal 

Fig. 58. table in two seconds ; at 6^ it en- 

counters a rough portion, and a consequent constant friction of 
2 oz. Required the distance described beyond 0, and the time 
occupied in coming to rest. Since we shall use 32.2 for g, 
times must be in seconds, and distances in feet ; as to the unit 



8 = ^-^;, . . (3), and5 = K^ + ^)^5 ... (4) 



KECTILINEAK MOTION OF A MATERIAL POINT. 55 

-t force, as that is still arbitrary, say ounces. Since AO was 
smooth, it must have been described with a uniform motion 
(the resistance of the air being neglected); hence with a veloc- 
ity = 4 ft. -^ 2 sec. = 2 ft. per sec. The initial velocity for 
the retarded motion, then, is c = -|- 2 at (9, At any point be- 
yond the acceleration = force -— mass = ( — 2 oz.) -r- (8 oz. 
-f- 32.2) = — 8.05 ft. per square second, i.e., p = — 8.05 = 
constant ; hence the motion is uniformly accelerated (retarded 
here), and we may use the formulae of § 56 with g = + 2, pi = 
— 8.05. At the end of the motion v must be zero, and the 
corresponding values of s and t may be found by putting v = 
in equations (3) and (1), and solving for s and t respectively : 
thus from (3), 5 =^(-4)-4- (— 8.05), i.e.,s = 0.248 +, which 
must be feet ; while from (1), t={—2)-^{— 8.05) = 0.248 +, 
which must be seconds. 

Examjple 2. (Algebraic.) — Fig. 59. The two masses J/", = 
G^ -~ g and M ^ G -^ g are connected by a flexible, inexten- 
sible cord. Table smooth. Required the acceleration common 
to the two rectilinear motions, and the tension in the string S,. 



i^s. 




Fig. 59. Fig. 60. 

there being no friction under G^, none at the pulley, and no 
mass in the latter or in the cord. At any instant of the mo- 
tion consider G^ free (Fig. 60), iV being the pressure of the 
table against G^. Since the motion is in a horizontal right line 
^(vert. compons.)= 0, i.e., iV— G^ = 0, which determines iV„ 
S, the only horizontal force (and resultant of all the forces) = 
M,p, i.e., 

S= G,p-^g. (i; 

At the same instant of the motion consider G free (Fig. 61);. 
the tension in the cord is the same value as above = S. The 
accelerating force is ^ — 8, and 

.*. = mass X ace, or G — 8 ^= {G -^ g)p. . (2) 



y 



56 MECHANICS OF ENGINEERING. 

j |g From equations (1) and (2) we obtain p — {Gg) ~ 

\ y, {G -^ G^ = a constant ; hence each motion is uniformly 

j r S accelerated, and we may employ equations (1) to (4) of 

i ^ § 56 to find the velocity and distance from the starting- 

^ points, at the end of any assigned time t, or vice versa. 

■ The initial velocity must be known, and may be zero. 

Also, from (1) and (2) of this article, 

S = {GG,) -^ ((9 + G,) = constant. 

Example 3. — A body of 2J (short) tons weight is acted on 
during ^ minute by a constant force P. It had previously de- 
scribed 316f yards in 180 seconds under no force ; and subse- 
quently, under no force, describes 9900 inches in -^ of an hour, 
Eequired the value of P. Ans. P = 22.1 lbs. 

. / Example 4. — A body of 1 ton weight, having an initial 

''velocity of 48 inches per second, is acted on for \ minute by a 

force of 400 avoirdupois ounces. Required the final velocity, 

Ans. 10.03T ft. per sec. 

Exa/mjple 5. — Initial velocity, 60 feet pei second ; body weighs 
0.30 pf a ton. A resistance of 112|- lbs. retards it for -^-^ of 
a ininute. Required the distance passed over during this time. 

Ans. 286.8 feet. 




Example 6. — ^Required the time in which a force of 600 avoir- 
dupois ounces will increase the velocity of a body weighing \\ 
tons from 480 feet per minute to 240 inches per second. 

Ans. 30 seconds. 

Example 7. — What distance is passed over by a body of (0.6) 
tons weight during the overcoming of a constant resistance 
(friction), if its velocity, initially 144 inches per sec, is reduced 
to zero in 8 seconds. Required, also, the friction. 

Ans. 48 ft. and 55 lbs. 

Example 8. — Before the action of a force (value required) a 
body of 11 tons had described uniformly 950 ft. in 12 minutes. 
Afterwards it describes 1650 feet uniformly in 180 seconds. 
The force acts 30 seconds. P — % Ans. P = 178 lbs. 



KliCTILINEAR MOTION OF A MATERIAL POINT. 



67 



58. Graphic Eepresentations. Unif. Ace. Motion. — With the 
initial velocity = 0, tlie equations of § 56 become 



V =pit,. 



= V 



2pi, 



(1) 

(3) 



s = ipit",. 



and 



(2) 

(4) 



Eqs. (]), (2), and (3) contain each two variables, which may 
graphically be laid off to scale as co-ordinates and thus give a 
curve corresponding to the equation. Thus, Fig. 62, in (I.), we 




A 

V 

J 


s 
. 0- 


t- 


!t 


MQ- 

0^ 


J^-y( 




(II.) 




(HI.) 






Fig. 63. 









have a right line representing eq. (I.) ; in (II.), a parabola with 
axis parallel to s, and vertex at the origin for eq. (2) ; also a 
parabola similarly situated for eq. (3). Eq. (4) contains tlii'ee 
variables, s, -y, and t. Tliis relation can be shown in (I.), s be- 
ing represented by the a?'ea of the shaded triangle = ^vt. 
(11.) and (III.) have this advantage, that the axis OS may be 
made the actual path of the body. [Let the student determine 
how the origin shall be moved in each case to meet the supposi^- 
tion of an initial velocity = -|- c or — c] (SeelSTotes, p. 120.) 

59. Variably Accelerated Motions. — We here restate the equa- 
tion s ( differential) 



9) = 



ds 
dt 



dv d^s 



•(i-);p = ^ = 



dt df 



. (XL); v<?'y=^^5..(III.) 



and resultant force 

= P=Mp,. , . . o . (lY.); 

which are the only ones for general use in rectilinear motion 
and involve the five variables, s, t, v, p, and P. 

Problem 1. — In pulling a mass M along a smooth, horizon- 
tal table, by a liorizontal cord, the tension is so varied that 
i? = 4:f {not a homogeneous equation / the units are, say, the 
foot and second). Required by what law the tension varies. 



58 MECHANICS OF ENGINEERING. 

From (L) , = ^^ = A_l ^ i^f ; from (IL), p = ^-^ = 

24:t', and (lY.) the tension = P = 2fp = ^^Mt, i.e., varies, 
directly as the time. 

Pkoblem 2. " Harmonic Motion," Fig. 63. — A small block 



Tig. 63 

on a smooth horizontal table is attached to two horizontal 
elastic cords (and they to pegs) in snch a way that when the- 
block is at (9, each cord is straight but not tense ; in any other 
position, as ^y?., one cord is tense, the other slack. The coi'ds 
are alike in every respect, and, as with springs, the tension 
varies directly with the elongation (= 5 in figure). If for an 
elongation s^ the tension is Tj, then for any elongation s it is- 
r = riSH-Si.v If the block be given an initial velocity =c 
at 0, it begins to execute an oscillatory motion on both 
sides of 0; m is any point of its motion. The tension 
T is the accelerating force ; variable and always pointing 
toward 0. The acceleration at any point m, then, is 
p — — {T -^ M) = — {T^s -f- J/Sj), which for brevity put 
^ = — as, a being a constant. Required the equations of 
motion, the initial velocity being = -[- c, at 0. From eq. (til.)- 

vdv = — asds ; .'. / vdv := — a I sds, 

i.e., ^(y' — c') = — ^as^ ; or, y' = 6'' — as\ . (1)- 

From (I.), dt = ds^v',\ CK^ r\, .- 

-, \ ,-,^ > I dt= I [ds^yc^—as^]: or, 

hence from (1), }Jo Jo 



1 . JsV^\ 
Va ^ 



= ~-^sm-^[-—j (2), 



RECTILINEAR MOTION OF A MATERIAL POINT. 



59 



Inverting (2), we have 5 = (c -r- Va) sin (t Va), ... (3) 
iLgain, by differentiating (3), see (I.), -y = c cos (tVa) (4) 
Differentiating (4), see (II.), 2^ = — cVa sin {t Va). . . (5) 

These are the relations required, each between two of the 
four variables, s, t, v, and p; but the peculiar property 
of the motion is made apparent by inquiring the time of pass- 
ing from (? to a state of rest ; i.e., put i) = in equation (4), 
we obtain i =z ^tt -i- Va, or ^tc — Va, or ^n -^ V~a, and so on, 
while the cori-esponding vakies of s (from equation (3)), are 
'\-{g -^ Va), — (c -^ Va), -f- (c -^ Va), and so on. This shows 
that tlie body vibrates equalW on both sides of in a cycle or 
period whose duration = 2;r -^ Va, and is indejpendent of the 
initial velocity given it at 0. Each time it passes the 
velocity is either -|- c, or — c, the acceleration = 0, and the 
time since the start is = nn -f- \/a^ in which n is any whole 
number. At the extreme point ^9 = :f c j/a, from eq. (5). 
If then a different amplitude be given to the oscillation by 
changing c, the duration of the period is still the same, i.e., 
the vibration is isochronal.'^ The motion of an ordinary pen- 
dulum is nearly, that of a cycloidal pendulum exactly, harmonic. 

If the crank-pin of a reciprocating engine moved uniformly 
in its circular path, the piston would have a harmonic motion 
if the connecting-rod were infinitely long, or if the design in 



•< 2r > 




Fig. 64. 

Fig. 64 were used. (Let the student prove this from eq. (3).) 
Let 2r = length of stroke, and c = the uniform velocity of the 
crank-pin, and M = mass of the piston and rod A£. Then 
the velocity of M at mid-stroke must = c, at the dead-points, 
zero; its acceleration at mid-stroke zero; at the dead-points 
the ace. = c Va, and s = r = c -7- Va (from eq. (3)) ; .*. V~a 
=: c -^ r, and the ace. at a dead point (the maximum ace.) 
* See illustrations and example on pp. 47, 48, of the "Notes." 



60 



MECHANICS OF ENGINEEEING- 



:= c* -4- r. Hence on account of the acceleration (or retarda- 
tion) of J!/^in the neighborhood of a dead-point a pressure will 
be exerted on the crank-pin, equal to mass X ace. = M& -^ r 
at those points, independently of the force transmitted due to 
steam-pressure on the piston-liead, and makes the resultant 
pressure on the pin at G smaller, and at D larger than it would 
be if the '"''inertia)'' of the piston and rod were not thus taken 
into account. We may prove this also by the free-body method, 
considering AB free immediately after passing the dead-point 



P, 



A 



Fig. 65. 



\ 



?' 



C, neglecting all friction. See Fig. 
65. The forces acting are : G, the 
weight ; W, the pressures of the 
guides ; P, the known effective steam- 
pressure on piston-head ; and P', the unknown pressure of 
crank-pin on side of slot. There is no change of motion ver- 
tically ; .-. iT' + i\^— 6^ = 0, and the resultant force is P — P' 
= mass X accel. = Mc" -^ r^ hence P' ^= P — M& — r. 
Similarly at the other dead-point we would obtain P' =^ P -{- 
Mc^ -^ r. In high-speed engines with heavy pistons, etc., 
Mg^ -^ r is no small item. [The upper half-revol., alone, is 
here considered.] (See example on p. 68, "Notes.") 

Problem 3. — Supposing the earth at rest and the resistance 
of the air to he null, a body is given an initial upward vertical 
velocity = c. Required the velocity at any distance s from 
the centre of the earth, whose attraction varies in- 
versely as the square of the distance s. 

See Fig. Q^. — The attraction on the body at the 
surface of the earth where s = r, the radius, is its 
weight G; at any point m it will be P = G(r2-=- s^),* 
while its mass = G -7- g. 

Hence the acceleration at m = j? = ( — P) -j- J/ 
= — ^(r° -=- s°). Take equation III., vdv = j)ds, 
and we have 

vdv = — gr^s ~^ds°, .'. 




Fig. 66. 



I vdv =z — gr^ I s 



'ds 



or. 



4^,« 



fv = — gr 



I.e. 



ii^' 



^')=-^r^{l-~ 



(1) 



* That is, the force of attraction, {P, lbs.) at any distance, s, from O 
is to the force at the surface (viz., G lbs.) as r'' is to s^. 



BECTILINEAR MOTION OF A MATERIAL POINT. 



61 



Evidently v decreases, as it should. Now inquire how small 
A value c may have that the body shall never return/ i.e., 
that V shall not = until 5 = oo. Put v = and s =s 00 in 
(1) and solve for g ; and we have 

c = V2ffr = V2 X 32.2 X 21000000, 

= about 36800 ft. per sec. or nearly 7 miles per sec. Con- 
versely, if a body be allowed to fall, from rest, toward the 
earth, the velocity with which it would strike the surface 
would be less than seven miles per second through whatever 
distance it may have fallen. 

If a body were allowed to fall through a straight opening in 
the earth passing through the centre, the motion would be har- 
monic, since the attraction and consequent acceleration now 
vary directly with the distance from the centre. See Prob. 2. . 
This supposes the earth homogeneous. 

Problem 4. — Steam working expansively and raising a weight. 
Fig. 67. — A piston works without 
friction in a vertical cylinder. Let 
S = total steam-pressure on the 
underside of the piston ; the weight 
G, of the mass G — g (which in- 
cludes the piston itself) and an 
atmospheric pressure = A^ con- 
stitute a constant back-pressure. 
Through the portion OB = s^, of 
the stroke. Sis constant = S^, while beyond B, boiler com- 
munication being " cut off," S diminishes with Boyle's law, i.e., 
in this case, for any point above JB, we have, neglecting the 
" clearance", 2^ being the cross-section of the cylinder, * 

S:S,::Fs,: Fs; or S=S,s,-^s. 

(Which gives *S as a function of s at any point above B.) 

Full length of stroke = ON" = s^. Given, then, the forces 

S^ and A, the distances s, and 5„, and the velocities at and 

at ^both = (i.e., the mass M = 6^^ -j- ^ is to start from rest at 

O. and to come to rest at iV), required the proper weight G to 

* See p. 627 for meaning of "clearance." 




Fig. 67. 



€2 MECHANICS OF ENGINEERING. 

fulfil these conditions, ^S* varying as already stated. The accel- 
eration at any point will be 

j9= [_S-A-G^~M. . . . . (1) 

Hence (eq. III.) Mvdv = [S — A — G'jds, and .*. for the 

whole stroke 

J/j^ vdv=J^ [S-A- G]ds; i.e., 

= Sj^ & + <Sa/ -^-A^de-oX ds, 

or 8A[l + iog/fJ = As,+ es,.. . . (2) 

Since S = /S^ ^ constant, from O to £, and variable, = 
^i^i -^ 5, from ^ to JV, we have had to write the summation 



X 



Sds in two parts. 





From (2), G becomes known, and .". J!f also {= G — g). 

Required, further, the time occupied in this upward stroke. 
From to B (the point of cut-off) the motion is uniformly 
accelerated, since p is constant {8 being = 8^ in eq. (1) ), 
with the initial velocity zero; hence, from eq. (3), § 56, 
the velocity 2it B = v^ z= V'i, \_8^ — A — G'\s^ -^ M'k known ; 
.'. the time ^j = 2Sj -i- v^ becomes known (eq. (4), § 56) of de- 
scribing OB. At any point beyond B the velocity v may be ob- 
tained thus : From (III.) vdv — _pds, and eq. (1) we have, 
summing between B and any point above, 

M^vd. = S,s.£ ~-iA + <?)/&; i.e., 

G (v" — V') S f A y n^ f N 

-- — 2—^ = ^1*1 log. --{A^G){s- 5,). 

This gives the relation between the two variables v and 8 
anywhere between B and iV"; if we solve for -y and insert its 
value in dt ^= ds -~- v, we shall have dt = a. function of s and 
ds, which is not integrable. Hence we may resort to approxi- 



KECTILINEAU aiOTION OF A MATERIAL POINT. 63 

mate methods for the time from B to iT. Divide the space 
BN'mio an imeveii nnuiber of equal parts, say five; the dis- 
tances of the points of division from will be s„ s,, s^, s^, s^, 
and Sn- For these values of s compute (from above equation) 
Wi (already known), v„ -y,, v,, v,, and v^ (knowTi to be zero). To 
the first four spaces apply Simpson's Eule,* and we have the 
time from JS to the end of 5,, 



t 



1 , 4 , 3 . 4: . 1 



: / - ; approx. = ^. - - -\ L- 



while regarding the motion from 5 toiTas uniformly retarded 
(approximately) with initial velocity = t\ and the final = zero, 
we have (eq. (4), § 56), 

-TV 

t = 2{S^ - S,) -r- V,. 

—6 

By adding the three times now found we have the whole time 
of ascent. In Fig. 67 the dotted curve on the left shows by 
horizontal ordinates the variation in the velocity as the distance 
s increases ; similarly on the right are ordinates showing the 
variation of jS. The point ^, where the velocity is a maximum 
= v,n, may be found by putting p = 0, i.e., for S ■= A-\- G, 
the acelerating force being = 0, see eq. (1). Below ^the ac- 
celerating force, and consequently the acceleration, is positive; 
above, negative (i.e., the back-pressure exceeds the steam- 
pressure). The horizontal ordinates between the line IIEKL 
and the right line HT hve proportional to the accelerating force. 
If by condensation of the steam a vacuum is produced be- 
low the piston on its arrival at iV^, the accelerating force is 
downward and ^ A-\- G. [Let the student determine how 
the detail of this problem would be changed, if the cylinder 
were horizontal instead of vertical.] 

60. Direct Central Impact. — Suppose two masses J/j and Jf, 
to be moving in the same right line so that their distance apart 
continually diminishes, and that when the collision or impact 
takes place the line of action of the mutual pressure coincides 
with the line joining their centres of gravity, or centres of 
* See p. 13 of the "Notes and Examples." 



64 MECHANICS OF ENGINEERING. 

mass, as they may be called in this connection. This is called 
a direct central impact, and the motion of each mass is varia- 
bly accelerated and rectilinear during their contact, the only 
force being the pressure of the other body. The whole mass 
of each body will be considered concentrated in the centre of 
mass, on the supposition that all its particles undergo simul- 
taneously the same change of motion in parallel directions. 
(This is not strictly true ; the effect of the pressure being 
gradually felt, and transmitted in vibrations. These vibrations 
endure to some extent after the impact.) When the centres 
of mass cease to approach each other the pressure between the 
bodies is a maxinmm and the bodies have a common velocity; 
after this, if any capacity for restitution of form (elasticity) 
exists in either body, the pressure still continues, but dimin- 
ishes in value gradually to zero, when contact ceases and the 
bodies separate with different velocities. Reckoning the time 
from the first instant of contact, let t' = duration of the first 
period, just mentioned ; t" that of the first -(- the second (resti- 
tution). Fig. 68. Let Jf^ and Jf^ be the masses, and at any 
<..?i. I — ^..3?2.^ instant during the contact let v^ and v^ 

~1)T be simultaneous values of the velocities 



-^ 



p 

Ml M2 of the mass-centres respectively (reckon- 

^^®- ^^- ing velocities positive toward the right), 

and f* the pressure (variable). At any instant the acceleration 

of J[fj is j?j = — (P -^ J!/i), while at the same instant that of 

M, is j)^ = -\- {P -^ M^ ; Jfj being retarded, M^ accelerated, 

in velocity. Hence (eq. 11.,^ = dv -f- dt) we have 

M,dv,= — Pdt\ and M^dv^— -\- Pdt. . . (1) 

Summing all similar terms for the first period of the impact, 
we have (calling the velocities before impact c^ and c^, and the 
common velocity at instant of maximum pressure G) 

^Jcy^^= - H'P^i^ ^-e-' ^^(^' -^a) = - S! Pdt ; (2) 



EECTILINEAK MOTION OF A MATERIAL POINT. 66 

The two integrals * are identical, numerically, term by term, 
since the pressure which at any instant accelerates J/, is nu- 
merically equal to that which retards J/, ; lience, though we do 
not know liow P varies with the time, we can eliminate the 
definite integral between (2) and (3) and solve for C. If 
the impact is inelastic (i.e., no power of restitution in either 
body, eitlier on account of their total inelasticity or damaging 
effect of the pressure at the surfaces of contact), they continue 
to move with tliis common velocity, which is therefore their 
final velocity. Solving, \Ne have 

^- M,+M, ^^^ 

Next, supposing that the impact \q partially elastic, ihid, the 
bodies are of the same material, and that the summation 

I Pdt for the second period of the impact bears a ratio, e, 

to that / Pdt, already used, a ratio peculiar to the material, 

if the impact is not too severe, we have, summing equations 
(1) for the second period (letting Y^ and Y^ = the velocities 
after impact), 

^^ X""^^^ = - S"^^*^ ^•^•' ^'( ^- ^) = - ^/*^^^; (s) 

M, / V = + X'Pdt, i.e., M,{ Y- 0) = + ej^dt. (6) 

6 is called the coefficient of restitution. 

Having determined the value of / Pdt from (2) and (3) in 

terms of the masses and initial velocities, substitute it and that 
of (7, from (4), in (5), and we have (for the final velocities) 

y. = W^: + M^c- eMlc-c:j\ - [if, + JfJ; . (7) 
and similarly 

F, = [JfA + ^.c.+ ^^,(^-^.)]-W + ^,]- • (8) 
For d = 0, i.e., for inelastic impact^ Y^= Y,= C m eq. (4) ; for 
* That is, the right-hand members of eqs. (2) and (3). 



66 MECHAISriCS OF ENGINEERING. 

<g= 1, or elastic impact^ (7) :ind (8) become somewhat simpli- 
fied. 

To deterinino e experimeiitallj, let a ball (-3/,) of the sub- 
stance fall upon a very large slab [M^ of the same substance, 
noting both the height of fall h^. and the height of rebomid H^. 
Considering M^ as = cc, with 



Ci= ^ ^yh^, V = —V ^H,, and c, = o, 
eq. (7) gives 



— ^/ "^gH^, =— eV "Igh, ; .-. e = V^, -^ h^. 

Let the student prove the following from equations (2), (3), 
(5), and (6) : 

{a) For any direct central impact whatever, 

[The product of a mass by its velocity being sometimes 
called its momentum^ this result may be stated thus : 

In any direct central impact the snm of the momenta before 
impact is eqnal to that after impact (or at any instant during 
impact). This principle is called the Conservation of Momen- 
tum. The present is only a particular case of a more genei'al 
proposition. 

It can. be proved that C, eq. (4), is the velocity of the centre 
of gravity of the two masses before impact ; the conservation 
of momentum, then, asserts that this velocity is unchanged by 
the impact, i.e., by the mutual actions of the two bodies.] 

(h) The loss of velocity of Jf,, and the gain of velocity of 
J/j, are twice as great when the impact is elastic as when in- 
elastic. 

(c) If e = 1, and M, = M„ then V, = + c^, and V^ = c,. 

Example. — Let Mi and M^ be perfectly elastic, having weights = 4 and 
6 lbs. respectively, and let Ci = 10 ft. per sec. and C2 = — 6 ft. per see. 
(i. e., before impact M^ is moving in a direction contrary to that of Mi). 
By substituting in eqs. (7) and (8), with 6 = 1, Mi — A-^ g, and Jfa = 5 -i- g, 
we have 

Vi = ir4 X 10 + 5 X (- 6) - 5 (lO - (- 6))]= - 7.7 ft. per sec. 

Vi = lr4 X 10 + 5 X (- 6) + 4 Ao - (- 6))]= + 8.2 ft. per sec. 
as the velocities after impact. Notice their directions, as indicated by thdr 



"virtual velocities." 



67 



CHAPTER II. 



"VIRTUAL velocities; 



61. Definitions. — If a material point is moving in a direction 
not coincident with that of the resultant force acting (as in 
cnrvihnear motion in the next chapter), and any element of its 
path, ds, projected npon this force;* the length of this projec- 
tion, du, Fig. 69, is called the "Virtual Yelocity" of the 
force, since du -^ dt uvAy be considered the veloc- 
ity of the force at this instant, just as ds -=- dt is ,7,, 
that of the point. The product of the force by q^ 
its du will be called its mrtucd moment, reckoned 
-f- or — according as the direction from (9 to Z^ is 
the same as that of the force or opposite. 

62. Prop. I. — The virtual moment of a force equals the 
algebraic sum of those of its components. Tig. YO. Take the 

p direction of ds as an axis JT; let P^ and P, 

'^ be components of P\ a^, a^, and a their 
angles with X. Then (§ 16) P cos a. =^ 





cos a^-\-P^ cos a^. Hence P{ds cos ar)= 
P^{ds cos a^) -\- PJids cos a^. But ds cos a 
= the projection of ds upon P, i.e., ^ du ; 

Fig. 70. ^^ gQg ^^ _ ^/^^^^ g^g^ . _._ J^^^^ _ p^g.^^ _j_ 

P^du^. If in Fig. 70 a^ M'ere > 90°, evidently we would 
have Pdu = — Pfi.u^ -|- P^du^^ i.e., Pflu^ would then be 
negative, and OD^ would fall behind 0; lience the definition 
of -\- and — in § 61. For any number of components tlie 
proof would be similar, and is equally applicable whether they 
are in one plane or not. 

63. Prop. II. — The sum of the mrtual moments equals zero, 
for concurrent forces in equilihrium. , 

* We should rather say " projected ou the line of action of the force ;* 
but the phrase used may be allowed, for brevity. 



68 MiCCHANICS OF ENGINEERINGo 

(If the forces are balanced, the material point is moving in 
a straight line if moving at all.) The resultant foi'ce is zero. 
Hence, from § 62, P^du^ -f- P^du^ -\- etc. = 0, having proper 
regard to sign, i.e., ^{Pdv) ■=■ 0. 

64. Prop. III. — The sum of the mrtual moinents equals zero 
for any small displacement or motion of a rigid hody in equi- 
librium under non-concurrent forces in a plane ^ all points oj 
the hody moving parallel to this plane. (Although the kinds 
of motion of a given rigid body which are consistent with 
balanced non-concurrent forces have not yet been investigated, 
we mav ima^ne any slio-ht motion for the sake of the alo;'e- 
braic relations between the different du^^ and forces.) 

• First, let the motion be a translation^ all 

points of the body describing equal parallel 
..^ lengths = ds. Take ^parallel to ds ; let aTj, 

J^.3.^^ \jr etc., be the angles of the forces with X. 
^^v Then (§ 35) '2{P cos «') = ; .-. ds^{P cos a) 
* = ; but ds cos a^ = du^ ; ds cos a^ = du^ ; 
^^ ^ etc. ; .-. :2{Pdu) ^ 0. Q. E. D. 

Secondly, let the motion be a rotation 

Fig. 71. through a small angle dd in the plane of the 

forces about any point in that plane, Fig. 72. With (9 as a pole 

let /Oj be the radius-vector of the point of application of P^. and 

«j its lever-arm from 0\ similarly for the p 



other forces. In the rotation each point of 

application describes a small arc, ds.^^ ds^, / / ' ' 

etc., propoi'tional to Pj, Pg, etc., since ds^ //'^^ ..--''^\ 

^ p,dd, ds, = p,dd, etc. From § 36, ^ ..^'-r^'.'-i^^ ^/^ 

P^a^ -\- etc. = ; but from similar triangles '^""-•-d 
ds^ : du^ :: Pi : a, ; .*. «, = p^d^i-^ -^ ds^ '"-■' 

= du^ -f- dd ; similarly a^ = du, -=- dO, etc. ^^'^' ^^" 

Hence we must have [P^du^ -\- P^du, -j- . . .] -f- dd = 0, i.e., 
^{Pdu) = 0. Q. E. D. 

Now since any slight displacement or motion of a body may 
be conceived to be accomplished by a small translation fol- 
lowed by a rotation through a small angle, and since the fore- 





"VIRTUAL VELOCITIES. 69 

going deals only with projections of paths, the proposition is 
established and is called the Principle of Virtual Velocities. 
[A simihxr proof may be used for any slight motion what- 
ever in space when a system of non-concurrent forces is bal- 
anced.] Evidently if the path {ds) of a point of application is 
perpendicnlar to the force, the virtual velocity {du), and con- 
sequently the virtual moment {Pdu) of the force are zero. 
Hence we may frequently make the displacement of such a 
character in a problem that one or more of the forces may be 
excluded from the summation of virtual moments. 

65. Connecting-Rod by Virtual Velocities. — Let the effective 
steam-pressure P be the means, througli the connecting-rod 
and crank (i.e., two links), of raising the weight G very slowly ^ 
neglect friction and the weight of the links themselves. Con- 
sider AB as free (see (5) in Fig. 73), BC also, at (c) ; let the 



Bi 




B/- \ N "^^X nN^^>\ 




Fig. 73. 



"small displacements" of both be simidtaneous small portions 
of their ordinary motion in the apparatus. A has moved to A^ 
througli dx ; B to ^i, through ds, a small arc ; C has not 
moved. The forces acting on AB are P (steam-pressure), N 
(vertical reaction of guide), and N' and :7^(the tangential and 
normal components of the crank-pin pressure). Those on BC 
are N' and T (reversed), the weight (7, and the oblique pressure 
of bearing P'. The motion being slow (or rather the accelera- 
tion being small), each of these two systems will be considered ais 
balanced. Now put 2{Pdu) = for AB, and we have 

Pdx -\-]^xO-\-]V' XO-Tds = 0. . . (1) 

For the simultaneous and corresponding motion of BC, 
^(Pdu) = gives 



70 MECHANICS OF ENGINEERING. 

iV^' X + Tds - Gdh + P' X = 0, . . (2) 

(Zh being the vertical projection of {j's motion. 

From (1) and (2) we Lave, easily, Pdx — Gdh = 0, . (3) 

./Bi which is the same as we mio'ht have 

I _^^>'--^k(^^ -?^ 'i obtained by putting 2{Pdu) = Ofor 

i^.-.-.-::."^-''' .... Jq\ p' the two links together^ regarded col- 

^ ' \ lectively as a free hody^ and describ- 

^i**- 74. iiig a small portion of the motion 

they really have in the mechanism, viz., (Fig. 74,) 

Pdx+WxO- Gdh-^P' X0 = 0. , o (4) 

We may therefore announce the — 



66. Generality of the Principle ofVirtual Velocities. — If ci^y 

mechanism of jtexible inextensible cords, or of rigid bodies 
jointed together, or hoth, at rest, or in motion with very son all 
accelerations, he considered free collectively {or any portion of 
it), and all the external forces put in ; then {disregarding 
mutual f'ictions) for a small portion of its prescribed motion, 
2{Pdu) must = 0, in which the du, or virtual velocity, of 
each force, P, is the projection of the path of the point of 
application upon the force (the product, Pdu, being -j- or — 
according to § 61). 

67. Example. — In the problem of § 65, having given the 
weight G, required the proper steam-pressure (effective) P to 
hold G in equilibrium, or to raise it uniformly, if already in 
motion, for a given position of the links. That is. Fig. 75, 
given a, r, c, a, and /?, re- \^^/\p--a \^ 
quired the ratio dh \ dx; for, ^^■■-'^^^■:'''^ ' 
from equation (3), § 65, P ^^.i^:::^^^^^''''''''''^ 

= G{dh : dx). The projec- p^_^^.^-::::^^^y^ /! "^^§^1 

tions of dx and ds upon AP dx A, ^ r -•>- 

will be equal, since A£ = ^^^- '^^■ 

A^B^, and makes an (infinitely) small angle with A^B^, i.e., 

6^ cos a = ds cos (/3 — a). Also, dh = {c '. r)ds sin J3. 



"virtual velocities." 



71 



Eliminating ds, we have, 

dfi c sin /? cos a 
dx r cos (/i — oc) ' 



P= 6^ 



<? sin /? cos a 
7" cos (/? — a)' 



68. When the acceleration of the parts of the mechanism is 
riot practically zero, 2{Pdu) will not = 0, but a function of 
the masses and velocities to be explained in the chapter on 
Work, Energy, and Power. If friction occurs at moving joints, 
enough '• free bodies" should be considered that no free body 
extend beyond such a joint ; it will be found that this friction 
cannot be eliminated in the way T and N' were, in § 65. 

69. Additional Problems; to be solved by "virtual velocities." Problem 
1. — Find relations between the forces acting on a straight lever in equi- 
librium; also, on a bent lever. 

Problem 2. — When an ordinary copying-press is in equilibrium, find 
■the relation between the force applied horizontally and tangentially at 
Lhe circumference of the wheel, and the vertical resistance under the 
screw-shaft. See Fig. 75a. 

Solution. — Considering free the rigid body consisting of the wheel and 
screw-shaft, let B be the resistance at the point of the shaft (poiuting 
along the axis of the shaft), and Pthe required horizontal tangential force 
at edge of wheel. Let radius of wheel be r. Besides R and P there are 
also acting on this body certain pressures, or "supporting forces," consist- 
ing of the reactions of the collars, and reactions of the threads of nut against 
the threads of screw. Denote by s the " pitcJi" of the screw, i.e., the dis- 
tance the shaft would advance for a full turn of the wheel. Then if we 
imagine the wheel to turn through a small angle dB, the corresponding 

advance, ds, of the shaft would be x<-, from the proportion s : ds :: %Tt: dB . 

The path of the point of application of P is AS, a small portion of 
a helix, the projection of which on the line of P is rdQ, while d& projects, 
in its full length on the line of the 
force R. In the case of each of 
the other forces, however, the path 
of the point of application is per- 
pendicular to the line of the force 
(which is normal to the rubbing 
surfaces, , friction being disre- 
garded). Hence, substituting in 
I{Pdu) = 0, we have 

+ P . rdd-R . ds + + 0=0; 
whence 

ds 



P= 



rdO 



R- 



'Inr 



R. 




Fig. 75a. 



72 



MECH ATTICS OF EWGINJfiEKINO. 



CHAPTER HI. 



CURYILINEAR MOTION OF A MATERIAL POINT. 



A° f ° 


D° 


Ln 


R'/7 


^ AC^-^ 


/'/ ° 


o A// bA 




1 -ii/ 1 \ kJ \ 



Fig. 76. 



[Motion in a plane, only, will be considered in this chapter.] 

70. Parallelogram of Motions. — It is convenient to regard 
the curvilinear motion of a point in a plane as compounded, or 
made up, of two independent rectilinear motions parallel 
respectively to two co-ordinate axes X. and 7^ as may be ex- 
plained thus : Fig. 76. Consider . the 
drawing-board CD as fixed, and let the 
head of a T-square move from A 
toward B along the edge according to 
any law whatever, while a pencil moves 
from M toward Q along the blade. The 
result is a curved line on the board, whose 
form depends on the character of the 

two ^ and IT component motions, ^^ they may be called. If 
m a time z5, the 2^-square head has moved an ^distance = J/7V, 
and the pencil simultaneously a Y distance = MP, hy com- 
pleting the parallelogram on these lines, we obtain li, the 
position of the point on the board at the end of the time t^. 
Similarly, at the end of the time t^ we find the point at R'. 

71. Parallelogram of Velocities. — Let the X and T" motions 
be uniform, required the resulting motion. Fig. 77. Let g„ 
and Cy be the constant uniform X and Y velocities. Then in 
any time, t, we have a? = c^,^ and y = v Y/ 
€yt ; whence we have, eliminating t, 
as ~r y = c„ -^ Cy =. constant, i.e., x is 
proportional to y, i.e., the path is a O-^, 

straight line. Laying off OA = c^, I — ^. /<*.- 

and AB = Cy, ^ is a point of the path, Fig. n. 

and OB is the distance described by the point in the first 




CUKVILINEAR MOTION OF A MATERIAL POINT. 73 



second. Since by similar triangles OR : x i: OB : c„ we 
have also OH = OB . t ; hence the resultant motion is uniform, 
and its velocity, OB = g, is the diagonal of the parallelograTn, 
formed on the two component velocities. 

Corollary. — If the resultant motion is curved, the direction 
and velocity of the motion at any point will be given by the 
diagonal formed on the component velocities at that instant. 
The direction of motion is, of course, a tangent to the curve. 

72. TJniformly Accelerated X and Y Motions. — The initial 
velocities of hath heing zero. Required the resultant motion. 
Fig. Y8. From § 56, eq. (2) (both c^, andCj, / ^ 
being = 0), we have x = ^pj^ and y = //'" ^<^ 
^yf, whence x -i- y = Px-^Py=^ constant, wVy'^^^i / 

and the resultant motion is in a straight j/^^ ^h -r- L — y^ 

line. Conceive lines laid off from 6^ on ^ ^* ^ 

and Y to represent j?a. and^j, to scale, and ^®' '^ ' 

form a parallelogram on them. From similar triangles {OB 

being the distance described in the resultant n^^otion in any 

time t), OB : x :: 'OB : p^ ; .-. Oir= ^OBf\ Hence, from the 
form of this last equation, the resultant motion is uniformly 
accelerated, and its acceleration is OB =pi, (on the same scale 
as^^ and^j,). 

This might be called the parallelogram of accelerations, but 
is really a parallelogram of forces, if we consider that a free 
material point, initially at rest at 0, and acted on simulta- 
neously by constant forces P^ and Py (so that p^. = P^ -^ M 
2a\^ Py — Py -=- J[/), must begin a uniformly accelerated recti- 
linear motion in the direction of the resultant force, {having 
no initial velocity in any direction.) 

73. In general, considering the point hitherto spoken of as a 
free material point, under the action of one or more forces, in 
view of the foregoing, and of Newton's second law, given the 
initial velocity in amount and direction, the starting-point, 
the initial amounts and directions of the acting forces and the 



74 



MECHANICS OF ENGINEERING-. 



laws of their variation if they are not constant, we can resolve 
them into a single ^ and a single T^ force at any instant, 
determine the ^ and T' motions independently, and afterwards 
the resultant motion. 

Note. — The resultant force is never in the direction of the tangent to the 
path (except at a point of infiection). The relations which its amount 
and position at any instant bear to the velocity, rates of change of 
that velocity (i.e., accelerations), both as to amount and position, and 
the radius of curvature of the path, will now be treated (§ 74). 

In Fig. 79, A, B, and C are three ' 'consecutive" positions of the moving 
point, AB and BC being two short chords of the curve. When dt is 
taken smaller and smaller (position B remaining unchanged) and finally 
becomes zero, the points A and C merge into B and the chords AB and 
BC becomes tangents at B; and hence the results to be obtained only 
apply to a single point, B. But note that before dt becomes zero each 
equation [except (7)] is divided through by dt (or dt^) and therefore the 
individual terms do not necessarily become zero also. 

74. General Equations for the curvilinear motion of a ma- 
terial point in a plane. — The motion will be considered result- 
,.K I ing from the composition of 

,H'''' "'•- independent JT and 1^ motions, 

..-0' r^ ^' \ C. - , 

' ' 5^ tv ^ X and Y being perpendicular to 
^2\L..--" eacli other. Fig. 79. In two 
consecutive equal times (each 
= dt), let dx and dx' = small 
spaces due to the X motion ; 
and dy and CK ^ dy\ due to 
the Y motion. Then ds and 
ds' are two consecutive elements 
of the curvilinear motion. Pro- 
long ds, making BE =^ ds; then EE = d'x, dF= d^y, and 
00 = d^s {EO being perpendicular to BE). Also draw CL 
perpendicular to BG and call CL d^n. Call the velocity of 
the JT motion v^.-, its acceleration 'p^; those of the J" motion, 
Vy and fy. Then, 

dx dy dv. 

For the velocity along the curve (i.e., tangent) 
V =■ ds -T dt, we shnll have, since ds^ = dx^ -\- dy'' 




Fig. 79. 



_ d'^x , _ d^y _ ^V 

~ df' ^^^^'>~~dt~ d¥° 



dsV _ fdxV [dy\ 
+ Vdl) 



dt 



dtl 



'Vx + V 



(1) 



CURVILINEAR MOTION OF A MATERIAL POINT. 75 

Hence v is the diagonal formed on v^^ and Vy (as in § 71). 
Let pf=thG acceleration of v, i.e., tlie tangential acceleration. 
then Pt = ^'"5 -J- ^^? ai^d, since d's = the sum of the projec- 
tions of ^^and CI^ on £C, i.e., d''s = d^x cos a -\- d^y sin a, 
we have 

d^s d^x , «^'2/ . . , . /^v 

^ = ^ cos «? +^ sin or; i.e.,_^j =^^ cos or +^^s]n o'. (2) 

By Normal Acceleration we mean the rate of cliaisge of the 
velocity in the direction of the normal. In describing the ele- 
ment AB = ds^ no progress lias been made in the direction of 
the normal JBHi.e.^ there is no velocity \\\ the direction of the 
normal; bnt in describing ^C' (on account of the new direc- 
tion of path) the point has progressed a distance GL (call it 
d^n) in the direction of the old iiormal BH (though none in 
that of the new normial (7/). Hence, just as the tang. ace. 

ds' — ds d^s ^ - , CL — zero d^7i 

= 775 = -m, so the normal accel. = ■ ^, = ^-,-. 

df df dt df 

It now remains to express this normal acceleration (^j^^) in 

terms of tlie X and Y accelerations. From the figure, CL 

= CM- ML, i.e., 

d^n = d'y cos a — d^x sin a | since EF = d?x\ ; 

dj'n d^y d^x . 

-df^df "^' ""-df '^" ''• 

Hence ^^^^j^cosor — ^a.sino' (3) 

The norm. ace. may also be expressed in terms of the tang. 
Telocity -y, and the radius of curvature r, as follows : 
ds' =. rda, or da = ds' ~ r ; also d^n = ds'da, = ds'''' ~ r, 
. d'n [ds'yi v' 

'■''■^df-\-dtl P ^^ ^- = r (^) 

If now, Fig. 80, we resolve the forces jf = Mp^ and Y 
Y = Mpy, which at this instant account for the 

JT and Y accelerations (M = mass of tlie 
\ /\.'^^ material point), into components along the 
■.\^-'''\a \ tangent and normal to the curved path, w^e 



■-Jt. 






,£-^^-^M\ ,.---'" shall have, as their eqioivalent, a tangential 



-^ 



T = Mj>x cos ^ + ^Pv sin or, 



76 



MECHANICS OF ENGINEERING. 



and a normal force 

J7 = Mjpy COS a — Mjpx sin or. 
But [see equations (2), (3), and (4)] we maj also write 

r = Jf?>, = J/-^; and N^M^^^m"^. . (5) 

Hence, if a free material point is moving in a curved path^ 
the sum of the tangential components of the acting forces must 
equal (the mass) X tang, accel.; that of the normal components, 
=^ (the mass) X normal accel. = (mass) X (square of veloc. ia 
path) H- (rad. curv.). 

It is evident, therefore, that the resultant force (= diagonal 
on T and N or on JTand Y, Fig. 80) does 7iot act along the tan- 
gent at any point, but toward the concave side of the path ; un- 
less r = oo. 

Hadius of curvature. — From the line above eq. (4) we 
have d'^n = ds'^ -~ r ; hence (line above eq. (3) ), ds''' -r- r = 
d^y cos a — d^x sin a ; but cos a-=:dx-^ ds, and sin a = dy ^ ds, 



T 



dx -,„ dy 

I-; dX-f 

as ds 

ds'^ds 



- = dVij7 - ^''^^o ' 



I.e. 



= dx^d 






ds ds , , 

or ' = dx 

r 

dy- 

_dx_ 

ds'''ds'^ 

w 



'dxd'y — dyd"x 



dx^ 

=■ dx^d (tan a-), 

~ I dx\' di2iCQ. a" 
'dij dT' 



or. 



y = -y -^ 



r 2 ^ *^" ^' 



(6) 



whicli is equally true if, for v^ and tan or, we put Vy and 

tan (90° — a;). I'espectively. 

Change in the velocity square. — Since the tangential accelera^ 

dv ^ ^ dv , . 

tion -.- ^Pf, we have ds-^- =pfd8\ i.e., 

-^-dv-=ptds, or vdv=.ptds and /. — r — = / pfds. (7) 

having integrated between any initial point of the curve where 
V = c, and any other point where v = v. This is nothing 
more than equation (HI.), of § 50. 



CURVILINEAR MOTION OF A MATERIAL POINT. 



77 




75. Normal Acceleration. Another Method. — Fig. 81. Consider a material 
point TO describing a circular path ABC, with constant velocity 
= v; the center of the curve being at 
and the radius = r. The velocity v is 
always tangent to the curve. Let the 
linear arc BC be described in the small 
time dt, the angle at the center being da. 
At B the velocity is directed along the 
tangent BT, while at C it is 1 to OC 
and makes an angle da with a line parallel 
to BT. As m moves along the curve 
from B to C, the point n, which is the 
foot of the T dropped from any position Fig. 81. 

of TO upon the normal BO, moves from B toward D; whUe the foot, a, 
of the T let fall from to upon the tangent BT, moves along BT with 
an average velocity = v', a little less than v. Now the motion of to may 
be regarded as compounded of these two motions, viz., that of n and 
that of a. The motion of n is called the "motion of to along the normal." 
The velocity of n is zero at B, where i" is T to the normal, and is v sin da 
at the point D; hence in the time dt the gain of n's velocity is v sin da — 0, 
and the rate of gain, or acceleration, is pn=v sin da-i-dt. But sin da 
= CD-^r and CD = BC' = v'dt. Substituting, we have pn=vv'^r. 

Now make dt equal to zero and we have v' = v; and finally pn=v^-i-r, 
as the value of the normal acceleration, just at the point B. 

76. TTniform Circular Motion. Centripetal Force. — The ve- 
locity being constant, j!?^ must be = 0, and .'. T{ov 2Tii there 
are severalforees) must = 0. The resultant of all the forces, 
therefore, must be a normal force = {Mc^ -i- r) = a con- 
stant (eq. 5, § 74). This is called the " deviating force," 
or " centripetal force ;" without it the body would continue 
in a straight line. Since forces always occur in pairs (§ 3), 
a " centrifugal force," equal and opposite to the " centri- 
petal" (one being the reaction of the other), will be found 
among the forces acting on the body to whose constraint the 
deviation of the first body from its natural straight course is 
due. For example, the attraction of the earth on the moon 
acts as a centripetal or deviating force on the latter, while the 
equal and opposite force acting on the earth may be called 
the centrifugal. If a small block moving on a 
smooth horizontal table is gradually turned from 
its straight course ^^ by a fixed circular guide, 
tangent to AB at ^, the pressure of the guide 
against the block is the centripetal force M&-^ r 
directed toward the centre of curvature, wliile 




78 



MECHANICS OF ENGINEERING. 



y/////////^,/// 




Fig. 83. 



the cent "■'^■ugal force Mc^ -^ r is tlie pressure of the bloct 

against th«. uide^ directed away from that centre. 

Note. — One is not justified, therefore, in saying that a body descrifeing 
a circular path is under the action of a "centrifugal force.' 

The Conioal Pendulum^ or governor- ball. — Fig. 83. If a 
material point of mass z= M =^ G ^ g^ suspended on a cord of 
leLgth ^ Z, is to maintain a uniform cir- 
cular motion in a horizontal plane, with a 
given radius r, under the action of gravity 
and the cord, required the velocity c to be 
given it. At B we have the body free. 
The only forces acting are G and the cord- 
tension P. The sum of their normal com- 
ponents, i.e., -5'i\^, must = Mc^ -r- r, i.e., P sin a = Md^ -f- r ; 
but, since -2" (vert, comps.) = 0, /^ cos a =. G. Hence 
G tan a = Gc^-^ gr; .•. c = Vgr tan a. Let u = number of 
revolutions per unit of time, then u = c -v- 27rr = Vg -i- 27r Vh ; 
i.e., is inversely proportional to the square root of the vertical 
projection of the length of cord. The time of executing one 
revolution is =1 -hw. 

JElevation of the outer rail on raih'oad curves (considera- 
tions of traction disregarded). — Consider a single car as a 
material point, and free^ having a given 
velocity = c. J^ is the rail-pressure r^ 

against the wheels. So long as the car $^_ -t — R-H;; 

follows the track the resultant P of P ' "■^■■ 

and 6r must point toward the centre of 

curvature and have a value = Md^ -^ r. 

But ^=:= 6^ tan a, whence tan or = c^-f- gr. 

If therefore the ties are placed at this 

angle a with the horizontal, the pressure 

will come upon the tread and not on the flanges of the wheels ; 

in other words, the car will not leave the track. (This is really 

the same problem as the preceding) 

Apparent weight of a body at the equator. — This is less than 
the true weight or attraction of the earth, on account of the 
uniform circular motion of the body with the earth in its 
diurnal rotation. If the body hangs from a spring-balancej 




Fig. 84. 



CURVILINEAR MOTION OF A MATERIAL POINT. 79 

whose indication is G lbs. (apparent weight), while the true 
attraction is G' lbs., we have G' — G == M& -^ r. For M 
we may use G ^ g (apparent values); for r about 20,000,000 
ft.; for c, 25,000 miles in 24 hrs., reduced to feet per second. 
It results from this that 6^ is < ^' by -^G' nearly, and 
(since 17^ = 289) hence if the earth revolved on its axis seven- 
teen times as fast as at present, G would = 0, i.e., bodies 
would apparently have no weight, the earth's attraction on 
them being jnst equal to the necessary centripetal or deviating 
force necessary to keep the body in its orbit. 

Centripetal force at any latitude. — If the earth were a ho- 
mogeneous liquid, and at rest, its form would be spherical ; but 
when revolving uniformly about the polar diameter, its form 
of relative equilibrium (i.e., no motion of the particles relatively 
to each other) is nearly ellipsoidal, the pohir diameter being an 
axis of symmetry. 

Lines of attraction on bodies at its surface do not intersect 
in a common point, and the centripetal force requisite to keep 
a suspended body in its orbit (a small circle of the ellipsoia), 
at any latitude /? is the resultant, iT, of the attraction or true 
weight G' directed (nearly) toward the centre, and of G tiie 
tension of the string. Fig. 85. ^ = the apparent weight, in- 
dicated by a spring-balaTice and MA is its ..^G 

line of action (plumb-line) normal to the y^r.:.....L.:'i:\.:^^U 
ocean surface. Evidently the apparent 
weiglit, and consequently g, are less than 
the true values, since N must be perpen- x eq^^j^— 
dicular to the polar axis, while the true 
values themselves, varying inversely as ^^igTssT 

the square of MC, decrease toward the equator, hence the ap- 
parent values decrease still more rapidly as the latitude dimin- 
ishes. The apparent g for any latitude /5, at h ft. above sea- 
level, is (Chwolson, 1901), for foot-second units,* 

^ = 32.lY23-0.083315cos2,5-0.000003/i. 
(The value 32.2 is accurate enough for practical purposes.) 
Since the earth's axis is really not at rest, but moving about 

* At the equator, g^ = 32.09 at sea-level but decreases to 32.06 at an ele- 
vation of 10,000 ft. above the sea. 




80 



MECHANICS OF ENGINEERING. 



the sun, and also about the centre of gravity of the moon and 
earth, the form of the ocean surface is periodically varied, i.e., 
the phenomena of the tides are produced. 

77. Cycloidal Pendulum. — This consists of a material point 
at the extremity of an imponderable, flexible, and inextensible 
cord of length = Z, confined to the arc of a cycloid in a ver^ 
tica] plane by the cycloidal evolutes shown in Fig. 86. Let 
the oscillation begin (from rest) at A, a height = h above 





*;he vertex. On reaching any lower point, as JS (height = 3 
above 0), the point has acquired some velocity v, which is at 
this instant increasing at some rate = Pf l^ow consider the 
point free, Fig. 87; the forces acting are I^ the cord- tension, 
normal to path, and G the M^eight, at an angle (p with the 
path. From § 74, eq. (5), ^T = Mpt gives 

6^ cos ^ + P cos 90° = {G -^ g)2Jt\ :. Jpt = ^ cos ^ 

Hence (eq. (7). § 74), 'vdv = p4s gives 

qidv = g cos cpds ; bnt ds 0,0% cp =. — dz\ .'. vdv = — gdz. 

Summing between A and _5, we have 



¥^' 



= - ^A^^; ^'' ^' = 2^(^^ ~ ^)5 



♦-.he same as if it had fallen freely from rest through the height 
h — z. {This result evidently applies to any form of path 
when, he sides the weight G, there is hut one other force, and 
that always norwal to the path, y^ 

From ^iV" = J/v" -^ r,, we have P — G sin q) = Mv* -j-ri, 
* Compare with lower part of p. 83. 



CURVILINEAR MOTION OF A MATElilAL POINT. 



81 



whence P^ the cord-tension at any point, may be found (here 
r{=- the radius of curvature at any point = length of straight 
portion of the cord). 

To find the time of passing from ^ to (9, a half-oscillation, 
substitute the above value of -y^ in v ^ ds -^ dt^ putting ds^ 
= dx" -j- ds"", and we have df = {dx" + dz'') -=- [2^(A — s)]. 
To find dx in terms of dz, differentiate the equation of the 
curve, which in this position is 



a? = r ver. sin.~^ (0 -^- r) -j- V'irz — z^ ; 



whence 



dx = 



.'. dx^ = 



rdz 



(r^— z)dz {2r — z)dz 



V2rz 
~2r 



V2rz — z^ 



V2: 



rz 



- 1 



dz' 



{r = radius of the generating circle). Substituting, we have 

^r (— dz) 
g ' \/hz - z"" ' 



dt = 



{> = V ^-/ 



dz 



rh 
ver. sin. 




^^0 y/is — s^ y g 

Hence the whole oscillation occupies a time = rr Vl ^ g 
(since I = 4r). This is independent of A, i.e., the oscillations 
are isochronal. This might have been proved by showing that 
'pt is proportional to OB measured^ along the curve j i.e., that 
the motion is harmonic. (§ 59, Prob. 2.) 

78. Simple Circular Pendulum. — If the material point oscil- 
lates in the arc of a circle, Fig. 88. proceeding 
as in the preceding problem, we have finally, 
after integration by series, as the time of a full ■ \ ■■< 

oscillation, in one direction,* L.t^ _ \§-''' ^ 



"\ 



gjL 



1 + 



9 ^ 

256' P 



225 h^ 
18432 *F 



+ .. 




Hence for a small h the time is nearly tt Vl -=- g. and the os- 
* See p. 651 of Coxe's translation of Weisbach's Mechanics. 



82 MECHANICS OF ENGINEERING.. 

dilations nearly isochronal. (For the Compound Pendulum, 
see § 117.) 

[Note. — While the simple pendulum is purely ideal, the conception is 
a very useful one. A sphere of lead an inch in diameter and suspended 
by a silk thread, or very fine wire, more than 2 ft. in length, makes a 
close approximation to a simple pendulum; the length I being measured 
from the point of suspension to the middle of the sphere (strictly it 
should be a little greater). The length of the simple pendulum beating 
seconds (small amplitude) is about 3.26 ft.; (see p. 120)]. 

79. Change in the Velocity Square. — From eq. (7), § 74, we 
have ^{v'^ — g'^) z=Jpfds. But, from similar triangles, du be- 
ing the projection of any ds of the path upon the resultant 
fo]-ce S at that instant, Hdu = Tds (or, Prin. of Yirt. Yels. 
§ 62, lidu = Tds + iV^ X 0). T and iV^are the tangential and 
normal components of ^. Fig. 89. Hence, finally, 

IMv' -\MG^^fRdu, («) 

for all elements of the curve between any two points. In ^gen- 

-^^ -v eneral R is different in amount and direc- 

-/ '""" -4. . '" tion for each ds of the path, but du is the 
~'--.jp distance through which R acts, in its own 



N 



Fig. 89. direction, while the body describes any ds ; 

Rdu is called tlie work done by R wlien ds is described by the 
body. The above equation is read : The difference hetween the 
initial and final hinetio energy of a hody = the work done hy 
the residtant force in that portion of the path. 

(These phrases will be fui-ther spoken of in Chap. YI.) 
Application of equation (a) to a planet in its orhit about 
the sun. — Fig. 90. Here the only force at any iTistant is the at 
traction of the sun R =^ O -^ u"^ (see Prob. 3, § 59), 
where (7 is a constant and u the variable I'adius Ns. 
vector. As u diminishes, v increases, therefore \ " 
dv and du have contrary signs ; hence equation i c??^X^® 
{a) gives {p being the velocity at some initial 1 / \ 
point 0) LJr X^ 



Ji 2} Juo U 



■1 1 

u, u„ 



\Q>) 



^Cri In ^su- 

•. -y, =A / c^'-f ijF — — — .which is independ- fig. 90. 
ent of the direction of the initial velocity c. 



CUKVlLlNEAli MOTION OF A MATKKIAL POINT. 



83 



Ajpplication of eq. (a) to a projectile in vacuo 
body's weight, is the only force acting, and 
therefore = i?, while M= G -^ g. Tliere- 
fore equation {a) gives 

■-■ . -^- = Gj dy = Gy,', 



■G, the 



dy-dii 




G=R 



"""^\}yl 


dy] 


.V ^r<i 






G^-' 


Fig 


93. 



,". -y. = \/c' ~\- 2^2/,' which is independent of fiq. 91 

the angle, a^ of projection. 

Application of equation (a) to a body sliding, wit/iovt fric* 
tion, on a iixed curved guide in a vertical plane, initial velo- 
city = c at 0. — Since there is some pressure at each point be- 
tween the body and the guide, to consider the bod)' free in 
space, we must consider the guide removed and that the body 

describes the given curve as a re- 
sult of the action of tlie two forces, 
its weight G, and the pressure /*, 
of the guide against the body. G 
is constant, wdiile jP varies from 
point to point, though always (since 
^ there is no friction) normal to curve. 
At any point, H being the resultant 
of G and jP, project ds upon i?, thus obtaining du ; on G, 
thus obtaining dy ; on I^, thus obtaining zero. But by the 
principle of virtual velocities (see § 62) we have Rdu = Gdy 
+P X zero"^ = Gdy, which substituted in eq. (a) gives 

~l{v,^ - 0^) =f''Gdy=Gfyy=Gy:., .-.v,^ VT^W, 

and therefore depends only on the vertical distance fallen 
througli and the initial velocity, i.e., is independent of the 
fo7")n of the guide. 

As to the value of P, the mutual pressui-e betw'een the guide 
and body at any point, since ^iVinust equal 3fv'^ ~ r, r being 
the variable radius of curvature, we have, as in §77, 

P — G&m q) = Mv" -^ r ; .. P = G\fix\ cp-^^v" ~ gr)\ 

As, in general, q) and r are different from point to point of 

* It is quite essential that the guide be fixed, as well as smooth, in order 
that this projection be zero; sinci if the guide were in motion, the 
*orce P, although 1 to the guide, would not be T to the ds or element 
of the path oi the body, for that path would then be different from the 
curve of the guide. 



84 MECHANICS OF ENGINEERING. 

the path, P is not constant. Should the curve at the point in 
question be convex upward (instead of concave upward as in 
Fig. 92) we must write G &m.(f)—P=Mv^-i-r; etc. 

80. Projectiles in Vacuo. — A ball is projected into the air 

(whose resistance is neglected, hence the 

'"""f!^ ,'--"'Tf plirase in vacuo) at an angle = a^ with the 

/•i" G'''^ horizontal ;* required its path ; assuming it 

\ i i „ coniined to a vertical plane. Resolve the 

'oil Cx ±\ ... . , -^ 

ZZTl^'ZIZl^""' motion into independent horizontal {X) 

Fig. 93. and vertical {Y) motions, G, the weight, 

the only force acting, being correspondingly replaced b}^ its 

horizontal component = zero, and its vertical component 

= — G. Similarly the initial velocity along X = 0^^-=^ c cos <x^, 

along y, = Cy = csin a^. The JT acceleration =j?a; = -^ J/ 

= 0, i.e., the X motion is uniform, the velocity v^. remains 

= c^ = c cos a^ at all points, hence, reckoning the time from 0^ 

at the end of any time t we have 

X = c(cos a^t (1) 

In the Y motion, j?y = ( — G) -^ M=. — g^ i.e., it is uniformly 

retarded, the initial velocity being Cy =^ c sin a^ ; hence, after 

any time t, the Y velocity will be (see § 56) v^ = c sin a^ — gt, 

while the distance 

y = c(sin a^)t - ^gf (2) 

Between (1) and (2) we may eliminate t, and obtain as the 

equation of the trajectory or path 

y =: X tan a. — —-z — . 

^ " 2c^ cos' o'o 

For brevity put c' = 2gh, h being the ideal height due to the 
velocity c, i.e., c^ -f- 2g (see § 53 ; if the ball were directed ver- 
tically upward, a height h = o^ -^ 2g would be actually at- 
tained, oTfl being = 90°), and we have 

y = xt^na,--^j^^^- (3) 

This is easily shown to be the equation of a parabola, with its 
axis vertical. 



* And with a velocity of c ft. per sec. 



CURVILINEAR MOTION OF A MATERIAL POINT. 



85 



The ho7'izontal range.- 
tion (3), we obtain 

a? 



X tan ar„ 



4A cos^ «„ 



-Fig. 94. Putting y = in equa- 



0, 




Fig. 94. 



which is satisfied both by a? = (i.e., at the 
origin), and by a? = 4:A cos a^ sin a^. Hence 
the horizontal range for a given g and a^ is 
x^ = 4A cos <arg sin a^ = 2A sin 2a^. 

For afg =: 45° this is a maximum (c remaining the same), 
being then = 2A. Also, since sin 2a^ = sin (180° — 2a^) = 
sin 2(90° — a^), therefore any two complementary angles of 
projection give the same horizontal range. 

Greatest height of ascent / that is, the value of y maximum, 
= y^. — Fig. 94. Differentiate (3), obtaining 

dy X 

dx ~ " 2A cos'' or ' 



which, put = 0, gives a? = 2A sin a^ cos ar„, and this value of 
X in (3) gives y^=i h sin" oc^. 

(Let the student obtain this more simply by considering the 
Y motion separately.) 

81. Actual Path of Projectiles. — Small jets of water, so long as 
they remain unbroken, give close approximations to parabolic 
paths, as also any small dense object, e.g., a ball of metal, hav- 
ing a moderate initial velocity. The course of a cannon-ball, 
however, with a velocity of 1200 to 1400 feet per second is 
much affected by the resistance of the air, the descending 
branch of the curve being much steeper than the ascending; 
see Fig. 96(2. The equation of this curve has not yet been 
determined, but only the expression for the slope (i.e., 
dy : dx) at any point. See Professor Bart- 
lett's Mechanics, § 151 (in which the body 
is a sphere having no motion of rotation). 
Swift rotation about an axis, as well as an 
unsymmetrical form with reference to the 
direction of motion, alters the trajectory 
still further, and may deviate it from a vertical plane, 
presence of wind would occasion increased irregularity 




Fig. 96a. 



The 

See 



Johnson's Encyclopaedia, article " Gunnery." (See p. 823.) 



86 



MECHANICS OF ENGINEERING. 




82. Special Problem (imaginary ; from Weisbach's Mechan- 
ics. The equations are not homogeneous). — Suppose a ma- 
terial point, mass = M^ to start 
from the point 0, Fig. 97, with 
a velocity = 9 feet per second 
along the — Y axis, beiiig snb- 
ilJ_J^,^ jected thereafter to a constant 
attractive JT force, of a valne X 
= 12M, and to a variable Y 
force increasino; with the time 
Fig. 97. (in scconds, reckoned from 0), 

viz., Y = 8Mi. Required the path, etc. For the JC motioa 
we Imvepx = X -^ If = 12, and hence 

dvy. = I ])Jit = 13 / dt\ i.e., 'o^ = 12^; 

dx ■=■ j v^dt ; i-.e., a? = 12 / tdt = 6f. . (1) 

For the Y motion ^j, = Y~3f=St, ..f 'd/Vy=%f tdt ; 

/y pt 

dy =^ I Vydtj 

.'. y = ^f fdt — 9^ dt, or y = |f — 9f. . . (2) 



Eliminate t between (1) and (2), and we have, as the equa- 
tion of the path, 

4:fx\^ (x\i 

which indicates a curve of the third order. 
The velocity at any jpoint is (see § 74, eq. (1) ) 



(3) 



-y=|/^,^ + 'U/=4if + 9 (4) 

The length of cu7've measured from will be (since v = 
ds -i- dt) 

s —Tds =f vdt = 4c f fdt -^9jdt = ^f + 9^. (5) 

The slojpe, tan a, at any point = -y^^ -^ -Va, = {Aff — 9) -^ 12^, 
d tan a U^ -^9 



and .*. 



dt 



l^f 



CUKVILINEAK MOTION OF A MATERIAL POINT. 87 

TTie radius of curvature at any point (§ 74, eq. (6) ), sub- 
stituting Vg. = 12f, also from (4) and (6), is 



r = v^ -i- \vj 



d tan 



a 



1 



^^j=i#^'+9r, . . (7) 

and the normal acceleration = v^ — r (eq. (4), § 74), becomes 
from (4) and (7)^^= 12 (ft, per square second), a constant. 
Hence the centripetal or deviating force at any point, i.e., the 
SW of tiie forces X and Y, is the same at all points, and = 
Mv' -^r = 12M. 

From equation (3) it is evident that the curve is sjunmetrical 
about the axis X. Negative vahies of ?! and s would apply to 
points on the dotted portion in Fig. 97, since the body may be 
considered as having started at any point whatever, so long as 
ill the variables have their proper values for that point. 

(Let the student determine how the conditions of this motioa 
could be approximated to experimentally.) 

83. Relative and Absolute Velocities. — Fig. 98. Let JUT be a 

material point having a uniform motion of velocity v^ along a 
straight groove cut in the deck of a steamer, which itself has 
a uniform motion of translation, of velocity v^, over the bed of 
a river. In one second 3f ad- \ / 

vances a distance v^ along the \ / 

groove, which simultaneously has z"^^-' iVi72^II^f^^^^----..._ 
moved a distance v, = AJB with I I ///\/ ll 
tlie vessel. The absolute path of — ~ d^r'[^~J^ r:::. 
M during the second is evidently fig. 98. 

w (the diagonal formed on -y^ and -y^), which may therefore be 
called the ahsolute velocity of the body (considering the bed 
of the river as fixed) ; while v^ is its relative velocity, i.e., rela- 
tive to the vessel. If the motion of the vessel is not one of 
translation, the construction still holds good for an instant of 
time, but V^ is then the velocity of that point of the deck over 
which JSTis passing at this instant, and v^ is Jff's velocity rela- 
tively to that point alone. 

Conversely, if M be moving over the deck with a given 
absolute velocity = lu, v^ being that oi the vessel, the relative 
velocity v,^ may be found by resolving w into two components^ 
one of which shall be v^ ; the other will be v^. 



'A„..-"B 



'S8 MECHAJMICS OF EJS'GlNEElllNG. 

If w is the absolute velocity and direction of the wind., the 
vane on rfie mast-head mmII be parallel to 3£T, i.e., to v^ the 
relative velocity; while if the vessel be rollins^ and the mast- 
head therefore describing a sinuous path, the direction of tha 
vane varies periodically. 

Evidently the effect of the wind on the sails, if any, will 
'depend on v^ the relative, and not directly on w the absolute, 
velocity. Similarly, if w is the velocity of a jet of water, and 
Vj that of a water-wheel channel, which the water is to enter 
without sudden deviation, or impact, the channel-partition 
■should be made tangent to v^ and not to w. 

Again, the aberration of light of the stars depends on the 
;Same construction ; v^ is the absolute velocity of a locality of the 
■earth's surface (being practically equal to that of the centre) ; 
w is the absolute direction and velocity of the light from a 
certain star. To see the star, a telescope must be- directed 
.2,\o\)g MT, i.e., parallel to v^ the relative velocity; just as in 
the case of the moving vessel, the groove must have the direc- 
tion MT. if the moving material point, having an absolute 
velocity w, is to pass down the groove without touching its 
sides. Since the velocity of light = 192,000 miles per second 
=: w, and that of the earth in its orbit = 19 miles per second 
= -Wj, the angle of aberration SMT, Fig. 98, Avill not exceed 
20 seconds of arc ; while it is zero when w and v^ are parallel. 

Returning to the wind and sail-boat,"^ it will be seen from 
Fig. 98 that when 'y, = or even > w, it is still possible for -y, 
to be of such an amount and direction as to give, on a sail 
properly placed, a small wind-pressure, having a small fore-and 
aft component, which in the case of an ice-boat may exceed 
the small fore-and-aft resistance of such a craft, and thus v^ will 
be still further increased ; i.e., an ice-boat may sometimes travel 
faster than the wind which drives it. This has often been 
proved experimentally on the Hudson Hiver. (See p. 819.) 

84. Examples. — 1. A platform-car on a straight I-evel track carries a 
vertical smooth pole loosely encircled by an iron ring weighing 30 lbs., 
and is part of a train having a uniform northward motion with velocity of 
20 miles/ hour. The ring, at first fastened at the top of pole, 10 ft. above 
floor, is set free. Find its absolute velocity just before striking the floor 
and the distance the car has progressed during the fall of ring. 

Solution. — The X-motion (horizontal) of the ring is that of the car and 
has a constant velocity Cx = 20X5280^3600 = 29.34ft./sec. Its F-motion 



CUKVILINEAR MOTION OF A MATERIAL POINT. 89 

■(i.e., along pole) has initial velocity =0 and a constant downward acceler- 
ation py=g (since the only force acting on ring is vertical and is its own 
weight) . Hence from § 56 the time of the 10-ft. fall = i/2X10h-32.2 = 0.788 
sec. and the F-velocity generated at end of that time is Vy~gt = 25A ft./sec. 
This is now combined with the simultaneous X- velocity of riag, i.e., 29.34, 
to give V, =l/ca;^ + %^ =38.8 ft./sec, for the required final absolute 
velocity of ring, which is therefore at this instant moving obliquely north- 
ward and downward at an angle of 40° 52' with the horizontal (since 
^j,-=-Ca;= 25.4-^29.34 = 0.8655 =tan 40° 52'). 

Example 2. — Pole and car, etc., as in example 1, but the train now has 
a uniformly accelerated motion, gaining 25 velocity-units (ft. per sec.) in 
each 5 sees, of time. The ring begins to drop when the train already has 
a velocity of 6 ft./sec. Find the final absolute velocity of ring; also the 
final pressure of pole on ring. 

Solution. — The ^''-motion of ring is the same as before, since the pressure 
on the ring from the pole (smooth vertical sides) must be horizontal and 
hence does not affect the F-motion. Hence the time of descent is, as before, 
€.788 sec. During this time the velocity of the train has increased to a 
value of i;x = 6-h (25^5)0.788 = 9.94 ft./sec, which is the velocity of the 
JC-motion of the ring at the final instant, whence its final absolute velocity, 
-w, =1/(9.94)2+ (25.4)2, =27.3 ft./sec, directed obliquely downward and 
northward at an angle of 68° 38' with horizontal (9.94-^27.3) =0.3642 
= cos 68° 38'. The pressure of the pole on ring is constant and =Px = Mpx 
= (30^32.2) X 5 = 4.65 lbs. 

Example 3. — Conical pendulum. Fig. 83, p. 78. Given G = 8 lbs. and 
1=2 ft., at what angle a will the cord finally place itself with the vertical 
if a steady rotation is kept up at the rate of 50 revs./min. ; and what will 
then be the tension in the cord? 

Solution. — ^With the ft., lb., and sec. as units we have w = 0.8333 revs. /sec, 
= 8, 1 = 2, a=? Hence from v? = g-^ {4:n%) , we find /i = 1.174 ft. and 
cos a, =h-^l, =0.5873; /. a = 54°0'. As for the tension in cord, 
P=G-^ cos a = 13. 62 lbs. 

\Note. — In this example, if the assigned value of u, or of the cord-length 
Z, had been so small as to make lu'''^g^-{i7i^), we should have obtained for 
cos a a value ^1.00; which is, of course, impossible for a cosine. That 
is, the value assigned for u must be ^i/g-^(27r]/Z), in order that the cord 
may depart at all from its original vertical position.] 

Example 3. — Compute the length Z of a simple pendulum which is to 
oscillate 4500 times in an hour. Amplitude small; 5°. 

Solution. — For small oscillations we have, from p. 81, t = Tt\/l-^g as the 
time of one oscillation ; that is, for the foot and second as units, 
3600-^500 = 7r-j/Z-^32.2; and therefore Z = 2.089 ft. 

Example 4. — A leaden ball weighing ^ ounce, and of diameter 0.53 in., 
is allowed to slide down the inside of a fixed and rigid hemispherical bowl, 
of perfectly smooth internal surface and with its upper edge in a horizontal 
plane. Its radius is 18 in. The ball is to start from rest at upper edge. 
Find the time occupied by the ball in reaching the lowest point, and the pressure 
under it as it passes that point; also the pressure in passing the 45° point. 



90 MECHANICS OF ENGINEERING. 

Solution. — Regarding the ball as a material point we note that its motion 
is practically that of a simple pendulum with ^ = [18 — ^(0. 53)] in., =1.478 ft., 
for which (see Fig. 88, p. 81) the ratio h-r-l=1.00. Hence (§ 78, p. 81) the 
time of a half oscillation (applicable here) will be (ft. and sec), 

(For a small amplitude this would be only -|7Z-(.02143)JL00] = 0.337 sec.) 

At the bottom the velocity will be (p. 83), v = '\/2gl, whence v'^-hgl = 2; 
and the pressure (see foot p. 83, with sin = 1.0, is P=-K1 +2) = 1.5 ou nces. 

As the ball passes the 45° point its velocity is v' = '\/2gX0.707l; i.e., 
^'^-|-3i = 1.414, while sin 45° = 0.707; whence, for the pressure, P', 
P' = i[0.707 + 1.414]= 1.06 ounces. 

Example 5. — A body at latitude 41° weighs apparently (i.e., by spring 
balance) 10 lbs.; what is the amount and direction of its real weight? 
(Fig. 85.) That is, we have given G = 10 lbs. and angle /? = 41°; and desire 
the value of force G' and of the angle which it makes with MA (plumb line) . 
(This angle, 0, =that at vertex G of the parallelogram in Fig. 85). 

Solution. — At the equator the earth's radius is r = 20,920,000 ft. and 
the velocity of objects at the surface is c=1521 ft. /sec. The radius of the 
small circle at M is r' = r cos 41° = 15,780,000 ft., and hence the velocity of 
the 10-lb. body at M is c', =(r'H-r)c, =1148 ft./ sec. Therefore the result- 
ant iV, = Mc^~r', = [(10 --32.2)(1148)2]H- 15,780,000 = 0.0259 lbs. 

Call the projection of N on GM prolonged, T, and its projection on a 
1 to GM, S; then T, =N cos /?, = 0.01954 lbs., and S, =Nsm /?, =0.01699 
lbs. We have also tan = 5-4- [G + T] = 0.0016957; hence = 0° 5' 48". 
Then G', ={G + T) sec 0, =10.01955 lbs. 

[By a somewhat more refined process we obtain 10.01964 lbs. (Du Bois).] 

Example 6. — A small compact jet of water (see Fig. 94, p. 85) issues 
obliquely from a nozzle. It strikes the horizontal plane of nozzle at 6 ft. 
from the latter, and its highest point is 26.4 in. above that plane. Find 
c, the velocity at nozzle, and the angle of projection a^. 

Solution. — From p. 85 (foot and second units) we have Ah cos ao sin «„ 
= 6 ft., and h sin^ q;o = 2.2 ft.; whence, by division (tan «(,-=- 4) = (2.2 -=-6), 
or tan ctg = 1 .4666 ; and therefore ao = 55°43'. a^ being now known we 
find from /i sin^ ao = 2.2 that A = 3.22 ft. But h simply stands for the ex- 
pression c^^2g, and hence, finally, we obtain for the velocity of the jet 
where it leaves the nozzle c = 14.4 ft. per sec. 

Example 7. — If in Fig. 98, the absolute velocity of the air-particles 
(wind) is w = 10 miles/ hour and directly from the northwest, the boat's 
velocity being =12 miles /hour toward the east, in what direction and with 
what velocity does the wind appear to come, to a man on the boat? 

Ans. From a direction 34° 52' east of north, and at 8.62 ft. /sec. 

Example 8. — If to a passenger on board a boat going eastward at 15 
miles /hour, the wind appears to come from the northeast and to have a 
velocity 10 miles/ hour, what is the true or "absolute" velocity of the wind, 
and what is its true direction (angle with north and south line)? 

Ans. 10.63 ft./ sec, and from a point 41° 44' east of north. 



MOMENT OF INEKTIA. 91 

CHAPTER lY. 

MOMENT OF INERTIA. 

[Note, — For the propriety of this term and its use in Mechanics, see 
■§§ 114, 216, and 229 ; for the present we deal only with the geometrical 
Jiature of these two kinds of quantity.] 

85. Plane Figures. — Just as in dealing with the centre of 
gravit}' of a plane figui-e (§ 23), we had occasion to suni tlie 
■&eY\ei fzdF, 3 being the distance of any element of area, dF, 
from an axis ; so in subsequent chapters it Mall be necessarj' to 
know the value of the seriesysW-^for plane figures of various 
shapes referred to various axes. This summation J'z'^dF of 
the products arising from multiplying each elementary area of 
the figure by the square of its distance from an axis is called 
the moment of inertia of the plane figure rcith respect to the 
■axis ill question / its symbol will be I. If the axis is perpen- 
dicular to the plane of the figure, it may be named the polar 
mom. of inertia (§94); if the axis lies in the plane, the rec- 
tangular mom. of inertia (§§ 90-93). Since the / of a plane 
figure evidently consists oi four dimensions of length., it inay 
always be resolved into two factors, thus /= Fk^^ in which 
i^= total area of the figure, while h = Vl-r- F, is called the 
Tadius of gyration, because if all the elements of area were 
situated at the sa^ne radial distance, Jc, from the axis, the 
moment of inertia would still be the same, viz., 

I = fk'dF = kfdF = Fh\ 

For example, if the moment of inertia of a certain plane figure about a 
specified axis is 248 biquadratic inches (i.e., four-dimension inches; or in.'*), 
while its area is 12 sq. in. (or in. 2), the corresponding radius of gyration is 
A; = 1/248-^12 = 4.55 in. 

86. Rigid Bodies. — Similarly, in dealing with the rotary 
motion of a rigid body, we shall need the sura of the series 
fp'^dM, meaning the summation of the products arising from 
multiplying the mass dM oi each elementary volume dY oi &. 



92 MECHANICS 0¥ ENGINEEKING. 

rigid body bj the square of its distance from a specified axis.. 
This will be called the moment of inertia of the hody with 
respect to the particular axis mentioned (often indicated by a. 
subscript), and will be denoted by /. As before, it can of tea 
be conveniently written Mh^^ in which 3£ is the whole muss, 
and h its "radius of gyration" for the axis used, h being 
= Vl -T- M. If the body is homogeneous, the heaviness, y, of 
all its particles will be the same, and we may write 

. I =fp''dM ={r~ g)fp^d V={y~g) Vl\ 

87. If the body is a homogeneous plate of an infinitely smaW. 
thickness = r, and of area = F, we have Z = (/ -f- g^fp'd K 

'"= {y -^ 9YfP^'^^'-, i-e-, = {y -^ g) X thickness X mom. iner- 
tia of the plane figure. 

88. Two Parallel Axes. Reduction Formula.* — Fig. 99. Let 

Z and Z' be two parallel axes. Tiien Ig 
=fp'dM, and I^.^fp'^dM. Bu t d being 
the distance between the axes, so that a^' 
-f lf= d\ we have p'^= {x - af-\-{y-hf 
= (£»' + y"") -\-d^ — 2aa? — %y, and .-. 
I^, =fp'dM-\-dYdM- ^afxdM 

-%fydM. . (1) 

Fig. m. V>\\ifp''dM = /^, fdM= M, and from the- 

theory of tlie centre of gravity (see§23, eq. (1), knowing that 
dJI =yd V~ g, and .-. that S^fyd F] -^ g=M) we \\2,YefxdM 
= Mx dindifydM = My\ hence (1) becomes 

/^, = I,-\- Mid' - 2«^ - %y\ .... (2)^ 

in which a and h are the x and y of the axis Z'\ x and y refer 
to the centre of gravity of the body. If Z is a. gravity-axis- 
(call it g), both x and y = 0, and (2) becomes 

I,.=I^ + Md\... or h,'^k;+d\ . . (3) 

It is therefore evident that the mom. of inertia about a grav- 
ity-axis is smaller than about any other parallel axis. 

Eq. (3) includes the particular case of a plane figure, by 

* The particle of mass = dM, shown in Fig. 99, is typical of the vast 
number of particles which form the rigid body. That is, o, 6, and d 
are constants, but x, y, z, p, and p' are variables. 




MOMENT OF INEKTIA. 



93 



writing area instead of mass, i.e., wlien Z (now g) is a gravitj- 
axis, 

I,,=I^-\-Fd\ (4) 

89. Other Reduction Formulse ; for Plane Figures. — (The axes 
here mentioned lie in the plane of the figure.) For two sets 
of rectangtdar axes, having the smne origin, the following holds 
good. Fig. 100. Since 

I^=fifdF, and ly^fx'dF, 

we have Ix + Iy =/(«;' + y')dF. 

Similarly, I^ + Iy =f{v' + %iyF. 

But since the x and y of any <^^have the same hypothennse as 
the u and v, we have v^ -{- v^ = ic^-J- y"; . ". -^x + -^r = A;- + -^r- 





Fig. 100. 



Fig. 100a. 



Let Xl)e an axis of symmetry ', then, given Ix and Iy {0 is 
anywhere on X). required Ijj, JJheing an axis through and 
maJcing any angle a with X. See Fig. 100a. 

I^ -^fv^dF^fiy co&oc —X sin dfdF\ i.e., 

Ijj = cos^ af/dF — 2 sin a. cos afxydF-\- sin* afx^dF. 

But since the area is symmetrical about X, in summing up the 
products xydF, for every term x{ -\- y)dF, there is also a term 
K — y)dF to cancel it ; which gives fxydF =: 0. Hence 

Ijj^ — cos* al^ -f- sin* aly. 

The student may easily prove that if two distances a and h 
be set ofE from on X and Y^ respectively, made inversely 
proportional to Vix and VTy, and an ellipse described on a and 
h as semi-axes ; then the moments of inertia of the figure about 



94 



MECHANICS OF ENGINEERING. 



im 



dz 



^n 



any axes through are inversely proportional to the squares 
of the corresponding semi-diameters of this ellipse ; called 
therefore the Ellijpse of Inertia. It follows therefore that the 
moments of inertia about all gravity-axes of a circle, or a 
regular polygon, are equal ; since their ellipse of inertia must 
be a circle. Even if the plane figure is not symmetrical, an 
" ellipse of inertia" can be located at any point, and has the 
properties already mentioned ; its axes are called t\\e principal 
axes for that point. 

90. The Rectangle. — First, ahout its hase. Fig. 101. Since 
all points of a strip parallel to the base 

-j,....^ ^ ?) -^ have the same co-ordinate, 0, we may take 

dz the area of such a strip for dF ■=^ hds\ 

.-. Ib= z'dF= I / z'dz 

L-o 
Secondly, about a gravity-axis parallel to hase. 

z'dz = -^W. 

-ih 

Hence the radius of gyration =k = h-^\' 12. 
Thirdly, about any other axis in its plane. Use the results 
already obtained in connection with the reduction-formulae of 

§§ 88, 89. 

90a. The Triangle. — First, about an axis through the vertex 

and parallel to the base ; i.e., 1-^ .^ .1, » ^ .j,, 

in Fig. 103. Here the length 

of the strip is variable ; call it y. ^ 

From similar triangles l 

_i. .\j^. _v AZ i V 

2/ = (& -i- h)z ; Fig. 103. Fig. 104. 



Fig 101. 



Fig. 102. 




Uh' 



.'. ly ^fz'dF^ fz'ydz = {h-^ h)f z'dz = I 

Secondly, about g, a gravity -axis parallel to the hase. 
104. From § 88, eq. (4), we have, since F=^ ^hh and 

d = |A, Ig = Iy- Fd' = IW - Ihh . ^h' = ^W. 



Fig 



MOMENT OF INEKTIA. 



95 



Thirdly^ Fig. 104, about the hase • Ijb = 1 From § 88, eq. 
(4), Ib=^ Ig-\- Fd^, with 6? = -JA ; hence 

I J, = ^^hh' + ^hh . \h' = -^hkK 

91. The Circle. — About any diameter, as g, Fig. 105. Polar 
co-ordinates, Ig =^ fz^dF. Here we take dF=^ area of an ele- 
mentary rectangle = pdq) . dp, while z=^ p sin cp. 

■» h — -• 

1 




f-i^r 



hi 
^1 



-- *•- 

c 



ibi- 



Fig. 105. 



Fig. 106. 



Ig= I I {p sin (pypdcpdp = I I sin' cpdcpj p^dp J 
= — / sin'' 9?<^^ = T / "^^"^ ~" ^^^ '^(p)dqi 

^* /.S'^fl 1 ~| 

= :f y^ 1^2^^ - J . cos 2(?)^(2^)J 
_o)-(0-0)_. 



1 .r^Vl 



= r 



= r 



2;r 

2" 



»^. = 4'^^*- 



Hence the radius of gyration =\r. 

92. Compound Plane Figures. — Since I =^ fz^dF is an in- 
finite series, it may be considered as made up of separate 
groups or subordinate series, combined by algebraic addition, 
corresponding to the subdivision of the compound figure into 
component figures, each subordinate series being the moment of 
inertia of one of these component figures ; but these separate 
moments Tnust all he referred to the same axis. It is con- 
venient to remember that the {rectangular) / of a plane 
figure remains unchanged if we conceive some or all of its 
elements shifted any distance parallel to the axis of refer- 
ence. E.g., in Fig. 106, the sum of the Is of the rectangle CE, 
and that of FD is = to the Ib of the imaginary rectangle 



96 



MECHANICS OF ENGINEERING. 



formed by shifting one of tliem parallel to B, until it touches 
the other ; i.e., I^ of CE-^ Ib of FD = ^hX (§ 90). Hence 
the Ib of the T shape in Fig. 106 will be = I^ of rectangle 
AD - Ib of rect. CE- Ib of rect. FIf. 

That is, /^ of T = i[5/''' - \Kl • • • (§ 90). . . (1) 

Ahoiit the gravity-axis, g, Fig. 106. To find the distance d 
from the base to the ceiirre of gravity, we may make use of 
eq. (3) of § 23, wu-iting areas instead of volumes, or, experi- 
mentally, having cut the given shape out of sheet-metal or 
card-board, we may balance ition a knife-edge. Supposing d 
to be known by some such method, we have, from eq. (4) of 
§ 88, since the area E= bh — hjt„ Ig=: Ib— Fd^ ; 



i.e., Ig = l\hk' - hji,'-] - {hh - lji,)d' 



(2) 




The doiihle-'Y (on), and the hox forms of Fig. 106a, if 

syminetrical about the gravity- 
axis g, have moments of inertia 
alike in form. Here the grav- 
ity-axis (parallel to base) of the 
compound figure is also a grav- 
FiG. 106a. ity axis (parallel to base) of each 

of the tw-o component rectangles, of dimensions h and A, h^ and 
Aj, respectively. 

Hence by algebraic addition we have (§ 90), for either com- 
pound figure, 

I,= i,\}h^-\h.n (3) 

(If there is no axis of symmetry parallel to the base we must 
proceed as in dealing with the T-form.) Similarly for the ring, 





Fig. 107. 



Fig. 108. 



Fig. 107, or space between two concentric circumferences, we 
have, about any diam-eter or ^ (§ 91), 

Io = \«-r:) (4) 



MOMENT OF INERTIA. 97 

The rhorribus about a gravity-axis, g, perpendicular to a 
diagonal, Fig. 108. — This- axis divides the figure into two 
equal triangles, symmetrically jplaced, hence the Ig of the 
rhombus equals double the moment of inertia of one triangle 
about its base ; hence (§ 90a) 

/, = 2 . ^li^Kf = -i^W (5) 

(The result is the same, if either vertex, or both, be shifted 
•any distance parallel to AB.) 

For practice, the student may derive results for the trapezoid ^ 
for the forms in Fig. 106, when the inner corners are rounded 
into equal quadrants of circles; for the double- "f, when the 
lower flanges are shorter than the upper; for the regular 
polygons, etc. (See table in the Cambria Steel Co.'s hand-book. ) 

93. If the plane figure be bounded, wholly or partial]}', by 
curves, it may be subdivided into an infinite number of strips,, 
and the moments of inertia of these (referred to the desired 
axis) added by integration, if the equations of the curves are 
hnown I if not, Simpson's Rule,* for a finite even number of 
strips, of equal width, may be employed for an approximate 
result. If these strips are parallel to the axis, the / of any one 
strip = its length X its width X square of distance from axis; 
while if perpendicular to, and terminating in, the axis, its 
/= -J its width X cube of its length (see § 90). 

A graphic method of determining the moment of inertia of 
any irregular figure will be given in a subsequent chapter.* 

94. Polar Moment of Inertia of Plane Figures (§ 85). — Since 
the axis is now perpendicular to the plane of the figure, inter- 
secting it in a point, (9, the distances of the ele- 
ments of area will all radiate from this point, 
and would better be denoted by p instead of s; 
hence, Fig. 109, fp^dF'^s the polar moment, of | 
inertia of any plane figure about a specified 
point ; this may be denoted by Ip. But p^ Fiq. 109. 

= a?" -|- 2/^ for each dJ^', hence 

4 =f{x' + f)dF=fx^dF+fy^dJ^= /^+ /^. 

7 

* See pp. 13, 79, 80, and 81 of the author's "Notes and Examples in 
Mechanics," and p. 454 of this book. 



98 MECHANICS OF ENGINEERING. 

i.e., the polar Tnoment of inertia ahout any gwen point i/n 
the plane equals the sum of the rectangular moine^its of iner- 
tia about any two axes of the plane figure^ which intersect ai 
right angles in the given point. We liave therefore for the 
circle about its centre 

7p = \7ir' -}- ^Ttr" = ^Ttr" ; 
For a ring of radii r^ and r,, 

4 = k7t{r: - r:) ; 
For the rectangle about its centre^ 

For the square, this reduces to 

-^p — 6" • 

(See §§90 and 91.) 

95. Slender, Prismatic, Homogeneous Rod. — Returning tcv the 

moment of inertia of rigid bodies, or solids, we begin with tliat 
of a material line, as it uiiglit be called, about 
^^^yy"^' an axis througli its extremity making some an- 
r y^^i ./"'' gle oi with the rod. Let I = length of the rod, 
y^--^" X i^its cross-section (very small, the result being 

y'' strictly true only when F = 0). Subdivide 

Fig. 110. the rod into an inlinite number of small prisms, 

each having _^as a base, and an altitude = ds. Let y = the 
heaviness of the material ; then the mass of an elementary 
prism, or dJif, = (r "^ 9)F'ds, while its distance from the axis 
Z \% p ^= s sin a. Hence the moment of inertia of the rod 
with respect to Z as an axis is 

Jz = fp'dM= {y -^ ^)i^sin^ afs^^ds = ^{y H- g)FT sin' a. 

But yFl -~ g =z mass of rod and I sin a z= a, the distance ot 
the further extremity from the axis ; lience Iz = ^Ma^ and 
tiie radius of gyration, or Jc, is found by writing-|-J!/a'''= Mh^ ; 
.-. ]c' = ^a\ or h = V^a (see § 86). If or = 90°, a = l. 

96. Thin Plates. Axis in the Plate. — Let the plates be homo- 
geneous and of small constant thickness = t. If the surface of 



MOMENT OF INEETIA. 



99 



the plate be = F, and its lieaviness y, then its mass = yFr — g. 

From § 87 we have for the plate, about anj axis, 

I ^^ {y -^ g)r X m,oin. of inertia of the j)lane fgure formed hy 

the shape of the plate (1) 

Rectangular 'plate. Gravity-axis parallel toJ)ase. — Dimen- 
sions h and h. From eq. (1) and § 90 we have 

Similarly, if the base is the axis, I^ = \MU, .'. Tc^ = -|A^ 
Triangular plate. Axis through vertex parallel to hase. — 
From eq. (1) and § 90a, dimensions being h and A, 

ly = {y -^ g)rlW = {yihhr ^ g)^h' = ^Mh'; .'. ¥ = \h\ 

Circular plate, with any diameter as axis. — From eq. (1) 
and § 91 we have 



Ig = {y -^ gy^Ttr" — {yytr'^t -^ g)^r^ = ^Mr^; 



¥ = ^\ 




Fig. 111. 



97. Plates or Right Prisms of any Thickness (or Altitude). 
Axis Perpendicular to Surface (or Base). — As before, the solid is 
homogeneous, i.e., of constant heaviness y, 
let the altitude = h. Consider an elementary 
pi'ism, Fig. Ill, whose length is parallel to the 
axis of reference Z. Its altitude = h = that 
of the whole solid ; its base = dF = an element \ 
of i^the area of the base of solid ; and each 
point of it has the same p. Hence we may 
take its mass, = yhdF -^ g, as the dMin summing the series 
l^=fp^dM', 
.'.Iz={yh^g)fp-'dF 

= {yh -^ g) X polar mom. of inertia of base. . . (2) 

By the use of eq. (2) and the results in § 94 we obtain the 
following: 

Circular plate, or right circular cylinder, about the geo- 
metrical axis, r ■= radius, h = altitude. 

Ig = {yh -^ g)^7tr' = (yhrrr' -i- g)ir' = ^Mr'; .'. ¥ — \r\ 

Right parallelopiped or rectangular plate. — Fig. 112, 

I. = (r^ - g)^M^' + ^1 = -^iV^'; ••• ^^ = ^^- 



loo 



MECHANICS OF EKGINEEKING. 



For a hollow cylinder^ about its geometric axis, 





— 


"A 




— ~] 


(n 




b:::::^- 




V 




Fig 112. Fig. 113. 

98. Circular Wire. — Fig. 113 (perspective). Let Z be a 
gravity-axis pei-pendicular to tlie plane of the wire ; X and Y 
lie in this plane, intersecting at right angles in the centre 0. 
The wire is hoitiogeneons and of constant (small) cross-section. 
Since, referred to Z, each dM has the same p — r, we have 
/^ ^fr''dM= Mr\ ]N"ow I^ must equal 7^, and (§ 94) their 
sum = Iz-, 

.-. 7x5 or Iy, = iMr\ and ^x^ ov Ity = ^^■ 

99. Homogeneous Solid Cylinder, dboxd a diameter of its base. 
— Fig. 114. 7x ^ ? Divide the cylinder into an infinite num- 
ber of larainse, or thin plates, parallel to the 
base. Each is some distance s from X, of 
thickness ds, and of radius r (constant). In 
each draw a gravity-axis (of its own) parallel to 

Fig. 114. X. We may now obtain the I^ of the whole 

cylinder by adding the IxS of all the laminae. The Ig of any 
one lamina (§96, circular plate) = its mass X i^''; hence its 
Ix (eq- (3), § 88) = its ^ -(- (its mass) X ^^ Hence for the 
whole cylinder 

Ix= f\{ydznT'^^g){\r'^^z-^)-\ 

I/O 

i.e., Jx = {jtr-'hy - g\lr^ + W) == M^kr' + W)- 

100. Let the student prove (1) that if Fig. 114 represent 
any right prism, and hp denote the radius of gyration of any 
one lamina, referred to its gravity-axis parallel to X^ then the 
Ix of whole prism = M{]i^ -|- \li^) ; and (2) that the moment 




MOMENT OF INERTIA. 



101 




of inertia of the cylinder about a gravitj-axis parallel to the 
base is = M{ir' + J^^')- 

101. Homogeneous Right Cone. — Fig. 115. First, about an 
axis F, through the vertex and jparallel to the base. As before, 
divide into laminae parallel to the base. Each is a 
circular thin plate, but its radius, x, is not = r, but, 1^ 
from proportion, is a? = (r -^ h)z. \ \i 'rX:^ 

The /of any lamina referred to its own gravity- ^ 
axis parallel to "Fis (§96) = (its mass) X ia?^, and |_ 
its Iv (eq. (3), §88) is .-. = its mass X i^i^ + fig. lis. 
its mass X s\ 

Hence for the whole cone, 

ly— I {nx^dzy -^ g)[iaf -j- s'] 

^ 

/Secondly, about a gravity-axis parallel to the hase. — From 
eq. (3), § 88, with d = ^h (see Prob. 7, § 26), and the result 
just obtained, we have /= J^-i-o[_r" -\- ^h^^. 

Thirdly, about its geometric axis, Z. — Fig. 116. Since the 
axis is perpendicular to each circular lamina through the centre, 
its Iz (§ 97) is 

= its mass X i-(rad.)° = {ynx^dz -^ g')^. 
Now a? = (r H- Ti)z, and hence for the w^hole cone 

Iz = \{yitr' - gU) t z'dz = {litr^hy - g)i-^r' = M^r\ 




Fig. 116. 





Fig. 118. 



102. Homogeneous Eight Pyramid of Rectangular Base. — 

About its geometrical axis. Proceeding as in the last para- 



102 



MECHANICS OF ENGHSTEERINft. 



graph, we deri^^e Iz = M^^d\ in which d is the diagonal of the 
base. 



103. Homogeneous Sphere. — About any diameter. Fig. 118. 
Iz = ? Divide into lamiuge perpendicular to Z. By § 97, and 
noting that a?' = r'— z% we have finally, for the whole sphere, 



Iz = {yTt ^ 2g) 



r+r 



{r'^ - ¥'^' + ¥1 = T^r^r^ - ^ 



For a segment, of one or two bases, put proper limits for s 
in the foregoing, instead oi -\- r and — r. 



104. 



Other Cases. 

Y 




Fig. 119. 



Fig. 120. 



-Parabolic plate, Fig. 119, homogeneous 

and of (any) constant thickness, about 

an axis through 0, the middle of the 

)-X chord, and perpendicular to the plate. 

This is 

The area of the segment is = f As. 

For an elliptic plate. Fig. 120, homogeneous and of any 
constant thickness, semi-axes a and h, we have about an axis 
through 0, normal to surface Iq = M^[a^ -f- h^'] ; while for a 
very small constant thickness 

I^=Mih% and Iy=Mia\ 

The area of the ellipse = rrah. 

Considering Figs. 119 and 120 as plane figures, let the 
student determine tlieir polar and rectangular moments of 
inertia about various axes. 

For numerous other cases Kent's Mechanical Engineers' 
Pocket-Book may be consulted ; also Trautwine's Civil Engi- 
neers' Pocket-Book. 

105. liTumerical Substitution. — The momsnts of inertia of 
'plane figures involve dimensions of length alone, and will be 
utilized in the problems involving flexure and torsion of beams, 
where the inch is the most convenient linear unit. E.g., the 



MOMENT OF INEKTIA. 



103 



polar moment of inertia of a circle of two inches radius about 
its centre is ^Ttr* = 25.1 o -[-Mquadralie, or four-dimension^ 
inches, as it may be called. Since this quantity contains iowv 
dimensions of length, the use of the foot instead of the inch 
would diminish its numerical value in the ratio of the fonrtli 
power of twelve to unity. 

The moment of inertia of a rigid hody, or solid, liowever, 
= MTc* = (G- -^ ffWi ill which G, the weight, is expressed in 
units oi force, g involves both time and space (length), while W 
involves length (two dimensions). Hence in any homogeneous 
formula in whicli the / of a solid occurs, we must be careful to 
employ units consistently ; e.g., if in substituting G -^ g for M 
(as will always be done numerically) we put g = 32.2, we 
should use the second as unit of time, and the foot as linear 
unit. 

106. Example. — Hequired the moment of inertia, about the 
axiS of rotation, of a pulley consisting of a rim, four parallelo- 
pipedical arms, and a cylindrical hub which may be considered 
solid, being filled by a portion of the shaft. 
Fig. 121. Call the weight of the hub G, 
its radius t\ similarly, for the rim, {r^, r^ 
and 7*2 ; the weight of one arm being = G^. 
The total / will be the sum of the /'s of 
the component parts, referred to the same 
axis, viz. : Those of the hub and rim will 
be {G ~ g)hy and {G, ~ ^)K^.= + r;), 
respectively (§ 97), while if the arms are ^^cj- 1-^- 

not very thich compared with their length, we have for them 
(§§ 95 and 88) 

4 (^1- g) [i(^. - ry ~ i(r, -. ry + [r + l{r, - r)]'], 
i.e., 4((7i-^g)[Kr2-r)2+rr2] .... (4) 

as an approximation (obtained by reduction from the axis at 
the extremity of an arm to a parallel gravity-axis, then to the 
required axis, then multiplying by four). In most fly-wheels, 
the rim is proportionally so heavy, besides being the farthest 
removed from the axis of rotation, that the moment of inertia 
of the other parts is only a small part of the whole. 

Numerically let us have given r = 4, r2 = 36, and r^ = ^7 inches; the 




104 MECHANICS OF ENGINEERING. 

respective weights being <?2 = 500 lbs. for the rim, (ri = 48 lbs. for each 
arm, and ^ = 120 lbs. for the hub. The quantity ^ will be retained as 
a mere symbol. Using the foot-pound-second system of units we then 
have for the moment of inertia of the huh (120-^gf)^[^]^= 6.66 -i-g'; 
for that of the four arms [by substitution in eq. (4) above] 

4 36' 



-S) 



2-/36 _£Y 
3\12 12/ 



'^ 12 12 



1 

while for the rim we obtain (500-f-gr)— 



Sry /36 
12J ^ll2 



1647.2 ^q; 



= 4627.0 ^g. 



^^=(^-^j^(^) = 7.91 sq.ft., 



These results are seen to be approximately in the ratio of the numbers 
1, 100, and 700; showing that the neglect of the hub and arms in com- 
puting the moment of inertia would give a result about ^ too small. 

Adding, we find for the total moment of inertia of the body about 
the axis of rotation the quantity / = 5280.8 -^gr, for the units foot and 
pound. The unit of time is still involved in the quantity g. 

We are now ready to compute the square of the corresponding radius of 
gyration, viz., k"^, by dividing / by the whole mass M, =668 -i-g (see § 86) ; 
whence 

and therefore k itself = 2.82 ft. 

This is seen to be a little less than the 3.04 ft. value for k which would 
be implied in the approximate assumption that the moment of inertia 
is the same as if the whole mass were concentrated at the mid-point 
of the thickness of the rim, which assumption would be very nearly 
true if the masses of the hub and arms could be neglected. 

107. Ellipsoid of Inertia. — The moments of inertia about 

all axes ptissing through any given point of any rigid body 
whatever may be proved to be inversely proportional to the 
squares of the diameters which they intercept in an imaginary 
ellipsoid, whose centre is the given point, and whose position 
in the body depends on the distribution of its mass and the 
location of the given point. The three axes which contain the 
three principal diameters of the ellipsoid are called the Princi- 
pal Axes of the body for the given point. This is called the 
ellipsoid of inertia. (Compare §89.) Hence the moments of 
inertia of any homogeneous reguhir polyedron about all gravity- 
axes are equal, since then the ellipsoid becomes a sphere. It 
can also be proved that for any rigid body, if the co-ordinate 
axes .X^ T", and .^, are taken coincident with the three principal 
axes at any point, we shall have 

fxydM = ; fyzdM = ; and fsxdM = 0. 

Note. — These three siimmations are called the "-products of inertia" and 
will occur in § 114 of this book. 



KINETICS OF A RIGID BODY. 105 

CHAPTER Y. 

KINETICS OF A RIGID BODY. 

108. General Method. — Among the possible* motions of a 
figid body the most important for practical purposes (and for- 
tunately the most simple to treat) are : a motion of translation, 
in which the particles move in parallel right lines with equal 
accelerations and velocities at any given instant; and rotation 
about a fixed axis, in which the particles describe circles in 
parallel planes with velocities and accelerations proportional 
(at any given instant) to their distances from the axis. Other 
motions will be mentioned later. To determine relations, or 
-equations, between the elements of the motion, tlie mass and 
form of the body, and the forces acting (which do not neces- 
sarily form an unbalanced system), the most direct method to 
be employed is that of two equivalent systems of forces (§ 15), 
one consisting of the actual forces acting on the body, con- 
sidered free, the otlier imaginary, consisting of the infinite 
number of forces which, applied to the separate material points 
composing the body, would account for their individual mo- 
tions, as if they were an assemblage of particles without mutual 
actions or coherence. If the body were at rest, then considered 
J'ree, and the forces referred to three co-ordinate axes, they 
would constitute a balanced system, for which the six summa- 
tions ^X, 2Y, ^Z, ^(mom.)x. ^''(mom.)y, and -^'(mom.)^. 
would each = ; but in most cases of motion some or all of 
these sums are equal (at any given instant), not to zero, but to 
the corresponding summation of the imaginary equivalent 
system, i.e., to expressions involving the masses of the particles 
(or material points), their distribution in the body, and the 
elements* of the motion. That is, we obtain six equations by 
putting the IX of the actual system equal to the IX of the 
imaginary, and so on ; for a definite instant of time (since some 
of the quantities may be variable), 

* Motions of such character that the particles of the body do not 
.change their relative positions. In other words, the body remains rigid. 



106 MECHANICS OF ENGINEERING. 

108a. The "Imaginary System." — In conceiving the imagi- 
nary equivalent system in § 108, applied to the material points 
or particles (supposed destitute of mutual action, and not 
exposed to gravitation), which make up the rigid body, we 
employ the simplest system of forces that is capable, by the 
Mechanics of a Material Point, of producing the motion, which 
the particles actually have. If now the mutual actions, co- 
herence, etc., were suddenly re-established, there would evi- 
dently be no change in the motion of the assemblage of parti- 
cles ; that is, in what is now a rigid body again, hence the imagi- 
nary system is equivalent to the actual system. 

In applying this logic to the motion of translation of a rigid 
body (see § 109 and Fig. 122,) we reason as follows : 

If the particles or elementary masses did not cohere together, 
being altogether without mutual action and not subjected to 
gravitation, their actual rectilinear motion in parallel lines, each 
having at a given instant the same velocity and also the sams 
acceleration, p, as any other, could be maintained only by the 
application, to each particle, of a force having a value = its mass 
X p, directed in the line of motion. In this way system (II.) is 
conceived to be formed and is evidently composed of parallel 
forces all pointing one way, whose resultant must be equal to "^eir 
sum, viz. I dMXp. But since at this instant p is common 
to the motion of all the particles, this sum can be written 
p i dM, =the whole mass Mxp. 

If now the mutual coherence of contiguous particles were sud- 
denly to be restored, system (II.) still acting, the motion of the 
assemblage of particles would not he affected (precisely as the fall- 
ing motion in vacuo of two wooden blocks in contact is just the 
same whether they are glued together or not) and consequently 
we argue that the imaginary system (II.) is the equivalent of 
whatever system of forces the body is actually subjected to, 
viz. system (I.), (in which the body's own weight belongs) 
producing the actual motion. 

Since the resultant of system (II.) is a single force, = ikfp, 
parallel to the direction of the acceleration, and in a line passing 
through the center of gravity of the body, it follows that th& 
resultant of the actual system is the same. 



KINETICS OF A RIGID BODY, 



107 



• 109. Translation. — Fig. 122. At a given instant all the par- 
ticles liave the same velocity = v, in parallel right lines (par- 
allel to the axis >^, say), and the 
same acceleration p. Required 
the 2^ of the acting forces, 
shown at (I.). (II.) shows the 
imaginary equivalent system, con- 
sisting of a force = mass X ace. 
= dMp applied parallel to 21 to 
each particle, since such a force 
would be necessary (from eq. {YY.) 
§ 55) to account for the accelerated rectilinear motion of the 
particle, independently of the others. Putting {'2X)i-={'2X)ii, 
we have 




Fig. 122. 



(^X)j =fpdM =j)fdM = Mp. 



(^•) 



It is evident that the resultant of system (II.) must be paral- 
lel to X; hence* that of (I.), which = (2X)j and may be de- 
noted by -S, must also be parallel to X; let a = perpendicular 
distance from H to the plane YX; a will be parallel to Z. 
Now put [-2'(mom.)y]j = \_2 (mom. y)]ii, (T'is an axis perpen- 
dicular to paper through 0) and we have — lia = —fdMjpz 
= —pfdMz = — pMz (§88), i.e., a := 2. A similar result 
may be proved as regards y. Hence, if a rigid hody has a 
motion of translation., the resultant force m,ust act in a line 
through the centre of gravity (here more ])roperly called the 
centre of mass), and parallel to the direction of motion. Or, 
practically, in dealing with a rigid body having a motion of 
translation, we may consider it concentrated at its centre of 
mass. If the velocity of translation is uniform, R =■ M X 
= 0, i.e., the forces are bnlanced. 

109a. Example. — The symmetrical rigid body in Fig. 122a weighs 
(G = ) 4 tons, and touches a smooth horizontal floor at the two points 
and B, symmetrically situated. Its center of gravity, C, is 6 ft. above 
the floor; and it is required to find the effect of applying a horizontal 
orce of P=l ton, pointing to the right and 4 ft. below the level of the 
center of gravity C. Evidently a motion of translation will ensue from 
left to right, with some acceleration p, unless the body should begin 
to overturn about or 5 as a pivot. The latter would be proved to 

* The forces of system (I.) cannot form a couple; since those of system 
(II.) do not reduce to a couple, all pointing one way. 



108 



MECHANICS OF ENGINEERING. 




J^G. 122a. 



be the case if either reaction, Vg or V, of the floor against the body at 
O and B, is found to be negative as the result of an analysis which 
assumes translation to occur. The actual forces acting on the body 

are only four, viz.: G and P, and the 
unknown vertical reactions V and Fq. 
A special device (very convenient for 
the present case) will now be used as a 
means of solution. The resultant of the 
"equivalent system," II, in this case 
of translation (see Fig. 122), is R, = Mp, 
lbs., acting through the center of gravity 
in a line parallel to that of the motion 
and in the direction of the acceleration, 
and hence is also the resultant of the 
actual system (just described). If, there- 
fore, we annex to the actual system its 
anti-resultant (which is a force, R', of the 
same value, Mp, as R, and in same line, but pointing in the opposite 
direction) we thereby form a system under which the body would be in 
equilibrium; which would justify our writing iX = 0, IY=0, i'(moms.) 
= 0, etc. (R' is called the "reversed inertia force" and is, of coiKse, 
fictitious). With this system, then, in view, putting IX = we obtain 
P—R' = 0; i.e., R', =Mp, =lton; whence the acceleration p=lH- (G-h^) 
= 1-^(4-^32.2) = 8.05 ft./sec.2 

By T(moms. about point A) = we find E'X4'-F(?X2'-FX4' = 
or 7 = 2.5 tons; and, by -Z = 0, ¥ + ¥^-0 = 0, or Fo = 4-2.5=1.5 tons. 
Since neither V nor Fq is found to be negative the body does not tend 
to overturn but moves parallel to itself (i.e., translation) with a uniformly 
accelerated motion, the value of the acceleration being p = 8.05 ft. /sec. ^; 
so that at the end of the first second the body would be 4.025 ft. from 
the start (no initial velocity); at the end of the second second, 16.1 ft. 
If P were zero, or if P Were applied horizontally through the center 
of gravity, F and Fq would each be one haK of G, i.e. 2 tons. It appears, 
therefore, that the effect of the eccentric application of P (viz. 4 ft. 
below the center of gravity C) is to increase F by 0.5 ton and diminish 
Fq by an equal amount. If P acted 4 ft. above C, F and Fq would 
change places in this respect. For F to be just zero, P (in its present 
position) would need to have a value of 2 tons, and the body would 
be on the point of overturning toward the left. Or, again, with P 
= 1 ton, its line of application would have to be 8 ft. below or above 
C, for one of the reactions to be just zero. In fact, in the fictitious 
equilibrated system which includes R', since P=R' (in this simple case) 
they form a couple; and hence the three forces G, F, and Fq are equiva- 
lent to a couple of equal and opposite moment (viz. 4 ft.-tons in Fig. 122a). 
From the above it is seen that in the case of the last car of a railroad 
train, when it has an accelerated motion (just leaving a station), the 
pressures under the front and rear trucks will be slightly different from 
their values when the motion is uniform or zero, if the pull in the coupling 
does not pass through the center of gravity of the car. 



KINETICS OF A RIGID BODY. 109 

110. Rotation about a Fixed Axis. — First, as to the elements 
of space and time involved. Fig. 123. Let be the axis of 
rotation (perpendicular to paper), OY d. fixed e ^"— --- x,w 
line of refei-ence, and OA a convenient line of (^ y^V\ 
the rotating body, passing through the axis and / X.^ \ 

perpendicular to it, accompanying the body in / — — 1~ 
its angular motion, Mdiich is the same as that of V_^_,,^-__^ — ^ 
OA. Just as in linear motion we dealt with ^^*^- ■'^■ 
linear space (.§), linear velocity (-y), and linear acceleration {^jp)y 
so here M'e distinguish at any instant ; 

a, the angular space between OY and OA, (radians; or de- 
grees, or revolutions) ; 

a) = -TT^ the angular velocity, or rate at which a is changing, 

(such as radians per sec. , or revolutions per minute, etc.); and 

^ = -^ = -77^, the angular acceleration, or rate at which 0/ 

is changing (radians per sec. per sec, e.g.) 

These are all in angular measure and may be + or — , ac- 
cording to their direction against or with the hands of a watch. 
da. is a small increment of a, while d^a is the difference be- 
tween two da^s, described in two consecutive small and equal 
time-intervals, each= dt. 

(Let the student interpret the following cases : (1) at a cer- 
tain instant gd is -|-, and — ; (2) go is — , and 6 -{-; (3) a is 
— , GO and 9 l)otli -f ; (4) a -{-, go and 6 both — .) For rotary 
motion we have therefore, in general, 

and .•. (by elimination) codco — 6da; (VIIL) 

corresponding to eqs. (I.), (II.), and (III.) in § 50, for rectilinear 
motion. . 

Hence, for uniform rotary motion, go being constant and 
^ = 0, we have a = Got, t being reckoned from the instant 
when a = 0. 



* See pp. 132, 133, of the "Notes," etc, for further illustration. 



110 MECHANICS OF ENGINEERING. 

For uniformly accelerated rotary motion Q is constant, and 
if (jjQ denote tlie initial angular velocity (when a and t = 0) we 
may derive as in § 5G, denoting the constant d by 6'i, 
,Go= Go^-\-dit; . . (1) a= Go^t + ^dxt'-', . . (2) 

a = — ^ — -"- ; . . (3) and a = i{oa^ + ^y- - • (4) 

If in an_y problem in rotary motion 0, go, and a have been 
determined for any instant, the corresponding linear vakies for 
any point of the body whose radial distance from tlie axis is p, 
will be 5= o'p (= distance described by the point measured 
along its circular path from its initial position), v = cop = its 
velocity, and j?^ = dp its tangential acceleration, at the instant 
in question, ii a, w and d, are expressed in radians. 

Example. — (1) What value of co, the angular velocity, is 
implied in the statemient that a pulley is revolving at the rate 
of 100 revolutions per minute if the radian is unit angle? 

100 revolutions per minute is at the rate of 2;rXl00 
= 628.32 radian units of angular space per minute = 10.472 
per second. .". (y = 628.32 radians per minute or 10.472 
radians per second. 

(2) A grindstone whose initial speed of rotation is 90 revo- 
lutions per minute is brought to rest in 30 seconds, the an- 
gular retardation (or negative angular acceleration) being con- 
stant; required the angular acceleration, di, and the angular 
.space a described. Use the second and radian as units. 
a»o = 27r|| = 9.4248 radians per second; .'. from eq. (1) 

/?!= — - — = — 9. 424 -=-30= —0.3141 radians per sec. per sec. 

t 

The angular space, from eq. (2) is 

a =a;o^ + J^ii2 = 30X9. 42-1(0.314)900 = 141. 3 
radians; that is, the stone has made 22.4 revolutions in 
<3oming to rest and a point 2 ft. from the axis has described a 
distance s = ap = 141. 3 X 2 = 282. 6 ft. in its circular path. 

111. Rotation. Preliminary Problems. Axis Fixed. — For 

clearness in subsequent matter we now consider the following 



KINETICS OF A RIGID BODY, 



111 




80 lbs. 



problem. Fig. 124 shows a rigid homogeneous right cylin- 
der A of weight G = 200 lbs. and radius r = 2 ft., mounted on 
a horizontal axle and concentric with the same. The center 
of gravity of the cyUnder is in the axis of rotation (Z). The 
axle carries a Ught and concentric drum, of 10 in. radius, from 
which a light inextensihle cord may unwind as the attached 
weight B descends, thus imparting an accelerated rotary motion 
to the cylinder. The weights and masses of the drum, cord, 
and axle, and all friction, will be neglected; and the two journals 
will be considered as one. The cylinder being originally at rest 

we wish to deter- 
mine its motion 
as produced by 
a constant down- 
ward pull or ten- 
sion of 80 lbs. in 
the vertical cord. 

(I) ^V.^L^--^B"(n> (T^' necessary 

y Fig. 124. \t weight, G', of the 

body to be used at B*, to secure this 80 lbs. tension in the 
cord, will be found later.) During this motion the real system 
of forces (system (I)) acting on a body A consists of the weight 
200 lbs., always acting through Z, the fixed axis of rotation; 
the downward pull of 80 lbs. at 10 in. from the axis; the ver- 
tical component V of the reaction of the bearing; and the 
horizontal component (if any), H. At (II), Fig. 124, is shown 
an imaginary equivalent system capable of producing the same 
motion in the particles, each of mass = dM, if they were inde- 
pendent. Since each particle is moving in a circle of some 
radius p with some linear (tangential) acceleration pt at any 
instant, the cylinder having at that same instant some an- 
gular velocity co and some angular acceleration 6, we have 
v = a)p atid pt = ^P- (^ ^nd 6 in radians.) 

This circular motion of each particle could be produced 
(see eq. (5), p. 76) by a tangential force dT lbs., =dMpt, 
= 6dMp, accompanied by a normal force dN lbs., =dMv^-^p, 
= co^dMp. Our equivalent system, then, in (II), consists of 
a dT and a dN of proper value applied to each particle of body 
A at a given instant. Axes X and Y are shown in Fig. 124, 



* The body B is not shown in the figure. 



112 MECHANICS OF ENGINEERING. 

axis Z, the axis of rotation, being i to the paper through 
origin 0. Let us now, for any instant of the motion, equate 
I (moms.)^ of the actual system, (I), to J (moms.)^ in sys- 
tem (II) ; using the integral sign to denote a summation which 
extends over all the particles of body A (for this instant; the 
integral might therefore be called an instantaneous integral). 
This gives, if we note that each of the normal forces dN of 
system (II) has no moment about axis Z, and that d is common 
to all the particles at this instant, (with ft.-lb.-sec. units), 
+ 80x10/12= +/dT.p=+d/dMp^=^ +61,. . (1) 

The summation (instantaneous) /dMp^ is seen to be the 
quantity called "moment of inertia," about axis Z, of the 
body A and remains constant, since the p's do not change 
in value as the motion proceeds. For a solid homogeneous 
cylinder Ig = ^Mr^ (p. 99), and hence 

800 = 6^[200-^32.2](2)2; i.e., ^ = 7.376 rads./sec.2 

That is, 6 is constant and the rotary motion of the cylinder 
is uniformly accelerated. 

(N. B. — From eq. (1) we note that, in general, in order to obtain the 
angular acceleration, 6, of the rotary motion [of a rigid body about a 
fixed axis Z we have only to treat the body as a "free body" and write 
J! (moms.) about axis of rotation = angul. accel.Xmom. of inertia about Z.) 

112. Further Results in Preceding Problem. — As to the necessary weight, 
G', of body B (suspended on the cord and causing the motion of both 
bodies), in order to produce the 80 lbs. tension in the cord, we note 
that body B has the same motion (only in a right line), as a point 

in the circumference of the drum, where the acceleration is p' = dX ^ 

= 4.48 ft. /sec. That is, the 'motion of B wUl be uniformly accelerated, 
with an acceleration of 4.48 ft. /sec. ^ Hence the weight of B must not 
only produce the 80 lbs. tension in the cord but also accelerate the mass 
of B, {M' = G'-^g) with an acceleration of 4.48 ft./ sec. ^ I.e., we have 
G' = 80+ {G'^g)p'; which is nothing more than saying that the net 
accelerating force, G' — 80, =massXaccel. ; whence we find, on solving, 
G' = 92.9 lbs. for the weight of the body to be used at B. 

For example, in the first 3 sec. of time, starting from rest, B will 
descend a distance (see p. 54), s^ = ip'(3y = 20.16 ft. and will have ac- 
quired a (linear) velocity of V3 = p'X3 = 13.44 ft. /sec. ; while body A 
will have turned through an angle of a3 = Ji9(3)^ = 33.19 radians, (or 
5.283 revolutions) and will possess an angular velocity of a>3 = 5X3 = 72.13 
rads./sec. or, (33.19 ^2;r = ), 3.525 revs./sec. 

Reaction of the bearing; (two journals considered as one). To find 
the two components H and V of this reaction, we again have recourse 
to the two equivalent systems of Fig. 124, acting on body A. (N.B. — 
The upward 80 lbs. and the force G' do not belong to system (I), since 
they act on body B.) During the motion, the coordinates x and y of each 
particle (of mass = dM) are continually changing, as also the angle ^ 



KINETICS OF A RIGID BODY. 



113 



between the p of the particle and axis Y (but not p itself). At any given 

instant we note that x = p sin 4> ^^d y — p cos (f>, for each particle. Let us 

now put i'F of system (I) equal io lY oi system (II). This gives us 

V -2m-m= fdT sin ^-/dN cos <^ . .... (2) 

As before, these integrals are "instantaneous integrals," being extended 
over all the particles of the body at a given instant of time, [so that in 
general the value of each may change with the progress of the motion. 

Substituting for dT and dN, etc., this may be written 

F - 200 - 80 = efdMp sin ^ - u?fdMp cos ^S ... (3) 
or, F-200-80=5/dM:c-w2/dMy, (4) 

Note that the value of (9, and also of w, at this single instant are 
common to all the particles and have been factored out, as shown. 

But the summation of fdMx is nothing more than Mx\ where M 
is the mass of the whole cylinder A ( = 200-^?) [see p. 18, eq. (1)], and 
X is the X coordinate of its center of gravity; and, similarly, J~dMy = My. 
We may therefore write 

V-2m-SQ = eMx-w'^My (5) 

But in the present case, since the center of gravity of body A is in 
the axis of rotation at all times, we have both x and 2/ = zero at all 
times; and hence finally 7-200-80=0; or F = 280 lbs. 

As to the horizontal component, H, of the bearing reaction, we place 
2X of system (I) equal to IX of system (II) and obtain 

H = -fdT cos ^ -fdN sm4>=- OfdMp cos ^ - oJ^fdMp sin 4>, (6) 
i.e., H=-efdMy-w^fdMx,= -eMy-w^Mx (7) 

But since x and y are zero at all times, H must be zero, from (7) ; and 
we therefore conclude that in this case the reaction of the bearing is 
purely vertical at all times and is V, = 280 lbs. 

113. Centre of Percussion of a Rod suspended from one End. — > 

Fig. 126. The rod is initially at rest (see (I.) in figure), is sti-aight, 

homogeneous, and of constant 

(small) cross-section. Neglect its 

weight. A horizontal force or 

pressure, P, due to a blow (and 

varying in amount during the 

blow), now acts upon it from the 

left, perpendicularly to the axis, 

Z, of suspension. An accelerated 

rotary motion begins about the fixed axis Z. 




°y° n 



^^dt 



(in.) 



(TI.) 
Fig. 126. 

(II.) shows the rod 
free, at a certain instant, with the reactions X^ and Y,, put in 
at 0„. (III.) shows an imaginary system which would produce 
the same effect at this instant, and consisting of a dT = dMOp, 
and a <^iV = oo^dMp applied to each dM, the rod being composed 
of an infinite number of dM^s, eacli at some distance p from 
tlie axis. Considering that the rotation has just begun, go, the 



114 



MECHANICS OF ENGINEERING. 



angular velocity is as yet small, and will be neglected. Re- 
quired Yo tlie horizontal reaction of the support at in terms 
of P. By putting lYji= lYm, we have 

P-Yo =/dT = e/pdM = eu'p. 
/. J^o = -P — OM p ; p is the distance of the centre of gravity 
from the axis (IST.B. J'pdM = M p is only true when all the 
p's are parallel to each other). But the value of the angular 
acceleration 6 at this instant depends on P and a, for 2 (mom.)> 
in (IL) = :2 (luom.)^ in (III.), whence Fa = dfp'dM^ diz, 
where Iz is the moment of inertia of the rod about Z, and from 
§ 95 = \Ml\ Now p = i^ ; hence, finally, 

"F — pfl _ ?- -' 

U. ± J. i) ' 1 

If now J^u is to = 0, i.e., if there is to be no shook between 
the rod and axis, we need only apply P at a point whose dis- 
tance a = f / from the axis ; for then Y^ = 0. This point is 
called the centre of percussion for the given rod and axis. It 
and the point of suspension are interchangeable (see § 118). 
(Lay a pencil on a table; tap it at a point distant one third of 
the length from one end ; it will begin to rotate about a vertical 
axis through the farther end. Tap it at one end ; it will begin 
to rotate about a vertical axis through the point first mentioned. 
Such an axis of rotation is called an axis of instantaneous rota- 
tion, and is different for each point of impact — ^just as the 
point of contact of a wheel and rail is the one point of the 
wheel which is momentarily at rest, and about which, therefore, 
all the others are turning for the instant. Tap the pencil at 
its centre of gravity, and. a motion of translation begins; see 
§ 109.) 

114. Rotation. Axis Fixed. General Formulae. — Consider 





Fig. 127. Fig. 128. 

|.ng now a rigid body of any shape whatever, let Fig. 12Y indi- 
cate the system of forces acting at any given instant.^ Z being 



KINETICS OF A RIGID BODY. 115 

the fixed axis of rotation, go and 6 tlie angular velocity and 
angular acceleration, at the given instant. X^ and IT are two 
axes, at right angles to each other and to Z^ fixed in space. At 
this instant eacii clM oi the body has a definite x, y, and q) 
(see Fig. 128), which will change, and also a p, and 0, which \v ill 
not change, as the motion progresses, and is pursuing a circu- 
lar path with a velocity = cop and a tangential acceleration 
= dp. Hence, if to each dM of the body (see Fig. 128) we 
imagine a tangential force dT =^ dMO p -Audi a normal force 

— dJf{oopy -^ p = QJ^dMp to be applied (eq. (5), §74), and 
these alone, we have a system comprising an infinite number of 
forces, all parallel to XJ^, and equivalent to the actual system 
in Fig. 127. Let ^JT, etc., represent the sums (six) for Fig. 
127, whatever they may be in any particular case, while for 
128 we shall write the corresponding sums in detail, looting 
that 

fdli cos cp = GoYdMp cos cp = coydMy = g9^J/^(§88); 
that/6^iV^sin (p = coydMp sin cp = coydMx = go'Mx; 
and similarly, that /dT cos cp = dfdMp cos q) = 6 My, and 
fdT sin q) = OMx; while in the moment sums (the moment 
of dT cos 9? about J^, for example, being — dT cos (p . z ■=■ 

— OdMp (cos (p)s= — 6dMyz, the sum of the moms, y of all the 
(^rcos 9>)'s = - QfdMyz) 

fdTeo% (pz = dfdMyz,fdN^m cpz = aoydMxz, etc., 
W6 have, since the systems are equivalent, 

:sX=-{-6My-Go'Mx; . . . . (IX.) 
:SY :=-6Mx-co'Ify, . . . . (X.^ 

2Z= 0; (XL) 

2 moms.x = - 0/dMxz — coydMyz ; . (XIL) 

:S moms.y = - e/dllyz + JfdMxz ; . (XIII.) 

:S moms.^ = OfdMp' = 61^. . • . (XIY.) 

These hold good for any instant. As the motion proceeds x 

and y change, as also the sums fdMxz and fdMyz. If the 

Ijody, however, is homogeneous, and symmetrical about the 

plane XY, fdMxz and fdMyz would always = zero ; since 



116 



MECHANICS OF ENGINEERING. 



the z of any <^JI/'does not change, and for every term dMy{-\-z\ 
there would be a term dMy{ — z) to cancel it ; similarly for 
fdMxz. The eq. (XIY,), ^ (moms, about axis of rotat.) = 
fdTp = QJdMff = {angular accel.) X {mom. of inertia oj 
hody about axis of rotat.), shows how the snvafdMp^ arises in 
problems of this chapter. That a iovce dT :=^ dMdp should 
be necessary to account for the acceleration (tang-ential) dfj of 
the mass dM, is due to the so-called inertia of the mass (§ 54), 
and its moment dTp, or OdMp^, might, with some reason, l>e 
called t\\e moment of inertia oi the dll, imdf6dMp^= OfdMp' 
that of the whole body. But custom has restricted the nanse 
to the snmfdMp^, Mdiich, being without the 0, has no term to 
suggest the idea of inertia. For want of a better the name is 
still retained, and is generally denoted by /. (See §§86, etc. ) 



115. Example of the Preceding. — A 



liomoofeneous rig 



lit par- 




FiG. 129. 



allelopiped is mounted on a vertical 
axle (no friction), as in figure. is 
at its centre of gravity, hence hoth 

X and y are zero. Let its henviness 
be y, its dimensions A, 5„ and b (see 
§ 97). XY is a plane of symmetry, 
hence both fdMxz and fdMyz are 
zero at all times (see above). The 
tension P in the (inextensible) cord 
is caused by the hanging weight P^ 
(but is not = /^j, unless the rotation is uniform). The figure 
shows both rigid bodies ^r^e. P^ will have a motion of trans- 
lation ; the parallelopiped, one of rotation about a fixed axis. 
No masses are considered except P^ -^ g. and bhb^y -^ g. The 
Iz = MTc^ of the latter = its mass X tV(^i' + ^')' § ^T. At 
any instant, the cord being taut, if ^ = linear acceleration of 
^., we have jp = da. eq. (<?) 

From (XIY.), Pa = 61^ ; .: P = Olz ~ a. . . . (1) 

For the free mass P^ -i- g we have (§ 109) P^ — P = 
mass X ace, 

= {P.'^9)p = {P.-^g)ea; .:P = P,{l-ea^g). (2) 
Equate these two values of P and solve for 6^, whence 

Mkl-^{P,-^g)a' ^^^ 



6 = 



KINETICS OF A RIGID BODY. 117 

All the terms here are constant, hence d is constant ; there- 
fore the rotary motion is uniformly accelerated, as also the 
translation of P,. The formulae of § 56, and (1), (2), (3), and 
(4) of §110, are applicable. The tension P is also constant; 
see eq. (1). As ior the five unknown reactions (components) 
at (?i and 0^, the bearings, we shall find that they too are con- 
stant ; for 

from (IX.) we have' Zi + Z2 = 0; (4) 

from (X.) we have P + ri+F2 = 0; (5) 

from (XI.) we have Z^-G = 0; (fi) 

from (XII.) we have P . AO + Y, .0,0-Y^ .0^0 = 0; (7) 

from (XIII.) we have -X^ . Ofi + X^ ."0^ = 0. (8) 

Numerical substitution in the above problem. — Let the parallelepiped 
be of wrought-iron ; let Pi = 48 lbs.; a = 6 in. = J ft.; 6 = 3 in. = J ft. 
(see Fig. 112); 6i = 2 ft. 3 in. = |^ft.; ^.nd /i = 4 in.=i ft. Also let 
0^0 = 020 = 18 in.= | ft., and AO = S in. = i ft. Selecting the foot- 
pound second system of units, in which g' = 32.2, the linear dimensions 
must be used in feet, the heaviness, ^, »of the iron must be used in lbs. 
per cubic foot, i.e., ^' = 480 (see § 7), and all forces in lbs., times in 
seconds. 

The weight of the iron will be G = Fr = ^f'i^r = i • I ■ iX480 = 90 lbs.; 
its mass = 90 -T- 32.2 = 2.79; and its moment of inertia about Z=/z = MA;2^ 
=Mxi^(V_-|-62) = 2.79X0.426 = 1.191. (That is, the radius of gyration, 
kz, ="i/0.426 = 0.653 ft.; or the moment of inertia, or any result depend- 
ing solely upon it, is just the same as if the mass were concentrated in 
a thin shell, or a line, or a point, at a distance of 0.653 feet from the 
axis.) We can now compute the angular acceleration, d, from eq. (3) ; 

48 X + 24 

1.191 + (48H-32.2)Xi'" 1.191+0.372" 

radians per sec. per sec. The linear acceleration of Pi is p = 0a = 7.68 
feet per sec. per sec. for the uniformly accelerated translation. 

Nothing has yet been said of the velocities and initial conditions of 
the motions; for what we have derived so far applies to any point of 
time. Suppose, then, that the angular velocity <y = zero when the time, 
t=0; and correspondingly the velocity, v = o^a, of translation of Pj, 
be also = when t = Q. At the end of any time t, a> = 9t (,§§ 56 and 110) 
and v = pt = 6at; also the angular space, a = ^dt^, described by the 
parallelopiped during the time t, and the linear space s = \pt'^ = ^Qat'^, 
through which the weight P^ has sunk vertically. For example, during 
the first second the parallelopiped has rotated through an angle a = ^9t' 
= iX 15.36 XI = 7.68 radians, i.e., (7.68 -^2;r) = 1.22 revolutions, while P^ 
has sunk through s = ^9at^ = 3.84: ft., vertically. 




118 MECHANICS OF ENGINEERING. 

The tension in the cord, from (2), is 

P = 48(l-15.36Xi-^?) =48(1-0.24) = 36.48 lbs. 

The pressures at the bearings will be as follows, at any instant: from 
(4) and (8), X^ and X2 must individually be zero; from (6) Z2 = G=Vj- 
= 90 lbs.; while from (5) and (7), Yi= -21.28 lbs., and 1^2= -15.20 lbs., 
and should point in a direction opposite to that in which they were 
assumed in Fig. 129 (see last lines of Jj- 39). 

117. The Compound Pendulum is any rigid body allowed to 
oscillate without friction under the action of gravity when 
mounted on a horizontal axis. Fig. 131 shows the 
body /"ree, in any position during the progress of 
the oscillation. C is the centre of gravitj^; let OG 
= s. From (XI Y.), § 114, we have 2 (mom. about 
fixed axis) 

= angul. ace. X mom. of inertia. 

.-. — Gs sin a = 61^, 
and = — Gs sin a -^ I^ = — Mgs sin a -f- MTcl, 

i.e., = — ^s sin «r -^ ^/ (1) 

Hence d is variable, proportional to sin a. Let us see what 
the length I = OJT, of a simple circular pendulum, must be, to 
have at this instant (i.e., for this value of a) the same angular 
acceleration as the rigid body. The linear (tangential) accelera- 
tions of ^ the extremity of the required simple pendulum 
would be (§ 77) Pt = — 9 sin a, and hence its angular accelera- 
tion* would = — gsina-^l. "Writing this equal to d in eq. 
(i), we obtain 

^ = ^0^^^ (2) 

Bat this is independent of a ; therefore the length of the sim- 
ple penduhim having an angular acceleration equal to that of 
the oscillating body is the same in all positions of the latter, 
and if the two begin to oscillate simultaneously from a position 
of rest at any given angle oc^ with the vertical, they will keep 
abreast of each other during the whole motion, and hence have 

* Most easily obtained by considering that if the body shrinks into a mere 
point at K, and thus becomes a simple pendulum, we have both ka and s 
equal to I ; which in (1) gives B — — g sin a -i- 1. 



KINETICS OF A RIGID BODY. 



119 



the same duration of oscillation ; which is .*. , for small ampli- 
tudes (§ 78), 

t' = 7t VT^ = 7t Vk; -^ gs, .... (3) 

j^is called the centre of oscillation corresponding to the given 
centre of suspension 0, and is identical with the cenl/re of per- 
cussion (§ 113). 

Example. — Required the time of oscillation of a cast-iron 
cylinder, whose diameter is 2 in. and length 10 in., if the axis 
of suspension is taken 4 in. above its centre. If we use 32.2 
for g, all linear dimensions should be in feet and times in 
seconds. From § 100, we have 



Jf(f 



From eq. (3), § 88, 
.-. i; = 0.170 sq. ft.; 



1 7,2\ — llffi 1 II 10 0\ — ]\ f 1 103 



M[^.\^^ + i-]=Mx0.m; 



t'= 7t VO.ITO -^ ^32.2 Xi) = 0.395 sec. 



118. The Centres of Oscillation and Suspension are Inter- 
changeable. — (Strictly speaking, these centres are points in the 
line through the centre of gravity perpendicular to the axis of 
suspension.) Refer the centre of oscillation K to the centre 
of gravity, thus (Fig. 132, at (I.) ) : 



= l-s = 



Ms 



s = 



MJ ic' + Ms' 
Ms 



s = —- (1) 
s ^ ^ 




!N'ow invert the body and suspend it at K', 
required CK^, or s^i to find the centre of 
oscillation corresponding to K as centre of 
suspension. By analogy from (1) we have s 
s^ = he -^ -Si ; but from (1). k^ -^ s^ ^ s .'. 
s^ = s\ in other words, ^j is identical with 
0. Hence the proposition is proved. 

Advantage may be taken of this to determine the length X 
of the theoretical simple pendulum vibrating seconds, and thus 
finally the acceleration of gravity from formula (3), § 117, viz.. 



(I.) (11.) 

Fig. 132. 



120 MECHANICS OF ENGINEEEING. 

when i! = 1.0 and I (now = Z) has been determined experi- 
mentally, we have 

g (in ft, per sq. second) = X (in ft.) X tt*. . . (2) 

This most accurate method of determining g at any locality 
requires the use of a bar of metal, furnished with a sliding 
weight for shifting the centre of gravity, and with two project- 
ing blocks provided with knife-edges. These blocks can also 
be shifted and clamped. By suspending the bar by one knife- 
edge on a proper support, the duration of an oscillation is com- 
puted by counting the total number in as long a period of 
time as possible; it is then reversed and suspended on the 
other with like observations. By shifting the blocks between 
successive experiments, the duration of the oscillation in one 
position is made the same as in the other, i.e., the distance be- 
tween the knife-edges is the length, I, of the simple pendulum 
vibrating in the computed time (if the knife-edges are not equi- 
distant from the centre of gravity), and is carefully measured. 
The I and t' of eq. (3), § 117, being thus known, g may be com- 
puted. The length, in feet, of the simple pendulum vibrating 
seconds, at any latitude /?, and at a height of h ft. above sea- 
level, is (Chwolson, 1902). 

L = 3.25974-0.008441 cos 2/3-0.0000003/i. 

119. Isochronal Axes of Suspension. — In any compound 
'pendulum., for any axis of suspension, there are always three 
others, parallel to it in the same gramty-plane, for which the 
oscillations are Tnade in the same time as for the first. For 
any assigned time of oscillation t', eq. (3), § 117, compute the 
corresponding distance CO = s oi O from O; 

. Mk: 7r\MkJ + Ms^ 
i.e.,from t = ^ ^^- = ^- , 



we have s= {gt"-^27r')± V{g'r-r-4:7r') — kJ. . . (1) 

Hence for a given f, there are two positions for the axis O 
parallel to any axis through C, in any gravity-plane, on both 
sides; i.e., four parallel axes of suspe?ision, in any gravity- 
plane, giving equal times of vibration ; for two of these axes 



KINETICS or A RIGID BODY. 



121 



we must reverse the body. E.g., if a slender, homogeneous, 
prismatic rod be marked off into thirds, tlie (small) vibrations 
will be of the same duration, if the centre of suspension is 
taken at either extremity, or at either point of division. 

Examjple. — Required the positions of the axes of suspension, 
parallel to the base, of a right cone of brass, whose altitude is 
six inches, radius of base, 1.20 inches, and weight per cubic inch 
is 0.304 lbs., so that the time of oscillation may be a half- 
second. (N.B. For variety, use the inch-pound-second system 
of units, first consulting § 51.) 

120. The Fly-Wheel in Fig. 133 at any instant experiences 
a pressure P' against its crank-pin from the connecting-rod 
and a resisting pressure P" from the teeth of a spur-wheel with 




Fi& 133. 

which it gears. * Its weight G acts through C (nearly), and 
there are pressures at the bearings, but these latter and G have 
no moments about the axis C (perpendicular to paper). The 
figure shows it free^ P" being assumed constant (in practice 
this depends on the resistances met by the machines which D 
drives, and the fluctuation of velocity of their moving parts). 
P\ aiivl therefore T its tangential component, are '.variable, 
depending on the effective steam-pressure on the piston at any 
instant, on the obliquity of the connecting-rod, and in high- 
speed engines on the masses and motions of the piston and con- 
necting-rod. Let r. = radius of crank-pin circle, and a the 
perpendicular from G on P" . From eq. (XIY.), § 114, we 
have 
Tr - P"a = Qlg, .: 6 = {Tr - P"a) -^ Ig, • (1 

*Bearings at C not shown. P is the thrust in the piston-rod due 
to steam pressure on piston. 



122 MECHANICS OF ENGHSTEEKING. 

as the angular accelei-atioii at any instant ; substituting wliicliin 
the general equation (VIII.), § 110, we obtain 

IqWcLoo = Trda — P"ada (2) 

From (1) it is evident that if at any position of the crank-pin 
the variable Tr is equal to the constant P"a^ 6 is zero, and 
consequently the angular velocity a) is either a maximum or a 
minimum. Suppose this is known to be the case both at m 
and n\ i.e., suppose T, which was zero at the dead-point A, 
has been gradually increasing, till at n, Tr = P"a\ and there- 
after increases still further, then begins to diminish, until at m 
Tr again = P"a^ and continues to diminish toward the dead- 
point P. The angular velocity go, whatever it may have been 
on passing the dead-point A, diminishes, since 6 is negative^, 
from A to n, where it is c»^, a minimum ; increases from n to- 
??^. where it reaches a maximum value, c»,^. n and m being 
known points, and supposing co^ known, let us inquire what 
Go^ will be. From eq. (2) we have 

IcJ^ Goda>=J^^ Trda-P"J^^ ada. . . (3> 

But rda = 6/s = an element of the path of the crank-pin, and 
also the " virtual velocity" of the force T, and ada = ds", an. 
element of the path of a point in the pitch-circle of the fly- 
wheel, the small space through which P" is overcome in dt. 
Hence (3) becomes 

/ci(c»J - GD^') =J^ Tds - P" X linear arc EF. (4> 

To determine / Tds we might, by a knowledge of the vary- 
ing steam-pressnre, the varying obliqnit}^ of the connecting-rod, 
etc., determine T for a number of points equally spaced along- 
the cnrve nm^ and obtain an approximate value of this sum by 
Simpson's Rule; but a simpler method is possible by noting 
(see eq. (1), § 65) that each term Tds of this sum = the corre- 

sponding term Pdx in the series / Pdx, in which P = the^ 



KINETICS OF A RIGID BODY. 123 

effective steam-pressure on the piston in the cylinder at any in- 
stant, dx the small distance described by the piston while the 
crank-pin describes any ds^ and n' and m' the positions of the 
piston (or of cross-head, as in Fig. 133) when the crank-pin is 
at n and m respectively. (4) may now be written 

Ic\{po^ - O =J^^, Pdx - P" X linear arc EF, (5) 

from which c»^ may be found as proposed. More generally, it 
is available, alone (or with other equations), to determine any 
one (or more, according to the number of equations) unknown 
quantity. This problem, in rotary motion, is analogous to that 
in §59 (Prob, 4) for rectilinear motion. Friction and the in- 
ertia of piston and connecting-rod liave been neglected. As 
to the time of describing tlie arc wm, from equations similar to 
(5), we may determine values of co for points along nm, divid- 
ing it into an even number of equal parts, calling them cw^, &?„ 

etc., and then employ Simpson's Rule* for an approximate value 
pm n,m g^ 

of the sum \ t= I — (from eq. (YL), § 110) ; e.g., with 
four parts, we would have 



f^l ri4241~ 
^ = T-=: (angle wC'w, in rads.) — I 1 1 1- — 

Ln 12 ^ ° Lo^n ' Ci5, C^j 0^3 <^m-^ 



.(6) 



121. Numerical Example. Fly-Wheel.— (See Fig. 133 and 
the equations of § 120.) Suppose the engine is non-condensing 
and non-expansive (i.e., that P is constant), and that 

P = 5500 lbs., r = 6in. = ift., a = 2ft., 

and also that the wheel is to make 120 revolutions per minute, 
i.e., that its inean angular velocity is to be 

oo' = ^^- X 27r, i.e., oa' = 47r " radians" per sec. 

First, required the amount of the resistance P" (constant) 
that there shall be no permanent change of speed, i.e., that the 
angular velocity shall have the same value at the end of a com- 
plete revolution as at the beginning. Since an equation of the 
form of eq, (5) holds good for any range of the motion, let 
* See p. 13 of the "Notes and Examples in Mechanics." 



134 MECHANICS OF ENGINEERING. 

that range be a complete revolution, and we shall have zero as 
the left-hand member ; fPdx = P X 2 f t. = 5500 lbs. X 2 ft., 
or 11,000 foot-pounds (as it may be called); while P" is un- 
known, and instead of lin. arc EF we have a whole circumfer- 
ence of 2 ft. radius, i.e., 4;r ft.; 

.-. = 11,000 - P" X 1 X 3.1416; whence P" = 875 lbs. 
Secondly, required the pi'oper mass to be given to the fly- 
wheel of 2 ft. radius that in the forward stroke (i.e., while the 
crank-pin is describing its upper semicircle) the max. angular 
velocity g?^ shall exceed the minimum go^ by only ^L-g?', assum- 
ing (which is nearly true) that ^{oj^ -j- go^) = go'. There be- 
ing now three unknowns, we require three equations, which 
are, including eq. (5) of § 120, viz.: 

J^^C i-(^m + COn){(^m " ^n) 

=J^^ Pdx - P" X linear arc EF\ (5) 

\{<^m-\- (^n)= co'=4:7r; (7) and go^- go^ = -^gj' = ^n. (8) 

The points n and m are found most easily and with sufficient 
accuracy by a graphic process. * Laying off the dimensions to 
scale, by trial such positions of the crank-pin are found that 
T, the tangential component of the thrust P' produced in the 
connecting-rod by the steam-pressure P (which may be resolved 
into two components, along the connecting-rod and a normal 
to itself) is =(a -^r)P'\ i.e., is = 3500 lbs. These points will 
be n and m (and two others on the lower semicircle). The 
positions of the piston n' and W, corresponding to n and m of 
the crank-pin, are also found graphically in an obvious manner. 
"We thus determine the angle nCm to be 100°, so that linear 
arc EF= \^7t X 2 ft. = ^tt ft., while 

nm' pin' 

/ Pdx = 5500 lbs. X / dx= 5500x^/iW=5500x 0.77 ft., 

n'm' being scaled from the draft. 

Xow substitute from (7) and (8) in (5), and we have, with 
Jcq = 2 ft. (which assumes that the mass of the fly-wheel is con- 
centrated in the rim), 

* See p. 85, "Xotes and Examples," etc. 



KINETICS OF A RIGID BODY. 



125 



{G-^g)X4:X4:7tx^7c = 5500 X C.77 - 875 X ^^, 
which being solved for G (with ff = 32.2 ; since we have used 
the foot and second), gives G = 600.7 lbs. 

The points of max. and min. angular velocity on the back- 
stroke may be found similarly, and their values for the fly- 
wheel as now determined ; they will differ but slightly from 
the Go^ and co^ of the f orwai'd stroke. Professor Cotterill says- 
that the rim of a fly-wheel should never have a max. velocity 
> 80 ft. per sec; and that if made in segments, not more than 
4rO to 50 feet per second. In the present example M^e have for 
the forward stroke, from eqs. (7) and (8), gl;^= 13.2 (7r-measure 
units) per second; i.e., the corresponding velocity of the wheel- 
rim is VJ,^ = co^a = 26.4 feet per second. 

122. Angular Velocity Constant. Fixed Axis. — If co is con- 
stant, the angular acceleration, 6, must be = zero at all times^ 
which requires 2 (mom.) about the axis of 
rotation to be = (eq. (XIY.), § 114). An 
instance of this occurs when the only forces 
acting are the reactions at the bearings on 
the axis, and the body's weight, parallel to 
or intersecting the axis ; the values of these ^"\2i2'EA^y' 
reactions are now to be determined for dif- /' ^ — ^ 
ferent forms of bodies, in various positions fig. 134. 

relatively to the axis. (The opposites and equals of these reac- 
tions, i.e., the forces with which the axis acts upon the bearings, 
are sometimes stated to be due to the " centrifugal forces^'' or 
" centrifugal action," of the revolving body.) 

Take the axis of rotation for Z, then, with = 0, the equa- 
tions of § 114 reduce to 







— 00" Mx ; 

— oo'My ; 
0: . . 



^ moms.x = — GofdMyz ; 
'2 moms.y = -{- a^fdMxz ; 
'2 moms.^ = 0. . . . 



(IX«.) 

(X«..). 

(XIa.) 

(X1I«.) 

(Xllla.) 

(XIYa.) 



126 MECHANICS OF ENGIITEEEIWa. 

For greater convenience, let ns suppose the axes ^ and Y 
{since tlieir position is arbitrary so long as tliey are perpen- 
dicular to each other and to Z) to revolve with the hody in its 
uniform rotation. 

122a. If a homogeneous hody have a plane of symmetry 

■and rotate uniformly about any axis Z perpendicular to that 

plane {intersecting it at 0)^ then the acting forces are equiva- 

. lent to a single force^ = co'^Mp, applied 

at and acting in a groA^ity-line, hut 

directed away from the centre of 

gravity, it is evident that such a 




/^- ^ . - . 

force P :3= ofMp, applied as stated 

^"'- '^''- (see Fig. 135), will satisfy all six con- 

ditions expressed in the foregoing equations, taking ^through 
the centre of gravity, so that x = p. For, from (IX«.), i^must 
■ = afMp, while in each of tlie other summations the left- 
hand member will be zero, since P lies in the axis of ^; and 
as their right-hand meinbers will also be zero for the present 
body (y = 0; and each of the sums fdMyz and fdlfxB is zero, 
since for each term dMy{ -\- z) there is another dMy{^ — z) 
to cancel it ; and similarly, for fdMxz), they also are satisfied ; 
Q.E.D. Hence a single point of support at will suffice to 
maintain the uniform motion of the body, and the pressui-e 
against it will be equal and opposite to P.* 

First Example. — Fig. 136. Supposing (for greater safety) 
that the uniform rotation of 210 revolutions 
per minute of each segment of a fly-wheel is .^--""' 

maintained solely by the tension in the cor- <f - f' 

responding arm, P ; required the value of P * P 
if the segment and arm together weigh -J^- of 
a ton, and the distance of their centre of ^^<^- ^3^- 

gravity from the axis is p = 20 in., i.e., = |- ft. "With the foot- 
ton-second system of units, with g = 32.2, we have 

P = co'Mp = [^ X 27cY X [^ ^ 32.2] X f = 0.83 tons^ 

or 1660 lbs. 

* That is, neglecting gravity. The body's weight, if considered, will 
take its place among the actual forces acting on the body. 



KINETICS OF A RIGID BODY. 



127 




Second Exmnple. — Fig. 137. Suppose the Tiniform rotation 
of the same ily-wheel depends solely on the tension in the rim, 
required its amount. The figure shows the half- 
rim fi-ee, with the two equal tensions, ]r*\ put in at 
the surfaces exposed. Here it is assumed tliat the 
arms exert no tension on the rim. Erom §122a we 
have 2P' = oo'^Mp^ where J/^is themass of the half- p' 
rim, and p its gravity co-ordinate, which may be ob- fig. 137. 
tained approximately by § 26, Problem 1, considering the rim 
as a circular wire, viz., p = 2r -^ rt. 

Let M = (180 lbs.) -^- g, with r = 2 ft. We have then 

P' = i(22)Xl80 ~ 32.2)(4 -^ n) = 1718.0 lbs. 

(In realit}^ neither the arms nor the rim sustain the tensions 
just computed ; in treating the arms we have supposed no duty 
done by the rim, and vice versa. The actual stresses are less, 
and depend on the yielding of the parts. Then, too, we have 
supposed the wheel to take no part in the transmission of mo- 
tion by belting or gearing, which would cause a bending of the 
arms, and have neglected its weight.) 

122b.' If a homogeneous hody have a line of symmetry and 
rotate uniformly dboxit an axis parallel to it [0 being the foot 
of the perpendicular from the centre of gravity on the axis)^ 
then the acting forces are equivalent to a single force P 
= Go'Mp, applied at O and acting in a gravity-line away 

from the centre of gravity. 
Taking the axis X^ tiirough the 
(JM centre of gravity, Z being the 
"H axis of rotation, Fig. 138, while 
j Z' is the line of symmetry, pass 
an auxiliary plane Z' IT' parallel 
to ZY. Then the sum fdMxz 
may be written fdM{p -\- x')z 
which = JfdMz + fdMx'z. 
Fi»- 138. ButfdMz = Mz = 0, since 1 

= 0, and every term dJif{-\- x')z is cancelled by a numerically 




128 



MECHANICS OF ENGINEERIISrG. 



equal term dM{— x')3 of opposite sign. 'H.eme fdMxz = 0. 
Also ydMyz = 0, since each positive product is annulled by an 
equal negative one (from symmetry about Z'). Since, also, 
3/ r= 0, all six conditions in § 122 ai-e satisfied. Q. E. D. 

If the lioiiiogeneous body is any solid of revolution whose 
geometrical axis is jparallel to the axis of rotation, the forego- 
ing is directly applicable. 

122c. If a homogeneous hody revolve uniformly about any 
axis lying in a plane of symmetry, the acting forces are equiv- 
alent to a single force P = oa'Mp, acting parallel to the grav- 
ity-line which is perpendicular to the axis (Z), and away 
from the centre of gravity, its distance from any origin in 
the axis Z being = [fdMxB] -^ Up {the plane ZX being a 
gravity plane). — Fig. 139. From the position of the body we 
have p z=z X, and y = \ hence if a 
value cD^Mp be given to P and it be 
made to act through Z and parallel to 
X, and away from the centre of gravity, 
all the conditions of § 122 are satisfied 
except {Xlla.) and (XIII«.). But 
symmetry about the plane XZ makes 
fdMyz = 0, and satisfies (XJI«.), and 
by placing P at a distance a =fdMxz -^ Mp from along Z 
we satisfy (XIII«.). Q. E. D. 

Example. — A slender, homogeneous, prismatic rod, of length 
= I, is to have a uniform motion, about a ver- q 
tical axis passing through one extremity, 
maintained by a cord-connection with a fixed p 
point in this axis. Fig. 140. Given oo, (p, I, 
(p =: \l cos cp), and F the cross-section of the 
rod, let s = the distance from to any dJif 
of the rod, dM being = Fyds -^ g. The x 
of any dM =^ s cos q); its s = s . sin qj ; 

.\fdMx3 = {Fy -T- g) sin (p cos q) / s^ds 

— -^{Fyl -r- g)r sin cp cos (p = iJifl' sin q> cos <p. 




Fig. 139. 




Fig. 140. 



KINETICS OF A RIGID BODY. 129 

HeiKje a, =^fdMxz -^ Mp, is = |Z sin 9?, and the line of ac- 
tion of P ( = oo'Mp = gl)' (i'VZ -^- ^) ^Z cos q)) is therefore 
higher up than the middle of the rod. Find the intersection 
D of G and the horizontal drawn tlirongh ^ at distance <zfroni 
0. Determine P' by completing the parallelogram GP', at- 
taching the cord so as to nudvc it coincide with P'^ for this will 
satisfy the condition of maintaining the motion, Mdien once be- 
gun, viz., that the acting forces G, and the cord-tension P', 
shall be equivalent to a force P = oo'^Mp, applied horizontally 
through Z at a distance a from 0. 

123. Free Axes. Uniform Rotation. — Referring again to § 122 
and Fig. 134, let us inquire under what circumstances the 
lateral forces, J^^. 1^„ ^^. Y^, with which the bearings pi'ess 
the axis, to maintain the motion, are individually zero, i.e., that 
the hearings are not needed, and may therefore he removed 
(except a sniooth horizontal plane to sustain the body's weight), 
leaving the motion undisturbed like that of a top "asleep." 
For this, not only must 2X and 2 Y both be zero, but also 
(since otherwise X^ and X^ might form a couple, or Y^ and Y^ 
similarly) ^ (moms.)^and 2 (moms.)y must each = zero. The 
necessary peculiar distribution of the body's mass about the 
axis of rotation, then, must be as follows (see the equations of 
§122): 

First, X and y each = 0, i.e., the axis must he a gravity-axis. 

Secondly, fdMyz — 0, KndfdMxz = 0, the origin being any- 
where on Z, the axis of rotation. 

An axis {Z) (of a body) fulfilling these conditions is called 
a Free Axis, and since, if either one of the three Principal Axes 
for the centre of gravity (see § 107) be made an axis of rotation 
(the other two being taken for X and Y), the conditions 
^ = 0, y = 0, fdMxz = 0, and fdMyz = 0, are all satisfied, 
it follows that every rigid hody has at least three free axes, 
which are the Principal Axes * of Inertia of the centre of 
gramit/y at right angles to each other. 

In the case of homogeneous hodies free axes can often be . 
determined by inspection : e.g.. any diameter of a sphere ; any 
* See § 107. p. 104. 



IBO 



MECHANICS OF ENGINEP:RI Nti. 




transverse diameter of a right circular cylinder througli its 
centre of gravity, as well as its geometrical axis; the geomet- 
rical axis of aTi}^ solid of revohition ; etc. 

124. Rotation about an Axis which has a Motion of Translation, 

— Take only the particular case where the moving axis is a 
fjj^ gravity-axis. At any instant, let the 

dM d? velocity and acceleration of axis Z be?; 
and p ; the angular velocity and accelera- 
tion about that axis, oo and 6. Then, since 
'1^ the actual motion of a dM in any dt is 
compounded of its motion of rotation 
about the gravity-axis and the motion of 
translation in common with that axis, 
Fig. 141. we may, in forming the imaginary equiva- 

lent system in Fig. 141, consider each dM as subjected to the 
simultaneous action of dP = dMjp parallel to ^, of the tan- 
gential dT = dMdp^ and of the normal dN ■= dMioopf -^ p 
= Go^dMp. Take ^in the direction of translation, Z (perpen- 
dicular to paper through 0) is the moving gravity-axis ; Y 
perpendicular to both. At any instant we shall have, then, the 
following conditions for the acting forces (remembering that 
/> sin 9? = y.fdMy = 3fy ~ ; etc.) : 

:2X = fdP - fdT sin cp - fdN Qo&<p = Mp; . (1) 

2Y=/dTcos(p -/dJ^siu (p =0;. . (2) 

^ moms.^ =/dTp -fdPy ^ dfdMp" = dl^ ^ OMlcz', (3) 

and three other equations not needed in the following example. 
Example. — A homogeneous solid of revolution rolls {with- 
out slipping) down a rough inclined 
plane. Investigate the motion. Con- 
sidering the \>o^j free., the acting forces 
are G (known) and N and P., the un- ^ ...- 
known normal and tangential compo- ''V 
nents of the action of the plane on the 
roller. If slipping occurs, then P is the ^'*^- ^^^ 

gliding friction due to the pressure Ni^ 156); here, however, it is 
less by hypothesis (perfect rolling). At any instant the four 
unknowns are found by the equations 




KINETICS OF A RIGID BODY. 131 

JSX, i.e., G sin fi - P, = {G-^ g)p ; . (1) 

:SY, i.e., (? cos ^ - if, = ; . . . . (2) 

^ moms.^, i.e., Pa, = OMkz ; • • (^) 

while on account of the perfect rolKng, da = p . . • (4) 

Solving, we have, for the acceleration of translation, 

_2? = ^ sin /i - [1 + (/.■/ - a')l 
(If the body slid wirhont friction, j9 would = p'sin /?.) Hence 
for a cylinder (§ 97), kz" beiiio- = -^a", we have^ = f^ sin /3 ; 
and for a sphere (§ 103) j> = ^g sin /3. 

(If the plane is so steep or so smooth that both rolling and 
slipping occur, then da no longer = _p, but the ratio of i-* to iV 
is known from experiments on sliding friction ; hence there are 
still four equations.) 

The motion of translation being thus found to be uniformly 
accelerated, we may use the equations of § 56. 

Numerically, if a homogeneous solid sphere took 1.20 sec. to descend 
(from rest() 10 ft. along a rough inclined plane, with /? = 30°, did any- 
slipping occur, or was the motion perfect rolling? From p. 54 we have 
s = -|-p/^ that is, 10 = -|- . ^ .^ sin 30° . t^, for perfect rolling; from which 
we obtain i=1.32 seconds, which is >1.20 sec. Hence some slipping 
must have occurred. (The time of descent would have been only 
i = "|/2s-=-^g = 1.114 sec, if the surfaces had been perfectly smooth; and 
the sphere would have had simply a motion of translation, the force P 
being zero). 

N.B. — A hollow sphere would occupy a longer time than a solid one in 
descending the plane (if rough) ; since the ratio kz^a is greater for the 
former. 

125. Parallel-Rod of a Locomotive. — "When the locomotive 
moves uniformly, each dJf of the rod between the two (or 
three) driving-wheels rotates with j \ ; 

uniform velocity about a centre of its 
own on the line j5i>, Fig. 143, and with 
a velocity -y* and radius r common ^._.^ 
to all, and likewise has a horizontal ( * js : , ; m t ) 
■M/l'^/or■7;^ motion ot transhition. Hence (ii.) 

if we inquire what are the reactions P ^^^- ^^3. 

* This velocity is that which the dM ?ias relatively to the frame of the 
locomotive, in a circular path. E.g., if the locomotive (frame; has a velocity 
of GO miles per hour and the radius r is one-third of the radius of the driver, 
then V is 20 miles per hour. 



132 MECHANICS OF ENGI]S'EEEING. 

of its supports, as induced solely T)y its weight and motion^ 
w'.ien in its lowest position (independently of any thrust along 
the rod), we put JSJT of (I.) = 2Y of (II.) (II. shows the 
imaginary equivalent system), and obtain 

2P - G =fdN =fdMo' -^r^iv'-^ r)fdM = Mv' :- r. 

Example. — Let the velocity of translation = 50 miles per 
hour, the radius* of the pins be 18 in. = f ft., and = half that 
of the driving -wheels, while the weight of the rod is 200 lbs. 
With g = 32.2, we must use the foot and second, and obtain 

V = i[60 X 5280 ^ 3600] ft. per second = 36.6; 

while Jf = 200 ^ 32.2 = 200 X .0310 = 6.20 ; 

and finally P = i[200 + 6.2(36.6)^-=- f] = 2868.3 lbs., 

or nearly If tons, about thirty times that due to the weight 
alone. 

126. So far in this chapter the motion has been prescribed, 
and the necessary conditions determined, to be fulfilled by the 
acting forces at any instant. Problems of a converse nature, 
i.e., where the initial state of the body and the acting forces 
are given while the resulting motion is required, are of much 
greater complexity, but of rare occurrence in practice. 

For further study in this direction the reader is referred to Routh's 
"Rigid Dynamics," Rankine's "Applied Mechanics," Sehell's " Theorie 
der Bewegung und der Kraefie," and Worthington's "Dynamics of Rota- 
tion" (this last being a small but clearly written and practical book). 

In Wood's "Analytical Mechanics" will be found the proof of "Euler's 
Equations," which are the basis of the treatment of the gyroscope in the 
book of that name by Gen. J. G. Barnard (Van Nostrand's Science Series, 
No. 90). The article on the gyroscope in Johnson's Cyclopaedia is by 
Gen. Barnard. Perry's "Spinning Tops" is an interesting popular book. 

The Brennan "Monorail Car" (model) is described in the Engineering 
Record for Aug. 31, 1907, p. 226, and depends for its stability (there 
being but one rail under the car) upon two gyroscope wheels revolving 
at 7000 revs, per min. in a vertical plane parallel to the rail. See also 
McClure's Magazine for Dec, 1907, p. 163. 

* Or, rather, the radius of the circular path of the pin-centre, whose 
velocity in this path is 25 miles per hour. 



WORK, ENEKGY, AND POWER. 18b 



CHAPTER YI. 

WORK, ENERGY, AND POWER. 

127. Remark. — These quantities as defined and developed 
in this cliapter, though compounded of the fundamental ideas 
of matter, force, space, and time, enter into theorems of such 
wide application and practical use as to more than justify their 
consideration as separate kinds of quantity, 

128, Work in a XJniform Translation. Definition of Work. — 

Let Fig. 144 represent a rigid body having a motion of trans- 
lation parallel to X, acted on by a yrp^ 
system of forces P^, P^, H^, and Ii^, ^f — ''~'&</<VY^ 
which remain constant.* ("'^y\,...s..-~^ ''•'2_s-..:.-\J_ 

Let s be any distance described by f ^^'"'''sa.., g \ ^ 

the body during its motion ; then ^ JT^+VP^" '"*"" '^'\\P/j 
must be zero (§ 109), i.e., noting that Rf- ^-v,^_^ /\y^^ 
R^ and R^ have negative X com- ^"^^ ^fi 

ponents (the supplements of tlieir ^^'^- '^^'^• 

angles with ^are used), 

P^ cos a-j -J- P^ cos a^ — R^ cos a^ — R^ cos o'^ = ; 

or, multiplying by s and transposing, we have (noting that 
s cos cfj = Sj the projection of s on P^^ that s cos a^ = s^, the 
•projection of s on P^^ and so on), 

P,s, + P,5, - R^s^ + R,s^ {a) 

The projections s^^ s^, etc., may be called the distances de- 
scribed in their respective directions by the forces Pj, P^, etc; 
Pj and P^ having \\\Q)\%^ forward^ since 5, and s^ fall in front 
of the initial position of their points of application ; R^ and R^ 
backward, since s^ and a\ fall behind the initial positions in 
their case, (By forward and backward we refer to the direc- 

* Constant in direction as well as amount. 



134 MECHANICS OF ENGINEERING. 

tion of each force in turn.) The name Work is given to tlie 
product of a force hy the distance described in the direction 
of the force l)y the point of application. If the force moves 
forward (see above), it is called a worhingforce, and is said ta 
do the work (e.g., P^s^ expressed by this prodnct ; while if 
hachioard, it is called a resistance, and is then said to have the 
worh (e.g., R^s^, done upon it, in overcoming it through the 
distance mentioned (it might also be said to have done nega- 
tive work). 

Eq. {a) above, then, proves the theorem that : In a uniform 
translation, the ivorhing forces do an amount of work which 
is entii'ely applied to overcoming the resistances. 

129. Unit of Work. — Since the work of a force is a product 
of force by distance, it may logically be expressed as so many 
foot-pounds, inch-pounds, kilogram-meters, according to the 
system of units employed. The ordinary English unit is the 
foot-pound, or ft.-lb. It is of the same quality as a force- 
moment. 

130. Power. — Work as already defined does not depend on 
the time occupied, i.e., the work P^s^ is the same whether per- 
formed in a long or short time ; but the element of time is of 
so great importance in all the applications of dynamics, as well 
as in such practical commercial matters as water-supply, con- 
sumption of fuel, fatigue of animals, etc., that the rate of worh 
is a consideration both of interest and necessit}'. 

Power is the rate at which work is done, and one of its 
units is one foot-pound per second in English practice ; a larger 
one will be mentioned presently. 

The power exerted by a working force, or expended upon a 
resistance, may be expressed symbolically as 

L = P,s, -■ t, or ^,^3 -^ t, 

in which t is the time occupied in doing the work P^s, or P^s, 
(see Fig. 144) ; or if v^ is the component in the direction of 
the force P^ of the velocity v of the body, we may also write 
L=P\V\, ft. -lbs. per sec {h} 




WOEK, ENERGY, AND POWER. 135 

131. Example. — Fig. 145, shows as a free hody a sledge 
which is being drawn uniformly up 
a rough inclined plane by a cord 
parallel to the plane. Required the 
total power exerted (and expended), 
if the tension in the cord is P^ = 100 
lbs., the weiglit of sledge ^3 = 160 Fig. 145. 

lbs., P = 30°, and the sledge moves 240 ft. each minute. iV^ 
and J^^ are the normal and parallel (i.e., P^ = friction) com- 
ponents of the reaction of the plane on the sledge. From eq. 
(1), § 128, the work done while the sledge advances through 
s = 240 ft. may be obtained either from the working forces, 
which in this case are represented by _Pj alone, or from the 
resistances JR, and P^. Take the former method first. Pro- 
jecting s upon ^j we have s^ = s. 
Hence P,s, or 100 lbs. X 240 ft. = 24,000 ft.-lbs. 
of work done in 60 seconds. That is, the power exerted hy the 
working forces is 

L = P,5, -=- ^ = 400 ft.-lbs. per second. 

As to the other method, we notice that ^g and R^^ are resist- 
ances, since the projections s^= s sin ^, and s^ — s, would fall 
back of their points of application in the initial position, while 
JV is neutral, i.e., is neither a working force nor a resistance, 
since the projection of s upon it is zero. 

From :SX = we have — £,— R3 sin /? + Pi = 0, 

■And from 2 T = (§ 109) JV — Ji, cos /3 =0; 

whence /^, the friction = 20 lbs., and JV = 138.5 lbs. Also,, 
since s, = 5 sin /6f = 240 X i = 120 ft., and s^ = s, = 240 ft., 
we have for the work done upon the resistances (i.e., in over- 
coming them) in 60 seconds 

B^s, + ^,6', = 160 X 120 + 20 X 240 = 24,000 ft.-lbs., 

and the power expended in overcoming resistances, 

L = 24,000 -^ 60 = 400 ft.-lbs. per second, 

as already derived. Or, in words the power exerted by the 
tension in the cord is expended entirely in raising the weight 
a vertical height of 2 feet, and overcoming the friction through 



136 MECHANICS OF EJ^iGlJ^EEKIlJJG. 

a distance of 4 feet along tlie plane, every second ; the motion 
heing a uniform translation. 

132. Horse-Power.- — As an average, a horse can exerts a trac- 
tive effort or pull of 100 lbs., at a uniform pace of 4 ft. per sec- 
ond, for ten hours a day without too great fatigue. This gives 
a power of 400 ft.-lbs. per second ; but Boulton & Watt in 
rating their eiigines, and experimenting with the strong drav- 
liorses of London, fixed upon 550 ft.-lbs. per second, or 33,000 
ft.-lbs. per minute, as a convenient large unit of power. (The 
Prench horse-power, or cheval-vapeur^ is slightly less than the 
English, being 75 kilogrammeters per second, or 32,550 ft.-lbs. 
per minute.) This value for the horse-power is in common 
use. In the example in § 131, then, the power of 400 ft.-lbs, 
per second exerted in raising the weight and overcoming fric- 
tion may be expressed as (400-^550 =) yjof a horse-power. A 
man can work at a rate equal to about J^ of a liorse-powe» , 
with proper intervals for eating and sleeping, 

133. Kinetic Energy. Retarded Translation. — In a retarded 
translation of a rigid body whose mass = Jf, suppose thei-e 
are no working-forces, and that the resistances are constant and 
their resultant is H. (E.g., Fig. 146 shows such a case ; a 

sledge, having an initial velocity c and slid- 

—7 ^^ ing on a rough horizontal plane, is gnidu- 

^^T ally retarded by the friction H.) i?is par- 

allel to the direction of translation (§ 109) 
Fig. 146. and the acceleration is j? = — M -^ M ] 

hence from vdv =-pds we have '; \ ) 

fvdv = - (1 -^ M)fRds. .... (1) 

But the projection of each ds of the motion upon R \q ■= ds 
itself ; i.e. (§ 128), Rds is the work done upon jR, in overcoin 
ing it through the small distance ds, and /Rds is the snm of 
,all such amounts of work throughout any definite portion ol 
the motion. Let the range of motion be between the points 
where the velocity = c, and where it = zero (i.e., the mass 
lias come to rest). "With these limits in eq. (1), (0 and s' be- 
ina; the corresponding 1 M(? C^'t^ . 
limits for s), we have J 2 *Jo ^ ' 



WORK, ENERGY, AND POWEK. 137 

Til at is, in giving up all its velocity c the hody has heen ahle 
to do the worh fRds (this, if R remains constant, reduces to 

JR.s') or its equal —^7—. If, then, bv energy we designate tlie 

ability to ^perform work, we give the name kinetic energy of 

a niuving body to the product of its mass hy half the square 

. fJ'fv^\ 
cf its velocity \~h~"); ,i-e., energy due to motion. 

Example. — If the sledge in Fig. 146 has initially a velocity oi c=^j 
ft./sec. and its weight is G = 322 lbs. (so that its mass in the ft.-lb.-sec. 
system is M = 10) its initial kinetic energy is ilfc^J_2=500 ft.-lbs. If 
the friction or resistance, R, is constant and has a value of 20 lbs., we 
compute s' = 25 ft. (from 500 = Rs') as the distance the sledge will go 
in overcoming this resistance; i.e., in giving up all its kinetic energy. 
If the sledge goes 40 ft. we conclude the average resistance to have been 
only 12.5 lbs.; since 500-^-40 = 12.5. Now suppose R variable, say 

= (20 + 4s) lbs., (s in ft.), and we have 500= / [20 + 4s]ds = 20s'+2s'2; 

.-. s' = 11.6ft. 

134, Work and Kinetic Energy in any Translation. — Let P 

be the resultant of the working forces at any instant, R that 
of the resistances ; they (§ 109) will both M 



u 



act in agravity-line* parallel to the di- <- 

rection of translation. The acceleration C— '_ -§■'. ----*o' 

at any instnnt is ^ = {^A. -^ M) fig. 147. 

=: {^P — R) -T- M\ hence from vdv ^ pds we have 

Mvdv = Pds — Rds (1) 

Integrating between any two points of the motion as and 0' 
where the velocities are 'o^ and v\ we have after transposition 



/ Pds= Rds 4- 



'Mv" Mv, 

. 2 ~ 2 J 



{d) 



But P being the resultant of P^, P^, etc., and R that oi 
^,, ^5,. etc., we may prove, as in § 62. that if dii^. du,, etc.. be 
the respective projections of any ds upon P^. P^, etc., while 
dw^, dv\, etc., are those upon R^, R^, etc., then 

Pds=.P,du^-\-P^du^-\- and Rds=R,dw^-\-R,dw, ; 

and (d) may be rewritten 

* That is, a line passing through the centre of gravity. 



Ig8 MECHANICS OF ENGINEERING. 

£ P^du, +y ' P^du, + . . . . 

P,dw, +y P,dw, + + I -^ ~2^ J 5 (^) 

° 

or, in words : In any translation, a portion of the worh done 
hy the working forces is applied in overcoming the resistances 
lohile the remainder equals the change in the kinetic energy of 
the l)ody. 

It will be noted that the bracket in {e) depends only on the 
initial and final velocities, and not upon any intermediate 
values ; hence, if the initial state is one of rest, and also the 
final, the total change in kinetic enei'gy is zero, and the work 
of the working forces has been entirely expended in the work 
of overcoming the resistances ; but at intermediate stages the 
former exceeds the work so far needed to overcome resistances, 
and this excess is said to be stored in the moving mass ; and as 
the velocity gradually becomes zero, this stored energy becomes 
available for aiding the working forces (which of themselves 
are then insufficient) in overcoming the resistances, and is then 
said to be restored. (The function of a fly-wheel might be 
stated in similar terms, but as that involves rotary motion it 
will be deferred.) 

Work applied in increasing the kinetic energ}' of a body is 
sometimes called " work of inertia," as also the work done by 
a moving body in overcoming resistances, and thereby losing 
speed. 

135. Example of Steam-Hammer. — Let us apply eq. {e) to 
determine the velocity v' attained by a steam-hammer at the 
lower end of its stroke (the initial velocity being = 0), just 
before delivering its blow upon a forging, supposing that 
the steam-pressure i-*^ ^^ "^ stages of the downward stroke is 
given by an indicator. Fig. 148. Weight of moving mass 
is 322 lbs.; .-. J!/' =10 (foot-pound-second system), Z = 1 foot. 
The working forces at any instant are P^^ O ^= 322 lbs.; P^, 
which is variable, but whose values at the s,&wQn equally spaced 



WORK, ENERGY, AND POWER. 



139 



fZs 



joints a, h, c, d, e, f, g, are 800, 900, 900, 800, 600, 500, 450 
lbs., respectively. R^ the exhaust-pressure (16 
lbs. per sq. inch X 20 sq. inches piston-area) = 
320 lbs., is the only resistance, and is constant. 
Hence fi*om eq. {e), since here the projections 
du^^ etc., of any ds upon the respective forces i 
are equal to each other and = ds, 

Pj ds +y P,ds = Rj ds + ^-. (1) 

'The term fP^ds can be obtained approximately * 
^Qj Simpson's Rule, using tlie above values for 
six equal divisions, vi^hich gives 

J^[800 + 4(900 + 800 + 500) 
-f 2(900 -(- 600) + 450] 
.= 725 ft.-lbs. of work. Hence, making all the substitutions. 




we have, since I ds =^1 ft., 



322 X 1 + 725 = 320 X 1 + IMv"; .-. ^Mv" = 727 ft.-lbs. 
of energy to be expended on the forging. (Energy is evi- 
dently expressed in the same kind of unit as work.) We may 
then say that the forging receives a blow of 727 ft.-lbs. 
•energy. The pressure actually felt at the surface of the ham- 
mer varies from instant to instant during the compression of 
the forging and the gradual stopping of the hammer, and 
-depends on the readiness with which the hot metal yields. 

If the mean resistance encountered is R^, and the depth of 
■compression s", we would have (neglecting the force of gravity, 
and noting that now the initial velocity is v', and the final 
zero), from eq. (c), 

^Mv" = Rrr^s"; i.e., R^ = [727 -^ s" (ft.)] lbs. 

E.g., if s" = I of an inch = -gL of a foot, R^ = 43620 lbs., 
and the maximum value of R would probably be about double 
this near the end of the impact. If the anvil also sinks during 
the impact a distance s'", we must substitute s'" -\- s" instead 
•of s" ; this will give a smaller value for ^^. 

* See p. 13 of "Notes and Examples in Mechanics." 



140 



MECHANICS OF ENGINEEKINa. 



By mean value for B, is meant [eq. (c)] that value, B,^^ which 
satisfies the relation 

BJ =f Bds. 

This may be called more explicitly a space-average, to dis- 
tinguish it from a time-average, which might appear in some 
problems, viz,, a value Bt^, to satisfy the relation {t' being the 
duration of the impact) 

I^tnt' = / Bdt, 

and is different from B^. 

From \Mv'^ = T2T ft.-lbs., we have v' = 12.06 ft. per sec, 
whereas for a free fall it would have been 4/2x32.2x1 = 8.03. 
(This example is virtually of the same kind as Prob. 4, § 59, 
differing chiefly in phraseology.) 



136. Pile-Driving.* — The safe load to be placed upon a pile 
after the driving is finished is generally taken us a fraction (from 
^ to ^) of the resistance of the earth to the passage of the pile as 
indicated by the effect of the last few blows of the ram, in ac- 
cordance with the following approximate theory : Toward the 
* end of the driving the resistance B encountered by 

! the pile is nearly constant, and is assumed to be that 

^ met by the I'am at the head of the pile; the distance 

i s' through which the head of the pile sinks as an 

M^^ effect of the last blow is observed. If G, then, is 
the weight of the ram, = 3Ig, and h the height of 
free fall, the velocity due to h, on sti-iking the pile, 
is c = V2gh (§ 52), and we have, from eq. (c), 

iMc\ i.e., GL = f Bds = Bs' . . (1) 

{B being considered constant) ; hence B = Gh -f- s' . 
and the safe load (for ordinary wooden piles), 

P = from \ to ^ oi Gh^s' (2) 

Maj. Sanders recommends |- from experiments made at Fort 



Fig. 149. 



* See also p. 87 of the author's Notes and Examples in Mechanics. 



WORK, ENERGY, AND POWER. 141 

Delaware in 1851; Molesworth, |-; General Barnard, ^, from 
extensive experiments made in Holland. 

Of course from eq. (2), given J*, we can compute s'. 

(Owing to the uncertainty as to how much of the resistance 
H is due to friction of the soil on the sides of the pile, and 
how mucli to the inertia of the soil around the shoe, the more 
elaborate theories of Weisbach and Rankine seem of little 
practical account.) 

137. Example. — In preparing the foundation of a bridge-pier 
it is desired that each pile (placing them 4 ft. apart) shall bear 
safely a load of 72 tons. If the ram weighs one ton, and falls 
12 ft., what should be the effect of the last blow on each pile? 
Using the foot-ton-second systein of units, and Molesworth 
factor \, eq. (2) gives 

s' = 1(1 X 12 -j- 72) = ^- of a foot = J of an inch. 

That is, the pile should be driven until it sinks only J incih 
^nder each of the last few blows. 

138. Kinetic Energy Lost in Inelastic Direct Central Impact. — 

Referring to § 60, and using the same notation as there given, 
we find that if the united kinetic energy possessed by two in- 
elastic bodies after their impact, viz., ^Jf^C -j" i^^^O^ C' hav- 
ing the value {M,c, + M^o^) -^ {M^ + i/,), be deducted from 
the amount before impact, viz., ^M^c^ -{- ^M^e^. the loss of 
Tcinetic energy dxiring iwijpact of two inelastic hodies is * 

An equal amount of energy is also lost by partially elastic 
bodies during the first period of the impact, but is partly re- 
gained in the second. If the bodies were perfectly elastic, we 
would find it wholly regained and the resultant loss zero, from 
the equations of § 60 ; but this is not quite the reality, on 
account of internal vibrations. 

The kinetic energy still remaining in two inelastic hodies 
after impact (they move together as one mass) is 

* See Eng, News, July, 1888, pp. 33 and 34. 



142 



MECHANICS OF ENGINEERING. 



•|^(J/"j + ^^C-, or, after inserting the value of 
C = {M,G, + M^c^) -- {M, + M^, we have 



2 






(2) 



M,| 



Exainjple 1. — The weight ^^ = M^g falls freely 
through a height A, impinging upon a weight 6^, 
= JI/2^, which was initially at rest. After their {in- 
elastic) impact they move on together with the com- 
bined kinetic energy just given in (2), which, since 
Cj and (?25 the velocities before impact, are respectively 
\^'2gh and 0, may be reduced to a simpler form. 
This energy is soon absorbed in overcoming the 
flange-pressure R^ which is proportional (so long as 
the elasticity of the ]-od is not impaired) to the 
elongation 6', as with an ordinary spring. If from 
Fig. 150. previous experiment it is known tliat a force R^ 

produces an elongation «„, then the variable R = (^„ -^ s^)s. 

Keglecting the weight of the two bodies as a working force, 

we now have, from eq. (d), 



+ 4^ 
s ids 



R. 



I.e. 



R, 



0=^ f sds + 









(3) 



When s = s\ i.e., when the masses are (momentarily) at rest 
in the lowest position, the flange-pressure or tensile stress in the 
rod is a maximum, R' = {R^ -^ s^)s', whence s' = R's^ ~- R^; 
and (3) may be written 

M:gh 



or 



R' 

2 

2R. 



s = 



M^gh 



(4) 



(5) 



Eq. (3) gives the final elongation of the rod. and (5) the greatest 
tensile force upon it, provided the elasticity of the rod is not 



W(3KK, ENERGY, AND POWKK. 143 

impaired. The forin ^R's' in (4) may be looked upon as a direct 

integration of / jRds, viz., the mean resistance {^H') multi- 

plied by the whole distance {s') gives the work done in over- 
coming the variable R through the successive ds's,. 

If the elongation is considerable, the working-forces G, and 
G^ cannot be neglected, and would appear in the term-|-(^i 
-f- G^s' in the right-hand members of (3"). (4), and (5). The 
upper end of the rod is firmly fixed, and the rod itself is of 
small mass compared with M^ and M^. 

Exmnple 2. — Two ears, Fig. 151, are connected by an elastic 
chain on a horizontal track. Yelocities before impact (i.e., 

before the stretching of the chain be- ^g^ o ^ci 

gins, by means of which they are l_-[~.,_____^ H 
brought to a common velocity at the M^ Mi 

instant of greatest tension R', and Fig. 151. 

elongation s' of tlie chain) are <?j = g^, and c^ = 0. 

During the stretching, i.e., the first period of the impact, the 
kinetic enei'gy lost by the masses has been expended in sti-etch- 
ing the chain, i.e., in doing the work ^i?V ; hence we may 
write (the elasticity of the chain not being impaired) (see eq. (1) ) 

M,M,e,^ _ 1 _ R, ^^_^.„ 

in which the different symbols have the same meaning as in 
Example 1, in which the rod corresponds to the chain of this 
example. 

In this case the mutual accommodation of velocities is due 
to the presence of the chain, whose stretching corresponds to 
the compression (of the parts in contact) in an ordinary impact. 

In numerical substitution, 32.2 for g requires the use of the 
units foot and second for space and time, while the unit of 
force may be anything convenient. 

139. Work and Energy in Rotary Motion. Axis Fixed. — 

The rigid body being considered free, let an axis through O 
perpendicular to the paper be the axis of rotation, and resolve 
all forces not intersecting the axis into components parallel 
and perpendicular to the axis, and the latter again into com- 
ponents tangent and normal to the circular path of the point 




144 MECHANICS OF ENGINEEKING. 

of application. These tangential com- 
ponents are evidently the only ones 
of the three sets mentioned which 
have moments about the axis, those 
having moments of the same sign as 
00 (the angular velocity at any instant) 
being called working forces^ T^, T„ 
etc. ; those of opposite sign, resist- 
ances, T^', T^', etc.; for when in time 
dt the point of application ^j, of T^, describes the small arc 
ds^ =: a^da, whose projection on T^ is = ds^, this projection 
falls ahead (i.e., in direction of force) of the position of the 
point at the beginning of dt, while the reverse is true for T/. 
From eq. (XIY.), § 114, we have for 6 (angul. accel.) 

6 = ' J , (1) 

which substituted in codco = Qda (from § 110) gives (remem- 
bering that a^doi ■=. ds^, etc.), after integration and transposition, 

T,ds,+J^ TA + etc. 

T^dsi-^j^ T:ds: ^^i^.-\-\_\oo^^i -koo:i\ (2) 

where and n refer to any two (initial and final) positions of 
the rotating body. Eq. (4), § 120, is an example of this. 

Now \oo^I— \Qo^fdMp' =f^dM{GOnpf, which, since go^P 
is the actual velocity of any dM ^i this (final) instant, is nothing 
more than the sum of the amounts of kinetic energy possessed 
at this instant by all the particles of the body ; a similar state- 
ment may be made for \oa^I. (a»o a^nd ojn in radians.) 
Eq. (2) therefore may be put into words as follows : 
Between any two positions of a rigid hody rotating about a 
fixed axis, the worh done hy the working forces is partly used 
in overcoming the resistances, and the remainder in changing 
the kinetic energy of the individual particles. If in any case 
this remainder is negative, the final kinetic energy is less than 
the initial, i.e., the work done by the working forces is less than 
that necessary to overcome the resistances through their respec- 
tive spaces, and the deficiency is made up by the restoring of 



WORK, ENERGY, AND POWER. 145 

some of the initial kinetic energy of the rotating body. A 

moving fly-wheel, then, is a reservoir of kinetic energy. 

Example. — The 668-lb. pulley of p. 104 was found to have a radius 
of gyration of |/7.91 ft., and a moment of inertia about its axis, Z, of 
ikfA;^ = (668-7-3) 7.91. Let us suppose it mounted on a short shaft of 
(ro = ) 2 in. radius (whose 7z may be neglected) supported in proper 
bearings. The pulley and shaft are in contact with nothing except 
the bearings, which offer a friction T/, tangent to outer surface of shaft, 
of 120 lbs. If the pulley has an initial rotary speed of 300 revs./min., 
in how many turns, n, will it be brought to rest? Evidently ^^ = 0, 
while o^Q, = 27r300 -;- 60, =31.41 rads./sec. That is, the initial kinetic 
energy is i^'Mk^ =a(31.41)2(668-h32.2) 7.91, =80,810 ft.-lbs.: and 
the final, zero. T'i' = 120 lbs., constant, and the work done on T/ is, 

T^' I "'ds^' = 120 .n{2n) .lr== 125. Qn ft.-lbs. Hence from eq. (2) we. 

JO 
have = 125.6n + [0-80,810]; i.e., n-643 turns, Ans. 

140. Work of Equivalent Systems the Same. — If two plane 
systems of forces acting on a rigid hody are equivalent (§ 1 oa), 
the aggregate worh done hy either of them during a given slight 
displacement or motion of the hody parallel to their plane is 
the same. By aggregate work is meant what Las ah'eady been 
defined as the sum of the " virtual moments" (§§ 61 to 64), iu 
any sniall displacement of tlie body, viz., the algebraic sum of 
the products, 2 [Pdu), obtained by multiplying eacli foi'ce by 
the pi'ojection {du) of the displacement of (or small space 
described by) its point of application upon the force. (We 
here class resistances as negative working forces.) 

Call the systems A and B] then, if all the forces of B were 
reversed in direction and applied to the body along Avith those 
of A. the compound system would be a balanced system, and 
lience we should have (§ 64), for a small motion parallel to the 
plane of the forces, 

:2{Pdii) = 0, i.e.. 2{Fdu) for A - :S{Fdu) for ^ = 0, 

or . + 2{Pdtc) for A = -{- 2{Fd'u) for B. 

But -f- 2 (Fdu) for A is the aggregate work done by the forces 
of A dui-ing the given motion, and -f 2(Fdu) for B is a 
similar quantity for the forces of B (not reversed) during the 
same small motion if B acted alone. Hence the theorem is 
p)-oved, and could easily be extended to space of three dimen= 
sions. 

10 - 



146 



MECHANICS OP ENGINEERING. 




Fig. 153. 

£>, of the body; a final, n 



141. Relation of Work and Kinetic Energy for any Extended 
Motion of a Rigid Body Parallel to a Plane. — (If at any instant 

any of the forces acting are not 
parallel to the plane mentioned, 
their components lying in or 
parallel to that plane, will be used 
instead, since the other compo- 
nents obviously would be neither 
working forces nor resistances.) 
Fig. 153 shows an initial position, 
and anj' intermediate, as q. The 
forces of the system acting may vary in any manner during 
the motion. 

In this motion each dM describes a curve of its own with 
varying velocity v, tangential acceleration j^t-, ^^^d radius of 
curvature r ; hence in any position ^, an imaginary system JB 
(see Fig. 154), equivalent to the actual system A (at q in Fig. 
153), would be formed by applying to each dM a 
tangential force dT =^ dMpt, and a normal force 
dN' = dMv'^ -V- r. By an infinite number of con- 
secutive small displacements, the body passes from 
o to n. In the small displacement of which q is the 
initial position, each 6? J/^ describes a space ds^ and 
dT does the work dTds = dMvdv, while dJV does the work- 
dJV X = 0. Hence the total work done by £ in the small 
displacement at q would be 



dN 



dT 



dM'v'dv' + dM"n}"dv" -f etc., 



(1) 



including all the dM^& of the body and their respective veloci- 
ties at this instant. 

But the work at q in Fig. 153 by the actual forces (i.e., of 
system A) during the same small displacement must (by § 140) 
be equal to that done by B. hence 

P,du, -f P,du^ + etc. = dM'v'dv' + dM"v"dv" -f etc. (^) 

Now conceive an equation like {g) written out for each of 



WORK, ENERGY, AND POWER. 147 

the small consecutive displacements between positions o and 
yi and corresponding terms to be added ; this will give 

P/hc^ -\- 1 P^du^ -\- etc. 

= dW / v'dii' + dM" / v"dv" + etc. 

= \dM'{v^^ - ^;=) + i^^Jf'X V - <'') + etc. 
The second member may be rewritten so as to give, finally, 

/ P,dit,+ P,du,-^etG.=:S{idMv,')-:S{^dMv,'),{XY.) 

or, in words, the worTi, done hy the acting forces {treating a re- 
sistance as a negative worhing force) between any two posi- 
tions is equal to the gain {or loss) in the aggregate Icinetic 
energy of the particles of the hody hetwee7i the tioo positions. 
To avoid confusion, 2 has been used instead of the sign y in 
one member of (XY.), in which v^ is the final velocity of any 
dM {not the same for all necessarily) and v^ the initial. 

(The same method of proof can be extended to three dimen- 
sions.) 

Since kinetic energy is always essentially positive, if an ex- 
pression for it comes out negative as the solution of a problem, 
some impossible conditions have been imposed. 

142. Work and Kinetic Energy in a Moving Machine. — 

Defining a mechanism or machine as a series of rigid bodies 
jointed or connected together, so that working-forces applied 
to one or more may be the means of overcoming resistances 
occurring anywhere in the system, and also of changing the 
amount of kinetic energy of the moving masses, let us for 
simplicity consider a machine the motions of whose parts are 
all parallel to a plane, and let all the forces acting on any one 
piece, considered free, at auy instant be parallel to the same 
plane. 

Now consider each piece of the machine, or of any series of 
its pieces, as a free body, and write out eq. (XY.) for it be- 
tween any two positions (whatever initial and final positions are 
selected for the first piece, those of the others must be corre- 
sponding initial and corresponding final positions), and it will 



348 



MEOIIA.MCS OF EXGINEEIilNG. 



be found, on adding np corresponding members of these equa- 
tions, that the terms involving tliose components of the mutual 
pressures (between the pieces considered) which are normal 
to the rubbing surfaces at any instant will cancel out, while 
their components tangential to the rubbing surfaces {i.e., fric- 
tion, since if the surfaces are perfectly smooth there can be 
no tangential action) will appear in the algebraic addition as 
resistances multiplied by the distances rubbed through, meas- 
ured on the rubbing surfaces. For example. Fig. 155, where 
one I'otating piece both presses and rubs on another. Let the 
normal pressure between them at A be R^ = P^ ; it is a work- 
ing force for the body of mass M" , but a resistance for M' , 
hence the separate symbols for the numerically equal forces 
(action and reaction). 

Similarly, the f liction at ^ is i?3 = ^Pg ; a resistance for M' , 
a working-force for M" . (In some cases, of course, friction 
may be a resistance for both bodies.) For a small motion, A 
describes tlie small arc AA' abont 0' in dealing with M\ but 
for M" it describes the arc AA" about 0" . A' A" being 
parallel to the surface of contact AD, while AB is perpen- 




FiG. 156. 



Fig. 157. 



Fig. 155. 

dicular to A' A" . In Figs. 156 and 15Y we see M' and M" 
free, and their corresponding small rotations indicated. During 
these motions the kinetic energy (K. E.) of each mass has 
clianged by amounts <f(K. E.)j,f/ and (i(K. E.)j/// respectively, and 
hence eq. (XY.) gives, for each free body in turn, 

P\a^' - R,AB - R,A^ = di^. E.)^. . (1) 

- RW+ P.AB + P^JJ^ = d(K. 'E.)m". . (2) 

Now add (1) and (2), member to member, remembering that 
P^ = P^ and P^ = P^ = P^ = friction, and we have 



P,aa' - F,A'A" - R^jb" = d{K. E.)^' + d{lL E.)m", (3) 



WUllK, ENERGY, AND POWER. 149 

in which the mutual actions of M' and M" do not appear, 
except the friction, the work done in overcoTning which, when 
the t'loo hodies are thus considered collectively, is the product 
of the friction hy the distance A' A" of actual nibbing meas- 
tired on the rubbing sttrface. For any number of pieces, then, 
considered free collectively, the assertion made at the beginning 
of this article is true, since any finite motion consists of an 
infinite number of small motions to each one of wliich an equa- 
tion like (3) is applicable. 

Summing the corresponding terms of all such equations, we 
have 

f" P,du, -{-fF,du,+ etc. = :^(K.E.),,- :^(K. E.)o.(XYI.) 

This is of the same form as (XY.), but instead of applying to a 
single rigid body, deals with any assemblage of rigid pai-ts 
forming a machine, or any part of a machine (a similar proof 
will apply to thi-ee dimensions of space); but it must be remem- 
bered that it excludes all the mutual actions* of the pieces con- 
sidered except friction, which is to l)e introduced in the manner 
just illustrated. A flexible inextensible cord may be considered 
as made up of a great number of short rigid bodies jointed 
M'ithout friction, and hence may form part of a machine with- 
out vitiating the truth of (XVI.). 

^(K. E.)„ signifies the sum obtained by adding the amounts 
of kinetic energy {^dMv^ for each elementary mass) possessed 
by all the particles of all the rigid bodies at their final posi- 
tions ; ^(K. E.)„, a similar sum at their initial positions. For 
example, the K. E. of a rigid body having amotion of transla- 
tion of velocity -y, =^ ^vfdM =^ ^Mv^ ', that of a rigid body 
having an angular velocity go about a fixed axis Z, = ^oo'^Iz 
(§ 139) ; while, if it has an angular velocity w about a gravity- 
axis Z, which has a velocity Vz of translation at right angles to 
itself, the (K. E.) at this instant may be j)roved to be (§ 143) 

the sum of the amounts due to the two motions separately. 

* These mutual actions consist only of actions by contact (pressure, ruo, 
etc.) . No magnetic or electrical attractions or repulsions are here considered. 




150 MECHANICS OF ENGINEERING. 

143. K. E. of Combined Rotation and Translation. — The last 

statement may be thus proved. Fig. 158. 
At a given instant the velocity of any dJf is 
V, the diagonal formed on the velocity Vz of 
translation, and the rotaiy velocity oop rela- 
tively to the moving gravity-axis Z (per- 
pendicnlar to paper) (see § Yl), 

Fig. 158. i-©., v' = Vz + {oopY — ^{Gop)vz COS 9? ; 

hence vv^e have K. E., at tliis instant, 

= f^dMv' = \v^fdM + WfdMp" - GovzfdMp cos ^, 

but p cos q) ^=:y, and fdMy = My = 0, since Z is a gravity- 
axis, 

.-. K. E. = iMvz' + WIz- Q. E. D. 

It is interesting to notice that the K. E. due to rotation, viz., 
\go^Tz = \M{w]tY^ is the same as if the whole mass were con- 
centrated in a point, line, or thin shell, at a distance ^,-the 
radius of gyration, from the axis. 

Example. — A solid homogeneous sphere of radius r = 6 in. and weight 
= 322 lbs. is rolling down an incline. At a certain instant the velocity 
of its centre is 10 ft. per sec. and hence, i'/ no slipping occurs, its angular 
velocity about its centre is co, ==Vz-^r, =10-7- J, =20 radians/ sec. Con- 
sequently, at this instant (see § 103, p. 102) its total kinetic energy is 
i(322 4_32.2)[(10)2 + (20)2 . f(i)2] = 700 ft.-lbs. 

144. Example of a Machine in Operation. — Fig. 159. Con- 
sider the four consecutive moving masses, M\ M'\ M"\ and 
M'^^ (being tlie piston ; connecting-rod ; fly-wheel, crank, drum, 
and chain ; and weiglit on inclined plane) as free, collectively. 
Let us apply eq. (XYI.), the initial and final positions being 
taken when the crank-pin is at its dead-points o and n\ i.e., we 
deal with the progress of the pieces made while the crank-pin 
describes its upper semicircle. Remembering that the mutual 
actions between any two of these four masses can be left out 
of account (except friction), the only forces to be put in are 
the actions of other bodies on each one of these four, and are 



WORK, ENERGY, AND POWER. 151 

shown in the figure. The only mutual friction considered will 
be at the crank- pin, and if tliis as an average — F'\ the work 
done on it between o and n = F"7tr" ^ where r" = radius of 
cranlv-pin. The work done by jP^ the effective steam- pressure 
(let it be constant) daring this period is = I^^l' ; that done in 
overcoming J^j, the friction between piston and cylinder, = I^^l' ; 
that done upo?i the weight G'oi connecting-rod is cancelled by 
the work done by it in the descent following ; the work done 




Fig. 159, 

upon G''', = G'^Tta sin /?, where a = radius of drum ; that 
upon the friction i^^, = J^^rra. The pressures JV, W, N'^, and 
N'", and weights G' and G'", are neutral, i.e., do no work either 
positive or negative. Hence the left-hand member of (XVI.) 
becomes, between o and n, 

P,V - F,V - F"7tr" - G'^Tta sin /? - Fjta, . . (1) 

provided the respective distances are actually described by 
these forces, i.e., if the masses have sufficient initial kinetic 
energy to carry the crank-pin beyond the point of minimum 
velocity, with the aid of the working force P^^ whose effect is 
small up to that instant. 

As for the total initial kinetic energy, i.e., ^(K. £.)„, lei; us 
express it in terms of the velocity of crank-pin at o, viz., Y^. 
The (K.E.)„ of M' is nothing ; that of M" , which at this in- 
stant is rotating about its right extremity {fixed ioix \\\& instant) 
with angular velocity oo" = F„ ^ l'\ is \(^"^1^' \ that oiM'" 
= \o!}"'^I^'\ in which oo'" = V^^r; that of M''' (translation) 
^ iJf ^X''"? in which v;^ = {a-^r) V,. 2(K. E.)„ is expressed 



152 



MECHANICS OF ENGINEERING. 



in a corresponding manner with F^ (final velocity of crank-pin) 
instead of Y^. Hence the right-hand member of (XYI.) will 
give (potting the radius of gyration of Jf about 0" = Jc", 
and that of Jf about G = Jc) 

i( K' - F;)[j/-|^ + M^-~ +M-y~']. . . (2) 

By writing (1) = (2), we have an equation of condition, capa- 
ble of solution for any one unknown quantity, to be satisfied 
for the extent of motion considered. It is understood that the 
chain is always taut, and. that its weight and mass are neg- 
lected. 

145. Numerical Case of the Foregoing. — (Foot-pound-second 
system of units for space, force, and time ; this 'requires g 
= 32.2.) 

Suppose the following data : 



Feet. 


Lbs. 


Lbs. 


Mass Units. 


V = 2.0 
I" = 4.0 
a = 1.5 
r = 1.0 
k = 1.8 
k" = 2.3 
r" = 0.1 


Pi = 

Fi = 
F" (av'ge) = 

F,= 


6000 
200 
400 
300 


0' = 60 
G" = 50 
0'" = 400 
0'^ = 3220 


(and .-.) 

M' = 1.86 
M" = 1.55 
M'" = 12.4' 
M^^ = 100.0 


Also let Fo = 4 


ft. per sec; /:/=30'' 



Denote (1) by TTand the large bracket in (2) by M (this l)y 
some is called* the total mass ^'- reduced''^ to the crank-pin). 
Putting (1) = (2) we have, solving for the unknown Vn, 



K = 



2Tf 



i-v:. 



(3) 



For above values, 

W = 12,000 - 400 - 125.T - Y590.0 - 141 Y.3 
= 2467 foot-pounds ; 
while ^ = 0.5 + 40.3 -f 225.0 = 265.8 mass-units; 
whence F„ = 4/18.56 + 16 = VsU^ = 5.88 ft. per second. 
As to whether the crank-pin actually reaches the dead-point 
n, requires separate investigations to see whether F becomes 
zero or negative between o and n (a negative value is inad- 



WORK, ENERGY, AND POWER. 153 

inissible, since a reversal of direction Implies a different 
value for W), i.e., whether the proposed extent of motion is 
realized ; and these are made by assigning some othei' inter- 
mediate position 771, as a final one, and computing F^, remem- 
bering that when m is not a dead-point the (K. E.),^ of M' is not 
zero, and must be expressed in terms of F^, ;uid that the 
(K. E.)to of the connecting-rod J/'''^raust be obtained from § 143. 

146. Eegulation of Machines. — As already illustrated in 
several exauiples (§ 121), a fly-wheel of sufficient weight and 
radius may prevent too great fluctuation of speed in a single 
•stroke of an engine ; but to prevent a permanent change, which 
must occur if the work of the working force or forces (such as 
the steam-pressure on a piston, or water-impulse in a turbine) 
exceeds for several successive strokes or revolutions the work 
required to overcome resistances (such as friction, gravity, re- 
sistance at the teeth of saws, etc., etc.) through their respective 
spaces, automatic governors are employed to diminish the 
working force, or the distance tlu-ough wliich it acts per stroke, 
until the normal speed is restored ; or vice versa, if the speed 
slackens, as when new resistances are temporarily brought into 
play. Hence when several successive periods, strokes (or other 
-cycle), are considered, the kinetic energy of the moving parts 
will disappear from eq. (XYI.), leaving it in this form : 

work of' worhing-forces = work done upon resistances. 

147. Power of Motors. — In a mill Avhere the same number of 
machines are run continuously at a constant speed proper for 
theii- work, turning out per hour tlie same number of barrels 
of flour, feet of himber, or other coujmodity, the motor (e.g., 
a steam-engine, or turbine) woi'ks at a constant rate, i.e., de- 
velops a definite horse-power (H.P.), which is thus found in 
the case of reciprocating steam-engines (doubie-actingj, 

II.P. = total mean effective \ l distance in feet ] 

steam-pressure on I X •< travelled by pis- I ~- 550, 
piston in lbs. ) ( ton per second. ) 

i.e., the work (in ft.-lbs) done per second by the working force 



164 MECHANICS OF ENGINEERING. 

divided by 550 (see § 132). The total effective pressure at auj" 
instant is the excess of the forward over the back-pressure^ 
and by its mean vakie (since steam is nsnally used expansively) 
is meant such a vahie^' as, multiplied by the length of stroke- 
I, shall give 



P'l=.J Pdx, 



where P is the variable effective pressure and dx an elemenfc 
of its path. If u is tlie number of strokes per second, we may 
also write {foot-jpound- second system) 



H.P. = P'lu -^ 550 = f 



Pdx 



u -^ 550. (XYII.)' 



Yery often the number of revolutions _^er minute, m, of th&: 
crank is given, and then 

H.P. = P' (lbs.) X 2Z (feet) X m ^ 33,000. 

II P= area of piston we may also write P' ^Pp', where j?' 
is the mean effective steam-pressure per unit of area. Evi- 
dently, to obtain P' in lbs., we multiply i^in sq. in. byj?' ia 
lbs. per sq. in., or P m sq. ft. hj p' in lbs. per sq. foot ; the 
former is customary, p^ in practice is obtained by measurements 
and computations from " indicator- cards " (see § 152, p. 159, 
where the value of P' is found by Simpson's Rule) ; or P'7, i.e., 

/ Pdx, may be computed theoretically as in § 59, Problem 4.. 

The power as thus found is expended in overcoming the- 
friction of all moving parts (which is sometimes a large item), 
and the resistances peculiar to the kind of work done by the ma- 
chines. The work periodically stored in the increased kinetic 
energy of the moving masses is restored as they periodically 
resume their minimum velocities. 

Example. — In the steady running of a (reciprocating) steam-engine 
operating a mill, the value of the mean total effective pressure on the 
piston is P' = 16,800 lbs. and the radius of the circle described by the 
crank-pin is 10 in. (so that the length of stroke is 1 = 20 inches). The^ 
fly-wheel turns at rate of 330 revs./min. Find the horse-power developed.. 
Substituting, we find H.P. = [16,800X2 XffXSSO]-^ 33,000 = 560 H.P. 



WORK, ENEIJGY, AND POWEU. 155 

148. Potential Energy. — There are other ways in which work 
or energy is stored and then restored, as follows : 

First. In raising a weight G through a height h, an amount 
of work = Gh is done ujyon G, as a resistance, and if at any 
subsequent time the weight is allowed to descend through the 
same vertical distance Ti (the form of path is of no account), G, 
now a worMng force, does the work Gh, and thus in aiding the 
motor repays, or restores, the Gh expended by tlie motor in 
raising it. If h is the vertical height tlirough which the centre 
of gravity rises and sinks periodically in the motion of the 
machine, the force G may be left out of account in i-eckoning 
the expenditure of the motor's work, and the body when at its 
liighest point is said to possess an amount Gh of potential 
energy, i.e., energy of jposition, since it is capable of doing the 
work Gh in sinking through its vertical range of motion. 

Second. So far, all bodies considered have been by express 
stipulation rigid, i.e., incapable of changing shape. To see 
the effect of a lack of rigidity as affecting the principle of 

work and energy in machines, -^---.JC^^^^^^^rr^ A 

take the simple case in Fig. 160. ^ Y^^'^''^^^ ? 

A helical spring at a given in- ^f---^^;^;^;|^^^^j~^ 
Btant is acted on at each end by f'^'^'i r ti>?' ^ 

a force jP in an axial direction ' ' ' / j 

(they are equal, supposing the Fig. leo. 

mass of the spring small). As the machine operates of which 
it is a member, it moves to a new consecutive position J^, 
suffering a further elongation dX in its length (if F is increas- 
ing). P on the right, a woi'king force, does the work Pdx'', 
how is this expended ? ^ on the left has the work Pdx done 
upon it, and the mass is too small to absorb kinetic energy or 
to bring its weight into consideration. The remainder, Pd'x' 
— Pdx = Pdx, is expended in stretching the spring an addi- 
tional amount dX, and is capable of restoration if the spring 
retains its elasticity. Hence the work done in changing the 
form of bodies if they are elastic is said to be stored in the 
form of potential energy. Tliat is, in the operation of ma- 
chines, the name potential energy is also given to the energy 
stored and restored periodically in the changing and regaining 
of form of elastic bodies. 



156 MECHANICS OF ENGINEERING. 

EKatiif)le. — A given helical spring, when held stretched s'=J ft. beyond 
its "natural" (or unstrained) length, exerts a pull of i2' = 1200 lbs. 
at its two ends; and the "potential energy" residing in it is — mean forceX 
distance, =^R's\ = (i) 1200 (^), =300 ft.-lbs. If such a stretched 
spring be placed on a car of 644 lbs. weight on a level track and properly 
connected with a driving-wheel, which does not slip on the track, its 
recovery of natural length may be made the means of starting the car 
into motion and causing it to attain a final velocity of v = 5A7 ft. /sec. 
(if no friction is met with); from i(644-f-32.2)i;2 = 300. 

149. Other Forms of Energy. — Numerous experiments witli 
various kinds of apparatus have proved that for every 7Y8 
(about) ft.-lbs. of work spent in overcoming friction, one British 
unit of heat is produced (viz., tlie quantity of lieat necessary to 
raise tlie temperature of one pound of water from 32° to 33° 
Fahrenheit); while from converse experiments, in which the 
amount of heat used in operating a steam-engine was all carefully 
estimated, the disappearance of a certain portion of it could only 
be accounted for by assuming that it liad been converted into 
work at the same rate of (about) 778 ft.-lbs. of vs^ork to each 
unit of heat (or 427 kilogrammetres to each French unit of 
lieat). This number 778 or 427, according to the system of 
units employed, is called the Mechanical Equivalent of Htot 

Heat then is energy, and is supposed to be of the kinetic 
form due to tJie rapid motion or vibration of the molecules of 
a substance. A similar agitation among the molecules of the 
(hypothetical) ether diffused through space is supposed to pro- 
duce the phenomena of light, electricity, and magnetism. 
Chemical action being also considered a method of transform- 
ing energy (its possible future occurrence as in the case of coal 
and oxygen being called potential energy), the well-known 
doctrine of the Conser nation of Energy^ in accordance with 
which energy is indestructible, and the doing of work is simply 
the conversion of one or more kinds of energy into equivalent 
amounts of others, is now an accepted hypothesis of physics. 

Work consumed in friction, though practically lost, still re- 
mains in the universe as heat, electricity, or some other subtile 
form of energy. 

150. Power Required for Individual Machines. Dynamome- 
ters of Transmission. — If a machine is driven by an endless 
belt from the main-shaft, A^ Fig. 161, being the driving-pulley 



WORK, ENERGY, AND POWER. 



167 




Fig. lUl. 



on the machine, the working force which drives the machine,, 
in other words tlie " grip" with which the 
belt takes hold of the pulley tangentially^ 
= /^ — P' ^ P and P' being the tensions 
in the "driving" and ''following" sides of 
the belt respectively. The belt is supposed 
not to slip on the pulley. If v is the ve- 
locity of the pulley -circumference, the 
work expended on the machine per second, i.e., \\\q lyower, is 

L = (P-POv, ft.-lbs. per sec. .... (1) 

To measure the force {P — P')^ an apparatus called a Dy- 
nainometer of Transmission may be placed between the main 
shaft and the machine, and the belt made to pass through it in 
such a way as to measure the tensions P and P' ^ or princi- 
pally their difference, without meeting any resistance in so do- 
ing ; that is, the power is transmitted^ not absorbed, by the 
apparatus. One invention for this purpose (mentioned in the 
Journal of the FranMin Institute some years ago) is shown 

{in principle) in Fig. 162. A ver- 
tical plate carrying four pulleys and 
a scale-pan is first balanced on the 
pivot C. The belt being then ad- 
justed,- as shown, and the power 
turned on, a sufficient weight G is 
placed in the scale-pan to balance 
Fig. 163. tlie plate again, for whose equilib- 

rium we must have Gh = Pa — P'a, since the P and P' on 
the right have action-lines passing through C. The velocity of 
belt, V, is obtained by a simple counting device. Hence 
(P —P') and V become known, and .'. L from (1). 

Many other forms of transmission-dynamometers are in use, 
some applicable whether the machine is driven by belting or 
gearing from the main shaft. Emerson's Hydrodynamics de- 
scribes his own invention* on p. 283, and gives results of meas- 
m-ements with it ; e.g., at Lowell, Mass., the power required 
to drive 112 looms, weaving 3 6 -inch sheetings, No. 20 yarn, 
60 threads to the inch, speed 130 picks to the minute, was 
found to be 16 H.P., i.e., \- H.P., to each loom (p. 335). 

* Prof . Flather's '^Dynamometers" is a standard book (1907). 




158 



MECHANICS OF ENGINEERING. 



Example. — The endless belt connecting the pulley (running at n=180 
revs./min., with a radius of r = 2 ft.) of an engine shaft with that of a 
planing machine is led over the idle pulleys of the apparatus in Fig. 162, 
as there shown (engine pulley on left, and that of machine on right; 
but neither shown in figure). To balance the plate in position shown 
(with a = 2 ft. and 6 = 4 ft.) is found to require a weight G = 210 lbs. 
We have, therefore, from {P-P')a = Gb, P-P' = 210X4 --2 = 420 lbs. 
as the net working force operating the machine; while the velocity of 
the belt is v, =n2;rr, =(180-^60) 2;r2= 18.85 ft. /sec. Hence the 
power transmitted through the dynamometer of transmission is L, 
= {P-P')v, =420X18.85 = 79,170 ft.-lbs. per sec, or 14.4 H.P 

151. Dynamometers of Absorption. — These are so named 
.since they furnish in themselves the resistance (friction or a 
weight) in tlie overcoming (or raising) of which the power is 
expended or absorbed. Of these the Prony Friction Brake 
is the most common, and is nsed for measuring the power 
developed by a given motor (e.g., a steam-engine or turbine) 
mot absorbed in the friction of the motor itself. Fig:. 163 




«hows one fitted to a vertical pulley driven by the motor. By 
tightening the bolt B^ the velocity i) of pulley-rim may be 
made constant at any desired value (within certain limits) by 
the consequent friction, -y is measni-ed by a counting appara- 
tus, while the friction (or tangential components of action be- 
tween pulley and brake), = I\ becomes known by noting the 
weight G which must be placed in the scale pan to balance the 
arm between the checks; then with G^''= weight of brake and 
h' =tlie horizontal distance of its center of gravity from the 
center of pulley, we have, for the equilibrium of the brake 
(moments about pulley center), 

Fa=Gb + GV', (1) 

and the work done on F per unit of time, or power, is 

L=i^y, ft.-lbs. per sec (2) 

(In case the pulley is horizontal, a bell-crank must be inter- 
posed between the arm and the scale-pan.) 



WORK, ENERGY, AND POWER. 



159 



Example. — A vertical pulley of a = 2 ft. radius and run by a turbine 
water-wheel, is gripped by a Prony brake, as in Fig. 163, with arm 
fe = 4 ft. 9 in. A load of G = lQO lbs. is placed in the scale pan, the water 
turned on, and the bolt B screwed up until the friction F of pulley-rim 
on brake is just sufficient to lift the weight and hold the brake in equi- 
librium Weight of brake is (r' = 40 lbs., with centre of gravity 6' = 1.5 ft. 
on right of pulley centre. The speed to which the pulley now adjusts 
itself is at rate of 210 revs./min. The friction is F ={Gb + Gb')^a = 
(160X4.75+40X1.5)-h2 = 410 lbs.; the velocity of pulley-rim is v = 
(210^60) 27rX2 = 44 ft. /sec; hence the power developed is F'?; = 410X44 
= 18,040 ft.-lbs. per sec. ; or 32.8 H.P. 

Note. — For an account of various modern designs of absorption and 
transmission dynamometers, the reader is referred to Prof. Flather's 
book, already mentioned in the foot-note on p. 157. This is a recent 
and a standard work. In the "Notes and Examples in Mechanics" by 
the present writer, brief descriptions are given (pp. 96 and 97) of the 
Appold and the Carpentier dynamometers of absorption, with the theory 
of the same; both of these are "automatic" or "self-adjusting." 

It must be carefully noted by the student that in the absorption dyna- 
mometer, which for purposes of test temporarily takes the place of useful 
machines, the power is absorbed and converted into heat, necessitating 
cooling devices if the parts are to run smoothly and lubricants are to 
remain unaffected; whereas in the dynamometer of transmission the 
power simply passes through without heating effect. 

152, The Indicator, used with steam and other fluid engines, 
is a special kind of dynamometer in which the automatic mo- 
tion of a pencil describes a curve 
on paper whose ordinates are 
proportional to the fluid pres- 
sures exerted in the cylinder at 
successive points of the stroke. 
Thus, Fig. 164, the back-pres- 
sure being constant and = P^, fig. lu. 

the ordinates P^, P^, etc.. represent the effective pressures at 
equally spaced points of division. The mean effective pressure 
P' (see § 147) is, for this figure, by Simpson's Rule (six equal 
spaces), 

P' = tV[^o + 4(P, + P3 + P.) + 2(P. + P.) + Pe]. 

This gives a near approximation. The power is now found by 
§147^ 

153. Mechanical Efficiency of a Steam or Vapor Engine (gas, 
petroleum, gasoline, or alcohol vapor, etc.). By the term 
*' mechanical efficiency " is meant the ratio of the power obtained 



< -- 




l 




! 






^\/ 


fk< 






P1 


P2 


P3 


P4 


Pa 


T 1 










Pe 












l^b 




2ERC 


3 LINE 




X 



160 MECHANICS OF ENGINEERING. 

at tlie rim of tlie pnlley or fly-wheel on the main shaft of the 
engine (where it would be connected with machinery to be 
operated or where in a test the resistance of brake -friction 
wonld be overcome) to the power exerted directly on the piston 
of the engine by the pressure of the fluid concerned. This 
latter item becomes known through the use of the ' ' indicator ' ' 
(see preceding paragraph) and is hence often called the 'indi- 
cated 'power ; " the power spent on friction provided by a Prony 
brake, for testing purposes, being called the " brake-power.^^ 

Example. — If from indicator-cards the value of P', or total mean 
effective pressure on the piston of an engine, is found to be 12,000 lbs., 
the piston speed being at the (mean) rate of 6 ft. per sec, the "indi- 
cated power" of the engine is = 72,000 ft.-lbs./sec. Now, when the 
engine is running under these same conditions of pressure and speed, 
if it is found by the use of a Prony friction brake that the power spent 
on brake friction consists of overcoming a friction of 6000 lbs. through 
10 ft. each second, and that therefore the power obtained at the brake, 
or "brake power," is equal only to 60,000 ft.-lbs./sec, the mechanical 
efficiency of the engine (in this test) is 60,000-^72,000, =0.833 or 83 J per 
cent. In other words, 16t per cent of the power exerted by the fluid 
pressure on the piston, or "indicated power," is lost in the overcoming 
of the friction of the engine itself, i.e., among the moving parts situated 
between the piston and the rim of the test pulley. 

153a. Efficiency of Power Transmission. — In transmitting 
power through a long line of shaftmg, or by ropes or belts, or 
water in pipes, or by electric current, the efficiency is the ratio 
of the power put in at the sending station to that obtained at 
at the receiving station. For example, 

Example. — An engine exerts power at the rateof (say) 600,000 ft.-lbs./sec, 
in running a "dynamo" at the sending or power station. The electric 
current so generated is conducted 60 miles through wires to a receiving 
station, where by operating an electric motor it enables a pulley to be 
run within a Prony brake from whose indications it is found that a power 
of 360,000 ft.-lbs./sec. is there obtainable. Hence the efficiency of 
transmission is 360,000-^600,000, =60 per cent. 

154. Boat-Rowing. —x^'ig. 166. During the stroke proper, 
let /* = mean pressure on one oar-handle ; hence the pressures 
on the foot-rest are 2P, resistances. Let J!f"= mass of boat 
and load, v^ and Vn its velocities at beginning and end of stroke. 
Pj = pressures between oar-blade and water. Ji = mean re- 
sistance of water to the boat's passage at this (mean) speed. 
These are the only (horizontal) forces to be considered as act- 
ing on the boat and two oars, considered free collectively. 
During the stroke the boat describes the space s^ = CD, the 
oar-liandle tlie space s„ = AB, wliile tlie oar-blade slips back- 



WORK, ENERGY, AND POWER. 



161 



ward through the small space (the " slip") = s^ (average). 
Hence by eq. (XYI.), § 142, 

i.e., 2P{s,-s,)=2FxAJi=2Fs =Bs,+2P,s,-\- iMiv^'-v,'); 

or, in words, the product of the oar-handle pressures into the 
distance described by them measured on the hoat, i.e., the work- 
done by these pressures relatively to the hoat, is entirely ac- 
counted for in the work of slip and of liquid-resistance, and in^ 

■■..si>^ 




Fig. 166. 

creasing the kinetic energy of the mass. (The useless work 
due to slip is inevitable in all paddle or screw propulsion, as 
well as a certain amount lost in machine-friction, not considered 
in the present problem.) During the '' recover" the velocity 
decreases again to v^. (See example on p. 9T, Notes, etc.) 

155. Examples. — 1. What work is done* on a level traci^, in 
bringing up the velocity of a train weighing 200 tons, from 
zero to 30 miles per hour, if the total frictional resistance (at 
any velocity, say) is 10 lbs. per ton, and if the change of speed 
is accomplished in a length of 3000 feet ? 

{Foot-ton-second system.) 30 miles per hour = 44 ft. per 
sec. The mass 

- 200 -^ 32.2 = 6.2 ; 
.'. the change in kinetic energy, 

(= Wv-' -iM X 0% 
= i(6.2) X 44* = 6001.6 ft.-tons. 

* That is, what work is done by the pull, or tension, P, in the draw-bar 
between the locomotive and the "tender." 



162 MECHANICS OF ENGlNEEKliMG. 

The work done in overcoming friction = Fs, i.e., 

= 200 X 10 X 3000 = 6,000,000 ft.-lbs. = 3000 ft.-tons ; 
.-. total work = 6001.6 + 3000 = 9001.6 ft.-tons. 

(If the track were an up-grade, 1 in 100 say, the item of 
200 X 30 = 6000 ft.-tons would be added.) 

Exmnjple 2. — Required the rate of work, or power, in Ex- 
ample 1. The power is variable, depending on the velocity of 
the train at any instant. Assume the motion to be uniformly 
accelerated, then the working force is constant ; call it P. 
The acceleration (§ 56) will be ^='y'-=- 2^=1936-^6000 = 0.322 
ft. per sq. sec; and since P — J^= Mjp^ we have 

P = 1 ton + (200 -=- 32.2) X 0.322 = 3 tons, 

which is 6000 -^ 200 = 30 lbs. per ton of train, of which 20 is 
due to its inertia, since when the speed becomes uniform the 
work of the engine is expended on friction alone. 

Hence when the velocity is 44 ft. per sec, the engine is 
working at the rate of P'o = 264,000 ft.-lbs. per sec, i.e., at the 
rate of 480 H. P.; 

At i of 3000 ft. from the start, at the rate of 240 H. P., half 
as much ; 

At a uniform speed of 30 miles an hour the power would be 
simply 1 X 44 = 44 ft. -tons per sec. = 160 H. P. 

Example 3. — The resistance offered by still water to the 
passage of a certain steamer at 10 knots an hour is 15,000 lbs. 
What power must be developed by its engines, at this uniform 
speed, considering no loss in " slip" nor in friction of ma- 
chinery ? A71S. 461 H. P. 

Example 4, — Same as 3, except that the speed is to be 15 
knots (i.e., nautical miles ; each = 6086 feet) an hour, assum- 
ing that the resistances are as the square of the speed (approxi- 
mately true). Ans. 1556 H. P. 

Example 5. — Same as 3, except that 12^ of the power is ab- 
soi-bed in the " slip" (i.e., in pushing aside and backwards the 
water acted on by the screw or paddle), and 8^ in friction of 
machinery. Ans. 576 H. P. 

Example 6. — In Example 3, if the crank-shaft makes 60 



"WORK, ENERGY, AND POWEE. 163 

revolutions per minute, the crank-pin describing a circle of 15 
incbes radius, required the average* value of the tangential 
component of the thrust (or pull) of the connecting-rod against 
the crank-pin. Ans. 26890 lbs. 

Example 7. — A solid sphere of cast-iron is rolling up an in- 
cline of 30°, and at a certain instant its centre has a velocity of 
36 inches per second. Neglecting friction of all kinds, how 
much further will the ball mount the incline (see § 143) % 

Ans. 0.390 ft. \ 

Example 8.— In Fig. 163, with J = 4 f t. and a = 16 inches, 
it is found in. one experiment that the friction which keeps the 
speed of the pulley at 120 revolutions per minute is balanced 
by a weight G — 160 lbs. Eequired the power thus measured.. 

Ans. 14.6 H. R 

Although in Examples 1 to 6 the steam cylinder is itself in 
motion, the work per stroke is still ■=■ mean effective steam- 
pressure on piston X length of stroke, for this is the final form 
to which the separate amounts of work done by, or upon, the 
two cylinder heads and the two sides of the piston will re- 
duce, when added algebraically. See § 154. 

* By " average value" is meant such a value, Tm, as multiplied into the 
length of path described by the crank-pin per unit of time shall give the 
power exerted. 



164 MECHANICS OF EWGINEEKING. 

CHAPTEK YIL 

FRICTION. 

156. Sliding Friction. — When the surfaces of contact of two 

bodies are perfectly smooth, the direction of the pressure or pair 

of forces between them is normal to these surfaces, i.e., to their 

; tangent-plane ; but when thej are rough, and 

'Y"'f \ moving one on the other, the forces or ac- 

pV 4N tions between them incline awaj from the 

i \ I ij -, P- normal, each on the side opposite to the di- 
WmJ^ S^ W//Jw///M ^^ction of the (relative) motion of the body 
/-. ^q/ ///M ^^ which it acts. Thus, Fig. 167, a block 
Fig. 167. whose weight is G, is drawn on a rough 

horizontal table by a horizontal cord, the tension in which is 
P. On account of the roughness of one or both bodies the ac- 
tion of the table upon the block is a force P^^ inclined to the 
normal (which is vertical in this case) at an angle = (p away 
from the direction of the relative velocity -y. This angle q) is 
called the angle of friction^ while the tangential component of 
P^ is called the friction = F. The normal component N^ 
which in this case is equal and opposite to G the weight of tlie 
body, is called the normal pressure. 

Obviously i^= iV^tan <p, and denoting tan cp hjf we have 

F=fJ^. (1) 

/"is called the coefficient of friction, and may also be defined 
as the ratio of the friction ^to the normal pressure JSf which 
produces it. 

In Fig. 167, if the motion is accelerated (ace. =J?), we have 
(eq. (lY.), § 55) P - i^ = J!/^ ; if uniform, P - F= ; from 
vv'hich equations (see also (I))/" may be computed. In the 
latter case/" may be found to be different with different veloci- 
ties (the surfaces retaining the same character of course), and 
then a uniformly accelerated motion is impossible unless P 
— P were constant. 

As for the lower block or table, forces the equals and op- 
posites of iV^andP(or a single force equal and opposite to P^) 
are comprised in the system of forces actirg upon it. 

As to whether i^ is a worMng force or a resistance, when 



FRICTION. 166 

either of the two bodies is considered free, depends on the cir- 
cumstances of its motion. For example, in friction-gearing 
the tangential action between the two pulleys is a resistance 
for one, a working force for the other. 

If the force P^ Fig. 167, is just sufficient to start the body, 
or is just on the point of starting it (this will be called impending 
fnotion), F\& called ihe friction of rest. If the body is at rest 
and P is not sufficient to start it, the tangential component will 
then be < the friction of rest, viz., just =^ P. As P increases, 
this component continually equals it in value, and P^^ acquires 
a. direction more and more inclined from the normal, until the 
instant of impending motion, when the tangential component 
=/'-ZV"= the friction of rest. When motion is once in prog- 
ress, the friction, called then the friction of motion., = fJV, 
in which/" is not necessarily the same as in the friction of rest. 

157. Variation of Friction and of tlie Coefficient of Friction, f. — Careful 
distinction must be made between the friction of dry surfaces and of 
those that are lubricated; though in the latter case as the supply of lubri- 
cant (oil, soap, graphite, etc.) is reduced from the extreme state of 
"flooding," the friction approximates in variation and magnitude to 
that of dry surfaces. Also, if the pressure is very great, the lubri- 
cant may be pressed out and the phenomena reduced to those of dry 
surfaces, which imder great pressures "seize," i.e., abrade, one another. 

With dry surfaces the amount of friction, F (lbs.), depends on the 
nature of the materials and their initial roughness, being somewhat 
reduced as they become more polished, when a sliding motion has been 
long continued. With the surfaces in a given condition it is found 
(unless the pressure is very low) that increase of velocity diminishes 
the friction, as is unfortunately the case with railroad brakes, the 
friction between a brake-shoe and the rim of the car-wheel being least 
^t the first application of the brakes, when the velocity of rubbing is 
greatest (see p. 168). The friction increases with the normal pressure N 
(the coefficient /, itself, increasing with N when N is large) and is some- 
what smaller after motion begins than when motion is "impending" 
(friction of rest). 

With well lubricated surfaces, however, the following may be said: 
The nature of the materials of the two bodies has but slight influence 
on the amount of friction, F, and when motion has begun, the friction 
is very much less than that of "impending motion." The friction is 
practically independent of the pressure when the lubrication is very 
■copious (bearings "flooded") (resembling, therefore, "fluid friction;" see 
p. 695), the coefficient / being as small as 0.001 or under (Tower); but 
with more scanty lubrication conditions may approach those of dry 
surfaces. As to the effect of velocity (Goodman), the frictional resistance 



166 



MECHANICS OF ENGINEERING. 



varies directly as the speed for low pressures. For high pressures, how- 
ever, it is relatively great at low velocities, a minimum at about 100 
ft./min., and afterwards increases approximately as the square root of 
the speed. A rise of temperature has a very important influence in 
diminishing the viscosity of the oil and enlarging the diameter of the 
bearing of a shaft more than that of the shaft itself. 

In the problems of this chapter the coefficient / will be considered as 
constant; so that where it really varies (as when the velocity changes) 
an average value will be understood. 

158. Experiments on Sliding Friction. — These may be made 
with simple aj)pai'atus. If a block of weight = G, Fig. 168, 
be placed on an inclined plane of uniformly rough surface, 
and the latter be gradually more and more inclined from the 
horizontal until the block begins to move,, the value of fS at 





Fig. 168. 



Fig. 169. 



this instant =; cp, and tan cp ■= f ■=^ coefficient of friction of 
rest. For from -2'X = we have F, \.%.^ fN^ = G sin ySj 
from ^1^= 0, iV^=: G cos fi ; whence tan /? =y, .-, /? must 
= cp. 

Suppose /? so great that the motion is accelerated, the body 
starting from rest at o, Fig. 169. If it is found that the 
distance x varies as the square of the time, then (§ 56) the 
motion is uniformly accelerated (along the axis X). (Notice 
in the figure that G is no longer equal and opposite to Pi, the^ 
resultant of N and P, as in Fig. 168.) We have, then 

lY = 0, which gives N- (? cos ^= ; 

JZ = Mpi, whicli gives G sin /?- fN = {G ■^g)pi ; 
while (from § 56) pi = '2x^t^. 

Hence, by elimination, x and the corresponding time t having 
been observed, we have for the coefficient of friction of motion, 

2x 
' '^ gt^ cos /? 

as an average (since the acceleration may not be uniform). 



FRICTION. 167 

In view of (3), § 157, it is evident that if a value /3^ has been 
found experimentally for /? such that the block, once started hy 
hand, preserves a uniform motion down the plane, then, since 
tan /3^ =.f for friction of motion, /?^ may be less than the /? 
in Fig. 168, for friction of rest. 

159. Another apparatus consists of a horizontal plane, apul- 
lej^, cord, and two weights, as shown in Fig. 59. The masses 
of the cord and pulley being small and hence neglected, the 
analysis of the problem when G is so large as to cause an ac- 
celerated motion is the same as in that example [(2) in § 57], 
except in Fig. 60, where the frictional resistance yW^ should be 
put in pointing toward the left. iT still = G^^ and .*. 

8-fG^^{G,^g):p', (1) 

while for the otlier free body in Fig. 61 we have, as before, 

G-8={G-^g):p (2). 

From (1) and (2), 8 the cord-tension can be eliminated, and 
solving for p, writing it equal to 2^ -^ f, s and t being the ob-^ 
served distance described (from rest) and corresponding time,, 
we have finally for friction of motion (as an average) 

'^- G, - G, ' gf ^^> 

If G, Fig. 59, is made just sufficient to start the block, or 
sledge, G^, we have for the friction of rest 

. /=|.. ....... ii) 

160. Results of Experiments on Sliding Friction. — For accounts of recent 
experiments (and others) and deductions therefrom, the reader may consult 
the Engineer's "Pocket-books" of Kent and Trautwine; also Thurston's 
"Friction and Lost Work," Barr and Kimball's "Machine Design," and 
" Lubrication and Lubricants, ' ' by Archbutt and Deely. The following 
table gives a few values for the coefficient of friction, /, for slow motion, 
taken from the results obtained by Morin and others. Small changes in 
the condition of the surfaces may produce considerable variation in the 
value of /. Our knowledge is still quite incomplete in this respect. 



168 MECHANICS OF ENGINEEKING. 

TABLE FOR FRICTION OF SLOW MOTION. 



No. 


Surfaces. 


Unguent. 


Angle <t). 


/ = tan ^. 


1 


Wood oil wood. 


None. 


14° to 36i° 


0.25 to 0.5U 


2 


Wood on wood. 


iSoap. 


2° to nr 


0.04 to 0.20 


3 


Mcilal on wood. 


None. 


26r to 3ir 


0.50 lo 0.60 


4 


Metal on wood. 


Water. 


15° to 20° 


0.25 10 0.35 


5 


Metal on wood. 


Soap. 


1U° 


0.20 


6 


Leather on metal. 


None. 


29;5r° 


0.56 


7 


Leather on metal. 


Greased. 


13° 


0.23 


8 


Leather on metal. 


Water. 


20° 


0.36 


9 


Leather on metal. 


Oil. 


sr 


0.15 


10 


Metal on metal 


Mone 


8*° to 18° 


0.15 to 0.30 


11 


Metal on metal 


Water 


18° (average) 0.30 


12 


Metal on metal 


Oil 




0.001 to 0.080 



For the coefficient of friction of rest, the above values might be in- 
creased from 20 to 40 per cent., roughly spealdng. 

As showing the effect of velocity in diminishing the friction of dry 
surfaces, we may note that in the Galton-Westinghouse experiments 
with railroad brakes (cast-iron brake-shoes on steel-tired car- wheels), 
values for / were found as follows: When the velocity of rubbing was 
10 miles per hour, / = 0.24; for 20 miles per hour, / = 0.19; for 30, 40, 
50, and 60 miles per hour / was found to be 0.164, 0.14, 0.116, and 
0.074, respectively. The foregoing values of / were obtained imme- 
diately on the application of the brake, but when the brake-shoe and 
wheel had been in contact some five seconds at a constant velocity, 
/ was reduced some 20 or 30 per cent. ; while for a contact of 15 seconds 
still further reduction ensued. The value of / for a "skidding" car- 
wheel (i.e., held fast by the brake pressure) sliding or "skidding" on the 
rail, was reduced from 0.25 for impending skidding, to 0.09 at a velocity 
of 7 miles per hour; and to 0.03 for 60 miles per hour. (See p. 190.) 

That increasing the velocity of lubricated surfaces diminishes the co- 
efficient of friction is well shown in the results obtained by Mr. Welling- 
ton, in 1884, with journals revolving at different speeds, viz., 
For velocity = 0.00 2.16 4.86 21.42 53.01 106ft /min. 

Coeff. / =0.118 0.094 0.069 0.047 0.035 0.026 

For a sledge on dry ground Morin found / = 0.66. For stone on stone, 
see p. 555 of this book. 

161. Cone of Friction. — Fig. 170. Let A and B be two 
a-ougli blocks, of which jB is immovable, and P the resultant 
■■.-.. 0/ of all the forces acting on A. except the pres- 
sure from £. jB can furnish any required 
normal pressure JV to balance P cos /?, but 
the limit of its tangential resistatice is /"iV^. 
So long then as /? is < cp the angle of fric- 
tion, or in other words, so long as the line of 
Fig. 170. action of P is within the " cone offriGtion" 




FRICTIOIT. 169 

generated by revolving 6^6* about ON^ the block A Mall not 
slip on B^ and the tangential resistance of B is simplj P sin 
/? ; but if ^ is > cp, this tangential resistance being oiiij fN 
and < P sin /?, A will begin to slip, with an acceleration. 

162. Problems in Sliding Friction. ^ — In the following prob- 
lemsy is supposed known at points where rubbing occurs, or 
is impending. As to the- pressure iV^ to which the friction is 
due, it is generally to be considered unknown until determined 
by the conditions of the problem. Sometimes it may be an 
advantage to deal with the single unknown force P\ (resultant 
of N 'Aw^fN^ acting in a line making the known angle 9? with 
the normal (on the side away from the motion). 

Problem 1. — Keqnired the value of the weight P^ Fig. lYl, 
the slightest addition to which will cause motion of the hori- 
zontal rod OB, resting on rough planes at 45°. The weight 
G of the rod may be applied at the ys^^y 

middle. Consider the rod free ; at 
€ach point of contact there is an un- 
known JSf and a friction due to it 
fN\ the tension in the cord will be 
= P, since there is no acceleration 
and no friction at pulley. Notice fig. 171. 

the direction of the frictions, both opposing the impending 
motion. Take axes OX and OF as shown, and note the inter- 
sections, A and C, of the line of G with axes OX. and OY . 
The student should not rush to the conclusion that, if G were 
transferred to A and there resolved into components along 
AO and AB, the value of 'N (and A^i) would be equal to that of 
one of these components, viz., mG (where m denotes sin 45°). 
Few problems in mechanics are so simple as to admit of an 
immediate mental solution, and guess-work should be care- 
fully avoided. 

It will be found of advantage to take C as a center of 
moments. The line of P at is considered as passing through 
point C , as also the lines of /AT and /iVi- For equilibrium 
(impending slipping) we have, therefore, 

i-Z^O; i.e.,/iVi+mG-iV-P = 0; . . . (1) 
27 = 0; i.e., i\ri + /Ar-mG = 0; .... (2) 
i'(moms.)c=0; i.e., iVa-A^'ia = 0. ... (3) 




170 



MECHANICS OF ENGINEERING. 



The three unknowns P, N, and Ni, can now be founds 
From (3) we have N = Ni, which in (2) gives A^ = t-71- 
Also from (1) we now find P=^^jN; and hence finally 



P 



1 +/"!+/■ 



G. 




Peoblem 2. — Fig. 1Y2. A rod, centre of gravity at middle,., 
leans against a rough wall, and rests on an equally rough floor; 
how small may the angle a become before it 
slips ? Let a = the half-length. The figure- 
p shows the rod free, and following the sugges- 
P tion of § 162, a single unknown force ^, 
p making a known angle (p (whose tan =/") 
P with the normal D^, is put in at D, leaning^ 
away from the direction of the impending 
motion, instead of an JV and /"iV; similarly 
7^2 acts at C. The present system consisting^ 
of but three forces, the most direct method of finding or, with- 
out introducing the other two unknowns jP^ and P*^ at all, is 
to use the principle that if three forces balance, their lines 
of action must intersect in a point. That is, P*^ must inter- 
sect the vertical containing G, the weight, in the same point 
as Pi, viz., A. 

ISTow, since CF is 1 to FD and the two angles <f> are equal, 
CA is T to DA and therefore DAC is a right triangle. "We 
also note that if CA be prolonged to meet DF in some point K, 
A must be the mid-point of CK, since B is the mid-point of 
CF ; and therefore it follows that in triangle DCK not only 
does DA bisect the side KC but is 1 to it. In other words, 
KDC is an isosceles triangle, with DK and DC as the two 
equal sides, and therefore the line DA bisects the angle KDC. 
Hence the angle KDC = '2(p. That is to say, the angle a, 
which was to be determined, is the complement of double the 
friction-angle, or 

a: = 90°- 20. 



PKICTION. 



171 



Pkoblem 3. — Fig. 173. Given the resistance Q, acting 
parallel to tlie fixed guide C, the angle a, and tlie (equal) co- 
efficients of friction at the rubbing surfaces, required the 




Fig. 173. 




amount of the horizontal force P, at the head of the block A 
(or wedge), to overcome Q and the frictions. D is fixed, and 
ah is perpendicular to cd. Here we have four unknowns, viz., 
JP, and the three pressures iV^, JV^, and JV^, between the blocks. 
Consider ^ and 5 as free bodies, separately (see Fig. 174), re- 
membering l^ewton's law of action and reaction. The full 
values {e.g.,f]V) of the frictions are put in, since we suppose 
.a slow uniform motion taking place. 
For A, 2X= and :S r = give 



i^i — iVcos a -|-y^sin a — J^ sin a = ; . 
f¥^ + N sin a -[-/iTcos a — P cos a = 0. . 
'FoyB, :SXand ^Zgive 
'Q-JSr,+fJV, = 0;....iS) and W,-fJ^, = 0. 
Solve (4) for JV^ and substitute in (3), whence 

j^.o^-r) = Q 



(1) 

(2) 

m 

(5) 

Solve (2) for JV, substitute the result m (1), as also the value 
of i\r^ from (5), and the resulting equation contains but one un- 
known, P. Solving for P, putting for brevity 



ycos a-\- sin a 
■we have P = 



m and cos a — /"sin a = n, 

{w,-\-fn)Q 



or 



{n . cos a -\- m . sin a)(l — yy 



{T} 



172 MECHANICS OF ENGINEERING. 

Numerical Examjple of Problem 3. — If Q = 120 lbs., /" 
= 0.20 {an abstract number, and .*. the same in any system of 
units), while a = 14°, whose sine = 0.240 and cosine = .970,, 
then 

m = 0.2 X. 97 + 0.24 = 0.43 and t^ = .97 — .2X.24 = 0.92,. 

whence P = OMQ = 76.8 lbs. 

While the wedge moves 2 inches P does the work (or exerts 
an energy) of 2 X 76.80 = 153.6 in.-lbs. = 12.8 ft.-lbs. 

For a distance of 2 inches described by the wedge horizon- 
tally, the block P (and .•. the resistance Q) has been moved 
through a distance = 2 X sin 14° = 0.48 in. along the guide- 
G, and hence the work of 120 X 0.48 = 57.6 in.-lbs. has been 
done upon Q. Therefore for the supposed portion of the 
motion 153.6 — 57.6 = 96.0 in.-lbs. of work has been lost in 
friction (converted into heat). 

For the "mechanical efficiency" of this machine (see § 153) 
we have 57.6-153.6=0.375. Also note that for / = 1.0(> 
P = '^ ; andfor a: = 90°, P = Q. 

Problem 4. Numerical. — With what minimum pressure 

P should the pulley A be held against P, which it drives by 

■^ n x riA " frictioiial gearing," to transmit 2 H. p.;. 

^ - p -(--- ^ > if a = 45°, f for impending (relative) 

«y 7>! ^ motion, i.e., for impending slipping = 
Fig. 175. 0.40, and the velocity of the pulley-rim; 

is 9 ft. per second ? 

The limit-value of the tangential " grip" 

T = 2/i\^= 2 X 0.40 X P sin 45°, 
2 H. P. = 2 X 550 = 1100 ft.-lbs. per second. 

Putting T X 9 ft. = 1100, we have* 

2 X 0.40 X 4/5 X P X 9 = 1100 ; .: P = 215 lbs.. 

Problem 6. — A block of weight G lies on a rough plane, 
inclined an angle ^ from the horizontal ; find the pull P, mak- 
ing an angle a with the first plane, which will maintain a uni- 
form motion up the plane. 

* In this problem the student should note that, in general, when a is not 45°, 
we have N = iP -v- cos a (since in such a case the parallelogram of. forces i& 
not a square). 



FRICTION. 173 

Pkoblem 7. — Same as 6, except that the pull P is to permit 
a uniform motion down the plane. 

Pkoblem 8. — The thrust of a screw-propeller is 15 tons. 
Tlie ring against which it is exerted has a mean radius of 8 
inches, the shaft makes ©ne revolution per second, andy = 0.06. 
Required the H. P. lost in friction from this cause. 

Ans. 13.7 H. P. 

163. The Bent-Lever with Friction. Worn Bearing. — Fig. 
176. Neglect the weight of the lever, and suppose the plumb- 
er-block so worn that there is 
contact along one element only of 
the shaft. Given the amount and 
line of action of the resistance R^ 
and the line of action of jP, re- 
quired the amount of the latter for 
impending slipping in the direction 
of the dotted arrow. As P grad- 
ually increases, the shaft of the 
lever (or gear-wheel) rolls on its fig. 176. 

bearing until the line of contact has reached some position Ay 
when rolling ceases and slipping begins. To find A^ and the 
value of P^ note that the total action of the bearing upon the 
lever is some force ^,, applied at A and making a known 
angle q) (^f =: tan q)) with the normal A C. P^ must be eqnal 
and opposite to the resultant of the known P and the unknown 
P, and hence graphically (a graphic is much simpler here than 
an analytical solution) if we describe about C 2i circle of radius 
= r sin 9?, r being the radius of shaft (or gudgeon), and draw 
a tangent to it from P, we determine PA as the line of action 
of P^. If PG is made = P, to scale, and ^^ drawn parallel 
to P . . . P, P is determined, being = PP, while P^ = PK 

If the known force P is capable of acting as a working force, 
by drawing the other tangent PP from P to the " friction- 
circle," we have P = PP, and P^ = PK, for impending 
rotation in an opposite direction. 

If P and P are the tooth-pressures upon two spur-wheels, 
keyed upon the same shaft and nearly in the same plane, the 




174 MECHANICS OF ENGINEERING. 



y = [Pj sin (p]27rr. 



same constructions hold good, and for a continuous uniform 
motion, since the friction = P, sin cp, 
the work lost in friction 
per revolution, 

It is to be remarked, that without friction Pj would pass 
through 0, and that the moments of i? and I^ would balance 
about C (for rest or uniform rotation) ; whereas with friction 
thej balance about the proper tangent-point of the friction- 
circle. 

Another way of stating this is as follows : So long as the 
resultant of I^ and P falls within the " dead-angle" BDA, 
motion is impossible in either direction. 

If the weight of the lever is considered, the resultant of it 
and the force M can be substituted for the latter in the fore- 
going. 

164. Bent-Lever with Friction. Triangular Bearing. — Like 
the preceding, the gudgeon is much exaggerated in the figure 

(1Y7). For impending rotation in 
direction of the force P, the total 
actions at A^ and A^ must lie in 
known directions, making angles = cp 
with the respective normals, and in- 
clined away from tlie shpping. Join 
the intersections D and L. Since 
the resultant of P and R i^i D must 
act along PL to balance that of P^ 
and P^^ having given one force, say 
Fig. 177. B, wc easily iind P = PE, wliile 

P^ and P^ = ZJf^and ZiV respectively, LO having been made 
= PP, and the parallelogram completed. 

(If the direction of impending rotation is reversed, the change 
in the construction is obvious.) If P^ = 0, the case reduces 
to that in Fig. 176 ; if the construction gives P^ negative, the 
supposed contact at A^ is not realized, and the angle A^OA^ 
should be increased, or shifted, until P^ is positive. 

As before, P and P may be the tooth-pressures on two 




FRICTION. 



175 



spur-wheels nearly in tiie same plane and on the same shaft ; 
if so, then, for a uniform rotation. 
"Work lost in fric. per revol. = [P^ sin cp -\- P^ sin cp\^7tr. 

165. Axle-Friction. — Tiie two foregoing articles are intro- 
ductory to the subject of axle -friction. When the bearing is 
new, or nearly so, the elements of the axle which are in contact 
with the bearing are iniinite in number, thus giving an infinite 
number of unknown forces similar to P^ and P^ of the last 
paragraph, each making an angle cp with its normal. Refined 
theories as to the law ox distribution of these pressures are of 
little use, considering tne uncertainties as to the value of 
y ( ^ tan <p) ; hence tor practical purposes axle-friction may be 
written 

in which f is a coejficieni of axle-friction derivable from 
experiments with axles, and JR the resultant pressure on the 
bearing. In some cases Jti may be partly due to the tightness 
of the bolts with which the cap of the bearing is fastened. 

As before, the work lost in overcoming axle-friction j)6r 
revolution is ■=.fR'^7tr^ in wliich r is the radius of the axle. 
/'', like y, is an abstract number. As in Fig. 176, a " friction' 
circle," of radius =fr, majr be considered as subtending the 
" dead-angle." 

166. Experiments with Axle-Friction. — Prominent among 
recent experiments liave been those 
of Professor Thurston (1872-73), 
who invented a special instrument 
for that purpose, shown (in princi- 
ple only) in Fig. 178. By means of 
an internal spring, tne amount of 
whose compression is reaa on a scale, 
a weighted bar or pendulum i5 oaused 
to exert pressure on a projecting axle 
from which it is suspended. Tlie 
axle is made to rotate at any desired 
velocity by some source of power, the axle-friction causing 




176 MECHANICS OF ENGINEERING. 

the pendulum to remain at rest at some angle of deviation 
from the vertical. The figure shows the pendulum free, the 
action of gravity upon it being (r, that of the axle consisting 
of the two pressures,* each = i?, and of the two frictions (each 
being F =^f R\ due to them. Taking moments about (7, we 
have for equilibrium 

^f'Rr = Gh, 

in which all the quantities except jT are known or observed. 
The temperature of the bearing is also noted, with reference 
to its effect on the lubricant employed. Thus the instrument 
covers a wide range of relations. 

General Morin's experiments as interpreted by Weisbach 
give the following practical results: (See also p. 192). 

0.054 for well-sustained 
lubrication ; 

0.07 to .08 for ordinary 
lubrication. 



For iron axles, in iron or 
brass bearings 



/' = 



By "pressure per square inch on the bearing" is commonly 
meant the quotient of the total pressure in lbs. by the area in 
square inches obtained by multiplying the width of the axle by 
the length of bearing (this length is quite commonly four times, 
the diameter) ; call it j?, and the velocity of rubbing m feet per 
minute, v. Then, according to Rankine, to prevent overheat- 
ing, we should have 

p{v + 20) < 44800 . . . (not homog.). 

Still, in marine-engine bearings pv alone often reaches 60,000^ 
as also in some locomotives (Cotterill). Good practice keeps 
P within the limit of 800 (lbs. per sq. in.) for other metals 
than steel (Thurston), for which 1200 is sometimes allowed. 

With ^ = 200 (feet per min.) Professor Thurston found that 
for ordinary lubricants p should 2iot exceed values ranging, 
from 30 to 75 (lbs. per sq. in.). 

The product pv is obviously proportional to the power ex- 
pended in wearing the rubbing surfaces, per unit of area. 

* The weight O being small compared with the compressive force Hia. 
the spring, each pressur*^ is practically equal to B. 



FRICTION. 



177 



167. Friction-Wheels — (Or, rather, anti-friction wheels). 
In Fig. 179, M and M (and two more behind) are the "fric- 
tion-wheels," with axles Ci and C\ in fixed bearings. 

G is the weight of a heavy wheel, Pi is a known vertical 
resistance (tootli-pressnre), and P an 
unknown vertical working force, 
whose value is to be determined to 
maintain a uniform rotation. The 
utility of the friction-wheels is also 
to be shown. The resultant of P^, 
G, and J-* is a vertical force P, pass- 
ing nearly through the centre C of 
the main axle which rolls on the four 
friction-wheels. J?, resolved along 
€A and CB, produces (nearly) equal 
pressures, each being J^ =: P -r- 2 cos (x, at the two axles of 
the friction- wheels, which rub against their fixed plumber- 
blocks. P ^ P -\- P^-\- 6^„ and .*. contains the unknown P, 
but approximately ■= G-{- 2P^, i.e., is nearly the same (in this 
case) whether friction-wheels are employed or not. 

When G makes one revolution, the friction /'''iV^ at each axle 
C^ is overcome through a distance = (r, : a^) 27rr, and 

Work lost per revol. \ 




Fig. 179. 



T T \ 

" n. n. cc 



«, COS oc 



fP^Ttr. 



with 
friction-wheels, 

Whereas, if C revolved in a fixed bearing, 

Work lost per revol. ) 

without V =f'P'i7tr. 

friction-wheels, ) 

Apparentl)^, then, there is a saving of work in the ratio r^ : 
a, cos o', but strictly the P is not quite the same in the two cases ; 
for with friction-wheels the force P is less than without, and P 
depends on P as well as on the known G and P^. By dimin- 
ishing the ratio r^ : a^^ and the angle or, the saving is increased. 
If a were so large that cos or < r, : a^, there would be no saving, 
but the reverse. 

As to the value of P to maintain uniform rotation, we have 
12 



178 MECHANICS OF ENGINEERING. 

foi' equilibrium of moments about (7, with fri-ction-wlieels (con^ 
sidering the large wheel and axleyV'ee), 

P5 = PA + 21>, ....... (1) 

in which T is the tangential action, or "grip," between one 
pair of friction-wheels and the axle C which rolls upon them. 
^ would noL equal yiV unless slipping took place or were im- 
])ending at E^ but is known bj considering a pair of friction- 
wheels free, when ^ (-P«) about C^ gives 



2 ' cos «' 
which in (1) gives finally 



b T T 

P = iP,^ ^-f'R. (2) 

' ' »! cos a*^ ^ ' 

Without friction-wheels, we would have 

P^\p,^fR\ : . (3) 

The last term in (2) is seen to be less than that in (3) (unless 
a is too large), in the same ratio as already found for the saving 
of work, supposing the jS's equal. 

If P^ were on the same side of C as P^ it would be of an 
opposite direction, and the pressure i? would be diminished. 
Again, if P were horizontal, R would not be vertical, and the 
friction-wheel axles would not bear equal pressures. Since P 
depends on Pj, G^ and the frictions^ while the friction depends 
on R^ and R on P^^ G, and P, an exact analysis is quite 
complex, and is not warranted by its practical utility. 

Example. — If an empty vertical water-wheel weighs 25,000 
lbs., required the force P to be applied at its circumference to 
maintain a uniform motion, with « = 15 ft., and r = 5 inches. 
Here P^ = 0, and R = G (nearly ; neglecting the influence of 
P on R), i.e., R = 25,000 lbs. 

Eirst, iDiihout friction-wheels (adopting the foot-pound-sec- 
ond system of units), withy =: .07 (abstract number). Frona 
eq. (3) we have 

P =: + 0.07 X 25,000 X (tV "^ 1^) = 48.6 lbs. 



FRICTION, 179 

The work lost in friction per revolution is 

f'B^Tir = O.or X 25,000 X 2 X 3.14 X A = ^580 ft.-lbs. 

Secondly, with friction-wheels, in which r^ '. a^ =: ^ and 
cos a = 0.80 (i.e., a = 36°). From eq. (2) 

J> = 0-^^.\^X 48.6 = only 12.15 lbs., 

while the work lost per revolution 

= 1. . jMi X 4580 = 1145 ft.-lbs. 

Of course with friction- wheels the wheel is not so steady as 
without. 

In this example the force J* has been simply enough to 
overcome friction. In case the wheel is in actual use, JP is the 
weight of water actually in the buckets at any instant, and does 
the work of overcoming I^^, the resistance of the mill machinery, 
and also the friction. By phicing J*^ pointing upward on the 
same side of C as P, and making h^ nearly ■=!), H will = G 
nearly, just as when the Avheel is running empty; and the 
foregoing numerical results will still hold good for practical 
purposes. 

168. Friction of Pivots. — In the case of a vertical shaft or 
axle, and sometimes in other cases, the extremity requires sup- 
port against a thrust along tlie axis of the axle or pivot. If 
the end of the pivot \%flat and also the surface 
against which it rubs, we may consider the 
pressure, and therefore the friction, as uniform 
over the surface. With a flat circular pivot, 
then. Fig, 180, the frictions on a small sector 
of the circle form a system of parallel foices 
whose resultant is equal to their sum, and is ^^'^' "^^°" 

applied a distance of -|r from the centre. Hence the sum of 
the moments of all the frictions about the centre =y!^|r, in 
which .^ is the axial pressure. Therefore a force P necessary 
to overcome the friction with uniform rotation must have a 
moment 

Pa =fR\r, 




180 MECHAl^ICS OF ENGINEERING, 

and the work lost in friction per revolution is 

^fR^Tt .\T = ^,7tfRr. . . . . (1) 

As the pivot and step become worn, the resultant frictioii* 
in the small sectors probably approach the centre; for the 
greatest wear occurs first near tlie outer edge, since there the 
product ^J)^> is greatest (see § 166). Hence for \r we may more 
reasonably put ^\ 

Exam]jle. — A vertical flat-ended pivot presses its step with 
a force of 12 tons, is 6 inches in diameter, and makes 40 revolu- 
tions per minute. Required the H. P. absorbed by the friction. 
Supposing the pivot and step new, and /"for good lubrication 
= 0.07, we have, from eq. (1) {foot-lb. -second), 

"Work lost per revolution 

= .07 X 24,000 X 6.28 X I • i = 1758.4 ft.-lbs., 

and .*. work per second 

= 1758.4 X |-t = 1172.2 ft.-lbs., 

which -i- 550 gives 2.13 H. P. absorbed in friction. If ordi- 
nary axle-friction also occurs its effect must be added. 

If the flat-ended pivot is hollow, with radii r^ and r^, we may 
put ^{i\-\-')\) instead of the fr of the preceding. 

It is obvious that the smaller the lever-arm given to the 
resultant friction in each sector of the rubbing surface the 
smaller the power lost in friction. Hence pivots should be 
made as small as possible, consistently with strength. 

For a conical pivot and step, Fig. 181, the resultant friction 
in each sector of the conical bearing surface has 
a lever-arm = f r, about tlie axis A, and a value 
>- than for a flat-ended pivot ; for, on account 
of the wedge-like action of the bodies, the 
pressure causing friction is greater. The sum of 
the moments of these resultant frictions about 
A is the same as if only two elements of the 
cone received pressure (each = iV= ^R -f- sin or). Hence the 




FRICTION. 181 

moment of friction of the pivot, i.e., the moment of the force 
necessary to maintain uniform rotation, is 

'^ 3 ' "^ sm or 3 " 

4 B 

and work lost per revolution = o'^f~ '^v 

o sin oi 

By making r^ small enough, these values may be made less 
than those for a flat-ended pivot of the same diameter = 2r. 

In Schiele's " anti-friction" pivots the outline is designed 
according to the following theory for securing uniform vertical 

wear. Let j!? = the pressure per r— — ^^ — _ — 

horizontal unit of area (i.e., Ip jA ip 

= -^ -r- horizontal projection of .^ ! ■ L<c^^ ^^^^ 

the actual rubbing surface) ; "^^^^ I "^^t^^^^ 

this is assumed constant. Let i!?C..'^^<n z^^^. 

tlie unit 01 area be small, for M ~ic ^ 

algebraic simplicity. The fric- fig. 182. 

tion on the rubbing surface, whose horizontal projection = unity, 
is = yiV =y (^ -f- sin a) (see Fig. 182; the horizontal com- 
ponent of J9 is annulled by a corresponding one opposite). The 
work per revolution in producing wear on this area = fN^ny. 
But the vertical depth of wear per revolution is to be the same 
at all parts of the surface ; and this implies that the same 
volume of material is worn away under each horizontal unit of 

area. HenceyiTSTr^/, i.e.,y-r^^ — B^ry, is to be constant for all 

values of y ; and since ^7^ and 27r are constant, we must have, 
as the law of the curve, 

y 

, i.e., the tangent BC = the same at all potnts. 



sm a 



This curve is called the " tractrixP Schiele's pivots give a 
very uniform wear at high speeds. The smoothness of wear 
prevents leakage in the case of cocks and faucets. 

169. Normal Pressure of Belting. — "When a perfectly flexible 
cord, or belt, is stretched over a smooth cylinder, both at rest, 



182 



MECHANICS OF ENGINEERING. 




the action between them is normal at every point. As to its 
j^ \_\ t t s ^^0^1^ ^3 i^j P^r linear unit of arc, the fol- 
C\\^-d:::^ > lowing will determine. Consider a semi- 
circle of the 2'2vd tree, neglecting its weight. 
Fig. 183. The forces holding it in equilib- 
rium are tlie tensions ar the two ends (these 
are equal, manifestly, the cylinder being 
„ smooth ; for tnev are the only two forces 
* / 7/1 having moments about c/, and each has the 

Fig. 183. smus lever-ariTi^. and the normal pressures, 

which are infinite in number, but nave an intensity, p^ per 
linear unit, which must be constant along the curve since 8 is 
the same at all points. The normal pressure on a single ele- 
ment, ds, of the cord is = //Jsr. aiid its JT component = 
pds cos 6 — prdd cos ^. Hence S.X = gives 

cos BdQ — 2/S' = 0. i.e., rjp\ sin d = 28; 



.-in 



.'. rp[l — (— 1)] = 26' or z> = 



S 



(1) 



170. Belt on Eough Cylinder. ImDending Slipping. — If fric- 
tion is possible between the two bodies, the tension may vary 
»k)ug the arc of contact, so that ^also varies, and consequently 




Fig. 184. 




Fig. 185. 



the friction on lui element (^s being =^ds =f{8-^ r)ds, also 
varies. If slipping is impendina. the law of variation of the 
tension S may be found, as follows ° Fig. 184, in which the 



ERICTIWN. 183 

impending slipping is toward the left, shows the cord free. 
For any element, ds, of the cord, we have, putting 2 (moms, 
about 0) = Q (Fig. 185), 

{8+ dSy = Sr + dFr ; i.e., dF= dS, 

or (see above) dS =f{S -^ r)ds. 

But ds = rdO ; hence, after ti-ansfonning, 

fde = §. (1) 

In (1) the two variables and S are separated ; (1) is there- 
fore ready for integration. 

fa = loge 8n — l0g« 8, = l0ge[_^J. (2) 

Or, by inversion, 8^ef"- — 8n, (3) 

<?, denoting the Naperian base, = 2.71828 -{-; a of course is in 
TT-measure. 

Since 8n evidently increases very rapidly as oc becomes 
larger, 8^ remaining the same, we have the explanation of the 
well-known fact that a comparatively small tension, 8^, exerted 
by a man, is able to prevent the slipping of a rope around a 
pile-head, when the further end is under the great tension 8^ 
due to the stopping of a moving steamer. For example, with 
^ = ^, we have (Weisbach) 

for or = J turn, or ^r = ^tt, 8^ = l-BQ-SI, ; 

= "I turn, or a = 7t^ 8n = '2.S68„ ; 

= 1 turn, or a- = 2;r, 8^ = S.lS^Su ; 

= 2 turns, or a = 4c7t, 8^ = 65.94^„ ; 

= 4 turns, or a = Stt, 8^ = 4348.56/S'„. 

If slipping actually occurs, we must use a value of f for fric- 
tion of motion. 

Examjple. — A leather belt drives an iron pulley, covering 
one half the circumference. What is the limiting value of the 



184 



MECHANICS OF ENGINEERING. 



ratio of Sn (tension on driving-side) to S^ (tension on follow- 
ing side) if tlie belt is not to slip, taking the low value of 
y = 0.25 for leather on iron ? 

We have given ya: = 0.25 X ^r = .T854, which by eq. (2) is 
the Naperian log. of {S^ '. /So) when slipping occurs. Hence the 
common log. of {S^ : /S,) = 0.7854 X 0.43429 = 0.34109 ; i.e., 
if 

(5;:/S;) = 2.193,say2.2, 
the belt will (barely) slip (for/= 0.25). 

(0.43429 is the modulus of the common system of loga- 
rithms, and = 1 : 2.30258. See example in § 48.) 

At very high speeds the relation^ = /S' -i- r (in § 169) is not 
strictly true, since the tensions at the two ends of an element 
ds are partly employed in furnishing the necessary deviating 
force to keep the element of the cord in its circular path, the 
remainder producing normal pressure. 

171. Transmission of Power by Belting or Wire Eope. — In the 

simple design in Fig. 186, it is required to find the motive 
weight Gy necessary to overcome the given resistance ^ at a 



DRIVING SIDE 




Fig. 186. 



uniform velocity = v^; also the proper stationary tension 
weight G„ to prevent slipping of the belt on its pulleys, and 
the amount of power, Z, transmitted. 
In other words. 



Given : 



j B, a, y, a^, r^; a z= n for both pulleys, ) 
( -y,; andy for both pulleys ; ) 



andy for both pulleys; 

-p • ^ . j Z ; ^, to furnish Z ; G^ for no slip ; 'y the velocity 
'\<A G\ v' that of belt ; and the tensions in belt. 



FRICTION. 



185 



Neglecting axle-friction and the rigidity of the belting, tlie 
power transmitted is that required to overcome i? through a 
distance = v^ every second, i.e., 

Z = Bv, ,. (1) 

Since (if the belts do not slip) 



we have 



a : r::v' : V, and a^ : r^iiv' : v^, 



V = —v., and v = v.. 



(2) 



I^egleeting the mass of the belt, and assuming that each pul- 
ley revolves on a gravity-axis, we obtain the following, by con- 
siderino^ the free bodies in Fig'. 187 : 




CA free) 



(B free) 
Fig. 187. 



(B and tr_u.ck fr.e.e) 



2 (moms.) = in A free gives Er^ = {8^ — S,)a, ; . (3) 
2 (moms.) = in ^ free gives Gr = (Sn — Sa)a ; . (4) 



whence we readily find 



r a. 



Evidently JR and G are inversely proportional to their velo- 
cities v^ and v ; see (2). This ought to be true, since in Fig. 
186 G is the only working-force, ^ the only resistance, and 
the motions are uniform ; hence (from eq. (XYI.), § 142) 

Gv - Ev, = 0. 
2J^ = 0, for _5 and truck free, gives 

G, = S^+S„ (5) 

while, for impending slip, 

^n = ^0^^' (6) 



186 MECHANICS OF ENGINEERING-. 

By elimination between (4), (5), and (6), we obtain 

and ^n = -/ • ^aZII: (8) 

Hence G^ and 8^, vary directly as the power transmitted and 
inversely as the velocity of the belt. For safety G^ should be 
made > the above value in (T) ; corresponding values of the 
two tensions may then be found from (5), and from the rela- 
tion (see § 150) 

{8^-8y = L (64 

These new values of the tensions will be found to satisfy the- 
condition of no slip, viz., 

(^,:xS'„)<^-(§170). 

For leather on iron, ef"" = 2.2 (see example in § lYO), as a. 
low value. The belt should be made strong enough to with- 
stand 8n safely. 

As the belt is more tightly stretched, and hence elongated,, 
on the driving than on the following side, it ^' creeps'^ back- 
ward on the driving and forward on the driven pulley, so that 
the former moves slightly faster than the latter. The loss of 
work due to this cause does not exceed 2 per cent with ordinary 
belting (Cotterill). 

In the foregoing it is evident that the sum of the tensions in 
the two sides = G„, i.e., is the same, whether the power is^ 
being transmitted or not ; and this is found to be true, both in 
theory and by experiment, when a tension-weight is not used, 
viz., when an initial tension S is produced in the whole belt 
before transmitting the power, then after turning on the latter 
the sum of the two tensions (driving and following) always 
= ^S, since one side elongates as much as the other contracts ; 
it being understood that the pulley-axles preserve a constant 
distance apart. 

172. Rolling Friction. — The few experiments which have- 
been made to determine the resistance offered by a level road- 



FRICTION. 187 

"Way to the uniform motion of a roller or wheel rolling upon it 
corroborate approximately the following theory. The word 
friction is hardly appropriate in this connection (except when 
the roadway is perfectly elastic, as will be seen), but is sanctioned 
by usage. 

Firsts let the roadway or track be compressible, but inelastic, 
O the weight of the roller and its load, and P the horizontal 
force necessary to preserve a uniform motion ^ — ~\ — > 

(both of translation and rotation). The track / |g \ 




(or roller itself) being compressed just in h"^ j'"*'^ 

front, and not reacting symmetrically from \ q[\ ,_^^ 
behind, its resultant pressure against the //////////mmW^y^-'^'^^^- 
roller is not at vertically under the centre, ^^**" ^^^' 

but some small distance, OD = h, in front. (The successive 
crushing of small projecting particles has the same effect.) 
Since for this case of motion the forces have the same relations 
as if balanced (see § 124), we may put 2 moms, about D = 0, 

.'.Fr=Gh; or, P = ~G (1) 

According to Professor Goodman we have the following 
values of b, approximately : 

Inches. 
Iron or steel wheels on iron or steel rails. . 6 = 0.007 to 0.015 

" " " " " wood 0.06 " 0.10 

" " " " " macadam 0.05 " 0.20 

" " " " " soft ground 3.0 "5.0 

Pneumatic tires on good road, or asphalt.. 0.02 " 0.022 

'' " heavy mud 0.04 " 0.06 

Solid rubber tires on good road, or asphalt 0.04 

" " heavy mud 0.09 " 0.11 

According to the foregoing theory, P, the " rolling fi-iction" 
(see eq. (1)), is directly proportional to G, and inversely to the 
radius, if h is constant. The experiments of General Morin and 
others confirm this, while those of Dupuit, Poiree. and Sauvage 
indicate it to be proportional directly to G, and inversely to the 
square root of the radius. 



188 MECHAlSriCS OF ENGINEEEING. 

Although J is a distance to be expressed in linear units, and 
not an abstract number like the /"and f for sliding and axle- 
friction, it is sometimes called a " coefficient of rolling fric- 
tion." In eq. (1), h and r should be expressed in the same 
unit. 

Of course if P is applied at the top of the roller its lever- 
arm about D is 2r instead of r^ with a corresponding change 
in eq. (1). 

With ordinary railroad cars the resistance due to axle and 
rolling frictions combined is about 8 lbs. per ton of weight on 
a level track. For wagons on macadamized roads & = |- inch, 
but on soft ground from 2 to 3 inches. 

Secondly^ when the roadway is jperfectly elastic. This is 
chiefly of theoretic interest, since at first sight no force would 
be considered necessary to maintain a uniform rolling motion. 
But, as the material of the roadway is compressed under the 
roller its surface is first elongated and then recovers its former 
state ; hence some rubbing and consequent sliding friction must 




occur. Fig. 189 gives an exaggerated view of the circum- 
stances, P being the horizontal force applied at the centre 
necessary to maintain a uniform motion. The roadway (rub- 
ber for instance) is heaped up both in front and behind the 
roller, being the })oint of greatest pressure and elongation 
of the surface. The forces acting are ^, P^ the normal 
pressures, and the frictions due to them, and must form a 
balanced system. Hence, since G and P^ and also the normal 
pressures, pass through C^ the resultant of the frictions must 
also pass through G\ therefore the frictions, or tangential 
actions, on the roller must be some forward and some backward 
(and not all in one direction, as seems to be asserted on p. 260 
of Cotterill's Applied Mechanics, where Professor Reynolds' 



FRICTION. 189 

explanation is cited). The resultant action of the roadwaj 
upon the roller acts, then, through some point J9, a distance 
OD = h ahead of (9, and in the direction DC, and we have as 
before, with 2? as a centre of moments, 

Pr=Gh, or P=-G. 

If rolling friction is encountered above as 
well as helow the rollers, Fig. 190, the 
student may easily prove, by considering 
three separate free bodies, that for uniform 
motion 



p = '-^<^' 




Fia. 190. 



where h and h^ are the respective " coefficients of rolling fric- 
tion ' ' for the upper and lower contacts. (See Kent's " Pocket- 
Book for Mechanical Engineers'' for "friction-rollers,'^ 
"ball-bearings," and "roller-bearings." 

Exa/mjple 1. — If it is found that a train of cars will move 
uniformly down an incline of 1 in 200, gravity being the only 
working force, and friction (both rolling and axle) the only 
resistance, required the coefficient, f\ of axle-friction, the 
diameter of all the wheels being 2f = 30 inclies, that of the 
journals 2a = S inches, taking h = 0.02 inch for the rolling 
friction. Let us use equation (XYI.) (§ 142), noting that while 
the train moves a distance s measured on the incline, its weight 

1 A 

G does the work G ^z s, the rolling friction — G (at* the axles) 

has been overcome through the distance s, and the axle-friction 

(total) through the (relative) distance — sin the journal boxes j 

whence, the change in kinetic energy being zero, 

1 ^ b ^ a 

Gs 



-^Ga-^Gs--fas = 0. 



Gs cancels out, the ratios h : r and a : r are = tAtt ^"<5 iV* 
respectively (being ratios or abstract numbers they have the 

* That is, the ideal resistance, at centre of axles and || to the incline, equiV' 
alent to actual rolling resistance. 



190 MECHANICS OF ENGINEERING. 

same numerical values, whether the inch or foot is used), an (J 
solving, we have 

/ = 0.05 - 0.0133 = 0.036. 
Examjple 2. — How many pounds of tractive effort per ton 
of load would the train in Example 1 require foi- uniform mo- 
tion on a level track ? Ans. 10 lbs. 

173. Eailroad Brakes.* — During the uniform motion of a 
railroad car the tangential action between the track and each 
wheel is small. Thus, in Example 1, just cited, if ten cars of 
eight wheels each make up the train, each car weighing 20 tons, 
the backward tangential action of the rails upon each wheel is 
only 25 lbs. When the brakes are applied to stop the train 
this action is much increased, and is the only agency by which 
the rails can retard the train, directly or indirectly : directly^ 
when the pressure of the brakes is so great as to prevent the 
wheels from turning, thereby causing them to "skid" (i.e., 
slide) on the rails ; indirectly^ when the brake-pressure is of 
such a value as still to permit perfect rolling of the wheel, in 
which case the rubbing (and heating) occurs between the brake 
and wheel, and the tangential action of the rail has a value 
equal to or less than the friction of rest. In the first case, 
then (skidding), the retarding influence of the rails is the/r^c- 
iion of motion between rail and wheel; in the second, a force 
which may be made as great as the friction of rest between rail 
and wheel. Hence, aside from the fact that skidding produces 
objectionable flat places on the wheel-tread, the brakes are 
more effective if so applied that skidding is impending, but 
not actually produced ; for the friction of rest is usually greater 
than that of actual slipping (§160). This has been proved 
experimentally in England. The retarding effect of axle and 
rolling friction has been neglected in the above theory. 

Example 1. — A twenty-ton car with an initial velocity of 80 
feet per second (nearly a mile a minute) is to be stopped on a 
level within 1000 feet ; required the necessary friction on each 
of the eight wheels. 

Supposing the wheels not to skid, the friction will occur 

* See statement on p. 168, as to diminution of the coefficient / with 
; speed. 



iJ'RICTION. 191 

between the brakes ana wheels, and is overcome through the 

(relative) distance 1000 feet. Eq. (XYI.), § 14:2, gives (foot- 
Ib.-second system) 

1 40000 
- 8i^'X 1000 = - 1 ^^(80)^ 

from which F { = friccion at circumference of each wheel) 

= 496 lbs. 

Note. — This result of 496 lbs. must be looked upon as only an average 
value. For a given pressure, A'^, of brake-shoe on wheel-rim on account 
of the variation of the coefficient /' with changing speed (see p. 168) 
the friction will be small at first and gradually increase. This same 
remark applies to Examples 3 and 4, also. 




1/ 



Examjple 2.— Supoose sisiddins^ to be impending in the fore- 
going, and tlie coefhcient of friction of rest (i.e., impending 
slipping) between ran ana wneel to be/'=0,20o In what 
-distance will the. car oe stopped? Ans. 496 ft 

Example 3. — Supoose tne car in Example 1 to be on an up» 
grade of 60 feet to tne mile. Qn applying eq. (XVI.) here, 
the weight 20 tons win enter as a resistance.) Ans. 439 lbs. 

Example 4. — In Ji,xample 3. consider all four resistances, 
viz., gravitj^, rolling triction. and brake and axle frictions, the 
distance being 1000 ft., and \F the unknown quantity. 

(Take the wheel dimensions of p. 189.) Ans. 414 lbs. 

174. Friction of Car-journals in Brass Bearings. — :(Prof. J. 
E. Denton, in Vol. xii Transac. Am. See, Mecli. Engs., 
p. 405; also Kent's Pocket-Book.) A new brass dressed 
with an emery wheel, loaded with 5000 lbs., may have an 
actual bearing surface on the journal, as shown by the polish 
of a portion of the surface, of only one sq. inch. "With this 
pressure of 5000 Ibs./sq.in. the coefficient of friction may be 
0.060 and the brass may be overheated, scarred, and cut ; or, 
on the contrary, it may wear down evenly to a smooth bearing, 
giving a highly polished area of contact of 3 sq. in., or more, 
inside of two hours of running, gradually decreasing the 
pressure per sq. in. of contact, and showing a coefficient of 
friction of less than 0.005. A reciprocating motion in the 
direction of the axis is of importance in reducing the friction. 
With such polished surfaces any oil will lubricate and the 



192 MECHANICS OF ENGINEERING. 

coefficient of friction then depends on the viscosity of the oil. 
With a pressure of 1000 lbs, per sq. in. , revolutions from 170 
to 320 per min., and temperature of 75° to 113° Fahr., with 
both sperm and parraffine oils, a coefficient as low as 0.0011 
has been obtained, the oil being fed continuously by a pad. 

175. Well Lubricated Journals. Laws of Friction. — In the 
Proc. Inst. Civ. Engs. for 1886 (see also Engineering News 
for Mar. 31, April 7 and 14, 1888) Prof. Goodman presents 
the conclusions arrived at by him as to the laws of friction of 
well lubricated journals as based on the experiments made by 
Thurston, Beauchamp Tower, and Stroudley. They are as 
follows : 

1 . The coefficient friction with the surfaces efficiently lubri- 
cated is from ^ to ^ that for dry or scantily lubricated surfaces. 

2. The coefficient of friction for moderate pressures and 
speeds varies approximately inversely as the normal pressure ; 
the frictional resistance varies as the area- in contact, the 
normal pressure remaining constant. 

3. At very low journal speeds the coefficient of friction is 
abnormally high, but as the speed of sliding increases from 
about 10 to 100 ft. per min. the friction diminishes; and 
again rises when that speed is exceeded, varying approximately 
as the square root of the speed. 

4. The coefficient of friction varies approximately inversely 
as the temperature, within certain limits, viz., just before 
abrasion takes place. 

In one of Mr. Tower's experiments it was found that when 
the lubrication was made by a pad under the journal (which 
received pressure on its upper surface) the coefficient was 
some seven times as large as when an ' ' oil bath, ' ' or copious 
supply of oil, was provided; (0.0090 as against 0.0014). 

176. Rigidity of Ropes. — If a rope or wire cable passes over 
a pulley or sheave, a force J-* is required on one side greater 
than the resistance Q on the other for uniform motion, aside 
from axle-friction. Since in a given time botli I^ and Q 
describe the same space s, if ^ is > Q, then I^sis > Qs, i.e., 
the work done by i^ is > than that expended upon Q. This 
is because some of the work J*s has been expended in bending 
the stiff rope or cable, and in overcoming friction between the 
strands, both where the rope passes upon and where it leaves 



FRICTION. 



198 



the pulley. With hemp ropes, Fig. 191, the material being 
nearly inelastic, the energy spent in bending it on at D is 
nearly all lost, and energy must also be spent in straightening 




Fig. 191. 

it at E\ but with a wire rope or cable some of this energy is 
restored by the elasticity of the material. The energy spent 
in friction or rubbing of strands, however, is lost in both cases. 
The iigure shows geometrically why P must be > ^ for a 
uniform motion, for the lever-arm, a, of P is evidently < h 
that of Q. If axle-friction is also considered, we must have 

Pa=qb^f{P-^Q)r, 

r being the radius of the journal. 

Experiments with cordage have been made by Prony, Cou- 
lomb, Eytelwein, and Weisbach, with considerable variation in 
the results and forraulse proposed. (See Coxe's translation of 
vol. i., "Weisbach's Mechanics.) 

With pulleys of large diameter the effect of rigidity is very 
slight. For instance, Weisbach gives an example of a pulley 
five feet in diameter, with which, Q being = 1200 lbs., P 
= 1219. A wire rope f in. in diameter was used. Of this 
difference, 19 lbs., only 5 lbs. was due to rigidity, the remainder, 
14 lbs., being caused by axle-friction. When a hemp-rope 1.6 
inches in diameter was substituted for the wire one, P — ^=27 
lbs., of which 12 lbs. was due to the rigidity. Hence in one 
case the loss of work was less than \ of \%. in the other about 
1^, caused by the rigidity. For very small sheaves and thick 
ropes the loss is probably much greater. 
13 



194 MECHANICS OF ENGINEERING. 

/Vl'^, Miscellaneous Examples. — Example 1. Tiie end of a 

\ /shaft 12 inches ill diameter and making 50 revolutions per min- 

1/ ute exerts against its bearing an axial pressure of 10 tons and 

/ a lateral pressure of 40 tons. With /" ^y = 0.05, required 

the H. P. lost in friction. Ans. 22.2 H. P. 

Example 2. — A leather belt passes over a vertical pulley, 
covering half its circumference. One end is held by a spring 
balance, which reads 10 lbs. vi'hile the other end sustains a 
vreight of 20 lbs., the pulley making 100 revolutions per min- 
ute. Required the coefficient of friction, and the H. P. spent 
in overcoming the friction. Also suppose the pulley turned 
in the other direction, the weight remaining the same. The 
diameter of the pulley is 18 inches. . {f = 0.22 ; 

^^' I 0.142 and .284 H. P. 
Example 3. — A grindstone with a radius of gyration = 12 
inches has been revolving at 120 revolutions per minute, and 
at a given instant is left to the influence of gravity and axle 
friction. The axles are 1|- inches in diameter, and the wheel 
makes 160 revolutions in coming to rest. Required the coeffi- 
(jient of axle-friction. (Average.) Ans. / = 0.039. 

Example 4. — A board A, weight 2 lbs., rests horizontally on 
another B:, coefficient of friction of rest between them being 
f = 0.30. B is now moved horizontally with a uniformly 
accelerated motion, the acceleration being = 15 f set per " square 
seco'id ;" will A keep company with it, or not ? Ans, " if o." 



y 



V 



STRENGTH OF MATERTALSo 

[Or Mechanics of Materials] = 



CHAPTEE I. 

ELEMENTARY STRESSES AND STRAINS. 

178. Beformation of Solid Bodies. — In tlie preceding por- 
tions of this work, what was called technically a " rigid 
body," was supposed incapable of changing its form, i.e., 
the positions of its particles relatively to each other, under 
the action of any forces to be brought upon it. This sup- 
position was made because the change of form which must 
actually occur does not appreciably alter the distances, 
angles, etc., measured in any one body, among most of 
the pieces of a properly designed structure or machine. 
To show how the individual pieces of such constructions 
should be designed to avoid undesirable deformation or 
injury is the object of this division of Mechanics of En- 
gineering, viz., the Strength of Materials, 




,^6 



'€5. 



D 
Fis. 193. § 178. 




AS perhaps tne simplest instance of the deformation or 
distortion of a solid, let us consider the case of a prismatic 
rod in a state of tension, Eig. 192 (eye-bar of a bridge 

195 



196 MECHANICS OF EI>J GINBERmG. 

truss, e.g.). The pull at each end is P, and the body is 
said to be under a tension of P (lbs., tons, or other unit), 
not 2P. Let ABGD be the end view of an elementary 
parallelopiped, originally of square section and with faces 
at 45° with the axis of the prism. It is now deformed, the 
four faces perpendicular to the paper being longer"^ than 
before, while the angles BAD and BCD, originally right 
angles, are now smaller by a certain amount d, ABC and 
ADG larger by an equal amount d. The element is said 
to be in a state of strain, viz.: the elongation of its edges 
(parallel to paper) is called a tensile strain, while the alter- 
ation in the angles between its faces is called a shearing 
strain, or angular distortion (sometimes also called a slid- 
ing, or tangential, strain, since BG has been made to slide, 
relatively to AD, and thereby caused the change of angle). 
[This use of the word strain, to signify change of form and 
not the force producing it, is of recent adoption among 
many, though not all, technical writers.] 

179. Strains. Two Kinds Only. — Just as a curved line may 
be considered to be made up of small straight-line ele- 
ments, so the substance of any solid body may be consid- 
ered to be made up of small contiguous parallelopipeds, 
whose angles are each 90° before the body is subjected to 
the action of forces, but which are not necessarily cubes. 
A line of such elements forming an elementary prism is 
sometimes called a> fibre, but this does not necessarily imply 
a fibrous nature in the material in question. The system 
of imaginary cutting surfaces by which the body is thus 
subdivided need not consist entirely of planes ; in the sub- 
ject of Torsion, for instance, the parallelopipedical ele- 
ments considered lie in concentric cylindrical shells, cut 
both by transverse and radial planes. 

Since these elements are taken so small that the only 
possible changes of form in any one of them, as induced 
by a system of external forces acting on the body, are 

* When a is nearly 0° (or 90°) BG and AD (or AB and DG) are shorter 
than before, on account of lateral contraction; see § 193. 



ELEMENTARY STRESSES, ETC, 197 

elongations or contractions of its edges, and alteration of 
its angles, there are but two kinds of strain, elongation 
(contraction, if negative) and shearing. 

180. Distributed Forces or Stresses. — In the matter preced- 
ing this chapter it has sufficed for practical purposes to 
consider a force as applied at a point of a body, but in 
reality it must be distributed over a definite area ; for 
otherwise the material would be subjected to an infinite 
force per unit of area. (Forces like gravity, magnetic at- 
traction, etc., we have already treated as distributed over 
the mass of a body, but reference is now had particularly 
to the pressure of one body against another, or the action 
•of one portion of the body on the remainder.) For in- 
stance, sufficient surface must be provided between the 
end of a loaded beam and the pier on which it rests to 
avoid injury to either. Again, too small a wire must not 
be used to sustain a given load, or the tension per unit 
of area of its cross section becomes sufficient to rupture 
it. 

Stress is distributed force, and its intensity at any point 
of the area is 

• o e (1) 



"where dF is a small area containing the point and dP the 
force coming upon that area. If equal dP^s (all parallel) 
act on equal dF'soi a plane surface, the stress is said to 
be of uniform intensity, which is then 



i>=p . . . . (2) 



where P=-= total force and ^the total area over which it 
acts. The steam pressure on a piston is an example of 
stress of uniform intensity. 



X98 MECHANICS OF ENGINEEKING. 

For example, if a force P= 28800 lbs, is uniformly dis- 
tributed over a plane area of ^=72 sq. inches, or ^ of a 
sq. foot, the intensity of the stress is 

28800 ,^^,, . , 

p= =400 lbs. Der sq. inch, 

(or jp = 28800^ >^ =57600 lbs. per sq. foot, or p=14400-j' 
^=28.8 tons per sq. ft,, etc... 

181. Stresses on an Element : of Two Kinds Only. — When a 
solid body of any material is in eauiiibrium under a sys- 
tem of forces which do not rupture it. not only is its shape 
altered (i.e. its elements are strained), and stresses pro- 
duced on those planes on which the forces act, but other 
stresses also are induced on some or all internal surfaces 
which separate element from element, f over and above the 
forces with which the elements mav have acted on each 
other before the application of the external stresses or 
" applied forces "). So long as the whole solid is the "free 
body " under consideration, these internal stresses, being 
the forces with which the portion on one side of an imag- 
inary cutting plane acts on the portion on the other side, 
do not appear in any equation of eauiiibrium (for if intro- 
duced they would cancel out); but if we consider free a 
portion only, some or all of whose bounding surfaces are 
cutting planes of the original bodv. the stresses existing 
on these planes are brought into the eauations of equilib- 
rium. 

Similarly, if a single element of the body is treated by 
itself, the stresses on all six of its faces, together with its 
weight, form a balanced system of forces, the body being 
supposed at rest. 




FiS. 138. 




ELEMENTAE.T STRESSES, ETC. 



199 



As an example of internal stress, consider again the case 
of a bar in tension ; Fig. 193 shows tlie whole bar (or eye- 
bar) free, the forces P being the pressures of the pins in. 
the eyes, and causing external stress (compression here) 
on the surfaces of contact. Conceive a right section made 
through BS, far enough from the eye, (7, that we may con- 
sider the internal stress to be uniform * in this section, and 
consider the portion BSG as a free body, in Fig. 194. The 
stresses on R8, now one of the bounding surfaces of the 
free body, must be parallel to P, i.e., normal to B8; 
(otherwise they would have components perpendicular to 
P, which is precluded by the necessity oi lY being = 0, 
and the supposition of uniformity.) Let .^ = the sec- 




FlG. 194, 




Fig. 195. 



tional area RS, and p = the stress per unit of area ; then. 

P 



IX-= gives P= Fp, i.e., p= 



F 



.(2). 



The state of internal stress, then, is such that on planes 
perpendicular to the axis of the bar the stress is tensile and 
normal (to those planes). Since if a section were made 
oblique to the axis of the bar, the stress would still be 
parallel to the axis for reasons as above, it is evident that 
on an oblique section, the stress has components both nor- 
mal and tangential to the section, the normal component 
being a tension. 

* As will be shown later (§ 295) the line of the two P's in Fig. 193 must 
pass through the centre of gravity of the cross-section RS (plane figure) of 
the bar, for the stress to be uniform over the section. 



200 



MECHANICS OF EISTGINEERLNG. 



The presence of the tangential or shearing stress in ob- 
lique Sections is rendered evident by considering that if an 
oblique dove-tail joint were cut in the rod, Fig. 195, the 
shearing stress on its surfaces may be sufficient to over- 
come friction and cause sliding along the oblique plane. 

If a short prismatic block is under the compressive ac- 
tion of two forces, each =P and applied centrally in one 
base, we may show that the state of internal stress is the 
same as that of the rod under tension, except that the nor- 
mal stresses are of contrary sign, i.e., compressive instead 
of tensile, and that the shearing stresses (or tendency to 
slide) on oblique planes are opposite in direction to those 
in the rod. 

Since the resultant stress on a given internal plane of a 
body is fully represented by its normal and tangential 
components, we are therefore justified in considering but 
iwo kinds of internal stress, normal or direct, and tangen- 
tial or shearing. 

182. Stress on Obliq[ue Section of Rod in Tension, — Consider 
free a small cubic element whose 
edge =a in lengthy it has two 
faces parallel to the paper, being 
taken near the middle of the rod 
in Fig. 192. Let the angle which 
the face AB, Fig. 196, makes with 
the axis of the rod be = a. This 
angle, for our present purpose, is 
considered to remain the same 
while the two forces P are acting, 
as before their action. The re- 
sultant stress on the face AB hav- 
ing an intensity p=P-h-F, (see eq. 
2) per unit of transverse section 
of rod, is = jp (a sin a) a. Hence 
its component normal to AB is 

■pa^ sin^ a ; and the tangential or shearing component along 

AB '=*pa^ sin a cos a. Dividing by the area, a^, we have 

the following : 

For a rod in simple tension we have, on a plane making 

an angle, a, with the axis : 




ELEMENTARY STRESSES, ETC. 201 

a Normal Stress =p Bin? a per unit of area . . (1) 
and a Shearing Stress =p sin a cos a per unit of area . (2) 

" Unit of area " here refers to the oblique plane in ques- 
tion, while p denotes the normal stress per unit of area of 
a transverse section, i.e., when a=90°. Fig. 194. 

The stresses on CD are the same in value as on AB, 
while for BG and AD wq substitute 90° — a for a. Fig. 
197 shows these normal and shearing stresses, and also, 
much exaggerated, the strains or change of form of the 
element (see Fig. 192). 

182a, Eelation between Stress and Strain. — Experiment 
shows that so long as the stresses are of such moderate 
value that the piece recovers its original form completely 
when the external forces which induce the stresses are re- 
moved, the following is true and is known as Hoohe's Law 
(stress proportional to strain). As the forces P in Fig. 
193 (rod in tension) are gradually increased, the elonga- 
tion, or additional length, of BK increases in the same 
ratio as the normal stress, p, on the sections BS and KI^^ 
per unit of area [§ 191]. 

As for the distorting effect of shearing stresses, considei 
in Fig. 197 that since 

p sin a cos a = p cos (90° — a) sm (90° — a) 

the shearing stress per unit of area is of equal value on all 
four of the faces (perpendicular to paper) in the elementary 
block, and is evidently accountable for the shearing strain, 
i.e., for the angular distortion, or difference, d, between 
90° and the present value of each of the four angles. Ac- 
cording to Hooke's Law then, as P increases within tlvg 
limit mentioned above, d varies proportionally to 

p sin a cos a, i.e. to the stress. 

182b. Example. — Supposing the rod in question were of 
a kind of wood in which a shearing stress of 200 lbs. per 
sq. inch along the grain, or a normal stress of 400 lbs. per 
8q. inch, perpendicular to a fibre-plane will produce rup- 
ture, required the value of a the angle which the grain 
must make with the axis that, as P increases, the danger 
of rupture from each source may be the same. This re- 



202 MECHANICS OE ENGIl>rEBEI2fG. 

quires that 200:400::p sin a cos a :p sin^a, i.e. tan. a must 
= 2.000.-.a=63i^°. If the cross section of the rod is 2 sq. 
inches, the force P at each end necessary to produce rup- 
ture of either kind, when a=63^°, is found by putting 
p sin a cos ^=^00. '.^=500.0 lbs. per sq. inch. "Whence, since 
p=P-^F, P=1000 lbs. (Units, inch and pound.) 

183. Elasticity is the name given to the property which 
most materials have, to a certain extent, of regaining their 
original form when the external forces are removed. If 
the state of stress exceeds a certain stage, called the Elastic 
Limit, the recovery of original form on the part of the ele- 
ments is only partial, the permanent deformation being 
called the Set. 

Although theoretically the elastic limit is a perfectly defi- 
nite stage of stress, experimentally it is somewhat indefi- 
nite, and is generally considered to be reached when the 
permanent set becomes well marked as the stresses are in- 
creased and the test piece is given ample time for recovery 
in the intervals of rest. 

The Safe Limit of stress, taken well within the elastic 
limit, determines the working strength or safe load of the 
piece under consideration. E.g., the tables of safe loads 
of the structural steel beams for floors, made by the Cambria 
Steel Co. , at Johnstown, Pa, , are computed on the basis that 
the greatest normal stress (tension -or compression) occurring 
on any internal plane shall not exceed 16,000 lbs. per sq. inch; 
and, again, by the building laws of Philadelphia, the greatest 
shearing stress to be permitted in ' ' web plates " of " mild 
steel" is 8750 lbs. /in. 2 

The tJltimate Limit is reached when rupture occurs. 

184. The Modulus of Elasticity (sometimes called co-efficient 
of elasticity) is the number obtained by dividing the stress 
per unit of area by the corresponding relative strain. 

Thus, a rod of wrought iron ^ sq. inch sectional area 
being subjected to a tension of 2^ tons =5,000 lbs., it is 



ELEMENTARY STRESSES, ETC. 203 

iound that a length whicli was six feet before tension is 
»= 6.002 ft. during tension. The relative longitudinal strain 
or elongation is then= (0.002)-^6= 1 : 3,000 and the corres- 
ponding stress (being the normal stress on a transverse 
plane) has an intensity of 

i?t=P^i^= 5,000-^ 1^=10,000 lbs., per sq. inch. 

Hence by definition the modulus of elasticity is (for ten- 
sion), if we denote the relative elongation by s, 

Bt-=Pt'^ £=10,000^ g-^ =30,000,000 lbs. per sq. inch, (the 

sub-script " t " refers to tension). 

It will be noticed that since £ is an abstract number, Et 
is of the same quality as p^, i.e., lbs. per sq. inch, or one di- 
mension of force divided by two dimensions of length. 
(In the subject of strength of materials the inch is the 
most convenient English linear unit, when the pound is 
the unit of force ; sometimes the foot and ton are used to- 
gether.) 

The foregoing would be called the modulus of elasticity 
of lorought iron in tension in the direction of the fibre, as 
given by the experiment quoted. • But by Hooke's Law p 
and £ vary together, for a given direction in a given ma- 
terial, hence ivithin the elastic limit E is constant for a given 
direction in a given material. Experiment confirms this 
approximately. 

Similarly, the modulus of elasticity for compression E^ 
in a given direction in a given material may be determined 
by experiments on short blocks, or on rods confined lat- 
erally to prevent flexure. 

As to the modulus of elasticity for shearing, E^, we 
divide the shearing stress per unit of area in the given 
direction by (? (in radians) the corresponding angular strain 
or distortion; e.g., for an angular distortion of 0.10° or 
,^ = .001T4, and a shearing stress of 15,660 lbs. per sq. inch, 
we have £;=^^ = 9,000,000 lbs. per sq. inch. 



204 MECHANICS OF ENGINEERING. 

184a. Young's Modulus is a name frequently given toEf and 
Ec, it being understood that in the experiments to determine 
these moduli the elastic limit is not passed, and also that the 
rod or prism tested is not subjected to any stress on the sides. 

See p. 507. 

185. Isotropes. — This name is given to materials which 
are homogenous as regards their elastic properties. In 
such a material the moduli of elasticity are individually 
the same for all directions. E.g., a rod of rubber cut out 
of a large mass will exhibit the same elastic behavior when 
subjected to tension, whatever its original position in the 
mass. Fibrous materials like wood and wrought iron are 
not isotropic ; the direction of grain in the former must 
always be considered. The " piling " and welding of nu- 
merous small pieces of iron prevent the resultant forging 
from being isotropic. 

186. Resilience refers to the potential energy stored in a 
body held under external forces in a state of stress which 
does not pass the elastic limit. On its release from con- 
straint, by virtue of its elasticity it can perform a certain 
amount of work called the resilience, depending in amount 
upon the circumstances of each case and the nature of the 
material. See § 148. 

187. General Properties of Materials. — In viev/ of some defi- 
nitions already made we may say that a material is ductile 
when the ultimate limit is far removed from the elastic 
limit ; that it is brittle like glass and cast iron, when those 
limits are near together. A small modulus of elasticity 
means that a comparatively small force is necessary to 
produce a given change of form, and vice versa, but implies 
little or nothing concerning the stress or strain at the 
elastic limit ; thus Weisbach gives E^, lbs. per sq. inch for 
wrought iron = 28,000,000= double the E^ for cast iron 
while the compressive stresses at the elastic limit are the 
same for both materials (nearly). 



ELEMENTARY STRESSES, ETC. 



205 




188. Element with Normal Stress on Sides as well as on End-Faces. 
Ellipse of Stiess.— In Fig. 193, p. 198, the parallelopiped RKNS is sub- 
jected to stress on the two end-faces only. Let us now consider a small 
square-cornered element of material subjected to a normal stress p^ 
(tension) on the two vertical end-faces, while on the horizontal side faces 
there acts a normal (also tensile) stress of pj Ibs./in.^; (but no stress 

on the vertical side ^- ^n'---^ / 

faces). In Fig. 197a '''' ' ^""- '^ 

is shown, as a free 
body in equilibrium, 
a triangular prism 
ABC, which is the 
upper right-hand 
half of such an ele- 
ment; obtained by 
passing the cutting 
plane AC along a 
diagonal of the side 
plane (plane of 
paper) on which 
there is no stress, 
and 1 to it. The 
angle 6 may have 
any value and it 
is desired to deter- 
mine the unit stress "iG- 197a. 
5o induced on the oblique plane AC by the normal stresses Pi and Pz 
acting respectively on the end face BC and on the side face AB. The 
unit stress q^ on the face AC is not 1 to that face but makes with it 
some angle ^. Let AB = h inches, BC = n in., and AC = c in.; each 
of the rectangular areas having a common dimension, =d in., T to 
the paper. Then the total (oblique) stress on face AC is q^cd lbs., that 
on AB is P'pd, and that on BC is p-jid lbs. Since the total stress on AC 
is the anti-resultant of the other two, and these are T to each other, 
we have 

{,q^cdy={,PTndy+{p^hdy; i.e., ?o^ = fpiyj + (pj- 
But, since ?i^c = sin d, and fe-^c = cos 6, this may be written 

q-'={p,smey+{p,coBdy (i) 

Eq. (1) gives the magnitude of q^ for any value of angle d; but both 
position and magnitude are best shown by a geometrical construction. 
being any point on AC, draw a circle with center at and radius, 
OH^, equal by scale to the unit stress p^. Similarly, with radius OH 2, 
equal (on same sea'"") to the unit stress P2, draw the circle H2E2. Through 
draw EfiN normal to the face AC on which the stress 50 is to be deter- 
mined, and note the intersections E^ and £'2 (both on left of 0) with 
the two circles respectively. A vertical line through E^ and a hori- 
zontal through E2 intersect at some point m. Cm is the magnitude 
and position of the stress q^; since mD2 = EiDi = pi sin 0, and ODj = 
P2COSO; hence from eq. (1) Om=5o. 



206 MECHANICS OF ENGINEERING. 

T he po int m is a point in an ellipse whose semi-principal axes are OH^ 
and OH 2, i.e., p^ and pj. This ellipse is called the Ellipse of Stress; 
Om being a semi-diameter, determined in the way indicated. (Similarly, 
if the elementary right parallelepiped is subjected to the action of three 
normal stresses, Pi, p2, and p^, on all three pairs of faces, respectively, 
the unit stress on any oblique plane is a semi-diameter of an Ellipsoid 
of Stress). 

The unit shearing stress on the oblique face AC is qs=^qo cos ,u; and 
the unit normal stress is q=qo sin /;. 

In case the normal stress P2 on the face AB were compressive, p^ being 
tensile, a horizontal would be drawn through E'2 on the circle of radius 
OH2, instead of through E2, to meet the vertical through E^, and would 
thus determine Om', instead of Om, as the stress on AC. If, in such 
a case, P2 were numerically equal to p^, and d were 45°, go would = Pi = pj, 
and would lie in the surface AC (pure shear; compare with Exam. 5, 
p. 242). With Pi = P2, and both tensUe, or both compressive, 50 would 
be equal to Pi, =P2, for all values of d. 

189. Classification of Cases. — Althougli in almost any case 
whatever of the deformation of a solid body by a balanced 
system of forces acting on it, normal and shearing stresses 
are both, developed in every element which is affected at 
all (according to the plane section considered,) still, cases 
where the body is prismatic, and the external system con- 
sists of two equal and opposite forces, one at each end of 
the piece a,nd directed away from each other, are commonly 
called cases of Tension; (Fig, 192); if the piece is a short 
prism with the same two terminal forces directed toward 
each other, the case is said to be one of Compression ; a case 
similar to the last, but where the prism is quite long 
(" long column "), is a case of Flexure or bending, as are also 
most cases where the " applied forces " (i.e., the external 
forces), are not directed along the axis of the piece. Rivet- 
ed joints and " pin-connections " present cases of Shearing; 
a twisted shaft one of Torsion. When the gravity forces 
due to the weights of the elements are also considered, a 
combination of two or more of the foregoing general cases 
may occur. 

In each case, as treated, the principal objects aimed at 
are, so to design the piece or its loading that the greatest 
stress,* in whatever element it may occur, shall not exceed 
a safe value ; and sometimes, furthermore, to prevent too 
great deformation on the part of the piece. The first ob- 
ject is to provide sufficient strength; the second sufficient 
stiffness. 

* See § 405b for mention of the "elongation theory" of safety. This 
is based on considerations of strain, or deformation, instead of stress. 



TMNSION. 



207 



te:nsion. 

191. Hooke's Law by Experiment. — As a typical experiment 
in the tension of a long rod of ductile metal sncli as 
wrought iron and the mild steels, the following table is quot- 
ed from Prof. Cotterill's " Applied Mechanics." The experi- 
ment is old, made by Hodgkinson for' an English Railway 
Commission, but well adapted to the purpose. From the 
great length of the rod, which was of wrought iron and 
0.517 in. in diameter, the portion whose elongation was 
observed being 49 ft. 2 in. long, the small increase in length 
below the elastic limit was readily measured. The succes- 
sive loads were of such a value that the tensile stress 
p=P^F, or normal stress per sq. in. in the transverse 
section, was made to increase by equal increments of 2657.5 
lbs. per sq. in., its initial value. After each application of 
load the elongation was measured, and after the removal 
of the load, the permanent set, if any. 

Table of Elongations of a Wrought Iron Rod, of a 
Length = 49 Ft. 2 In. 



p 


X 


JA 


e^X^l 


X' 


Load (lbs 
square ii 


. per Elongation, 
ich.) (inches.) 


Increment 

of 
Elongation. 


s, the relative 
elongation, (ab- 
stract number.) 


Permanent 

Set. 
(inches.) 


1X266 


7.5 .0485 


.0485 


0.000082 




2X ' 


. 1095 


.061 


.000186 




3X ' 


. 1675 


.058 


.000283 


0.0015 


4X ' 


.224 


.0565 


.000379 


.002 


5X ' 


.2805 


.0565 


.000475 


.0027 


6X ' 


.337 


.0565 


.000570 


.003 


7X ' 


.393 


.056 




.004 


8X ' 


.452 


.059 


.000766 


.0075 


9X ' 


.5155 


.0635 




.0195 


lOX ' 


.598 


.0825 




.049 


IIX ' 


.760 


.162 




.1545 


12X ' 


1.310 


.550 




.667 


etc. 











208 



MECHANICS OF E]SrGIl!fEEEIl!fG. 



Referring now to Fig. 198, the notation is evident. P 
is the total load in any experiment, F the cross section of 
the rod ; hence the normal stress on the transverse section 
is p=P-r-F. When the loads are increased by equal in- 
crements, the corresponding increments of the elongation 
a should also be equal if Hooke's law is true. It will be 
noticed in the table that this is very nearly true up to the 
8th loading, i.e., that JX, the difference between two con- 
secutive values of }., is nearly constant. In other words the 
proposition holds good ; 



if P and Pi are any two loads below the 8th, and X and ki 
the corresponding elongations. 

The permanent set is just perceptible at the 3d load, and 
increases rapidly after the 8th, as also the increment of 
elongation. Hence at the 8th load, which produces a ten- 
sile stress on the cross section of j9= 8x2667.5= 21340.0 
lbs. per sq. inch, the elastic limit is reached. 

As to the state of stress of the individual elements, if 

we conceive such sub-division 
of the rod that four edges of 
each element are parallel to the 
axis of the rod, we find that it 
is in equilibrium between two 
normal stresses on its end faces 
''^^ (Fig. 199) of a value ^pdF== 
{P^F)dF where dF is the hor- 
izontal section of the element. 
If dx was the original length, 
and dX the elongation produced by pdF, we shall have, 
since all the dx's of the length are equally elongated at the 

dX X 
same time, w" ^ T 

where Z = total (original) length. But dX^dx is the relative 
elongation e, and by definition (§ 184) the Modulus of Elas- 
ticity for Tension, Ei, = jp^e, {Young's Modulus, § 184a). 




TENSION. 209 

.'.E.=-4rr or E,=^^ .... (1) 




Jblq. (1) enables us to solve problems involving the elonga- 
tion of a prism under tension, so long as the elastic limit 
is not surpassed. 

The values of E^ computed from experiments like those 
just cited should be the same for any load under the elas- 
tic limit, if Hooke's law were accurately obeyed, but in 
reality they differ somewhat, especially if the material 
lacks homogeneity. In the present instance (see Table) 
we have from the 

2d Exper. ^=^-^£=28,680,000 lbs. per sq. in. 
5th " Ec= " =28,009,000 
8th " ^t= " =27,848,000 

192. Stress-Strain Diagrams. — If the relative elongations 
or "strains " (s) corresponding to a series of values of the 
tensile unit-stresses (p) (Ibs./in.^) to which a rod of metal 
has been subjected in a testing machine, are plotted as 
abscissae, and the unit-stresses themselves (p) as ordinates, 
we have in the curve joining these points a useful graphic, 
representation of the results of experiment. 

Fig. 200 shows some of these curves, giving average re- 
sults for the principal "ferrous " metals. On the left, in 
(I), the scale adopted (horizontal) for the "strain " (e) or 
"unit-elongation " is one hundred times as great as that 
used in the right-hand diagram, (II) ; while the vertical 
scale for stress (p) in (I) is only twice as great as that in 
' (II) . The change of form within the elastic limit is so 
small compared with that beyond, that this difference in 
scale is quite necessary in order that diagram (I) may show 
what occurs within the elastic limit and a Httle beyond. 
Diagram (II) shows the remainder of the curves of wrought 
iron and soft steel, up to the point of rupture. 

We have here the means of comparing the properties of the 
four typical metals represented, as to elasticity and tenacity. 
Up to the respective elastic hmits, B, B' , B" , and B'" , stress 
is fairly proportional to strain, and a straight line is the result; 



210 



MECHANICS OF ENGINEERING. 



the "true elastic limit " being regarded as the point where 
such proportionahty ceases. In the case of wrought iron and 
soft steel there is a point Y, called the "yield point," a little 
above the true elastic limit, and sometimes called the "apparent 
elastic limit/' or "commercial elastic limit/' immediately be- 
yond which further slight increments of stress produce rela- 
tively great increments of strain, permanent set becoming 
very marked; i.e., the part YD of the curve is almost hori- 
zontal. Beyond D the curve rises again, more steeply, but 
just before rupture [see (II)] may descend somewhat; since, 



50,000 r r--y^ — ^ 



40,000 




10,000 
Ibs./iv} 



Fig. 200. 

on account of the lateral contraction mentioned in the next 
paragraph, here plotted, the stress being computed by dividing 
the total pull by the original sectional area, is less toward 
rupture than at stages closely preceding. 

If at any point beyond the elastic limit, as at C (see curve 
for wrought iron) in (I), the stress be gradually removed, the 
relations of stress and strain during this gradual diminution 
of stress, are shown by the straight line CC. The position of 
the point C indicates that there is in the rod (now under no 
stress) a permanent set, or relative elongation, of £ = 0.0015, 
or 15 parts in 10,000, an elastic recovery having occurred from 
0.0028 to 0.0015 (see horizontal scale). 

Since by definition the modulus of elasticity E = p^s, the values 
of the respective moduli for the metals in diagram (I) are propor- 



TENSION. 



211 



tional to the tangent of the angle which the corresponding 
straight portions OB, OB' , etc., make with the horizontal axis. 
From the various ordinates and abscissae for the points B, B', 
etc., we find E for cast iron to be 14,000,000 Ibs./in.^ and for 
the other three metals 28,000,000, 30,000,000, and 40,000,000, 
respectively. The curve for the "harder steel" is not 
shown in (II), being beyond the limits of the diagram, as 
to stress; and the complete curve for cast iron is contained 
within the limits of diagram (I), since the elongation at 
rupture is very small in the case of this metal, only about 
3/10 of one per cent, or 3 parts in 1000; whereas that for 
wrought iron or soft steel is 300 parts in 1000 (or 30 per 
cent). In the case of cast iron the elastic limit is very ill- 
defined and the proportion of carbon and the mode of manu- 
facture have much influence on its behavior under test. 

"Soft steel" is another name for "structural steel," used 
in construction on a large scale, as in buildings and bridge 
trusses; "medium steel " being a somewhat harder grade 
of the same. Many grades of steel are made which are 
much stronger and harder than these, such as tool steel, 
nickel steel, and piano wire (whose rupturing stress may 
be as high as 300,000 Ibs./in.^). Wrought iron in the form 
of wire is much stronger than in bars. 

Note. — Such a line as CC, showing the relation of stress and strain 
as the stress is gradually removed, will be called an "elasticity line" 
on p. 241. In § 206 some mention will be made of the phenomena of 
"overstraining" a test-piece of iron or steel, showing that on re-applying 
stress after a certain period of rest the plotted results of stress and 
strain relations show that the line C'C is retraced to C and continues 
in the same straight line prolonged, to a new elastic limit higher than C, 
before curving off to the right. 

193. Lateral Contraction. — In the stretching of prisms of 
nearly all kinds of material, accompanying the elongation 
of length is found also a diminution of width whose rela- 
tive amount in the case of the three metals just treated is 
about ^ or i^ of the relative elongation (within elastic 
limit). Thus, in the third experiment in the table of § 191, 
this relative lateral contraction or decrease of diameter 
~ H ^^ /i ^^ ^> ^■^•> about 0.00008. In the case of cast 
iron and hard steels contraction is not noticeable ex- 




212 MECHANICS OF ENGINEEBING 

csept by very delicate measurements, both within and with- 
out the elastic limit ; but the more ductile metals, as 
wrought iron and the soft steels, when stretched beyond 
the elastic limit show this feature of their deformation 
in a very marked degree. Fig. 201 shows by dotted lines 
the original contour of a wrought iron rod, while the con- 
tinuous lines indicate that at rupture. At the cross section 
of rupture, whose position is determined by some 
local weakness, the drawing out is peculiarly 
pronounced. 

The contraction of area thus produced is some- 
times as great as 50 or 60% at the fracture. 

194. "Flow of Solids." — When the change in re- 
lative position of the elements of a solid is ex- 
treme, as occurs in the making of lead pipe, 
I drawing of wire, the stretching of a rod of duc- 
j tile metal as in the preceding article, we have 
Fig. 201. instances of what is called the Flow of Solids, in- 
teresting experiments on which have been made by 
Tresca. 

195. Moduli of Tenacity. — The tensile stress per square 
inch (of original sectional area) required to rupture a 
prism of a given material will be denoted by T and called 
the modulus of ultimate tenacity ; similarly, the modulus oj 
safe tenacity, or greatest safe tensile stress on an element, 
by T' ', while the tensile stress at elastic limit may be 
called T". The ratio of T' to T" is not fixed in practice 
but depends upon circumstances (from j4, to ^). 

Hence, if a prism of any material sustains a total pull 
or load P, and has a sectional area=jP, we have 

P= FT for the ultimate or breaking load. \ 

P'=FT' " " safe load. f ' ' (^^ 

P"=FT" " " load at elastic limit. ) 

Of course T' should always be less than T". (The hand-, 
book of the Cambria Steel Co. , in quoting from the building 
laws of various cities of the U. S., gives allowable unit- 
stresses for ordinary materials, both in tension and com- 
pression.) 



TENSION. 213 

196. Resilience of a Stretched Prism. — ^Fig. 202. In the 
gradual stretcliing of a prism, fixed at one extremity ^ the 
value of the tensile force P at the other necessarily de- 
pends on the elongation A at each stage of the lengthening, 
according to the relation [eq. (1) of § 191.] 

' ^^ (8) 



FE, 



within the elastic limit. (If we place a weight G on the 

^^ flanges of the unstretched prism and then leave 

^ it to the action of gravity and the elastic action 

of the prism, the weight begins to sink, meeting 

an increasing pressure P, proportional to l, from 

the flanges). Suppose the stretching to continue 

until P reaches some value P" (at elastic limit 

\\ say), and I a value X'. Then the work done so 

N^ far is (see p. 155) 

Fig 802 ^7= mean force X space = ^ P" /I" . . (4) 

But from (2) F'=FT", and (see §§ 184 and 191) 

K"=e"l 
.-. (4) becomes XJ=y2 T e". Fl=}^ T e" V . . (6) 

where Fis the volume of the prism. The quantity }4T"£", 
or work done in stretching to the elastic limit a cubic 
inch (or other unit of volume) of the given material, may 
be called the Modvlus of Resilience for tension. From (5) 
it appears that the amounts of work done in stretching to 
the elastic limit prisms of the same material but of differ- 
ent dimensions are proportional to their volumes simply. 
The quantity }4T"e" is graphically represented by the 
area of one of the triangles such as OA'B, OA'B" in Fig. 
200 ; for (in the curve for wrought iron for instance) the 
modulus of tenacity at elastic limit is represented by ^'P, 
and e" (i.e., e for elastic limit) by OA'. The remainder of 



214 MEOHAE^ICS OF ENGINBERI^a. 

the area OBG included between the curve and the hori- 
zontal axis, i.e., from B to G, represents the work done in 
stretching a cubic unit from the elastic limit to the point 
of rupture, for each vertical strip having an altitude =p 
and a width =de, has an area ^pde, i.e., the work done by 
the stress p on one face of a cubic unit through the dis- 
tance de, or increment of elongation. 

If a weight or load = (r be " suddenly "applied to stretch 
the prism, i.e., placed on the flanges, barely touching 
them, and then allowed to fall, when it comes to rest again 
it has fallen through a height X^, and experiences at this 
instant some pressure P\ from the flanges; Pi=:?. Apply- 
ing to this body the "Work and Energy" method (p. 138), 
noting that its initial and final kinetic energy are each zero 
and that the force G is constant, while the upward force P 
(from the flanges) is variable, with an average value of JPi, 
we have 

GAi = iPiAi + 0-0; whence Pi = 2(?. 
Since Pi = 2G, i.e., is >(t, the body does not remain in 
this position but is pulled upward by the elasticity of the 
prism. In fact, the motion is harmonic (see §§ 59 and 
138). Theoretically, the elastic limit not being passed, the 
oscillations should continue indefinitely. 

Hence a load O " suddenly applied " occasions double the 
tension it would if compelled to sink gradually by a sup- 
port underneath, which is not removed until the tension is 
just = Q, oscillation being thus prevented. 

If the weight G sinks through a height —h before strik- 
ing the flanges. Fig. 202, we shall have similarly, within 
elastic limit, if ^i= greatest elongation, (the mass of rod 
being small compared with that of G). 

G{h^K)=%P,K .... (6) 

If the elastic limit is to be just reached we have from eqs. 
(5) and (6), neglecting ^ compared with h, 

Gh=%T"B"V . . . (7> 



TBNSIOX. 



215 



nn equation of condition that tlie prism shall not be in- 
jured. 

Example. — If a steel prism have a sectional area of i/ 
eq. inch and a length ^=10 ft. =120 inches, what is the 
greatest allowable height of fall of a weight of 200 lbs., 
that the final tensile stress induced may not exceed T" = 
30,000 lbs. per sq. inch, if e" z=.002 ? From (7), using tha 
inch and pound, we have 



h= 



T"e"V 



30,000 X. 002x1^x120. 



2^ 



2x200 



:4.5 inches. 



197. Stretching of a Prism by Its Own Weight. — In the case 
of a very long prism such as a mining- 
pump rod, its weight must be taken into 
account as well as that of the terminal 
load P , see Fig. 203. At (a.) the prism 
is shown in its unstrained condition ; at 
(&) strained by the load P^ and its own 
weight. Let the cross section be =jP, the 
heaviness of the prism =y. Then the rela- 
tive extension of any element at a distance 




Fig. 203. 



jy from o is "^ 



^_dX {P,+rFx) 



dx' 



FE, 



(1) 



{See eq. (1) § 191) ; since P^-^-Fjx is the load hanging upon 
the cross section at that locality. Equal c?a?'s, therefore, 
are unequally elongated, x varying from to I. The total 
elongation is 



=/*=jk/f^''"+''^"'"^=/i; 



'/2GI 



FE, 



Le., k= the amount due to Pi, plus an extension which 
half the weight of the prism would produce, hung at the 
lower extremity. 
PI 

* In A. = put dX for A, dx for I, and (P, + j^^.r) for P, 

FEt 



216 MECHANICS OF ENGINEERING. 

The foregoing relates to tlie deformation of the piece, 
and is therefore a problem of stiffness. As to the strength 
of the prism, the relative elongation e=dA-h-dx [see eq. (1)], 
which is variable, must nowhere exceed a safe value e'= 
T'^E, (from eq. (1) § 191, putting P=FT', and X=X), 
Now the greatest value of the ratio dX : dx, by inspecting 
eq. (1), is seen to be at the upper end where x=l. The 
proper cross section F, for a given load Pj, is thus found. 

Putting ^]^-~^ ^e have F =^^ . (2) 

198. Solid of Uniform Strength in Tension, or hanging body 
of minimum material supporting its own 
weight and a terminal load Pj. Let it be a 
solid of revolution. If every cross-section 
P at a distance =x from the lower extrem- 
ity, bears its safe load FT', every element 
of the body is doing full duty, and its form 
is the most economical of material. 

The lowest section must have an area 

Fi(j.204. Fo=P^-^T', since Pi is its safe load. Fig. 

204. Consider any horizontal lamina ; its weight is yFdx, 

(j= heaviness of the material, supposed homogenous), and 

its lower base Pmust have Pi-\-G for its safe load, i.e. 

G+F,=FT' ... a) 

in which G denotes the weight of the portion of the solid 
below F. Similarly for the upper base F-\-dF, we have 

G+P,+r^dx={F+dF)T' . , (2) 

By subtraction we obtain 

rFdsc=T'dFi ie. -l.dx^ ^ 
T F 




TENSION. 217 

in whicli the two variables x and F are separated. By in- 
tegration we now have 






;or^,=log.e^ . . (3) 



. \x p yx 

1.6., F=Foer =-1, e~ (4) 

from which i^may be computed for any value of x. 

The weight of the portion below any F is found from (1) 
and (4); i.e. 

while the total extension ^ will be 

^=^"^1 (6) 

"the relative elongation dX-i-dx being the same for every dx 
and bearing the same ratio to e" (at elastic limit), as T' 
does to T". 

199. Tensile Stresses Induced by Temperature. — If the two 
ends of a prism are immovably fixed, when under no strain 
and at a temperature t, and the temperature is then low- 
ered to a value t', the body suffers a tension proportional 
to the fall in temperature (within elastic limit). If for a 
rise or fall of 1° Fahr. (or Cent.) a unit of length of the 
material would change in length by an amount t^ (called 
the co-efficient of expansion) a length =1 would be con- 
tracted an amount X=-fjl(t-t') during the given fall of tem- 
perature if one end were free. Hence, if this contraction 
is prevented by fixing both ends, the rod must be under a 
:tension P, equal in value to the force which would be 



218 MECHANICS OF exgi]s:eeeing. 

necessary to produce the elongation X, just stated, under 
ordinary circumstances at the lower temperature. 

!From eq. (1) §191, therefore, we have for this tension 
dtite to fall of temperature 

For 1° Cent, we may write 

For Cast iron -f] = .0000111 ; 
« Wrought iron = .0000120 ; 
« Steel = .0000108 to .0000114 J 

« Copper yj = .0000172 ; 

« Zinc 7^ = .0000300, 



COMPRESSION OF SHORT BLOCKS. 

200, Short and Long Columns. — In a prism in tension, its- 
own weight being neglected, all the elements between thl 
jocaiities of application of the pair of external forces pro- 
ducing; the stretching are in the same state of stress, if the 
external forces act axially (excepting the few elements in the 
immediate neighborhood of the forces ; these suffering 
local stresses dependent on the manner of application of 
the external forces), and the prism may be of any length 
without vitiating this statement. But if the two external 
forces are directed toivard each other the intervening ele- 
ments will not all be in the same state of compressive 
stress unless the prism is comparatively short (or unless 
numerous points of lateral support are provided). A long 
prism will buckle out sideways, thus even inducing tensile 
stress, in some cases, in the elements on the convex side. 

Hence the distinction between sTiort UocTcs and long 
columns. Under compression the former yield by crush- 
ing or splitting, while the latter give way by flexure (i.e. 
bending). Long columns, then will be treated separately 



COMPRESSION OF SHORT BLOCKS. 219 

In a subsequent chapter. In the present section tlie blocks 
treated being about tliree or four times as long as wide, 
^11 tlie elements will be considered as being under -equal 
compressive stresses at tbe same time. 

201. Notation for Compression. — By using a subscript c, 
we may write 

E^= Modulus of Elasticity;* i.e. tlie quotient of the 
compressive stress per unit of area divided by the relative 
shortening. (Young's Modulus; no stress on sides); 

C= Modulus of crushing ; i.e. the force per unit of sec- 
tional area necessary to rupture the block by crushing ; 

G'= Modulus of safe compression, a safe compressive 
stress per unit of area ; and 

G"= Modulus of compression at elastic limit. 

For the absolute and relative shortening in length we 
may still use X and e, respectively, and within the elastic 
limit may write equations similar to those for tension, F 
being the sectional area of the block and F one of the ter- 
minal forces, while p = compressive stress per unit of area 
of Ff viz.: 



. (1) 







v-F _ 

-dx 


X- 


rF_ 

■A ~ 


.PI 
~FX 


ithin the elastic limit. 










Also for 


a short block 











Crushing force =FG 

Compressive force at elastic limit =iFG" }• . (2]f 

Safe compressive force =FG' 



limit r=FG" \ . 

7' ) 



202. Remarks on Crushing. — As in § 182 for a tensile 
stress, so for a compressive stress we may prove th<at a 

*[NoTE. — It must be remembered that the modulus of elasticity, 
whether for normal or shearing stresses, is a number indicative of stiff- 
ness, not of strength, and has no relation to the elastic limit (except 
that experiments to determine it must not pass that limit).] 



220 MECHANICS OF ENGINEERING. 

shearing stress = p.sin. a cos a is produced on planes at an 
angle a with the axis of the short block, p being the com- 
pressive stress per unit of area of transverse section. Experi- 
ment shows, however, that, although the above value for the 
shea,riug stress is a maximum for a = 4: 5°, in the crushing of 
shoi^t blocks or rather brittle materials like cast iron and stone, 
the surface along which separation takes place makes an angle 
smaller than 45° with the axis (35° for cast iron, according to 
Hodgkinson's experiments) ; but the block must be two or 
three times as long as wide to enable this phenomenon to take 
place. This seems to show that the presence of the com- 
pressive stress on the 45° plane is sufficient to strengthen the 
material against rupture by shearing on that plane, causing 
the separation to occur along a plane on which the compressive 
stress is considerably less. Crushing by splitting into pieces 
parallel to the axis sometimes occurs. 

Blocks of ductile material, however, yield by swelling 
out, or bulging, laterally, resembling plastic bodies some- 
what in this respect. 

The elastic limit is more difficult to locate than in ten- 
sion^ but seems to have a position corresponding to that 
in tension, in the case of wrought iron and steel. With 
cast iron, however, the relative compression at elastic 
limit is about double the relative extension (at elastic 
limit in tension), but the force producing it is also double. 
For all three metals it is found that E^^=E^ quite nearly, 
so that the single symbol U m.aj be used for both. 



EXAMPLES IIS" TENSION AND COMPRESSION. 

203. Tables for Tension and Compression. — The round 

numbers in the following tables are to be taken as rude aver- 
ages only ; the scope and design of the present work admitting 
of nothing more. For abundant details of the more import- 
ant experiments and researches of recent years, the reader 
is referred to Professor J. B. Johnson's "Materials of Con- 
struction ' ' and the works of Professors Thurston, Burr, and 
Lanza ; also to ' ' Testing of Materials ' ' by Unwin, and 
Martens' work of similar title. Another column might 
have been added giving the Modulus of Resilience, viz. : 
ie"T'\ {~^T"^^2E; see §196). e is an abstract num- 



EXAMPLES Ilf TEirSIOlS' A^D COMPRESSION 



221 



ber, and =X-^l, while E^, T", and T are given in pounds 
per square incli: 

TABLE OF THE MODULI, ETC., OP MATEEIALS IN TENSION. 





e" 


£ 


K 


rpn 


T 


Material. 


(Elastic limit.) 


At Rupture. 


Mod. of Elast. 


Elastic limit. 


Eupture. 




abst. number. 


abst. number. 


lbs. per sq. in. 


lbs. per sq. in. 


lbs. per sq. in. 


Soft Steel, 


.00120 


.3000 


30,000,000 


35,000 


60,000 


Hard Steel, 


.00200 


.0500 


40,000,000 


60,000 


120,000 


Cast Iron, 
Wro't Iron, 

Brass, 


.00066 
.00080 

.00100 


.0020 
.3000 


14,000,000 
28,000,000 

10,000,000 


9,000 

22,000 

f 7,000 

to 
L 19,000 


18,000 
• ■ 45 000 

to 
60,000 
16.000 

to 
50,000 


Glass, 






9,000,000 




3,500 


Wood, with 
the fibres. 


( .00200 
K to 
( .01100 


.0070 

to 
.0150 


200,000 
to 
2,000,000 


3,000 

to 

19,000 


6,000 

to 

28,000 


Hemp rope, 










7,000 



[N.B.— Expressed in kilograms per square centim., E^, T and T" would be nu 
merically about V]4 as large as above, while € and e" would be unchanged.] 



TABLE OF MODULI, ETC.; COMPEESSION OP SHORT BLOCKS. 





e" 


£ 


E, 


G" 


C 


Material. 


Elastic limit. 


At lupture- 


Mod. of Elast. 


Elastic limit. 


Rupture. 




abst. number 


abst. number. 


lbs. per sq. in. 


lbs. per sq. in. 


lbs. per sq. in. 


Soft Steel, 0.00100 




30,000,000 


30,000 




Hard Steel. 


0.00120 


0.3000 


40,000,000 


50,000 


200,000 


Cast Iron, 


0.00150 




14,000,000 


20,000 


90,000 


Wro't Iron, 


0.00080 


0.3000 


28,000,000 


24,000 


40,000 


Glass, 










20,000 


Granite, 

Sandstone, 

Brick, 






See 
§213a 




10,000 
5,000 
3,000 


Wood, with 
the fibres. 




I 0.0100 
< to 
1 0.0400 


350,000 

to 

2,000,000 




2,000 

to 
10,000 


Portland 1 
Cement, f 






(§ 213a) 




4,000 



"222 MECHANICS OF ENGINBEEING. 

204. Examples. No. 1. — A bar of tool steel, of sectional 
area =0.097 sq. inches, is ruptured by a tensile force of 
14,000 lbs. A portion of its length, originally ^ a foot, 
is now found to have a length of 0.532 ft. Required T, 
and e at rupture. Using the inch and pound as units (as 
in the foregoing tables) we have T= 1^=144326 lbs. per 
Bq. in.; (eq. (2) § 195) ; while 

e=(0.532— 0.5)x12h-(0.50x12)=0.064. 

Example 2. — Tensile test of a bar of " Hay Steel " for 
the Glasgow Bridge, Missouri. The portion measured was 
originally 3.21 ft. long and 2.09 in. X 1.10 in. in section. 
At the elastic limit P was 124,200 lbs., and the elongation 
was 0.064 ins. Required E^, T''^ and e" (for elastic limit). 

e"=^ =,-M54^=.00166 at elastic limit. 
I 3.21x12 

r"=124,200--(2.09xl.l0)=54,000 lbs. per sq. in. 

Nearly the same result for E^ would probably have been 
^obtained for values of p and e below the elastic limit. 

The Modulus of Resilience of the above steel (see § 196) 
would be ^2 e" :!r"= 44.82 inch -pounds of work per cubic 
inch of metal, so that the whole work expended in stretch- 
ing to the elastic limit the portion above cited is 

Cr= y^ e" T" r=3968. inch -lbs. 

An equal amount of work will be done by the rod in re- 
-covering its original length. 

Example 3. — ^A hard steel rod of ^ sq. in. section and 
:20 ft. long is under no stress at a temperature of 130' 



EXAMPLES IN TENSION AND COMPEESSION. 223 

Cent., and is provided witli flanges so that tlie slightest 
contraction of length will tend to bring two walls nearer 
together. If the resistance to this motion is 10 tons how 
low must the temperature fall to cause any motion ? tj be- 
ing =.0000110 (Cent, scale). From § 199 we have, ex- 
pressing P in lbs. and F in sq. inches, since E^= 40,000,000 
^hs. per sq. inch, 

10x2,000 =40,000,000 x }4 X(1304') x 0,000011 ; whence 
^'=39.0° Centigrade. 

Example 4. — If the ends of an iron beam bearing 5 tons 
at its middle rest upon stone piers, required the necessary 
bearing surface at each pier, putting C for stone =200 
lbs. per sq. inch. 25 sq. in., Ans. 

Example 5. — How long must a wrought iron wire* be, 
supported vertically at its upper end, to break with its 
own weight ? 216,000 inches, Ans. 

Example 6, — One voussoir (or block) of an arch-ring 
presses its neighbor with a force of 50 tons, the joint hav- 
ing a surface of 5 sq. feet ; required the compression per 
sq. inch. 138.8 lbs. per sq. in., Ans. 

205. Factor of Safety. — When, as in the case of stone, the 
value of the stress at the elastic limit is of very uncertain 
determination by experiment, it is customary to refer the 
value of the safe stress to that of the ultimate by making 
it the w'th portion of the latter, n is called a factor o/ 
safety, and should be taken large enough to make the safe 
stress come within the elastic limit. For stone, n should 
not be less than 10, i.e, C'^G-^n; (see Ex. 6, just given). 



206. Practical Notes. — It was discovered independently by 
Commander Beardslee and Prof. Thurston, in 1873, that 
if wrought iron rods were strained considerably beyond 
the elastic limit and allowed to remain free from stress 

* Take T = 60,UU0 lbs. per square incli. 



224 MECHANICS OF ENGINEERING. 

for at least one day thereafter, a second test would sliow 
higher limits both elastic and ultunate. 

In 1899 Mr. James Muir discovered that this recovery of 
elasticity and raising of both the yield-point and ultimate 
strength, in the case of iron and steel after "overstraining," 
may be brought about by sim23ly heating the metal for a few 
minutes in a bath of boiling water. In one experiment a bar 
of a kind of mild steel which under ordinary tests broke at 
39 tons/in.2 with 20% elongation on 8 in., was stretched just 
to its yield-point, then relieved and heated for a few minutes 
to 100° Cent., then stretched just to its new yield-point, 
then relieved and heated as before; and so on, for three 
times more. The first yield -point was at 27, the others at 
33, 38, 43i, and 47 tons/in.2 The bar was then broken at 
49 tons/in.2 with total extension of 12%. The diminished 
ultimate extension shows the hardening effect of the treatment. 
(See Prof. Ewing's ''Strength of Materials,'' pp. 38 and 40.) 

'Bj fatigue of metals we understand the fact, recently dis- 
covered by Wohler in experiments made for the Prussian 
Ci-overnment, that rupture may be produced by causing the 
stress on the elements to vary repeatedly between two 
limiting values, the highest of which may be considerably 
below T (or G), the number of repetitions necessary to 
produce rupture being dependent both on the range of 
variation and the higher value. 

For example, in the case of Phoenix iron in tension, 
Tupture was produced by causing the stress to vary from 
to 52,800 lbs. per sq. inch, 800 times ; also, from tc 
44,000 lbs. per sq. inch 240,858 times ; while 4,000,000 va- 
liations between 26,400 and 48,400 per sq. inch did not 
cause rupture. Many other experiments were made and 
the following conclusions drawn (among others): 

Unlimited repetitions of variations of stress (lbs. per 
^q. in.) between the limits given below will not injure the 
paetal (Prof. Burr's Materials of Engineering). 

^ , , . j From 17,600 Comp. to 17,600 Tension, 

roug iron. | ^^ ^ ^^ ^^^^^^ 

( From 30,800 Comp. to 30,800 Tension. 
Axle Cast Steel. ^ " to 52,800 

( " 38500 Tens, to 88,000 « 
(See p, 5i3'2 for an addendum to this paragraph.) 



SHEARING. 



225 



SHEARING. 

207. Rivets. — The angular distortion called shearing 
strain in the elements of a body, is specially to be provided 
for in the case of rivets joining two or more plates. This 
distortion is shown, in Figs. 205 and 206, in the elements 
near jhe plane of contact of the plates, much exaggerated. 



jT^ 



i > 



>r* 



T 



Fig. 205. 




Fig, 206. 



In Fig. 205 (a lap-joint) the rivet is said to be in single 
shear ; in Fig. 206 in double shear. If P is just great 
enough to shear off the rivet, the modulus of ultimate shear- 
ing, which may be called S, (being the shearing force per 
unit of section when rupture occurs) is 



F iTid^ 



(1) 



in which i^==the cross section of the rivet, its diameter 
being =d. For safety a value S'= }{ to ^ of >S' should 
be taken for metal, in order to be within the elastic limit. 

As the width of the plate is diminished by the rivet 
hole the remaining sectional area of the plate should be 
ample to sustain the tension P, or 2P, (according to the 
plate considered, see Fig. 206), P being the safe shearing 
force for the rivet. Also the thickness t of the plate 
should be such that the side of the hole shall be secure 
against crushing ; P must not be > C'td, Fig. 205. 

Again, the distance a, Fig. 205, should be such as to 
prevent the tearing or shearing out of the part of the 
plate between the rivet and edge of the plate. 



226 



MECHANICS OF E:NGINEEKING. 



For economy of material tlie seam or joint sliould be 
no more liable to rupture by one tban by another, of the 




o o ^ o 



■.t 




Fig. 307. 



four modes just mentioned. The relations which must 
then subsist will be illustrated in the case of the " butt= 
joint " with two cover-plates, Fig. 207. Let the dimen- 
sions be denoted as in the figure and the total tensile force 
on the joint be = Q. Each rivet (see also Fig. 206) is ex- 
posed in each of two of its sections to a shear of I2 Q} 
hence for safety against shearing of rivets we put 



12 Q- 



-% Tides' 



(1) 



Along one row of rivets in the main plate the sectional 
area for resisting tension is reduced to {b — ^d)t,, hence for 
safety against rupture of that plate by the tension Q, we 
put 

Q=(h—3d)t,T' , (2) 

Equations (1) and (2) suffice to determine d for the rivets 
and ^1 for the main plates, Q and b being given ; but the 
values thus obtained should also be examined with refer- 
ence to the compression in the side of the rivet hole, i.e., 
J^ Q must not be > C't.d. [The distance a, Fig. 205, to the 
edge of the plate is recommended by different authorities 
to be from d to 3d.] 

Similarly, for the cover -plate we must have 



and 12^ not > G'td, 



^4QoT(b—dd)tT' 
< 



(8) 



SHEARi:srG. 227 

If the rivets do not fit their holes closely, a large margin 
should be allowed in practice. Again, in boiler work, the 
pitch, or distance between centers of two consecutive rivets 
may need to be smaller, to make the joint steam-tight, than 
would be required for strength alone. 

208, Shearing Distortion. — The change of form in an ele- 
ment due to shearing is an angular deformation and will 
be measured in tt -measure. This angular change or dif- 
ference between the value of the corner angle during strain 
and ^4i'^, its value before strain, will be called d, and is 
proportional (within elastic limit) to the shearing stress 
per unit of area, p^, existing on all the four faces whose 
angles with each other have been changed. 

Fig. 208. (See § 181). By § 184 the Modulus of Shearing 
Elasticity is the quotient obtained by dividing p^hj d \ i.e. 
{elastic limit not passed)^ 

^s=^ . . . . (1) 

or inversely, d=p^-^E^ (1)' 

The value of E^ for different substances is most easily 
I determined by experiments on torsion 

in which shearing is the most promi- 
/ nent stress.* (This prominence depends 
y\ on the position of the bounding planes 
j^ of the element considered ; e.g., in Fig. 
_ __.L 208, if another element were considered 

/ '---cix->, within the one there shown and with 

Fig. 208. its plaues at 45° with those of the first, 

we should find tension alone on one pair of opposite faces, 
compression alone on the other pair.) It will be noticed 
that shearing stress cannot be present on two opposite 
faces only, but exists also on another pair of faces (those 
perpendicular to the stress on the first), forming a couple 
of equal and opposite moment to the first, this being 
necessary for the equilibrium of the element, even when 

* For instance, see numerical example on p. 237, giving a value of 
Es as resulting from a torsion test made by students in the Civil Engi- 
neering Laboratory at Cornell University, April, 1904. 



228 



MECHAIflCS OF ENGINBERIKG. 



tensile or compressive stresses are also present on the 
faces considered. 

209. Shearing Stress is Always of the Same Intensity on the 
Four Faces of an Element. — (By intensity is meant per unit 
of area ; and the four faces referred to are those perpen- 
dicular to the paper in Fig. 208, the shearing stress being 
parallel to the paper.) 

Let dx and dz be the width and height of the element 
in Fig. 208, while dy is its thickness perpendicular to the 
paper. Let the intensity of the shear on the right hand 
face be =q^, that on the top face =Ps. Then for the ele- 
ment aw a free body, taking moments about the axis per- 
pendicular to paper, we have 

q^ dz dy X dx — ^g dx dy x dz =0 .•. qs =p^ 
{dx and dz being the respective lever arms of the forces 
q^ dz dy and p^ dx dy.) 

Even if there were also tensions (or compressions) on 
one or both pairs of faces their moments about would 
balance (or fail to do so by a differential of a higher order) 
independently of the shears, and the above result would 
still hold. 



210. Table of Moduli for Shearing. 





d" 


^s 


8" 


s 


Material. 


i.e. 6 at elastic 
limit. 


Mod. of Elasticity 
for Shearing. 


(Elastic limit.) 


(Rupture.) 




arc in radians. 


lbs. per sq. in. 


lbs. per sq. in 


lbs. per sq. in. 


Soft Steel, 




9,000,000 


30,000 


70,000 


Hard Steel, 


0.0033 


14,000,000 


45,000 


90,00C 


Cast Iron, 


0.0021 


7,000,000 


15,000 


30,00C 


Wrought Iron, 


0.0022 


9,000,000 


20,000 


50,000 


Brass, 




5,000,000 






Glass, 










Wood, across ( 
fibre, 1 








1,500 

to 

8,000 


Wood, along ( 
fibre, ( 








500 

to 

3,200 



SHEAKIISG. 



229 



As in tlie tables for tension and compression, tlie above 
values are averages. The true values may differ from 
these as mucli as 30 per cent, in particular cases, accord-* 
ing to the quality of the specimen. 

211. Punching rivet holes in plates of metal requires the 
overcoming of the shearing resistance along the convex 
surface of the cylinder punched out. Hence ii d = diam- 
eter of hole, and t= the thickness of the plate, the neces- 
sary force for the punching, the surface sheared being 
F= tjvd^ is 



P=8t^d 



(2)- 



Another example of shearing action is the " stripping " 
of the threads of a screw, when the nut is forced off lon- 
gitudinally without turning, and resembles punching in 
its nature. 

212. EandEgj Theoretical Relation. — In case a rod is in 
iension within the elastic limit, the relative (linear) lateral 
contraction (let this =m) is so connected with E^ and E^ 
ihat if two of the three are known the third can be de- 
duced theoretically. This relation is proved as follows, 
by Prof. Burr. Taking an elemental cube with four of its 
faces at 45° with the axis of the piece, Fig. 209, the axial 
half-diagonal AD becomes of a length AD'=AD-\-s.AD 
under stress, while the transverse half diagonal contracts 
to a length B'D'=AD — m.AD, The angular distortion d 







.-\U.S . 


• 




X<'>^' 






/^<\^^^ 




A< 


^ D D/^ ^^' 


,. 




\>6-^ 




Fig. 209. § 212. 



Fig. 210. 



230 MECHAIflCS OF EI^GINEEKING. 

is supposed very small compared with 90° and is due to 
the shear j9g per unit of area on the face BG (or BA\ 
From the figure we have 

tan(45°— -) = __^,=__=1— m— s, approx. 

[But, Fig. 210, tan(45° — x)=l — 2x nearly, where a; is a 
small angle, for, taking CA=unitj=AE, ian AD=AF= 
AE—EF. Now approximately EF= EG, y2 and EG= 
BDa^^=x^^ .'. AF= 1 — 2a7 nearly.] Hence 

1 — d= 1 — m — £ ; or d= m+e . . (2) 

Eq. (2) holds good whatever the stresses producing the 
deformation, but in the present case of a rod in tension, 
if it is an isotrope, and if ^ = tension per unit of area on 
its transverse section, (see § 182, putting «=45°), we have 
^t==p-=-£ and E^={psOJx BG')-^d=}^p-^d. Putting also 
(m : £)= h, whence m=k£, eq. (2) may finally be written* 

>4-==(^ + l)4-; i.e., ^s=-^^ . . (3) 

Prof, Bauschinger, experimenting with cast iron rods,, 
found that in tension the ratio m: £was =m} as an average, 
which in eq. (3) gives 

^=12^^,= !^, nearly. , , . (4) 
246 5 ^ ^ ^ 

His experiments on the torsion of cast iron rods gave 
^,= 6,000,000 to 7,000,000 lbs. per sq. inch. By (4), then, 
E, should be 15,000,000 to 17,500,000 which is approxi- 
mately true (§ 203). 

Corresponding results may be obtained for short blocks 
in compression, the lateral change being a dilatation in- 
stead of a contraction. 

* This ratio, m-^e, denoted by k, is called Poisson's Ratio. For metals 
its value lies approximately between 0.20 and 0.35. See also p. 507. 




313. Examples in Shearing. — Example 1. — Kequired the^ 
proper length, a, Fig. 211, to 
guard against the shearing off, 
along the grain, of the portion 
ah, of a wooden tie-rod, the force 
P being = 2 tons, and the width 
of the tie = 4 inches. Using a 
value of S' = 100 lbs. per sq. irL.s 
we put 6a/S''= 4,000 cos 45° ; i.e. 
■Fi^m. a= (4,000x0. 707) --(4x100)= 7.07 

inches. 

Example 2.— A ^ in. rivet of wrought iron, in single 
shear (see Eig. 205) has an ultimate shearing strength 
P= FS=}(7T(PS= %7t{ Yq y X 50,000= 30,050 lbs. For safety, 
putting aS"= 8,000 instead of aS',P'=4,800 lbs. is its safe 
shearing strength in single shear. 

The wrought iron plate, to be secure against the side- 
crushing in the hole, should have a thickness t, computed 
thus I 

P'=tdC' ; or 4,800=^.^ x 12,000 ,-. ^=0.46 in. 

If the plate were only 0.23 in. thick the safe value of P 
would be only ^ of 4,800. 

Example 3. — Conversely, given a lap-joint, Fig. 205, in 
which the plates are ^ in. thick and the tensile force on 
the joint = 600 lbs. per linear inch of seam, how closely 
must ^ inch rivets be spaced in one row, putting jS"=8,000 
and 6" =12,000 lbs. per sq. in. ? Let the distance between 
centres of rivets be =x (in inches), then the force upon 
each rivet =600a7, while its section P=0.44 sq. in. Having 
regard to the shearing strength of the rivet we put 600cc= 
0.44x8,000 and obtain a?=5.86 in.; but considering that the 
safe crushing resistance of the hole is =1^-^.12,000= 
2,250 lbs., 600aj=2,250 gives a;=3.75 inches, which is the 
pitch to be adopted. What is the tensile strength of the. 
reduced sectional area of the plate, with this pitch '? 



232 MECHAXIC3 OF EKGIISTEBIilNG. 

Example 4 — Double butt-joint ; (see Fig. 207) ; ^s iiich 
plate; ^ in. rivets; F =C' =12,000 ; S' =8,333; width of 
plates=14 inches. Will one row of rivets be sufficient at 
each, side of joint, if ^=30,000 lbs.? The number of rivets 
^^ ? Here each rivet is in double shear and has therefore 
a double strength as regards shear. In double shear the 
safe strength of each rivet =2i^>S"= 7,333 lbs. Now 30,000^ 
7,333=40 (saj). With the four rivets in one row the re- 
duced sectional area of the main plate is =[14 — 4x ^] X^s 
=4,12 sq. in., whose safe tensile strength is =i^J"=4„12x 
12,000=49,440 lbs.; which is > 30,000 lbs. .% main plate is 
safe in this respect. But as to side-crushing in holes 
in main plate we find that G't^d (i,e, 12,000 X Vs >^ M^^'^'^^ 
lbs.) is <.%Q i,e. <7,500 lbs., the actual force on side of 
hole. Hence four rivets in one row are too few unless 
thickness of maiiL plate be doubled. Will eight in one 
row be safe ? 

213a, (Addendum to § 206.) Elasticity of Stone and Cements. 
— Experiments by Gen. Gill more with the large Watertowi» 
testing-machine in 1883 resulted as follows (see p. 221 for 
notation) : 

With cubes of Haverstraw Freestone (a homogeneous brown- 
stone) from 1 in. to 12 in. on the edge, E^ was found to be 
from 900,000 to 1,000,000 lbs. per sq. in. approximately ; and 
C about 4,000 or 5,000 lbs. per sq. in. Cubes of the same 
range of sizes of Djckerman's Portland cement gave E^ from 
1,350,000 to 1,630,000, and G from 4,000 to 7,000, lbs. per sq. 
in. Cubes of concrete of the above sizes, made with the 
Newark Cc.'s Rosendale cement, gave E^ about 538,000^ while 
cubes of cement-mortar, and some of concrete, both made with 
National Portland cement, showed E^ from 800,000 to 2,000,- 
OOO lbs. per sq. in. 

The compressibility of hrick jpiers 12 in. square in section 
and 16 in. high was also tested. They were made of common 
North River brick with mortar joints f in. thick, and showed 
a value for E„ of about 300,000 or 400,000, while at elastic 
limit C" was on the average 1,000, lbs. per sq. in. 



TORSIOM. 



233 



CHAPTER IL 



TOKSION. 



S14. Angle of Torsion and of Helix. "When a cylindrical 
beam or shaft is subjected to a twisting or torsional action, 
I. e. when it is the means of holding in equilibrium two 
couples in parallel planes and of equal and opposite mo- 
ments, the longitudinal axis of symmetry remains straight 

and the elements along it exper- 
lience no stress (whence it may be 
I called the "line of no twist"), 
while the lines originally parallel to 
Fig. 212. i^ assume the form of helices, each 

element of which is distorted in its angles (originally 
right angles), the amount of distortion being assumed pro- 
portional to the radius of the helix. The directions of the 





faces of any element were originally as follows : two radial, 
two in consecutive transverse sections, and the other two 
tangent to two consecutive circular cylinders whose com- 
mon axis is that of the shaft. E.g. in Fig. 212 we have 
an unstrained shaft, while in Fig. 213 it holds the two 



234 



MECHANICS OF ENGINEEHmG. 



couples (of equal moment Pa = Qh) in equilibrium. These 
couples act in parallel planes perpendicular to the axis of 
the prism and a distance, ?, apart. Assuming that the 
transverse sections remain plane and parallel during tor- 
sion, any surface element, m, which in Fig. 212 was entire- 
ly right-angled, is now distorted. Two of its angles have 
been increased, two diminished, by an amount d, the angle 
between the helix and a line parallel to the axis. Suppos- 
ing m to be the most distant of any element from ihe axis, 
this distance being e, any other element at a distance s 

from the axis experiences an angular distortion =- <§„ 

If now we draw B' parallel to 0' A the angle B B', 
=a, is called the Angle of Torsion, while d may be called the 
helix angle', the former lies in a transverse plane, the latter 
in a plane tangent to the cylinder. Now 

tan d = (linear arc B B')-t-1; but lin. arc B B' =' ea; hence, 
putting d for tan d, (3 being small) 






(1) 



(d and « both in radians). 

215. Shearing Stress on the Elements. The angular distor- 
tion, or shearing strain, d, of any element (bounded as al- 
ready described) is due to the shearing stresses exerted on 
it by its neighbors on the four faces perpendicular to the 

tangent plane of the cylindri- 
cal shell in which the element 
is situated. Consider these 
neighboring elements of an 
outside element removed, and 
the stresses put in ; the latter 
are accountable for the dis- 
^®- ^*^ tortion of the element and 




-pdF 



TOKSiou". 235 

hold it in equilibrium. Fig. 214 shows this element 
"free." Within the elastic limit ^ is known to be propor- 
tional to jOg, the shearing stress per unit of area on the 
faces whose relative angular positions have been changed. 
That is, from eq. (1), § 208, S -^^p^-r-Us; whence, see (1) of 
§ 214, 

In (2) Ps, and e both refer to a surface element, e being 
the radius of the cylinder, and p^ the greatest intensity of 
shearing stress existing in the shaft. Elements lying nearer 
the axis suffer shearing stresses of less intensity in pro- 
portion to their radial distances, i.e., to their helix-angles. 
That is, the shearing stress on that face of the element 
which forms a part of a transverse section and whose dis- 
tance from the axis is z, is p, =— p^, per unit of area, and 

the total shear on the face is pdF, c?^ being the area of the 

face. 

216. Torsional Strength. — ^We are now ready to expose tlia 
full transverse section of a shaft under torsion, to deduce 
formulae of practical utility. Making a right section of 
the shaft of Fig. 213 anywhere between the two couples 
and considering the left hand portion as a free body, the 
forces holding it in equilibrium are the two forces P of 
the bft-hand couple and an infinite number of shearing 
forces, each tangent to its circle of radius s, on the cross 
section exposed by the removal of the right-hand portion. 
The cross section is assumed to remain plane during tor- 
sion, and is composed of an infinite number of dF's, each 
being the area of an exposed face of an element | see !Fig. 



236 







elementary shearing force = S p^dF, and s is its 
lever arm about the axis Oo . For equilibrium, S (mom.), 
about the axis Oo must =0 ; i.e. in detail 



_p^o^p^a+ f ( -£ p,dF)%^ii 



©r, redncing. 



h rz^dF=Pa\ or, A.^Pa 
eJ e 



(3) 



Eq, (3) relates to torsional strength, since it contains ^s, tha 
greatest shearing stress induced by the torsional couple, 
whose moment Pa is called the Moment of Torsion, the 
stresses in the cross section forming a couple of equal and 
opposite moment. Pa is also called the "torque." 

Ip is recognized as the Polar Moment of Inertia of the cross 
section, discussed in § 94 ; e is the radial distance of the 
outermost element, and = the radius for a circular shafto 

217. Torsional Stiffness. — In problems involving the angle 
of torsion, or deformation of the shaft, we need an equa- 
tion connecting Pa and a, which is obtained by substitut- 
ing in eq. (3) the value of p^ in eq. (2), whence 



I 



=Pa. 



(4) 



From this is appears that the angle of torsion, a, is pro- 
portional to the moment of torsion, or " torque," Pa inch -lbs., 
within the elastic limit; a must be expressed in radians. 



TORSION. 237 

Example. — A portion 3.4 ft. long, of a solid cylindrical shaft of soft 
steel, of diam. = 1.5 in., is found by the use of "Torsion Clinometers" 
(see frontispiece) to be held at an angle of torsion of a = 5.41°, =0.0944 
radians, just before the elastic limit is reached, by a "torque," =Pa, of 
10,200 in. -lbs. Compute the Modulus of Elasticity for Shearing. 

Substituting in eq. (4), with I-p=Tzr^l2, (§94), =7r(0.75)^^2, =0.497 
in.*, and Z = 3.4X 12 = 40.8 in., we have 

10,200X40 .8 Qg^nnnniv. 
' ^ 0.0944 X 0.4 97 ^ 8,870,000 lbs. per sq. m. 

218. Torsional Resilience is the work done in twisting a 
shaft from an unstrained state until the elastic limit is 
reached in the outermost elements. If in Fig. 213 we 
imagine the right-hand extremity to be fixed, while the 
other end is gradually twisted through an angle each 
force P of the couple must be made to increase grafdually 
from a zero value up to the value Pj, corresponding to ai. 
In this motion each end of the arm a describes a space 
= ^aai, and the mean value of the force = }4Pi (compare 
§ 196). Hence the work done in twisting is 

Ui=}4FiX}4aaiX2=}4Piaaj^ . . (5) 
By the aid of preceding equations, (5) can be written 

If for ps "^e write 8' (Modulus of safe shearing) we have 
for the safe resilience of the shaft 

U'=4r^ -. . . . (7) 

If the torsional elasticity of an originally unstrained shaft 
is to be the means of arresting the motion of a moving 
mass whose weight is O, (large compared with the parts 
intervening) and velocity =v, we write (§ 133) 

g 2' 
as tlie condition that the shali shall not be injnrecL 



238 mecha:^ics of engineering. 

21U. roiar Moment of Inertia. — ^For a shaft of circular 
cross section (see § 94) /p=i^7rr*; for a hollow cylinder 
/p=i^7r(ri* — r^) ', while for a square shaft If=yih^, h being 
"the side of the square ; for a rectangular cross-section 
sides h and li, I^=lJbh{lf-\-¥). For a cylinder e=r; if hoi- 
low, e=r , the greater radius. For a square, e=i^6y'2. 

220. Ifon-Circular Shafts. — If the cross-section is not cir- 
cular it becomes warped, in torsion, instead of remaining 
plane. Hence the foregoing theory does not strictly ap- 
ply. The celebrated investigations of St. Tenant, how- 
ever, cover many of these cases. (See § 708 of Thompson 
and Tait's Natural Philosophy ; also, Prof. Burr's Elas- 
ticity and Strength of the Materials of Engineering). His 
results give for a square shaft (instead of the 

ab'E.^ Pa of eq. (4) of § 217), 



Ql 



Pa=OMl^t . . . . (1) 



and Pa=Jffi^p^f instead of eq. (3) of § 216, 2?s being the 
greatest shearing stress. 

The elements under greatest shearing strain are found 
at the middles of the sides, instead of at the corners, when 
the prism is of square or rectangular cross-section. The 
warping of the cross-section in such a case is easily veri ' 
fied by the student by twisting a bar of india-rubber in 
his fingers. 

221. Transmission of Power. — Fig. 216. Suppose the cog- 
wheel B to cause A, on the 
same shaft, to revolve uni- 
formly and overcome a resis- 
tance Q, the pressure of the 
teeth of another cog-wheel, 
P 5 being driven by still another 
Fig. 216. wheel. The shaft AB is un- 

der torsion, the moment of torsion being =Pa= Qh. (Pi 
and ^1 the bearing reactions have no moment about the 
axis of the shaft). If the shaft makes u revolutions per 
unit-time, the work transmitted {transmitted ; not expend^ 




TOESION. 239 

ed in twisting the shaft whose angle of torsion remains 
constant, corresponding to Fa) per unit-time, i.e. the Power, 
is 

X/=P.27ra.u=27ruPa . , , (8) 

To reduce L to Horse Power (§ 132), we divide by N, 
the number of units of work per unit -time constituting 
vOne H. P. in the system of units employed, i.e., 

Horse Power =H. p=?!E!^ 

JSl 

For example JSf =33,000 ft. -lbs. per minute, or =396,000 
inch -lbs. per minute ; or = 550 ft. -lbs. per second. Usually 
the rate of rotation of a shaft is given in revolutions per 
minute. 

But eq^. (8) happens to contain Fa the moment of torsion 
acting to maintain the constant value of the angle of tor- 
sion, and since for safety (see eq. (3) § 216) Fa= S' I.^-^ e, 
with -ZJ,= y^TiT^ and e=r for a solid circular shaft, we have 
for such a shaft 

(Safe),H.P.=?f^ . . . (9) 

N 

which is the safe H. P., which the given shaft can trans- 
mit at the given speed. S' may be made 7,000 lbs. per sq. 
inch for wrought iron ; 10,000 for steel, and 5,000 for cast- 
iron. If the value of Pa fluctuates periodically, as when 
a shaft is driven by a connecting rod and crank, for (H. P.) 
we put toX(H. p.), m being the ratio of the maximum to 
the mean torsional moment; m= about 172 under ordi- 
nary circumstances (Cotterill). 

With a hollow cylindrical shaft, of outer radius = rj, and inner = r 2 
the r^ of eq. (9) must be replaced by (?*i*— /•2*)-^'"i- If> furthermore, the 
thickness of metal is small, we may proceed thus, taking numerical data: 
Let the radius to the middle of the thickness be ro = 10 in., the thickness 
t=\ in., and the (steel) shaft make m=120 revs./min. ; with *S' = 5000 
Ibs./in.^; then the total safe shearing stress in the cross-section is- 
12' = 27rroi5;' = 27rlOXiX 5000 = 78,540 lbs., whHe the velocity of the 
mid-thickness is v' = 27rroU = 27: 10X2 = 125.6 in./sec. = 10.47 ft. /sec. Hence 
the (safe) power that may be transmitted at given speed is L = R'v' 
= 78,540X10.47 = 822,100 ft.-lbs. per sec; or, (-^550), =1495 H.P. 



240 



MECHAITICS OF ElifGINEEEING. 



222. Autographic Testing Machine. — Tlie principle of Prof 
Thurston's invention bearing this name is shown in Fig 




• Fie. 217. 

217. The test-piece is of a standard shape and size, its 
central cylinder being subjected to torsion. A jaw, carry- 
ing a handle (or gear-wheel turned by a worm) and a drum 
on which paper is wrapped, takes a firm hold of one end 
of the test-piece, whose further end lies in another jaw 
rigidly connected with a heavy pendulum carrying a pen- 
cil free to move axially. By a continuous slow motion of 
the handle the pendulum is gradually deviated more and 
more from the vertical, through the intervention of the 
test-piece, which is thus subjected to an increasing tor- 
sional moment. The axis of the test-piece lies in the axis 
of motion. This motion of the pendulum by means of a 
proparly curved guide, WH, causes an axial (i.e., parallel 
to axis of test-piece) motion of the pencil A, as well as an 
angular deviation /9 equal to that of the pendulum, and 
this axial distance CF,=^sT, of the peiicil from its initial 
position measures the momenr of torsion =i^«=:^(? sin )5.. 
As the piece twists, the drum and paper move relatively 
to the pencil through an angle sUo equal to the angle 
of torsion a so far attained. The abscissa so and ordinate 
sT oi the curve thus marked on the paper, measure,, 
when the paper is unrolled, the values of a and Pa through. 



TOESrOK 



241 



all the stages of the torsion. Fig. 218 shows typical 




Fig. 218. 

Jurves thus obtained. Many valuable indications are 
given by these strain diagrams as to homogeneousness of 
composition, ductility, etc., etc. On relaxing the strain 
at any stage within the elastic limit, the pencil retraces 
its path ; but if beyond that limit, a new path is taken 
called an " elasticity-line," in general parallel to the first 
part of the line, and showing the amount of angular re- 
CO very, BC, and the permanent angular set, OB. 

2222i.. Torsion Clinometers. — ^When the test-piece used in the 
Thurston testing machine is short, the indicated angles of 
torsion below the elastic limit are far in excess of the actual 
values, on account of the initial yielding of the wedges in the 
jaws. By the use of " torsion clinometers," however (see 
frontispiece) the angle of torsion can be measured accurately 
within one minute of arc. 

223. Examples in Torsion. — The modulus of safe shearing 
strengtn. S', as given in § 221, is expressed in pounds per 
square inch ; hence these two units should be adopted 
throughout in any numerical examples where one of the 
above values for S' is used. The, same statement applies 
to the modulus of shearing elasticity, E^, in the table of 
§ 210. 
- Example 1.— Fig. 216. With P = 1 ton, a = 3 ft., I ^ 
10 ft. , and the radius of the cylindrical shaft r=2.5 inches, 
required the max. shearing stress per sq. inch, ps, the 
shaft being of wrought iron. From eq. (3) § 216 

Pae 2,000x36x2.5 o oomi, • -u 

^----T^' V..X(2.5)^ =2,930 lbs. per sq. inch, 

which is a safe value for any ferrous raetaL 



242 



MECHANICS OF EIN GINEEEIXG. 



Example 2. — What H. P. is the shaft in Ex. 1 transmit- 
ting, if it makes 50 revolutions per minute ? Let u = 
number of revolutions per unit of time, and N = the num- 
ber of units of work per unit of time constituting one 
horse-power. Then H. V.^Pu^na-^N, which for the foot» 
pound-minute system of units gives 

H. P.=2,000x50x27rx3--33,000=57i4: H. P. 

Example 3. — What different radius should be given t(- 
the shaft in Ex. 1, if two radii at its extremities, originally 
parallel, are to make an angle of 2° when the given moment 
of torsion is acting, the strains in the shaft remaining con- 
stant. From eq. (4) § 217, and the table 210, with a=i|^c;r=* 
0.035 radians (i.e. ;7-measure), and I^=^j^T^^ we have 



y^ 



2,000x36x120 



>^7r0.035x 9,000,000 



-—=17.45 .-. r=2.04 inches. 



(This would bring about a different p,, but still safe.) The 
foregoing is an example in stiffness. 

Example 4. — A working shaft of steel (solid) is to tran:^- 
mit 4,000 H. P. and make 60 rev. per minute, the maximum 
twisting moment being 1^ times the average; requireil 
its diameter. • c^=14.74 inches. Ans. 

Example 5. — In example 1, p = 2,930 lbs. per square 
inch ; what tensile stress does this imply on a plane at 45° 
with the pair of planes on which Ps acts ? Fig. 219 shows 



p,dx 





dx^'Ps 



'dx^ps 



Via. 220. 



TOESiOE^. 243 

a small cube, of edge =dx, (taken from the outer helix of 
Fig. 215,) free and in equilibrium, tbe plane of the paper 
being tangent to the cylinder ; while 220 shows the portion 
BD 0, also free, with the unknown total tensile stress jorfa;^,^/^ 
acting on the newly exposed rectangle of area =dxxdx^% 
p being the unknown stress per unit of area. From sym- 
metry the stress on this diagonal plane has no shearing 
component. Putting 2' [components normal to^-D]=0, 
we have 

pdx^^2=2dx'^p^Gos4:5°=dx^p^^/2.'.p=ps . (1) 

That is, a normal tensile stress exists in the diagonal 
plane BD of the cubical element equal in intensity to the 
shearing stress on one of the faces, i.e., =2,930 lbs. per sq. 
in. in this case. 

Similarly in the plane AG will be found a compressive 
stress of 2,930 lbs. per sq. in. If a plane surface had been 
exposed making any other angle than 45° with the face of 
the cube in Fig. 219, we should have found shearing and 
normal stresses each less than p^ per sq. inch. Hence the 
interior dotted cube in 219, if shown " free " is in tension 
in one direction, in compression in the other, and with 
no shear, these normal stresses having equal intensities. 
Since S' is usually less than T' or C, ii Ps is made = S' 
the tensile and compressive actions are not injurious. It 
follows therefore that when a cylinder is in torsion any 
helix at an angle of 45° with the axis is a line of tensile, 
or of compressive stress, according as it is a right or left 
handed helix, or vice versa. 

Example 6. — A solid and a hollow cylindrical shaft, of 
equal length, contain the same amount of the same kind 
of metal, the solid one fitting the hollow of the other. 

Compare their torsional strengths, used separately. 
The solid shaft has only ^ the strength of the hollow 
one, Ans. 

Example 7. — Compare the shafts of Example 6 as to tor- 
sional stiffness (i. e. , the angles of torsion due to equal moments) . 
The solid shaft is only one-third as stiff as the other ; an equal 
moment produces three times the angle. Ans. 



244 MECHANICS OF El!fGrNEi:KrN:G. 



CHAPTER in. 

FliEXURE OF HOMOGENEOUS PRISMS UHDEK 
PERPENDICULAK FORCES IN ONE PLANE. 

224. Assumptions of tlie Common Theory of Plexure. — Wlien 
a prism is bent, under tlie action of external forces per- 
pendicular to it and in tlie same plane witli each otlier, it 
may be assumed tliat the longitudinal fibres are in tension 
on the convex side, in compression on the concave side, 
and that the relative stretching or contraction of the ele- 
ments is proportional to their distances from a plane in- 
termediate between, with the understanding that the flex- 
ure is slight and that the elastic limit is not passed in any 
element. i 

This " common theory " is sufficiently exact for ordinary 
engineering purposes if the constants employed are prop- 
erly determined by a wide range of experiments, and in- 
volves certain assumptions of as simple a nature as possi- 
ble, consistently with practical facts. These assumptions 
are as follows, (for prisms, and for solids with variable cross 
sections, when the cross sections are similarly situated as 
regards a central straight axis) and are approximately 
borne out by experiment : 

(1.) The external or " applied " forces are all perpendicu- 
lar to the axis of the piece and lie in one plane, which may 
be called the force-plane ; the force-plane contains the 
axis of the piece and cuts each cross-section symmetri- 
cally ; 

(2.) The cross-sections remain plane surfaces during 
flexure ; 

(3.) There is a surface (or, rather, sheet of elements) 
which is parallel to the axis and perpendicular to the 
force-plane, and along which the elements of the solid ex- 



FLEXURE. 



245 



perience no tension nor compression in an axial direction, 
this being called tlie Neutral Surface; 

(4.) The projection of the neutral surface upon the force 
plane (or a || plane) being called the Neutral Line or Elastic 
Curve, the bending or flexure of the piece is so slight that 
an elementary division, ds, of the neutral line may be put 
^dx, its projection on a line parallel to the direction of 
the axis before flexure ; 

(5.) The elements of the body contained between any 
two consecutive cross-sections, whose intersections with 
the neutral surface are the respective Neutral Axes of the 
sections, experience elongations (or contractions, accord- 
ing as they are situated on one side or the other of the 
neutral surface), in an axial direction, whose amounts are 
proportional to their distances from the neutral axis, and 
indicate corresponding tensile or compressive stresses ; , 

(6.) E,=E,; 

(7.) The dimensions of the cross-section are small com- 
pared with the length of the piece ; 

(8.) There is no shear perpendicular to the force plane 
on internal surfaces perpendicular to that plane. 

In the locality where any one of the external forces is 
Applied, local stresses are of course induced which demand 
separate treatment. These are not considered at present. 

225. Illustration. — Consider the case of flexure shown in 
Fig. 221. The external forces are three (neglecting the 




Fig. 22i. 



246 



MECHANICS OF EXGIXEBRING. 



weight of the beam), viz.: P^, Pg, and P3. P^ and P3 are 
loads, P2 the reaction of the support. 

The force plane is vertical. N^L is the neutral line or 
elastic curve. NA is the neutral axis of the cross-section 
at m / this cross-section, originally perpendicular to the 
sides of the prism, is during flexure ~| to their tangent 
planes drawn at the intersection lines ; in other words, the 
side view QNB, of any cross-section is perpendicular to 
the neutral line. In considering the whole prism free we 
have the system Pj, P2, and P3 in equilibrium, whence 
from 2^=0 we have P2=Pi+P3j and from 2" (mom. about 
P) =0, P3?3=Pi?i. Hence given Pi we may determine the 
other two external forces. A reaction such as Pg is some- 
times called a supporting force. The elements above the 
neutral surface NiOLS Sive in tension ; those below in com- 
pression (in an axial direction). 

226. The Elastic Forces. — Conceive the beam in Fig. 221 
separated into two parts by any transverse section such 
as QA, and the portion NiOJSf, considered as a free body 
in Fig. 222. Of this free body the surface QAB is one of 



^^dx 




T«e. 222. 



FLEXURE. 247 

i 

tlie bounding surfaces, but was originally an internal sur- 
face of tlie beam m Fig. 221. Hence in Fig. 222 we must 
put in the stresses acting on all the dF^^ or elements of area 
of QAB. These stresses represent the actions of the bodj 
taken away upon the body which is left, and according to 
assumptions (5), (6) and (8) consist of normal stresses (ten- 
sion or compression) proportional per unit of area, to th© 
distance, z, of the cZi^'s from the neutral axis, and of shear- 
ing stresses parallel to the force-plane (which in most 
cases will be vertical). 

The intensity of this shearing stress on any dF varies 
with the position of the dF with respect to the neutral 
axis, but the law of its variation will be investigated later 
(§§ 253 and 254). These stresses, called the Elastic Forces 
of the cross-section exposed, and the external forces Pj and 
P2, form a system in equilibrium. We may therefore ap- 
ply any of the 3onditions of equilibrium proved in § 38. 



227. The Neutral Axis Contains the Centre of Gravity of the 
Cross-Section. — Fig. 222. Let e— the distance of the outer- 
most elem.ent of the cross-section from the neutral axis, and 
the normal stress per unit of area upon it be =p, whether 
tension or compression. Then by assumptions (5) and (6), 
§ 224, the intensity of nprmal stress on any dF is = -1 p 
and the actual 

normal stress on any dFis= — pdF , {1} 

This equation is true for dF's having negative «'s, i.e. 
on the other side of the neutral axis, the negative value 
of the force indicating normal stress of the opposite char- 
acter ; 'for if the relative elongation (or contraction) of two 
axial fibres is the same for equal g's, one above, the other 
below, the neutral surface, the stresses producing the 
changes in length are also the same, provided ^t=:^^; see§§ 
184 and 201. 

For this free body in equilibrium put 2'X=0 (Xis a 
horizontal axis). Put the normal stresses equal to their 
X components, the flexure being so slight, and the X com- 



248 MECHANICS OF ENGINEEEIKG. 

ponent of the shears = for the same reason. This gives 
(see eq. (1) ) 

r± pdF= ; i.e. Z PdFz^ ; or, ^ i^i=0 (2) 



Ih which z— distance of the centre of gravity of the cross- 
section from the neutral axis, from which, though un- 
known in position, the g;'s have been measured (see eq. 
(4) § 23). 

In eq. (2) neither p-^e nor F can be zero .•. z must = ; 
i.e. the neutral axis contains the centre of gravity. Q. E. D. 
[If the external forces were not all perpendicular to the 
beam this result would not be obtained, necessarily.] 

228. The Shear. — The " total shear," or simply the 
'* shear," in the cross-section is the sum of th.e vertical 
shearing stresses on the respective dF's. Call this sum 
J, and we shall have from the free body in Fig. 222, by 
putting ^y=0 (F being vertical) 

P,—F,—J=0.:J=F,—Pi . . (3) 

That is, the shear equals the algebraic sum of the ex- 
ternal forces acting on one side (only) of the section con- 
sidered. This result implies nothing concerning its mode 
of distribution over the section. 

229. The Moment. — By the "Moment of Flexure" or 
simply the Moment, at any cross- section is meant the sum 
of the moments of the elastic forces of the section, taking 
ihe neutral axis as an axis of moments. In this summa- 
tion the normal stresses appear alone, the shear taking no part, 
having no lever arm about the axisiVA. Hence, Fig. 222, the 
moment of flexure (or "moment of resistance") 

=J(ipdF).=f/dF.^=£^ (*) 

This function, CdFz^, of the cross-section or plane figure 



FLEXURE. 249 

is the quantity called Moment of Inertia of a plane figure, 
§ 85. For the free body in Fig. 222, by putting 2'(mom.3 
about the neutral axis NA)=0, we have then 

^ — PiX^-]rP^X2=Q, or in general^ lL=zM . (5) 
e e 

in which M signifies the sum of moments,* about the neutral 
axis of the section, of all the forces acting on the free body 
considered, exclusive of the elastic forces of the exposed 
section itself. M is also called the "Bending Moment." 

Example. — In Fig. 222 let Pi = 3 and P2 = 4 tons, Xi = l ft. 8 in. and 
^2 = 5 in.; the section of the beam being a rectangle, with NA=b = 3 in. 
and QB=^h = Q in. Then I about axis NA is, (p. 94), fo/i^n- 12 = 54 in.*; 
and e=3 in. Hence the "bending moment," M, =3X20-4X5 = 40 
in. -tons. Equating M to the "moment of resistance" [or moment of 
the "stress couple" (see § 230)] we obtain, from eq. (5), p = Me-^I = 
40 X 3 H- 54 = 2.22 tons/in.^ for the unit normal stress in the outer 
fibre at Q, or B. We find also, for the shear at section QB, / = 4— 3 = 1 ton. 

230. Strength in Flexure. — Eq. (5) is available for solving 
problems involving the Strength of beams and girders, since 
it contains p, the greatest normal stress per unit of area to 
be found in the section. 

In the cases of the present chapter, where all the exter- 
nal forces are perpendicular to the prism or beam, and 
have therefore no components parallel to the beam, i.e. to 
the axis X, it is evident that the normal stresses in any 
section, as QB Fig. 222, are equivalent to a couple ; for the 
condition I!X=0 falls entirely upon them and cannot be 
true unless the resultant of the tensions is equal, parallel, 
and opposite to that of the compressions. These two equal 
and parallel resultants, not being in the same line, form a 
couple (§ 28), which we may call the stress -couple. The 
moment of this couple is the " moment of flexure " '~ , and 
it is further evident that the remaining forces in Fig. 222, 
viz.: the shear J and the external forces Pj and Pg* are 
equivalent to a couple of equal and opposite moment to 
the one formed by the normal stresses, 

* It is evident, therefore, that J!f (ft.-lbs., or in. -lbs.) is numerically equal 
to the "moment of flexure," or moment of the " stress couple " ; so that 
occasionally it maybe convenient to use "Jf" to denote the value of the 
latter momeut also. 



250 



MECHAKICS OF EXGHirBEIiriirG. 



231. Flexural Stiffness. — Tiie neutral line, or elastic curvo^ 
containing the centres of gravity of all tlie sections, was 
originally straight ; its radius of curvature at any point, 
as N, Fig. 222, c'uring flexure may be introduced as fol- 
lows. QB and U'V are two consecutive cross -sections, 
originally parallel, but now inclined so that the intersec- 
tion G, found by prolonging them sufficiently, is the centre 
of curvature of the ds (put =dx) which separates them, at 
JSf, and CG=p= the radius of curvature of the elastic 
curve at N. From the similar triangles U' TIG and GNG we 
have dk'.dx'.:e;Pf in which dX is the elongation, U' U^ of a 
portion, originally =c?cc, of the outer fibre. But the rela- 
tive elongation £=-t— of the latter is, by §184, within the 

elastic limit, =^.\ -:^ =— and eq. (5) becomeL 
E E p ^ ^ 



EI 



=M 



(6) 



AXIS X 



From (6) the radius of curvature can be computed. E~ 
the value of E^—E^, as ascertained from experiments in 
bending. 

~ To obtain a differential equation of the elastic curve, (6) 
may be transformed thus, Fig. 223. The curve being very 

flat, consider two consecutive 
(is's with equal dx's ; they may 
be put = their c^x's. Produce 
the first to intersect the dy of the 
second, thus cutting off the d^y^ 
i'e. the difference between two 
^■^Jfy consecutive dy'^. Drawing a per- 
pendicular to each ds at its left 
extremity, the centre of curva- 
ture G is determined by their in- 
tersection, and thus the radius 
of curvature p. The two shaded 
Fig. 223. triangles have their small angles 

equal, and d^y is nearly perpen- 
dicular to the prolonged ds ; 
hence, considering them sim- 
ilar, we have 




\p,dx:'.dx'.d^y :.-^J^^, 



FLEXURE. 251 

and hence from eq. (6) we ) , , ^A^V 

may write | (approx.) ±EI^=M . (7) 

as a differential equation of the elastic curve. From this 
the equation of the elastic curve may be found, the de- 
flections at different points computedj and an idea thus 
formed of the stiffness. All beams in the present chap- 
ter being prismatic and Jiomogeneous both jE' and / are the 
same (i.e. constant) at all points of the elastic curve^ In 
using (7) the axis Xmust be taken parallel to the length 
of the beam before flexure, which must be slight ; the 
minus sign in (7) provides for the case when d^y-r-dx^ ises"= 
sentialiy negative. 

232. Resilience of Flexure. — If the external forces are made 
to increase gradually from zero up to certain maximum 
Yalues,. some of them may do work, by reason of their 
points of application moving through certain distances 
due to the yielding, or flexure, of the body. If at the be- 
ginning and also at the end of this operation the body is 
at rest, this work has been expended on the elastic resis- 
tance of the body, and an equal amount, called the work 
of resilience (or springing-back), will be restored by the 
elasticity of the body, if released from the external forces, 
provided the elastic limit has not been passed. The energy 
thus temporarily stored is of the potential kind; see §§ 
148, 180, 196 and 218, 

232a. Distinction. Between Simple, and Continuous, Beams (or 
■** Girders "). — The external forces acting on a beam consist 
generally of the loads and the " reactions " of the sup* 
ports. If the beam is horizontal and rests on two supports 
only, the reactions of those supports are easily found by 
elementary statics [§ 36] alone, without calling into ac- 
count the theory of flexure, and the beam is said to be a 
Simple Beam, or girder ; whereas if it is in contact with 
more than two supports, being " continuous,'* therefore, 
over some of them, it is a Continuous Girder (§ 271). The 
Temainder of this chapter will deal only with simple 



252 



MECHANICS or E:NGi:tfEEIl]:NG. 



ELASTIC CURVES. 

233. Case I. Horizontal Prismatic Beam, [Supported at Both 
Ends, With a Central Load, Weight of Beam Neglected. — Fig. 
224. First considering the whole beam free, we find eack 



k- 



-Vd- 



% 



.—x- 



-I- 



:^ 



Fig. 324. § 233. 



reaction to be =%P. AOB is the neutral line ; required 

the equation of the portion OB referred to as an origin, 
and to the tangent line through as the axis of X To 
do this consider as free the portion mB between any sec- 
tion, m on the right of and the near support, in Fig. 
225 The forces holding this free body in equilibrium 




Fig. S25. 



Fis. S26. 



nre the one external force ^P, and the elastic forces act- 
ing on the exposed surface. The latter consist of J, the 
shear, and the tensions and compressions represented in 
the figure by their equivalent " stress-couple." Selecting 
N, the neutral axis of m, as an axis of moments (that J 
may not appear in the moment equation) and putting 
2 (mom) =0 we have 



P (I 

2V"2 



— X j 



rd^y-i 



'y -P (I 



dx" dx' 2 \2 / 



(1) 



Fig. 226 shows the elastic curve OB in its purely geomet- 
rical aspect, much exaggerated. For axes and origin as in. 
figure d^y-^doc^ is positive. 



ELASTIC CURVES. 253 

Eq. (1) gives the second a;-deriYative of y equal to a 
function of x. Hence tlie first fl?-derivative of y will be 
equal to tlie a?-anti-derivative of tliat function, plus a con- 
stant, (7. (By anti -derivative is meant tlie converse of de- 
rivative, sometimes called integral though not in the sense 
of summation). Hence from (1) we have (^/ being a con- 
stant factor remaining undisturbed) 

M^=~(Lx — -\+G . . (2)* 
dx 2 V2 2r 

(2)' is an equation between two variables c?2/-i-c?a; and a?, and 
holds good for any point between and B; dy-^dx de- 
noting the tang, of a, the slope, or angle between the tan- 
gent line and X At the slope is zero, and x also zero ; 

nencs at (2)' becomes 

^7x0=0— 0+C 

which enables us to determine the constant C, whose value 
must be the same at as for all points of the curve. 

Hence C=0 and (2)' becomes 



EI 



ay _ r ( I xf\ ..^ 

^~2"i-2'^2j • • ' ^^' 



from which the slope, tan. «, (or simply a, ir jt -measure: 
since the angle is small) may be found at auy point. Thus 
at B we have x=}4l and dy-^dx=ai, and 

. _ 1 PI' 
••"^""16" M 

Again, taking the cp-anti-derivative of both, members of eq, 
(2) we have 

^i2/=-f-(^-^)+C" . . . (3)' 

and since at both x and y are zero, G' is zero. Hence 
the equation of the elastic curve OB is 



254 MECHANICS OF ENGIXEEEIFG. 

^^^=f (^f) • • • <«' 

To compute the deflection of from the right line joiii'^ 
ing A and ^ in Fig. 224, i.e. BK, =c?, we put x^}^lm{^), a 
being then =d, and obtain 

^^=■^=©•5 • • • <** 

Eq. (3) does not admit of negative values for x ; for if 

the free body of Fig. 225 extended to the left of 0, the ex- 
ternal forces acting would be P, aownward, at ; and y^P, 
upward, at B, instead of the latter alone ; thus altering 
the form of eq. (1). From symmetry, however, we know 
that the curve AO, Fig. 224, is symmetrical with OB about 
the vertical through Q. 

Numerical Illustration. — Let [the beam shown in Fig. 224, resting 
on two unyielding supports at the same level, be of white oak timber 
and bear a load of P = 200 lbs. at the middle, its length being Z=12 ft. 
and cross-section rectangular with a width (horizontal) of 6 = 2 in. and 
height /i = 6 in. The modiilus of elasticity E will be taken as 1,600,000 
lbs./ in. ^ Required the radius of curvature, p, or the elastic curve at 
a point 4 ft. from the right-hand pier (or left). 

From the free body in Fig. 225 we have, using the form El-i-p for 
the moment of the stress-couple in the section, and putting i'(moms.)j\r 
= 0, with x = 2 ft., £7 -^J0= 100X48, the inch and pound being selected 
as units. Now I=bh^-irl2 (p. 94) which = 36 iu.^j whence, solving, 
(0 = 1,600, 000 X 36 -H 4800 = 4000 in. The cin-ve is evidently very flat. 
The smallest radius of curvature is found at the middle of the beam 
and is 2666 in.; at either extremity, A or B, it is infinite, since at each 
of these points the moment of the stress-couple is zero. 

At the same point (4 ft. from B) the "slope" of the elastic curve, 
viz., dy^dx, is found by putting x = 2 ft. = 24 in!, in eq. (2) from which 
is derived tan a = dy I dx = Q.Q025, corresponding to an angle of 0° 8' 36". 
At the extremity B we find, from ai = PP-7-16£'/, the slope of the tangent 
line to be ai = 0.0045; which is the tangent of 0° 15' 29". 

The deflection of the middle point is known from eq. (4), viz., 
d = PP^48EI; i.e., d=(200X 144 X 144 X 144) -v- (48X1,600,000X36) = 
0.216 in. 

It now remains to ascertain if the elastic limit is passed in any fibre 
of the beam. If we put the form p/-^e (for moment of stress-couple) 
in place of the present left-hand member of eq. (1), and solve for the 
unit (normal) stress in outer fibre, we findp=JPe(J Z— 2;)-f-/, which 
shows that p is greatest in the outer fibre of the section for which ^l—x 
is greatest, within the limits of the half-length; and this occurs at the 
middle of the beam, where x = 0. With this substitution we obtain 
p(max.) = pm = Pie h- (47) ; or pm = (200 X 12 X 12 X 3) -f- (4 X 36) = 600 
lbs./ in. ^, which is well within the elastic limit, for tension or compression 
in white oak. 



ELASTIC CURVES. 255 

233a. Load Suddenly Applied. — Eq. (4) gives the deflection 
d corresponding to the force or pressure P applied at the 
middle of the beam, and is seen to be proportional to it 
If a load G hangs at rest from the middle of the beam, 
P=G', but if the load G, being initially placed at rest 
.upon the unbent beam, is suddenly released from the ex- 
ternal constraint necessary to hold it there, it sinks and 
dehects the beam, the pressure P actually felt by the beam 
varying with the deflection as the load sinks. What is 
the maximum deflection d^ ? and what the pressure P^^ 
between the load and the beam at the instant of maximum 
deflection? In this motion of the body, or "load," it is 
acted on by two forces, the constant downward force G (its 
weight) and the variable upward force P, whose average vahie 
is |Pni ; while its initial and final kinetic energy are each zero. 
G does the work Gd^, while the work done upon P is ^Pmdm \ 
hence, by the theorem of '' Work and Energy " (p. 138), v/e 
have 

Gd^^hPrJra + 0-Q (5) 

That is, Pm = 2(r. Since at this instant the load is sub- 
jected to an upward force of 2 (r and to a downward force 
of only G (gravity) it immediately begins an upward mo- 
tion, reaching the point whence the motion began, and 
thus the oscillation continues. We here suppose the elas- 
ticity of the beam unimpaired. This is called the " sud- 
den " application of a load, and produces, as shown above, 
double the pressure on the beam which it does when grad- 
ually applied, and a double deflection. The work done 
by the beam in raising the weight again is called its re- 
silience. 

Similarly, if the weight G is allowed to fall on the mid- 
dle of the beam from a height Ji, we shall have 

Gx(h+d^, or approx., Gh,= ^P^d^i 

and hence, since (4) gives d,^ in terms of P^, 

e;i=i .^P oreA=2i^^ . (6) 



256 MECHANICS OIT EXGINEEEIXG. 

This theory suppcs5es the mass of the beam small com- 
pared with the falling weight. 

234. Case II. Horizontal Prismatic Beam, Supported at Both 

End? Bearing a Single Eccentric Load. Weight of Beam Neg- 

p p lected. — Fig. 227. The reactions 

4 t . of the -points of support, Pn and 

O I AXIS X I B "■ . X i ' >^ 

y'^^^^i^^J?^ Jvm ^i^^^^^^^^fe -^1' ^^^ easily found by consider- 

j, — -]Ei^._-_J^^^^^^'^ I ing the whole beam free, and put- 

j Ip j ting first 2'(mom.)o=0, whence i'l 

^\ '^' 1 =PZh-Zi, and then J(mom.)B=0, 

ri«227. whence Po= An— O^^i- i'o and 

Pi will now be treated as known quantities. 

The elastic curves 0(7 and OP, though having a comm on 
tangent line at (and hence the same slope a^, and a com- 
mon ordinate at 0, have separate equations and are both 
referred to the same origin and axes, as shown in the 
figure. The slope at 0, «o> and that at P,«i, are unknown 
constants, to be determined in the progress of the work. 

Eq[uation of OC. — Considering as free a portion of the 
beam extending from P to a section made anywhere on 
OC, X and y being the co-ordinates of the neutral axis of 
that section, we conceive the elastic forces put in on the 
exposed surface, as in the preceding problem, and put 
2'(mom. about neutral axis of the section) =0 which gives 
(remembering that here d?y-~dx^ is negative.) 

Ei^^=p{y-x)—p,{k—x)', , . (1) 

(X OC 

whencO;. by taking the x anti-derivatives of both members 

■ M ^ =P(lx—^)-F,{lx— -^)+ C 
ax 2 2 

To find 0, write out this equation for the point 0, where 
dy-^dx=aQ and a;=0, and we have 0=P/«o> hence the 
equation for slope is 



FLEXURE ELASTIC CUKVES. 257 

EI^=P{lx—^)-P,{l,x-^)+EIa^ .. (2) 

Again taking the x anti-derivatives, we have from (2) 

Ely =P (^^|1_|^._P,^^|^_^ YEIa,x+{C'=0) (3) 

'^at Oboth X and 2/ are —0 .°. C'=0). In equations (1), (2), 
and (3) no value of x is to be used <0 or >Z, since for 
points in CB different relations apply, thus 

Equation of CB. — Fig. 227. Let the free body extend 
from ^ to a section made anywhere on (7^.2'(moms.), as 
before, =0, gives (see foot-note on p, 322) 

^^^=-^^^1-^) . . . (4) 

(N.B. In (4), as in (1), Eld^y—dx^ is written equal to a neg- 
ative quantity because itself essentially negative ; for the 
curve is concave to the axis X in the first quadrant of the 
co-ordinate axes.) 

From (4) we have in the ordinary way (aj-anti-deriv.) 

EI"^ =-Pil,x -J^)+C" . . (5X 
ax 2 

To determine C", consider that the curves CB and OG 
have the same slope (dy-r-dx) at G where x=l; hence put 
x~l in the right-hand members of (2) and of (5)' and 
equate tha results. This gives C" = %PV-\-EIaQ and .-. 

^it-^ + ^I'^PS.^t^ . (5) 



A A o 



258 MECHANICS OE E2fGlNEEIlIN&. 

At C, where J5 = ?, botli curves have the same ordinate; 
hence, by putting x — l in the right members of (3) and (6)' 
and equating results, we obtain C'"— — }iPl^. .'. (6)' bo 
comes 



Mly = y2Pl'x+EIa^7>—P^ 



~2 6" 



~6" 



(6) 



as the Equation of CB, Fig. 227. But ag is still an unknown 
constant, to find which write out (6) for the point B where 

X = li, and y = 0, whence we obtain 

- 1 ^Fl'—3Fl\-{-2P,l,^] . . , (7) 



" 6m\ 

«!= a similar form, putting Pq ^^t P,, and (l^ — I) for I. 

235. Maximum Deflection in Case II — Fig. 227. The or- 
dinate ?/„ of the lowest point is thus found. Assuming 
^> /4 k> it will occur in the curve G. Hence put the 
dy-h-dx of that curve, as expressed in equation (2), =0. 
Also for O.Q write its value from (7), having put Pi=P?-r-Zij 
and we have 

whence [a? for max. y] = ^yi(2k—l) ■ 

Now substitute this value of x in (3), also ao from (7), and 
putPi =P?-T-Zi, whence 

Max. Deflec.=2/max=^% . -^ ll'—3l%+W,'] ^M^^O- 

236. Case III. Horizontal Prismatic Beam Supported at Both 
Ends and Bearing a Uniformly Distributed Load along its Whole 
Length. — (The weight of the beam itself, if considered, 



FLEXUEE. ELASTIC CUltVES. 



259 



constitutes a load of this nature.) Let 1= the length 
of the beam and w= the weight, per unit of length, 
of the loading ; then the load coming upon any length x 
will be =ivx, and the whole load ^=ui. By hypothesis w 
is constant. Fig. 228. From symmetry we know that the 



W=«)? 



Ulli I 1 1 1 lU 





Fig. 228. 



reactions at A and B are each =}4iol, that the middle of 
the neutral line is its lowest point, and the tangent line at 
is horizontal. Conceiving a section made at any point 
m of the neutral line at a distance x from 0, consider as 
free the portion of beam on the right of m. The forces 
holding this portion in equilibrium are yz'^h ^^^ reaction 
at B ; the elastic forces of the exposed surface at m, viz.: 
the tensions and compressions, forming a couple, and J 
the total she?r ; and a portion of the load, iv(^/2l — x). The 
sum of the mc ments of these latter forces about the neu- 
tral axis of m, is the same as that of their resultant; (i.e., 
their sum, since they are parallel), and this resultant acts in 
the middle of the length ^Z — x. Hence the sum of these 
moments =w(}4l — x)^[}4l — x). Now putting 2' (mom. 
about neutral axis of w)=0 for this free body, we have 



BI 



dx^ 



}4wl{}4l—x)—}^w(}41^xy 



i.e.,^/g- = > 



^t^(l^?2_^) 



(1) 



260 MECHANICS OF ENGINEEEIN&. 

Taking tlie cc-anti-derivative of both sides of (1), 

^^Tx =yM}{l^'^—}i^')+{G=0) (2) 

as the equation of slope. (The constant is =0 since at 
both dy-i-dx and x are =0.) From (2), 

my=-^i}il'x'-%x')+[C'=0] . . (3) 

which is the equation of the elastic curve ; throughout, 
i.e., it admits any value of x from x=-\-y2^ to x= — yil. 
This is an equation of the fourth degree, one degree high- 
er than those for the Curves of Cases I and II, where 
there were no distributed loads. If w were not constant, 
but proportional to the ordinates of an inclined right line, 
eq. (3) would be of the fifth degree ; if lo were propor- 
tional to the vertical ordinates of a parabola with axis 
vertical, (3j would be of the sixth degree ; and so on. 

By putting x=y^l in (3) we have the deflection of be- 
low the horizontal thro ugh A and B, viz.: (with W=^ total 
load ^wl) 

384 ' m S84: ' FI ' ' ^ ^ 

237. Case IV. Cantilevers. — A horizontal beam whose only 
support consists in one end being built in a wall, as in 
Fig. 229(a), or supported as in Fig. 
229(&) is sometimes called a canti- 
lever. Let the student prove that in 
Fig. 229(a) with a single end load P, 
the deflection of^ below the tangent 
at Ois d=j/Pl^-i-£^I;the same state- 
ment applies to Fig. 229(&), but the 
tangent at is not horizontal if the 
beam was originally so. It can also 
be proved that the slope at B, Fig. 
229(a) (from the tangent at 0) is 




FLEXUEE ELASTIC CURVES. 261 



«i= 



2^7" 



The greatest deflection of the elastic curve from the right 
line Joining AB, in Fig. 229(6), is evidently given by the 
equation for y max. in § 235, by writing, instead of P of 
that equation, the reaction at in Fig. 229(&). This assumes 
that the max. deflection occurs between A and 0. If it 
occurs between and B put (li—l) for I. 

If in Fig. 229(a) the loading is uniformly distributed 
along the beam at the rate of w pounds per linear unit, 
the student may also prove that the deflection of B below 
the tangent at is 

238. Case V. Horizontal Prismatic Beam Bearing Equal Ter- 
minal Loads and Supported Symmetrically at Two Points.— 
Fig. 231. Weight of beam neglected. In the preceding 
cases we have made use of the approximate form Eld'^y-r-dx^ 
in determining the forms of elastic curves. In the present 



ris~T% 




p\ 

1-^ '- < 1- 

Pig. 231. Fig. )i^Z. 

case the elastic curve from to (7 is more directly dealt 
with by employing the more exact expression EI-^f> (see 
§ 231) for the moment of the stress-couple in any sectioUo 
The reactions at and Care each =P, from symmetry. 
Considering free a portion of the beam extending from A 
to any section m between and C (Fig. 232) we have, by 
putting 2 (mom. about neutral axis of m)=0, 

P{i+x)- ^~Px==o .-. p^ 4r 



262 MECHANICS OF EXGINEBKIJfG. 

That is, the radius of curvature is the same at all points 
of OG ] in other words 0(7 is the arc of a circle with the 
above radius. The upward deflection of F from the right 
line joining and G can easily be computed from a knowl- 
edge of this fact. This is left to the student as also the 
value of the slope of the tangent line at (and G). The 
deflection of D from the tangent at G=^l^Pf-^EL as ip 
Fig, 229(a), 



SAFE LOADS IIS^ FLEXUKE. 

239. Maximum Moment. — As we examine the different sec- 
tions of a given beam undar a given loading we find differ- 
ent values oi p, the normal stress per unit of area in the 
outer element, as obtained from eq. (5) § 229, viz.: 

^=il/. . . . , (1) 

e 

in which I is the " Moment of Inertia " (§ 85) of the plane 
figure formed by the section, about its neutral axis, e the 
distance of the most distant (or outer) fibre from the neu< 
tral axis, and ilf the sum of the moments, about this neu- 
tral axis, of all the forces acting on the free body of which 
the section in question is one end, exclusive of the stresses 
on the exposed surface of that section. In other words 
Jf is the sum of the moments of the forces which balance 
the stresses of the section, these moments being taken 
about the neutral axis of the section under examination. 
For the prismatic beams of this chapter e and /are the 
same at all sections, hence p varies with M and becomes a 
maximum when J/ is a maximum. In any given case the 
location of the " dangerous section" or section of maximum 
M, and the amount of that maximum value may be deter- 
mined by inspection and trial, this being the only method 
(except by graphics) if the external forces are detached. 



FLEXUEE SAFE LOADS. 263 

If, however, the loading is continuous according to a de- 
finite algebraic law the calculus may often be applied, 
taking care to treat separately each portion of the beam 
between two consecutive reactions of supports^ or detached 
loads. 

As a graphical representation of the values of 31 along 
the beam in any given case, these values may be conceived 
laid off as vertical ordinates (according to some definite 
scale, e.g. so many inch-lbs. of moment to the linear inch 
of paper) from a horizontal axis just below the beam. If 
the upper fibres are in compression in any portion of the 
beam, so that that portion is convex downwards, these or- 
dinates will be laid off below the axis, and vice versa ; for 
it is evident that at a section where ilf=0, p also =0, i.e., 
the character of the normal stress in the outermost fibre 
changes (from tension to compression, or vice versa) when 
if changes sign. It is also evident from eq. (6) § 231 that 
the radius of curvature changes sign, and consequently the 
curvature is reversed, when J/ changes sign. These mo- 
ment ordinates form a Moment Diagram, and the extremities 
a Moment Curve. 

The maximum riioment, ilf^, being found, in terms of 
the loads and reactions, we must make the p of the " dan- 
gerous section," where M= M^^ equal to a safe value R', 
and thus may write 

^=M^ . . . o (2) 
e 

Eq. (2) is available for finding any one unknown quanti- 
ty, whether it be a load, span, or some one dimension of 
the beam, and is concerned only with the Strength, and not 
with the stiffness of the beam. If it is satisfied in any 
given case, the normal stress on all elements in all sections 
is known to be = or <i?', and the design is therefore safe 
in that one respect. 

As to danger arising from the shearing stresses in any 



264: 



MECHAXrCS OF ENGINEEKING. 



section, the consideration of the latter will be taken up in 
11 subsequent chapter and will be found to be necessary 
only in beams composed of a thin web uniting two flanges. 
The total shear, however, denoted by J, bears to the mo- 
ment ilf, an important relation of great service in deter- 
mining M^. This relation, therefore, is presented in the 
next article. 



pd? 



t-pdF ^ 



240, The Shear is the First x-Derivative of the Moment. — ^ 
Fig. 233. {x is the distance of any section, measured parallel 
wdx ij' to the beam from an arbitrary 

p'dp origin). Consider as free a ver- 
tical slice of the beam included 
between any two consecutive 
vertical sections whose distance 
apart is dx. The forces acting 
are the elastic forces of the two 
internal surfaces now laid bare, 
and, possibly, a portion, tvdx, 
of the loading, which at this 
part of the beam has some intensity =w lbs. per running 
linear unit. Putting 2'(mom. about axis .iV)=0 we have 
(noting that since the tensions and compressions of section 
JSf form a couple, the sum of their moments about N' is 
just the same as about N,) 




i- ^^ — + Jdx+ivdx 



:0 



But P^=M, the Moment of the left hand section,^ =31% 

6 e 

that of the right ; whence we may write, after dividing 
through by dx and transposing. 



M'—M 
dx 



-r , (jjdu 



. dM r 
dx 



(3) 



for w -2 vanishes when added to the finite J, and M*^ — M= 
d3l= increment of the moment corresponding to the incre- 
ment, dx, of X. This proves the theorem. 



FLEXURE. SAFE LOADS, 265 

Now the value of a? wliich renders M a maxininm or 
minimum would be obtained by putting the derivative 
dM ~ dx = zero; hence we may state as a 

Corollary. — At sections whe7'e the 'tnoment is a maximum 
■or Tninimiwrn the shear passes through the value zero. 

The shear J at any section is easily determined by con° 
sidering free the portioiL of beam from the section to either 
end of the beam and putting 2'( vertical components) = 0. 

In this article the words maximum and minimum are 
used in the same sense as in calculus ; i.e., graphically, 
they are the ordinates of the moment curve at points 
where tie tangent line is horizontal. If the moment curve be 
reduced to a straight line, or a series of straight lines, it 
ias'no maximum or minimum in the strict sense just 
stated ; nevertheless the relation is still practically borne 
out by the fact that at the sections of greatest and least 
ordinates in the moment diagram the shear changes sign 
suddenly. This is best shown by drawing a shear diagram, 
whose ordinates are laid off vertically from a horizontal 
axis and under the respective sections of the beam. They 
will be laid off upward or downward according as J" is 
found to be upward or downward, when the free body con- 
sidered extends from the section toward the right. 

In these diagrams the moment ordinates are set off on 
an arbitrary scale of somany inch-pounds, or foot-pounds, 
to the linear inch of paper ; the shears being simply 
pounds, or some other unit oi force, on a scale of so many 
pounds to the inch of paper. The scale on which the 
beam is drawn is so many feet, or inches, to the inch of 
(paper. 

241. Safe Load at the Middle of a Prismatic Beam Support- 
ed, at the Ends. — Fig. 234. The reaction at each support 
is ^P. Make a section n at any dis.tance cc<-L from B. 
Consider the portion nB free, putting in the proper elas- 
tic and external forces. The weight of beam is neglected. 
From i'(mom. about %)=0 we have 



2G6 MECHAIN'ICS OF EXGrSTEEKING. 

pL=^x; i.e., M=%Px 
e 2 

Evidently Tlfis proportional to x, and tlie ordinates repre^ 
senting it will 4;lierefore be limited by the straight line 




Fig. 234. 



B'Bt forming a triangle B'BA'. From symmetry, another 
triangle OR A' forms the other half of the moment dia- 
gram. Frqm inspection, the maximum iHf is seen to be in 
the middle where cc= }4l, and hence 



(il/max.)=7!/;„=i^P? 



. (1) 



Again by putting 2'(vert. compons.)=0, for the free body 
nB we have 

and must point downward since ~ points upward. Hence 
the shear is constant and = i^P at any section in the right 
hand half. If n be taken in the left half we would have, 
nB being free, from J(vert. com.)=0, 



FLEXUKE. SAFE LOADS, 267 

tlie same numerical value as before ; but J" must point up- 
ward, since | at 5 and J at n must balance tbe downward 
P at A. At A, then, the shear changes sign suddenly, 
that is, passes through the value zero; also at A, Mis a 
maximum, thus illustrating the statement in § 240. Notice 
the shear diagram in Fig. 234. 

To find the safe load in this case we write the maximum 
value of the normal stress, p,^=R% a safe value, (see table 
in a subsequent article) and solve the equation for P. 
But the maximum value of p is in the outer fibre at A, 
since Jf for that section is a maximum. Hence 

S^^%Pl (2) 

is lh.e equation for safe loading in this case, so far as the 
normal stresses in any section are concerned. 

Example. — If the beam is of wood and has a rectangu- 
lar section with width &= 2 in., height h-= 4 in., while its 
length 1= 10 ft., required the safe load, if the greatest nor- 
mal stress is limited to 1,000 lbs. per sq. in. Use the 
pound and inch. From § 90 1=^1^ M^=Vi2X 2x64= 10.66 
biquad. inches, while e=l=2 in. 

.-. P- ifiZ-ixiM^O^^lTT.T lbs. 
le 120x2 

■^ 242. Safe Load Uniformly Distributed along a Prismatic Beam 
Supported at the Ends.— Let the load per lineal unit of the 
length of beam he =w (this can be made to include the 
weight of the beam itself). Fig. 235. From symmetry, 

each reaction = yiwl. For the free body wO we have, put'' 
ting 2'(mom. about n)=Q, 

pi wl / X a? ii/r w ,1 9\ 

^ = -^x— (tax) - .-. Jf= j-ilx-^3ty) 



268 



MECHANICS OF ENGINEEKIN^ 



wliicli gives Jf for any section by making x vary from (F 
to I. Notice tliat in this case tlie law of loading is con- 
tinuous along tlie wliole length, and that hence the mo- 
ment curve is continuous for the whole length. 



W=«;? 




Fig. 235. 



To find the shear J, at n, we may either put 2'(vert. com 
pons.)=0 for the free body, whence e7= YiWl — wx^ and mus 
therefore_be downward for a small value of x ; or, employ 
ing § 240, we may write out dM-~dx, which gives 



J= 



dM 



dx 



(l—2x) 



(1/ 



the same as before. To find the max. 31, or Jfn,, put J- O.- 
which gives cc^^L This indicates ajnaximum, forwliaB 
substituted in d^3I-^dx\ i.e., in — iv, a negative result TB 
obtained. Hence ilf^ occurs at the middle of the beam and 
its value is 



= iiwl'; .'. ^=yiwV=%Wl 



m 



the equation of safe loading. W= total load=tyl- 

It can easily be shewn that the moment curve is 2 por« 



FLEXURE. SAFE LOADSo 



26& 



lion of a parabola, whose vertex is at A" under the mid- 
Jls of the beam, and axis vertical. The shear diagram 
consists of ordinates to a single straight line inclined to 
its axis and crossing it, i.e., giving a zero shear, under the 
middle of the beam, where we find the max. 31. 

If a frictionless dove-tail joint with vertical faces were 
introduced at any locality in the beam and thus divided 
the beam into two parts, the presence of J" would be made 
manifest "by the downward slipping of the left hand part 
oji the right hand part if the joint were on the right of the 
middle, and vice versa if it were on the left of the middle. 
This shows why the ordinates in the two halves of the 
shear diagram have opposite signs. The greatest shear 
is close to either support and is Jj^=^wl. 

243, Prismatic Beam Supported at its Extremities and Loaded 

in any Manner. Equation for Safe Loading. — Fig. 236. Given 

p .p^ p the loads Pj, P^, and P3, whose 

g I ' I I Q distances from the right sup- 

^^.. lLi port are l^, l^, and ^ ; ,required 
the equation for safe loading ; 
i.e., find ilf^ and write it = 

If the moment curve were 
continuous, i.e., if M were a 
continuous function of x from 
end to end of the beam, we 
could easily find Jf^ by making 
Fig. 236. dM-^dx=0, i.e., J=0, and sub- 

stitute the resulting value of x in the expression for M. 
But in the present case of detached loads, J is not zero, 
necessarily, at any section of the beam. Still there is 
sore J one section where it changes sign, i.e., passes sud- 
denly through the value zero, and this will be the section 
of greatest moment (though not a maximum in the stric^j 
sense used in calculus). By considering any portion n '^ 
as free, «/is found equal to the Reaction at Diminished by 
the Loads Occurring Between n and 0. The reaction at B is 




270 MECHAls'ICS or ENGi:JfEEEING. 

obtained by treating the whole beam as free (in which case 
no elastic forces come into play) and putting 2'(mom. 
about O)=0; while that at 0,=Pf,=Py-^P^-\-P^—Ps 
If n is taken anywhere between and E, J=Pq 

E " F,J=Po-Pi 
F " H,J^P^-P^-P^ 

H " B, J=Po-P\-P2-Ps 

This last value of j/also = the reaction at the other 
support,^. Accordingly, the shear diagram is seen to 
consist of a number of horizontal steps. The relation 
J=dM-^dx is such that the dope of the moment curve is 
proportional to the ordinate of the shear diagram, and 
that for a sudden change in the slope of the moment curve 
there is a sudden change in the shear ordinate. Hence in 
the present instance, J being constant between any two 
consecutive loads, the moment curve reduces to a straight 
line between the same loads, this line having a different 
inclination under each of the portions into which the beam 
is divided by the loads. Under each load the slope of the 
moment curve and the ordinate of the shear diagram change 
suddenly. In Fig. 236 the shear passes through the value 
zero, i.e., changes sign, at E; or algebraically we are sup- 
posed to find that Pq—P^ is + while PQ—P1—P2 is — , in 
the present case. Considering EO, then, as free, we find 
Jf;„ to be 

Mai=Poh~Pi{h~^i) and the equation for safe loading is 

?^-Pol-P.{k-k) (1) 

(i.e., if the max. il/is at F). It is also evident that the 
greatest shear is equal to the reaction at one or the other 
support, whichever is the greater, and that the moment 
at either support is zero. 

The student should not confuse the moment curve, which 



FLEXUKE. SAFE LOADS. 



271 



is entirely imaginary, with the neutral line (or elastic 
curve) of the beam itself. The greatest moment is not 
necessarily at the section of maximum deflection of the 
neutral line (or elastic curve). 

For the case in Fig. 236 we may therefore state that the 
max. moment, and consequently the greatest tension or 
compression in the outer fibre, will be found in the sec- 
tion under that load for which the sum of the loads (in- 
cluding this load itself) between it and either support first 
equals or exceeds the reaction of that support. The 
amount of this moment is then obtained by treating as free 
either of the two portions of the beam into which this 
section divides the beam. 

244. Numerical Example of the Preceding Article. — Fig. 237. 
Given Pi, Pg* -Ps* equal to i/^ ton, 1 ton, and 4 tons, re- 




spectively ; <i =5 feet, ^2= 7 feet, and ^3= 10 feet ; while ike 
total length is 15 feet. The beam is of timber, of rectan- 
gular cross-section, the horizontal width being b=10 
inches, and the value of B' (greatest safe normal stress), 
= ^ ton per sq. inch, or 1,000 lbs. per sq inch. 



272 ' MECHANICS Of^ ENGINEERING. 

Requirea the proper deptli k lor the beam, for safe load- 
ing. 

Solution. — Adopting a definite system of units, viz., the 
inch-ton-second system, we must reduce all distances such 
as I, etc., to inches, express all forces in tons, write K'= ^^ 
(tons per sq. inch), and interpret all results by the same sys- 
tem. Moments will be in inch-tons, and shears in tons. 
[N. B. In problems involving the strength of materials 
the inch is more convenient as a linear unit than the foot, 
since any stress expressed in lbs., or tons, per sq. inch, is 
. numerically 144 times as small as if referred to the square 
foot.] 

Making the whole beam free, we have from moms, about 
O, Pb~ [>^X 60+1x84+4x120] ^3.3 tons .-. Po=5.5— 
3.3=2.2 tons. 

The shear anywhere between O and ^is J= Po=2.2 tons. 

^ and i^ is e/ =2.2— 1^=1.7 

tons. 
The shear anywhere between i^ and His J =2.2 — ^ — 1 = 

0.7 tons. 
The shear anywhere between H and B is J = 2.2 — }4 — 1 

—4 =—3.3 tons. 
Since the shear changes sign on passing H, .-. the max. 
moment is at ^; whence making HO free, we have 
M at H=M,,, =2.2 x 120— ^^ x 60—1 x 36 =198 inch -tons. 

For safety M,„ must = , in which B'='^ ton per sq. 

inch, e = }4^ — }4 of unknown depth of beam, and /, §90, = 

I bM, with & = 10 inches 

,vi. >^ .|-Xl0.¥^198; or 71^-237.6 .-. h=15A inches. 



245. Comparative Strength of Rectangular Beams. — For such 
a beam, under a given loading, the equation for safe load- 
ing is 

^=3i;„ i. e. ye E bh'=M^ .... (1) 

« 



FLEXUEE. SAFE LOADS. 273 

whence the following is evident, (since for the same length, 
mode of support, and distribution of load, M^ is propor- 
tional to the safe loading.) 

For rectangular prismatic beams of the same length, 
same material, same mode of support and same arrange- 
ment of load : 

(1) The safe load is proportional to the width of beams 
having the same depth (A). 

(2) The safe load is proportional to the square of the 
depth of beams having the same width (h). 

(3) The safe load is proportional to the depth of beams 
having the same volume (i. e. the same hh] 

(It is understood that the sides of the section are hori- 
zontal and vertical respectively and thai the materia] \^ 
homogeneous.) 

246. Comparative Stiflfness of Rectangular Beams.— Taking tli*. 
deflection under the same loading as an inverse me^-sure 
of the stiffness, and noting that in §§ 233, 235, and 236, 
this deflection is inversely proportional to I—k hh^ = 
the " moment of inertia "of the section about its neutral 
axis, we may state that : 

For rectangular prismatic beams of the same length, 
same material, same mode of support, and same loading .• 

(1) The stiffness is proportional to the width for beams 
of the same depth. 

(2) The stiffness is proportional to the cube of the 
height for beams of the same width (&). 

(3) The stiffness is proportional to the square of the 
ciepth for beams of equal volume (hhl), 

(4) It the length alone vary, the stiffness is inversely 
proportional to the cube of the length. 

247. Table of Moments of Inertia. — These are here recapitu- 
lated for the simpler cases, and also the values of *?. the 
distance of the outermost fibre from the axis. 

Since the stiffness varies as /(other things being equal). 



274 



MECHANICS OF ENGINEERING. 



while tlie strength, varies* as I~-e, it is evident that a 
square beam has the same stiffness in any position (§89), 
while its strength is greatest with one side horizontal, for 
then e is smallest, being —^6. 

Since for any cross-section 1= j dF z^^ in which «=the 

distance of any element, dF, of area from the neutral axis, 
a beam is made both stiffer and stronger by throwing 
most of its material into two flanges united by a vertical 
web, thus forming a so-called " I-beam " of an I shape. But 
not without limit, for i;he web must be thick enough to 
cause the flanges to act together as a solid of continuous 
substance, and, if too high, is liable to buckle sideways^ 
thus requiring lateral stiffening. These points will be 
treated later. 




SECTION. 


/ 


e 


Rectangle, width = b, depth = h (vertical) 


Vm bh^ 


%h 


Bollow Rectaiigle, symmet. about neutral axia. See 1 
Fig. 238 (a) f 


Vi» [6i h,»-b^ h\^ 


%h, 


•Triangle, width =6, height = h, neutral axis parallel ^_ 
to base (horizontal). ) 


Vse M3 


%h 


Circle of radius r 


%^r^ 


r 


Eing of concentric circles. Fig. 238 (b) 


}in(r\~r*^) 


Ti 


Ehombus; Fig. 238 (c) h = diagonal which is vertical. 


V4e 5AS 


%h 


Square with side b vertical. 


Via b* 


%b 


" " " 6 at 45° with horiz. 


Vn ** 


HbVS 



248. Moment of Inertia of I-beams, Box-girders, Etc. — In 
common with other large companies, the Cambria Steel 

* This function, /-r-e, of the plane figure formed by the cross-section 
of a beam is evidently of three dimensions of length (cubic inches, for 
example), and is tabulated in the handbooks of the steel companies for 
different shapes of section; it is called the "section-modulus." See 
next page. 



FLEXURE. SAFE LOADS. 



275 



Co. of Johnstown, Pa., manufactures prismatic rolled beams 
and other "shapes," of structural steel, which are variously 
called I-beams, deck-beams (or " bulb -beams "), rails, angles, 
T-bars, channels, Z-bars, etc., according to the form of their 
sections. See Fig. 239 for some of these forms. The company 



d- 



^ 




T 



CHANNEL. DECK-BEAM. RAIL. 

Fie. 239. 



publishes a pocket-book giving tables of quantities rela- 
ting to the strength and stiffness of beams, such as the 
safe loads for various spans, moments of inertia of their 
sections in various positions, etc., etc„ The moments of 
inertia of /-beams and deck -beams are computed accord- 
ing to §§ 92 and 93, with the inch as linear unit. The 
/-beams range from 4 in. to 24 inches deep, the deck- 
beams being about 7 and 8 in. deep. (See foot-note, p. 274.) 
For beams of still greater stiffness and strength com- 
binations of plates, channels, angles, etc., are riveted to- 
gether, forming " built-beams,"' or " plate girders." The 
proper design for the riveting of such beams will be ex- 
amined later. For the present the parts are assumed to 
act together as a continuous mass. For example, Fig. 240 
shows a " box -girder," formed of two " channels " and 
two plates riveted together. If the axis of symmetry, JV, 

h h \ is to be horizontal it becomes the neu- 

3 ff"* tral axis. Let (7= the moment of iner- 
tia of one channel (as given in the 
pocket-book mentioned) about the axis 
iV perpendicular to the web of the chan- 
nel. Then the total moment of inertia oj 
the combination is (nearly) 



^ 



m 



^ 



Fig. 340. 



4 =W^'iUd?—ld't'{d^y2if 



(1> 



276 



MECHANICS OF EKGIXEBRING. 



In (1), &, t, and d are the distances given in Fig. 240 {d ex- 
tends to the middle of plate) while d' and t' are the length 
and width of a rivet, the former from head to head 
(i.e., d' and t' are the dimensions of a rivet-hole). 

For example, a box-girder of structural steel is formed of 
two 15 -in. channels (35 lbs, per foot) and two plates 10 in. 
wide and 1 in. thick ; the rivet-holes | in, wide and If in. 
long. That is, 5-10; i = l; d=8; f = f; and d' = l| in. 
Also from the hand-book we find that for the channel in 
question C==320 in,-* (i.e., biquad. in.). Hence, eq. (1), 
7^=640 + 2X10X1X64-4x1. 1(8-^)^ = 1625 in.4^ 

In this instance e=8^in, ; and if 15,000 lbs, /in.^ ( = 7,5 
tons/in.2) be taken as the value of R' (greatest safe normal 
stress in the extreme fibre of any section) as used by the 
Cambria Steel Co. for box-girders in buildings, we have 

" R'l 15000X1625 . . ., 

— = ^^ = 2, 867, 500 inch-lbs. 

e 8.5 ' ' 

That is, the max, safe ^'moment of resistances^ of the box- 
girder is M^ = 2,867,00O inch-lbs. = 1433.7 inch-tons; this 
quantity having to do with normal stresses in the section. The 
greatest " hending-momenf^ due to the amount, and mode, of 
loading on the beam, must not exceed this. Proper provision 
for the shearing stresses in the section, and in the rivets, wiU 
be considered later,) 



249. Strength of Cantilevers. — In Fig. 241 with a single 
, I ^ w^w? concentrated load P at the 

i::!::^:::::::;;^ Q_i_J_J_J_J_Jl projecting extremity, we 
r° easily find the moment at 
71 to be Jf =Px, and the 
/ max. moment to occur at 





the section next the wall, 

j ^f its value being M^=Pl. 

The shear, J", is constant, 

Fi«-24i. Fig. 242. ^mj = P at all sections. 

The moment and shear diagrams are drawn in accordance 

with these results. 



FLEXUKE. SAFE LOADS. 



277 



If the load W= wli^ uniformly distributed on tlie can- 
tilever, as in fig. 242, by making nO free we have, putting 
-2'(mom. about n) = 0, 



pi 



^-^=ivx . I .'.M=}iwaP.'. J4=;^w;Z2= ^ 



Wl. 



Hence the moment curve is a parabola, whose vertex is at 
0' and axis vertical. Putting I (vert, compons.) = we 
obtain J = ivx. Hence the shear diagram is a triangle, 
and the max. J = wl = TV. 

250. Resume'ofthe Four Simple Cases. — The following table 
shows the values of the deflections under an arbitrary 
load P, or W, (within elastic limit), and of the safe load ; 



Deflection 

J Safe load (from ?!I 

I = Mm) 

Relative strength 

j Relative stiffness 
1 under same load 

j Relative stiffness 
I under safe load 

J Max. shear = Jm,(.wa.d 
t location, 



Cantilevers. 



With one end 
loadP 
Fii?. 241 



EI 
B'l 



P, (at wall) 



With unif . load 
Fig. 242 



Beams with two end supports. 



Load P in 
middle 
Fig. 234 



EI 
B'l 



J. 
3 

W, (at wall) 



iS'EI 
,B'l 



ViP, (at supp). 



Unif. loaa 
W=wl 
Fig. 235 



5 y\/fi 

384' EI 



8 
128 

6 
16 
5 



W, (at suppi) 



also the relative strength, the relative stiffness (under the 
same load), and the relative stiffness under the safe load, 
for the same beam. 

The max. shear will be used to determine the proper 
web-thickness for /-beams and " built-girders." The stu- 
dent should carefully study the foregoing table, noting 
especially the relative strength, stiffness, and stiffness 
under safe load, of the same beam. 

Thus, a beam with two end supports will bear a double 



278 MECHANICS OF ENGINEERING. 

load, if uniformly distributed instead of concentrated in 
the middle, but will deflect ^;( more ; whereas with a given 
load uniformly distributed the deflection would be only 
5/^ of that caused by the same load in the middle, provided 
<-he elastic limit is not surpassed ii? either case. 

261. E', etc. For Various Materials.— The formula& = Jf^, 

e 

from which in any given case of flexure we can compute 
the value of p^, the greatest normal stress in any outer 
element, provided all the other quantities are known, 
holds good theoretically within the elastic limit only. 
Still, some experimenters have used this formula for the 
rupture of beams by flexure, calling the value of p^ thus 
obtained the Modulus of Rupture, B. R may be found to 
differ considerably from both the ^ or C of § 203 with 
some materials and forms, being frequently much larger. 
This might be expected, since even supposing the relative 
extension or compression (i.e., strain) of the fibres to be 
proportional to their distances from the neutral axis as 
the load increases toward rupture, the corresponding 
bLi'esses, not being proportional to these strains beyond the 
elastic limit, no lono^'er vary directly as tlie distances from the 
neutral axis ; and the neutral axis does not pass through the 
centre of gravity of the section, necessarily. 

The following table gives average values for R, R', R", 
and E for the ordinary materials of construction.'^ E, the 
modulus of elasticity for use in the formulsB for deflection, 
is given as computed from experiments in flexure, and is 
nearly the pame as E^^ and E^. 

In any example involving R', e is usually written equal 
to the distance of the outer fibre from the neutral axis, 
whether that fibre is to be in tension or compression ; 
since in most materials not only is the tensile equal to the 
compressive stress for a given strain (relative extension 
or contraction) but the elastic limit is reached at about 
the same strain both in tension and compression. 

* Wet, or unseasoued, timber is very cousiderably weal^er than that (such as 
ordinary " dry" timber) containing only 12 per cent, of moisture. Large pieces 
of timber talie a much longer time to season than small ones. (Johnson.) 



FLEXUKE. SAFE LOADS. 



279 



Table foe Use in Examples in Flexure. 





Timber. 


Cast Iron. 


Wro't Iron, 


Structural 
Steel. 


Max. safe stress in outer fi- ) 
bre— if'dbs. per sq. inch). ) 


600 

to 

1,200 


6,000 in tens. 
12,000 in comp. 


12,000 


15,000 


Stress in outer fibre at Elas. ) 
limit = j?''(lbs. per sq. in.) ) 






17,000* 

to 
35,000 


30,000 
and upward. 


" Modul. of Rupture " 1 
=i?=lbs. per sq. inch. ) 


4,000 

to 

10,000 


40,000 


50,000 


60,000 


E^Mod. of Elasticity, j 
=lbs. per sq. inch. ) 


1,000,000 

to 
2,000,000 


17,000,000 


25,000,000 


29,000,000 



In the case of cast iron, however, (see § 203) the elastic 
limit is reached in tension with a stress =9,000 lbs. per 
sq. inch and a relative extension of ^^ of one per cent., 
while in compression the stress must be about double to 
reach the elastic limit, the relative change of form (strain) 
being also double. Hence with cast iron beams, once 
extensively used but now largely replaced by rolled beams 
of structural steel, an economy of material was effected 
by making the outer fibre on the compressed side twice 
as far from the neutral axis as that on the stretched side. 
Thus, Fig. 243, cross-sections with unequal flanges were 
used, so proportioned that the centre of 
gravity was twice as near to the outer 
fibre in tension as to that in compression, 
i.e., e2=2ej; in other words more material 
J is placed in tension than in compression. 
The fibre A being in tension (within elas- 
tic limit), that at B, since it is twice as far from the neu- 
tral axis and on the other side, is contracted twice as much 
as A is extended ; i.e., is f.nder a compressive strain 
double the tensile strain at A, but in accordance with the 
above figures its state of stress is proportionally as much 
within the elastic limit as that of A. 

* In the tests by U. S. Gov. in 1879 with I-beams, B" ranged from 25,000 
to 38,000, and tlie elastic limit was reached with less stress in the large 
than in the smaller beams. Also, for the same beam, U" decreased with 
larger spans. 



1 

N 




i 
1 ^ 








A 




f. 



Fig. 243. 



280 MECHANICS OP ENGIJfEERING. 

The great range of values of R for timber is due not 
only to the faet that the various kinds of wood differ 
widely in strength, while the behavior of specimens of 
any ona kind depends somewhat on age, seasoning, etc., 
but also to the circumstance that the size of the beam un- 
der experiment has much to do with the result. The ex- 
periments of Prof. Lanza at the Mass. Institute of Tech- 
nology in 1881 were made on full size lumber (spruce), of 
dimensions such as are usually taken for floor beams in 
buildings, and gave much smaller values of R (from 3,200 
to 8,700 lbs. per sq. inch) than had previously been ob- 
tained. The loading employed was in most cases a con- 
centrated load midway between the two supports. 

These low values are probably due to the fact that in 
large specimens of ordinary lumber the continuity of it& 
substance is more or less broken by cracks, knots, etc., 
the higher values of most other experimenters having 
been obtained with small, straight-grained, selected pieces, 
from one foot to six feet in length. See footnote p. 278* 

Yaluable information and tables relating to timber beams 
may be found in the hand-book of the Cambria Steel Co. 

The value R' = 1^,000 lbs. per sq. inch is employed by the 
Cambria Steel Co. in computing the safe loads for their 
rolled I-beams of structural steel ; but with the stipulation 
that the beams (which are high and of narrow width) must 
be secure against yielding sideways. If such is not the 
case the ratio of the actual safe load to that computed with 
i2' = 16,000 is taken less and less as the span increases. 
The lateral security referred to may be furnished by the 
brick arch-filling of a fire-proof floor, or by light lateral 
bracing with the other beams. 

252. Numerical Examples. — Example 1. — A square bar of 
wrought iron, 1^ in. in thickness is bent into a circular :|: 
arc whose radius is 200 ft,, the plane of bending being par- 
allel to the side of the square. Bequired the greatest nor^ 
mal stress p^ in any outer fibre. 

Solution. From §§ 230 and 231 we may write 

—t =£— .'. p=eU-T-p, i.e., is constant. 



FLEXURE. SAFE LOADS. 281 

For the units incli and pound (viz. tliose of the table in § 
251) we have e=% in., /> =2,400 in., and ^=25,000,000 lbs. 
per sq. inch, ani .•. 

i>=i>m=^x 25,000,000-^2,400 =7,812 lbs. per sq. in.^ 

which is quite safe. At a distance of ^ inch from tne 
neutral axis, the normal stress is =\_%-^}i.^Pm — %Pm— 
5,208 lbs. per sq. in. (If the force-plane (i.e., plane of 
bending) were parallel to the diagonal of the square, e 
would =}4x 1.5^^2 inches, giving p^ = \l ,S12x ^/2 ] H^s. 
per sq. in.) § 238 shows an instance where a portion, 0C7, 
Fig. 231, is bent in a circular arc. 

Example 2. — A hollow cylindrical cast-iron pipe of radii 
3 y2 and 4 inches* is supported at its ends and loaded in 
middle (see Fig. 234). Eequired the safe load, neglecting 
the weight oi the pipe. From the table in § 250 we have 
for safety 

P=4^ 

le 

From § 251 we put i?'= 6,000 lbs. per sq. in.; and from § 

247/=^(ri* — rf)\ and with these values, r2 being =4> '"'i — 
4l, e=ri=4, 7i=-B- and Z=144 inches (the inch must be the 
unit of length since i?' = 6,000 lbs. per sq. inch) we have 

7>=4x6,000x;^- ^(256-150)-r-[144x4] .-. P=3,470 lbs. 
The weight of the beam itself is (r= Vy, (§ 7), i.e., 

Q^^^r,'^ri)lr= f-(16-12i^)144Xjg=448 lbs. 

(Notice that y, here, must be lbs., per cubic inch). This 
weight being a uniformly distributed load is equivalent to 
half as much, 221 lbs., applied in the middle, as far as the 
strength of the beam is concerned (see § 250), .*. P must be 
taken =3,249 lbs. when the weight of the beam is consid- 
ered. 

* And length of 13 feet, should be added. 



282 MECHANICS OF ENGINEERING. 

Example 3. — A Cambria I-beam, of structural steel, is to 

be placed horizontally on two supports at its extremities and 

is to be loaded imiformly (Fig. 235), the span being ^ = 20 ft. 

5g„_^ Its cross-section. Fig. 244, has a depth 

T ^ V^ parallel to the web, of 15 in. In the 



lA 



W f\ 



^ 



Fig. 244. 



handbook of the Cambria Steel Co. it 
-\ n" is designated as B 53, 15 in. in depth, 
and weighing 42 lbs. per foot of length; 
its section having a moment of inertia 
/i = 442 in.4 about a gravity axis per- 
pendicular to the web (for use when the web is vertical; the 
strongest position) and 72 = 14.6 in.^ about a gravity axis 
parallel to the web (i.e., when the web is placed horizontally). 
First, placing the web vertically, we have from § 250, 

7->/7 

Wi = safe load, distributed, = 8-^. With R' = 16,000, 

1 1 = 442, I = 240 inches, and ei --^ 7 J inches, this gives * 
TFi = (8 X 16,000 X 442) - (240 X 7.5) - 31,430 lbs. 

But this includes the weight of the beam, =20x42 = 840 
lbs.; hence a distributed load of 30,600 lbs., or 15.3 tons, 
may be placed on the beam (secured against lateral yielding). 
The handbook of the Cambria Steel Co. referred to gives 15.7 
tons as the safe load.) 

With the web placed horizontally, we find as safe load 

Tf 2 = 8^-^^ = (8 X 16,000 X 14.6) ^ (^240 X ^) = 2830 lbs. ; 

or less than 1/10 of Wi. Hence in this position the beam 
could carry safely only 1990 lbs. above its own weight. 

Example 4. — Required the deflection at the middle in the 
first case of Ex. 3. From § 250 this deflection is 

, _A 

^'"384 

* The handbook of the Cambria Steel Co. also gives in a separate 
column the quantity /j^ej, called the "section-modulus," S, (cub. in. 
or in.^); so that the formula for the safe load would be TFj = 8JS'*S -e- i, 
S having the value 58.9 in.' in the present instance. 



Wil^ 5 8R'h 


l^ 


5 


R' Z2 


Ell "384" lei 


'Ell 


"48 


E ' ei 



FLEXUliE. SAFP] LOADS. 283 

. , 5 16,000 (240)2 

'-'•' ^^^48 • 29,000,000 ' V ^ ^'^^ 

.-. d,=0. 4:4:1 in. 
Example 5. — A rectangular beam of yellow pine, of widtli 
6=4 inches, is 20 ft. locg, rests on two end supports, and is 
to carry a load of 1,200 lbs. at the middle ; required the 
proper depth h. From § 250 

le T 12 ' ^h 

.: h?=6Pl-^4:B'b. For variety, use the inch and ton. For 
this system of units P=0.60 tons, i?'=0.50 tons per sq. in., 
1=24:0 inches and 6= 4 inches. 

.-. /i2=(6x0.6x240)4-(4x0.5x4)=108sq. in. .-. ^=10.4 in. 

Example 6. — Suppose the depth in Ex. 5 to be deter- 
mined by the conditien that the deflection shall be = Ygoo 
•f the span or length. "We should then have from § 250 

d= k 1=1 ^' 
600 48 EI 

Using the inch and ton, with ^=1,200,000 lbs. per sq. in., 
which = 600 tons per sq. inch, and /=Yl2^^^ we have 

;^3^ 500x0.60x240x240x12 ^^^ _. ^^^^^ ^ 
48x600x4 

As this is > 10.4 the load would be safe, as well. 

Example 7. — Required the length of a wro't iron pipe 

supported at its extremities, its internal radius being 2}^ 

in., the external 2.50 in., that the deflection under its oivn 

weight may equal Yioo of the length. 579.6 in. Ans. 

Example 8. — Fig. 245. The wall is 6 feet high and one 

foot thick, of common brick work 

(see § 7) and is to be borne by an 

7-beam in whose outer fibres no 



n 



greater normal stress than 8,000 

*£ lbs. per sq. inch is allowable. If 
Pio. 245. a number of I-beams is available, 



284 MECHANICS OF ENGINEERLNG. 

ranging in height from 6 in. to 15 in. (by whole inches), 
which one shall be chosen in the present instance, if their 
cross-sections are Similar Figures, the moment of inertia of 
the 15-inch beam being 800 biquad. inches ? 

The 12-inch beam. Ans. 



SHEARIXa STRESSES IN FLEXURE. 

253. Shearing Stresses in Surfaces Parallel to the Neutral 
Surface. — If a pile of boards (see Fig. 246) is used to sup- 
port a load, the boards being free to slip on each other, it 
is noticeable that the ends overlap, although the boards 



Fig. 246. 




are of equal length (now see Fig. 247) ; i.e., slipping has 
occurred along the surfaces of contact, the combina- 
tion being no stronger than the same boards side by 
side. If, however, they are glued together, piled as in the 
former figure, the slipping is prevented and the deflection 
is much less under the same load P. That is, the com- 
pound beam is both stronger and stiifer than the pile of 
loose boards, but the lendency to slip still exists and is 
known as the " shearing stress in surfaces parallel to the 
neutral surface." Its intensity per unit of area will now 
be determined by the usual " free-body " method. In Fig. 
248 let AN' be a portion, considered free, on the left of any 




N N 



FiQ. S48. 



SHEAB, US FLEXURE, 



285 



section N', of a prismatic beam slightly bent under forces 
in one plane and perpendicular to the beam. The moment 
equation, about the neutral axis at JSf', gives 



■^=M' ; whence «'= — =^ 
e 1 



(1) 



Similarly, with AN as a free body, NN' being =^dx, 

t—=M; whence ^=- . 
e I 



. (2) 



p and p' are the respective normal stresses in the outer 
fibre in the transverse sections N and N' respectively. 

Now separate the block NN', lying between these two 
consecutive sections, as a free body (in Fig. 249). And 



^W 




%f^^ I 




BART OF J 



furthermore remove a portion of the top of the latter block, 
the portion lying above a plane passed parallel to the neu- 
tral surface and at any distance z" from that surface. This 
latter free body is shown in Fig. 250, with the system of 
forces representing the actions uj)on it of the portions taken 
away. The under surface, just laid bare, is a portion of a sur- 
face (parallel to the neutral surface) in which the above men- 
tioned slipping, or shearing, tendency exists. The lowfer por- 
tion (of the block NN') which is now removed exerted this 



286 MECHANICS OF ENGINEEEIKG. 

rubbing, or sliding, force on the remainder along the under 
surface of the latter. Let the unknown intensity of this 
shearing force be X(per unit of area) ; then the shearing 
force on this under surface is =Xy"dx, (y",= oa in figure, 
being the horizontal width of the beam a.t this distance z" 
from the neutral axis of N') and takes its place with the 
other forces of the system, which are the normal stresses 

between , and portions of J and J', the respective 

_z=z" 

total vertical shears. (The manner of distribution of J 
over the vertical section is as yet unknown ; see next arti- 
cle.) 

Putting 2 (horiz. compons.) = in Fig. 250, we have 

P -p'dF— r -pdF—Xy"dx=0 

,'.Xy"dx=P'—P fzdF 
^ z" 

But from eqs. (1) and (2), p'— p = (iHf — Jf)J-=^ dM, 
while from § 240 dM = Jdx ; 

.■.Xrd.=^-^jlaF.:X=^fliF (3) 

z " z 

as the required intensity per unit of area of the shearing 
force in a surface parallel to the neutral surface and at a 
distance z" from it. It is seen to depend on the " shear " J 
and the moment of inertia I of the whole vertical section; 
upon the horizontal thickness* y'' of the beam at the sur- 
face in question ; and upon the integral / zdF, 

^«" 
which (from § 23) is the product of the area of that part of 
the vertical section extending from the surface in question to 
the outer fibre, by the distance of the centre of gravity of that 
part from the neutral surface. 

* Thickness of actual substance. 



SHEAR IN FLEXURE. 



287 



It now follows, from § 209, that the intensity (per unit 
area) of the shear on an elementary area of the vertical 
cross section of a bent beam, and this intensity we may call 
Z, is equal to that X, just found, in the horizontal section 
which is at the same distance (z") from the neutral axis. 

254. Mode of Distribution of J, the Total Shear, over the Verti- 
cal Cross Section. — The intensity of this shear, Z (lbs. per 
sq. inch, for instance) has just been proved to be 



Z=X=^, CzdF 



(4) 



ly 

To illustrate this, required the 
value of Z two inches above the neu- 
tral axis, in a cross section close to 
the abutment, in Ex. 5, § 252. Fig. 
251 shows this section. From it we 
have for the shaded portion, lying 
above the locality in question, y" = 

4 inches, and C ~ ' sdF = (area 
^ z"= 2 

of shaded portion) X (distance of 

its centre of gravity from J^A) = Fia.aoi. 

(12.8 sq. in.) x (3.6 in.) = 46.08 cubic inches. 

The total shear J = the abutment reaction = 600 lbs., 
while 1= L bM = ^ X 4 X (10.4)^ = 375 biquad. inches. 
Boih J Siud I refer to the whole section. 

600x46.08 ift.oiK 

= 18.42 lbs. per sq. m.. 




..Z- 



375x4 

•qui+e insignificant. In the neighborhood of the neutral 
axis, where z" = 0, we have y'" = 4 and 



r^ zdF= r^ zdF=^ 20.8 X 2.6=54.8, 
J z"=0 J 



z''=0 J 

wh__e J and I of course are the same as before, 
for z" =0 



Hence 



288 



MECHANICS OF ENGINEERING. 




Z=^o^ 21.62 lbs. per sq. in. 
At the outer fibre since f^ zdF^O, %" being = e, ^ is = 

for a beam of any shape. 

For a solid rectangular section like the 
above, Z and 2" bear the same relation to 
each other as the co-ordinates of the para- 
bola in Fig. 252 (axis horizontal). 

Since in equation (4) the horizontal 
thickness, y" , from side to side ef the sec- 
tion of the locality where Z is desired, ^iq, 25? 

occurs in the denominator, and since / %dF 

increases as g" grows numerically smaller, the following 
may be stated, as to the distribution of J, the shear, in 
any vertical section, viz.: 

The intensity (lbs. per sq. in.) of the shear is zero at 
the outer elements of the section, and for beams of ordi- 
nary shapes is greatest where the section crosses the neu- 
tral surface. For forms of cross section having thin webs 
its value may be so great as to require special investiga- 
tion for safe design. 

Denoting by Z^ the value of ^at the neutral axis, (which 
=Xo in the neutral surface where it crosses the vertica 
section in question) and putting the thickness of the sub- 
stance of the beam = &o at the neutral axis, we have, 



Zq—Xq — 



J 



Iho 



X 



area above 
neutral axis 
(or below) 



X 



the dist. of its cent, 
grav.from that axis 



(5) 



255. Values of Zo for Special Forms of Cross Section. — From 
the last equation it is plain that for a prismatic beam the 
value of Zo is proportional to J, the total shear, and hence 
to the ordinate of the shear diagram for any particular 
case of loading. The utility of such a diagram, as obtain- 



SHEAR ll>r FLEXURE. 



289 



ed in Figs. 234-237 inclusive, is therefore evident, for by- 
locating tlie greatest shearing stress in the beam it 
enables us to provide proper relations between the load- 
ing and the form and material of the beam to secure safety 
against rupture by shearing. 

The table in § 210 gives safe values which the ^^s-^ 
maximum Zq in any case should not exceed. It is \ 
only in the case of beams with thin webs (see Figs. 
238 and 240) however, that Zq is likely to need at- 
tention. 

For a Rectangle we have, Fig. 253, (see eq. 5, § 



! 
l4 



Fig. 253. 



254) 6o=&, I=>/xib'h\ and C\dF=%'h'h , yih^yiW 



.'.Zn — Xn- 



2 



~ (total shear) -7- (whole area) 



Hence the greatest intensity of shear in the cross-section 
is A as great per unit of area as if the total shear were 
uniformly distributed over the section. 




Fig. 254. Fig. 233. Fig. ^5 

For a Solid Circular section Fig. 254 



Fig. 257. 



Z,= -^f^dF = 

IhffJo l^nr^ . 2r 



Stt 3 Tir^ 



[See § 26 Prob. 3]. 

For a Hollow Circular section (concentric circles) • Fig. 
255, we have similarly, 



290 



Zo=- 



MECHATTICS OF BKGIKEERIIJ^G. 
J 



Xri*-r2*)2(ri-n) 






J{r{'—r.i') 



3 ;r(ri*— r2*)(ri— rg) 



A-pplying this formula to Example 2 § 252, we first have 
as the max. shear J„, = ^P =1,735 lbs., this being the abut- 
ment reaction, and hence (putting tc = (22 -=- 7)) 



^0 max. = 



_ 4x7xl735[64-42.8] 



3x22[256-150](4-3.5) 



= 294 lbs. per sq. in. 



which cast iron is abundantly able to withstand in shear- 
ing. 

For a Hollow Rectangular Beam, symmetrical about its 
neutral surface, Fig. 256 (box girder) 

;7_ 3}i{hJh'-hJi i) _3 J[5,V-5,V ] 
' %ihji,'-hjh%h-h,) 2* [6A-^-M/][6;-&,] 

The same equation holds good for Fig. 257 (I-beam with 
square corners) but then &2 denotes the sum of the widths 
of the hollow spaces. 



256. Shearing Stress in the Web of an I-Beam. — It is usual to 
consider that, with I-beams (and box- 
beams) with the web vertical the shear J, 
in any vertical section, is borne exclusively 
by the web and is uniformly distributed 
■ over its section. That this is nearly true 
may be proved as follows, the flange area 
being comparatively large. Fig. 258.- Let 
Fi be the area of one flange, and F^ that of 
the half web. Then since 



_i = 

N 



b< 



Fig. 258. 



/=^ W+2i^, (f ) 



SHEAE IK PLiEXUEE. 291 

(tlie last term approximate, ^ /iq being taken for the radi- 
us of gyration of jPi,) while 

r zdF=Fi ±.-{-Fq-^, (the first term approx.) we have 

J r\dF 
Jo _ J%K{^F,+F,) J 

if we write (2i^i+i^o) ^ (6i^i+2i^o)=K • ^ut &o^o is the 
area of the whole web, .'. the shear per unit area at the 
neutral axis is nearly the same as if J were uniformly dis- 
tributed over the web. E. g., with jPi = 2 sq. in., and Fq 
= 1 sq. in. we obtain Zq = 1.07 (J-r-hoh^. 

Similarly, the shearing stress per unit area at n, the 
upper edge of the web, is also nearly equal to e7-f- JqAo (see 

eq.,4(§254) for then \ T {zdF)'\ ^ F^.y^K nearly, 

while / remains as before. 

The shear per unit area, then, in an ordinary I-beam 1& 
obtained by dividing the total shear J by the area of the 
web section.* 

Example. — It is required to determine the proper thick- 
ness to be given to the web of the 15- inch structural steel 
rolled I-beam of Example 3 of p. 282, the height of web 
being 13 in., and the maximum safe shearing stress being 
taken as 8750 lbs. /in. ^ (as prescribed by the Philadelphia 
building laws for mild steel). The web is vertical. 

The greatest total shear, J^^ which occurs at either support, 
is equal to half the load, i.e., to 15,715 lbs.; and hence, 
with 6o denoting the thickness of web, we have 

J 15 715 

Zomax.=^; i.e., 8750= ,-^—j^ ; .-. 6o = 0.138 in. 

* That, is, for the vertical, or horizontal, section of web. The shear 
on bome oblique plane may be somewhat larger than this. . See §§ 270a 
and 314. 



292 



MECHANICS OF ENGINEERING. 



(Units, inch and pound). The 15-incli I-beam in question of 
the Cambria Steel Co.), weighing 42 lbs. to the linear foot, 
has a web 0.41 in. thick, which provides a very ample resist- 
ance to shearing stress. 

In the middle of the span, Zo = 0, since J = 0. ' 

257. Designing of Riveting for Built Beams. — The latter are 
generally of the I-beam and box forms, made by riveting 
together a number of continuous shapes, most of the ma- 
terial being thrown into the flange members. E. g., in fig. 
259, an I-beam* is formed by riveting together, in the 
manner shown in the figure, a " vertical stem plate " or 
web, four "angle-bars," and two "flange-plates," each of 





Fig. 259. 



Fig. 260. 



these seven pieces being continuous through the whole 
length of the beam. Fig 260 shows a box-girder. If the 
riveting is well done, the combination forms a single rigid 
beam whose safe load for a given span may be found by 
foregoing rules ; in computing the moment of inertia, how- 
ever, the portion of cross section cut out by the rivet 
holes must not be included. (This will be illustrated in 
a subsequent paragraph.) The safe load having been com- 
puted from a consideration of normal stresses only, and 
the web being made thick enough to take up the max. 
total shear, J",,, with safety, it still remains to design the 
riveting, through whose agency the web and flanges are 
caused to act together as a single continuous rigid mass. 
It will be on the side of safety to consider that at a given 

* Such a built I-beam is usually designated a " plate-girder. '\ See 
handbook of the Cambria Steel Co. 



SHEAK m FLEXURE. 293 

locality in the beam the shear carried by the rivets con- 
necting the angles and flanges, per unit of length of beam, 
is the same as that carried by those connecting the angles 
and the web ("vertical stem -plate"). The amount of this 
shear may be computed from the fact that it is equal to 
that occurring in the surface (parallel to the neutral sur- 
face) in which the web joins the flange, in case the web 
and flange were of continuous substance, as in a solid I- 
beam. But this shear must be of the same amount per 
horizontal unit of length as it is per vertical linear unit in 
the web itself, where it joins the flange ; (for from § 254 Z 
=X) But the shear in the vertical section of the web, 
being uniformly distributed, is the same per vertical linear 
unit at the junction with the flange as at any other part 
of the web section (§ 256,) and the whole shear on the ver- 
tical section of web = J, the " total shear " of that section 
of the beam. 

Hence we may state the following : 

The riveting connecting the angles with the flanges, (or 
the web with the angles) in any locality of a built beam, 
must safely sustain a shear equal to J on a horizontal length 
eqtial to the height of web. 

The strength of the riveting may be limited by the re- 
sistance of the -rivet to being sheared (and this brings 
into account its cross section) or upon the crushing resist- 
ance of the side of the rivet hole in the plate (and this in- 
volves both the diameter of the rivet and the thickness of 
the metal in the web, flange, or angle. In its hand-book, the 
Cambria Steel Co. gives tables for the safe strength of rivets, 
and compressive resistance of plates ; based on unit shearing 
stresses from 6,000 to 10,000 Ibs./sq. in. for shearing stress in 
the rivets, and 12,000 to 20,000 Ibs./sq. in. compressive re- 
sistance, in the side of the rivet hole, the axial plane section of 
the hole being the area of reference. 

In fig. 259 the rivets connecting the web with the angles 
are in double shear, which should be taken into account in 
considering their shearing strength, which is then double ; 
those connecting the angles and the flange plates are in 



294 MECHANICS OF ENGINEERING. 

single shear. In fig. 260 (box-beam) where the beam is 
built of two webs, four angles, and two flange plates, all 
the rivets are in single shear. If the web plate is very high 
compared with its thickness, vertical stiffeners in the form 
of "angles " may need to be riveted upon them laterally 
[see § 314]. 

Example. — ^A built I-beam of structural steel (fig. 259) 
is to support a uniformly distributed load of 40 tons, its ex- 
tremities resting on supports 20 feet apart, and the height 
and thickness of web being 20 ins. and J in. respectively. 
How shall the rivets, which are | in. in diameter, be spaced 
between the web and the angles which are also ^ in. in thick- 
ness? Let the unit stresses taken be 7500 for shearing, and 
12,500 for side compression (Ibs./in.^). Referring to fig. 235 
we find that J = | W = 20 tons at each support and diminishes 
regularly to zero at the middle, where no riveting will there- 
fore be required. Each rivet, having a sectional area of J7r(|)2 
= 0.60 sq. inches, can bear a safe shear of 0.60x7500 = 4500 lbs. 
in single shear, and hence of 9000 lbs. in double shear, which is 
the present case. But the safe compressive resistance of the 
side of the rivet hole in either the web or the angle is only 
I in. Xj in. X 12500 = 5470 lbs., and thus determines the spacing 
of the rivets as follows : 

Near a support the riveting must sustain a shear equal 
to 40,000 lbs. on a horizontal length equal to the height of 
web, i.e., to 20 ins., and the safe compression for each rivet 
is 5470 lbs. Hence 4000 h- 5470, or 7.2, rivets will be needed 
for the 20-inch length. In other words, they must be spaced 
20-^7.2 = 2.7 in. apart, center to center, near the supports; 
5.4 in. apart at ^ the span from a support; none at all in the 
middle. By the Cambria handbook, this distance apart 
should never be less than 3 diameters of the rivet; and, in 
connecting plates in compression, should not exceed 16 times 
the thickness of the plate. 

As for the rivets connecting the angles and flange plates, 
being in two rows and opposite (in" pairs) the safe shear- 



FLEXURE, BUILT BEAM. 



295 



ing resistance of a pair (eacli in single shear) is 9,000 lbs., 
while the safe compressive resistance of the sides of the 
two rivet holes in the angle bars (the flange plate being 
much thicker) is =10,940 lbs. Hence the former figure 
(9,000) divided into 40,000, gives 4.44 as the number of 
pairs of rivets for 20 in. of length of the beam; i.e., the 
rivets in one row should be 20^-4.44 = 4.5 in. apart, centre 
to centre, near a support ; the interval to be increased in 
inverse ratio to the distance from the middle of span, 
(^bearing in mind the practical limitation just given). 

If the load is concentrated in the middle of the span, 
instead of uniformly distributed, e/is constant along each 
half-span, (see fig. 234) and the rivet spacing must accord- 
ingly be made the same at all localities of the beam. 



SPECIAL, PROBLEMS IN FLEXURE. 

258. Designing Cross Sections of Built Beams. — The last par- 
agraph dealt with the riveting of the various plates ; we 
now consider the design of the plates themselves. Take 
for instance a plate-girder, fig. 261 ; one vertical stem= 




Fig. 261. 



296 MECHANICS OF ENGINEERING, 

plate, four angle bars, (each of sectional area = A, re- 
maining after the holes are punched, with a gravity axis 
parallel to, and at a distance = a from its base), and two 
flange plates of width = h, and thickness = t. Let the 
whole depth of girder = 7^, and the diameter of a rivet 
hole =f. To safely resist the tensile and compressive 
forces induced in this section by M,^ inch-lbs. (itf^ being 
the greatest moment in the beam which is prismatic) we 
have from § 239, 

ifn. = — - (1) 

e 
E' for mild steel = 15,000 lbs. per sq. inch, e is = ^ ^ 
while i, the moment of inertia of the compound section, 
is obtained as follows, taking into account the fact that 
the rivet holes cut out part of the material. In dealing 
with the sections of the angles and flanges, we consider 
them concentrated at their centres of gravity (an approx-. 
imation, of course,) and treat their moments of inertia 

about N as single terms in the series fdF z^ 

(see § 85). The subtractive moments of inertia for the 
rivet holes in the web are similarly expressed ; let 6o = 
thickness of web. 

j It, for web = ^h, (h—2tf—2b,t' [^—t—a'Y 
„•, ■} I^ for four angles = 4 A ['l — t — aY 

( In for two flanges = 2(6— 2f) t ('^f 

the sum of which makes the ^ of the girder. Eq. (1) may 

now be written 

which is available for computing any one unknown quan- 
tity. The quantities concerned in /^ are so numerous and 
they are combined in so complex a manner that in any 
numerical example it is best to adjust the dimensions of 
the section to each other by successive assumptions and 



FLEXUEE BUILT BEAM. 297 

trials. (The hand-book of the Cambria Steel Co. gives tables 
of safe loads of beam box-girders and plate-girders for a large 
variety of sizes and distances between supports ; but attention- 
is called to the fact that the loads given in the tables are based 
on the assumption that the girder is supported laterally, and 
that otherwise a proper reduction in the allowable safe load 
must be made, as explained elsewhere in the hand-book. The 
value of 15,000 Ibs./sq. in. for R' has been used in computing 
these tables.) 

Example. — (Units, inch and pound) . A plate-girder with 
end supports, of span = 20 ft. = 240 inches, is to support 
a uniformly distributed load of 45 tons = 90,000 lbs. If f 
inch rivets are used, angle bars 3" X 3'' X 2'% vertical 
web I" in thickness, and plates 1 inch thick for flanges, how 
wide (6 = ?) must these flange-plates be ? taking h = 22 
inches = total height of girder. 

Solution. — From the table in § 250 we find that the max. 
31 for this case is ^ Wl, where W = the total distributed 
load (including the weight of the girder) and I = span- 
Hence the left hand member of eq. (2) reduces to 

Wl h 90000 X 240 X 22 

16 • R' - 16 X 15000 "■^^^^• 

That is, the total moment of inertia of the section must 
be = 1,980 biquad. inches, of which the web and angles 
supply a known amount, since &o = >^", t = l''> t'= )'i" , 
a' = 1^", A= 2.0 sq. in., a = 0.9', and h = 22", are 
known, while the remainder must be furnished by the 
flanges, thus determining their width b, the unknown 
quantity. 

The elective area. A, of an angle bar is found thus : 
The full sectional area for the size given, = 3 X ^ + 
2>^ X % = 2.75 sq. inches, from which deducting for two 
rivet holes we have 

A= 2.75—2 X ^ X >^_ 2.0 sq. in. 

The value a = 0.90" is found by cutting out the shape 



298 MECHANICS OF ENGINEEKING. 



X3 



of two angles from sheet iron, tlius : I 

and balancing it on a knife edge* (The 

gaps left by the rivet holes may be ignored, 

without great error, in finding or). Hence, fig!^ a 

substituting we have 

Ih for web =A. . 1^x20^— 2x>^ . ^ l8}(Y=282.d 
In for four angles =4x2x [9.10]2=662.5 
In for two flanges=2(6— |)xlx(10>^)2=220.4(&— 1.5) 
.-. 1980=282.3+662.5+(Z^1.5)220.4 
whence b = 4.6 + 1.5 = 6.1 inches 

the required total width of each of the 1 in. flange plates. 
This might be increased to 6.5 in. so as to equal the 
United width of the two angles and web. 

The rivet spacing can now be designed by § 257, and 
the assumed thickness of web, )4 in., tested for the max, 
total shear by § 256. The latter test results as follows ; 
The max. shear J^„ occurs near either support and = 
)4 ^=45,000 lbs. .-., calling 6'o the least allowable thickness 
of web in order to keep the shearing stress as low as 8, 000 
lbs. per sq. inch, 

6'o X 20" X 8000 = 45000 .-. 6'o=0.28 in. 

showing that- the assumed width of )4 in. is safe. 

This girder will need vertical stiffeners near the ends, 
as explained subsequently, and is understood to be sup- 
ported laterally, f Built beams of double web, or box- 
form, (see Fig. 260) do not need this lateral support, 

259. Set of Moving Loads. — When a locomotive passes over 
a number of parallel prismatic girders, each one of which 
experiences certain detached pressures corresponding to 
the dijfferent wheels, by selecting any definite position of 
the wheels on the span, we may easily compute the reac- 
tions of the supports, then form the shear diagram, and 
finally as in § 243 obtain the max. moment, Jf^s and the 

* The Cambria hand-book gives values of / and a for sections of angle- 
bars. 

t See § 314. 



FliEXUEE. MOVING LOADS. 



299 



max. shear J^, for this particular position of the wheels. 
But the values of Jf^ and J^ for some other position may 
be greater than those just found. We therefore inquire 
"which will be the greatest moment among the infinite 
number of {M^Js, (one for each possible position of the 
wheels on the span). It is evident from Fig. 236 from the 
nature of the moment diagram, that when the pressures or 
loads are detached, the 31^ for any position of the loads, 
which of course are in this case at fixed distances apart, 
must occur under one of the loads (i.e. under a wheel). 
We begin .*. by asking : What is the position of the set of 
moving loads when the moment under a given wheel is 
greater than will occur under that wheel in any other po- 
sition? For example, in Fig. 262, in what position of the 




Fig 263. 



loads Pi, P2> stc. on the span will the moment ilfa* i-6., 
under Pn, be a maximum as compared with its value under 
Pg in any other position on the span. Let P be the resultant 
of the loads which are now on the span, its variable distance 
from be = cc, and Unfixed distance from Pg = a'', while 
a, h, c, etc., are the fixed distances between the loads 
(wheels). For any values oi x , as the loading moves 
through the range of motion within which no wheel of the 
set under consideration goes off the span, and no new 

wheel comes on it, we have Pi =--^ P, and the mament 

under Pg 

=M^=RS-(x-a'y]—P^h-Pih-\-c) ' 

i.e. M2=j(l^—'^^a'x)—P,b—P,(b-\-c) 



(1) 



300 MECHANICS OF EXGINEBRING. 

In (1) we liave M2 as a function of x, all tlie other quan- 
tities in tlie right hand member remaining constant as the 
loading moves ; x may vary from x=^a-\-d to 
x=l—{c-\-h—a). For a max. M2, we put dMi-h-dx=0, i. e. 

m 

j{l-2x+a)=0 .'. x{ioT Max M.,)=}4l+}^(^' 

(For this, or any other value of x, (FM^-^daf is negative, 
hence a maximum is indicated). For a max. M2, then, B 
must be as as far {%Oj') on one side of the middle of the 
span as P2 is on the other ; i.e., as the loading moves, the 
moment under a given wheel becomes a max. when that 
wheel and the centre of gravity of all the loads {then on 
the span) are equi-distant from the middle of the span. 

In this way in any particular case we may find the- 
respective max. moments occurring under each of the 
wheels during the passage, and the. greatest of these is the 
3I„^ to be used in the equation ilt/,„ =^R'I-^e for safe loading:.* 

As to the shear J, for a given position of the wheels this 
will be the greatest at one or the other support, and 
equals the reaction at that support. When the load moves 
toward either support the shear at that end of the beam 
evidently increases so long as no wheel rolls completely 
over and beyond it. To find J" max., then, dealing with 
each support in turn, we compute the successive reactions 
at the support when the loading is successively so placed 
that consecutive Avheels, in turn, are on the point of roll- 
ing ofi^ the girder at that end ; the greatest of these is the 
max. shear, J^^. As the max. moment is apt to come under 
the heaviest load it may not be necessary to deal with 
more than one or two wheels in finding M,„. 

Example. — Given the following wheel pressures, 

^< . . 8' . . >B< . . 5' . . >C< . . 4 . . <D 
4 tons. 6 tons. 6 tons. 5 tons, 

on one rail which is continuous over a girder of 20 ft. span, 
under a locomotive. 



* Since this may be regarded as a case of " sudden application" of a load, it is 
customary to make R' much gmaller than for a dead load; from one-third to one-half 
smaller. 



FLEXURE. MOVIl!lG LOADS. 



301 



1. Required the position of the resultant of A, B, and (7' 

2. " " " " A, B, C, and I) ; 

3. " " " " B, G, and I). 
4.. In what position of the wheels on the span will the 

moment under ^ be a max. ? Ditto for wheel C? Required 
the value of these moments and which is M^^ ? 

5. Required the value of J^, (max. shear), its location and 
the position of loads. 

Results.— (1.) 7.8' to right of A. (2.) 10' to right of A. 
(3.) 4.4' to right of B. (4.) Max. M^ = 1,273,000 inch lbs. 
with all the wheels on ; Max. Jfc = 1,440,000 inch-lbs. with 
wheels B, C, and JD on. (5.) J^ = 13.6 tons at right sup- 
port with wheel I) close to this support. 

260. Single Eccentric Load. — In the following special cases 
of prismatic beams, peculiar in the distribution of the 

loads, or mode of support, or both, 
the main objects sought are the 
values of the max. moment M^] for 
use in the equation 

il4^:?y(see§239); 
e 

and of the max. shear J,^, from 
which to design the web riveting 
in the case of an I or box-girder. 
The modes of support will be such 
that the reactions are independent 
of the form and material of the 
beam (the weight of beam being 
neglected). As before, the flexure is to be slight, and the 
forces are all perpendicular to the beam. 

The present problem is that in fig. 263, the beam being 
prismatic, supported at the ends, with a single eccentric 
load, P. We shall first disregard the weight of the beam 
itseli Let the span = ?i4-?2- First considering the whole 
beam free we have the reactions Bi = PI, ^ I and B2 = 
PI, -f- Z. 

Making a section at m and having Om free, x being < I2, 
S (vert, compons,) == gives 




Fig. 263. 



302 MECHANICS OF ENGINEEELNG. 

i?2 — J=0, i.e., e7=i?2 ; 
while from H (moni.),„=0 we have 

P-^-R.,x= .-. Jf = B,x=?}}x 
e I 

These values of J and M hold good between and 0, J 
being constant, while 31 is proportional to x. Hence for 
C the shear diagram is a rectangle and the moment dia^ 
gram a triangle. By inspection the greatest M for C is 
for X =■ I2, and = FI1I2 -4- I. This is the max. M for the 
beam, since between G and B, M is proportional to the disr 
tance of the section from B. 

.'.M^=?}^a.Tid.^=I}i' ... (1) 

lei 

is the equation for safe loading. 

J = A\ ill any section along OB, and is opposite in sign 
to what it is on 0(7; i.e., practically, if a dove-tail joint 
existed anywhere on 0(7 the portion of the beam on the 
right of such section would slide downward relatively to 
the left hand portion ; but vice versa on GB. 

Evidently the max. shear «/„, = ^x 01* ^2> as I2 or Z^ is the 
greater segment. 

It is also evident that for a given span and given beam 
the safe load P', as computed from eq. (1) above, becomes 
very large as its point of application approaches a sup- 
port ; this would naturally be expected but not without 
limit, as the shear for sections between the load and the 
support is equal to the reaction at the near support and 
may thus soon reach a limiting value, when the safety of 
the web or the spacing of the rivets, if any, is considered. 

Secondly, considering the weight of the heam, or any 
uniformly distributed loading, weigliing w lbs. per unit of 
length of beam, in addition to P, Fig. 264, we have the 
reactions ' 

iJ,=^+|'; and B,=i^+^ 

Let 1-2 he >?i ; then for a portion Om of length a?<^ 
moments about m give 



FLEXURE. SPECIAL PROBLEMS. 



303 



^ — BoX + ivx.~ =0 

e " 2 

Le., on 00, M—B-iX — y^ lox^ . . . , (2) 

Evidently for x = (i.e. at 0) M = 0, while for x = Iz (i.e. 
at (7) we have, putting w = W -r- I 



Mr=B>,l,— y: 



wll- 



Z 2 ^^ J 



(3) 




(4) 



it remains to be seen whether a value of M may not exist 
in some section between- and G, (i.e., for a value of x 
<l2 in eq. (2)), still greater than Mq. Since (2) gives Ji" as 
a continuous function of x between and C, we put 
dM-r- dx = 0, and obtain, substituting the value of the con- 
stants B2 and w, 

( max, 

B^—ivx^O .'. Xa -< for M or 

( min. 

This must be for 31 max., since d^M -^ dx^ is negative 

when this value of x is sub- 
stituted. If the particular 
-—J value of X given by (4) is 
'P <l2, the corresponding vahie 
of 31 (call it iSiJ from eq. 
(2) will occur on 00 and will 
be greater than 3Iq (Dia- 
grams II. in fig. 264 show 
this case) ; but if x„ is> h, 
we are not concerned with 
the corresj)onding value of 
31, and the greatest 31 on 00 
would be 3Ic. 

For the short portion BG, 
which has moment and shear 
diagrams of its own not con- 
tinuous with those for 00, it 
"' may easily be shown that 
3£c is the greatest moment of 
P,gj_ge4. any section. Hence the 31 




304 



MECHANICS OF EXGIXEERING. 



max., or Ji|„, of tlie whole beam is either Mq or J^, 
according as x^ is > or < I2. This latter critPT.ion may be 
expressed thus, [with h — yi I denoted by l^, the distance 
of P from the middle of the span] : 



From (eq. 4) 
and since from (4) and (2) 



£ii_L.i/n>7 IS equiv 
?F^^'7<- alent to L 



tf)<OjJ 



if. 



-Pk 



W 



■PI, 

w 






(5) 



The equation for safe loading is 



and 



e JV li 



— =Jf„,when^is < ^ 
e W 



k 



. . . . (6) 

Seeeqs. (3) and (5) 
for M, and Jf„ 



If either P, W, \, or \ is the unknown quantity sought, the 
criterion of (6) cannot be applied, and we .•. use both equa- 
tions in (6) and then discriminate between the two results. 
The greatest shear is J^=Bi, in Fig. 264, where l^ is 

281. Two Equal Terminal Loads, Two Symmetrical Supports 

Fig. 265. [Same case as in Fig. 231, § 238]. Neglect 
weight of beam. The reaction at each support being =P, 
(from symmetry), we have for a free body Om with a; < Z, 

.Pl .0 



Px—^. 
e 



M=Px 



while where a? > Zi and < ?i+?2 



Px-P {x--\)—^=0 .: M=P\ 



(1> 



(2) 



That is, see (1), ilf varies directly with x between and C, 
while between G and D it is constant. Hence for safe 
loading 

i.e., — ^Pl , . (3) 






FLEXURE. SPECIAL PROBLEMS. 



305 




a( 


i 


11 


illlllllk.i 




MOMS. 




illlllllllllllllllll 


1 






SHEARS 











Tlie construction of the 

B moment diagram is evident 

^l^ ^1 ^^ p from equations (1) and (2). 

\ ! As for J", tlie shear, the 

same free bodies give, from 

I, (vert. forces)=0. 

On OG . J=^P ... (4) 
On CD . J=P—P==zerol5) 

(4) and (5) might also be ob- 
Pig 265 tained from (1) and (2) by- 

writing J=d M-T-dx, but the 
former method is to be preferred in most cases, since the 
latter requires M to be expressed as a function of x while 
the former is applicable for examining separate sections 
without making use of a variable. 

If the beam is an I-beam, the fact that J is zero any- 
where on G D would indicate that we may dispense with 
a web along G D io unite the two flanges ; but the lower 
flange being in compression and forming a " long column " 
would tend to buckle out of a straight line if not stayed by 
a web connection with the other, or some equivalent brac- 
ing. 

282. XTniform Load over Part of the Span. Two End Supports. 
Fig. 266. Let the load= W, extending from one support 
over a portion =c, of the span, (on the left, say,) so that 
W= IOC, w being the load per unit of length. Neglect 
wei ght o f beam. For a 'free body Dm of any length 
X <, B (i.e. < c), 2" moms^=0 gives 



pi 



wx- 



2 



-^icc=0 .'.M= 



(1) 



which holds good for any section on B. As for sections 
on B (7 it is more simple to deal with the free body m'G, 

of leiigth 

' x' < G B from which we have M^R^ x' . (2) 



MECHANICS OF ENGINE EEH^TG. 




Fig. 2G6. 



wMch shows the moment 
curve for B G tohe a. straight 
line DC, tangent at D to the 
parabola 0' D representing 
eq. (1.) (If there were a con- 
centrated load at B, CD 
would meet the tangent at 
D at an angle instead of co- 
inciding with it ; let the stu- 
dent show why, from the 
shear diagram). 

The shear for any value of 
ic on -S is : 



On 5 
while on B C 



. e/= Bo= constant 



(3) 



The shear diagram is constructed accordingly. 



To find the position of the max. ordinate of the para- 
bola, (and this from previous statements concerning the 
tangent at the point D must occur on B, as will be seen 
and will .'. be the M^ for the whole beam) we put e/=--0 in 
eq (3) whence 



X (for JC)= 



i?i_JF[?— |] ^_(? 



w 



tu 



(5) 



W 



and is less than c, as expected. [The value oi Bi^--j- (l — '^\ 

—[wc ~-T) (I — 2), (the whole beam free) has been substi- 
tuted]. This value of x substituted in eq. (1) gives 



is the equation for safe loading. 

The max. shear J^ is found at and is 
evidently >i?2j at C. 



Bx, which is 



FLEXUEE. SPECIAL PROBLEMS. 



30^ 



263. XTniforin Load Over Whole length With Two Symmetricj 
Supports. Fig. 267. — With the notation expressed in the fig- 
ure, the following results may be obtained, after having 
divided the length of the beam into three parts for sepa- 
rate treatment as necessitated by the external forces, which 
are the distributed load W, and 
and the two reactions, each = 
}^ W. The moment curve is 
made up of parts of three dis- 
tinct parabolas, each with its 
axis vertical. The central par- 
abola maj sink below the hori- 
zontal axis of reference if the 
supports are far enough apart, 
in which case (see Fig.) the elas- 
tic curve of the beam itself becomes concave upward be- 
tween the points E and F of " contrary flexure." At each 
of these points the moment must be zero, since the radius 
of curvature is co and M = EI ^ p (see § 231) at any sec- 
tion ; that is, at these points the moment curve crosses its 
horizontal axis. 

As to the location and amount of the max. moment M^, 
inspecting the diagram we see that it will be either at H, 
the middle, or at both of the supports B and C (which from 
symmetry have equal moments), i.e., (with I = total length,) 




Fig. 267. 



w 



Mr 



.[and.-.— J= 



( either ~ \ %li-l,^-] at ^ 



or 



Ell' at ^ and a 

2Z 



according to which is the greater in any given case ; i.e. 
according as I2 is > or < l^ y'g. 

The shear close on the left oi B = ivl^, while close to the 
right oiBit=}4 W — id^. (It will be noticed that in this 
case since the beam overhangs, beyond the support, the 
shear near the support is not equal to the reaction there, 
as it was in some preceding cases.) 



308 



MECHANICS OF ENGINEEEHsTG. 



Hence (/„= 



wli 



/2 ^-t^Zi P^^^^^^^g ^^ ^1 <^ 



264, Hydrostatic Pressure Against a. Vertical Plank. — From 
elementary hydrostatics we know that the pressure, per 
unit area, of quiescent water against the vertical side of a 
tank, varies directly with the depth, x, below the surface, 
and equals the weight of a prism of water whose altitude 
= X, and whose sectional area is unity. See Fig. 268. 




Fig. 26S. 

*Tt& plank is of rectangular cross section, its constant 
breadth, — b, being r~ to the paper, and receives no sup- 
port excepi at its two extremities, and B, being level 
with the water surface. The loading,' or pressure, per unit 
of length of the beam, is here variable and, by above defini- 
nition, is = w= yxb, where ;' = weight of a cubic unit 
(i.e. the heaviness, see § 7) of water, and x = Om =■ depth 
of any section m below the surface. The hydrostatic pres- 
sure on dx = ivdx. These pressures . for equal dx's, vary 
as the ordinates of a triangle ORiB. 

Consider Onti free. Besides the elastic forces of the ex- 
posed section m, the forces acting are the reaction Bq, and 
the triangle of pressure OEm. The total of the latter is 



W. 



0(? 



= I wdx = yb I xdx = 'fb-^ 



(1) 



and the sum of the moments of these pressures about m is 
equal to that of their resultant ( = their sum, since they 

are parallel) about m, and .% ==: jb -^ , ^, 

A o 



rLEXHRB. SPECIAL PEOBLEMS. 309 

[From (1) wh«n x==1,wq have for tlie total water pres- 

sure on the beam Wi = jb ^ and since one-third of this 

will be borne at we have i?o =^}i T^^^-~\ 

Now putting i'( moms, about the neutral axis of wi)=0, 
for Om free, we have 



Box—JK . ^—^=0 .-. 31= /eyb {Vx—:j(?) 

O 6 



(2) 



(which holds good from x = Oto x — I). From I (horiz. 
forces) = we have also the shear 

J=R,— W^=% yh {P—Sx') .... (3) 

as might also have been obtained by differentiating (2), 
since J = dM -^ dx. By putting e7 = (§ 240, corollary) 
we have for a max. M, x = I -i- V3, which is less than I 
and hence is applicable to the problem. Substitute this 
in eq. 2, and reduce, and we have 

Efl ,, . R'l 1 1 

-^=Ji^, i.e. — =g "^^•rbV' . (4) 

as the equation for safe loading. 

265. Example. — If the thickness of the plank is h, re- 
quired 7i = ?, if B' is taken = 1,000 lbs. per sq. in. for 
timber (§ 251), and I = 6 feet. For the inch-pound-second 
system of units, we must substitute B' = 1,000 ; ? = 72 
inches ; y = 0.036 lbs. per cubic inch (heaviness of water 
in this system of units); while I =h¥ -4- 12, (§ 247), and e 
— }i h. Hence from (4) we have 

1000 &A3 0.0366x723 ,„ ..^ . n 07 • 
^rs 7T— — n /- 1 ••• ^^=5.16 .'. h = 2.27 m. 

It will be noticed that since x for ilfm = I -^ Vs, and not 
^ I, ifm does not occur in the section opposite the resul- 
tant of the water pressure ; see Fig. 268. The shear curve 
is a parabola here ; eq. (3). 



310 



MECHANICS OF ENGINEERING. 




^ ^ ^ ^'^ ^T 

£r ^ 1 wTtr- 



Fig. 289. 



Fig. 269a. 



266. Flat Circular Plate, Homogeneous and of Uniform Thick- 
ness, Supported all Eound its Edge and Subjected to Uniform Fluid 
Pressure of w lbs. per sq. in. A strict treatment of this case 
being very complicated, an approximate method, due to Prof. 
C. Bach, will be presented."* Fig. 269 shows a top view of the 
circular plate, in a horizontal position and covering a circular 

I opening, its edge 

being supported 
C^) continuously on 
the edge of the 
opening (but not 
clamped to it). 
Let the radius of 
the plate be r and 
its thickness h. 
Under the plate 
is the atmosphere, 
while on its upper surface, acting uniformly over the whole of 
that surface, is a fluid pressure whose excess over that of the at- 
mosphere is w Ibs./sq. in. The particles near the upper surface 
are under compressive stress, which is obviously greater near the 
center of the circle ; those near the lower surface are in tension. 
Let now the half-plate, CODE, (cutting along the diameter 
CD) be considered as a "free body" in Fig. 269a; the tensile 
and compressive stresses in the section COD being assumed to 
form a stress-couple, as in previous case of flexure, the unit- 
stress varying as the distance from the middle of the thick- 
ness, with the stress in the outermost fiber denoted by p. 
Then the moment of this couple will be written pi — e, as 
before, where e = ^h and I =2rh^ -i- 12. On the upper surface 
of the free body we find a total pressure of | W7rr^ lbs., covering 
a semicircular surface ; so that (p. 22) the distance of the re- 
sultant from is 4 r -^ Stt. The upward reaction from the 
supporting edge is also | wirr'^ lbs., but its resultant acts 2r/7r 
in. from (center of gravity of a semicircular "wire," p. 20). 
Taking moments, then, about we have 
wrrr^ r2 r 4 rl _ prh^ 
~2~ [V~3^J^ ""3" 

* Elasticitaet und Festigkeit, by C. Bach ; Berlin, 1898. 



or, tv = -,« 



(0) 



FLEXUEE. PLATES. 



311 



Notwithstanding the imperfections of this analysis, the 
experimental work of Prof. Bach shows that a modification of 
eq. (0), viz. : — - 

(1) 



5 K" „, 
r 



may be used with safety for the design of a plate under these 
circumstances ; R% a safe unit working stress for the material, 
having been substituted for p. 

For example, let the plate (e.g., cylinder-head of a loco- 
motive) be of mild steel with h = 1 in. and r = 8 in. Putting 
R' = 16,000 lbs. /sq. in., we have from eq. (1) a safe w 
= 1 (16000) X (1 H- 8)' =. 208 lbs. per sq. in. 

[N.B. If the plate is clamped all round the edge, we may 
write f instead of the |. (Bach.) ] 

266a. Homogeneous Circular Plate of Uniform Thickness, h, 
Supported all Round the Edge, with Concentrated Load (P lbs.) in 
Center. By the same method as before we may here derive 
P = 1 Trh^p ; but from his experiments in this case Prof. Bach 
concludes that the formula for safe design should be written 



P 



lirh'R'. 
o 



(2) 



It is seen from eq. (2) that the value of P is independent of 
the radius of the plate; depending only on the material and 
the thickness, h. 

266b. Homogeneous Elliptical and Rectangular Plates of Con- 
stant Thickness, h , Supported all Round the Periphery. According 
to Prof. Bach's approximate analysis, as supplemented by his 
experimental researches, we may use the following formulae for 




Fig. 269&. 

safe design of elliptical and rectangular plates, supported (not 
clamped) around the whole periphery. See Fig. 2696 for 
notation of dimensions ; h being the uniform thickness in each 



312 MECHANICS OF ENGINEERING. 

case, and a being > 6. R' = max. safe unit stress for the 
material. 

For the elliptical plate under unifornily distributed pressure 

(over whole area) of w lbs. / sq. in., denoting the ratio 6 -r- a by 
m, we have 

w = ^ {1 +m')J^,.R'; .... (3) 

and under a central concentrated load of P lbs., 

3^ 3 + 2m^ + 3w^ , ... 

(N.B. If the edge is clamped down all round we may use 
values of w and P about 50 per cent, greater than the above.) 
With rectangular plates under a uniformly distributed pressure, 
denoting the ratio 6 -;- a by m, we have 

w = l{l +m')f,.R' ..... (5) 

and for a concentrated central load P, with n denoting the ratio 

P = i- (1 + n') . h'R' .'.... (6) 
on 

In the particular case of the square plate, the side of the 
square being a, eqs. (5) and (6) reduce to 

7 2 

(uniform pressure) w = S.6 — .R'; (7) 

(central load) P =o h'R' (8) 

o 

266c. Homogeneous Flat Circular Plate, of TTniforin Thickness, used as Piston 
of an Engine. In such a case we have fluid pressures ou both sides of the plate 
or disc, neither of which is necessarily one atmosphere ; while at the center 
we have acting the concentrated pull or thrust, P lbs., of the piston rod. (Fric- 
tional forces around the edge may be disregarded.) If w denote the greatest 
difference between the (uniform) fluid pressures (per sq. in.) on the two sides, 
we may write (according to Grashof's analysis, as quoted by Unwin), for safe 
design in this case : — 

^-If.-^' . (9) 

(E', h, and r, have the same meaning as before.) 



FLEXUEE SPECIAL PROBLEMS. 313 



267. ResilienceofBeamWithEndSupports.— Fig. 270. If a 

9g mass whose weight is G {G large com- 

!^ I pared with that of beam) be allowed to 

^^ J __^^'- l_p fall freely through a height = h upon 

g J I - 1^ ^j^^ centre of a beam supported at its 

a-.y TPm extremities, the pressure P felt by the 

Fig. 270. beam increases from zero at the first 

instant of contact up to a maximum P^, as already stated 

in §233a, in which the equation was derived, d^ being 

small compared with h, 

The elastic limit is supposed not passed. In order that 
the maximum normal stress in any outer fibre shall at most 
be=^', a safe value, (see table §251) we must put 

=-7^ [according to eq. (2) §241,] i.e. in equation (a) 

above, substitute F^= [4 Ii'l]-^Ie, which gives 

having put I==FJi? {h being the radius of gyration §85) 
and Fl= V the volume of the (prismatic) beam. From 
equation (&) we have the energy, Gh, (in ft. -lbs., or inch- 
lbs.) of the vertical blow at the middle which the beam of 
Pig. 270 will safely bear, and any one unknown quantity 
can be computed from it, (but the mass of G shaiili not 
be small compared with that of the beam.) 

The energy of this safe impact, for two beams of the 
same material and similar cross-sections (similarly placed), 
is seen to be proportional to fheii volumes; while if further- 
more their cross-sections are the same and similarly 
placed, the safe Gh is proportional to their lengths. (These 
same relations hold good, approximately, beyond the elas' 
tic limit.) 

It will be noticed that the last statement is just the re- 



314 



MECHANICS OF ENGINEEEING. 



verse of wliat was found in §245 for static loads, (the 
pressure at tlae centre of the beam being then equal to 
the weight of the safe load) ; for there the longer the beam 
(and .°. the span) the less the safe load, in inverse ratio. 
As appropriate in this connection, a quotation will be 
given from p. 186 of " The Strength ' of Materials an^ 
Structures," by Sir John Anderson, London, 1884, viz.: 

"It appears from the published experiments and state- 
ments of the Railway Commissioners, that a beam 12 feet 
long will only support )4 of the load that a beam 6 feet 
long of the same breadth and depth will support, but that 
it will bear double the weight suddenly applied, as in the 
case of a weight falling upon it," (from the same height, 
should be added) ; " or if the same weights are used, the 
longer beam will not break by the weight falling upon it 
unless it falls through twice the distance required to frac- 
ture the shorter beam." 

268. Combined Flexure and Torsion. Crank Shafts. Fig. 271. 
Let OiB be the crank, and NOi the portion projecting 

beyond the nearest bearing 
N. P is the pressure of the 
connecting-rod against the 
crank-pin at a definite in- 
stant, the rotary motion be- 
ing uniform. Let a= the 
perpendicular dropped from 
the axis OOi of the shaft 
upon P, and 1= the distance 
of P, along the axis Oj from 
the cross-section iV^TwiV^' of the 
Let NW be a diameter of this 
In considering the portion 
NOiB free, and thus exposing the circular section iVmZV^, 
we may assume that the stresses to be put in on the ele- 
ments of this surface -are the tensions (above NN') and 
the compressions (below NN') and shears "| to NN', due 
to the bending action of P ; and the shearing stress tan= 




shaft, close to the bearing, 
section, and parallel to a. 



FLEXURE. SPECIAL PROBLEMS. 



315 



gent to tlie circles which have as a common centre, and 
pass through the respective dF's or elementary areas, 
these latter stresses being due to the twisting action of P. 
In the former set of elastic forces let p = the tensile 
stress per unit of area in the small parallelopipedical ele- 
ment m of the helix which is furthest from NN' (the neu- 
tral axis) and /= the m.oment of inertia of the circle about 
NN'-, then taking moments about NN' for the free body, 
(disregarding the motion) we have as in cases of flexure 
(see §239) 

pT T.7 . .. . .. Plr 



.= PI 



; i.e., p- 



(«) 



[None of the shears has a moment about iVW.] Next 
taking moments about OOi, (the flexure elastic forces, both 
normal and shearing, having no moments about OOi) we 
have as in torsion (§216) 



■^^^-i^= Pa ; i.e., p^= 



Par 
~I7 



Q>) 



in which p^ is the shearing stress per unit of area, in the 
torsional elastic forces, on any outermost dF, as at m ; 
and 7p the polar moment of inertia of the circle about its 
centre 0. 

Next consider free, in Fig. 272, a small parallelopiped 
taken from the helix at m (of Fig. 271.) The stresses [see 
§209] acting on the four faces p" to the paper in Fig. 272 
are there represented, the dimensions (infinitesimal) being 
n " to NN, & II to 00,, and d -] io the paper in Fig. 272. 



/pnd 



pM' 



^Pgticl 

— "pM 



p^na 



pnd 



- H 

-J) M 




P,M^, 



-p/id 

Fig. 272. 



qcd- 



./"" 



Fiff. 273. 



pnd 



Sl(i MECHANICS OF EXGINEEKllJTG. 

By altering the ratio of 6 to % we may make the angle 6 
what we please. It is now proposed to consider free the 
triangular prism, GUT, to find the intensity of normal 
stress q, per unit of area, on the diagonal plane GH, (oi 
length — c,) which is a bounding face of that triangulai' 
prism. See Fig, 273. By writing 2" (compons. in direc-^ 
tion of normal to GII)=0, we shall have, transposing, 

qcd=pnd sin d+pjbd sin d+pjid cos d ; and solving for q 
q=jp -- sin d+p, -sin<?+-. cos 6j ; . (1; 

but n : c= sin d and b : c= cos 6 .*. 

q=p sin^^+Ps2 sin d cos d . . (2) 

This may be written (see eqs. 63 and 60, O. W. J. Trigo- 
nometry) 

q^}4pO—Gos2d)+p,sm2d . . (3) 

As the diagonal plane GH is taken in different positions 
(i.e., as d varies), this tensile stress q (lbs. per sq. in. for 
instance) also varies, being a function of d, and its max, 
value may be >^. To find 6 for q max. we put 

tJ, =j9sin2^4-2^sCos2(?, . . (4) 

= 0, and obtain: tan[2(^ for q max)]=> — ~ . . • (5) 

Call this value of 6, 6'. Since tan 2d' is negative, 2d' lies 
either in the second or fourth quadrant, and hence 

sin2^^=± , ^" and cos 2^'= rp— 7=^= (6) 

[See equations 28 and 29 Trigonometry, O. W. J.] The 



FliEXUBE. CEANK SHAFT. 317 

apper signs refer to tlie second quadrant, the lower to tlie 
fourth. If we now differentiate (4), obtaining 

^=2i)cos2^-4p,sin2^ . . . (7) 

W8 laote that if the sine and cosine of the [2^'] of the 2nd 
quadrant [upper signs in (6)] are substituted in (7) the re- 
sult is negative, indicating a maximum ; that is, g is a max- 
imum for 6= the d' of eq. (6) when the U2>per signs are taken 
(2nd quadrant). To iind q max., then, put 6' for 6 in (3) 
substituting from (6) (upper signs). We thus find * 

g-max =;^[p+Vy+4^] . . (8) 

A similar process, taking components parallel to GH, 
Fig. 273, will yield q^ max., i.e., the max. shear per unit of 
area, ^hich for a given p and p^ exists on the diagonal 
plane GH in any of its possible positions, as d varies. 
This max. shearing stress is 

g^max =^yp2_|_4^^2 ^ ^ (9j 

In the element diametrically opposite to m in Fig. 211, p 
is compression instead of tension ; q maximum will also 
be compression but is numerically the same as the q max. 
of eq. 8. 

269. Example.— In Fig. 271 suppose P=2 tons = 4,000 
lbs., a=Q in., 1=5 in., and that the shaft is of wrought 
iron. Required its radius that the max. tension or com- 
pression may not exceed i^' = 12,000 lbs. per sq. in.; nor the 
max. shear exceed /S" = 7,000 lbs. per sq. in. That is, we 
put 5'=12,000 in eq. (8) and solve for r : also ^,,=7,000 in 
(9) and solve for r. The greater value of r should be 
taken. From equations (a) and (5) we have (see §§ 219 and 
247 for /p and i) 

* According to the conceptions of § 405&, safe design would require 
that we put the max. '^ strain" in this case equal to a safe value, as 
determined by simple tensile or compressive tests. Here the max. 
strain (tensile) is £=[|p + |-\/p^ + 4ps^]-^-E'- (Grashof's method.) 



318 MECHANICS OF EKGINEEEIITG. 

\P= r ^^cl p^= -. 

•which in (8) and (9) give 

mas. g=>^~ [4?+|/(4^)=+4(2«)"] ^ . . (8^ 

p 

and max. g's=^— 3A/(4!)H4(2a)2 . . • (9«) 

With max. g= 12,000, and the values of P, a, and Z, already 
given, (units, inch and pound) we have from (8a), r^=2.72 
cubic inches .*. r=1.39 inches. 

Next, with max. 5's=7,000; P, a, and I as before; from 
(9a), r^=2.84 cubic inches .*. r=1.41 inches. 

The latter value of r, 1.41 inches, should be adopted. It 
is here supposed that the crank-pin is in such a position 
(when P= 4,000 lbs., and a=Q in.) that q max. (and q^ 
max.) are greater than for any other position ; a number 
of trials may be necessary to decide this, since P and a are 
different with each new position of the connecting rod. If 
the shaft and its connections are exposed to shocks, H and 
S' should be taken much smaller. 

270. Another Example of combined torsion and flexure is 

shown in Fig. 274. The 
'" ' /^^ "^^ "^i< ^B 'wo^k of the working force 

Pi( vertical cog-pressure) is 
B expended in overcoming the 
resistance (another vertical 
cog-pressure) Q^. 
^la- 27'4. That is, the rigid body 

consisting of the two wheels and shaft is employed to 
transmit power, at a uniform angular velocity, and since 
it is symmetrical about its axis of rotation the forces act- 
ing on it, considered free, form a balanced system. (See 
§ 114). Hence given Pi and the various geometrical quan- 




TLEXUEE. CEAXK SHAFT. 319 

titles «!, 5i, etc., we may obtain Q^, and the reactions Pq and 
Pr, in terms of Pj. The greatest moment of flexure in the 
shaft will be either FJi, at G; or PJ3, at B. The portion 
CD is under torsion, of a moment of torsion =Piai= Qih^. 
Hence we proceed as in the example of § 269, simply put- 
ting Poll (or Pb4, whichever is the greater) in place of Fl, 
and PiCTi in place of Pa. We have here neglected the 
weight of the shaft and wheels. If Qi were an upuard ver- 
tical force and hence on the same side of the sh:it as Pj, 
the reactions Pq and Pg would be less than before, and on© 
or both of them might be reversed in directioji. 

270a. Web of I-Beam. Maximum Stresses on an Oblique 
Plane. — The analysis of pp. 315, 316, etc., also covers the 
case of an element of the web of a horizontal I-beam under 
stress, when this element is taken near the point of junction 
with the flange. Supposing that the thickness of web has 
already been designed such that the shearing stress on the 
vertical (and therefore also on the horizontal) edges of such 
an element is at rate of 8000 lbs. per sq. inch ; and that the 
horizontal tension at each end of this element (since it is 
not far from the outer fibre of the whole section) is at rate 
of 10,000 lbs. per sq. in.; we note that Fig, 272 gives us a. 
side view of this element, with p^ = 8000, and p = 14,000, 
lbs. per sq. inch. GTis the lower edge of the upper flange, 
corresponding (in an end view) to the point n in Fig. 258 on 
p. 290. (We here suppose the upper flange to be in tension ; 
of course, an illustration taken from the compression side 
would do as well.) 

Substitution in equations (8) and (9) of p. 317 results in 
giving as maximum stresses on internal oblique planes : 

q max. = 17,630 lbs. per sq. in. tension; 
and g^ max. =10,630 " " " " shearing. 

These two values are seen to be considerably in excess of 
the respective safe values for shearing and tensile stresses in 
the case of structural steel, and the necessity is therefore em- 
phasized of adopting values for shearing stress in webs some- 
what lower than the 8000 lbs./in.2 used above ; to avoid the 
occurrence of excessive stress on internal oblique planes. See 
p. 291. 



320 MECHANICS OF ENGINEEHIiN^a. 



CHAPTER IV. 



FLEXURE, CONTINUED. 



CONTINUOUS GIRDERS. 



271. Definition. — A continuous girder, for present pur« 
poses, may be defined to be a loaded straight beam sup- 
ported in more than two points, in which case we can no 
longer, as heretofore, determine the reactions at the sup- 
ports from simple Statics alone, but must have recourse 
to the equations of the several elastic curves formed by its 
neutral line, which equations involve directly or indirect- 
ly the reactions sought ; the latter may then be found as 
if they were constants of integration. Practically this 
amounts to saying that the reactions depend on the man- 
ner in which the beam bends ; whereas in previous cases, 
with only two supports, the reactions were independent of 
the forms of the elastic curves (the flexure being slight, 
however). 

As an Illustration, if the straight beam of Fig. 275 is placed 
on three supports 0, B, and (7, at the same level, the 
reactions of these supports seem at first sight indeterm- 
inate ; for on considering the p -i ^ 

whole beam free, we have three \'^_~^~ 1. '^ j;* $ 

unknown quantities and only bZT""^ Z^° ^ — -^ 

two equations, viz : S (vert. fig. 275. 

compons.) = and S (moms, about some point) = 0. If 
now be gradually lowered, it receives less and less pres- 



FLEXUBE. CONTIJJUOUS GIKDEBS. 321 

sure, until it finally readies a position where the beam 
barely touches it ; and then O's reaction is zero, and B and 
C support the beam as if were not there. As to how 
low must sink to obtain this position, depends on the 
stiffness and load of the beam. Again, if be raised 
above the level of B and C it receives greater and greater 
pressuTt., until the beam fails to touch one of the other 
supports. Still another consideration is that if the beam 
were tapering in form, being stiffest at 0, and pointed at 
B and (7, the three reactions would be different from their 
values for a prismatic beam. It is therefore evident that 
for more than two supports the values of the reactions de- 
pend on the relative heights of the supports and upon the 
form and elasticity of the beam, as well as upon the load. 
The circumstance that the beam is made continuous over 
the support 0, instead of being cut apart at into two 
independent beams, each covering its own span and hav- 
ing its own two supports, shows the significance of the 
term " continuous girder." 

All the cases here considered will be comparatively 
simple, from the symmetry of their conditions. The 
beams will all be prismatic, and all external forces (i.e. 
loads and reactions) perpendicular to the beam and in the 
same plane. All supports at the same level, 

272. Two Equal Spans; Two Concentrated Loads, One in the Mid- 
^e of Each Span. Prismatic Beam. — Fig. 275. Let each half- 
Bpan = i^ /i. Neglect the weight of the beam. Required 
the reactions of the three supports. Call them P^, Pq and 



p 



\.. From symmetry P^ = Pc, and the tangent to the 
elastic curve at is horizontal ; and since the supports 
are on a level the deflection of C (and B) below O's tangent 
is zero. The separate elastic curves OD and DC have a 
common slope and a common ordinate at D. 

For the equation of OD, make a section n anywhere be- 
tween and Z>, considering n(7 a free body. Fig. 276 (a) 



322 



MECHANICS OF ENGINEERIXG. 




Y 



—X ^>| 



(6) 



Fig. 276. 



•with origin and axes as there indicated. * From H (moms 
about neutral axis oi n) = we have (see § 281) 



Ei'^^=p{y2i—x)—Pc{i—x) 



eA =F{y2ix—% 



dx 



(1 
(2) 



The constant = 0, for at both x, and dy -^ dx, = 0. 
Taking the x-anti-derivative of (2) we have 

^/2/=P(^_^')-Pe[^-|'] . . (3) 

Here again the constant is zero since at 0,x and y both =0. 
(3) is the equation of OD, and allows no value of cc <0 
or>^. It contains the unknown force P^. 

For the equation of BC, let the variable section n be made 
anywhere between D and C, and we have (Fig. 276 ih\ j x 
may now range between J^Z and T) 



^^^^— ^^(^-) 



^jdy_ 

dx 



Ix-t^+C 



(4) 

(5)' 



To determine C\ put x = }4l both in (5)' and (2), and 
equate the results (for the two curves have a common 
tangent line at D) whence C" = ^ PV 

Elp.^yiP¥—Pjl 



0[?\ 

2~j 



(5) 



* These are such that XOY is our "first quadrant"; in which, for points 
in a part of a curve convex toward the axis of X, d^yldx^ is essentially 
positive; and vice versa. It will be seen that both eqs. (1) and (4) are 
on this basis. They must be on the same basis; otherwise, later com- 
parisons of equations would result in error. 



FLBXUEE. CONTINUOUS GIRDEES. 323 

Hence Ely ^ % PT?x-PA^^—^'\j^O" . . (6)' 

At D tlie curves have the same y, hence put a? = i- in the 
right hand member both of (3) and (6)', equating results, 
and we derive C"= — ^ Pf 

EIy = y,PVa>~P^\^__p^^XPf . . (6) 

which is the equation of DC, but contains the unknown 
reaction P^. To determine P^ we employ the fact that O's 
tangent passes through G, (supports on same level) and 
hence when a? = Hn (6), y is known to be zero. Making 
these substitutions in (6) we have 

Q=y,pf-y,p^f-±pi^ ... P=^P 

From symmetry P^ also = —P, while Pq must = ~P, 
since P^ + P^ + P<7 = 2 P (whole beam free). [Note. — 
If the supports were not on a level, but if, (for instance) 
the middle support were a small distance = Ag below 
the level line joining the others, we should put x = I and 
y = — Iiq in eq. (6), and thus obtain P^ = Pc= -^^ P + 

SET—, which depends on the material and form of the 

prismatic beam and upon the length of one span, (whereas 
with supports all on a level, P^ — P^ = -| P is independent 
of the material and form of the beam so long as it is ho- 
mogeneous and prismatic.) If Pq, which would then = 
?| P — 6 EI {Jiq-^F'), is found to be negative, it shows that 
requires a support from above, instead of below, to 
cause it to occupy a position 7^o below the other supports, 
i.e. the beam must be " latched down " at 0.] 

The moment diagram, of this case can now be easily con- 
structed ; Fig. 277, For any free body nC, n lying in BG, 
we have 



324 



MECHANICS OF ENGINEEEING. 



i.e., varies directly as cc, un- 
c til X passes D wlien, for any 
point on DO, 



I wliicli is =0, (point of in- 
flection of, elastic curve) 
I for .T=yii? (note that x is 
Fig. 277. measured from C in this 

figure) and at 0, where x= I, becomes — ^^Pl 

•'• K=—lPl; M^^O; 3I^=LPl; andif„=0 

Hence, since if max. =^Ply the equation for safe loading 

is 

B'l 6 




-PI 



(7) 



The shear at (7 and anywhere on CD=~Pf while on DO 
it =^^P in the opposite direction 

.•.j;„=;ip . " . . . (8) 

The moment and shear diagrams are easily constructed, 
as shown in Fig. 277, the former being svmmetrical about 
a vertical line through 0, the latter about the point 0" 
Both are bounded by right lines. 

273. Two Equal Spans. IJniformly Distributed Load Over 
„, , Whole Length. Prismatic Beam. 

^ y\ ^_:^ ^ c —Fig. 278. Supports B, 0, 

bQIM I I 1 \ [\\ 111 C, on a level. Total load 

V~- — o| 1 ^ -- ^ ^---1 = 2 W= ^icl and may include 

I |po j ^'^k^'^ j that of the beam 



P. 



^ , , "w IS con- 

I I li I I 11 stant. Asbefore, from sym- 
metry P^=P^, the unknown 
i Pc| reactions at the extremi- 

VMt. 278. ties. 



FLBXUKE. CONTINUOUS GIRDERS. 



125 



Let On=x ; then with wC free, 2" moms, about n= gives 



rdy 



IV 



EI'^^- lFx-lx'+^]-F,[lx- ^]+[Coiist=0] (2) 

[Const. =0 for at both dy-i-dx the slope, and x, are =0] 
... EIy= '^[}4Fc^-j4M+Vi2^]-P.[}4lx^-y6^]+{G=0) (3) 

[Const. =0 for at both x and y are =0]. Equations (1), 
(2), and (3) admit of any value of x from to I, i.e., hold 
good for any point of the elastic curve OC, the loading on 
■which follows a continuous law (viz. : w= constant). But 
when x=l, i.e., at G, y is known to be equal to zero, since 
0, B and G are on the axis of X, (tangent at 0). "With 
these values of x and y in eq. (3) we have 

0= |L . t-y^pj? ... p,=y8wi=yQW 

.-. PB=^^and Po=27r— 2Pe=? W 



The Moment and Shear Diagrams can now be formed since 

;j jovv all tli6 external forces are 

Lw^ known. In Fig. 279 meas- 
ure X from G. Then in any 
section n the moment of the 
"stress-couple " is 




M^yQWoo- 



wxr 



. (1) 



j which holds good for any 
value of x on GO, i.e., from 
07=0 up to x=l. By inspec- 
PiG S79. tion it is seen that for 07=0, 

M=0 ; and also for x=yi, M=0, at the in/lection point' G, 
beyond which, toward 0, the upper fibres are in tension 



326 MECHANICS or engixeert^tg. 

the lower in compression, whereas between C and G the;^ 
are vice versa. As to the greatest moment to be found on 
CG, put dM-i-dx—0 and solve for x. This gives 

^ W—wx=0 .'. [X for if max.]=^Z . (2) 

i^rhich in eq. (1) gives 

Jfu(at JV, seefigure)=+^jr? . . (2) 

But this is numerically less than Mo{=—}i Wl) hence the 
stress in the outer fibre at being 

T/ Wle /Q\ 

Po=%—j-, ... (3) 



the equation for safe loading is 

B'l _., 



Wl . . . . (4) 



the same as if the beam were cut through at 0, each half, 
of length I, retaining the same load as before [see § 242 eq. 
(2)]. Hence making the girder continuous over the mid- 
dle support does not make it any stronger under a uni- 
formly distributed load ; but it does make it considerably 
stiffer. 

As for the shear, J, we obtain it for any section by tak- 
ing the x-derivative of M in eq. (1), or by putting ^(ver- 
tical forces) =0 for the free body nG, and thus have for 
any section on GO 

J=z/qW—wx ... (5) 

j/is zero for x='^l (where M reaches its calculus maxi- 
mum M^ ; see above) and for x=l it = — Yq fF" which is nu 
merically greater than yi W, its value at G. Hence 

Jm=y8w . . . ". (6) 



FLEXURE. CONTINUOUS GIRDERS. 327 

The moment curve is a parabola (a separate one for each 
span), the shear curve a straight line, inclined to the hor- 
izontal, for each span. 

Problem. — How would the reactions in Fig. 278 be 
changed if the support were lowered a (small) distance 
Iiq below the level of the other two ? 

274. Prismatic Beam Fixed Horizontally at Both Ends (at 
Same Level). Single Load at Middle. — Fig. 280. [As usual 

j /^~x p-| the beam is understood to 

1^^ — ■ V^py ■ — — - ' I be homogeneous so that E 

E ^P~ I '-' ^ is the same at all sections]. 

IJ I * j The building in, or fixing, 

i lyij ji of the two ends is supposed 

objr^-Jt:;;™--— 1^ — --—^^^ — (c* to be of such a nature as to 
Yi ' Br — —- — Vi-T--^ cause no 'horizontal con- 

FiG. 280. straint ; i.e., the beam does 

not act as a cord or chain, in any manner, and hence the 
sum of the horizontal components of the stresses in any 
section is zero, as in all preceding cases of flexure. In 
other words the neutral axis still contains the centre of 
gravity of the section and the tensions and compressions 
are equivalent to a couple (the stress-couple) whose mo- 
ment is the " moment of flexure." 

If the beam is conceived cut through close to both wall 
faces, and this portion of length=Z, considered free, the 
forces holding it in equilibrium consist of the downward 
force P (the load) ; two upward shears J^ and J^ (one at 
each section) ; and two " stress-couples " one in each sec- 
tion, whose moments are 31^ andJ/g. From symmetry we 
know that J, — J„ and that M^=M,. From I Y=Q for the 
free body just mentioned, (but not shown in the figure), 
and from symmetry, we have «/„= % P and J^-= % P ', but 
to determine M^ and M„ the form of the elastic curves 
B and B G must be taken into account as follows : 
Equation of OB, Fig. 280. I [mom. about neutral axis 
of any section n on 5] = (for the free body nC which 



528 



MBCHAXics or EXGi:^rEEEi:srG. 



lias a section exposed at each end, n being tlie variable 
section) will give 



BI^y-= P(y2 l-x) + M,- 



-y2P{i~x) 



(1) 



J^Note. In forming this moment equation, notice that 
M^ is the sum of the moments of the tensions and com- 
pressions at G about the neutral axis at n, just as much as 
about the neutral axis of 0', for those tensions and com- 
pressions are equivalent to a couple, and hence the sum of 
their moments is the same taken about any axis whatever 
"I to the plane of the couple (§32).] 

Taking the a:-anti-derivative of each member of (1), 



EI^=P(% I x—% a^)-f- if, x—y PQ x—% x") 
ax 



(2) 



(The constant is not expressed, as it is zero). Now from 
symmetry we know that the tangent-line to the curve B 
s>i B is horizontal, *.e., for x^y^l, dy-^dx^Q, and these 
values in eq. (2) give us 

0=yi Pf+ y^I^l—f^PV; whence M,=M,=}i PI , (3) 

Safe Loading. Fig. 281. Having now all the forces which 
act as external forces in straining the beam 00, we are 
ready to draw the moment diagram and find M^^. For con- 
venience measure x from 0. For the free body nO, we 
have [see eq. (3)] 

y2Px-M, + P~=0.'.M=}iPl-}4Px ... (4) 

p Eq. (4) holds good for any 
J section on OB. By put- 
f7^ ting x=0 we have M=M^= 
y% PI; \&j oEEO'=M, to 
scale (so many inch-pounds 
moment to the inch of pa- 
per). At B, for x^y I, 
M^= — y^ PI ; hence lay 
offB'I)=ys PI on theop- 
FiG. 281. posite side of the axis O'O' 



c 




FLEXUEE. CONTIGUOUS GIEDERS. 



329 



from HG', and join DH. DK, symmetrical witli i>^ about 
B'D, completes the moment curves, viz.: two riglit lines. 
The max. iHf is evidently =yi Fl and the equation of safe 
loading 



?Li=upi 



(5) 



Hence the beam is twice as strong as if simply supported 
at the ends, under this load ; it may also be proved to be 
four times as stiff. 

The points of inflection of the elastic curve are in the 
iniddles of the half-spans, while the max. shear is 



J.n = y2P 



(8) 



275. Prismatic Beam Fixed Horizontally at Both Ends [at Same 
Level]. Uniformly Distributed Load Over the Whole Length. 

Pig. 282. As in the preceding problem, we know from 
symmetry that e/o=^c=/^ ^=/^ *^^> ^^^ tl^s-^ Mq=M^, and 
determine the latter quantities by the equation of the 
curve OG, there being but one curve in the present in- 
stance, instead of two, as there is no change in the law of 
loading between and C. "With nO free, I (mom^)=0 
.gives 






ax 2 o 



X 

9 



(1) 



(2) 



^N^wl 



& 



i 



C \n 

J 



opr 



J i I 11 I I H i L 




Fig. 282. 



MBgHAITlCS OF ENGIXEEKIlira. 



The tangent line at being horizontal we have for x=0, 



dx 



0, .'. C=0. But since the tangent line at C is also hori- 
zontal, we may for x=l put dy-^dx=0, and obtain 



O^—i^Wl'+lIol+yewV; whence Mo=^Wl 



(3) 



a.s the moment of the stress-couple close to the wall at 
and at 0. 

Hence, Fig. 283, the equation of the moment curve (a 
single continuous curve in this case) is found by putting 
2' (moma)=0 for the free body nO, of length x, thus 
obtaining 



w 



y^=wl 



lUi I ] j lull 




Fig. 283. 



lL+i4Wx-Mo 



iva^ 



=0 



I.e. 



M=lWl+ ^-}4Wx 



, .(4) 



an equation of the second degree, indicating a conic. At 0, 
M=Mq of course,= 4- ^^/ ati?by putting a; = i^ Z in (4), we 
have M^— — }^ Wl, which is less than Jig, although M^ is the 
calculus max. (negative) for 31, as may be shown by writ- 
iijg the expression for the shear {J=% W — wx) equal to 
zero, etc. 



FiiEXUKE. coxTrsruous giudees. 331 

Hence 31^=^ Wl, and tlie equation for safe loading is 

--^Wl (5) 



B'l __^ 



Since (with this form of loading) if the beam were not 
built in but simply rested on two end supports, the equa- 
tion for safe loading would be \_R'I-^e\ = yi Wl,isee §242), 
it is evident that with the present mode of support it is 50 
per cent, stronger as compared with the other ; i.e., as re- 
gards normal stresses in the outer elements. As regards 
shearing stresses in the web if it has one, it is no stronger, 
since t/m = j^ JFin both cases. 

As to stiffness under the uniform load, the max. deflec- 
tion in the present case may be shown to be only i- of that 
in the case of the simple end supports. Eiieiit 

It is noteworthy that the shear diagram in Fig. 283 is 
identical with that for simple end supports §242, under 
uniform load ; while the moment diagrams differ as fol- 
lows : The parabola KB'A^ Fig. 283, is identical with tha*- 
in Fig. 235, but the horizontal axis from which the ordi- 
nates of the former are measured, instead of joining the 
extremities of the curve, cuts it in such a way as to have 
equal areas between it and the curve, on opposite sides 

i.e., areas [^C"^'+^i6^'0']=area R'G'B' 

In other words, the effect of fixing the ends horizontally 
is to shift the moment parabola upward a distance = 3Ic 
(to scale), = i Wl, with regard to the axis of reference, 

0'^', in Fig. 235. 

276. Remarks. — The foregoing very simple cases of con- 
tinuous girders illustrate the means employed for deter- 
mining the reactions of supports and eventually the max. 
moment and the equations for safe loading and for deflec- 
tions "When there are more than three supports, with 
spans of unequal length, and loading of any description 
the analysis leading to the above results is much more 
complicated and tedious, but is considerably simplified 



332 MECHAXTCS OF ENGINEERING. 

and systematized by tlie use of tlie remarkable theorem of 
three moments, the discovery of Clapeyron, in 1857. By 
this theorem, given the spans, the loading, and the vertical 
heights of the supports, we are enabled to write out a rela- 
tion between the moments of each three consecutive sup- 
ports, and thus obtain a sufficient number of equations to 
determine the moments at all the supports [p. 641 Eankine'a 
Applied Mechanics.] From these moments the shears 
close to each side of each support are found, then the 
reactions, and from these and the given loads the moment 
at any section can be determined ; and hence finally the 
max. moment ilf^,,, and the max. shear J^„. 

The treatment of the general case of continuous girdera 
hy algebraic methods founded on the properties of familiar geo- 
metrical figures, however, is comparatively simple ; and will 
be developed and applied in another part of this book. (See 
Chap. XII, pp. 485, etc.) 



THE DANGEROUS SECTIOIS^ OF ]S^O]?^-PRIS- 
MATIC BEAMS. 

277. Eemarks. By " dangerous section " is meant that sec- 
tion (in a given beam under given loading with given mode 
of support) where p, the normal stress in the outer fibre, 
at distance e from its neutral axis, is greater than in the 
outer fibre of any other section. Hence the elasticity of 
the material will be first impaired in the outer fibre of 
this section, if the load is gradually increased in amount 
(but not altered in distribution). 

In all preceding problems, the beam being prismatic, I, 
the moment of inertia, and e were the same in all sections, 

hence when the equation P—=M [§289] was solved for », 

e 

Me .... 

giving i>=— . . . . (1) 

we found that p was a max., = p^, for that section whose 
ili" was a maximum, since p varied as M, or the moment 



FLEXUEE NON-PEISMATIC BEAMS. 



333 



of the stress-couple, as successive sections along the beam 
were examined. 

But for a non-prismatic beam Zand e change, from sec- 
tion to section, as well as 31, and the ordinate of the 
moment diagram no longer shows the variation of p, nor 
is ^ a max. where ilf is a max. To find the dangerous 
section, then, for a non-prismatic beam, we express the ilf, 
the I, and the e of any section in terms of x, thus obtain- 
ing ^=func. {x), then writing dp-~dx=0, and solving for x. 

278. Dangerous Section in a Double Truncated Wedge. Two 
End Supports. Single Load in Middle. — The form is shown in 
Fig. 284. Neglect weight of beam ; measure x from one sup- 
port 0. The 
r e a c tion a t 
each support 
is i^ F. The 
width of the 
beam == 5 at 

all sections, while its height, v, varies, being = h at 0. 
To express thee = }4 v, and the /= 1 hv^ (§247) of any 
section on 0(7, in terms of x, conceive the sloping faces 
of the truncated wedge to be prolonged to their intersec- 
tion A, at a known distance = c from the face at 0. We 
then have from similar triangles 




[Tpl 



Fig. 284. 



V : X -{- c : : h 



3, .: V = ~ (x -\- c) 
c 



and 



e = 



h 



(x-{-G) and I = K^ 



-^x-^rcf 



For the free body nO, H (moms.^) 



Px—tL± 

e 



[That is, the M = )4 Fx.] 



p=SF 



and^= 3F ^ 
ax 



hH ' {x+cf ' 
By putting dp -t- dx = 



(1) 

(2) 

(3) 

we find X = + c\ showing a 



gives 
Fxe 

'~w • • • 

But from (2), (3) becomes 
dp_ o-p & {x-\-cy — 2a;(a;-t-c) 



p- 



334: 



MECHAXICS OF ENGINEERING. 



maximum for p (since it will be found to give a negative 
result on substitution in d^p 4- dotf). 

Hence tlie dangerous section is as far from tlie support 
0, as the imaginary edge^ A, of tlie completed wedge, but 
of course on the opposite side. This supposes that the 
half -span, )4l, is, > ^; if not, the dangerous section will 
be at the middle of the beam, as if the beam were 
prismatic. 

tx -xi, ) the equation for safe { ■Dfh'h'2 

Hence, with L ^^.^1^ .^^ ^f^>^^,^^^,\ ^-%Pl (5) 

A'' <^ ) at middle) ( ^ 

while with )tl^e equation for safe j ^,j ^A]^ , , p ,.. 

1/7 ^ r h loading is : (put x=c ■{ ^-^= V2 Pc (6) 

/^^ > ^ ) and_p=i?'in [3]) ( ^ 

(see §239.) 

279. Double Truncated Pyramid and Cone. Fig. 285. For 




Fig. 285. 



the truncated pyramid both width = u, and height = v^ 
are variable, and if h and Ji are the dimensions at 0, and 
c = QJ[ = distance from to the imaginary vertex A, we 

shall have from similar triangles u=~ (a;+c)and v= ~ (x-\-c). 

G c 

Hence, substituting 6=^^ and 7=1 uv^, in the moment 
equation 

£^_^=0.weW^=34,.^-|-,. . (7) 



. dp ^ op <^ (x+c)^ — 3x [x+cy 
' * dx bW" ' {x-\-cf 



(8) 



FLEXURE NON-PRISMATIC BEAMS. 335 

Putting the derivative =0, for a maximum p, we liave x = 
-h ^ c, hence the dangerous section is at a distance a? = ^ c 
from 0, and the equation for safe loading is 

either :?^= 14: PZ if >^ Z is < >^ c . . . , (9) 

(in which V and h' are the dimensions at mid-span) 



or 



MM)lhlf=y^P,,iy^i 



6 



is > >^ c ... (10) 



For the truncated cone (see Fig. 285 also, on right) where 
e = the variable radius r, and / = i^ ;r r*, we also have 

/=[Const.] .,—^.3 (11) 

and hence j9 is a max. for a? = ^ c, and the equation for 
safe loading 

either 5£i^ = % Fl,iox %l <% c , . , , . (12) 
(where r' = radius of mid-span section) ; 

^^ ^-R' (l^o)' ^%Fc,ioxy2l> %c (13) 

(where r^ = radius of extremity.) 



IS^ON-PMSMATIC BEAMS OF "UNIFORM 
. STRE]?^GTH." 

380. Eemarks. A beam is said to be of " uniform 
strength " when its form, its mode of support, and the dis- 
tribution of loading, are such that the normal stress ^ has 
the same value in all the outer fibres, and thus one ele- 
ment of economy is secured, viz. : that all the outer fibres 
may be made to do full duty, since under the safe loading, 
p will be = to B' in all of them. [Of course, in all cases 
of flexure, the elements between the neutral surface and 



336 MECHANICS OF EIS^GIJ^JEEEmG. 

fclie outer fibres being under tensions and compressions 
less than R' per sq. inch, are not doing full duty, as 
tegards economy of material, unless perhaps with respect 
to shearing stresses.] In Fig. 265, §261, we have already 
had an instance of a body of uniform strength in flexure, 
viz. : the middle segment, CD, of that figure ; for the 
moment is the same for all sections of CD [eq. (2) of that 
§], and hence the normal stress p in the outer fibres (the 
beam being prismatic in that instance). 

In the following problems the weight of the beam itself 
is neglected. The general method pursued will be to find 
an expression for the outer -fibre-stress p, at a definite sec- 
tion of the beam, where the dimensions of the section are 
known or assumed, then an expression for p in the varia- 
ble section, and equate the two. For clearness the figures 
are exaggerated, vertically. 

' 281. Parabolic Working Beam. UnsymmetricaL Fig. 286 




i. 



Pig. 286, 



CBO is a working beam or lever, B being the fixed fulcrum 
or bearing. The force P^ being given we may compute P^ 
from the mom. equation Pq^o = PJ^u while the fulcrum 
reaction is P^^P^-^-P^^. All the forces are ~\ to the beam. 
The beam is to have the same width h at all points, and is 
to be rectangular in section. 

Ilequir6*d first, the proper height hx, at B, for safety. 
From the free body BO, of length = l^, we have I (momss) 
= ; i.e., 

-^ rX,oxp^- -— ... (1) 



FLEXUEE. IfON-PIlISMATIC BEAMS. 337 

Hence, putting jo^ = B', h^ becomes known from (1). 

Required, lecondly, tlie relation between the variable 
height V (at any section n) and the distance x ot n from 0. 
For the free body nO, we have (2 momSu = 0) 

3iL=F,x ; or ^" ^^ ^^' =P,x and .-. p, =^l^ (2) 

But • for " uniform strength " p^ must = p^ \ hence 
equate their values from (1) and (2) and we have 

^ = — 1, which may be written {% vj = .>'^p' x (3) 

so as to make the relation between the abscissa x and the 
ordinate }4 v more marked; it is the equation of a para- 
bola, whose vertex is at 0. 

The parabolic outline for the portion BC is found simi- 
larly. The local stresses at G, B, and must be proper- 
ly provided for by evident means. The shear J = Pq, at 
0, also requires special attention. 

This shape of beam is often adopted in practice for the 
working beams of engines, etc. 

The parabolic outlines just found may be replaced by 
trapezoidal forms, Fig. 287, without using much more ma- 
terial, and by making the slop- 
ing plane faces tangent to the 
parabolic outline at points Tq 

and Ti, half-way between and ^^"^^b^^"'^'^ ° 

B, and C and B, respectively. It fis. 287. 

can be proved that they contain minimum volumes, among 
all trapezoidal forms capable of circumscribing the given 
parabolic bodies. The dangerous sections of these trape- 
zoidal bodies are at the tangent points Tq and Ti. This is 
as it should be, (see § 278), remembering that the subtan- 
gent of a parabola is bisected by the vertex. 




338 



MECHANICS OF ENGINEEKING. 



283. Rectang. Section. Height Constant. Two Supports (at Ex- 
tremities). Single Eccentric Load. 
— Fig. 289. h and Ji are tlie 
dimensions of the section at 
B. "With BO free we have 



Pah 



■Folo=0.\pj, 



■ 6Po?o 



(1) 




Fig. 289. 

At any other section on BO, as n, where the width is u, 
the variable whose relation to x is required, we have for 
wOfree 



P^^.F,x;ovPll/^=P,x 



QP,,x 

Pn = 



Equating pu and ^„ we have u :h :: x :Iq „ 
That is, BO must be wedge-shaped ; edge at 0, vertical. 



(2) 
(3) 



■k- 1 ^k-^wl 




Fig. 289 a. 



283a. Sections Rectangular and Similar. Otherwise as Before. — Fig. 289a. 
The dimensions at B are b and h; at any other section n, on BO, the 
height V, and width u, are the variables whose relation to x is desired, 
and by hypothesis are connected by the relation u:v::b:h (since the 
section at m is a rectangle similar to that at B). By the same method 
as before, putting pB = Pn, we obtain lf^^bh'^ = x-i-uv^; in which placing 
u^bv^h, we have finally 

v^=(h^^lf))x; and similarly, u^={b^-^lQ)x; . . . (4) 

i.e., the width u, and height v, of the different sections are each pro- 
portional to the cube root of the distance x from the support. (The 
same relation would hold for the radii, in case all sections were circular.) 
283b. Beam of Uniform Strength under Uniform Load. Two End Supports. 
Sections Rectangular with Constant Width. — Fig. 289&. WeigM of beam 
neglected. How should the height v vary, (the height and width 
at middle being h and b) ? As before, we equate pB and pn ', whence 
finally 

(ivy = [h^^P](lx-x') (5) 

This relation between the half-height ^v (as ordinate) and the abscissa. 
X is seen to be the equation to an ellipse with origin at vertex. 



FLEXURE OF BEINFOKCED CONCRETE BEAMS. 339 

CHAPTER V. 
Flexure of Reinforced Concrete Beams. 

284. Concrete and "Concrete-Steel" Beams. Concrete is an 
artificial stone composed of broken stone or gravel (sometimes 
cinders), cement and sand, properly mixed and wet beforehand 
and then rammed into moulds or " forms " and left to harden or 
" set." This material, after thorough hardening or " setting," 
thouglT fairly strong in resisting compressive stress is compara- 
tively weak in tension. When it is used in the form of beams 
to bear transverse loads (i. e., under " transverse stress ") the 
side of the beam subjected to tensile stress is frequently " re- 
inforced " by the imbedding of steel rods on that side. In this 
way a composite beam may be formed which is cheaper than a 
beam of equal strength composed entirely of concrete or one 
composed entirely of steel. 

Of course the steel rods are placed in the mixture when wet, 
and previous to the ramming and compacting, and their aggre- 
gate sectional area may not need to be more than about one per 
cent, of that of the concrete. 

No reliance being placed on the tensile resistance of the 
concrete (on the tension side of the beam) it is extremely 
important that there should be a good adhesion, and consequent 
resistance to shearing, between the sides of the steel rods and 
the adjacent concrete, for without this adhesion the -rods and 
the concrete would not act together as a beam of continuous 
substance.* 

In some specifications, for instance, it is required that the 
shearing stress, or tendency to slide, between the steel rods and 
the concrete shall not exceed 64 lbs. per sq. in. Sometimes 
the steel rods are provided with projecting shoulders, or ridges, 
or corrugations, along their sides, to secure greater resistance to 
sliding. * 

* For an account of tests of this adhesion see Engineering News, Aug. 15, 
1907, p. 169, and also p. 120 of the Engineering Record for Aug. 3, 1907. 



340 



MECHANICS OF ENGINEERBsTG. 



Fig. 290 gives a perspective view of a concrete-steel beam 
of rectangular section, placed in a horizontal position on two 




supports at its extremities, and thus fitted to sustain vertical 
loads or weights ; while Fig. 291 shows a concrete-steel beam 



flange- 



teeb, or stem- 




of T-section, in which the flange is intended to resist compres- 
sion, while the steel rods in the lower part of the " stem " are 
to take care of the tension. These two shapes of beam will be 
the only ones to be considered here, in a theoretical treatment. 

The ratio of the Modulus of Elasticity of steel (viz. — about 
30,000,000 lbs. per sq. in.) to that of concrete (say, from 
1,000,000 to 4,000,000 lbs. per sq. in., according to the propor- 
tions of ingredients used) is of great importance in the theory, 
since in general the stresses induced in two materials for a given 
percentage change of length are directly proportional to the 
modulus of elasticity (for same sectional area). 

Generally the diameter of a steel rod is so small compared 
with the full height of the beam that the stress in the rod is 
taken as uniform over the whole of its section. 

285. Concrete-Steel Beam of Rectangular Section. Flexure 
Stresses. — As in the common theory of flexure of homogeneous 
beams, it will be assumed that cross-sections plane before 
flexure are still plane when the beam is slightly bent, so that 
changes of length occurring in the various fibers are propor- 



FLEXUEE OF EEINFORCED CONCRETB BEAMS. 



541 



tional to the distances of those fibers from a certain neutral axis 
of the cross-section, and upon the amount of any such change 
of length (relative elongation) can be based an expression for 
the accompanying stress. Now in the case of concrete it is not 
strictly true that stresses are proportional to changes of length 
(" strains " or deformations) ; in other words its modulus of 
elasticity, E, is not constant for different degrees of shortening 
under compressive stress. Nevertheless, since this modulus 
does not vary much, within the limits of stress to which the 
concrete is subjected in safe design, it' will be considered con- 
stant, the resulting equations being sufficiently accurate for 
practical purposes. 

Let us now take as a " free body " any portion, ON, of the 
beam in Fig. 290, extending from the left-hand support to any 
section, at any distance x from that support. In the plane 
section terminating this body on the right, BNS (see now Fig. 
292, in which we have also, at the right-hand, an end-view of 






< 6 > 


t 

1 

] 

h 

\ 

1 

-4-- 

1 


AS-;>:-iy:V;-:;?il! 


N. axis 

— • — • — •— 



End View.. " 



the body), we note that the fibers of concrete from Z> down to a 
neutral axis iV^are in a state of compression, while below iVthe 
steel rod alone is considered as under stress, viz., a total tensile 
stress of F'p', where F' is the aggregate sectional area of all 
the steel rods, these rods being at a common distance a' above 
the lower edge of the section, and p' is the unit (tensile) stress 
in the steel rods. 

The distance BN, or " ^," of the neutral axis N below the 
" outer fiber " i), is to be determined. Let p denote the unit 
compressive stress in the fiber at I) (outer fiber) of the concrete ; 
then the unit stress in any fiber of the concrete at distance z 

from iV will be - j9, lbs. per sq. in., and the total stress on any 



342 MECHANICS OF ENGLNEERING. 

such fiber is — »c?i^, lbs. (where c?jPis the sectional area of the 

e 

fiber). All the (horizontal) fibers between the two consecutive 
cross sections DS and D' 8' were originally dx inches long, but 
now (during stress), we find that the fiber at D has been shortened 
an amount d\ and the steel rod "fibers " elongated an amount 
d\'^ so that we have the proportion d\ : d\' : :e: a — e; 

d\ e , ,». 
or, ■-— = (0) 

dV a- e ^ ^ 

For the free body in Fig. 292 we have, for equilibrium, the 
sum of horizontal components of forces = (the shear J" has no 
horizontal component); that is, remembering that below iVno 
tensile forces are considered as acting on the concrete, but 
simply the total tensile stress F'p' in the steel rods, 



t/n 



- pdF - F'p' = 0. 
e 



But here =^ is a constant ; and for the rectangular cross -section, 
dF = b . dz, and 

'iI>'^-'T4-'-'T-^y • • • W 

But from the definition of modulus of elasticity (F for the 
concrete and E^ for the steel), we have (§ 191) 

F = p —- (relat. elongation), or F = p -i- [dX/dx) ; and 
similarly, F' = p' -^ {dV /dx) ; whence 

d\' p'' E ^ ^ 

But, from eq. {\.)^ p -— p' = IF' -^ he, combim'ng which 
with eqs. (0) and (2), and denoting the ratio F' -^ Fhj n, 

we mid = — ^ — ■. . (3) 

a — e be 

The ratio n may have a value from 10 to 25 for " rock- 
concrete," and still higher for " cinder-concrete ; " see § 284. 
Now solve eq. (3) for the distance e, obtaining 



F'nfj2ab ^ A 



FLEXURE OF REINFORCED CONCRETE BEAMS. 



34S 



This locates the neutral axis, iV". [See, later, eq. (29), § 
291.] 

Returning to the free body OJV in Fig. 292 above, we note 
that the resultant compression in the concrete between iV^ 
and D, viz., ^ p .he, lbs. [see eq. (1)], is equal in value to 
the total tension F'p', lbs., in the steel rods at G', and that they 
are parallel. Consequently they form a couple (the " stress- 
couple " of the section) whose moment is equal to the product 
of one of these forces, say 
F'p\ by the perpendicular | j^ \ 

distance = a", between Gr' 
and a point Cr (see now Fig. 
293) whose distance from the 
" outer fiber " i> is one-third 
of e. The "arm" of this 



couple is a' 



a- -■ For 




Fig. 293. 



equilibrium of the free body ON in Fig. 292 the shear J and 
the two forces V (reaction) and P^ (load) must be equivalent 
to a couple of opposite and equal moment to that of the stress 
couple. Call this moment M [in this case it has a value of 
Yx — P^{x — a;J]; it is the " bending moment " of the section 
at DS. We may therefore write (see Fig. 293) : 



M = F'p' [« — i e] ; and .-. p' = 



M 



F\a-\e) 



(5) 



which will give the unit-stress p\ induced in the steel rods at 
section BS. It is seen to depend on the position of the neutral 
axis N (i.e., upon e); upon the bending moment, M, at that 
section; upon the sectional area F' of the steel rods (aggre- 
gate); and on the distance, a, at which they are placed from 
the compression edge, B, of the beam. 

But since the resultant compression, h p -he, is equal to the 
resultant tension, .F'p', we may also write 

2M 



M= ^p .be [« — i e] and .-. p = 



he (a — I e) 



(6> 



which gives the unit-stress (compression) in the outer " fiber " 
at i), of the concrete, for this section BS. 



1 
e 

..i 


G 








■^— 


N 


N" 


V'(p + clp'l 




s" 



344 MECHANICS OF ENGINEERING. 

286. Horizontal Shear in the Foregoing Case (Rectangular Sec- 
tion). The shear per sq. in. along the sides of the steel rods, 
I ^^ and also along the horizontal 

,/' " Neutral Surface,'' NN" (see 

Fig. 294), may be obtained as 
follows : — Let dx be the length 
of a small portion of the beam 
,, , , „ (of Fig'. 290) situated between 

■^ r^^^^n '- — two vertical sections Do and 

D"S". Fig. 294 shows this 
^^*^- 294. portion as a " free body." The 

forces acting consist of the tension F'p' on the left-hand end of 
the steel; the tension on the right-hand end of these rods 
[being something greater (say) and expressed by F' (^p' -\- dp'^ 
in which dp is the difference between the unit-tensions at the 
two ends of the steel " re-inforcement "] ; the resultant com- 
pression, i he.p, in the concrete on the left; and that, 
1 5g . (j? -f dp^, on the right ; and, finally, the two vertical shears, 
J'and J". Here p is the unit compressive stress (lbs. per sq. in.) 
in outer fiber of concrete at the left-hand extremity of the same, 
while p -\- dp expresses the unit compressive stress in the same 
outer fiber at the right-hand extremity. 

Evidently the difference between the total tensile stresses at 
the extremities of the steel rods will give the total horizontal 
shearing stress on the sides of those rods and this may be 
written pjl^dx (lbs.), where pj = unit shearing stress between 
the steel and concrete and l^ == aggregate perimeter of the steel 
rods (so that l^dx = total area of the outside surface of rods in 
Fig. 294); 

hence p/l^dx ^ F' {p' j^- dp') — F'p' .... (7) 

But if, for the free body of Fig. 294, we put 2 moms. = 
about the point Cr (a distance ^ e from upper fiber) 

we find Jdx = [F' (p' + dp') — F'p'']{a -\ e)\ . . (8) 

and hence ) , J" /q\ 

see (7), \ ^' ^ l^ (a -^ e) ^ ^ 



FLEXURE OP EElNFOliCED CONCRETE BEAMS. 



346 



p bdx 



Also, if we let p^ denote the unit shearing stress (or tendency 
to slide) along the horizontal sur- 
face NN" or neutral surface, the 
total amount is pjbdx (lbs.). 

In Fig. 295, which shows as a 
free body the portion NN"S"S of 
Fig. 294, we see this horizontal 
force (of concrete on concrete) act- 
ing toward the left. The other ^^°- ^^s. 
forces acting on the free body are as shown in Fig. 295 and, by 
putting 2 horiz. compons. = 0, 




we find 

and finally, 
see eq. (8), 



F' {f + df) — F'p' = pJbdx; 
J 



Ps 



(9a) 
(10) 



5(a-ie) • 

This (unit) shearing stress in the concrete along NN", the 
"neutral surface," should nowhere exceed a certain value \Q.g., 
64 lbs. per sq. in.). For horizontal planes above NN" it is 
smaller than along NN". Similarly, the unit stress pj should 
not exceed a proper limit. 

287. Numerical Example of a Concrete-Steel Beam of Bectangular Section. 

(See foregoing equations.) 

Fig. 296 sliows the section [8 by 11 inches] of the beam. Four round steel 
rods are imbedded near the under (tension) side, their centers being 10 in. from 



W=600 Ihs. 




.- d=0.45 



4L 



p-^? 



Fig. 296. 

the top of section (a = 10 in.). This beam is to be placed on two supports at 
the same level and 8 feet apart, and is to support a concentrated load P, lbs., 
at the middle of the span as well as its own weight, which is K = 600 lbs. 

P is to be determined of such a safe value that the greatest stress in the 
steel rods shall not exceed 16,000 lbs. per sq. in. The compressive stress in 
concrete is not to exceed 700 lbs. per sq. in., nor the greatest shear either in the 
concrete or between the steel and the concrete, 64 lbs. per sq. in. 

Each steel rod is continuous throughout the whole span and has a diameter 
of 0.45 in., from which we easily compute the aggregate perimeter of the rods 



346 MECHANICS OF ENGHSTEERING. 

to be 5.65 inches (=io)> ^^^ ^^^ aggregate sectional area to be 0.64 sq. in 

(= Fy 

The ratio of the modulus of elasticity for steel to that of the concrete will 
be taken as 15 to 1 ; i.e., n = 15. 

The first step is to locate the neutral axis by finding the value of e from 
eq. (4), thus: — 

64 15 / / 2 X 10 X 8 , , , \ „ J, . . 
^-iOQ-T(V .64X15 +l-l)=3.84m. 

Next, if for p' we write 16,000 (using inch and pound) and substitute in 
eq. (5), solving for M, we obtain the greatest bending moment to which any 
section of the beam should be exposed, so far as the steel is concerned, viz: — 



M = p'F'la- I) = 16,000 x 0.64 (10 - 1.28) = j 



e\ ..nnn.. n«..in 1 oe^ _ i 89,000 

in.-lbs. 



i.e., max. moment is to be 89,300 in.-lbs. 

For the mode of loading of the present beam the max. moment occurs at 

the section at the middle of the span and has a value (with I denoting the span, 

PI Wl 
or 96 in.) of — + -tt- • We therefore write 

PX96 ^ 6_00x9_6 ^ gg 3^^_ ^^^^^ ^ ^ g^^^O lbs. 
4 o 

To find the accompanying maximum compressive stress in the concrete, 
eq. (6) gives (for outer "fiber") 

2Jf 2 x 89,300 __ -, 

P = T-, ; — -s = 5 — FTTH s-^ = 666 lbs. per sq. m., 

•^ le{a-\e) 8 x 3.84 x 8.72 ±' ^ > 

which is within the limit set (700 lbs. per sq. in.). 

As for the max. shearing unit stresses Ps and ps, they are greatest where 
the vertical shear, J, is a max., which is close to one of the supjjprts. Here we 
note that J is equal to J of 3,420 + i of 600 = 2,010 lbs. Hence, from eq. (9), 

2,010 2,010 

^' - 5.65 X (10 - 1.28) = 5.65 x 8.72 " = ^^"^ ^^'- ^^^ "I- '''■' 

while ) 2,010 

from (10) i^^ = 8"3r8^ = ^^'^ ^^^- P^^ ^*1- ^"•' 

These shearing stresses are seen to be well within the limit set, of 64 lbs. per 
sq, in. As to compressive stress, the building laws of most cities put 500 lbs. 
per sq. in. as max, safe limit forp, the compressive stress in concrete. 

288. Concrete-Steel Beam of T-Form Section. See Fig. 297. 
In this form of beam, to secure simplicity in treatment, it will 
be considered that the flange {TK') alone is subjected to com- 
pressive stress [although strictly a small portion of "stem" 
between the flange and the neutral axis of a section is under 
that kind of stress] . The part of stem below the neutral axis 
(as before) is not considered to offer any tensile resistance, all 



FLEXUEE OF HEINFOECED CONCEETE BEAMS. 



347 



tension being borne by the steel rods or " re-inforcement." Fig. 
297 shows a side view and also an end-view of a portion of the 
beam in Fig. 291 extending from the left-hand support np to 
any section DjS (or up to W in Fig. 291). - As before, sections 
plane before flexure are considered to be still plane during 
flexure, so that the elongations or shortenings of any horizontal 
*' fiber," w^hether steel or concrete, are proportional to the dis- 




Fig. 297. 

tances from a neutral axis iV, at some distance e from the top 
fiber of the flange, where the unit compressive stress has some 
value p. 

Also, since the U for concrete in compression is to be taken 
as constant the stresses in the concrete will also be proportional 
to the distances of the " fibers " from H the neutral axis. Let 
p'^ denote the unit-stress in the concrete at H, the bottom fiber 
of the flange ; then, by proportion, p :p^' : : e: e — d, where d is 
the thickness of the flange. Since the compressive stresses in 
the concrete between H and D are distributed over a rectangle 
their average unit-stress is (p-\-p'')/2, and their resultant, 
which acts horizontally through some point Gr, has a value of 
bd.(p -\- p")l2\ or, as it may be written (see above for y ), 
'{^p {1 e - d) .Id) -^ {1 e\ . 

The total tensile stress in the steel rods will be ¥'p', as 
before, where ¥' is the aggregate sectional area of the rods and 
p' the unit stress in them at section DS. Besides the stresses 
just mentioned the other forces acting on the free body in Fig. 
297 are all vertical ; viz., the shear J'and the pier reaction and 
certain loads between and D ; hence by summing the horizon- 
tal components we note that the compressive stress in the con- 
crete is equal to the tensile stress F'p' in the steel, so that this 



348 MECHANICS OF ENGINEERING. 

tensile force F'p' and the resultant compressive stress form a 
couple (^^ stress-couple '^ of the section; with an arm = 6r6r', 
= a"), and we have 

pi2e-d)bd ^ • 

2e -^ ^ ^ 

Consequently the shear and the other vertical forces acting on 
the free body form a couple also, and the moment of this couple 
(equal to that of the " stress-couple ") will be called M. In the 
figure these vertical forces are not shown, but simply an equiva- 
lent couple (on the left). 

If at this part of the beam a length dx of the steel has 
stretched an amount dX' and an equal length, dx, of the outer 
fiber at D has shortened an amount dX, we have from eq. (2) 
of previous work 

dX'-p'"E' ^^^^' 

where E' and E are the moduli of elasticity of the steel and 
concrete, respectively. But, from (11), 

p' (e - ^d)bd' ^ ^ 

and from similar triangles dX : dX' : : e : (a — e) . . . .(14) 
Eqs. (33), (14), and (12), with E' -i- E = n, give 

F-n.a+—- 
'= bd + En ' ^^^^' 

and thus the neutral axis, iV, is located. 

It will now be necessary to locate the point of application,, 
between E and E, of the resultant compressive stress on EE ; 
that is, the point (r in Fig. 298 which gives a side view of 
these stresses alone, forming, as they do, a trapezoidal figure 
whose center of gravity, U, projected horizontally on EE gives- 
the desired point, G: The lower base EC^^ of this trapezoid 



FLEXURE OF JREINFOECED CONCRETE BEAMS. 



349 



represents the unit stress jw"; the upper, DC", represents the 

unit stress p. The distance, call it c, 

of G- from N, is to be determined. 

Let the trapezoid be divided into a 

rectangle BD'" C" H and a triangle 

D'"G"'Q". The center of gravity of 

the latter is at a vertical distance of \ d 

from a line WW" drawn horizontally 

at distance \ d from D . H" H'" passes 

through the center of gravity of the 

rectangle. Let us now find the distance GrR" by writing the 

moment of the resultant stress about point W equal to the 

sum of those of its two parts, or components, represented by 

the rectangle and the triangle ; whence we have 




Fig. 298. 



\{p^p").uy. aw = o + ^P-^P^'^ .'^ 



(16) 



Noting that ji?'' = 

d-" 



- d 



p, we have, solving, 



aw=\ 



6 2e- d 

NCr, i.e., c, = e 



, and therefore, measuring from N, 



d 1 

2 "^ 6 ■ 2e - cZ 



(17) 



Now that both e and c have been determined in any given 
case it remains to find expressions for the unit stresses p' and j? 
(in steel and fiber I) of concrete}. 

Since (r is the point of application of the resultant compres- 
sion in flange, the arm of the stress-couple, a" (Fig. 297), is 
the distance from G- to Gr' (see Fig. 297); that is, 

a" = c -{- [a— e); and hence we may write 

Pf(c + a-e)~M;.:f = ^,^/^^_^^ .(18) 

Also, by eliminating the ratio d\:d\' from eqs. (12) and 
(14) we have, solving for^, 



p = 



p e 
n (a — e) 



(19) 



350 MECHANICS OF ENGINEERING. 

289. Shearing Stresses in T-Form Concrete-Steel Beams. As 

regards the unit shearing stress, p'^ induced on the sides of the 
steel rods, in this case of the concrete steel beam of T-form 
section, an analysis similar to the corresponding one in the case 
of the beam of rectangular section leads to the result 

J ( where «7is the total vertical shear at ) ,oa\ 
Iq{c -\- a— e) I the section BS, and l^ the aggregate ) 

perimeter of the steel rods. 

And, similarly, for the unit shearing stress on the horizontal 
surface separating the flange from the "web" or "stem" (see 
Fig. 297 ) at H, where the width of the web is h", we find for this 
unit horizontal shear, p^, 

P'= V-(c + a-e) (^^' 

290. Deflections of Concrete-Steel Beams. The deflection of a 
loaded prismatic concrete-steel beam resting on two supports at 
its extremities, may be obtained for the cases dealt with in 
§§ 233-236 inclusive, in connection with homogeneous beams ; 
provided the product EI occurring in the expressions for these 

deflections be replaced by ■ [ a— -^), for concrete-steel beams 

of rectangular section; and by E' F' {a — e) (c + a — e), for 
those of T-form section. 

291. Practical Formulae and Diagrams for use with Concrete-Steel Beams 
of Rectangular Section. The equations of the foregoing theory will now be put 

into convenient form for practical use in designing these 
beams. Let us denote the ratio of p' (stress in steel at 
section of max. moment) to p (stress in outer fiber of 
concrete) by r ; i.e., r = p'/p ; while n = E'/E, as before. 
Also let ?n, = M -^h, denote the max. bending moment per 
inch of width (6) of beam ; and let F' (area of steel) -h & be 
called/, i.e., steel areajjer mc/i of width {b). In other words, 
we have the notation 

r =-?:_; n= -=-;™ = ^; and /'= -=-; . . . . (22) 
Fig. 299. p E 

If we now substitute e = 2 r/, from eq. (1), in eq. (3), 

we have 2Tf {r + n)=an (23) 

Now e = 2 r/, which from (23) =an-^{r + n) ; Hence eq. (5) 

-will give 3?n(r + Ti) = a/'p'(3r + 2n) (24) 




REINFORCED CONCRETE BEAMS 

DIAGRAM 1 


^ rectangular'). 

20 25 30 40 


L/ 1 r^ M 1 ir^ III .!,• 




; 


- -r -> 




1 H - 


■i:: 




:-_!_: '.!.'. 




.-H ::f. 


i-t.. 60 


s = 


/«- 


» 




:;-r-:-?: 


- T-- ^^ 


~ «->«./'„ . „ \ ' 


- _l- _ 


:.-^y-. 




in which ^c 




::z5:::t: 






,?-i- -i- 


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- -\-^ 




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y 




y 






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-■^•- 50 






— 


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1 ^1 




- 






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^ y J 




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.._,.(« 








y 




/'^ 1 


.,2^ 








y 




■'-y r- 


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\ d ah 1 


xc 






/ 




: _Ly 




1 "[ 


\ J 

^5 ^ ^ n Q 




'"i"" 


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/ 


- 




-y-- 


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y 




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p°^ 

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in 

<• 


o / O "^ 




/I 








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~"i~ 


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1 


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y 


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y. 




y 


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y 




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^ 


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. _L Its 


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X 


y 


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/> 


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/.. 


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y 


y 


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y 


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y 


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1 1 


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/ 


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^ 


> 


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y 


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1 


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1 1 


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i 


> 6 7 8 9 10 12 15 20 25 30 
VALUES OF n, =E'^E 


ijj 1 ly 

40 



\ To face page 350. 



REINFORCED CONCRETE BEAMS 
10 12 15 20 


(rectangular ^ 
« nil 1 » 


.75 

.80 

^ i 

.85 

O 
.90 

CO 

u 

D ' 
_l 
< 
> 

.95 












1 

1 
1 , , 




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\ 


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1 

1 




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V 


t 


\ 


s 


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1 

■-+-- 

1 

1 


-i 


3 (r -^n) 


\ 


sj 


\ 


s. 


\ 


^, 


1 25 30 


in the foftiiula j — , 

- Q />'« 

50 60 




s 


k 


> 


\ 


.p. 


1 


1 
..J... m 


^ 


V ' 


N 


s 




^^^ 


H 


1 or 


\ 


\ 


\ 


\ 


S 




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, 1 s 


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1 

. 1 




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+ ..,x.._l 1 




\ 


\ 


s. 




^s 


Pv 


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1 




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\ 


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1 

1 s 


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1 


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1 


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L_. 
















1 
1 


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1 

-1 — 
i 

1 


1 




> 1 T ^N -• 


1 


12 15 20 25 30 4( 
VALUES OF r^ 


D 50 60 



[To face page 351, 



FLEXURE OF EEINFOECED CONCRETE BEAMS. 351 

If now S denote the quotient /'-=-a (i.e., /S=area of steel section j3er unit 
area of concrete section above the steel), (23) may be written in the form 

S = n-h[2r{r+n)] (25) 

3 J'4-2 n 
Asain, let ^— r- be denoted by Q - . . . (26) 

° 3(r + n) ^ ' 

Prom (24), (25) and (26) we may then derive 

M=qSp'a^b .... (27); andi m = QSp'a^ ...... (28) 

for practical use in design; in connection with Diagrams I and II (opp. pp. 
350 and 351) from which we may obtain values of Q and 8, respectively, for 
any given values of r and n. 

As to unit shearing stresses, which should not exceed 64 lbs. per sq. in. 
(say), use is made of eqs. (9) and (10), § 286, in which the maximxim total vertical 
shear Jm should be substituted. In computing "e " for use in (9) and (10), it is 
simplest to employ the relation [derived from eqs. (1), (23) and (25)] 

e = 2 Sra (29) 

292. Numerical Examples. Rectangular Section. Let us suppose that in the 
cross-section of maximum moment the stresses in both materials are to attain 
their greatest safe values; viz., 16,000 lbs. per sq. in. tension for the steel, and 
600 for the compressive stress in outer fiber of the (rock) concrete. Also suppose 
that ^' = 30,000,000 and £' = 2,000,000 lbs. per sq. in. That is, we have n=15 
and r = 16,000 H- 500, = 32 ; and from Diagram I find S = 0.005. In other words 
the necessary area of steel section is J of one per cent, of the area of the concrete 
(above steel rods). "We also find, from Diagram II, that Q = 0.894, 

If now " a " be taken as 10 in., eq. (28) gives : — 

m = 0.894 X 0.005 X 16,000 X 100, = 7,152 inch-lbs. bending moment that 
could safely be withstood by each inch in the width &; so that if "&" were 8 
inches we should have M = mb = 7,152 x 8 = 57,216 inch-lbs., safe bending 
moment for the section of the beam. 

Again, with r and n still equal to 32 and 15, respectively, and hence with 
Q and /S as before, viz., 0.894 and 0.005, if b is assumed as 10 in., and the max. 
bending moment to be sustained is If = 80,000 inch-lbs. (so that m = 8,000), 
we find from eq. (28) 



= \/^ 



« = ^/ -.894x0^07x16,000 = ^^''^ ^^^^^^' 



as necessary value of a ; while the total area of steel section needed is 

F', = S . a &= 0.005 X 10. 58 X 10 = 0. 5290 sq. in. 

which is seen to be one half of one per cent, of the area [10x10.58] of the 
concrete above steel (i.e., S = 0.005, as found from Diagram I originally). 

293. Cost of Beams of Rectangular Section. While in a general sense 
economy in cost is favored by having the width " &" of the rectangular section 
small compared with the height " a," a limit to narrowness of width is set by 
the unit shearing stress in the neutral surface which would be found to exceed 
a safe limit if the beam were too narrow. The thickness "a'" of concrete 
below the steel rods (see Fig. 292) might be made -^5 of " a." 



362 



MECHANICS OF ENGINEERING. 



CHAPTER VI. 

Flexure. Columns and Hooks. Oblique Loads. 

294. Oblique Prismatic Cantilever. In Fig. 301, at (a), 
(on p. 354) we have a prismatic beam built in at K, projecting- 
out obliquely, and carrying -a vertical load P at upper end ; the 
line of action of P passing through the center of gravity of the 
upper base of the prism. In such a case the fibers of the beam 
where they cross any transverse plane mg will evidently be 
subjected to compressive stress (called a ^'■thrust'''') due to the 
component of P parallel to the axis OKoi the prism ; to a shear 
J" due to the component of P at right angles to that axis ; and 
also to additional stresses, both tensile and compressive, formings 
a " stress-couple^^ due to the moment of P (i.e., Pu) about ^, the 
center of gravity of the cross-section m'm. 

More in detail, consider in Fig. 300 a portion AB of the 
prism, being the part lying above a cross-section mm' near the 

top, so that the portion gO of 
axis is practically perpen- 
dicular to the section mm' 
which is a plane both before 
and after flexure, g being 
the center of gravity of the 
plane figure formed by the 
cross-section. 

Let the unit stress on the 
end of the extreme fiber at 
m be represented by the 
length sm and that [also com- 
pression (say)] on the other 
extreme fiber, at m', by s'm'. 
Draw the straight line ss' ; 
then by the common theory of flexure the stress on any inter- 
mediate fiber, at c, would be the intercept, or ordinate, ac to this 
line. Now the unit stress p^, on the fiber g at the center of 
gravity of cross-section, being gr, draw through r a line t'rt 




Fia. 300. 



FLEXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 353 

parallel to m'm, and we now have the stress on any fiber as o 
divided into two parts be, or p^, the same for all the fibers ; and 
ab, different for the different fibers but proportional to the 
distance z of the fiber from g. Hence we have : 
the unit stress on any fiber c is 

P = Pi -\ P2 (ibs. per sq. in.) . . . . (1) 

where p^ is st and e the distance of the extreme fiber ??? from g ; 
and hence the total stress on fiber c is pdF =p^dF H — Pnli^F, lbs. ; 

where dF is the area (sq. in.) of section of fiber, or element of 
area of the cross-section, F being the total area of the cross- 
section, mm' . Geometrically, we note that while the system of 
normal stresses on all the fibers forms a trapezoid, m's'sm in 
this side-view, and that they are all compressive, they are 
equivalent to a rectangle, m't'tm, of stress of uniform compressive 
unit-stress p^ ; and two triangles, one, rst, of compressive stress, 
and the other, rs't', of tensile stress.* It will now be shown that 
the sum of the moments of the stresses of the rectangle about 
center g is zero, and that the two triangles of stress form a couple. 

^(moms.) of stresses in triangle = I (p^dF)z = pi dFz 

=p^Fz = zero ; since z =zero, the 2's being measured from the 
center of gravity, g, of section mm' [§ 23, eq. (4)]. 

Again, if we sum (algebraically) the stresses of the two 
triangles, 

we have / - p dF = — I zdF = ^Fz, = zeYo 

Jz = -e' e e J e 

that is, the resultant of the compressive stresses in rts equals 
that of the tensile stresses in rs't' ; hence they form a couple. 

If, therefore, we have occasion to sum the moments about 
g, of all the stresses acting on the fibers in section wm' we are 
to note that this moment-sum involves the stresses of the triangles 
alo7ie (that is, of the couple), and is 

in.-lbs. ; where I^ is the " moment of inertia " of the cross-section 
* These plane figures are the side views of geometric solids. 



354 



MECHANICS OF ENGINEERING 



referred to an axis through g (its center of gravity) and perpen= 
dicular to the "force plane " (plane of paper here). 

If, again, we sum the components of all the stresses (on plane 
mm') parallel to the axis gO we note that this sum is zero for 
the couple and also for the shear J and hence reduces simply to 

fp^dF =_Pj CdF=p^F, lbs. (the Thrust) . . (3) 

(corresponding to the rectangle, fm). 

The sum of components perpendicular to axis ^ is of course 
simply the shear, J, lbs. 

Evidently the unit stress (normal) in fiber at m is expressed 

e' 
as jp,„ = j9j 4-^2' ^^^^ ^^^^ ^^ *^' '^^^ Pm'=Pi P2' ^^ ^^ ^^^J 

case the latter is negative it indicates that the actual stress in 

this fiber is tension. 

295. Oblique Cantilever. Fig. 301, (a) and (5). At (h) is shown 

as a " free body," a portion [of the cantilever at (a)] of any 

length X from top. The 
forces acting are the vertical 
load P at 0, and the stresses 
on the ends of the fibers in 
the section m'm.; and these 
stresses are now indicated 
as consisting of a thrust, T, 
of uniform intensity p^, the 
total thrust being ^9^^', lbs., 
(where Y is the total area of 
section) ; of a stress-couple, 

' '" C, whose moment is -^ in.- 

FiG. 301. e 

lbs., in which pj = Pm — P^ ^^^ I is the " moment of inertia " 
of the cross-section (about an axis through its center of gravity 
g at right angles to the plane (" force plane ") containing Og 
and force P ; the same I that has been used in previous cases of 
flexure) ; and the total shear, J, lbs., parallel to force plane and 
perpendicular to gO. The lever arm of P about g i^ u which 
practically = x sin a (unless the beam is considerably bent or 
is nearly vertical). 




FLEXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 355 



For this free bjdy (in order to find p^, p^ and J ) 

Xxr A • Ti -n f\ Pcosa 

Jl = Ogives: /^ cos a— p^i^ =0 ; .•. p^ = 






F 



X^ X A Vol r> A -^^^ 

(moms.)^ = . .-^ Fu = 0; .-. p^ = — — • 



(4) 
. (5) 



and 



Xi'=o 



P sin a — J" = (9 ; .-.J — P sin a, (6) 



As X varies, from to Z, we note that p^ and J remain 
unchanged but tliat p., increases ivith u ; so that the maximum 
value of the unit stress jo,,,, Avhicli = p^ + p,' will be found in 
the section at K, where x = I ; and if this stress is not to exceed 
a safe value, R', for the material, we put p^i,^^ K) +p^ = R', 
(as the equation of safe loading) ; 

^^nan ^, .... (7) 



or, 



P 



"cos a 

~'f~ 



Pn 



(N. B. For a cross-section of unusual shape the stress 

e' 
, = p^ P2, at K, might happen to be numerically greater 



than Pj^„ and thus govern the design). 

296. Experimental Proof of Foreg^oing. A 

stick or test piece of straight-grained pine 
wood, 12 inches in length and of square 
cross-section (one inch square), originally 
straight and planed smooth and with bases 
perpendicular to ^the length, was placed in 
a testing machine ; steel shoes, with (outside) 
spherical bearing surfaces, being centered 
on the ends. See Fig. 302, where AB is 
the stick and S, S\ the two steel shoes. The 
stick was gradually compressed between 
the two horizontal plates B,B'^ of the machine 
and bent progressively in a smooth curve 
under increasing force. From the nature of 
the "end conditions," as the stick changed 
form, the line of action of the two end 
pressures P,P, always passed through the 
centers of gravity, a and h, of the respective bases. 

When the force P had reached the value 4500 lbs. a fine 
wrinkle was observed to be forming on the right-hand surface 




Fie. 302. 



356 MECHANICS or ENGINEERING. 

of the stick at the outside fiber m of the middle section gm. The 
other tibers of this section were evidently uninjured. At m then, 
the unit-stress must have been about 8000 lbs. per sq. in., the 
crushing stress (as known from previous experiments with sticks 
of similar material and equal section but only three or four 
inches long; these were too short to bend, and wrinkles formed 
around the whole 'perimeter^ showing incipient crushing in all the 
fibers). The distance gc at this time was found to be \ in. ; 
i.e., the lever arm, w, of the force P about g, the center of gravity 
of the section. In this case, then, it is to be noted that the 
value of %i was entirely due to the bending of tit e piece. 

Substituting, in eqs. (4) and (5) of § 295, the values w = |- in., 
a = 0, cos a = l, e = e', =i inch, i^=l sq. in., 

hh^ 1x1^ 1 
and i; = — , = ^^ = — in/, we find p^ = 4500 

lbs. per sq. in. and p^= 3375 lbs. per sq. in. 
Hence stress at m, = Pi-\- P21 == 7875 lbs. per sq. in., which is 
about 8000, as should be expected. On the fiber at 0, how- 
ever, we find a stress of p^ — p., or of only 1125 lbs. per sq. in. 
compression. 

We find, then, that in the section om, when P reached the 
value of 4500 lbs., there was a total-thrust (p,F^ of 4500 lbs.; 
a unit-thrust (w^) of 4500 lbs. per sq. in. ; and a stress-couple 

pi 
having a moment of Pw, = ^— , = 562.5 in .-lbs., (implying a 

separate stress oi p^^^'^l^ lbs. per sq. in. in the outer fibers, 
to be combined with that due to the thrust). Also that /, the 
shear, was zero. 

297. Crane-Hooks. First (Imperfect) Theory. Fig. 303 shows 
a common crane-hook of iron or steel. Early writers (Brix and 
others) treated this problem as follows : — 

The load being P, if we make a horizontal section at AB^ 
about whose gravity axis, gr, P has its greatest moment, and con- 
sider the lower portion C as a free body, in Fig. 304), we find, 
using the notation and subdivision of stresses already set forth 
in § 294 for an oblique prism, that the uniformly distributed 
pull (or " negative thrust ") on the fibers is p^F = P, lbs. ; 



FI^EXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 357 



P 




while the moment of the stress-couple is ^- = Pa ft.-lbs.; and 

e 
that the shear, /, is zero. 

Hence on the ex- 
treme fiber at B we 
have a total unit 
tensile stress of 
Pae 

T' 

which for safe de- 
sign must not ex- 
ceed the safe unit- 
stress for the ma- 
terial, R' lbs. per sq. in. ; whence we should have 

p[l + f] = iS' . . 

as the equation of safe loading.* 

Example: Safe P = ?, if section AB is a circle of radius 
2 in., while a = 4 in. ; the material being mild steel for which 
(in view of the imperfection of the theory) a low value, say 
6000 lbs. per sq. in., should be taken for R'. With these data 
we obtain: — 

1 4x2" 



Fig. 303. 



Fig. 304. 



Fig. 305. 



(8) 



P = 6000 



-[ 



12.56 ^ 50.24 



h 



25130 lbs. 



The simple crane in Fig. 305, being practically an inverted 
hook, may be treated in the same manner. 

298. Crane-Hooks. Later, More Exact, Theories. The most 
exact and refined theory of hooks yet produced is that of 
Andrews and Pearson,! but it is very complicated in practical 
application and far too elaborate and extended to be given 
here. 

The next best (and fairly satisfactory) treatment is that of 
Winkler and Bach, of which the principal practical features 
and results will now be presented. 

* See experiments by Prof. Goodman, in Engineering, vol. 72, p. 537. Re- 
sults are irregular, due probably to the use of this imperfect theory. 

t Drapers' Company Research Memoirs. Technical Series I. London, 
1904. 



358 



MECHANICS OF ENGINEERING. 



In AB, Fig. 306, we have again the free body of Fig. 304, 
but the vertical stresses acting on the cross-section m'm are 
proportional to the ordinates of a curve instead of a straight 
line. The imperfection of the early theories lies in the fact 
that the sides of a hook are curved, and not straight and par- 
allel as in the prismatic body of Fig. 301 ; and the variation of 
stress from fiber to fiber on the cross-section must follow a dif- 
ferent law, as may thus be illustrated : 

As preliminary, the student should note, from the expres- 

P EX 
sion— = — -of p. 209, that in the case of two fibers under ten- 
F I 

sion, with the same sectional area F, the unit-stress P -^ F (or 
p) is not proportional to the elongation }. of the fiber unless the 
two lengths I are equal. In Fig. 306 the center of gravity of 
the cross-section is g, and is the center of curvature of the 
curved axis gk of this part of the hook (or other curved body). 
The two consecutive radial sections m'm and ft are assumed 
to remain plane during stress, and hence the changes of length, 

due to stress, of the (verti- 
cal) fiber lengths between 
them are proportional to 
the ordinates of a straight 
line ; and if these fiber 
lengths were equal in 
length (as would be the 
case for a prismatic beam) 
the unit-stresses acting 
would also be proportional 
to the ordinates of a 
straight line (this is the 
case in Fig. 301). 

But in the present case 
these fiber-lengths are un- 
equal, so that the unit- 
stresses in action are (in 
general) proportional to 
the ordinates of a curved line. Such a curved line we note in 
vCi Fig. 306, the ordinates between which and the horizontal 
line hi represent the unit-stresses, p, acting on the upper ends 




Fig. 306. 



FLEXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 359 

of the vertical fibers from m' to m. Tlius, the stress on the 
fiber mt is p^ = ei (tension); and that on the other extreme 
fiber, (at m') is p^, = hv (compression). 

If now we compute the average unit-stress p^ =^ P -i- F and 
lay it off, == is, upward from hi, and draw the horizontal 6s, we 
thereby re-arrange the stresses into a uniformly distributed pull 
(or " negative thrust ") p^F ^hs., represented by the rectangle 
hsih, and a stress-couple formed by the ordinates lying between 
the curve and the axis hs. 

It will be noted in Fig. 306 that there is a fiber at some 
point n (on right of g) where the stress is zero ; i.e., the " neu- 
tral axis " of the section is at n, ~1 to paper. Also, at some 
point n', the actual stress is equal to the average, p^, and an 
axis ~] to paper through this point would be the neutral axis if 
the forces acting on this free body, other than the fiber stresses, 
consisted, not of a single force P, but of a couple, with a mo- 
ment = Pa. This axis through n' might be called the neutral 
axis for " pure bending ", since then the whole system of fiber 
stresses would reduce to a couple and the stresses would be 
measured by the ordinates between hs and the curve. 

299. Crane-Hooks. Winkler-Bach Theory. Formula for Stress. In Fig. 
306, let F be the area of the plane figure formed by the section m'm, dF an 
element of this area, and z its distance (reckoned positive toward the right) 
from the gravity axis, g^ of the section. The radius of curvature of gk is r, 
and a is the lever arm of P, the load, about g. Let gm = e and gm' = e' (dis- 
tances of extreme fibers) and let 

/Vr r'=+^ I dF 
S denote the quantity ( t, / ( 

an abstract number depending on the area, shape, and position, of the cross- 
section m'ni ; and upon the radius of curvature r. Its value may be obtained 
by the calculus (or Simpson's Eule) for ordinary cases. For instance, if the 
section is a rectangle of width b, and altitude = A, = m'm, we find 



-1 ; ■ (1) 



'-jh-'^l)-' (^: 



From the Winkler-Bach theory it results that the unit-stress on any fiber 
between m and m', at a distance z from the gravity axis g (on the right, toward 
the center of curvature, 0; if on the left, z is negative) is 

a /. Z 1"' 



J" 

^^F 

I 



r\ r — z S/. 



(3) 



lbs. per sq. inch. A positive result from (3) indicates tension; a negative, com- 
pressive stress. Of course, for P -=- Pwe might write the symbol p^, or " aver- 
age stress." If p were set = zero, a solution of (3) for z would locate the neu- 



360 MECHANICS OF ENGINEERING. 

tral axis, n, of Fig. 306; while by placing p — Pj = 0, a solution for z would 
locate the point n', or neutral axis for "pure bending." 

300. Numerical Example. Let the cross-section be a trapezoid, of base 
6 = 3 in. at m, and upper base 6' = 1 in. at ?/i', both | to paper ; the altitude 
/i, =• m/m, being 4 in. This brings g f in. (= e ) from m and | in. (= e') from 
m'. Let N be in the same vertical as and Om = 2 in. Hence r=a = 2 + |.= 
y in. The material is mild steel and the load P is 8 tons ; find p^ and -pm' . 

From above dimensions we find area ^=8 sq. in,, while from eq. (1), 
(using the calculus), S= 0.0974. For p^ w^ put z = + f in. in eq. (3) ; and for 
Pm', 3 = - I in. ; obtaining, finally, p,« = 17,120 lbs. per sq. in. (tension) ; and 
Pm' = — 7,980 (compression). Evidently the elastic limit is not passed. 

Using the imperfect theory of § 297, we should have obtained pm = 12,000 
lbs. per sq. in., only ; which is seen to be about 30 per cent, in error, compared 
with the above value of 17,120. The reason for taking a low value for the safe 
unit-stress, B', in the example of § 297 is now apparent, an additional reason 
being the fact that loads are sometimes "suddenly applied " on hooks. 

301. By "column" or "long column" is meant a straight 
beam, usaally prismatic, which is acted on by two com- 
pressive forces, one at. each extremity, and whose length 
is so great compared with its diameter that it gives way 
(or " fails ") by buckling sideways, i.e. by flexure, instead 
of by crushing or splitting like a short block (see § 200). 
The pillars or columns used in buildings, the compression 
members of bridge-trusses and roofs, the " bents " of a 
trestle work, and the piston-rods and connecting-rods of 
steam-engines, are the principal practical examples of long 
columns. That they should be weaker than short blocks 
of the same material and cross-section is quite evident, but 
their theoretical treatment is much less satisfactory than 
in other cases of flexure, experiment being very largely 
relied on not only to determine the physical constants 
which theory introduces in the formulae referring to them, 
but even to modify the algebraic form of those formulae, 
thus rendering them to a certain extent empirical. 

302. End Conditions. — The strength of a column is largely 
dependent on whether the ends are free to turn, or are 
fixed and thus incapable of turning. The former condi- 
tion is attained by rounding the ends,' or providing them 
with hinges or ball-and-socket-joints ; the latter by facing 
off' each end to an accurate plane surface, the bearing on 
which it rests being plane also, and incapable of turning. 
In the former condition the column is spoken of as having 



FLEXURE. LONG COLUMNS. 



361 



round ends ; * Fig. 311, (a) ; in the latter as having fixed ends, 
(ov flat bases ; or square ends), Fig. 311, (&). 




Fig. 312. 

Sometimes a coliimn is fixed at one end while the othei 
end is not only round but incapable of lateral deviation from 
the tangent line of the other extremity ; this state of end 
conditions is often spoken of as "Pin and Square," Fig. 
311, (c). 

If the rounding * of the ends is produced by a hinge or 
** pin joint," Fig. 312, both pins lying in the same plane 
and having immovable bearings at their extremities, the 
column is to be considered as round-ended as regards flex- 
ure in the plane 1 to the pins, but as square-ended as re- 
gards flexure in the plane containing the axes of the pins. 

The " moment of inertia " of the section of a column will 
be understood to be referred to a gravity axis of the sec- 
tion which is "I to the plane of flexure (and this corres- 
ponds to the " force-plane " spoken of in previous chap- 
ters), or plane of the axis of column when bent. 

303. Euler's Formula. — Taking the case of a round-ended 
column, Fig. 313 (a), assume the middle of the length as 
an origin, with the axis X tangent to the elastic curve at 
that point. The flexure being slight, we may use the form 
EI (Py-^dx^ for the moment of the stress-couple in any 

* With round ends, or pin ends, it should be understood tliiit the force 
at each end must be so applied as to act through the centre of gravity of the 
base (plane figure) of the prismatic column at that end ; and continue to do 
so as the column b^ nds. 



562 



MECHANICS OF ENGINEERING. 






dp 


dy 


dy 


dx- 


—'r- 


y 


J/c 


dx- 


" / 


/ 




dx 

7 


f 


-a-y. 


1 

if 


4+- 


-Or— 


1 
"1 




\x 

1 

1 

1 
1 

1 

J 




1 

I 

Y 



Fig. 313. 



Fig. 314. 



^section w, remembering tliat with this notation the axis X 
must be || to the beam, as in the figure (313). Considering 
the free body nC, Fig, 313 (h), we note that the shear is 
zero, that the uniform thrust =P, and that 2'(moms.n)=0 
gi'ves (a being the deflection at 0) 



EI 



d'y 



dx^ 



--F{ar-y) 



Multiplying each side by dy we have 
El 



dx" 



dy (Fy=Pa dy — Fy dy 



(1) 



(2) 



' Since this equation is true for the y, dx, dy, and d^y of any 
element of arc of the elastic curve, we may suppose it 
written out for each element from where ?/=0, andc''y=0, 
up to any element, (where dy=dy and y=y) (see Fig. 314) 
and then write the sum of the left hand members equal to 
itliat of the right hand members, remembering that, since 
dx is assumed constant, l-^dsc^ is a common factor on the 
left. In other words, integrate between and any point 
of the curve, n. That is. 



f[dy]d[dy] =Fa f dy—P T ydy (3) 
The product dy d^y has been written {dy)d(dy\ (for d^y m 



EI 

da? 



FLEXURE. LONG COLUMNS. 363 

the differential or increment of dy) and is of a form like 
xdx, or ydy. Performing the integration we have 

EI d_l y^ .... (-1) 

dx' 2^2 ■ ^ 

which is in a form applicable to any point of the curvej 
and contains the variables x and y and their increments 
dx and dy. In order to separate the variables, solve for dxy 
and we have 

di 



dx=l^-JL==^OTdx=^ I EI, \aJ ... 



d(y) 

'^ (X \C(/ / 

i.e.,a!=±y-p- (vers, sin ^^j , , . (6) 

(6) is the equation of the elastic curve DOG^ Fig. 313 (a), 
and contains the deflection a. If P and a are both given, 
y can be computed for a given cc, and vice versa, and thus 
the curve traced out, but we would naturally suppose a to 
depend on P, for ineq. (6)whena7=^Z, y should —a. Mak- 
ing these substitutions we obtain - ^ 

'A^= V^ (^^^"- ^^^ "' ^-^^^ ' ^•^- >^^= 7^ I ^^^ 

Since a has vanished from eq. (7) the value for P ob- 
tained from this equation, viz.: 

i\=EI ^ .... (8) 

is independent of a, and 

is ,\ to be regarded as that force (at each end of the round' 

ended column in Fig. 313) which will hold the column at 

any small deflection at which it may previously have been 

set. 



364 



MECHANICS OF ENGINEERING. 



In other words, if the force is less than Pq no flexure at 
all will be produced, and hence P,, is sometimes called the 
force producing " incipient flexure." [This is roughly ver- 
ified by exerting a downward pressure with the hand on 
the upper end of the flexible rod (a T-squai e-blade for in- 
stance) placed vertically on the floor of a room ; the pres- 
sure must reach a definite value before a decided buckling 
takes place, and then a very slight increase of pressure oc- 
casions a large increase of deflection.] 

It is also evident that a force slightly greater than P^ 
would very largely increase the deflection, thus gaining for 
itself so great a lever arm about the middle section as to 
cause rupture. For this reason eq. (8) may be looked 
upon as giving the Breaking Load of a column with round 
ends, and is called Euler^ s fornfiula. 

Referring now to Fig. 311, it will be seen that if the three 
parts into which the flat-ended column is di- 
vided by its two points of inflection A and B 
are considered free, individually, in Fig. 315, 
the forces acting will be as there shown, viz.: 
At the points of inflection there is no stress- 
couple, and no shear, but only a thrust, =P, 
and hence the portion AB is in the condition 
of a round-ended column. Also, the tangents 
to the elastic curves at and G being pre- 
served vertical by the f rictionless guide-blocks 
and guides (which are introduced here simply 
as a theoretical method of preventing the ends 
from turning, but do not interfere with verti- 
cal freedom) OA is in the same state of flex- 
ure as half of AB and under the same forces. 
Hence the length AB must = one half the 
total length I of the flat-ended column. In 
other words, the breaking load of a round- 
ended column of length =^Z, is the same as 
that of a flat-ended column of length —I. 
Hence for the I oi eq. (8) write %l and we 
have as the breaking load of a column with 
flat-ends and of length =1. 




}il 



f/MWM 



Fig. 315. 



TLEXURE. LONG COLUMNS. 365 

r.^4.m^ .... (9) 

Similar reasoning, applied to tlie " pin-and-square " 
mode of support (in Fig. 311) where the points of inflec- 
tion are at B, approximately y^ I from G, and at the 
extremity itself, calls for the substitution of ^ I for I in 
eq. (8), and hence the breaking load of a ^'pin-and-square " 
column, of length = I, is 

P^=l ^/^ . . . (10) 

Comparing eqs. (8), (9), and (10), and calling the value of 
Pi (flat-ends) unity, we derive the following statement : 
The breaking loads of a given column are as the numbers 



1 

flat-ends 



9/16 
pin-and-square 



y^ j according to the 

round-ends \ mode of support. 

These ratios are approximately verified in practice. 

Euler's Formula [i.e., eq. (8) and those derived from it, 
(9) and (10)] when considered as giving the breaking load 
is peculiar in this respect, that it contains no reference to 
the stress per unit of area necessary to rupture the material 
of the column, but merely assumes that the load producing 
" incipient flexure ", i.e., which produces any bending at 
all, will eventually break the beam because of the greater 
and greater lever arm thus gained for itself. In the canti- 
lever of Fig. 241 the bending of the beam does not sensibly 
affect the lever-arm of the load about the wall-section, but 
with a column, the lever- arm of the load about the mid- 
section is almost entirely due to the deflection produced. 

It is readily seen, from the form of eqs. (8), (9) and (10), 
that when I is taken quite small the values obtained for Po, Pi, 
and P2 become enormous, and far exceed what would be 
found from the formula for crushing load of a short block, 
viz., P = FC (see p. 219), with F denoting the area of section 
of the prism and C the crushing unit-stress of the material. 
The degree of slenderness a column must have to justify the 
use of Euler's relations will appear in the next paragraph. 



36G 



MECHANICS OF ENGINEERING. 



304. Euler's Formula Tested by Experiment. — Since the 
"moment of inertia," /, (referred to a certain axis) of the cross- 
section of the column may be written I = Fk^, where k is the 
"radius of gyration " (see p. 91), and F the area of the plane 
figure, eq. (8), for ''round 1 Pq tz^E 

ends," may be written \ F~~(l^ky ' * 

Here Po^F is the average unit-stress (compressive) on the 
cross-section and l^k is a ratio measuring the slenderness of 
the column. (Of course, when the column actually gives way 
by buckhng, the unit-stress on the concave side at the middle 
of the length is much greater than the average). In the ex- 
periments by Christie, described on p. 112 of the Notes and 
Examples, the value of the ratio l-i-k ranges from 20 to 480. 

As an example consider a 3"x3"Xi" angle-bar (or "angle") 
of wrought iron, with Z = 15 ft., to be 
used as a column. Fig. 315a shows 
the cross-section of this shape, with di- 
mensions. Q is the center of gravity 
of this plane figure. Let the force be 
applied at each end of the column 
according to Christie's mode of "round 
ends," i.e., by a ball-bearing device. Fig. 3l5a. 

the force always passing through the point C of the section at each 
extremity of the column. Since the ends are free to turn in any 

plane, the axis 
of the column 
will deflect in 
the plane CN 1 
to the axis 2 ... 2 
(of the plane 
figure) about 
which the values 
of / and of k are 
least. For this 
shape, we find 
from the hand- 
book of the Cam- 
bria Steel Co., 
that k about 
axis 2 ... 2 is the least radius of gyration and =0.58 in. ; also that 





FLEXURE. LONG COLUMNS. 367 

the area of the figure is F = 2.75 sq. in. Hence the "slender- 
ness-ratio-;' l^k, is 180" -^ 0.58" = 310; and from eq. (11) we 
have, with E for wrought iron taken as 25,000,000, lbs./in.2 
(p. 279), 

I (Po -rF)=Ti^X 25,000,000 -^ (310)2 = 2570 lbs. / in.2 ; 
while from the Christie experiments we find (Po^P)=2650 
lbs. /in.2 as the average unit-stress at rupture; a fairly close 
agreement with the Euler result. The total rupturing load 
would then be Po = 2570X2.75 = 7070 lbs., and the safe load, 
with the "factor of safety " of 8 recommended in the Christie 
report, would be 884 lbs. 

In this way it may be ascertained that for values oi l^k 
from 200 to 400 for "round ends " and from 300 to 400 for 
fixed ends there is an approximate agreement between 
Euler's equations and the Christie experiments. But most 
of the columns used in engineering practice involve values 
oil^k less than 200, so that Euler's formulae are not adapted 
to actual columns (though used to some extent in Germany). 
A formula of such nature as to be available for all degrees 
of slenderness has therefore been established (Rankine's, 
see next paragraph), based partly on theory and partly on 
experiment, which has obtained a very wide acceptance 
among engineers. 

In Fig. 3156 is shown a curve, Er, resulting from plotting as abscissa 
and ordinate the values of Po-Hi^ and Z-h A:, as related in Euler's formula 

(8) for columns with round ends, for "medium" structural steel; with 
£' = 30,000,000 lbs./ in- ^ Ej is a similar curve plotted from Euler's formula 

(9) for fixed ends for the same material. Each of these Euler curves is 
tangent to both axes at infinity. The other curves wUl be referred to later. 

305. Rankine's Formula for Columns. — The formula of this 
name (some times called Gordon's, iu some of its forms) has 
a somewhat more rational basis than Euler's, in that it in- 
troduces the maximum normal stress in the outer fibre and 
is applicable to a column or block of any length, but stili 
contains assumptions not strictly borne out in theory, thus 
introducing some co-efficients requiring experimental de- 
termination. It may be developed as follows : 

Since in the flat-ended column in Fig. 315 the middle 
portion AB, between the inflection points A and B, is 
acted on at each end by a thrust = P, not accorapanied by 
any shear or stress-couple, it will be simpler to treat thai 



368 



MECHANICS OF ENGINEERING. 



p., 



portion alone Fig. 316, (a), since the thrust and stresa- 
couple induced in tlie section at 
R, the middle of AB, will be equal 
to those at the flat ends, and G, 
in Fig. 315. Let a denote the de- 
flection of R from the straight line 
AB. Now consider the portion 
AR as a free body in Fig. 316, (b), 
putting in the elastic forces of the 
section at R, which may be clas- 
sified into a uniform thrust = 
PiF, and a stress couple of moment Fiq. sie. 

294). (The shear is evidently zero, from 





= L:_, (see 
e 



I (hor comps.) = 0). Here p^ denotes the uniform pres- 
sure (per unit of area), due to the uniform thrust, and jpg 
the pressure or tension (per unit of area), in the elastic 
forces constituting the stress-couple, on the outermost 
element of area, at a distance e from the gravity axis (~| 
to plane of flexure) of the section. F is the total area of 
the section. / is the moment of inertia about the said 
gravity axis, g 

1 (vert, comps.) = gives P == p^F , . (Tj 
2' (moms.j,) = gives Pa = ■^— .... (2) 



For any section, n, between A and R, we should evidently 
have the same^j as at R, but a smaller pi, since Py < Pa 
while e, /, and F, do not change, the column being pris- 
matic. Hence the max. (pi+JJa) is oil the concave edge at 
R and for safety should be no more than G -^ n, where G 
is the Modulus of Crushing (§ 201) and w is a " factor of 
safety." Solving (1) and (2) for j9i and^gj and putting their 
sum = C -V- %, we have 



P.Pae G 



(3) 



We might now solve for P and call it the safe load, biat a§ 



FLEXURE. LONG COLUMNS. 369 

is customary to present the formula in a form for giving 
the breaking load, the factor of safety being appHed after- 
ward. Hence, we shall make n=l, and solve for P, calling 
it then the breaking load. Now the deflection a. is unknown, 
but may be expressed approximately, as follows, in terms 
of e and l. 

If we now consider ARB to be a circular arc, of radius = |0, 
we have from geometry (similar triangles) a=(Z-^- 4)^^2/9; 
and if we equate the two expressions for the moment of the 

EI V2I 
stress-couple at R there results — = — (see pp. 249 and 250) . 

A combination of these two relations gives ae=(p2^S2E)P, 
Now under a safe load the total stress, pi + p2, in the outer 
fibre (concave side) at R will have reached a safe value, R', 
for the material, and is therefore constant for this material, 
and if the rude assumption is made that the portion p2 of 
this stress is also constant, it follows that the fraction (p2 h- S2E) 
= a constant; which may be denoted by /?, (an abstract number). 
Let us also write, for convenience, I = Fk^, (k being the radius 
of gyration of the cross-section about a (gravity) axis through 
^ 1 to paper). Hence finally, we have, from eq. (3), 

Breaking load 1 FC 
forflatends J ^^r+^5(m)2 • • • (4) 

By the same reasoning as in § 303,' for a round-ended 
column we substitute 21 for I; for a column with one end round 
and the other '^fiat " or ''fixed " (i.e., for a " pin-and-square " 
column), ^l for I; and obtain 

Breaking load for a round- 1 FC ^ 

ended column \^^^TTW(hW' • • • (^) 

Breaking load for a ''pin- 1 FC 

and-square " column J ^^^l + 1.78/9(Z-^/c)^' ' ' • ^^^ 

Each of these equations (4), (5), and (6), is known as Ran- 
kine's Formula, for the respective end-conditions mentioned. 
They find a very extended use among engineers in English- 
speaking countries; with some variation, however, in the 



370 



MECHANICS OF ENGINEERING. 



numerical values used for quantities C and /?, which are con- 
stants for a given material; and also in the fraction of the 
breaking load which should be taken as the safe, or working, 
load (the reciprocal of this fraction being called the "factor 
of safety,") =n. A set of fair average values for these con- 
stants, as recommended by Rankine and others, is here pre- 
sented : 





Hard 

steel. 


Medium 
Steel. 


Soft 
Steel. 


Wrought 
Iron. 


Cast 
Iron. 


Timber, 


C (lbs./in.2) 


70,000 


50,000 


45,000 


36,000 


70,000 


7,200 


/? (abstract number) .... 


1 


1 


1 


1 

36,000 


1 


1 


25,000 


36,000 


36,000 


6,400 


3,000 



The factor of safety, n, usually employed with the fore- 
going formulae and constants, is n = 4 for wrought iron and 
steel in quiescent structures; and 5 under moving loads, as 
in bridges; while n = 10 should be used for timber and 8 for 
cast iron. 

In Fig. 315?) are two dotted curves, plotted for round ends 
(Rr) and fixed ends (Rj) in the case of medium steel; the above 
equations (Rankine), with the above values of C and /?, having 
been used. The ''slenderness ratio," l^k, i& the abscissa; 
and Po^F, or Pi^F, (the average breaking unit-stress), is 
the ordinate, of any point. These curves may now be com- 
pared with the Eule'r curves, E^. and £/, (in the same figure), 
.already mentioned as having been plotted for structural steel 
(of modulus of elasticity £? = 30,000,000 lbs./in.2) 

306. Examples; under the Rankine Formulae. — Example 1. 
Let it be required to compute the breaking load of awrought- 
iron solid cylinder, used as a column, of length 1 = 8 ft. and 
diameter, =d, =2.4 inches; with round ends, i.e., the pressure 
acting at each end at the center of the circular base, the ends 
being free to turn in any direction. 

The "end conditions " call for the employment of the 
''least k," but here k is the same for any gravity axis of the 
circular section. That is we have 

A;2 = / ^^ = 17,^4 ^ ;,^2_i ^2 _ 1(1, 2)2 _ 0.36 in.2; .-. /(; = 0.6 in. 



FLEXURE. LONG COLUMNS. 371 

and (Z-^A;) = "slenderness-ratio " = 96-j-0.6 = 160. Hence from 
eq. (5) 

■KT^C 1 

Po = j^pX^Tieop' ^^^^ '^^ 36 000 ^^^ <^ = 36,000 lbs./in.2; Le., 
;r( 1.2)236,000 162,800 ,<, .nn ik 

It is seen that, on account of the degree of slenderness of 
the column, the breaking load is about one quarter of what 
it would be for a short prism of same section. 

With a factor of safety of 5 we should take 5 of 42,300, 
i.e., 8460 lbs., as safe load. 

Example 2. — It is required to compute the diameter, d, . 
of a solid cast-iron cylinder, 16 ft. in length, to serve as a 
column with fiat ends, whose safe load is to be 6 tons, the 
factor of safety being 6. This calls for the use of eq. (4) in 
which we put Pi = 6x12,000 = 72,000 lbs., the required break- 
ing load ; with C = 70,000 lbs. / in.2 and /3 = 1 -^ 6400. The least 
radius of gryation should be used, but in this case the k"^ is 
constant for all axes of the section, viz., k'^ ^\7tii^ ^nr^ ^d^ ^IQ. 

Hence from eq. (4) we have (for inch and pound) 

„ \Kd^C 54,980^2 ^onnmu 

^' = l + 7ra-/cl2 =- r^ = 72,000 lbs. 

This on reduction leads to the bi-quadratic equation 
#-1.309^2 = 120.7; 
which being solved for d^ gives d2 = o.645± 11.01. The upper 
sign being taken we have, finally, c? = 3.41 in. as the required 
diameter. 

The "slenderness ratio," therefore, proves to be 192-^0.85 
= 225, which though seemingly high is not extreme for a flat- 
ended column; corresponding, as it does, to 112 for a round-'^ 
ended column. 

Example 3. — A prism of medium steel, of uniform rec- 
tangular section (solid) with dimensions 6 = 3 in. and /i = l in., 
is to be subjected to a thrust (connecting-rod of a steam- 
engine). Its ends are provided with pins (see Fig. 312) capable^ 
of turning in firm bearings, the axis of each pin being T to 



372 MECHANICS OF ENGINEERING. ' 

the "b" dimension of the rectangular section. The length 
between axes of pins, is Z = 6 ft. It is required to find the 
breaking load by the Rankine formula). 

Since the end conditions would be ''round-ends" if the 
axis of the column were to bend in a plane T to the axes of 
the pins (as in Fig. 312), but ''flat-ends" [Fig. 311(6)] in 
case it bent in the plane containing the axes of the pins; 
and since the k of the section is different for the two cases, 
it will be necessary to make each supposition in turn and 
take the smaller of the two results for breaking load (i.e., 
as the one to which the factor of safety should be applied). 

For round-ended buckling the value of k^ is I^F = 
[hh^^l2]^hh = 0.75 in.^; and, with the values of C and ^ 
for medium steel, we have from eq. (5), 

p 50,000X3.0 150,000 ^,^^^,. 

^36,000' 0.75 
while for flat-ended buckling, in the other plane, the P 
to be used would be A;^ = [6/^3^ 12] -6/i = 0.0833 in.2, and 
hence from eq. (4) 

p 50,000X3.0 150,000 -, ^.q lu 

^^36,000' 0.0833 
It is seen that Pi is smaller than Pq, so that with a factor 
of safety of 6 we have for the safe, or working, load, I of 
54,933, =9,155 lbs. 

307. Radii of Gyration. — The following table, taken from 
p. 523 of Eankine's Civil Engineering, gives values of ^'^, 
the square of the least radius of gyration of the given cross- 
eection about a gravity-axis. By giving the least value oi 

h^ it is implied that the plane of flexure is not determined 
by the end-conditions of the column (i. e., it is implied 
that the column has either flat ends or round ends). If 
either end (or both) is a pin-Joint the column may need to 
be treated as having a flat-end as regards flexure in a plane 
containing the axis of the column and the axis of the pin, 
if the bearings of the pin are firm ; while as regards flexure 
in a plane perpendicular to the pin it is to be considered 
round-ended at that extremity. 



FLEXURE. LONG COLUMNS. 



373 



In the case of a " thin cell " the value of h"^ is strictly 
true for metal infinitely thin and of uniform thieJcness ; still, 
if that thickness does not exceed ^ of the exterior diame- 
ter, the form given is sufficiently near for practical pur- 
poses; similar statements apply to the branching forms. 



f f^mm 



h i 



h 



<—-h > 



(a) (5) 



wmmm, 




Fig. 317. 








*— -.. 







\W 



(6) (cj 

Fia. 818. 




Solid Eectangle. 

%— least side. 

Thin Squfire Cell. 

Side— In. 

Thin Kectangular Cell. Yia 317 fc> -" /i^ /i+36 

h^=- least side. 

Solid Circular Section. 

Diameter —d. 

Thin Circular Cell. 

Exterior diam. = d. 



Fig. 317(a). ]z''=l^}i^ 
Fig. 317(6). A;2 = |/i2 

p = 

12'/i+6 

Fig. 317 (c?). p^i^2 

Id 

Fig. 317(e). A;2 = Jd2 



Angle-Iron of Equal j^ig. 317^^) A;2=^-62 

ribs 



F: 



62^* 



A^ngle-Iio-n of unequal , -^-g g^g^^^^ ^= 12(F+3?) 

Cross of equal arms. Fig. 318 (6). ^=4^' 

I-Beam as a pillar. 

Let area of web =5, j.- 313 (c). F= -. . -^-j-^ 

« « &o^A flanges & ^ ^ 12 A-^-H 

=A. 
Channel ^ig. 318(^). ^=^^^ [l2T^)+iT^^] 

Let area of web =B; of flanges =A (both). ^ extends 
from edge of flange to middle of web. 



374 



MECHANICS OF ENGINEERING. 



308. Built Columns. — The "compression members" of 
bridge trusses, and columns in steel framework buildings are 
generally composed of several pieces of structural steel riveted 
together, each column being thus formed of a combination of 
plates, channels, angles, Z-bars, etc. In Figs. 319 and 320 





PHCENIX COLUMN. 



Fig. 319. ^^**- 

are shown examples of these compound shapes. The Phoenix 
column is seen to consist of four quadrantal segments riveted 
together. In Fig. 319 is a combination of two channels and 
one plate, these three pieces being continuous along the whole 
length of the column. On the side opposite to the plate are 
seen lattice bars, arranged in zig-zag, which serve to stiffen 
the column on that side. The center of gravity of the cross- 
section of this column is nearer to the edge carrying the plate 
than to the lattice edge; and if the ends of the column are 
provided with pins 1 to the webs of the channels the axis of 
each of these pins should be so placed as to contain the center 
of gravity of the cross-section of the column at that point. 

The handbooks of the various steel companies present 
formulae and tables enabling the breaking loads to be found 
for their various designs of built columns, and for single I-beams 
used as columns. For example, the tables given in the hand- 
book of the Cambria Steel Co. for built columns of "medium 
steel " are stated to be computed from the following formulae 
(which are evidently of the Rankine type). 

The breaking load for a column of length I and with cross- 
section of area F and least radius of gyration k is (in pounds) : 

Square Bearing, Pin and Square Bearing. Pin Bearings. 

50,000i^ „ 50,000i^ _ 50,000i^ 



Pi = 



1 + 



36,000 



W 



P2 = 



1 + 



24,000 



ar 



Po = 



1+ 



18,000VA: 



FLEXURE. LONG COLUMNS. 375 

In these formulae I and k should be in the same unit (both 
feet, or both inches; since (l^k) is a ratio) and the proper 
k to be used for the case of "pin and square bearing " (i.e., 
one end provided with a pin and the other with a square 
bearing) should be ascertained as in example 3, p. 371. 
To obtain the total safe load for the column: "For quiescent 
loads, as in buildings, divide by 4. For moving loads, as in 
bridges, divide by 5." 

Considerable variety will be found among the formulae 
of the Rankine type proposed by different engineers as best 
satisfying the results of experiment. For accounts of ex- 
periments beyond those already quoted in the author's 
"Notes and Examples in Mechanics," the reader is referred 
to special works. Kent's Pocket Book for Mechanical 
Engineers contains much valuable matter on the subject 
of columns. The handbooks of the Carnegie Steel Co., the 
Pencoyd Iron Works, and the Phoenix Iron Co., give ex- 
tensive data relating to steel columns. Osborne's Tables of 
moments of inertia and radii of gyration of compound sec- 
tions is a valuable book in this connection. 

309. Moment of Inertia of Built ColumiL Example.— It is pro- 
posed to form a column by joining two I-beams by lattice- 
work, Fig. 321, (a). (While the lattice-work is relied upon 
to cause the beams to act together as one piece, it is not 
regarded in estimating the area F, or the moment of iner- 
tia, of the cross section). It is also required to find the 
proper distance apart = x, Fig. 321, at which these beams 
must be placed, from centre to centre of webs, that the 
liability to flexure shall be equal in all axial planes, i.e. 
that the 1 of the compound section shall be the same 
about all gravity axes. This condition will be ful- 
filled if Iy can be made ~i^* (§89), being the centre 
of gravity of the compound section, and X perpendicular 
to the parallel webs of the two equal I-beams. 

Let F' = the sectional area of one of the I-beams, Fx 
Tsee Fig. 321(a) its moment of inertia about its web-axis, 
that about an axis ~[ to web. (These quantities can 1)8 

* That is, with flat ends or ball ends ; but with pin ends, Fig. 313, if the 
pin is II to X. put 4/y = Ixi if II to Y, put 47x = Ir . 



376 



MECHANICS OF ENGINEERING. 



found in tlie hand-book of the iron company, for each size 
of rolled beam). 
Then the 

total 7x = 2rx ; and total I^ = 2ri'v -f W-Yl 

(see §88 eq. 4.) If these are to be equal, we write them so 
and solve for a?, obtaining 



X 



V jr, 



(1) 



310. Numerically; suppose each girder to be a 10}4 inch 
light I-beam, 105 lbs. per yard, of the N. J. Steel and Iron 
Co., in whose hand-book we find that for this beam I'x = 
185.6 biquad. inches, and I'r = 9.43 biquad. inches, while 
F' = 10.44 sq. inches. "With these values in eq. (1) we 
have 



„^J± (185.6-9.43) Vera = 8.21 inchea. 

V 1 0.4-4. 






n^ 



V 

7^ 



■^ 



r'l^ 



-a;— H 



^ 



iai 



^ 



^ 



^^ 



-P- 



^11:=. 



(6) 



Fig. 331. 



The square of the radius of gyration will be 

F=2Px-^2i^'= 371.2 -^20.88=17.7 sq. in. . (2) 

and is the same for any gravity axis (see § 89). 

As an additional example, suppose the two I-beams united 
by plates instead of lattice. Let the thickness of the plate 
■= t, Fig. 321, (&). Neglect the rivet-holes. The distance 
a is known from the hand-book. The student may derive 
a formula for x, imposing the condition that (total /x)= /y- 



FLEXURE. LONG COLUMNS. 377 

310a. Design of Columns. — General considerations governing 
economy and efficiency in the design of built columns are 
that the various pieces, besides being continuous for the whole 
length, should be placed as far from the axis of the column 
as possible, in order to increase the value of k the (least) radius 
of gyration, thus leading to a larger value of the safe load for 
a given amount of material, or to a minimum amount of material 
for a given required safe load; and that the parts should be well 
fastened together by rivets, preventing all relative motion. The 
economy secured by placing the material as far from the center 
as possible also holds, of course, for single pieces used as columns. 
For example, if the safe load of a hollow cylindrical cast-iron 
flat-ended column, 20 ft. long, is to be 40 tons, i.e., 80,000 lbs., 
and the thickness of metal is not to be less than \ in., we find, 
after a few trials with Rankine's formula eq. (4), p. 369, taking 
a factor of safety of 8 (so that the breaking load would be 
640,000 lbs.) that an outside diameter of d = 8 in. is the largest 
permissible. Thus, taking the least k"^, {^(F^8), from p. 373, 
for a thin cylindrical cell, with Z = 240 in., with the sectional 
area, F, as the quantity to be solved for, we have 

\^'^^^.94QN2 =640,000 lbs.; .-. i^ = 19.43 sq. in. 

1 + : 



6400 [82-8] 
Let ^2 denote the internal diameter of the section; then 

j(82 — d22) = 19.43; whence ^2 = 6.26 in.; i.e., the thickness 

of metal==^(d — ^2) =0.87 in., or practically | in. 

310b. The Merriman-Ritter-Formula for Columns was de- 
rived independently by Professors Merriman and Ritter (see 
Engineering News, July 19, 1894) and has a mathematical 
basis as follows. In Fig. 315& curves have been plotted for 
the Euler and Rankine formulae for medium steel, both for 
flat and round ends; and it is seen that each of the Rankine 
curves is tangent to the horizontal line through V and is roughly 
parallel to, and not very distant from, the corresponding 
Euler curve on the extreme right. Professor Merriman de- 
rives the equation (of the same form as Rankine's) for a curve 
which has a horizontal tangent at V, and is exactly tangent 



378 MECHANICS OF ENGINEERING. 

to the Euler curve at some point on the extreme right (at 
infinity, in fact) and thus secures a more rational value for 
the constant called ^ in Rankine's formula. 

With P' denoting the safe load for the column and C the 
safe compressive unit-stress for the material, this 

formula may be written . . . P' = — Tvrpp^i • • '^ • (M) 



where C" denotes the unit compressive stress at elastic limits 
E the modulus of elasticity, F the sectional area, and n an 
abstract number whose value (as before, in the Rankine for- 
mulae) is 1, 16/9 (or 1.78), and 4, for flat ends, pin-and-square, 
and round ends, respectively. 

If for Q we write P, the breaking load, and correspondingly 
C for C, and plot values oi P-^F and l-^k, the curve would 
not differ greatly from the Rankine curve in Fig. 120 for medium 
steel; and similarly for wrought iron; but for timber and cast 
iron the variation is considerable, and hence Prof. Merriman 
does not recommend the use of his formula for the latter two 
materials. (Crehore's formula differs from the above only 
in replacing C" by C.) 

310c. The "Straight-Line Formula."— It will be noticed 
that in Fig. 315& the straight line connecting points A and C 
(medium steel, round ends) or A' and C (medium steel, flat 
ends) would not vary widely from the Rankine curve, so that 
on account of its simpKcity, when restricted to proper Hmiting 
values of the ratio l-^k,& straight line, or hnear relation, between 
the quantity P-^F and ratio l-i-k was proposed by Mr. T. H. 
Johnson (see Transac. Am. Soc. C. E., 1886, p. 530) for the 
breaking loads of columns of various materials. Among them 
are the following : 

Wrought iron: Hinged ends, Po = [42,000- 157/'^]li^; 

" Flat ends. Pi = [42,000- 128(|-)lP; 

Mild steel : Hinged ends, Po = [52,000 - 220(77)]^; 

" " Flat ends, Pi = [52,000 -179(^)lp. 



FLEXURE. LONG COLUMNS. 379 

In these formulae Pq, or Pi, is breaking load in lbs., F= 
sectional area (in sq. in.), Z = tlie length, and k is the least 
radius of gyration of the cross-section for flat ends (as for 
hinged ends, see example 3, § 306) ; I and k in same unit. 

310d. The J. B. Johnson Parabolic Formula for Columns. 
— If in. Fig. 315a a parabola be plotted with its axis vertical 
(and downward) and vertex at the point V of the two Rankine 
curves, and also made tangent to the Euler curve for the end 
conditions concerned, the points on such a curve for values of 
l^k between zero and the point of tangency to the Euler curve 
are found to agree fairly well with experiment; and the corre- 
sponding formula, or the equation to the curve, is of much 
simpler form than that of the Rankine types, being almost 
as simple as the straight line formula. Such a formula was 
proposed by the late Prof. J. B. Johnson, those for mild steel 
and wrought iron being given below (breaking load in lbs.). 
Mild steel: 

Pin ends, Pq = [42,000 -0.97(^HF; U not >150 

Flat ends. Pi = [42,000- 0.62(^ MP; 1^ not >190 
Wrought iron: 

Pin ends, Po = [ 34,000 -O.erQHp; (^ not >170 

Flat ends. Pi = [34,000 -0.43(^ MP; U not >210 

The notation is the same as in the preceding article. The 
limiting values mentioned for l^k refer to the points of tan- 
gency with the Euler curve. In Fig. 3156 the curve FTT^A^ 
is a parabola fulfilling the above mathematical condition for 
medium steel, with flat ends. 

311. Solid Wooden Columns and Posts. Formula of U. S. 
Dept. of. Agriculture, Division of Forestry. — This formula was 
derived by Johnson from the results of experiments sonducted 
by the Division of Forestry and appUes to solid wooden columns 
provided with '' square ends," the constraint due to which, 
however, is not to be considered as fully equivalent to that 
of "fixed ends." The breaking load being denoted by Pi, 



380 MECHANICS OF ENGINEERING. 

the sectional area by F, the ratio of length I to the ^^ least 
dimension,'^ d,' oi the cross section, by m (i.e., l^d = m), and 
the unit crushing stress for the material by C, the formula is 

J700 + 15m)FC 
^ 700 + 15m + m2 ^^ 

The values of C to be used for different kinds of timber 
are given as follows : 

White oak and Georgia yellow pine 5000 Ibs./in.^ 

Douglas fir and short-leaf yellow pine 4500 '' 

Red pine, spruce, hemlock, cypress, chestnut, CaU- 

fornia redwood, and Cahfornia spruce 4000 ' ' 

White pine and cedar 3500 ' ' 

The fraction of Pi to be taken as the safe load depends 
on the wood and the degree of moisture present, four classes 
being designated in this respect; from Class A (18 per cent 
of moisture; timber exposed to weather), to Class D (10 per cent;, 
timber at all times protected from the weather). For yellow 
pine the safe load should be from 0.20Pi for Class A to O.SlPi 
for Class D. For all other timbers, from 0.20Pi for- Class A 
to 0.25Pi for Class D. 

312. Column under Eccentric Loading. — In Fig. 322 let the load P be 
applied at i, at a distance or "eccentricity" =c from the center of gravity 
1^1 A oi the upper base of the column, the reaction at 

j^\ j the. other end (at k) having an equal eccentricity 

yj i from B; the ends of the column being free to turn. 

^^ / j (In an extreme case Ai and Bk might be brackets 

/ [ "^ fastened to the ends of the column.) 

/ . I AOB is the elastic curve, or bent condition of the 

1^1 \ j^ axis of the column, originally straight. With as 

)"r" I I origin, any point n in the elastic curve has a vertical 

I I I co-ordinate x and a horizontal co-ordinate y. The 

\ I I unknown lateral deflection of the point from AB 

\ I T is a. With n cs any point in the elastic curve, and 

\i I nAi as free body, we have for the moment of the 

■^ Bj I stress couple in section at n £'/[d2?/^dx^] = P(c -I- a—?/);. 

!^_c_J which is seen to differ from eq. (1) of p. 362 only in 

I having the constant c + a in place of the constant a. 

Fig. 322. y^^ ^^j therefore use eq. (6) of p. 363 for the present 

case, after replacing a by c + a; and hence, denoting 's/P^EI by h, remem- 
bering that vers. sin. = 1 — cos, we may write, as the equation to the elastic 

curve, y=(c + a)[l-cos (6x)] (1) 

For x = ^l, y should = the deflection o; on substituting which values in. 
(1) there results finally 



FLEXURE. LONG COLUMNS. 



381 



o=c sec (-^1—1 . . (2); and c+o=c- sec (-^j. • 
Hence the moment of the stress couple at is M^ = P{c+ a) = Pc- sec I -^\ 



(3) 

bl\ 



and the unit stress in outer fibre on concave side at is 



p- 



P_ M,e 



P Pc secQftZ) 

"y^ I ■ 



(4) 



(In this case of eccentric loading, then, the deflection a is not indeter- 
minate as was the case in deriving Euler's formula on p. 363. Note that 
^bl is an angle in radians.) 

Example. — Let the value of P be 10,000 lbs., the length of the colunin 
be Z=20 ft. = 240 in., and the cross-section be a square cell [see Fig. 317 (6)] 
4 inches being the side of the outer square ; area F = 7 m? and / = 14.58 in.^ Let 
the eccentricity be c = 2 in., each force P being applied in the middle of a side 
of the 4 in. square. Let £' = 30,000,000 Ibs./in.^; material, medium steel. 
"With this position of the force plane, e = 2 in. 



Here we have hhl 



=*(/^ 



10,000 



) X 240 = 0.5736 radians, corre- 

V 



30, 000, 000 X 14. 58^ 

spending to 32° 52', whose sec. = 1.190; and therefore a = 2X (1.190-1) 
= 0.380 in., and Mo=10,OOOX2X 1.190 = 23,800 in.-lbs. Finally 



■p-. 



10,000 23,800X2 

■ + - 



= 1430 + 3265 = 4695 lbs./ in. ^ 



7 14.58 

With P= 20,000 lbs., we should obtain iW= 0.811 radians (46° 30'), 
a=0.906 in., Mo = 57,120 in.-lbs., and p = 2860 + 7835 =10,695 Ibs./in.^ 

This latter unit stress is seen to be only moderate in value for the mate- 
rial, leading to the conclusion that 20,000 lbs. for P is a safe load; but on 
account of the possible original lack of straightness in the column, and of 
lack of homogeneity, both of which causes might increase a and Mq, it 
would be better to limit the load to 15,000 lbs. ; considering, also, the fact 
that Rankine's Formula for round ends (with a safety factor of 4) applied 
to this column for the case of no eccentricity would give about 22,000 lbs. 
as safe load. 




Fig. 322a. 

318. Beam or Column with Eccentric End Pressures and also under Uniform 
Transverse Loading. — For example, in Fig. 322a let AB he the -bent axis 
of a beam, or column (originally straight), the longitudinal forces P and 
P being applied at an eccentricity c from A and B, while there is at the 
same time a vertical loading W, =wl, uniformly distributed along the 



382 MECHANICS OF ENGINEERING. 

whole length at rate of w lbs. per running inch. The reactions of the two 
end supports will therefore be each ^W. The ends of the column are free 
to turn. It is required to find the deflection a, the moment Mq of the couple 
at middle section 0, and the unit stress p on the concave side at O. Take 
the free body iAn, n being any point of the elastic curve AOB, with co- 
ordinates X and y referred to the horizontal and vertical axes through 
as an origin, as shown. Then the moment of stress couple at n is 

EI{d'y^dx') = P{c + a-y) + {i)w{P-4x^) .... (5) 

Since (d^y-r-dx^) is a variable, let us denote i—EI-i-P)(d^y-T-dx^) by u, 
as an auxiliary variable; and eq. (5) wUl now read 

y-u=c+a+[{i)w{P-4:X^)]^P (6) 

Differentiating (6) twice, with respect to x, we have 

d^y d^u w . d^u P w 

= ; that is, — = u-\ — (7) 

dx' dx' P' ' dx' EI P ^^ 

Multiplying (7) by 2du, and denoting P-i-EI by b^ and 2w-^P by h, we 

have by integration, {dx'^ is a constant, x being the independent variable), 

(dM)^-7-(dx)^= — 6^M^ + /iM+C, where C is a constant of integration; and 

hence dx = dM-f- (VC + Zim— 6V), which integrates into 

a;=A.sm-M , — +C^ (8) 

where C is a constant. Transfo rmation of (8) gives 

(Vh' + ACb') sin [b(x-C')]+h = 2b^u (9) 

Eliminating u by aid of eqs. (6) and (9) we have 

2b'y==\/h' + 4Cb'-sin [b(x-C')] + h + 2b\c+a) + (i)b^h{P-4x^) (10) 
from which 

2b\dyldx) = bVh^ + 4C6^ • cos [b {x- C')]-b^hx . . . (11) 
To determine the three constants C, C", and a, we now make use of the 
facts that in (10) when x = 0, y also =0, and for x = ^l, y = a; and that in 
(11) for x=0, dy/dx, must =0. The three equations thus obtained, con- 
taining constants only, enable us to determine C, C, and a, and insert their 
values in (10) ; thus giving us as the equation to the elastic curve AOB, 



2/ = 



(, h \ri-cos (6a: "1 ,, , ,^„^ 

^+26-^jL^os(i60-J"*^'^' ^^'^ 

as also the value of the deflection 

a^[c+{h^2b^)lsec{^bl)-l]-{i^)hV .... (13) 
To find the moment of stress couple, Mq, at 0, we have now only to sub- 
stitute a; = and y = in eq. (5), and for a its value from (13); and thus 
obtain 

[fc+^Ysec(i60-^j (14) 

With F as the sectional area of the cross-section of the (prismatic) column 
(or beam, as it might also be called in this connection), and e as the dis- 
tance of the outer fibre from the gravity axis of the section, we now have 
for p, the stress in outer fibre on concave side at 0, 

V-l^^f (15) 



M„=P 



FLEXURE. LONG COLUMNS. 



383 



Since 6 and h denote VP-i-EI and 2w-^P, respectively, it is seen 

that when w is zero, h is zero and eq. (13) reduces to eq. (2) of the 

previous article. Again, if the two forces P are central, i.e., [applied at 

A and B, we put c = 0; in which case an approximate result may be 

reached by writing for the deflection a the value it would have if the 

5 WP 
end forces P were not present, i.e., ^^ • -^j, as due to the uniform load 

W alone (see p. 260). On this basis the value of M^ is Pa+ (,^)Wl. 

(In case the vertical load on the beam or column in Fig. 322a is a 
single load Q concentrated in the middle at 0, a treatment similar to 
the foregoing may be applied, but is somewhat more complicated. For 
details of such a case the reader is referred to Mecanique AppUquee, 
by Bresse, Tome I. p. 384.) 

314. Buckling of Web-Plates in Built Girders. — In §257 men- 
tion was made of the fact that very high web plates in 
built beams, such as /beams and box-girders, might need 
to be stiffened by riveting " angles " on the sides of the web. 
(The girders here spoken of are horizontal ones, such as 
might be used for carrying a railroad over a short sj^an of 
20 to 50 feet. 

An approximate method of determining whether such 
stiffening is needed to prevent lateral buckling of the web, 
may be based upon Rankine's formula for a long column 
and will now be given. 

In Fig. 323 we have, free, a portion of a bent I-beam, 
between two vertical sections at a distance apart= Ai = 
the height of the web. In such a beam under forces L ^o 
its axis it has been proved (§256) that we may consider 
the web to sustain all the shear, J, at any section, and the 
flanges to take all the tension and compression, which 
form the " stress -couple" of the section. These couples 
and the two shears are shown in Fig. 323, for the two 
exposed sections. There is supposed to be no load on this 
portion of the beam, hence the shears at the two ends are 




Fig. 324. 



384 MECHANICS OF ENGINEERING, 

equal. Now tlie shear acting between eacli flange and tlie 
horizontal edge of the web is equal in intensity per square 
inch to that in the vertical edge of the web ; hence if the 
web alons, of Fig. 323, is shown as a free body in Fig. 324,, 
we must insert two horizontal forces = J, in opposite 
directioii^,, on its upper and lower edges. Each of theM 
^ J since we have taken a. horizontal length hi = height 
of web. In this figure, 324, we notice that the effect oi 
the acting forces is to lengthen the diagonal BD ancj 
shorten the diagonal AG, both of those diagonals making 
an angle of 45° with the horizontal. 

Let us now consider this buckling tendency along ^(7, 
by treating as free the strip ^(7, of small width = \. This 
is shown in Fig. 325. The only forces acting in the direc- 
tion of its length AG&ie the components along AG oi the 
four forces J' at the extremities. "VVe may therefore treat 
the strip as a long column of a length I = hi ^2, of a sec- 
tional area F = bb^, (where b is the thickness of the web 
plate), with a value of F = Yjg 6^ (see § 309), and with 
fixed (or flat) ends. Now the sum of the longitudinal 
components of the two J'.'s &t A is. Q = 2 J' ]/?, V2 

= J' V2 ; but J' itself =■ rr. b j4 bi \'% since the small 

rectangle on which J' acts has an area = b )4 h^ ^2, and 
ihe shearing stress on it has an intensity of (J -r- bh{) per 
unit of area. Hence the longitudinal force at each end of 
this long column iis 

'i-r/ w 

According to eq. (4) and the table in § 305, the safe load 
(factor of safety = 4) for a medium steel column of this form, 
with flat ends, would be (pound and inch) 

ib6i50,000 _ 12,500&&i 
1 1 2/ii2 1 h^ • • (2) 

^36,000* V1262 1 + 1,500' 62 

If, then, in any particular locality of the girder (of medium 
steel) we find that Q is >Pi, i.e. 



FLEXUEE. LONG COLUMNS. 



385 






12,500& 



l+rl,.-'^ 



(pound and inch). 



(3) 



W =40 TONS 



1,500 62 

then vertical stiffeners will be required laterally. 

When these are required, they are generally placed at inter- 
vals equal to hi, (the depth of web), along that part of the 
girder where Q is >Pi. 

Example Fig. 326.— Will stiffening pieces be required in 
a plate 'girder of 20 feet span, bearing a uniform load of 
40 tons, and having a web 24 in. deep 
and I in. thick? 



From § 242 we know that the -|t 
greatest shear, J max., is close to 
either pier, and hence we investigate 
that part of the girder first. 

J max. = iTF = 20 tons -40,000 lbs. 

.'. (inch and lb.), see (3), 

J _ 40,000 
hi 



-10^ — -^ 



Trn 




Fig. 326. 



24 



= 1666.6 



while, see (3), (inch and pound), 
12,500X1 



1 + . 



242 



1270 



(4) 



(5) 



1,500 (1)2 

which is less than 1666.66. 

Hence stiffening pieces will be needed near the extremities 
of the girder. Also, since the shear for this case of loading 
diminishes uniformly toward zero at the middle they will 
be needed from each end up to a distance of ^ of 10 ft. 
from the middle. 



^86 MECHANICS OF ENGLNEEBIlfG. 



CHAPTER Vn. 



UJTEAK ARCHES (OF BLOCKWOKI^. 



815. A Blockwork Arch is a structure, spanning an openicg 
or gap, depending, for stability, upon the resistance to 
compresssion of its blocks, or voussoirs, the material ot 
which, such as stone or brick, is not suitable for sustain- 
ing a tensile strain. Above the voussoirs is usually 
placed a load of some character, (e.q. a roadway,) whose 
pressure upon the voussoirs will be considered as vertical, 
only. This condition is not fully realized in practice, 
unless the load is of cut stone, with vertical and horizontal 
joints resting upon voussoirs of corresponding shape (see 
Fig. 327), but sufficiently so to warrant 
its assumption in theory. Symmetry 
of form about a vertical axis will also 
be assumed in the following treatment. 




316. Linear Arches. — For purposes of 
theoretical discussion the voussoirs of 
Fig. 327 may be considered to become 
Fig. 327. infinitely small and infinite in number, 

thus forming a " linear arch," while retaining the same 
shapes, their depth "1 to the face being assumed constant 
that it may not appear in the formulae. The joints 
between them are "1 to the curve of the arch, i.e., adjacent 
voussoirs can exert pressure on each other only in the 
direction of the tangent-line to that curve. 



LINEAR AKCHES. 



387 



317. Inverted Catenary, or Linear Arch Sustaining its Own 
Weight Alone. — Suppose tlie infinitely smalJ voussoirs to 
have weight, uniformly distributed along the curve, weigh- 
ing q lbs. per running linear unit. The eqiii]ibrium of 
such a structure, Fig. 328, is of course unstable but theo- 
retically possible. Required the form of the curve when 
equilibrium exists. The conditions of equilibrium are, 
obviously : 1st. The thrust or mutual pressure T between 
any two adjacent voussoirs at any point. A, of the curve 
must be tangent to the curve ; and 2ndly, considering a 
portion BA as a free body, the resultant of Hq the pres- 




FiQ. 328. 



Fig. 329. 



Fig. 330. 



sure at B the crown, and T &i A, must balance R the re- 
sultant of the il vertical forces (i.e.,weights of the elementary 
voussoirs) acting between B and A. 

But the conditions of equilibrium of a flexible, inexten- 
sible and uniformly loaded cord or chain are the very 
same (weights uniform along the curve) the forces being 
reversed in, direction. Fig. 329. Instead of compression 
we have tension, while the || vertical forces act toward in- 
stead of away from, the axis X. Hence the curve of equi- 
librium of Fig. 328 is an inverted catenary (see § 48) whose 
equation is 



y+c= 



e -\- e 



. (1) 



See Fig. 330. e = 2.71828 the Naperian Base. The "par- 
ameter " c may be determined by putting x = a, the half 
span, and y= Y, the rise, then solving for c by successive 



388 



MEOHAJSriCS OF ENGINEBKING. 



approximations. The " horizontal thrust" or H^^, is = yc, 
while if s = length, of arch OA, along the curve, the thrust 
T at any point A is 



T=^I Riffs' (2..) 

From the foregoing it may be inferred that a series ot vcui»' 

soirs of finite dimensions, arranged 

so as to contain the catenary curve, 

with joints "I to that curve and of 

equal weights for equal lengths of 

arc will be in equilibrium, and 

moreover in stable equilibrium on 

account of friction, and the finite 

width of the joints ; see Fig. 331. 




FIG. 331. 



318. Linear Arches under Given Loading. — The linear arches 
to be considered further will be treated as without weight 
themselves but as bearing vertically pressing loads (each 
voussoir its own). 

Problem. — Given the form of the linear arch itself, it is 
required to find the law of vertical depth of loading under 
which the given linear arch will be in equilibrium. Fig. 
332, given the curve ABC, i.e., the linear arch itself, re- 
quired the form of the curve MON, or upper limit of load- 
ing, such that the linear arch ABC shall be in equilibrium 
under the loads lying between the two curves. The load- 
ing is supposed homogeneous and of constant depth "^ to 
paper ; so that the ordinates z between the two curves are 
proportional to the load per horizontal linear unit. Assume 
a height of load z^ at the crown, at pleasure ; then required 
the z of any point m as a function of ^ and the curve 
ABC. 




LINEAB ARCHES. 



389 



Practical Solution. — Since a linear arcli under vertical 
pressures is nothing more than the inversion of the curve 
assumed by a cord loaded in the same way, this problenj 
might be solved mechanically by experimenting with a 
light cord, Fig. 333, to which are hung other heavy cords, 
or bars of uniform weight per unit length, and at equal 
horizontal distances apart ivhen in equilibrium,. By varying 
the lengths of the bars, and their points of attachment, we 
may finally find the curve sought, MON. (See also § 343.) 

Analytical Solution. — Consider the structure in Fig. 334 
A number of rods of finite length, in the same plane, are in 
equilibrium, bearing the weights P, P^ etc., at the con- 




FiG. 334. 



Tig. 335. 



necting joints, each piece exerting a thrust T against the 
adjacent joint. The joint A, (the " pin " of the hinge), im- 
agined separated from the contiguous rods and hence free, 
is held in equilibrium by the vertical force P (a load) and 
the two thrusts T and T', making angles = d and d' with 
the vertical ; Fig. 335 shows the joint -4 fi'^e. From 2'(hor«^ 
izontal comps.)=0, we have. 

That is, the horizontal component of the thrust in any ro J 
is the same for all ; call it H^. ,\ 



T^ 



H. 



Bin 



(1) 



390 



MECHANICS OF ENGINEEKING. 



Now draw a line As *i to T' and write 2* ( compons. I to 
As)=0; whence F sin ^'=2^ sin ^, and [see (1)] 



. p_ jgp sin /? 

sm 6 



sm 



(2) 



Let the rods of Fig. 334 become infinitely small a,nd infi- 
nite in number and the load continuous. The length of 
each rod becomes =ds an element of the linear arch, fi is 
the angle between two consecutive ds's, d is the angle be- 
tween the tangent line and the vertical, while P becomes 
the load resting on a single dx, or horizontal distance be- 
tween the middles of the two cZs's. That is, Fig. 336, if 
Y= weight of a cubic unit of the 
loading, P-=yzdx. (The lamina of 
arch and load considered is unity, 
1 to paper, in thickness.) -Ho=a 
constant = thrust at crown ; 
6=6', and sin /3=ds-^p, (since the' 
angle between two consecutive tan- 
gents is = that between two con- 
secutive radii of curvature). Hence 
eq. (2) becomes 



Yzdx = 



Kds 



p BID? 6 



but dx—ds sin 6y 




Fia. 336. 



,\yz- 



H. 



p siii^/? 



(3) 



Call the radius of curvature at the crown. /?o» and since 
there z=Zq and ^*=90°, (3) gives x^qPq~3^', hence (3) may 
be written 






sin^ d 



(4) 



This is the law of vertical depth of loading required. For 
a point of the linear arch where the tangent line is verti- 
cal, sin 6 =0 and z would == oo ; i.e., the load would be in- 



LINEAR ARCHES. 



391 



finitely high. Hence, in practice, a full semi-circle, for in- 
stance, could not be used as a linear arcli. 

319. Circular Arc as Linear Arch. — ^As an example of the 

preceding problem let us ap- 
ply eq. (4) to a circular arc, 
Fig. 337, as a linear arcb. 
Since for a circle p is con- 
stant — r, eq. (4) reduces 
to 




sin^ 6 






(5) 



Fig. 337. 



Hence tlie deptb of loading 
must vary inversely as the cube of tbe sine of the angle d 
made by the tangent line (of the linear arch) with the ver- 
tical. 

To find the depth z by construction.— Having z^ given, C 
being the centre of the arch, prolong Ga and make ob = 
go ; at 5 draw a 1 to Gb, intersecting the vertical through a 
at some point d ; draw the horizontal dc to meet Ga at 
some point c. Again, draw ce "| to Gc, meeting ad m e\ 
then ae= z required ; a being any point of the linear arch. 
For, from the similar right triangles involved, we have 

z„=ab=ad sin 0=ac sin d. sin ^=ae sin d sin d sin d 



ae= — ^— ; i.e., ae=2. Q.E.D. 

mn'd [-gee (5.)] 



320. Parabola as Linear Arch. — To apply eq. 4 § 318 to a 

parabola (axis vertical) as linear arch, we must find values 
of p and po the radii of curvature at any point and the 
crown respectively. That is, in the general formula. 



-M 



dy\ 

dx) _ 



dx 



we must substitute the forms for the first and second dif- 
ferential co-efficients, derived from the equation of the 



392 



MECHANICS OF EI*f G [2f EERIN^G. 







Fig. 338. 



Fig. 339. 



curve (parabola) in Fig. 338, i.e. from x^ =^ 2 py; whence 
we obtain 

~2.,or cot 0,= — ana-^=— 
ax p dor p 



Hence ^=i3^°M=^ 

^ l^P 



cosec. 



I.e. p 



sin^^ 



. (6) 



At tbe vertex d = 90** ,*. />„ = p. Hence by substituting 
for p and p^ in eq. (4) of § 318 we obtain 

g=s^= constant [Fig. 339) (7) 

for a parabolic linear arcli. Therefore tbe depth of homo- 
geneous loading must be the same at all points as at the 
crown ; i.e., the load is uniformly distributed with respect 
to the horizontal. This result might have been antici- 
pated from the fact that a cord assumes the parabolic 
form when its load (as approximately true for suspension 
bridges) is uniformly distributed horizontally. Sae § 46 
in Statics and Dynamics. 



321. Linear Arch for a Given Tipper Contour of Loading, the 
arch itself being the unknown lower contour. Given the 
upper curve or limit of load and the depth z^ at crown, re- 
quired the form of linear arch which will be in equili- 
brium under the homogenous load between itself and that 
upper curve. In Fig. 340 let MON be the given upper 
contour of load, z^ is given or assumed,s' and z" are the 
respective ordinates of the two curves -S^ (7 and MON, 
Required the eqation of BAG. 



LINEAR ARCHES. 



393 




Fig. mo. 



Fig. 341. 



As before, tlie loading is homogenous, so that the 
weights of any portions of it are proportional to the 
corresponding areas between the curves. (Unity thick- 
ness "I to paper.) Now, Fig. 341, regard two consecutive 
ds's oi the linear arch as two links or consecutive blocks 
bearing at their junction w the load dP =y (^z -\- z"} dx in 
which Y denotes the heaviness of weight of a cubic unit of 
the loading. If T and T' are the thrusts exerted on these 
two blocks by their neighbors (here supposed removed) 
we have the three forces dP, T and T', forming a system 
in equilibrium. Hence from IX =0, 



T cos <p = T' cos cp' 



(1) 



and 



1*7=0 gives T' sin cp'— T sin <p = dP ... (2) 



From (1) it appears that T cos f is constant at all points 
of the linear arch (just as we found in § 318) and hence 
.= the thrust at the crown, = Jff, whence we may write 

T=H-^ cos <p and r^E~ cos q)' . . . (3) 

Substituting from (3) in (2; we obtain 

H (tan <f' — tan <p)=dP (4) 

■But tan <p =-^ and tan ip' = ^ "J ^ , {dx constant) 

while dP = y {z' -\- z") dx. Hence, putting for convenience 
H = yo?, (where a = side of an imaginary square of the 



394 MECHANICS OP ENGINEERING. 

loading, whose thickness = unity and whose weight = IT) 
we have. 

^=^'+'"'> <^> 

as a relation holding good for any point of the linear arch 
which is to be in equilibrium under the load included 
between itself and the given curve whose ordinates are «", 
Fig. 340. 

322. Example of Preceding. Tipper Contour a Straight Line.— 
Fig. 342. Let the upper contour be a right line and hor- 
izontal ; then the a" of eq. 5 becomes zero at all points of 
ON. Hence drop the accent of z' in eq. (5) and we have 

dot? €^ 
Multiplying which by dz we obtain 
dz dh 1 



do(? 



zdz (6) 



This being true of the z, dz, d?z and dx of each element of 
the curve O'B whose equation is desired, conceive it writ- 
ten out for each element between 0' and any point m, and 
put the sum of the left-hand members of these equations 
= to that of the right-hand members, remembering that 
a,^ and dx'^ are the same for each element. This gives 

dz=dz z=z 

d^ I « / ** 2 al2 2j 

nJ (te— %/ z=Zo 



adz ^ [ zj .... (7.) 



d 



.'. da; = -T^==«« 






LINEAR ARCHES. 



395 




Fig. 342. Pis 343 

Integrating (7.) between 0' and any point m 

/, 



f =«[:iog..(^+,/(-i)-i) . . (8) 



i.e., fl?=a log. 






D-^]= 



or %= 



gg r i -E.-1 



(8.) 



(9.) 



This curve is called the transformed catenary since we may 
obtain it from a common catenary by altering all the ordi- 
nates of the latter in a constant ratio, just as an ellipse 
may be obtained from a circle. If in eq. (9) a were = z^ 
the curve would be a common catenary. 

Supposing Sj and the co-ordinates x^ and gj of the point 
B (abutment) given, we may compute a from eq. 8 by put- 
ting X =Xi and z = g„ and solving for a. Then the crown- 
thrust H = ya^ becomes known, and a can be used in eqs. 
(8) or (9) to plot points in the curve or linear arch. From 
eq. (9) we have 

(10) 



area 
00' mn 






Fig. 343. 

Call this area, A. As for the thrusts at the different 
joints of the linear arch, see Fig. 343, we have crown- 
thrust = ZT = ^a' . . . ; • - . • (11) 
and at any joint m the thrust 

T^VH'+irAf =rV^^^' .... (12} 



396 MECHANICS OF ENGINEERING. 

323. Remarks. — The foregoing results may be utilized 
with arches of finite dimensions by making the arch-ring 
contain the imaginary linear arch, and the joints 1 to the 
curve of the same. Questions of friction and the resist- 
ance of the material of the voussoirs are reserved for a 
succeeding chapter, (§ 344) in which will be advanced ^ 
more practical theory dealing with approximate linear 
arches or " equilibrium polygons " as they will then be 
called. Still, a study of exact linear arches is valuable on 
many accounts. By inverting the linear arches so far pre- 
sented we have the forms assumed by flexible and inexten- 
sible cords loaded ini the same way- 



GliAPHICAIi STATICS, tS9? 



CHAPTER VrX 



ELEMENTS OE GRAPHICAIi STATICS. 



324. Definition. — In many respects graphical processes 
titve duvantages over tlie purely analytical, whicli recom- 
mend their use in many problems where celerity is desired 
without refiiied accuracy. One of these advantages is that 
gross errors are more easily detected, and another that 
the relations of the forces, distances, etc., are made so 
apparent to the eye, in the drawing, that the general effect 
of a given change in the data can readily be predicted at 
a glance. 

Graphical Statics in the system of geometrical construc- 
tions by which prt^blems in Statics may be solved by 
the use of drafting iixsiruments, forces as well as distances 
being represented in amount and direction by lines on the 
paper, of proper length and position, according to arbi- 
trary scales ; so many fest of distance to the linear inch of 
paper, for example, for distances ; and so many pounds or 
tons to the linear inch of paper for forces. 

Of course results should be interpreted by the same 
scale as that used for the data. The parallelogram of 
forces is the basis of all constructions for combining and 
resolving forces. 

325. Force Polygons and Concurrent Forces in a Plsuae. — If a 

material point is in equilibrium under three forces Pi P, 
P3 (in the same plane of course) Fig. 344, any one of them, 



398 



MECHANICS OF ENGIXEERING. 




as Pi, must be equal and opposite to B the resultant of 
the other two (diagonal of their parallelogram). If now 
we lay off to some convenient scale a line in Fig. 345 = 
Pi and II to Pi in Fig. 344 ; and then from the pointed end 

of Pi a line equal and || to Pg and 
laid off pointing the same ivay, we 
note that the line remaining to 
p close the triangle in Fig. 345 must 
be = and || to Pg, since that tri- 
angle is nothing more than the 
left-hand half -parallelogram of 
Fig. 345. Fig. 344. Also, in 345, to close 
the triangle properly the directions of the arrows must 
be continuous Point to Butt, round the periphery. Fig. 
345 is called a force polygor ; of three sides only in this 
case. By means of it, given any two of the three forces 
which hold the point in equilibrium, the third can be 
found, being equal and 1| to the side necessary to " close " 
the force polygon. 

Similarly, if a number of forces in a plane hold a mate- 
rial point in equilibrium, Fig. 346, their force polygon. 




FiG.344. 





Fig. 347, must close, whatever be the order in which its 
sides are drawn. For, if we combine Pj and P2 into a re- 
sultant Oa, Fig. 346, then this resultant with P3 to form a 
resultant Oh, and so on ; we find the resultant of Pi, P2, Ps? 
and P4 to be Oc, and if a fifth force is to produce equilib- 
rium it must be equal and opposite to Oc, and would close 
the polygon OdabcO, in which the sides are equal and par- 



GEAPHICAL STATICS. 



399 



allel respectively to the forces mentioned. To utilize tliis 
fact we can dispense witli all parts of tlie parallelograms in 
Pig. 346 except tlie sides mentioned, and tlien proceed as 
follows in Fig. 347 : 

If P5 is the unknown force which is to balance the other 
four (i.e, is their anti-resultant), we draw the sides of the 
force polygon from A round to B, making each line paral- 
lel and equal to the proper force and pointing the same 
way ; then the line BA represents the required F^ in 
amount and direction, since the arrow BA must follow 
the continuity of the others (point to butt). 

If the arrow BA were pointed at the extremity B, then 
it gives, obviously, the amount and direction of the result^ 
ant of the four forces Pj . . . P4. The foregoing shows 
that if a system of Concurrent Forces in a Plane is in equi- 
librium, ii^ force polygon must close. 



326. Non-Concurrent Forces in a Plane. — Given a system of 
non-concurrent forces m a plane, acting on a rigid body, 
required graphic means of finding their resultant and anti- 
resultant ; also of expressing conditions of equilibrium. 
The resultant must be found in amount and direction ; and 
also in position (i.e., its line of action must be determined). 
E. g., Fig. 348 shows a curved rigid beam fixed in a vise 
at T, and also under the action of forces Pi P2 P3 and P^ 
{besides the action of the vise); required the resultant of 

By the ordinary 
parallelogram of 
forces we com- 
bine Pi and P2 at 
a, the intersection 
of their lines of 
PjQ 34g action, into a re- 

sultant Pa, ; then Pa with Pg at b, to form PbJ and finally P,, 
with P4 at c to form B^ which is .*. the resultant required, 
ie., of Pi . . . . P4 ; and c . , . P is its line of action. 




400 



MSCHAXICS OF EXGIXEEKIXG. 




Fig. 349. 



The separate force triangles (half-parallelograms) by 
wliich. the successive partial resultants B^^, etc., were found, 
are again drawn in Fig. 349. Now since B^ acting in the 

line C..F, Fig. 348, 
is the resultant of 
Pi . . Fi, it is plain 
that a force FJ 
equal to B,. and act- 
ing along c . . i^.but 
in the opposite di- 
rection, would balance the system Pi . . . P4, (is their anti- 
resultant). That is, the forces Pi P2 P3 P4 and BJ would 
form a system in equilibrium. The force B^' then, repre- 
sents the action of the vise T upon the beam. Hence re- 
place the vise by the force B/ acting in the line . . . F . . .c • 
to do which requires us to imagine a rigid prolongation of 
that end of the beam, to intersect F . . . c. This is shown in 
Fig. 350 where the whole beam is free, in equilibrium, under 
the forces shown, and in precisely the same state of stress, 
part for part, as in Fig. 348. Also, by combining in one 
force diagram, in Fig. 351, all the force triangles of Fig. 349 
(by making their common sides coincide, and putting B/ 
instead of B^., and dotting all forces other than those of 
Fig. 350), we have a figure to be interpreted in connection 
with Fig. 350. 



A "poL^^iQH J' 




SPACE DIAGRAM 
Fig. 350. 



FORCE DIAGRAM 
Fig. 351. 



Here we note, first, that in the figure called a force-dia- 
gram, P1P2P3P4 and R/ form a closed polygon and that 



Gr^APHlCAli STATICS. 401 

their arrows follow a continuous order, point to butt, 
around the jperimeter ; which proves that one condition of 
equilibrium of a system of non-concurrent forces ir^ a, plane 
is that its force polygon must close. Secondly, note that ah 
is II to Oa', and be to Oh' ; hence if the force-diagram has 
been drawn (including the rays, dotted) in order to deter- 
mine the amount and direction of HJ, or any other one force, 
we may then find its line of action in the space-diagram, as 
follows: (N. B. — By space diagram is meant the figure show- 
ing to a true scale the form of the rigid body and the lines 
of action of the forces" concerned). Through a, the intersec- 
tion of Fi and F-j, draw a line || to Oa' to cut P3 in some point 
b ; then through b a line || to Ob' to cut F^ at some point c; cF 
drawn || to Oc' is the required line of action of RJ, the anti- 
resultant of Pi, F2, P3, and P4. 

abc is called an equilibrium polygon; this one having but 
two segments, ab and bo (sometimes the lines of action of F^ 
and RJ may conveniently be considered as segments.) The 
segments of the equilibrium polygon are parallel to the respect- 
ive rays of the force diagram. 

Hence for the equilibrium of a system of no;ti-conciirrent 
forces in a plane not only must its force polygon close, 
but also the first and last segments of the corre- 
sponding equilibrium polygon must coincide with 
the resultants of the first two forces, and of the last 
two forces, respectively, of the system. E.g., ab coin- 
cides with the line of action of the resultant of F^ and F^, I 
he with that of F^ and E'c- Evidently the equil. polygon 
"will be different with each different order of forces in 
the force polygon or different choice of a pole, 0. But if 
the order of forces be taken as above, as they occur along 
the beam, or structure, and the pole taken at the " butt " of 
the first force in the force polygon, there will be only one j 
(and this one will be called the special equilibrium polygon 
in the chapter on arch-ribs, and the " true linear arch " in 
dealing with the stone arch.) After the rays (dotted in 
Fig. 351) have been added, by joining the pole to each 



402 



MEOn Allies OF ENGrKEEEi:srG. 



vertex with wliicli it is not already connected, tJbe finai 
figure may be called the/brce diagram. 

It may sometimes be convenient to give tlie name of 
rays to tlie two forces of tlie force polygon which, meet 
at the pole, in which case the first and last segments of 
the corresponding equil. polygon will coincide with the 
lines of action of those forces in the space-diagram (as we 
may call the representation of the body or structure on 
which the forces act). This " space diagram " shows the 
real field of action of the forces, while the force diagram, 
which may be placed in any convenient position on the 
paper, shows the magnitudes and directions of the forces 
acting in the former diagram, its lines being interpreted 
on a scale of so many lbs. or tons to the inch of paper ; in 
the space-diagram we deal with a scale of so many/ee^ to 
the inch of paper. 

We have found, then, that if any vertex or corner of the 
closed force polygon be taken as a pole, and rays drawn 
from it to all the other corners of the polygon, and a cor- 
responding equil. polygon drawn in the space diagram., the 
first and last segments of the latter polygon must co-incide 
with the first and last forces according to the order 
adopted (or with the resultants of the first two and last 
two, if more convenient to classify them thus). It remains 
to utilize this principle. 



327. To Find the Resultant of Several Forces in a Plane. — This 
might be done as in § 326, but since frei^^uently a given set 
of forces are parallel, or nearly so, a special method will 
now be given, of great convenience in such cases. Fig. 352. 

Let Pi Pg and 
Pa be the given 
forces whose 
resultant is re- 
quirsd. Let us 
first find their 
and -' resultant, 
or force which 
Fig. 352. Pm. 353. will balance 




GEAPHICAL STATICS. 403 

them. This anti-resultant may be conceived as decom- 
posed into two components P and P' one of which, say P, 
is arbitrary in amount and position. Assuming P, then, 
at convenience, in the space diagram, it is required to lind 
F'. The live forces must form a balanced system ; hence 
if beginning at Oi, Fig. 353, we lay off a line O^A = P by 
scale, then Al = and || to P,, and so on (point to butt), the 
line POi necessary to close the force polygon is = P' re- 
quired. Now form the corresponding equil. polygon in 
the space diagram in the usual way, viz.: through a the 
intersection of P and P^ draw ab || to the ray 0, . . . 1 
(Avhich connects the pole Oi with the point of the last force 
mentioned). From h, where ab intersects the line of Pg* 
draw he, || to the ray O^ . . 2, till it intersects the line of Pg. 
A line mc drawn through c and || to the P' of the force 
diagram is the line of action of P'. 

Now the resultant of P and P' is the anti-resultant of 
Pi, P2 and P3; .'. d, the intersection of the lines of P and 
P', is a point in the line of action of the anti-resultant re- 
quired, while its direction and magnitude are given by the 
line BA in the force diagram ; for BA forms a closed poly- 
gon both with Pi P2 P3, and with PP'. Hence a line 
through (i || to BA, viz., de, is the line of action of the anti- 
resultant (and hence of the resultant) of Pj, P2, P3. 

Since, in this construction, P is arbitrary, we may first 
choose Oi, arbitrarily, in a convenient position, i.e., in such 
a position that by inspection the segments of the result- 
ing equil. polygon shall give fair intersections and not 
pass off the paper. If the given forces are parallel the 
device of introducing the oblique P and P' is quite neces- 
sary. 

328. — The result of this construction may be stated as 
follows, (regarding Oa and cm as segments of the equil. 
polygon as well as ah and he): If any tivo segments of an 
equU. polygon he prolonged, their intersection is a point in 
the line of action of the resultant of those forces acting at 



404 



MECHANICS OF ENGINEERING. 



the vertices intervening between the given segments, 
the resultant of Pi P2 P3 acts through d. 



Here, 



329. Vertical Reaction of Piers, etc. — Fig. 354. Given the 
vertical forces or loads Pj P2 and P3 acting on a rigid body 
(beam, or truss) which is supported by two piers having 
smooth horizontal surfaces (so that the reactions must be 
vertical), required the reactions Fq and V^ of the piers. 
For an instant suppose V^ and V^ known ; they are in 



iVn 




Fig. 354. 

equil. with Pi Pg and P3. The introduction of the equal 
and opposite forces P and P' in the same line will not dis- 
turb the equilibrium. Taking the seven forces in the 
order P Vq Pj Pg P3 V^ and P', a force polygon formed with 
them will close (see (h) in Fig. where the forces which 
really lie on the same line are slightly separated). With 
Oy the butt of P, as a pole, draw the rays of the force dia- 
gram OA, OB, etc. The corresponding equil. polygon 
begins at a, the intersection of P and V^ in {a) (the space 
diagram), and ends at n the intersection of P' and V^. 
Join an. Now since P and P' act in the same line, an 
must be that line and must be || to P and P' of the force 
diagram. Since the amount and direction of P and P' are 
arbitrary, the position of the pole is arbitrary, while 
Pi, P2, and P3 are the only forces known in advance in the 
force diagram. 

Hence Vq and V^ may be determined as follows : Lay off 
the given loads Pi, P2, etc., in the order of their occur- 
rence in the space diagram, to form a " load-line " AD 



GRAPHICAL STATICS. 



405 



(see (h.) Fig. 854) as a beginning for a force-diagram ; take 
any convenient pole 0, draw the rays OA, OB, 00 and 
OD. Tlien beginning at any convenient point a in the 
vertical line containing the unknown Vq, draw ab || to OA, 
be II to OB, and so on, until the last segment [dn in this 
case) cuts the vertical containing the unknown V„ in some 
point n. Join an (this is sometimes called a closing line) 
and draw a || to it through 0, in the force-diagram. This 
last line will cut ths " load-line " in some point n', and 
divide it in two parts n' A and i>w', which are respectively 
Vq and Va required. 

Corollary. — Evidently, for a given system of loads, in given 
vertical lines of action, and for two given piers, or abut- 
ments, having smooth horizontal surfaces, the location of the 
point n' on the load line is independent of the choice of a 
•pole. 

Of course, in treating the stresses and deflection of the 
ligid body concerned, P and P' are left out of account, as 
being imaginary and serving only a temporary purpose. 

330. Application of Foregoing Principles to a Roof Truss- 
Fig. 355. Wi and W.^ are wind pressures. Pi and P. are 
loads, while the lemaining external forces, viz., the re- 




406 MECHANICS OF ENGINEERING. 

actions, or supporting forces. To, F'„ and H^i niay be fonnd 
by preceding §§. (We here suppose that the right abut- 
ment furnishes all the horizontal resistance ; none at the 
left). 

Lay off the forces (known) Wi, W2, Pi, and P2 in the 
usual way, to form a portion of the closed force polygon. 
To close the polygon it is evident we need only draw a 
horizontal through 5 and limit it by a vertical through 1. 
This determines H^ but it remains to determine ?^' the 
point of division between F^ and V^. Select a convenient 
pole Oi, and draw rays from it to 1, 2, etc. Assume a con- 
venient point a in the line of V„ in the space diagram, and 
through it draw a line || to Oil to meet the line of W^ in 
some point b ; then a line || to Oi2 to meet the line of W2 
in some point c ; then through c || to OjS to meet the line 
of Pi in some point d ; then through d || to Oi4 to meet the 
line of P2 in some point e, (e is identical with d, since Pi 
and P2 are in the same line) ; then efW to Oi5 to meet Hj^ 
in some point/; then fg \\ to OS to meet V^ in some 
point g. 

abcdefg is an equilibrium polygon corresponding to the 
pole Oj. 

Now join ag, the " closing-line," and draw a || to it 
through Oi to determine n', the required point of division 
between Vo and V„ on the vertical 1 6. Hence F^ and V^ 
are now determined as well as H^^. 

[The use of the arbitrary pole Oi implies the temporary 
employment of a pair of opposite and equal forces in the line 
ag, the amount of either being = Oiti']. 

Having now all the external forces acting on the truss, 
and assuming that it contains no " redundant parts," i.e., 
parts unnecessary for rigidity of the frame-work, we proceed 
tc find the pulls and thrusts in the individual pieces, on 
the following plan. The truss being pin-connected, no 
piece extending beyond a joint, and all loads being con- 
sidered to act at joints, the action, pull or thrust, of each 
piece on the joint at either extremity will be in the direction 
of the piece, i.e., in a knoivn direction, and the pin of each 



GKAPHICAL STATICS. 407 

joint is in equilibrium under a system of concurrent forces 
consisting of the loads (if any) at tlie joint and the pulls 
or thrusts exerted upon it by the pieces meeting there.. 
Hence we may apply the principles of § 325 to each joint 
in turn. See Fig. 356. In constructing and interpreting 
the various force polygons, Mr. E.. H Bow'g convenient 
notation will be used; this is as follows: In the space 
diagram a capital letter [ABC, etc.] is placed in each tri- 
angular cell of the truss, and also in each angular space in 
the outside outline of the truss between the external forces 
and the adjacent truss-pieces. In this way we can speak of 
the force Wi as the force BC, of W2 as the force C-E, the 
stress in the piece a/3 as the force QI), and so on. That 
is, the stress in any one piece can be named from the 
letters in the spaces bordering its two sides. Corresponding 
to these capital letters in the spaces of the space-dia~ 
gram, small letters will be used at the vertices of the closed 
force-polygons (one polygon for each joint) in such a way 
that the stress in the piece CD, for example, shall be thQ 
forc3 cd of the force polygon belonging to any joint in 
which that piece terminates ; the stress in the piece FO 
by the force fg, in the proper force polygon, and so on. 

In Fig. 356 the whole truss is shown free, in equili- 
brium under the external forces. To find the pulls or 
thrusts (i.e., tensions or compressions) in the pieces, con- 
sider that if all but two of the forces of a closed force 
polygon are known in magnitude and direction,, while the 
directions, only, of those two are known, the wliole force 
polygon may he drawn, thus determining the amounts of 
those two forces by the lengths of the corresponding 
sides. 

We must .'. begin with a joint where no more than two 
pieces nieet, as at a ; [call the joints a, /9, y, d, and the cor- 
corresponding force polygons a', /9' etc. Fig. 356.] Hence 
at a' (anywhere on the paper) make oh \\ and = (by scale) 
to the known force AB (i.e., V^) pointing it at the upper end, 
and from this end draw he — and || to the known force BG 
(i.e., Wj) pointing this at the lower end. 



iU8 



MECHANICS OF E:i^GINEEBraG. 




Fig. 356. 

To close the polygon draw througli c a || to the piece 
CD, and through a a || to AD ; their intersection deter- 
mines d, and the polygon is closed. Since the arrows 
must be point to butt round the periphery, the force with 
which the piece CD acts on the pin of the joint a is a 
force of an amount = cd and in a direction from c toward 
d ; hence the piece CD is in compression ; whereas the 
action of the piece DA upon the pin at a is from d toward 
a (direction of arrow) and hence DA is in tension. Notice 
that in constructing the force polygon «' a right-handed 
(or clock-wise) rotation has been observed in considering 
in turn the spaces ABC and D, round the joint a. A 
similar order will be found convenient in each of the other 
joints. 

Knowing now the stress in the piece GD, (as well as in 
DA) all but two of the forces acting on the pin at the joint 
/? are known, and accordingly we begin a force polygon, /3', 
for that joint by drawing dc,= and || to the dc of polygon 
a', hut pointed in the opposite direction, since the action of 
OD on the joint /? is equal and opposite to its action on 
the joint a (this disregards the weight of the piece). 
Through c draw ce = and || to the force CE (i.e., W^ and 



GRAPHICAL STATICS. 409 

pointing tlie same way ; tlien ef, = and || to tlie load EF 
(i.e. Pj) and pointing downward. Througli f draw a || to 
tlie piece FG and through d, a || to the piece OB, and the 
polygon is closed, thus determining the stresses in the 
pieces FG and GT>. Noting the pointing of the arrows, 
we readily see that FG is in compression Avhile GD is in 
tension. 

Next pass to the joint (5, and construct the polygon o' , 
thus determining the stress gli in GB. and that ad in AD ; 
this last force ad should check with its equal and oppo- 
site ad already determined in polygon a'. Another check 
consists in the proper closing of the polygon y', all of 
whose sides are now known. 

[A compound stress-diagram may be formed by super- 
posing the polygons already found in such a way as to 
make equal sides co-incide ; but the character of each 
stress is not so readily perceived then as when they are 
kept separate]. 

In a similar manner we may find the stresses in any pin- 
connected frame-work (in one plane and having no redun- 
dant pieces) under given loads, provided all the support- 
ing forces or reactions can be found. In the case of a 

braced-arch (truss) as 
shown in Fig. 357, hinged 
to the abutments at both 
ends and not free to slide 
laterally upon them, the 
reactions at and B de- 
FiG. 357. pend, in amount and direc- 

tion, not only upon the equations of Statics, but on the 
form and elasticity of the arch-truss. Such cases will be 
treated later under arch -ribs, or curved beams. 

332. The Special Equil. Polygon. Its Relation to the Stresses 
in the Rigid Body. — Eeproducing Figs. 350 and 351 in Figs. 
358 and 359, (where a rigid curved beam is in equilibrium 
under the forces P^ Pg, P3, P4 and B'^) we call a . . b . , 




MECHANICS OF Eis^GINEERIKG. 



tiie special equil. polygon because it corresponds to a force 
diagram in which the same order of forces has been ob- 
S3rved as that in which they occur along the beam (from 
left to right here). From the relations between the force 




SPACE DIAGRAM 

Fig. 358. 



FORCE DIAGRAM 
Pig, 359. 



diagram and equil. polygon, this special equil. polygon in 
the space diagram has the following properties in connec- 
tion with the corresponding rays (dotted lines) in the force 
diagram. 

The stresses in any cross-section of the portion O'A of 
the beam, are due to P^ alone ; those of any cross-section 
on AB to Pi and P2, i.e., to their resultant R , whose mag- 
nitude is given by the line Oa' in the force diagram, while 
its liLe of action is ah the first segment of the equil. poly- 
gon. Similarly, the stresses in BC are due to P^, P^ and 
P., i.e., to their resultant R^, acting along the segment &c, 
its magnitude being =^0h' in the force diagram. E.g., if 
the section at m be exposed, considering O'ABm as a free 
body, we have (see Fig. 360) the elastic stresses (or inter- 







P3 

Fig. 360. Fig. 361. 

nal forces) at m balancing the exterior or " applied forces " 
Pi, Pj and P3. Obviously, then, the stresses at m are just 



GEAPHIOAL STATICS. 411 

the same as if B^, tlie resultant of Pj, P^ and Pg, acted upon 
an imaginary rigid prolongation of tlie beam intersecting 
he (see Fig. 361).i?i, might be called the " anti-stress-resuU- 
ant" ior the portion PC of the beam. We may .•. state 
the following : If a rigid body is in equilibrium under a sys- 
tem of Hon-Concurrent Forces m a plane, and the special equi- 
librium polygon has been draivn, then each ray of tlie force 
diagram is the anti-stress-resultant of that portion of the beam 
which corresponds to the segment of the equilibrium polygon 
to which the ray is parallel ; and its line of action is the seg- 
ment just mentioned. 

Evidently if the body is not one rigid piece, but com- 
posed of a ring of uncemented blocks (or voussoirs), it may 
be considered rigid only so long as no slipping takes place 
or disarrangement of the blocks; and this requires that the 
" anti-stress-resultant " for a given joint between two 
blocks shall not lie outside the bearing surface of the 
joint, nor make too small an angle with it, lest tipping or 
slipping occur. For an example of this see Fig. 362, show- 
ing a line of three blocks in equilibrium under five forces. 

The pressure borne at the 
s^2 joint MN, is = Pa ^^ the 
force -diagram and acts in 
the line ab. The con- 
struction supposes all 
the forces given except 
Fig. 362. oue, in amount and posi- 

tion, and that this one could easily be found in amount, as 
being the side remaining to close the force polygon, while 
its position would depend ox^ I;he equil. polygon. But in 
practice the t^m forces Pj and B\ are generally unknown, 
hence the point 0, or pole of the force diagram, can not 
be fixed, nor the special equil. polygon located, until other 
considerations, outside of those so far presented, are 
brought into play. In the progress of such a problem, as 
will be seen, it will be necessary to use arbitrary trial po- 
sitions for the pole 0, and corresponding ^rmZ equilibrium 
polygons. 




412 



MECHANICS OF BJUGmBBlWSGc, 



CHAPTER IX. 



GRAPHICAL. STATICS OF VERTICAL. FORCES, 



333. Remarks. — (Witli the excoption of § 378 a) in prob- 
lems to be treated subsequently (either the stiff arch-rib, 
or the block-work of an arch-ring, of masonry) when the 
body is considered free all the forces holding it in equil. 
will be vertical (loads, due to gravity) except the reactions 
at the two extremities, as in Eig, 363 ; but for convenience 
each reaction will be replaced by its horizontal and verti- 
cal components (see Fig. 364). The two fi^'s are of course 
pqual, since they are the only horizontal forces in the 
system. Henceforth, aU equil. polygons under discussion will 
he understood to imply this kind of system of forces. Pi, Pz, 



r r f 

A t\ 



U 



v„ 





Fig. 363. 



Fig. 364a. 



Fig. 364. 

etc. , will represent the ' ' loads " ; Vq and F„ the vertical 
components of the abutment reactions; H the value of 
either horizontal component of the same. (We here sup- 
pose the pressures To and Tn resolved along the horizontal 
and vertical.) 



JRAPHICAIi STATICS. 



413 



334. Concrete Conception of an Equilibrium Polygon. — Any 
equilibrium polygon has this property, due to its mode 
of construction, viz.: If the ab and be of Fig. 358 were im- 
ponderable straight rods, jointed at b without frictioiij they 
would be in equilibrium under the system of forces there 
given. (See Fig. 364a). The rod ah suffers a compression 
equal to the H^ of the force diagram, Fig. 359, and be a 
-^compression = B^^. In some cases these rods might be in 
tension, and would then form a set of links playing the 
part of a suspension-bridge cable. (See § 44). 

335. Example of EcLuilibrium Polygon Drawn to Vertical Loads 
— Fig. 365. [The structure bearing the given loads is not 
shown, but simply the imaginary rods, or segments of an 
equilibrium polygon, which would support the given loads 
in equilibrium if the abutment points A and B, to which 
the terminal rods are hinged, were firm. In the present 
case this equilibrium is unstable since the rods form a 
standing structure ; but if they were hanging, the equilibri- 
um would be stable. Still, in the present case, a very light 
bracing, or a little friction at all joints would make the 
equilibrium stable. 




2 FT. TO aNOH 



Fig. S65. 



Given three loads Pi, F2, and P3, and two " abutment 
verticals " A' and B', in which we desire the equil. poly- 
gon to terminate, lay off as a "load-line," to scale, Pj, P2, 
and P« end to end in their order. Then selecting any pole. 



414 MBCHAXICS OF EjS^GI:N^EEIII]^G. 

0, draw the rays 01, 02, etc., of a force diagram (tlie F's 
and P's, though really on the same vertical, are separated 
slightly for distinctness ; also the H's, which both pass 
through and divide the load-line into V^ and F^). We 
determine a corresponding equilibrium polygon by draw- 
ing through A (any point in A') a line || to . . 1, to inter- 
sect Pi in some point b ; through 6 a || to . . 2, and so ou> 
until B'' the other abutment-vertical is struck in some 
point B. AB is the " abutment-line " or " closing -line." 

By choosing another point for 0, another equilibrium 
polygon would result. As to which of the infinite 
number (which could thus be drawn, for the given loads 
and the A' and B' verticals) is the special equilibrium poly' 
gon for the arch-rib or stone-arch, or other structure, on 
which the loads rest, is to be considered hereafter. In 
any of the above equilibrium polygons the imaginary 
series of jointed rods would be in equilibrium. 

336. Useful Property of an Equilibrium Polygon for Vertical 
Loads. — (Particular case of § 328). See Fig. 366. In any 
equil. polygon, supporting vertical loads, consider as free 
any number of consecutive segments, or rods, with the 
loads at their joints, e. g., the 5th and 6th and portions of 
C/r^.^ the 4th and 7th which, we sup- 

/g i ,> ^ ^6"--. ^ -^^ pose cut and the compressive 

— ~S<^^ forces in them put in, T^ and 
^^ Tj, in order to consider 4 5 6 7 
"^^ ^ as a free body. For equil,, 

~^~[-^\ according to Statics, the lines 

' ' 'Pe "\ of action of Ti and Ty (the com- 
i^iG. 366. pression in those rods) must in- 

tersect in a point, C, in the line of action of the resultant 
of Fi, P5, and Pq ; i.e., of the loads occurring at the inter- 
vening vertices. That is, the point C must lie in the ver- 
tical containing the centre of gravity of those loads. Since 
the position of this vertical must be independent of the 
particular equilibrium polygon used, any other (dotted 
lines in Fig. 366) for the same loads will give the same re- 



GKAPHICAL STATICS. 415 

suits. Hence tlie vertical CD, containing the centre of 
gravity of any number of consecutive loads, is easily found 
by drawing tlie equilibrium polygon corresponding to 
any convenient force diagram having the proper load-line. 
This principle can be advantageously applied to finding 
a gravity -line of any plane figure, by dividing the latter 
into parallel strips, whose areas may be treated as loads 
applied in their respective centres of gravity. If the strips 
are quite numerous, the centre of gravity of each may be 
considered to be at the centre of the line joining the mid- 
dles of the two long sides, while their areas may be taken 
as proportional to the lengths of the lines drawn through 
these centres of gravity parallel to the long sides and lim- 
ited by the end-curves of the strips. Hence the " load- 
line " of the force diagram may consist of these lines, or 
of their halves, or quarters, etc., if more convenient (§ 376). 



USEFUL, RELATIONS BETWEEN FORCE DIA- 
GRAMS AND EQUILIBRIUM POLYGONS,, 

(for vertical loads,) 

237. Il6sum6 of Construction.— Fig. 367. Given the loads 
Pi, etc., 'their verticals, and the two abutment verticals ^4' 
and B', in which the abutments are to lie ; we lay off a 
load-line 1 ... 4, take any convenient pole, 0, for a force- 
diagram and complete the latter. For a corresponding 
equilibrium polygon, assume any point A in the vertical" 
A', for an abutment, and draw the successive segments 
Al, 2, etc., respectively parallel to the inclined lines of the 
force diagram (rays), thus determiDiDg finally the abut- 
ment B, in B', v/hich {B) will not in general lie in the hor- 
izontal through A. 

Now join AB, calling AB the abutment-line, and draw a 
parallel to it through 0, thus fixing the point n' on the 



416 



MECHANICS or ENGINEERIKG. 

Pi 




P. t 



l' 



Fig. 



load-line. This point %' , as above determined, is indepen' 
dent of the location of the pole, 0, (proved in § 329) and 
divides the load-line into two portions ( V'o = 1 . . . n\ and 
V'n = n' .. .4:) which are the vertical pressures which two 
supports in the verticals A' and B' would sustain if the 
given loads rested on a horizontal rigid bar, as in Fig. 368. 

See § 329. Hence to find the point n' we may use any 
convenient pole 0. 

[N. B.— The forces V^ and V^ of Fig. 367 are not identi- 
cal with F'o and V'„, but may be obtained by dropping a 
"I from to the load-line, thus dividing the load-line 
into two portions which are V^ (upper portion) and F^. 
However, if A and B be connected by a tie-rod, in Fig. 
367, the abutments in that figure will bear vertical press- 
ures only and they will be the same as in Fig, 368, while 
the tension in the tie -rod will be = On'.^ 

338. Theorem. — The vertical dimensions of any two equili- 
brium polygons, drawn to the same loads, load-verticals, and 
abutment -verticals, are inversely proportional to their H^s {or 
"pole distances "). We here regard an equil. polygon and 
its abutment-line as a closed figure. Thus, in Fig. 369, 
we have two force-diagrams (with a common load-line, for 
convenience) and their corresponding equil. polygons, for 
the same loads and verticals. From § 337 we know that 
On' is II to AB and OqW' is || to A^B^. Let CD be any ver- 
tical cutting the first segments of the two equil. polygons. 



GRAPHICAL STATICS. 



417 



SB| 



Denote the intercepts thus determined by z' and %\, respect- 
gC r ively. From the 

parallelisms just 
mentioned, and 
others more famil~ 
,/ iar, we have the 
triangle \n' sim* 
ilar to the triangle 
Az' (shaded), and 
the triangle O^n' 
similar to the tri- 
angle Ajz,^,. Hence 



c 

1 


P. 




u-^ 


/ 


hi 

1 . 


p^ ^. 


A. 




^y 


1 



—- 1-.. 



-^-N^ 



Fig. 369. 




the proportions between ( \n' 
bases and altitudes ( 



H h 



and 









.*. z' : z\ : : H^ '■ H. The same kind of proof may easily 

be applied to the vertical intercepts in any other segments, 
e. g., 2" and z'\. Q. E. D. 

339. Corollaries to the foregoing. It is evident that : 
(1.) If the pole of the force-diagram be moved along a 
vertical line, the equilibrium polygon changing its form 
in a corresponding manner, the vertical dimensions of the 
equilibrium polygon remain unchanged ; and 

(2.) If the pole move along a straight line which con- 
tains the point n\ the direction of the abutment-line 
remains constantly parallel to the former line, while the 
vertical dimensions of the equilibrium polygon change in 
inverse proportion to the pole distance, or H, of the force- 
diagram, [^is the "1 distance of the pole from the load- 
line, and is called the pole-distance]. 

§ 340. Linear Arch as Equilibrium Polygon. — (See § 316.) 
If the given loads are infinitely small with infinitely small 
horizontal spaces between them, any equilibrijim polygon 
becomes a linear arch. Graphically we can not deal with 
these infinitely small loads and spaces, but from § 336 it 
is evident that if we replace them, in successive groups. 



418 



MECHAXICS OF ENGINEERING. 




Fig. 370. 



bj finite forces, eacli of wliicli = tlie STim of those com^. 

I I I I I I I I P°''^^ """^ ^^""^P ^""^ ^' 
.M i V ..M M. .^^ M I ,M I + /, applied tlirougli the cen- 
tre of gravity of that 
group, we can draw an 
equilibrium polygon 
whose segments will be 
tangent to the curve of 
the corresponding linear 
arch, and indicate its posi- 
tion with siifficient exactness for practical purposes. (See 
Fig. 370), The successive points of tangency A, m, n, etc.. 
lie vertically under the points of division between the 
groups. This relation forms the basis of the graphical 
treatment of voussoir, or blockwork, arches. 

341. To Peas an Equilibrium Polygon Through. Three Arbitrary 
Points. — (In the present case the forces are vertical. For 
a construction dealing with any plane system of forces see 
construction in § 378a.) Given a system of loads, it is re- 

^ quired to draw 
r /^l ^^ equilibrium 
polygon for 
t h e m through 
-anythree points, 
two of which 
may be consid- 
ered as abut- 
ments, outside of the load-verticals, the third point being 
between the verticals of the first two. See Fig. 371. The 
loads Pi, etc., are given, with their verticals, while A, p, 
and B are the three points. Lay oft the load-line, and 
with any convenient pole, Oj, construct a force-diagram, 
then a corresponding preliminary equilibrium polygon 
beginning at A. Its right abutment P,, in the vertical 
through B, is thus found. Oj n' can now be drawn || to AB^, 
to determine n\ Draw n'O \\ to BA. The pole of the 
required equilibrium polygon must lie on n'O (§ 337} 




Fig. 371. 



GEAPHICAL STATICS. 



Draw a vertical throiigli jp. The E. of tlie required equili- 
brium polygon must satisfy the proportion H : H^ : : rs i 
pm. (See § 338). Hence construct or compute H from 
the proportion and draw a vertical at distance H from 
the load-line (on the left of the load-line here) ; its inter- 
section with n' gives the desired pole, for which a 
force diagram may now be drawn. The corresponding 
equilibrium polygon beginning at the first point A will 
also pass through p and B ; it is not drawn in the figure. 

342. Symmetrical Case of the Foregoing Problem.— If two 

points A and B are on a level, the third, p, on the middle 
vertical between them ; and the loads (an even number) 
symmetrically disposed both in position and magnitvde, about 
p, we may proceed more simply, as follows : (Fig. 372). 

From symmetry n' 
must occur in the mid- 
dle of the load-line, of 
which we need lay off 
only the upper half. 
Take a convenient pole 
■piG. 372. Oi, in the horizontal 

through n', and draw a half force diagram and a corres- 
ponding half equilibrium polygon (both dotted). The up- 
per segment he of the latter must be horizontal and being 
prolonged, cuts the prolongation of the first segment in a 
point d, which determines the vertical CD containing the 
centre of gravity of the loads occurring over the half -span 
on the left. (See § 336). In the required equilibrium poly- 
gon the segment containing the point p must be horizon- 
tal, and its intersection with the first segment must lie in 
CD. Hence determine this intersection, C, by drawing the 
vertical CD and a horizontal through p ; then join CA, 
which is the ^rst segment of the required equil. polygon. 
A parallel to GA through 1 is the Jirst ray of the corres- 
ponding force diagram, and determines the pole on the 
horizontal through n'. Completing the force diagram foi 




420 



MECHAXIOS OF El^J^GINEBEIN^G. 




Fig. 373. 



this pole (half of it only here), the required equil. poly- 
gon is easily finished afterwards. 

343. To Find a System of Loads Under Which a Given Equi- 
librium Polygon Would be in Equilibrium, — Fig. 373. Let AB 
he the given equilibrium polygon. Through any point 
as a pole draw a parallel to each 
segment of the equilibrium polygon. 
Any vertical, as V, cutting these 
lines will have, intercepted upon it, 
a load-line 1, 2, 3, whose parts 1 . . 2, 
2 . . 3, etc., are proportional to the 
successive loads which, placed on 
ih@ corresponding joints of the equilibrium polygon would 
be supported by it in equilibrium (unstable). 

One load may be assumed and the others constructed. A 
hanging, as well as a standing, equilibrium polygon may be 

dealt with in 
hke manner, 
but will be 
in stable 
equilibrium. 
The problem 
in § 44 may 
be solved in 
this way, the various steps and final re- 
sults being as follows (Fig. 50 is here re- 
peated) : — 

Let weight Gi be given, =66 lbs., and 
the positions of the cord segments be as in 
Fig. 50. We first lay of! (see Fig. 373a) 
vertically, a6 = 66 lbs., by some convenient 
scale, and prolong this vertical fine indefi- d 
nitely downward. aO is then drawn parallel to 
and bO parallel to 1 ... 2. Their intersection determines a 
pole, 0, through which Oc and Od, parallel respectively to 
2 ... 3 and 3 . . . n, are drawn, to intersect ad in c and d. 

We also draw the horizontal On, in Fig. 373a. By scaling, 
we now find the results: — G2 = bc = 42 lbs.; G3 = cd = 50 lbs.; 
H = 58.5 lbs., ( = Ho and'^„ of Fig. 50); while 70 = ^^=100 
lbs. and y„ = 58 lbs. 





Fig. 373a. 

.1 of Fig. 50, 



AECHES OF MASOifBT. 



421 



CHAPTER X. 



RIGHT ARCHES OF MASONRY. 

Note. — The treatment given in this chapter is by many engineers 
considered sufficiently exact for ordinary masonry arches, the mOie 
refined methods of the "elastic theory" being reserved for arches of 
fairly continuous material, such as those of metal and of concrete (re- 
inforced or otherwise); and is accordingly retained in this revised 
edition. 

844. — In an ordinary "right" storce-arcli (i.e., one in 
which the faces are "[ to the axis of the cylindrical soffit, 
or under surface), the successive blocks forming the arch- 
ring are called voussoirs, the joints between them being 
planes which, prolonged, meet generally in one or more 
horizontal lines; e.g., those of a three-centred arch in three 
II horizontal lines ; those of a circular arch in one, the axis 
of the cylinder, etc. Elliptic arches are sometimes used. The 
inner concave surface is called the soffit, to which the radiat- 
ing joints between the voussoirs are made perpendicular. 
The curved line in which the soffit, is intersected by a plane 




Fig. 374. 



H to the axis of the arch is the Intrados. The curve in the 
same plane as the intrados, and bounding the outer ex- 
tremities of the joints between the voussoirs, is called the 
Extrados. 

Fig. 374 gives other terms in use in connection with, a 



422 



MECHAKICS OF ElSGIXEEIxiKG, 



stone arch, and explains those already given. 
" springing-line." 



AB is the 



345o Mortar and Friction. — As 'common mortar hardens 
very slowly, no reliance should be placed on its tenacity 
as an element of stability in arches of any considerable 
size ; though hydraulic mortar and thin joints of ordinary 
mortar can sometimes be depended on. Friction, however, 
between the surfaces of contiguous voussoirs, plays an 
essential part in the stability of an arch, and will there- 
fore be considered. 

The stability of voussoir-arches must .•. be made to 
depend on the resistance of the voussoirs to compresssion 
and to sliding upon each other ; as also of the blocks 
composing the piers, the foundations of the latter being 
firm. 

346. Point of Application of the Eesultant Pressure between 

two consecutive voussoirs ; (or pier blocks). Applying 
Navier's principle (as in flexure of beams) that the press- 
ure per unit area on a joint varies uniformly from the 
extremity under greatest comj)ression to the point of least 
compression (or of no compression); and remembering 
that negative pressures (i.e., tension) can not exist, as they 
might in a curved beam, we may represent the pressure 
per unit area at successive points of a joint (from the intra- 
dos toward the extrados, or vice versa) by the ordinates of 
a straight line, forming the surface of a trapezoid or tri- 
angle, in which figure the foot of the ordinate of the cen- 
tre of gravity is the point of application of the resultant 
pressure. Thus, where the least compression is supposed 










Fig. 575. 



Fig. 376. 



Fig. 377. 



Fig. 378, 



MASONEY ARCHES. 



423 



to occur at the intrados A, Fig. 375, tlie pressures vary as 
tlie ordinates of a trapezoid, increasing to a maximum value 
at B, in the extrados. In Fig. 376, where the pressure is zero 
at B, and varies as the ordinates of a triangle, the result- 
ant pressure acts through a point one-third the joint- 
length from A. Similarly in Fig. 377, it acts one-third 
the joint-length from B. Hence, when the pressure is not 
zero at either edge the resultant pressure acts within the 
middle third of the joint. Whereas, if the resultant press- 
ure falls loitliout the middle third, it shows that a portion 
-4m of the joint, see Fig. 378, receives no pressure, i.e., the 
joint tends to open along Am. 

Therefore that no joint tend to open, the resultant press- 
ure must fall within the middle third. 

It must be understood that the joint surfaces here dealt 
with are rectangles, seen edgewise in the figures. 

347. Friction. — By experiment it has been found the 
angle of friction (see § 156) for two contiguous voussoirs 
of stone or brick is about 30° ; i.e., the coefficient of fric- 
tion is / = tan. 30°. Hence if the direction of the press- 
ure exerted upon a voussoir by its neighbor makes an 
angle a less than 30° with the normal to the joint surface, 
there is no danger of rupture of the arch by the sliding 
of one on the other. (See Fig. 379). 

348. Resistance to Crushing. — When the resultant pressure 
falls at its extreme allowable limit, viz. : the edge of the 
middle third, the pressure per 
unit of area at n, Fig. 380, iy 
double the mean pressure per 
unit of area. Hence, in de- 
signing an arch of masonry, 
we must be assured that at 
every joint (taking 10 as a 
factor of safety) 

( Double the mean press- | ^^^^ ^^ j^^^ ^^^^ y g 
I ure per unit oi area \ ' 




Fig. 379. 



Fig. 380. 



424 MECHA]srics of engineekikg. 

C being tlie ultimate resistance to crushing, of tlie material 
emj)loyed (§ 201) (Modulus of Crushing). 

Since a lamina one foot thick will always be considered 
in what follows, careful attention must be paid to the units 
employed in applying the above tests. 

Example. — If a joint is 3 ft. by 1 foot, and the resultant 
pressure is 22.5 tons the mean pressure per sq. foot is 

p=22.5-^3=7.5 tons per sq. foot 

.'. its double=15 tons per sq. foot=208.3 lbs. sq. inch, 
which is much less than '/lo of C for most building stones ; 
see § 203, and below. 

At joints where the resultant pressure falls at the middle, 
the max. pressure per square inch would be equal to the 
mean pressure per square inch ; but for safety it is best to 
assume that, at times, (from moving loads, or vibrations) 
it may move to the edge of the middle third, causing the 
max. pressure to be double the mean (per square inch). 

Gem Gillmore's experiments in 1876 gave the following 
results, among many others : 

NAME OF BUILDING STONE. C IN LBS. PER SQ. INCH. 

Berea sand-stone, 2-inch cube, - - - - 8955 

4 " " - - - - 11720 

Limestone, Sebastopol, 2-inch cube {chalk)^ - - 1075 

Limestone from Caen, France, - - . . 3650 

Limestone from Kingston, ]^. Y., - - . - 13900 

Marble, Vermont, 2-inch cube, - - 8000 to 13000 

Granite, New Hampshire, 2-inch cube, 15700 to 24000 

349. The Three Conditions of Safe E(iiiilibriiim for an arch of 
uncemented voussoirs. 

Recapitulating the results of the foregoing paragraphs, 
we may state, as follows, the three conditions which must 
be satisfied at every joint of arch -ring and pier, for each 
of any possible combination of loads upon the structure : 

(1). The resultant pressure must pass within the middle- 
third, 

(2). The resultant pressure must not make an angle > 
30° with the normal to the joint. 

(3). The m'^.an pressure per unit of area on the surface 



AKCH OF MASOK^RT. 



425 




of tlie joint must not exceed Ygo of the Modulus of crush- 
ing of the material. 

350. The True Linear-Arch, or Special Equilibrium Polygon; 

and the resultant pressure at any joint. Let the weight 
of each voussoir and its load be represented by a vertical 
force passing through the centre of gravity of the two, as 
in Fig. 381o Taking any 
two points A and JB, A 
being in the first joint and 
B in the last ; also a third 
point, p, in the crown 
joint (supposing such to 
be there, although gener- 
ally a key-stone occupies | 
the crown), through these fig. ssi. 

three points can be drawn [§ 341] an equilibrium polygon 
for the loads given ; suppose this equil. polygon nowhere 
passes outside of the arch-ring (the arch-ring is the por- 
tion between the intrados, mn, and tha (dotted) extrados 
m'n') intersecting the joints at h, c, etc. Evidently if such 
be the case, and small metal rods (not round) were insert- 
ed at A, h, c, etc., so as to separate the arch-stones slight- 
ly, the arch would stand, though in unstable equilibrium, 
the piers being firm ; and by a different choice of A, p, and 
B, it might be possible to draw other equilibrium poly- 
gons with segments cutting the joints within the arch- 
ring, and if the metal rods were shifted to these new inter- 
sections the arch would again stand (in unstable equilib- 
rium). 

In other words, if an arch stands, it may be possible to 
draw a great number of linear arches within the limits of 
the arch-ring, since three points determine an equilibrium 
polygon (or linear arch) for given loads. The question 
arises then : luMch linear arch is the locus of the actual re- 
sultant pressures at the successive joints ? 

[Considering the arch-ring as an elastic curved beam 
inserted in firm piers (i.e., the blocks at the springing-line 



426 MEOHAKIOS OF ENGIKEEEING. 

are incapable of turning) and Jbaving secured a close fit at 
all joints before the centering is lowered, the most satisfac- 
tory answer to this question is given in Prof. Greene's 
" Arches," p. 131 ; viz., to consider the arch-ring as an 
arch rib of fixed ends and no hinges ; see § 380 of next 
chapter;, but the lengthy computations there employed 
(and the method demands a simple algebraic curve for the 
arch) may be most advantageously replaced by Prof. 
Eddy's graphic method (" New Constructions in Graphical 
Statics," published in Van Nostrand's Magazine for 1877)„ 
which applies to arch curves of any form. 

This method will be given in a subsequent chapter, on 
Arch Eibs, or Curved Beams ; but for arches of masonry a 
much simpler procedure is sufficiently exact for practical 
purposes and will now be presented]. 

If two elastic blocks 
of an arch-ring touch at 
one edge. Fig. 382, their 
adjacent sides making a 
small angle with each 
•"iG- ^82. Fig. 383. other, and are then grad- 

ually pressed more and more forcibly together at the edge 
m, as the arch-ring settles, the centering being gradually 
lowered, the surface of contact becomes larger and larger, 
from the compression which ensues (see Pig. 383); while 
the resultant pressure between the blocks, first applied at 
the extreme edge m, has now probably advanced nearer the 
middle of the joint in the mutual adjustment of the arch- 
stones. With this in view we may reasonably deduce the 
following theory of the location of the true linear areh 
(sometimes called the " line of pressures " and " curve of 
pressure ") in an arch under given loading and with ^rm 
piers. (Whether the piers are really unyielding, under the 
oblique thrusts at the springing-line, is a matter for sub- 
sequent investigation. 

351. Location of the True Linear Arch. — Granted that the 
voussoirs have been closely fitted to each other over the 





ARCH OF MASOXKT. 427 

■centering (sheets of lead are sometimes used in tlie joints 
to make a better distribution of pressure); and tliat the 
piers are firm ; and that the arch can stand at all without 
the centering ; then we assume that in the mutual accom- 
modation between the voussoirs, as the centering is low- 
ered, the resultant of the pressures distributed over any 
joint, if at first near the extreme edge of the joint, advances 
nearer to the middle as the arch settles to its final posi- 
tion of equilibrium under its load ; and hence the follow- 
ing 

352. Practical Conclusions. 

I. If for a given arch and loading, with firm piers, an 
•equilibrium polygon can be drawn (by proper selection of 
the points A, p, and B, Fig. 381) entirely within the mid^ 
die third of the arch ring, not only will the arch stand, but 
the resultant pressure at every joint will be within the 
middle third (Condition 1, § 349) ; and among all possible 
equilibrium polygons which can be drawn within the mid- 
dle third, that is the " true " one which most nearly coin- 
•cides with the middle line of the arch-ring. 

II. If (with firm piers, as before) no equilibrium poly- 
rgon can be drawn Avithin the middle third, and only one 
within the arch-ring at all, the arch may stand, but chip- 
ping and spawling are likely to occur at the edges of the 
joints. The design should .*. be altered. 

III. If no equilibrium polygon can be drawn within 
the arch-ring, the design of either the arch or the loading 
.must be changed ; since, although the arch may standi 
from the resistance of the spandrel walls, such a stability 
must be looked upon as precarious and not countenanced 
in any large important structure. (Very frequently, in 
small arches of brick and stone, as they occur in buildings, 
the cement is so tenacious that the whole structure is vir- 
tually a single continuous mass). 

When the " true " linear arch has once been determined, 
the amount of the resultant pressure on any joint is given 
by the length of the proper ray in the force diagram. 



428 



MECHANICS OF EXGIXEEHING. 



ARRANGEMENT OF DATA FOR GRAPHIC 
TREATMENT. 

353. Cli-aracter of Load. — In most large stone arch bridges 
the load (permanent load) does not consist exclusively of 
masonry up to tlie road-way but partially of earth filling 
above the masonry, except at the faces of the arch where 
the spandrel walls serve as retaining walls to hold the 
earth. (Fig. 384). If the intrados is a half circle or half- 




Fig. 385. 



Fig. 384. 

ellipse, a compactly -built masonry backing is carried up 
beyond the springing-line to AB about 60° to 45° from the 
crown. Fig. 385 ; so that the portion of arch ring below 
AB may be considered as part of the abutment, and thus 
AB is the virtual springing-line, for graphic treatment. 

Sometimes, to save filling, small arches are built over 
the haunches of the main arch, with earth placed over 
them, as shown in Fig. 386. In any of the preceding cases 




Fig. 387. 



it is customary to consider that, on account of the bond- 
ing of the stones in the arch shell, the loading at a given 
distance from the crown is uniformly distributed over the 
width of the roadway. 



AECHES OF MASONET. 429 

354, Reduced Load-Contour. — In the graphical discussioa 
of a proposed arch we consider a lamina one foot thick, 
this lamina being vertical and "| to the axis of the arch ; 
i.e., the lamina is || to the spandrel walls. For graphical 
treatment, equal areas of the elevation (see Fig, 387) of 
this lamina must represent equal weights. Taking the 
material of the arch-ring as a standard, we must find for 
each point "p of the extrados an imaginary height z of the 
arch-ring material, which would give the same pressure 
(per running horizontal foot) at that point as that due to 
the actual load above that point. A number of such or- 
dinates, each measured vertically upward from the extra- 
dos determine points in the "Reduced Load-Contour," i.e., 
the imaginary line, AM, the area between which and the 
extrados of the arch-ring represents a homogeneous load 
of the same density as the arch -ring, and equivalent to the 
actual load (above extrados), vertical hy vertical. 

355. Example of Reduced Load-Contour. — Fig, 388. Given 
an arch-ring of granite (heaviness = 170 lbs. per cubic 
foot) with a dead load of rubble (heav. = 140) and earth 
(heav. = 100), distributed as in figure. At the point p, of 
the extrados, the depth 5 feet of rubble is equivalent to a 
depth of [^^ x5]=4.1 ft. of granite, while the 6 feet of earth 
is equivalent to [l°?x6]=3.5 feet of granite. Hence the 
Reduced Load-Contour has an ordinate, above p, of 7.6 feet. 
That is, for each of several points of the arch -ring extrados- 
reduce the rubble ordinate in the ratio of 170 : 140, and 
the earth ordinate in the ratio 170 : 100 and add the re- 
sults, setting off the sum vertically from the points in the 
■extrados*. In this way Fig. 389 is obtained and the area 

*TUs is most conveniently done by graphics, thus : On a right-line set off 17 equal. 
parts (of any convenient magnitude.) Call this distance OA. Through t> draw another 
right line at any convenient angle (30° to 60°) with OA, and on it from O 

set off OB equal to 14 (for the ruhble ; or 10 for the earth) of the eame egaal 
parts. Join AB. From O toward A set off* all the rubble ordiaates to be reduced^ 
(each being set off from 0} and through the other extremity of each draw a Bne par- 
allel to AB. The reduced ordinates will be the respective lengths, from O, along OB, 
to the intersections of these parallels ynth OB. 

* Witli the dividers. 



430 



MBCHAISICS OF ENGINEERING. 



:/EART.HV; Av,*//%V;*i*»'v5i;'i!lV;?V/;i;;*'uVf/-^;;-';^^ 





there given is to be treated as representing liomogeneous 
granite one foot thick. This, of course, now includes the 
arch-ring also. AB is the " reduced load- contour." 

356. Live Loads. — In discussing a railroad arch bridge 
the " live load " (a train of locomotives, e.g., to take an ex- 
treme case) can not be disregarded, and for each of its po- 
sitions we have a separate Reduced Load-Contour. 

Example. — Suppose the arch of Fig. 388 to be 12 feet 
wide (not including spandrel walls) and that a train of lo- 
comotives weighing 3,000 lbs. per running foot of the track 
covers one half of the span. Uniformly * distributed later- 
ally over the width, 12 ft., this rate of loading is equiva- 
lent to a masonry load of one foot high and a heaviness of 
250 lbs. per cubic ft., i.e., is equivalent to a height of 1.4 
ft. of granite masonry [since ^[|- X 1.0 — 1.4] over the half 
span considered. Hence from Fig. 390 we obtain Fig. 391 
in an obvious manner. Fig. 391 is now ready for graphic 
treatment. 




Fig. 390. Fig. 391. 

357. Piers and Abutments. — In a series of equal arches 
the pier between two consecutive arches bears simply the 
weight of the two adjacent semi-arches, plus the load im- 



* If the earth-filling is sLallcw, the Icminse directly under the track prob* 
aibly receive a greater pressure than the others. 



AKCHES OF MASONRY. 



431 



mediately above the pier, and .-. does not need to be as 
large as the abutment of the first and last arches, since 
these latter must be prepared to resist the oblique thrusts 
of their arches without help from the thrust of another on 
the other side. 

In a very long series of arches it is sometimes customary 
to make a few of the intermediate piers large enough to 
act as abutments. These are called " abutment piers," and 
in case one arch should fall, no others would be lost except 
those occurring between the same two abutment piers as 
the first. See Fig. 392. A is an abutment-pier. 






Fig. 39;^. 



GRAPHICAL. TREATMENT OF ARCH. 

358. — Having found the " reduced load-contour," as in 
preceding paragraphs, for a given arch and load, we are 
ready to proceed with the graphic treatment, i.e., the first 
given, or assumed, form and thickness of arch-ring is to be 
investigated with regard to stability. It may be necessary 
to treat, separately, a lamina under the spandrel wall, and 
one under the interior loading. The constructions are 
equally well adapted to arches of all shapes, to Gothic as 
well as circular and elliptical. 

359.— Case I. Symmetrical Arch and Symmetrical Loading.— 
(The " steady " (permanent) or " dead " load on an arch is 
usually symmetrical). Fig. 393. From symmetry we need 




Fig. 



Fig. 394. 



Fis. 395. 



i33 MECHAJTICS OF ENGINEERmG. 

deal witli only one half (say the left) of tlie arch and load. 
Divide this semi-arch and load into six or ten divisions 
by vertical lines ; these divisions are considered as trape- 
zoids and should have the same horizontal width = 6 (a 
convenient whole number of feet) except the last one, LKN, 
next the abutment, and this is a pentagon of a different 
v\ridth hy, (the remnant of the horizontal distance LC). The 
weight of masonry in each division is equal to (the area 
of division) X (unity thickness of lamina) x (weight of a cu- 
bic unit of arch-ring). For example for a division having 
an area of 20 sq. feet, and composed of masonry weighing 
160 lbs. per cubic foot, we have 20x1x160=3,200 lbs., 
applied through the centre of gravity of the division. 
The area of a trapezoid. Fig. 394, is^&(7ii+7i2), audits cen- 
tre of gravity may be found. Fig. 395, by the construction 
of Prob. 6, in § 26 ; or by § 27a. The weight of the pen- 
tagon LN, Fig, 393, and its line of application (through 
centre of gravity) may be found by combining results for 
the two trapezoids into which it is divided by a vertical 
through K. See § 21. 

Since the weights of the respective trapezoids {excepts 
ing LN) are proportional to their middle vertical in- 
tercepts [such as ^(^1+7^2) Fig- 394] these intercepts (trans- 
ferred with the dividers) may be used directly to form the 
load-line, Fig. 396, or proportional parts of them if more 
convenient. The force scale, which this implies, is easily 
computed,, and a proper length calculated to represent the 
weight of the odd division LN ; i.e., 1 ... 2 on the load- 
line. 

Now consider A, the middle point of the abutment joint. 
Fig. 396, as the starting point of an equilibrium polygon 
(or abutment of a linear arch) for a given loading, and re- 
quire that this equilibrium polygon shall pass through j>, 
the middle of the crown joint, and through the middle of 
the abutment joint on the right (not shown in figure). 

Proceed as in § 342, thus determining the polygon Ap 
for the half-arch. Draw joints in the arch-ring through 
those points where the extrados is intersected by the ver- 



ABCHES OF MASONRY. 



433 




Jig. 396. Fig. 397. 

Heal separating tlie divisions (not the gravity verticals), 
Tlie points in which these joints are cut by the segments 
of the equilibrium polygon, Fig. 397, are (very nearly, if 
th« joint is not more than 60° from jp, the crown) the points 
of application in these joints, respectively, of the resultant 
pressures on them, (if this is the " true linear arch " for 
this arch and load) while the amount and direction of each 
such pressure is given by the proper ray in the force -dia- 
gram. 

If at any joint so drawn the linear arch (or equilibrium 
polygon) passes outside the middle third of the arch-ring, 
the point A, or p, (or both) should be judiciously moved 
(within the middle third) to find if possible a linear arch 
which keeps within limits at all joints. If this is found 
impossible, the thickness of the arch -ring may be increased 
at the abutment (giving a smaller increase toward the 
crown) and the desired result obtained ; or a change in the 
distribution or amount of the loading, if allowable, may 
gain this object. If but one linear arch can be drawn 
within the middle third, it may be considered the " true " 
one ; if several, the one most nearly co-inciding with the 
middles of the joints (see §§ 351 and 352) is so considered. 



360.— Case II Unsymmetrical Loading on a Symmetrical Arch,' 
(e.g., arch with live load covering one half -span as in Figs. 
390 and 391). Here we must evidently use a full force 
diagram, and the full elevation of the arch -ring and load* 



434 



MBCHAXICS OF EXGINEEELNG. 



See Fig. 398. Select three points A, p, and B, as follows, 
to determine a trial equilihriu'm ])6lygon : ' 

Select A at the Joicer limit of the middle third of tLa 




Fig. 398. 



abutment-joint at the end of the span -vihich is the more 
heavily -loaded ; in the other abutment-joint take B at tht 
upper limit of the middle third ; and take p in the middle 
of the crown-joint. Then by § 341 draw an equilibrium 
polygon (i.e., a linear arch) through these three points for 
the given set of loads, and if it does not remain within the 
middle third, try other positions for A, p, and B, within 
the middle third. As to the " true linear arch " alterations 
of the design, etc., the same remarks apply as already 
given in Case I. Very frequently it is not necessary to 
draw more than one linear arch, for a given loading, for 
even if one could be drawn nearer the middle of the arch- 
ring than the first, that fact is almost always apparent on 
mere inspection, and the one already drawn (if within 
middle third) will furnish values sufiiciently accurate for 
the pressures on the respective joints, and their direction 
angles. 

360a. — The design for the arch-ring and loading is not 
to be considered satisfactory until it is ascertained that for 
the dead load and any possible combination of live-load 
'(in addition) the pressure at any joint is 



ARCHES OF MASONRY. , 435 

1.) Witliin the middle third of that joint ; 

{2.) At an angle of < 30° with the normal to joint- 
SYirface. 

(3.) Of a mean pressure per square inch not > thanVa) 
of the ultimate crushing resistance. (See § 348.) 

§ 361. Abutments. — The abutment should be compactly 
and solidly built, and is then treated as a single rigid mass. 
The pressure of the lowest voussoir upon it (considering 
a lamina one foot thick) is given by the proper ray of the 
force diagram (0 .. 1, e. g., in Fig. 396) in amount and direc- 
tion. The stability of the abutment will depend on the 
amount and direction of the resultant obtained by com- 
bining that pressure P^ with the weight G of the abutment 
and its load, see Fig. 399. Assume a probable width RS 
for the abutment and compute the weight G 
of the corresponding abutment OBRS and 
MNBO, and find the centre of gravity of the 
whole mass G. Apply G in the vertical 
through C, and combine it with P„ at their in- 
tersection D. The resultant P should not cut 
the base R8m. a point beyond the middle third 
p/^ / " (or, if this rule gives too massive a pier, take 

I / / ° such a width that the pressure per square 

I// inch at 8 shall not exceed a safe value as 

^ Fis. 399. computed from § 362.) After one or two 
trials a satisfactory width can be obtained. 
We should also be assured that the angle PD G is less 
than 30°. The horizontal joints above RS should also be 
tested as if each were, in turn, the lowest base, and if 
nscessary may be inclined (like mn) to prevent slipping. 
On no joint should the maximum pressure per square inch 
be > than y^o the crushing strength of the cement. Abut- 
ments of firm natural rock are of course to be preferred 
where they can be had. If water penetrates under an 
abutment its buoyant effort lessens the weight of the lat- 
ter to a considerable extent. 



436 



MECHANICS OF ENGINEEKIifG. 



362. Maximum Pressure Per Unit of Area When the Resultant 
Pressure Falls at Any Given Distance from the Middle ; according 
to Navier's theory of the distribution of the pressure ; see 
§ 346. Case I. Let the resultant pressure P, Fig. 400, (a). 




Fig. 400. 



Fig. 401. 



fall within the middle third, a distance = wc? ( < ^ d) 
from the middle of joint [d = depth of Joint.) Then we 
have the following relations : 

p (the mean press, per.- sq. in.,),^,,, (max. press, persq. in.), 
and p^ (least press, per sq. in.) are proportional to the lines 
h (mid. width), a (max. base), and c (min. base) respectively, 
of a trapezoid. Fig. 400, (&), through whose centre of gravity 
P acts. But (§ 26) • 

nd=---. i.e., n = y^ — = — or a=h (6w+l) 

6 a-\-c n . 

''• Pm—JP (6w+l). Hence the following table: 



If 7id= j4> d Ya d 
press. Pn,= 2 y^ 



Vs 



then the max. 

times the mean pressure. 



Case II. Let P fall outside the mid. third, a distance = 
"nd {^ )4 d) from the middle of joint. Here, since the 
joint is not considered capable of withstanding tension, 
we have a triangle, instead of a trapezoid. Fig. 401. First 
compute the mean press, per sq. in. 



V - 



P (lbs.) 



(1— 2w) 18 d inches 
foot thick). 



or from this table : (lamina ona 



AKOHES OF MASOlfEY. 



437 



For nd = 


^d 


■hd 


■hd 


T\d 


^d 


^d 


P = 


1 P 

10* d 


1 P 

8 * d 


1 P 

6 * d' 


1 P 

4 <^ 


1 P 

2 6^ 


infinity. 



{d in inches and P in lbs.; with arch lamina 1 ft. in thickness.) 

Then the maximum pressure (at A, Fig. 401) />,„, = 2p, 
becomes known, in lbs. per sq. in. 

362a. Arch-ring under Uon- vertical Forces. — An example of 
this occurs when a vertical arch-ring is to support the pressure 
of a liquid on its extrados. Since water-pressures are always 
at right angles to the surface pressed on, these pressures on the 
extradosal surface of the arch-ring form a system of non-paral~ 
lei forces which are normal to the curve of the extrados at; 
their respective points of application and lie in parallel 
vertical planes, parallel to the faces of the lamina. "We here 
assume that the extradosal surface is a cylinder (in the most 
general sense) whose rectilinear elements are 1 to the faces of 
the lamina. If, then, we divide the length of the extrados, 
from crown to each abutment, into from six to ten parts, the 
respective pressures on the corresponding surfaces are obtained 
by multiplying the area of each by the depth of its centre of 
gravity from the upper free surface of the liquid, and this 
product by the weight of a unit of volume of the liquid ; and 
each such pressure may be considered as acting through the 
centre of the area. Finally, if we find the resultant of each 
of these pressures and the weight of the corresponding portion 
of the arch-ring, these resultants form a series of non-vertical 
forces in a plane, for which an equilibrium polygon can then 
be passed through three assumed points by § 378a, these three 
points being taken in the crown-joint and the two abutment- 
joints. As to the " true linear arch" see § 359. 

As an extreme theoretic limit it is worth noting that if the 
extrados and intrados of the arch-ring are concentric circles ; if 
the weights of the voussoirs are neglected ; and if the rise of 
tb« arch is very small compared with the depth of the crown 
^/elow the water surface, then the circularGentre-line of the 
wrch-ring is the " true linear archP 



4:38 MECHAIiflCS OF ENGLNiiJEJiLNG. 



CHAPTER XI. 



ARCH-RIBS. 

Note. — The methods used in this chapter for the treatment of the 
"elastic arch" are practically the equivalent of those based on the theory 
of "Least Work." 

364. Definitions and Assumptions. — An arcli-rib (or elastic- 
arch, as distinguished from a block-work arcb) is a rigid 
curved beam, either solid, or built up of pieces like a 
truss (and then called a braced arch) the stresses in which, 
under a given loading and with prescribed mode of sup- 
port it is here proposed to determine. The rib is sup- 
posed symmetrical about a vertical plane containing its 
axis or middle line, and the Moment of Inertia of any cross 
section is understood to be referred to a gravity axis of 
the section, which (the axis) is perpendicular to the said 
vertical plane. It is assumed that in its strained condi- 
tion under a load, the shape of the rib differs so little 
from its form when unstrained that the change in the ab- 
scissa or ordinate of any point in the rib axis (a curve) 
may be neglected when added (algebraically) to the co- 
ordinate itself ; also that the dimensions of a cross-section 
are small compared with the radius of curvature at any 
part of the curved axis, and with the span. 

365. Mode of Support. — Either extremity of the rib may be 
hinged to its pier (which gives freedom to the end-tangent- 
line to turn in the vertical plane of the rib when a load is 
applied); or may 'hef,xed, i.e., so built-in, or bolted rigid- 
ly to the pier, that the en^-tangent-line is incapable of 
changing its direction when a load is applied. A hinge 
may be inserted anywhere along the rib, and of course 



ARCH BIBS. 



439 



destroys the rigidity, or resistance to bending at that 
point. (A hinge having its pin horizontal "1 to the axis of 
the rib is meant). Evidently no more than three such 
hinges could be introduced along an arch- rib between two 
piers ; unless it is to be a hanging structure, acting as a 
suspension-cable. 

366. Arch Rib as a Free Body. — In considering the whole 
rib free it is convenient, for graphical treatment, that no 
section be conceived made at its extremities, if fixed ; hence 
in dealing with that mode of support the end of the rib 
will be considered as having a rigid prolongation reach- 
ing to a point vertically above or below the pier junction, 
an unknown distance from it, and there acted on by a force 
of such unknown amount and direction as to preserve the 
actual 'extremity of the rib and its tangent line in the same 
position and direction as they really are. As an illustra- 
tion of this Fig. 402 
shows free an arch rib. 
ONB, with its extremi- 
ties and BJixed in the 
piers, with no hinges, q 
and bearing two 
loads P. and P^. The 
other . :ces of the sys- 
tem holding it in equi- 
librium are che horizontal and vertical components, of the 
pier reactions {H, V, H,„ and V^), and in this case of fixed 
ends each .of these two reactions is a single force not in- 
tersecting the end of the rib, but cutting the vertical 
through the end in some point F (on the left ; and in G on 
the right) at some vertical distance c, (or d), from the end. 
Hence the utility of these imaginary prolongations OQF, 
and BRG, the pier being supposed removed. Compare 
Figs. 348 and 350. 

The imaginary points, or hinges, F and G, will be called 
ctbutments being such for the special equilibrium polygon 




Fig. 402. 



440 MBCHAlSriCS OF ENGINEEKING. 

(dotted line), while and B are the real ends of the curved 
beam, or rib. 

In this system of forces there are five unknowns, viz.: V, 
V,„ H = H^, and the distances c and d. Their determina- 
tion by analysis, even if the rib is a circular arc, is ex-, 
tremely intricate and tedious ; but by graphical statics 
(Prof, Eddy's method ; see § 350 for reference), it is com- 
paratively simple and direct aiid applies to any shape of 
rib, and is sufficiently accurate for practical purposes. 
This method consists of constructions leading to the loca- 
tion of the " special equilibrium polygon " and its force 
diagram. In case the rib is hinged to the piers, the re- 
actions of the latter act through these hinges, Fig. 403, 
i e., the abutments of the special 
equilibrium polygon coincide with 
the ends of the rib and B, and for 
a given rib and load the unknown 
quantities are only three V, F'n, and 
H; (strictly there are four ; but IX "^^ 
= gives H^ = H). The solution fig. 403. 

by analytics is possible only for ribs of simple algebraic 
curves and is long and cumbrous ; 'whereas Prol Eddy's 
graphic method is comparatively brief and simple and ia 
applicable to any shape of rib whatever. 

367. Utility of the Special Equilibriun Polygon and its force 

diagram. The use of locating these will now be illustrated 
[See § 832]. As proved in §§ 332 and 334 the compres- 
sion in each " rod " or segment of the '* special equilibrium 
polygon" is the anti-stress resultant of the cross sections in 
the corresponding portion of the beam, rib, or other struc- 
ture, the value of this compression (in lbs. or tons) being 
measured by the length of the parallel ray in the force 
diagram. Suppose that in some way (to be explained sub- 
sequently) the special equilibrium polygon and its force 
diagram have been drawn for the arch -rib in Fig. 404 hav- 
ing fixed ends, and B, and no hinges ; required the elastic 
stresses in any cross-section of the rib as at m. Let the 



ARCH RIBS. 



Ml 




FiG. 404. 

of the force-diagram on the right be 200 lbs. to the 
inch, say, and that of the space-diagram (on the left) 30 ft. 
to the inch. 

The cross section m lies in a portion TK, of the rib, cor- 
responding to the rod or segment he of the equilibrium 
polygon; hence its anti-stress-resultant is a force R2 acting 
in the line 6c, and of an amount given in the force-diagram. 
Now i?2 is the resultant of V, H, and Pj, which with the 
elastic forces at m form a system in equilibrium, shown in 
?ig. 405 ; the portion FO Tm being considered free. Hence 




Pig. 405. Fio. 406. 

taking the tangent line and the normal at m as axes we 
should have I (tang, comps.) = ; -T (norm, comps.) = j 
and 2* (moms, about gravity axis of the section at w) = Oj 
and could thus find the unknowns pi, "p^, and J", which ap- 
pear iu the expressions 'p^F the thrust, ^ the moment* of 



442 MECHANICS OF ENGINEERING. 

the stress-couple, and J the shear. These elastic stresses 
are classified as in § 295, which see. p^ and jpa are ^hs. per 
square inch, J is lbs., e is the distance from the horizontal 
gravity axis of the section to the outermost element of 
area, (where the compression or tension is p^ lbs. per sq. 
in., as due to the stress-couple alone) while I is the " mo- 
ment of inertia " of the section about that gravity axis. 
[See §§ 247 and 295 ; also § 85]. Graphics, however, gives 
us a m.ore direct method, as follows : Since i?2> i^ the line 
he, is the equivalent of V, H, and Pj, the stresses at m will 
be just the same as if ^2 acted directly upon a lateral pro- 
longation of the rib at T (to intersect ScFig. 405) as shown 
in Fig, 406, this prolongation Tb taking the place of TOF 
in Fig. 405. The force diagram is also reproduced here. 
Let a denote the length of the "] from m's gravity axis 
upon he, and 2 the vertical intercept between m and Jc. 
For this imaginary free body, we have, 

from I (tang. compons.)=0, i?2 cos a=^piF 

and from 2' (norm. compons.)=0,i?2 sin «=«/ 

while from J' (moms, about) ) ^ ^ rj P2I 

,-, ., ^- 4; \ A }-we have ixott =^ -^ • 

the gravity axis 01 to)=0, j ^ e 

But from the two similar triangles (shaded ; one of them 
is in force diagram) a :z :; Zf:i?2 .•. R^a^ Hz, wIh&tigq we 
may rewrite these relations as follows (with a general state- 
ment), viz.: 

If the Special Equilibrium Polygon and Its Force Diagram Have 
iBeen Drawn for a given arch-rib, of given mode of support, 
p.nd under a given loading, then in any cross-section of the 
J ib, we have {F = area of section): 

The projection of the proper 

i \ \ rri.^T>.,m.f i^ ^ rr- J ^«2/ (of tlie force diagram) up- 
(L) The Thrust. i.e.,i>,i^-^ ^^^^^ ^^^^^^^ ^.^^ ^^ ^-^^ ^i^, 

drawn at the given section. 



ARCH KIBS. 443 



(2.) The Shear, i.e., J", = C 

/ 1-11 1 J.1 The proieetion of me proper 

(upon which dependstne , » ,i « -,. ^ 

^, . , . ,1 ray (oi the lorce diaarram) up- 

shearmg stress m the-^ "^ .; ? , i^i -i 

web). (See §§ 253 and 



256). 



on the normal to the rib curve 
at the given section. 



(3.) The Moment of the 
stress couple, i.e.,-^ , = " 



6 



The product {Hz) of the fl 
(or pole-distance) of the force- 
diagram by the vertical dis- 
tance of the gravity axis of the 
section from the spec, equilib- 
rium polygon. 

By the ** proper ray " is meant that ray which is parallel 
to the segment (of the equil. polygon) immediately under 
or above which the given section is situated. Thus in 
Fig. 404, the proper ray for any section on TK is B2 ; on 
KB, i?3 ; on TO, Bi. The projection of a ray upon any 
given tangent or normal, is easily found by drawing through 
each end of the ray a line ^ to the tangent (or normal) ; 
the length between these "I's on the tangent (or normal) is 
the force required (by the scale of the force diagram). We 
may thus construct a shear diagram, and a thrust diagram 
for a given case, while the successive vertical intercepts 
between the rib and special equilibrium polygon form a 
moment diagram. For example if the s of a point m is ^ 
inch in a space diagram drawn to a scale of 20 feet to the 
inch, while Zf measures 2.1 inches in a force diagram con- 
structed on a scale of ten tons to the inch, we have, for the 
moment of the stress-couple at m, J!f=^s= [2.1x10] tons 
X [ ^ X 20] ft. =210 ft. tons. 

368. — It is thus seen how a location of the special equili 
Ibrium polygon, and the lines of the corresponding iovoi 
diagram, lead directly to a knowledge of the stresses in al 
the cross-sections of the curved beam under consideration, 
bearing a given load; or, vice versa, leads to a stateme?^^ 
of conditions to be satisfied by the dimensions of the rir 
for proper security. 

It is here supposed that the rib has sufficient latei 



444 



MECHANICS OF ENGlNEElil^rG, 



bracing (witli others wliicli lie parallel witli it) to prevent 
buckling sideways in any part like a long column. Before 
proceeding to the complete graphical analysis of the differ- 
ent cases of arch-ribs, it will be necessary to devote the 
next few paragraphs to developing a few analytical rela- 
tions in the theory of flexure of a curved beam, and to 
giving some processes in " graphical arithmetic." 

369. Change in the Angle Between Two Consecutive Rib Tan- 
gents when the rib is loaded, as compared with its value 
before loading. Consider any small portion (of an arch 
rib) included between two consecutive cross- sections ; Fig. 
407. KHG W is its unstrained form. Let EA, = ds, be 
the original length of this portion of the rib axis. The 
length of all the fibres (ii to rib-axis) was originally =ds 
(nearly) and the two consecutive tangent-lines, at E and Ay 
made an angle = dO originally, with each other. While 
under strain, however, all the fibres are shortened equally 
an amount dX^, by the uniformly distributed tangential 
thrust, but are unequally shortened (or lengthened, accord- 
ing as they are on one side or the other of the gravity axis 
E, or A, of the section) by the system of forces making 
what we call the " stress couple," among v/hich the stress 
at the distance e from the gravity axis A of the section is 
called p-i per square inch ; so that the tangent line at A' 
now takes the direction A'D ~j to H'A'G' instead of A'G 
(we suppose the section at E to remain fixed, for coUTezii- 



^6! =^ 




t/"/^'^' cIp.^' 



.-r- 



"'v-^.^f** 



^pd? 



AECH lilBS. 445 

ence, since tlie change of angle between tlie two tangents 
depends on the stresses acting, and not on tlie new posi- 
tion in space, of this part of the rib), and hence the angle 
between the tangent-lines at E and A (originally = dd) is 
now increased by an amount GA'D = d(p (or O'A'R = dip); 
G'H' is the new position of GH. We obtain the value of 
d(p as follows : That part {dk^ of the shortening of the 
fibre at Q, at distance e from A due to the force p.^dFy is 

§ 201 eq. (1), dX.2 = ^ft' But, geometrically, J^ also ~edf, 

Eedcp-^pzds ■ (1.) 

But, letting ilf denote the moment of the stress-couple 
at section A (ilf depends on the loading, mode of support, 
etc., in any particular case) we know from § 295 eq. (6) that 

Jf=-^j and hence by substitution in (1) we. have 



•, Mds r^x 

'^^^I . • . . (2) 

[If the arch-rib in question has less than three hinges, 
the equal shortening of the due to the thrust (of 

the block in last figure) p^F, will have an indirect effect on 
the angle d(p. This will be considered later.] 

370. Total Change i.e. CcU in the Angle Between the End 

Tangents of a Rib, before and after loading. Take the ex- 
ample in Fig. 408 of a rib fixed at one end and hinged at 



Fig. 408. 



446 



MECHAT^ICS OF ENGINEEEIJIG. 



the other. When the rib is unstrained (as it is supposed 
to be, on the left, its own weight being neglected ; it is not 
supposed sprung into place, but is entirely without strain) 
then the angle between the end-tangents has some value 

6' = j dd— the sum of the successive small angles dd for 

do 

each element ds of the rib curve (or axis). After loading., 
[on the right, Fig. 408], this angle has increased having 
now a value 



d'-\- r d(p, i.e., a value = d'+ C -—jr 



(I.) 




Fig. 409. 



There must oe no hinge between 
and B. 

§ 371. Example of Eq[iiation (I.) in Anal" 
ysis. — ^A straight, homogeneous, pris- 
matic beam, Fig. 409, its own weight 
neglected, is fixed obliquely in a wall. 
After placing a load P on the free end, 
required the angle between the end- 
tangents. This was zero before load- 
ing .'. its value after loading is 



=o+f'=o+ 4r r'-^^^^ 



UIJo 



By considering free a portion between and any da of the 
beam, we find that M=Fx=mom.. of the stress couple. 
The flexure is so slight that the angle between any ds and 
its dx is still practically =a (§ 364), and .*. ds=dx sec a. 
Hence, by substitution in eq. (I.) we have 



^'=A rms= l^rxdx= 

^ EI Jo EI Jo 



P sec ar'*'^* 
L2 ' 



EI 



... ^'=?^^^l [Compare with § 237]. 



ARCH EIBS. 



447 



It is now apparent that if hoth ends of an arcli rib are 
fixed, wlien unstrained, and the rib be then loaded (within 
elastic limit, and deformation slight) we must have 



r {Mds-^Eiy 



zero, since (p'=0. 



372. Projections of the Displacement of any Point of a Loaded 
Uib Relatively to Another Point and the Tangent Line at the Lat- 
ter. — (There must be no hinge between and B). Let 
be the point whose displacement is considered and B the 
other point. Fig. 410. If ^'s tangent-line is fixed while 
the extremity is not supported in any way (Fig. 410) 
then a load P put on, is displaced to a new position 0^, 




Fig. 410. 



Fig. 411, 



Fig; 413. 



With as an origin and OB as the axis of X, the projec- 
tion of the displacement OOj, upon X, will be called Ja?, 
that upon Y, Ay. 

In the case in Fig.. 410, O's displacement with respect to 
B and its tangent-line BT, is also its absolute displacement 
in space, since neither B nor BT has moved as the rib 
changes form under the load. In Fig. 411, however, the 
extremities and B are both hinged to piers, or supports, 
the dotted line showing its form when deformed under a 
load. The hinges are supposed immovable, the rib being 
free to turn about them without friction. The dotted line 
is the changed form under a load, and the absolute dis- 
placement of is zero ; but not so its displacement rela- 
tively to B and j5's tangent BT, for BT has moved to a 
new position BT'. To find this relative displacement con- 
ceive the new curve of the rib superposed on the old in 
a way that B and BT m&j coincide with their original po- 



448 



MECHANICS OF ENGIIsTEEEING. 



sitions. Fig. 412. It is now seen that O's displacement 
relatively to B and BT is not zero but =00„, and lias a 
small Jx but a comparatively large zly. In fact for this 
case of hinged ends, piers immovable, rib continuous be- 
tween 'them, and deformation slight, we shall write Jx = 
zero as compared with Jy, the axis X passing through OB). 

373. Values of the X and Y Projections of O's Displacement Rela- 
tively to Band B's Tangent; the origin being taken at 0. 
Fig. 413. Let the co- 
ordinates of the dif- 
ferent points jE, I), G, 
etc., of the rib, re- 
ferred to and an 
arbitrary X axis, be 
X and y, their radial 
distances from be- 
. ing u (i.e., u for G, u' 
for D, etc.; in gener- 
al, ^0- OEDG is the J 
unstrained form of the 
rib, (e.g., the form it 
would assume if it lay flat on its side on a level platform, 
under no straining forces), while 0,,E"B'GB is its form 
under some loading, i.e., under strain. (The superposi- 
tion above mentioned (§ 372) is supposed already made if 
necessary, so that BT i^ tangent at B to both forms). 
Now conceive the rib OB to pass into its strained condi- 
tion by the successive bending of each ds in turn. The 
straining or bending of the first ds, BC, through the small 
angle d(p (dependent on the moment of the stress couple 
at G in the strained condition) causes the whole finite piece 
OG io turn about (7 as a centre through the same small 
angle d(p ; hence the point describes a small linear arc 
00'=ov, whose radius = u the hypothenuse of the x and 
y of G, and whose value .*. is dv=ud(p. 

Next let the section B, now at D\ turn through its 
proper angle d(p' (dependent on its stress-couple) carrying 




Fig. 413. 



AECH KIBS. ' 449 

with it the portion D'O', into the position D'O", making 
0' describe a linear arc O'O" = {dvy =u'd(p', in which u' = 
the hypothenuse on the x' and y' (of D), (the deformation 
is so slight that the co-ordinates of the different points 
referred to and X are not appreciably affected). Thus, 
each section having been allowed to turn through the an- 
gle proper to it, finally reaches its position, 0„, of dis- 
placement. Each successive dv, or linear arc described by 
0, has a shorter radius. Let dx, {dx)', etc., represent the 
projections of the successive (^v)'s upon the axis X; and 
similarly dy, (dy)' etc., upon the axis Y. Then the total X 
projection of the curved line . . . . 0^ will be 

Jx= / (5j? and similarly J?/ = / dy , , , (1) 

But d V = u d (f, and from similar -right-triangles, 
3 x: dv : : y : u and dy : 8v :: x : u .'. 8x = yd<p and dy=xd<p ; 
whence, (see (1) and (2) of §369) 

Ax = fS. = fyd^=£JMl... (IL) 

and Ay = C dy = C xd(p = C ^^ . . . , (III.^ 

If the rib is homogeneous E is constant, and if it is of 
constant cross-section, all sections being similarly cut by 
the vertical plane of the rib's axis (i.e., if it is a " curved 
prism ") /, the moment of inertia is also constant. 

374. Hecapitulation of Analytical Relations, for reference* 
(Not applicable if there is a hinge between and E) 

Total Change in Angle between ) _ ^^Mds 



tai Lfiiaiige lu Aiigie oerween / _ p>"m.as .j ^ 

tangent-lines and ^ [ ~ Jo ^ ' o • • W 

The X-Projection of O's Displacement "] 

Relatively to B and B's tangent- I A^Myds /tt ^ 

line ; {the origin being at 0) I — / — ^fjr- • • • (JJ-J 
and the axes X and F 1 to [ ^o i^i 
each other) • 



450 



MECHAKICS OF EKGINEEKING. 



The Y-Prqjection of O's Displacement, | _ 

etc., as above. 



. (m.) 



Hviie X anv.. y are the co-ordinates of points in the rib- 
curve, ds an element of that curve, M the moment of the 
stress-couple in the corresponding section as induced by 
the loading, or constraint, of the rib. 

(The results already derived for deflections, slopes, eto„, 
for straight beams, could also be obtained from these 
formulae, I., 11. and III. In these formulae also it must 
be remembered that no account has been taken of the 
shortening of the rib-axis by the thrust, nor of the effect 
of a change of temperature.) 

374a. R^sumfe of the Properties of Ec^mlibiiiim Polygons and 
their Force Diagrams, for Systems of Vertical Loads. — See §§ 335 
to 343. Given a system of loads or vertical forces, P^, P2, 

1 etc., Eig. 414, and 
two abutment verti- 
cals, F' and G' ; if 
"we lay off, vertically, 
to form a " load- 
line," 1 .. 2 - P^, 2. . . 
8=P2> etc., select any 
Pole, Oi, and join 0^ 
... 1, Oi . . . 2, etc. ; 
also, beginning at 
any point F^ in the 
vertical P', if we draw i^i . . . a I| to Oj . . 1 to intersect the 
line of Pi ; then ah \\ to Oi . . 2, and so on until finally a 
point G]y in G', is determined; then the figure Pj ,ahc G^iis 
an equilibrium polygon for the given loads and load verti- 
cals, and Oi . . . 1234 is its ". force diagram." The former 
is so called because the short segments PjCt oh, etc., if 
considered to be rigid and imponderable rods, in a vertical 
plane, hinged to each other and the terminal ones to abut- 
ments Pi and G^, would be in equilibrium under the given 
loads hung at the joints. An infinite number of equilil> 




Fig. 414. 



AECH-HIBS. 451 

rium polygons may be drawn for tlie given loads and 
abntment-verticals, by choosing different poles in the force 
diagram. [One other is shown in the fignie ; O2 is its 
pole. {Fi Gi and F2 U-^ are abutment lines.)] For all of 
these the following statements are true : 

(1.) A line through the pole, i| to the abutiifint line cutii 
the load-line in the same point n', whichever equilibrium 
polygon be used ( /. any one will serve to determine n' 

(2.) If a vertical GI) he drawn, giving an intercept z' in 
each of the equilibrium polygons, the product Hz' is the 
same for all the equilibrium polygons. That is, (see Fig., 
414) for any two of the polygons we have 

H,:H,:: z/ : z,' ; or H,z,' = H, z,'. 

(3.) The compression in each rod is given by thai 
" ray " (in the force diagram) to which it is parallel. 

(4.) The " pole distance " H, or ~| let fall from the pole 
upon the load-line, divides it into two parts which are the 
vertical components oi the compressions in the abutment- 
rods respectively ( the other component being horizontal) ; 
H is the horizontal component of each (and, in fact, of 
each of the compressions in all the other rods). The 
compressions in the extreme rods may also be called the 
abutment reactions (oblique) and are given byti^e extreme 
rays. 

(5.) Three Points [not all in the same segment (or rod)] 
determine an equilibrium polygon for given loads. Hav- 
ing given, then, three points, we may draw the eaailibrium 

polygon by §341. 

« 

375. Summation of Prcducts. Before proceedini^ to treat 
graphically any case of arch-ribs, a few processes in 
graphical arithmetic, as it may be called, must be pre- 
sented, and thus established for future use. 

To make a summation of products of two factors in each 
by means of an equilibrium polygon. 



452 



MECHANICS OF ENGINEERING. 



Construction, Suppose it required to make the summa- 
tion I {x z) {. e., to sum the series 



Xi %+ X2 Z2 + x^z^ 4. 



bj graphics. 




Having first arranged the terms in the order of magnf- 
tude of the ic's, we proceed as follows : Supposing, for 
illustration, that two of the s's (% and z^ are negative 
(clotted in figure) see Fig. 415. These quantities x and z 
may be of any nature whatever, anything capable of being 
represented by a length, laid off to scale. 

First, in Fig 
416, lay off the 
s's in their 
order, end to 
end, on a ver- 
tical load-line 
taking care to 
"^ lay off % and 
.. % upuard in 
^« their turn. 
Take any con- 
FiG. 416. venient j-ola 

; draw the rays ... 1, ... 2, etc.; then, having pre- 
viously drawn vertical lines whose horizontal distances 
from an extreme left-hand vertical F' are made = x^, x-, 
Xs, etc., respectively, we begin at any point F, in the verti- 
cal F', and draw a line 11 to ... 1 to intersect the Xi ver- 
tical in some point ; then 1' 2' II to . . . % and so on, fol- 
lowing carefully the proper order. Produce the last seg- 
ment (6' ... (x in this case) to intersect the vertical F' in 
some point K. Let KF =k (measured on the same scale 
as the i»'s), then the summation required is 

J/ {xz) = m. 

H is measured on the scale of the 2's, which need, not be 
the same as that of the aj's ; in fact the 2's may not be the 
same kind of quantity as the a;'s. 

[Peoof. — From similar triangles H: z^v.x^^: h^, .'. x^z^ — IHc^ \ 
and " " " H\Zo :: x^ : ^2> •*• x.^Zi=Hki . 



Fig. 415. 



AECH-KIBS. 



453 



and so on. But H {h,+h+eiG.)^HxFK=Hh']. 

376. Gravity Vertical. — From the same construction in 
Fig. 415 we can determine tlie line of action (or gravity 
vertical) of tlie resultant of the parallel vertical forces 2i, 
Z2, etc. (or loads); by prolonging the first and last segments 

to their intersection at 
0. The resultant of the 
system of forces or loads 
acts through C and is 
vertical in this case ; its 
value being = ^ (2), 
that is, it = the length 
1 ... 7 in the force dia- 
gram, interpreted by the 
proper scale. It is now 
supposed that the 2's 
represent forces, the x'b, 
being their respective 
lever arms about F. If 
the ?'s represent the 
areas of small finite por- 
tions of a large plane 
figure, we may find a 
gravity -line (through C) 
of that figure by the 
above construction; each 
z being-applied through 
the centre of gravity of 
its own portion. 

Calling the distance 
X between the verticals 
through C and F, we 
have also x . I [z) = 
I (xz) because I (z) is 
the resultant of the || z's. 
^' This is also evident from 
the proportion (similar 
triangles) 
H : (1. .7)::x:Jc. 




454 MECHANICS OF ENGINEERING. 

376a. Moment of Inertia (of Plane Figure) by Graphics.* — 

rig. 416a. /n= ? First, for the portion on right. Divide OR 
into equal parts each = ^x. Let «i, Z2, etc., be the middle 
ordinates of the strips thus obtained, and x^, etc. their 
abscissas (of middle points). 
Then we have approximately 



/n for 0R=Ax.ZyX^-\-Ax.Z2xi-{- 

=Ax[{z^Xi)x^+{z2X^X2-\- ...]..(!) 

But by §375 we may construct the products ZyXi,Z2X2, etc., 
taking a convenient H\ (see Fig. 416, (&)), and obtain \, ki, 
etc., such that z^x^ = H'ki, z^x^ = H'k2i etc. Hence eq. (1) 
becomes : 

li,ioT OR a.-p-prox.=II'^x[kiXi-\-k2X2-{- ...]... (2) 

By a second use of § 375 (see Fig. 416 c) we construct Z, 
such that kiX^ + kzXz +....= £["l \^H" taken at con- 
venience]. .'. from eq. (2) we have finally, (approx.), 

In for OR=H'H"lAx (3) 

For example if OR — 4 in., with four strips. Ax would = 
1 in.; and if ^' = 2 in., H" = 2 in., and I = 5.2 in., then 

Jn for OR = 2x2x5.2x1.0=20.8 biquad. inches. 

The 7x for OL, on the left of N, is found in a similar 
manner and added to 7^ for OR to obtain the total I^. The 
position of a gravity axis is easily found by cutting the 
shape out of sheet metal and balancing on a knife edge ; or 
may be obtained graphically by § 336 ; or 376. 

377. Construction for locating a line vw (Fig. 417) at (a), in 
the polygon FG in such a position as to satisfy the two 
following conditions with reference to the vertical inter- 
cepts at 1, 2, 3, 4, and 5, between it and the given points 
1„ 2, 3, etc., of the perimeter of the polygon. 

* Another graphic method for this purpose will be found in § 76 (p. 80), 
of the author's Notes and Examples in Mechanics. 



ARCH-EIBS. 



455 



Condition I. — (Calling these intercepts u^, u^, etc., and tlieir 
horizontal distances from a given vertical F, x^, x.^, etc.) 

2" (u) is to = ; i.e., the sum of the positive u's must be 
numerically — - that of the negative (which here are at 1 
and 5). An infinite number of positions of vm will satisfy 
condition I. 



Condition II. — 2* (ux) is to = ; i.e., the sum of the 

ij 1 1 . r-— ;^?n moments of the positive u'^ 

• — ■' ■ — "^^^^ ' about F must = that of the 
'' negative -m's. i.e., the moment 
of the resultant of the posi- 
tive w's must = that of the 
resultant of the negative ; 
and .*. (Condit. I being 
already satisfied) these two 
resultants must be directly 
opposed and equal. But the 
ordinates u in (a) are indi- 
vidually equal to the difiFer- 
ence of the full and dotted 
ordinates in (&) with the 
same cc's .'. the conditions 
may be rewritten : 

I. 2 (full ords. in (6))= 
2" (dotted ords. in (&)) 

II. 2 [each full ord. in (h) 
X its £c] = - [each dotted 
ord. in (b) x its x] i.e., the 

Fig. 4ir. Centres of gravity of the full 

and of the dotted in (6) must lie in the same vertical 

Again, by joining ^(x, we may divide the dotted ordi- 
nates of (b) into two sets which are dotted, and broken, re- 
spectively, in (c) Then, finally, drawing in (d), 

B, the resultant of full ords. of (c) 
T, " " " broken " " " 

T', " " " dotted " " " 




456 MECHANICS OF ENGINEERING. 

we are prepared to state in still another and final form tlie 
conditions wliicli vm must fulfil, viz. : 

(I.) T+T must = i?; and (II.) The resultant of T 
and T' must act in the same vertical as R. 

In short, the quantities T, T', and R must form a bal- 
anced system, considered as forces. All of which amounts 
practically to this : that if the verticals in which T and T' 
act are known and R be conceived as a load supported by 
a horizontal beam (see foot of Fig. 417, last figure) resting 
on piers in those verticals, then T and T' are the respec- 
tive reac^'^ons o/" ^Aose jjiers. It will now be shown that the 
verticals of T and T' are easily found, being independent of 
the position of vm; and that both the vertical and the mag- 
nitude of R, being likewise independent of vm, are deter- 
mined with facility in advance. For, if v be shifted up 
or down, all the broken ordinates in (c) or {d) will change 
in the same proportion (viz. as vF changes), while the 
dotted ordinates, though shifted along their verticals, do 
not change in value ; hence the shifting of v affects neither 
the vertical nor the value of T', nor the vertical of T. 
The value of T, however, is proportional to vF. Similar- 
ly, if m be shifted, up or down, T' will vary proportionally 
to mG, but its vertical, or line of action, remains the same. 
T is unaffected in any way by the shifting of m. R, de- 
pending for its value and position on the full ordinates of 
(c) Fig. 417, is independent of the location of vm. We 
may .*. proceed as follows : 

1st. Determine R graphically, in amount and position, 
by means of § 376. 

2ndly. Determine the verticals of T and T' by any trial 
position of vm (call it v-im.,), and the corresponding trial 
values of T and T' (call them T, and T',). 

3rdly. By the fiction of the horizontal beam, construct 
(§ 329) or compute the true values of T and T', and then 
determine the true distances vF and w6^ by the propor- 
tions 

vF : v.F :: T : T. and mG : m,G : : T' i T^. 



AECH-EIBS. 



457 



Example of this. Fig. 418. (See Fig. 417 for s and t.) 

From A tovi^ard B in (e) Fig. 418, lay off the lengths (or 
lines proportional 
to them) of the full 
ordinates 1, 2, etc., 
of (/). Take any 
pole Oi, and draw the 
equilibrium poly- 
gon {/y and pro- 
long its extreme seg- 
ments to find C and 
thus determine ^'s 
vertical. JR is repre- 
sented by AB. In 
(g) [same as (/) but 
shifted to avoid 
complexity of lines] 
draw a trial VoWi and 
join V2 G2. Deter- 
mine the sum T2 of 
the broken ordi- TFig. 418. 

nates (between V2G2 ana ^^2^2) and its vertical line of ap- 
plication, precisely as in dealing with B ; also T'2 that of 
the dotted ordinates (five) and its vertical. Now the true 
T=Btj-{s+t) and the true T'=Bs^(s+t). Hen ce com- 
pute vF={T^T2) ^2 and ?^^=(T'-^^^) m^G^., and by 
laying them off vertically upward from F and G respec- 
tively we determine v and m, i.e., the line vm to fulfil the 
conditions imposed at the beginning of this article, rela- 
ting to the vertical ordinates intercepted between vm and 
given points on the perimeter of a polygon or curve. 

Note (a\ If the verticals in which the intercepts lie are 
equidistant and quite numerous, then the lines of action 
of T2 and T'2 will divide the horizontal distance between 
F and G into three equal parts. This will be exactly true 
in the application of this construction to § 390. 

Note (b). Also, if the verticals are symmetrically placed 
about a vertical line, (as will usually be the case) VjWg is 




458 



MECHANICS OF ENGINEERING. 



best drawn parallel to FG, for then T^ and T'^ will be 
equal and equi-distant from said vertical line. 

378, Classification of Arch-Eibs, or Elastic Arches, according 
to continuity and modes of support. In the accompany^ 
ing figures Htxefull curves show the unstrained form of the 
rib (before any load, even its own weight, is permitted to 
come upon it) ;the dotted curve shows its shape (much ex- 
aggerated) when bearing a load. For a given loading 
Three Conditions must be given to determine the special 
equilibrium polygon (§§ 366 and 367). 

Class A. — Continuous rib, free to slip laterally on the 
piers, which have smooth horizontal surfaces. Fig. 420. 

This is chiefly of theoretic interest, its consideration 
being therefore omitted. The pier reactions are neces- 
sarily vertical, just as if it were a straight horizontal 
beam. 

Class B. Rib of Three Hinges, two at the piers and one 
intermediate (usually at the crown) Fig. 421. Fig. 36 also 
is an example of this. That is, the rib is discontinuous 
and of two segments. Since at each hinge the moment of 
tlie stress couple must be zero, the special equilibrium 
polygon must pass through the hinges. Hence as three 
points fully determine an equilibrium polygon for given 
load, the special equilibrium is drawn by § 341. 




Fig. 420. 



Fig. 421. 



[§ 378a will contain a construction for arch-ribs of three 
hinges, when the forces are not all vertical.] 

Class C. Rib of Two Hinges, these being at the piers, the 
rib continuous between. The piers are considered im- 
movable, i.e., the span cannot change as a consequence of 
loading. It is also considered that the rib is fitted to its 



AKCH RIBS. 



459 



hinges at a definite temperature, and is tlien under no con- 
straint from the piers (as if it lay flat on the ground), not 
even its own weight being permitted to act when it is fi- 
nally put into position. When the " false works " 
or temporary supports are removed, stresses are in- 
duced in the rib both by its loading, including its 
own weight, and by a change of temperature. Stresses 
due to temperature may be ascertained separately and 
then combined with those duo to the loading. [Classes 
A and B are not subject to temperature stresses.] Fig. 

422 shows a rib of two hinges, 
at ends. Conceive the dotted 
curve (form and position un- 
der strain) to be superposed 
on the continuous curve 
(form before strain) in such 
a way that B and its tangent 
line (which has been dis- 
placed from its original position) may occupy their pre- 
vious position. This gives us the broken curve O^B. 00,^ 
is .*. O's displacement relatively to B and -S's tangent, 
Now the piers being immovable OqB (right line) =05 ; i.e., 
the X projection (or Jx) of OOn upon OB (taken as an axis 
of X) is zero compared with its Jy. Hence as one condi- 
tion to fix the special equilibrium polygon for a given load- 
ing we have (from § 373) 




Fig. 422. 



r^[Myds-^EI^=0 



(1) 



The other two are that the [ must pass through . (2) 
special equilibrium polygon ) " " " B . (3) 

Class D. Bill with Fixed Ends and no hinges, i.e., continu- 
ous. Piers immovable. The ends may be ^xed by being 
inserted, or built, in the masonry, or by being fastened to 
large plates which are bolted to the piers. [The St. Louis 
Bridge and that at Coblenz over the Rhine are of this 
class.] Fig. 423. In this class there being no hinges we 



460 



MECHANICS OF ENGINEERING. 




Fig. 423. 



have no point given in advance througli whicli the special 
equilibrium polygon must pass. However, since O's dis- 
placement relatively (and absolutely) to B and ^'s tangent 
is zero, both z/a:; and z/^[see § 373] = zero, AIsq the tan- 
gent-lines both at and B being 
fixed in direction, the angle be- 
tween them is the same under 
loading, or change of temperature, 
as when the rib was first placed 
in position under no strain and at 
a definite temperature. 
Hence the conditions for locating the special equilibrium 
polygon are 

p^ Mds _ Q . p Myds ^ ^ . n^ Mxds _ q 

Jo ^jT ' Jo '~m~ ' Jo EI 

In the figure the imaginary rigid prolongations at the 
ends are shown [see § 366]. 

Other designs than those mentioned are practicable 
(such as : one end fixed, the other hinged ; both ends fixed 
and one hinge between, etc.), but are of unusual occur- 
rence. 

378a. Eib of Three Hinges, Forces not all Vertical,* If the 
given rib of three hinges upholds a roof, the wind-press- 
ure on which is to be considered as well as the weights of 
the materials composing the roof-covering, the forces will 
not all be vertical. To draw the special equil, polygon in 

such a case the following 
construction holds : Re- 
quired to draw an equilib- 
rium polygon, for any 
plane system of forces, 
through three arbitrary 
xs^ points. A, p and B ; Fig, 
B 423a. Find the line of 
action of B^, the resultant 
of all the forces occurring 
between A and p; also, 




Fig. 423a. 



* See p. 117 of the author's "Notes and Examples io Mechanics" for > 
detailed example of the following construction. 



ARCH-EIBS. 4C1 

that of R,, tlie resultant of all forces between ^p and B ; 
also the line of action of B, tlie resultant of B^ and B.2, [see 
§ 328.] Join any point iH^ in ^ witli A and also witli B, 
and join the intersections iVand 0. Then A iV will be the 
direction of the first segment, B that of the last, and 
NO itself is the segment corresponding to p (in the de- 
sired polygon) of an equilibrium polygon for the given 
forces. See § 328. If A N' p 0' B are the corresponding 
segments (as yet unknown) of the desired equil. polygon, 
we note that the two triangles MNO and M'N' O, having 
their vertices on three lines which meet in a point [i.e., B 
meets Bi and B^ in C], are homological [see Prop. YII. of 
Introduc. to Modern Geometry, in Chauvenet's Geometry,] 
and that . • . the three intersections of their corresponding 
sides must lie on the same straight line. Of those inter-r 
sections we already have A and B, while the third must be 
at G, found at the intersection of AB and NO. Hence by 
connecting C and p, we determine N and 0'. Joining 
N'A aiid O'B, the first ray of the required force diagram will 
be II to NA, while the last ray will be || to O'B, and thus 
the pole of that diagram can easily be found and the cor- 
responding equilibrium polygon, beginning at A, will pass 
through p and B. 
(This general case includes those of §§ 341 and 342.) 

379. Arch-Rib of two Hinges; by Prof. Eddy's Method.* 
[It is understood that the hinges are at the ends.] Re- 
quired the location of the special equilibrium polygon. "VVe 
here suppose the rib homogeneous (i.e., the modulus of 
Elasticity E is the same throughout), that it is a " curved 
prism " (i.e., that the moment of inertia / of the cross- 
section is constant), that the piers are on a level, and that 
the rib-curve is symmetrical about a vertical line. Fig. 

424. For each point m of the rib 
curve we have an x and y (both 
known, being the co-ordinates of 
the point), and also a z (intercept 
between rib and special equilib- 
FiG. 424. ^ rium polygon) and a z' (intercept 

*P. S5 of Prof. Eddy's book ; see reference in preface of this work. 




462 



MECHANICS OF ENGINEEEING. 



between tlie spec. eq. pol. and the axis X (whicli is OB). 
The first condition given in § 378 for Class C may be 
transformed as follows, remembering [§ 367 eq. (3)] that 
M = Hz at any point m of the rib (and that EI is con- 
stant). 



1^ 

EI 



H 



r Myds = 0, i.e., — C zyds = . • . f zyds 

do El c/o t/o 







^^y _^, \-''J^(y~ ^')yds=0', i.e., J^ yyds =J^ yz'ds . (1) 

In practical graphics we can not deal with infinitesimals ; 
hence we must substitute As a small finite portion of the 
rib-curve for ds', eq. (1) now reads I^ yy As = 2'^ yz' As. 
But if we take all the As's equal, As is a common factor 
and cancels out, leaving as a final form for eq. (1) 

I^\yy) = I^^{yz') . . . (1/ 

The other two conditions are that the special equilibrium 
polygon begins at and ends at B. (The subdivision of 
the rib-curve into an even number of equal As's will be ob- 
served in all problems henceforth.) 

379a. Detail of the Construction. Given the arch-rib B, 
Fig. 425, with specified loading. Divide the curve into 




Fig. 425. 



ARCH RIBS. 



4G; 



eight equal ^s's and draw a vertical through the middle 
of each. Let the loads borne by the respective ^s's be 
Pi, P2, etc., and with them form a vertical load-line A C to 
some convenient scale. With any convenient pole 0" 
draw a trial force diagram 0" AC, and a corresponding 
trial equilibrium polygon F G, beginning at any point in 
the vertical F. Its ordinates %", 22", etc., are propor- 
tional to those of the special equil. pol. sought (whose 
abutment line is OB) [§ 374a (2)]. We next use it to de- 
termine n^ [see § 374a]. We know that OB is the " abut- 
ment-line " of the required special polygon, and that . ' . 
its pole must lie on a horizontal through n'. It remains 
to determine its H, or pole distance, by equation (1)' just 
given, viz. : IJ^ yy = Sfyz'. First by § 375 find the value 
of the summation Ii{yy), which, from symmetry, we may 

write = 2i'/(2/2/) =2 [2/12/1+2/22/2+2/32/3+2/42/4] 

Hence, Fig. 426, we obtain 

11 {yy)=2 [HM 

Next, also by § 375, see Fig. 
427, using the same pole dis- 
tance Ho as in Fig. 426, we 
find 







I\{yz")=HA"; i.e., 

+2/22:2'' + 2/3%" +2/4^'.= 



Again, since II {yz") = ysz/' 
+ 2/7^7" + 2/6^6" + 2/5^5" which 
from symmetry (of rib) 

=2/i%"+2/2^7"+2/326"+2/'<', 

we obtain, Fig. 428, 

l"! (2/O = ^oV', (same /^,); 

and .-. 

Jf {yz")=ff, {\"+hJ'). If now we find that /fc/'+/b/'=2yfc. 



464 MECHANICS or engineeking. 

the condition 2^1 (yy) = II {y^'') is satisfied, and the pole 
distance of our trial polygon in Fig. 425, is also that of 
the special polygon sought; i.e., the z" 's.are identical in 
value with the s"s of Fig. 424. In general, of course, we 
do not find that ky'~{-k/' = 2A;. Hence the z" 's must all 
be increased in the ratio 2k: {lc^"-\~k/') to become equal to 
the g"s. That is, the pole distance H of the spec, equil* 
polygon must be 

7j-_ ki'-\-'k/' jT,, (in which W = the pole distance of the 
2^c trial polygon) since from §339 the ordi- 

nates of two equilibrium polygons (for the same loads) 
are inversely as their pole distances. Having thus found 
the if of the special polygon, knowing that the pole must 
lie on the horizontal through n', Fig. 425, it is easily 
drawn, beginning at 0. As a check, it should pass through 
B. 

For its utility see § 367, but it is to be remembered that 
the stresses as thus found in the different parts of the 
rib under a given loading, must afterwards be combined 
with those resulting from change of temperature and the 
shortening of the rib axis due to the tangential thrusts, 
before the actual stress can be declared in any part. 

Note. — Variable Moment of Inertia. If the / of the rib section is dif- 
ferent at different sections we may proceed as follows: For eq. (1); we 

PB ds i'B ds 
now write I yy y = I yz'—-. Taking the I of the crown section (say) 

Jo i Jo I 
as a standard of reference, denoting it by /', we may write for any other 
section I = nl', where n is a variable ratio, or abstract number; whence 

eq. (1) becomes, after putting Js for ds, y / 2/J/— ="77 / V^—- 

If now the length of each successive Js, from the crown down, be made 
directly prop>ortional to the number n at that part of the rib, the quantity 
^ s^n will have the same value in all the terms of each summation and 
may be factored out ; and we then have a relation identical in form with 
eq. (1)', but with the understanding that the j/'s and 2"s concerned are 
those in the successive verticals drawn through the mid-points of the 
unequal -s's, or subdivisions along the rib, obtained by following the 
above plan that each As is proportional to the value of the moment of 
inertia at that part of the rib. For instance, if the / of a section near 
the hinge 0, or B, is three times that (/') at the (3rown, then the length 
of the Js at the former point must be made three tim,es the length of 
the As first assumed at the crown when the subdivision is begun. By 
a little preliminary investigation, a proper value for this crown , s may 
be decided upon such that the total .number of As's shall be sufficient 
for accuracy (sixteen or twenty in all) 



ABCn-KIBS. 



465 



f-$BO. Arch Rib of Fixed Ends and no Hinges,— Example of 
Class D. Prof. Eddy's Method.* As before, E and / are 
constant along tlie rib Piers immovable. Rib curve 
symmetrical about a vertical line. Fig. 429 shows such a 
rib under any loading. Its span is OB, wliicli is taken as 
an axis X. The co-ordinate of any point m' of the rib 
curve are x and y, and z is the vertical intercept between 
w' and the special equilibrium polygon (as yet unknown, 
but to be constructed). Prof. Eddy's method will now be 

■given for finding tha spe- 
cial equil. polygon. The 
three conditions it 
must satisfy (see § 378, 
Class D, remembering 
that E and /are constant 
and that M — ITz from 
§ 367) are 




H-« ^^ 



Fig. 429. 



/ zds=^ ; / xzds— ; and / yzds =0 
e/o e/o e/o 



(1) 



Now suppose the auxiliary reference line (straight) vm 
to have been drawn satisfying the requirements, with 
respect to the rib curve that 



/ z'ds—0 ; and / xz'ds=Q 
e/o c/o 



(2) 



in which z' is the vertical distance of any point m' from 
vm and x the abscissa of mf from 0. 

From Fig. 429, letting z" denote the vertical intercept 
(corresponding to any m') between the spec, polygon and 
the auxiliary line vm, we have z=z'—z", hence the three 
conditions in (1.) become 

r{z'-z")ds=0; i.e., see eqs. (2) C^. z"ds=0 , .. (3) 



* p. 14 of Prof. Eddy's book ', see reference in preface of this work. 



466 mecha:nics of engineeeing. 

B B 

fx {z'—^')ds=0 ; i.e., see eqs. (2) f xz"ds=o (4/ 
^nifh'-^)ds=;0,^7^^Zl-f}.'ds=fkds . (5) 

provided vm has been located ^s prescribed. 

For graphical purposes, having subdivided the rib curve 
into an even number of small equal J.s's, and drawn a verti- 
cal through the middle of each, we first, by § 377, locate 
vm to satisfy the conditions 

ll{z')=0 and l^,{xz')=0 . . (6) 

(see ec[. (2) ; the di cancels out) ; and then locate the 
special equilibrium polygon, with vm as a reference-line, 
by making it satisfy the conditions. 

:EI{z'')=0 . (7); Il{xz")=Q . (8); I^Xyz")^l^Xyz') . (9) 

(obtained from eqs. (3), (4), (5) by putting ds = ^s, and can- 
celling). 

Conditions (7) and (8) may be satisfied by an infinite 
number of polygons drawn to the given loading. Any one 
of these being drawn, as a trial polygon, we determine for it 
the value of the sum l'f^(yz") by § 375, and compare it with 
the value of the sum l'^,{yz') which is independent of ihe 
special polygon and is obtained by § 375. [N.B. Itmist 
be understood that the quantities (lengths) x, y, z, z\ and z" , 
kere dealt with are thosa pertaining to the verticals drawn 
through the middles of the respective ^s's, which must be 
sufficiently numerous to obtain a close result, and not to 
the verticals ia which the loads act, necessarily, since these 
latter may be few or many according to circumstances, see 
Fig. 429]. If these sums are not equal, the pole distance 
of the trial equil. polygon must be altered in the proper 
ratio (and thus change the 2;'"s in the inverse ratio) neces- 
sary to make these sums equal and thus satisfy conditicn 
(9). The alteration of the 2'"s, all in the same ratio, will 



AECH-EIBS. 



4G7 



aot interfere with conditions (7) and (8) whicli are alreadj^ 
satisfied. 



381. Detail of Construction of Last Problem. Symmetrical Arch- 
Rib of Fixed Ends. — As an example take a span of the St. 
Louis Bridge (assuming /constant) "with. " live load'' cov- 
sring the half span on the left. Fig. 430, where the verticaJ 




feME:^^d==Ui 



Fig. 430. 



scale is much exaggerated for the sake of distinctness*. 
Divide into eight equal Js's. (In an actual example sixteen 
or twenty should be taken.) Draw a vertical through the 



* Each arch -rib of the St. Louis bridge is a built up or trussed, rib of steel about 53i 
ft. span and 52 ft. rise, ia the form of a segment of a circle . Its moment of inertia, 
however, is not strictly constant, the portions near each pier, of a length equal to one 
twelfth of the span, having a value of / one-half greater than that of the remainder at 
the arc. 



468 MECHANICS OF ENGINEERING. 

middle of each ^s. P^ , etc., are the loads coming upon 
the respective Js's. 

First, to locate vm, by eq. (6) ; from symmetry it must 
be horizontal. Draw a trial vm (not phown in the figure), 
and if the (-{- 8')'^ exceed the (— 2')'s by an amount z^, the 

true vm will lie a height —z' above the trial vm (or below, 

if vice versa) ; n = the number of z/s's. 

Now lay off the load-line on the right (to scale), 
take any convenient trial pole 0'^' and draw a correspond- 
ing trial equil. polygon F'"G"\ In r"G"', by §377, 
locate a straight line v"'m!" so as to make 2^(2'") = and 
^l(xz!") = (see Note (&) of § 377). 

[We might now redraw F'" G'" in such a way as to bring 
v"'m!" into a horizontal position, thus : first determine a 
point n'" on the load-line by drawing 0"'n"' \ to v"'m"' , 
take a new pole on a horizontal through n'" , with the same 
II'" , and draw a corresponding equil. polygon ; in the lat- 
ter v"'m"' would be horizontal. We might also shift this 
new trial polygon upward so as to make v"'m!" and vm. 
coincide. It would satisfy conditions (7) and (8), having 
the same %'"'''& as the first trial polygon ; but to satisfy con- 
dition (9) it must have its 2""s altered in a certain ratio, 
which we must now find. But we can deal with the individ- 
ual 2""s just as well in their present positions in Fig. 430.] 
The points ^and L in vm, vertically over E'" and L'" in 
v"'m'", are now fixed ; they are the intersections of the special 
polygon 7'equired, ivith vm. 

The ordinates between v"'m"' and the trial equilibrium 
polygon have been called z'" instead of z" ; they are pro- 
portional to the respective g"'s of the required special 
polygon. 

The next step is to find in what ratio the (s'")'s need to 
be altered (or H'" altered in inverse ratio) in order to be- 
come the {z"^^ ; i.e., in order to fulfil condition (9), viz. : 



AUCH-EIBS. 



469 




^.{yz")=I\{yz') . (9) 

This may be done pre- 
cisely as for tlie rib with 
two hinges, but the nega- 
tive (s'")'s must be prop- 
erly considered (§ 375) 
See Fig. 431 for the de- 
tail. Negative 2;"s or g""s 
point upward. 

From Fig. 431a 

[j ' .*. from symmetry 

I\{yz')=2H^h 
From Fig. 4315 we have 

riyz"')=HX 

Pig. 431. 

and from Fig. 431c 

Il{yn=Ho^ 
[The same pole distance H^ is taken in all these construc- 
tions] .♦. I\yz")=H,{k,-\-\). 

If, then, Ho {\-\-k,) = 2HJc condition (9) is satisfied by the 
z""s. If not, the true pole dista-nce for the special equil. 
polygon of Fig. 430 will be 

2k ' 

With this pole distance and a pole in the horizontal through 
n'" (Fig. 430) the force diagram may be completed for the 
required special polygon ; and this latter may be con- 
structed as follows : Beginning at the point E, in vm, 
through it draw a segment || to the proper ray of the force 
diagram. In our present figure (430) this " proper ray " 
would be the ray joining the pole with the point of meet- 
ing of P2 and Pi on the load-line. Having this one seg- 



470 MECHANICS OF EXGIXEEEING. 

ment of the special polygon the others are added in an 
obvious manner, and thus the whole polygon completed. 
It should pass through L, but not and B. 

For another loading a different special equil. polygon 
would result, and in each case we may obtain the tkrusty 
shear, and moment of stress couple for any cross-section of 
the rib, by § 367. To the stresses computed from these, 
should be added (algebraically) those occasioned by change 
of temperature and by shortening of the rib as occasioned 
by the thrusts along the rib. These " temperature 
stresses," and stresses due to rib-shortening, will be con- 
sidered in a subsequent paragraph. They have no exist- 
ence for an arch-rib of three hinges. 

Note. — If the moment of inertia of the rib section is 
variable, instead of dividing the rib axis into equal Js's, 
we should make them unequal, following the plan indicated 
in the note on p. 464, the As being made proportional to 
the values of the moment of inertia along the rib. After such 
subdivision is made, and a vertical drawn through the mid- 
point of each Js, the various ^'s, z^'s, etc., in these verticals 
are dealt with in the same manner as just shown for the 
case of constant moment of inertia. 

381a. Exaggeration of Vertical Dimensions of Both Space and 
Force Diagrams. — In case, as often happens, the axis of the 
given rib is quite a flat curve, it is more accurate (for find- 
ing M) to proceed as follows : 

After drawing the curve in its true proportions and pass- 
ng a vertical through the middle of each of the equal 
z/s's, compute the ordinate (y) of each of these middle points 
from the equation of the curve, and multiply each y by 
four (say). These quadruple ordinates are then laid off 
from the span upward, each in its proper vertical. Also 
multiply each load, of the given loading, by four, and then 
with these quadruple loads and quadruple ordinates, and 
the upper extremities of the latter as points in an exagge- 
rated rib-curve, proceed to construct a special equilibrium 
polygon, and the corresponding force diagram by the 
proper method ( for Class B, C, or D, as the case may be) 
for this exaggerated rib -curve. 

The moment, Hz, thus found for any section of the ex- 



AECn-KIBS. 471 

aggerated rib-curve, is to be divided by four to obtain the 
moment in tlie real rib, in tlie same vertical line. To find 
the thrust and shear, however, for sections of the real rib, 
besides employing tangents and normals of the real rib W9 
must draw, and use, another force diagram, obtained from 
the one already drawn (for the exaggerated rib) by re- 
ducing its vertical dimensions (only), in the ratio of four 
to one. [Of course, any other convenient number besides 
four, may be adopted throughout.] 

382. Stress Diagrams. — Take an arch -rib of Class D, § 378,. 
i.e., of fixed ends, and suppose that for a given loading (in- 
cluding its own weight) the special , ^^^^ ^thrust 
equil. polygon and its force diagram 

have been drawn [§ 381]. It is re- "^^^^ " —coopte- 
quired to indicate graphically the 
variation of the three stress-elements 
for any section of the rib, viz., the 
thrust, shear, and mom. of stress- 
couple. / is constant. If at any 
point TO of the rib a section is made, then the stresses in 
that section are classified into three sets (Fig. 432). (See 
§§ 295 and 367) and from § 367 eq. (3) we see that the ver- 
tical intercepts between the rib and the special equil. 
polygon being proportional to the products Hz or 
moments of the stress-couples in the corresponding sec- 
tions form a moment diagram, on inspection of which we 

can trace the change in this moment, Hz = ^ , and 

e 

hence the variation of the stress per square inch, jJjj (as. 
due to stress couple alone) in the outermost fibre of any 
section (tension or compression) at distance e from the 
gravity axis of the section), from section to section along 
the rib. 

By drawing through lines On' and OV parallel re- 
spectively to the tangent and normal at any point m of the 
rib axis [see Fig. 433] and projecting upon them, in turn, 
the proper ray (B^ in Fig. 433) (see eqs. 1 and 2 of § 367) 




tJ 

Fig. 432. 



472 



MECHANICS OF ENGINEElimG. 



we obtain the values of the thrust and shear for the sec- 
tion at m. When found in this way for a number of points 
along the rib their values may be laid off as vertical lines 
from a horizontal axis, in the verticals containing the re- 
spective points, and thus a thrust diagram and a shear dia- 
gram may be formed, as constructed in Fig. 433. Notice 
that where the moment is a maximum or minimum the 
shear changes sign (compare § 240), either gradually or 




Fig. 433. 



suddenly, according as the max. or min. occurs between 
two loads or in passing a load ; see m', e. g.' 

Also it is evident, from the geometrical relations involv- 
ed, that at those points of the rib where the tangent-line 
is parallel to the " proper ray " of the force diagram, the 
thrust is a maximum (a local maximum) the moment (of 



ARCH RIBS 47S 

stress couple) is either a maximum or a minimum and the 
shear is zero. 

From the moment, Hz = ^, p2 — — - 

e 1 

may be computed. From the thrust = Fp^^, pi=- , (F 

= area of cross-section) may be computed. Hence the 
greatest compression per sq. inch (Pi+p^) may be found in 
each section. A separate stress-diagram might be con- 
structed for this quantity (pi+p^)- Its max. value (after 
adding the stress due to change of temperature, or to rib- 
shortening, for ribs of less than three hinges), wherever it 
occurs in the rib, must be made safe by proper designing 
of the rib. The maximum shear J,,^ can be used as in §256 
to determine thickness of web, if the section i?^ I-shaped, 
or box-shaped. See § 295. 



383. Temperature Stresses.— In an ordinary bridge truss 
and straight horizontal girders, free to expand or contract 
longitudinally, and in Classes A and B of § 378 of arch- 
ribs, there are no stresses induced by change of tempera- 
ture ; for the form of the beam or truss is under no 
constraint from the manner of support ; but with the arch- 
rib of two hinges (hinged ends, Class C) and of fixed ends 
(Class D) having immovable piers which constrain the dis- 
tance between the two ends to remain the same at all tem- 
peratures, stresses called " temperature stresses '* are in- 
duced in the rib whenever the temperature, t, is not the 
same as that, t^, when the rib was put in place. These 
may be determined, as follows, as if they were the only 
ones, and then combined, algebraically, with those due to 
the loading. 

384. Temperature Stresses in the Arch-Rib of Hinged Ends,— 
(Class C, § 378.) Fig. 434. Let E and /be constant, with 



i74 



MECHANICS OF ENGINEBErNG. 




Fig. 434. 



oilier postulates as in § 379. 
Let t^, = temperature of 
erection, and i — any other 
temperature ; also let I = 
length of span = OB (in- 
variable) and 7^ "CO -efficient 
of linear expansion of the 
material of the curved beam or rib (see § 199), At tempeia- 
fcure t there must be a horizontal reaction H at each hinge 
to prevent expansion into the form O'B (dotted cuive), 
which is the form natural to the rib for temperature t and 
without constraint. We may /. consider the actual form 
OB as having resulted from the unstrained form O'B by 
displacing 0' to 0, i.e., producing a horizontal displace- 
ment O'O =1 {t-Q-/j. 

But O'O = Jx (see §§ 373 and 374) ; (KB. B'% tangent 
has moved, but this does not affect Jx, if the axis X is 
horizontal, as here, coinciding with the span ;) and the 
ordinate y of any point m of the rib is identical Avith its 
z or intercept between it and the spec, equil. polygon, 
which here consists of one segment only, viz. : OB, Its 
force diagram consists of a single ray Oi n' • see Fig. 434. 
Now (§ 373) 

.J B 

Ja? = -A j3Iyds ; and M=Hz = in this case, Hy 



H 



.:l{t-Qrj=—Jfds; 



hence for graphics, and 
equal Js's, we have 



Ell{t-t,)y^=HJs I^y' . . . . (1) 

From eq. (1) we determine H, having divided the rib-curve 
into from twelve to twenty equal parts each called Js . 

For instance, for wrought iron, t and t^,, being expressed 
In Fahrenheit degrees, -/j = 0.0000066. If E is expressed 
in lbs. per square inch, all linear quantities should be la 
inches and H will be obtained in pounds. 

2'o?/^ may be obtained by § 375, or may be computed. B 
being known, we find the moment of stress-couple = Hy, 



AKCH-KIBS. 



475 



at any section, while the thrust and shear at that section 
are the projections of //, i.e., of O^n' upon the tangent and 
normal. The stresses due to these may then be determined 
in any section, as already so frequently explained, and 
then combined with those due to loading. 

385. Temperature Stresses in the Arch-Ribs with Fixed Ends,— 
See Fig, 435. (Same postulates as to symmetry, E and J 
constant, etc., as in § 380.) t and t^ have the same meaning 
as in § 384. 

Here, as before, we 
consider the rib to 
have reached its ac- 
tual form under tem- 
perature t by having 
had its span forcibly 
shortened from the 
length natural to 
temp, t, viz. : O'B', 
to the actual length OB, which the immovable piers compel 
it to assume. But here, since the tangents at and B are 
to he the same in direction under constraint as before, the two 
forces H, representing the action of the piers on the rib, 
must be considered as acting on imaginary rigid prolonga- 
tions at an unknown distance d above the span. To find 
H and d we need two equations. 

From § 373 we have, since M=Hz=H {y—d), 

Ax, i.e., WO-VBW, i.e., \t-t:)r^,=-^J{y-^yds . (2) 
or, graphically, with equal As's 




Fig. 435. 



EIl{t- 



-Qr- 



-HAs 



I-f-dSly 



(3) 



Also, since there has been no change in the angle betweeij 
end-tangents, we must have, from § 374, 

^_rMds=0; i.e., — / 2c?s=0;i.e., ny-d)ds=0 



476 MECHANICS OF ENGINEEEING. 

or for graphics, witli equal jU 's, I'^y = nd . , • (^\ 

in wMcli n denotes tli6 number of J.s's. From (4) wq 
determine d, and tlien from (3) can compute U. Drawing 
the horizontal F G, it is the special equilibrium polygon 
(of but one segment) and the moment of the stress-couple 
at any section = Hz, while the thrust and shea\' are the 
projections of H=^0{ii' on the tangent and normal respect- 
ively of any point m of rib. 

For example, in one span, of 550 feet, of the St. Louis 
Bridge, having a rise of 55 feet and fixed at the ends, the 
force H of Fig. 435 is = 108 tons, when the temperature is 
80° Fahr. higher than the temp, of erection, and the en- 
forced span is 3^ inches shorter than the span natural to 
iliat higher temperature. Evidently, ;f the actual temp- 
■erature I is lower than that ^„, of erection, ^must act in a 
direction opposite to that of Figs. 435 and 434, and th& 
"'thrust " in any section will be negative, i.e., a pull. 

386. Stresses Due to Rib-Shortening — In § 369, Fig. 407, the 
shortening of the element AE to a length A'E, due to the 
uniformly distributed thrust, PiF, was neglected as pro- 
ducing indirectly a change of curvature and form in the 
rib axis ; but such will be the case if the rib has less than 
three hinges. This change in the length of the different, 
portions of the rib curve, may be treated as if it were due 
to a change of temperature. For example, from § 199 we- 
see that a thrust of 50 tons coming upon a sectional area. 
of i^ = 10 sq. inches in an iron rib, whose material has a 
modulus of elasticity — E = 30,000,000 lbs. per sq. inch, 
and a coefficient of expansion yj = .0000066 per degree 
Fahrenheit, produces a shortening equal to that due to a 
fall of temperature {to—t) derived as follows: (See § 199) 
(units, inch and pound) 

^° ^ FEri 10 X 30,000,000 X. 0000066" 
Fahrenheit. 

Practically, then, since most metal arch bridges of 
©lasses G and D are rather flat in curvature, and the thrusts. 



AECH-RIBS. 477 

due to ordinary modes of loading do not vary more than 20 
or 30 per cent, from each other along the rib, an imagin- 
ary fall of temperature corresponding to an average thrust 
in any case of loading may be made the basis of a con- 
struction similar to that in § 384 or § 385 (according as the 
ends are hinged, ov fixed) from which new thrusts, shears, 
and stress-couple moments, may be derived to be combin- 
ed with those previously obtained for loading and for 
change of temperature. 

387. Resume — It is now seen how the stresses per square 
inch, both shearing and compression (or tension) may be 
obtained in all parts of any section of a solid arch-rib or 
curved beam of the kinds described, by combining the re- 
sults due to the three separate causes, viz.: the load, 
change of temperature, and rib-shortening caused by the 
thrusts due to the load (the latter agencies, however, com- 
ing into consideration only in classes G and D, see § 378). 
That is, in any cross-section, the stress in the outer fibre 
is, [letting J',/, T-^", T^"', denote the thrusts due to the 
ihree causes, respectively, above mentioned ; {H&)', {Hz)"y 
{Hz)'", the moments] 

^T}}.^I}l'^I^±tUHz)'±{Hzy'±{Hzy"'\ . . . (1) 

i.e., lbs. per sq. inch compression (if those units are used). 
The double signs provide for the cases 
where the stresses in the outer fibre, due 
to a single agency, may be tensile. Fig. 
436 shows the meaning of e (the same 
used heretofore) /is the moment of in- 
ertia of the section about the gravity 
axis (horizontal) (7. i^ = area of cross- 
section. [Ci = e ; cross section symmet- 
rical about (7]. For a given loading we 
may find the maximum stress in a given rib, or design the 
rib so that this maximum stress shall be safe for the ma- 
terial employed. Similarly, the resultant shear (total, not 




478 



MECHANICS OF ENGINEERrNG. 



per sq. inch) = «/' ± J" ± 3'" is obtained for any section 
to compute a proper thickness of web, spacing of rivets, 
etc. 

388 The Arch-Truss, or braced arch. An open-work 
truss, if of homogeneous design from end to end, may be 
treated as a beam of constant section and constant moment 
of inertia, and if curved, like the St. Loi*is Bridge and the 
Coblenz Bridge (see § 378, Class D), may be treated as an 
arch-rib.* The moment of inertia may be taken as 



r=2i^, 



A 



(I) 



where F^ is the sectional area of one of the pieces II to the 
curved axis midway between them. Fig. 437, and h = dis« 
fcance between them. 




Fig. 438. 



Fig. 437. 



Treating this curved axis as an arch-rib, in the usual 
way (see preceding articles), we obtain the spec, equil. pol. 
and its force diagram for given loading. Any plane ~| to 
the rib -axis, where it crosses the middle m of a " web- 
member," cuts three pieces, A^ B and 6', the total com- 

*The St Louis Bridge 13 not strictly of constant moment of inertia, being somewha* 

strengthened near eaoli pier, 



ARCH-EIBS. 479 

pressionB (or tensions) in which are thus found : For the 
point m, of rib-axis, there is a certain moment = Hz, a 
thrust = Th, and a shear = J, obtained as previously ex- 
plained. We may then write Psin/9 = J . . . • (1) 
and thus determine whether P is a tension or compres- 
sion ; then putting P'+P" ± P cos /? = T,, 2 

(in which P is taken with a plus sign if a compression, and 
mlQus if tension), and 

(P'-P")^=Rz ...... (3) 

we compute P* and P", whi(5h are assumed to be both com-' 
pressions here. /9 is the angle between the web member 
and the tangent to rib-axis at m, the middle of the piece. 
See Fig. 406, as an explanation of the method just 
adopted. 



Circular Ribs akd Hoops. 

389. Deflections and Changes of Slope of Curved Beams. Analyt- 
ical Method. For finding these quantities we may use eqs. (I.), 
(II.) and (III.) of § 374. For example, we have in Fig. 439, 
a curved beam of the form of the ox k — 

quadrant of a circle, fixed vertically ^ 
at lower extremity p, and carrying 
a single concentrated load, P, at 
the free end 0. [Its own weight 
neglected.] 

As a consequence of the load- i /^^^' \q \ 

ing, the extremity is displaced to kfl__ 1__^J^ 

some position, ^\, but the bending M^^^ 

is slight. Required, the projections ^^^' ^^^' 

Ax and Ay of this displacement and also the angle OKOn or <^, 
which the tangent-line at 0„ makes with its former fhorizon- 
tal) position OX. The beam is homogeneous and of constant 
cross-section ; i.e., E and / are constants. 

To use the equations for Ax and Ay we must take as an 
origin (since is the point whose displacement is under con- 



480 MECHANICS OF ENGINEERING. 

sideration). Hence the co-ordinates x and y of any point, m, of 
the axis of the beam are as shown in the figure. Taking now 
polar co-ordinates, as shown, we note tliat x = r cos_fi-| 
y = r ( 1 — sin ^) ; and ds = rdO. We must also put down 
the following integral forms for reference ; viz. : — 

fsin ^ . d^ = — cos ^ ; f^ . cos 6* . di9 = e . sin ^ + cos ^ ; 

fcos e.de = + sin d ; fsin^ 6 .dd = ^ 6 ~ \s,m2 d ; 

fsin e.cos e .de = \ sin^ e ; C c,o%^ 6 ,de = ^ 6 + \sva.20. 

Taking the portion 0„m {m being any point on curved axis 
of beam) as a free body, we have, for the moment of the stress 
couple a.t m, M — Px, = Pr cos 6, and hence derive, for the 
angle ^, 



Also 5 

and z z 

1 r^ Pr^ r /'^ z*^ 1 Pr^ 

^^'=^X^-^''^' = ^[Jo «<^«^-^^-X -^^•--^•^^] = 2^z- (3>- 

It must be understood that the elastic limit is not passed in 
any fiber and that the bending is very slight. A simple curved 
crane and a ship's davit are instances of this problem, provided 
the cross-section has the same moment of inertia, /, about a 
gravity axis perpendicular to the plane of the paper in Fig. 
439, at all parts of the beam. 

390. Semi-Circular Arch-Rib. Hinged at the Two Piers or Sup- 
ports, and Continuous Between. Fig. 440. The supports are at the 
same level. The arch-rib, or curved beam, is homogeneous 
and has a constant I at all sections. (It is a " curved prism".) 
It is stipulated that no constraint is necessary in fitting the rib 
upon the hinges at the piers before any load is placed on the 
rib ; that is, that the distance apart of the piers (which are 
unyielding') is just equal to the distance between the ends of 
the rib when entirely free from strain. In other words, after 
the rib is in position it is under no stress until a load is put 
upon it. Its own weight is neglected and the load is a concen- 
trated one of 2 P lbs. placed at the " crown ", B. As a conse- 



CIRCULAR RIBS AND HOOPS. 



481 




quence of the gradual placing of the load the crown B settles 
slightly, but on account of symmetry the tangent-line to the 
curved axis at B remains horizontal. Also the extremities 
and A tend to spread further apart, but this is prevented by 
the fact that the piers are immovable (or we may express it 
"the span is invariable"). Hence the reaction at each hinge 
support will have a horizontal component ^as well as a ver- 
tical component, V, lbs. Fig. 2 shows the axis of the rib. 
Taking the whole rib as a 
free body we easily find (by 
putting ^ vert, comps. = zero, 
and from symmetry) that 
each y=P; the whole load 
being called 2 P ; but for de- 
termining the value of H 
(same at each hinge ; from 
2 (horiz. comps.) =zero) we 
must have recourse to the 
theory of elasticity ; i.e., must depend on the following fact, 
viz. : — that in the gradual settling of point B under the load, 
B remains in the same vertical, and the tang, line at B remains 
horizontal, and hence (since moves neither horizontally nor 
vertically in actual space) the horizontal projection of O's dis- 
placement relatively to B and ^B's tangent is zero (or Ax^O), 
while the vertical projection of O's displacement relatively to 
B and 5's tangent (A?/) equals the distance B has settled in 
actual space. Here we must take as origin for x and y (as 
in figure) for any point m between and B\ and note that 
the X = r {1 — cos 6), and y = r mi 6; while ds = r. d6. 

With Om as a free body (m being any point between and 
5) we have for the moment of the stress couple at m, 
M.^Vx- Hy, = Px- Hy. 

1 rB r^ 

Ax, = — j Myds, =0 ; .'. I [P(X- rco^ 6) - Hr sm e'jr^ sm Odd =^0', 



Fig. 440. 



.-. P C sin0.de - P C smdcosddd-H C sin'6'.d^ = 0: 
Jo Jo Jo 

and hence, (see integrals in § 389), 



482 



MECHANICS or ENGINEERING. 



(-[- 



COS 6 



Slll^ 



H 



■0 _ sin 2 6h\"^ 
2"" 4 



= 0. 



Inserting the limits, we have 

pr_0 + l-i + 0J-//|^|-0-0-(-0)" 



= 0; 



.:H = 



2 P load 



Also we may obtain, for the settlement of the crown, at B 

1 C^ Pr'^ r3 TT 1 "I 
Aw of relatively to 5, = =7 / Mxds = -77^ -. — 2 

^ "^ EI Jo EI I 4: TT j 

while the tangent-line at 0, originally vertical, now makes with 
the vertical (on the outside) 

'eT 

This is a '■'statically indeterminate structure " ; that is, one in 
which a solution is impossible by ordinary statics but must 
depend on the theory of the elastic change of form of the beam 
or body in question. 

If the load were not placed at the crown, or highest point, 
we should be obliged to put 



an angle 



EI Jo 



Mds = 



- +1- 



jj£Myds^O 



EI 

for the Ax of relatively to A (instead of to jB). 

391. Cylindrical Pipe Loaded on Side. A cylindrical pipe of homogeneous 
material and small uniform thickness of pipe-wall, i, and length I, (so that 
the moment of inertia of the cross-section of wall is for present purposes 
I = It^ -i- 12) rests in a horizontal position on a firm horizontal floor and bears 
a concentrated load of 2P at the highest point, or crown, jB. See Fig. 441. 
It is to be considered as a continuous curved beam or " hoop ", without hinges. 
We neglect the weight of the pipe itself. The dotted circle shows the original 
unstrained form of the pipe-wall, or hoop, while the full line is its (slightly 
deformed) shape when it bears the load. The elastic limit is of course not to 
be passed. The upward force 2P at iV"is the reaction of the floor. Required, 
the maximum moment of stress-couple ; and also the increase in the length of 
the horizontal diameter, and the decrease in that of the vertical diameter. 

Consider as a free body the upper left-hand quadrant of the hoop, viz., 
OB, in Fig. 442, cutting just on the loft of the load at L', a horizontal section 
being made at 0. At each end of this body we must indicate a stress-couple, 
a shear, and a thrust. But at it is evident, after a little consideration, that 
the shear (which would be horizontal) must be zero ; there being at 0, .•. only 



CIECULAE EIBS AND HOOPS. 



483 



a thrust Tg and a stress-couple of unknown moment M^. At the other section 
the shear must he equal to one half of the load 2P (from considerations of 
symmetry) i.e., J" at B = P ; while the thrust at B is soon shown to be zero 
(since S (horiz. compons.) must = zero, and this thrust if it existed would b» 




Fig. 441. 



Fig 442. 



Fig. 443. 



the only horiz. force besides those formhig the stress-couple at B). At B, 
therefore, we find only a stress-couple, of an unknown moment Ms, and a 
shear Jb of direction shown in Fig. 442. By writing S (vert, compons.) 
= zero for this free body we find that the thrust, Tg, at 0, must have a value P. 
To determine M^ we make use of the fact (evident from Fig. 441) that in 
the deformed condition of the ' ' hoop ' ' the tangent-lines at points and B are 
still vertical and horizontal, respectively ; in other words that the angle be- 
tween them has not changed, i.e., is still' 90°. Hence the value of 0, or change 
of angle between tangents at and B is zero. Apply this fact to Fig. 442. Take 
as origin for the x and y of any point m on OB (using 6 later). From a 
consideration of the free body Om shown in Fig. 443 we have for the stress- 
couple-moment M at any section m the value M = Px — M^. We have also 
a; = r (1 — cos d) ; y = r sin 6 ; and ds = rdd. 



Since 



1 r^ r^ 

0, = ^ j Jfds, = 0, .-. j \_Pr^ - Pr^ cos 6 - Mf\ dd = ; 



i.e., {Pr^ [0 - sin e\ - M^rd) 2 = ; or, Pr^ f^ ~ ^ 1 ~ -^^^l ^ ^ '' 



whence, finally, we have Jlfg = Pr 1 I 



(1) 



Now that Jlfg is known, we may find Mb by taking moments about the 
lower section, 0, in Fig. 442, with OB as fj-ee body whence Mb = (2 -=- tt) Pr, 
which is greater than M^. Hence the equation for safe loading is {R'l -=- e) 
= 2Pr -^ TT, where R' is the maximum safe unit-stress for the material, and e 
the distance of the extreme fiber from the gravity axis of a section. (If, how- 
ever, the radius, r, of the cylinder is not large compared with the radial thick- 
ness of the section, see §§ 298 and 299.) 

Evidently the horizontal diameter has been lengthened by an amount 2 Ax, 
if Ax denote the horiz. proj. of O's displacement relatively to B and 5's tan- 
gent ; and similarly, the shortening of the vertical diameter is 2 Ay, if Ay denote 
the vert. proj. of O's displacement with regard to B and JS's tangent-line. 



484 MECHANICS OF ENGINEERING. 

Hence ^ 

ix =. =y j Myds = J- j [Pr^ sin 6 .dO - Pr^ cos 6 sin edd — M^r'' sin edd]; 

from which we have, with M^ = Pr [1 — {2 -t- tt)], 

TT 

^y = ^C -^a^s, = -^ r ^ [Pr3 (1 - COS ey de - M,r^ {dd - COS edd)-], 

Pr^ ^2 _ 8 

It will be noted that the results obtained in this problem apply also to the 
case where the hoop is a circular link of a chain under a tension 2 P, except 
that the moments will be of opposite character and shears and thrusts of oppo- 
site direction. Also, the change of length 2 Ax of the horiz. diameter will be a 
shortening, that of the vertical diameter, a lengthening. (See Prof. Filkins' 
article on p. 99 of Vol. IV of the Transac. of Assoc. C. E. of Cornell Univ. 
and Engineering News, Dec. 1904, p. 547.) 

BTumerical Example. Fig. 441. The length of a cast iron pipe is 10 ft., the 
thickness of wall ^ inch, and the radius of the pipe (measured* to the middle 
of the thickness) is 6 inches. Kequired, the value of the safe load at crown, 2 P 
when the pipe is supported horizontally on a firm smooth bed or floor; the 
max. safe unit-stress being taken at the low figure 2000 lbs. per sq. inch. 

Solution. "We have only to substitute these values in M^ = R' I -i- e and 

*9000 V 120 V C^y 
Obtain (since I =1. P ^ 12), frrr_iL^Lil^== (0.6366 P X 6) ; hence safeload 

(^) X (2) X 1^ 
= 2 P, = 5236. lbs. ; that is, 43.63 lbs. per running inch of pipe length. 

If now the thickness be doubled, i.e., t = 1", with other data unchanged, 
we find the safe load to be four times as great, i.e., 2 P = 20,944 lbs. ; or 174.5 
lbs. per running inch of pipe-length. 

Although the load is called "concentrated" as regards the end-view of the 
pipe, it must be understood to be uniformly distributed along the length. 



FLEXURE OF BEAMS: GEOMETEICAL TREATMENT. 485 



CHAPTER XII. 

Flexure of Beams ; both Simple and Continuous. 
Geometrical Treatment. 

392. By Geometrical Treatment is meant making use of the 
properties of geometrical figures to deduce algebraic relations. 
This does not necessitate the use of drafting instruments ; but the 
graphic ideas involved greatly simplify the algebraic detail of 
finding deflections, angles, moments, shears, etc., in the case of 
horizontal beams originally straight and slightly bent under 
vertical loads and reactions. In the case of " continuous 
beams", or "girders", (p. 320), this mode of treatment leads to 
conceptions and methods which are remarkably clear and simple. 

393. Angle Between End-Tangents of a Portion of a Bent Beam. 
If the cantilever beam of Fig. 443a (slender and originally 



o 



|C B 



.ax I 



norvuilto CD '_ ^^ 

— T" ^. ,, 

\d(p ^■'' » 

' -* 1 
^1 







Fig. 443a. Fig. 4436. 

straight) be loaded as shown, and the beam thus slightly bent, 
the two cross-sections, AH and CD, at the two ends of any dx 
of the axis of beam, are no longer parallel but become in- 
clined at a small angle d<^ which is also the angle between the 
normals to these sections, in their new (relative) position (see 
now Fig. 4436). AZT now occupies the position A'W (relatively 
to (7Z)). The outer fiber AC (originally of length = dx) is 
longer by some amount dX ; and evidently the value of angle 
d^ may be written = dX -j- g. But, by the definition of the 
modulus of elasticity of the material, J5', we have also 



E = p H- — , (p. 209) ; whence d^ = '^-^ 
dx hiQ 



(1) 



486 MECHANICS OF ENGINEERING. ' 

Now if M denote the moment of the stress-couple to which 
the tensions and compressions on the ends of the fibers in the 
section A^ H' are equivalent (M would equal Px in this simple 
case) we may combine the relation, (§ 229), M = pi -r- e with 
eq. (1) and thus derive, as 

a fundamental relation : , . . d4> = — — (2) 

EI 

for the angle between two tangents to the elastic curve, 
one at each end of the elementary length, dx, of the curve ; 
since the two normals to the sections A'H' and C D in Fig. 
4436 are tangents to the ends of the short length dx of the 
elastic curve. (This value of the angle d^ is in 7r-measure ; 
i.e., radians.) 

It follows, therefore, that when the cantilever of Fig. 443a 
is gradually bent from its original condition (in which the 
tangent lines at the two extremities and B Avere coincident, 

i.e., made with each other an angle of 
zero) into its final form, by the gradual 
application of the load P at 0, the 
angle between the tangent to elastic 
curve at 0^ (the final position of 0) 
and that at B (which tangent, in this 
,,. case, has not moved) will have a value 

Fig. 444. \ _ -^ 

obtained by summing up all the small 
values of d^, one for each of the dx'^ between and B (these 
dx'^ making up the length of curve between those points). 

Or, in general, if 0„ and B are any two points of an elastic 
curve (of axis of bent beam, originally straight and now only 
slightly bent, x beiag measured along the beam) we have for the 
angle between the ^ _ ., _ T^ Mdx ,on 

tangents at 0„ and B\^ Jo EI 

(See Fig. 444 for case of cantilever.) This may. be called 
the angle hstween end-tangents of any portion of such elastic 
curve. The beam must be continuous between these two 
j)oints and only slightly bent. Usually the beam in question 
is homogeneous and then E may be taken outside of the integral 
sign. Also, if the beam be prismatic in form (i.e., sides par- 
allel to a central axis, originally straight) the moment of inertia, 



FLEXURE OF BEAMS ; GEOMETEICAL TREATMENT, 



487 



/, of the cross-section is the same for each dx, and may be 

placed outside of the / sign. 

394. (Relative) Displacement of any Point, 0, of Elastic Curve 
of a Bent Beam. In the case of the simple cantilever of 
Fig. 443a let us consider that the axis of the beam, originally 
straight and in position OB, passes gradually into its final form 
or elastic curve 0„ A'" A'' . . . B hj the successive change of 
form of each small block, or elementary length dx ; beginning 




Successive bending of each 
dx of Cantilever. 

Fig. 444a. 

at the end B. When the section at A^ turns through its angle 
d(f)^, as due to the lengthening and shortening of the fibers 
forming the block (i.e., to the stress-couple in section A', of 
moment M') it carries with it all the portion OA' (still straight) 
into position O^A' so that the extremity describes a small 
distance (practically vertical) 00 ^ = OA' . d(f>^. Similarly 
when, next in order, the section at A^' turns through its small 
angle d(f>^, the left-hand end of the beam describes a further 
small distance Ofi^ ^ ^^" • ^^2 ' ^^^ ^° ^^^ ' "i^^til finally the 
extremity has arrived at its final position 0„, having executed 
a total (vertical) displacement OOn-, which will be called Ay. 

If, now, any one of the elementary vertical displacements 
(like OjOj? as typical) be called 8y, we note that Ay is the sum 
of all these small 8i/'s, each of which is practically a small cir- 
cular arc described with a radius x swinging through a small 
angle d(fi, (the successive x's being successively smaller for the 
Sy's lower in the series), so that By = xd(}>; hence 

Ay, =- fSy, = Cxd(j>. But, from eq. (2), d(f> = Mdx -^ AT; 

( Displacement of point 0) _ _ T ^ Mxdx 
I relatively to 5's tangent ) Jo -^-^ 



(4) 



488 MECHANICS OF ENGINEEEING. 

(N.B. In the use of this relation the x of each dx must he 
measured from the point whose displacement is desired.) 

Although the special case of the cantilever has been in mind 
in the figure used in this connection, this result in eq, (4) may- 
be generalized by stating that it gives the displacement A?/ of 
any point from the tangent-line drawn at any other point, B, 
of the elastic curve formed by the axis of a beam originally 
straight and slightly bent under the action of vertical forces 
and reactions. In order to use it, the value of the moment M 
of the stress-couple in each successive dx must be expressed as 
a function of x. If, in addition, the beam has a constant moment 
of inertia, /, of the cross-sections, the " I " may be taken outside 
of the sign of integration. An integration is then generally 
possible. (For example, in the above cantilever, for M we 
should write Px.) 

395. Deflections and Slopes of Straight Homogeneous Prismatic 
Beams Slightly Bent under Vertical Loads and Eeactions. (Beam 
Horizontal.) 

If the beam is a prism and homogeneous, both U and I are 
constant along its length and may be taken outside of the in- 

D^ tegral sign in eqs. (3) 

B,.-*':^^'' and (4), and these two 

J \ '^T^v----^c'""°''^ equations may now be 







I [dxi 



applied to a portion of a 
beam situated between 
any two points and B 
Fig. 4446. of the elastic curve as- 

sumed by the (originally straight) axis of the beam (Fig. 4446) 
under some load. The tangent-lines at 0„ and B were origi- 
nally coincident, and hence the angle between these tangents 
when the beam is bent is the total change in angle between the 

1 r^ 

tangents, and consequently may be written (^ = __ / Mdx and 

is 0„ 00 in the figure. Again, if a vertical be drawn through 
the point 0„ to B's tangent-line OB, the length 00,^ is evi- 
dently O's displacement relatively to 5's tangent-line, since 
originally the point 0„ was situated in 5's tangent itself. 

1 r^ 

That is, 00„, or A?/, = -— I Mxdx,\n which M is the mo- 

hil Jo 



PLEXUBE OF BEAMS : GEOMETRICAL TREATMENT. 



489 



ment of the stress-couple in the cross-section at any distance, x, 
from 0. Note that in general Mis a variable; also that the x 
must be measured from the point whose displacement is 
under consideration. 

Example I. Simple cantilever (Fig. 444,), huilt Inliorizontally at B and bear- 
ing a concentrated load = P lbs. at the free extremity. Both E and / are con- 
sfcant (homogeneous prism). Pind the deflection 00« and the slope (p. 

Solution. From the free body OhVI (m being any point between and B) 
we have M = Px as mom. of stress-couple at ?n. 

PP 
2ET 



1 /.5 p p! 



the "slope" at On- For Jy 



we have 



Pl^ 




Fig. 444i. Fig. 4442- 

Example II. Prismatic beam on two end-supports. Concentrated load P, 
V)S., in middle, Fig. 4442- The two supports being at same level we note that 
from symmetry the tangent at the middle point B of the elastic curve is hori- 
zontal. Hence the displacement OOn of the extremity from this tangent is 
«qual to the deflection of B itself below the horizontal line 0„G. To find OOn 
or Jy, 

Pl^ 
40/ 

Example III. Prismatic beam on two end-supports at same level, the load 
being uniformly distributed over the whole span, I. Fig. 4443. That is, W = wl, 



1 r-B 1 /-S pP -| P /•■!■= 2 

Jy = -=-:- I Mx dc = -r— 1 TT X \ xdx = -— , ^ I x^dx ^ 

^ ElJo EI Jo L2 J 2ElJx=^ 



W = ivl 





Fig. 4443. Fig. 4444. 

w being the load per running inch. As before, the tangent-line at middle 
point B of the elastic curve must be horizontal, so that the displacement of 
extremity On from this tangent will also give the deflection of B from the 
horizontal OnC. Measuring x from (as must always be done in these cases) 



we note that M at any point m 



W 



wx- 



I pBr-WX'^ WX3 -| 1 r WX^ WXn 5 

°^« = MJo I ^^"- -2-^^J= ^[-6-- Xjo = 



5 
584 



EI 



490 MECHANICS OF ENGINEEEIKG. 

Example IV. Prismatic beam on end-supports, hearing two equal loads, each 
= P, symmetrically placed on the span. Fig. 444^. Required, tlie deflection 
of the middle point, B, of the elastic curve, below the horizontal OnC. 
Length = 4a. 

Solution. In previous problems of this article the expression for M, the 
mom. of stress-couple for any point m betvreen the points O and B, has been a 
single function of x, applying to all such points m. But in the present problem, 
having found the reaction at O to be = P lbs., we note by considering a free 
body Onin (where m is any point between On and B) that the value of 31 is 
M= Pz ; whereas if the free body extends into the portion DB the expression 
for M (the free body being now Onm') is M=Px —P{x — a) which reduces to 
M=Pa, a, different function of x; (in fact a constant). Therefore, in making 

the summation 00« = (l-=-^J) ( Mx dx for all the dx^s between and B, this 

summation must be divided into two parts, viz. : one from to D, involving for 
X the limits x=0 and x=a; and the other from D to B, for which the limits for 
X are x=a and x=2a. Hence 

(The student should verify all details of this operation, noting that each sum- 
mation or integi'al contains the proper value of M, as a function of x, for the 
proper portion of the elastic curve. As before, it should be said that on ac- 
count of symmetry the tangent-line at the middle point B is horizontal, and 
parallel to OnC. Otherwise OnO would not he equal to the deflection of B.) 

396. Non-prismatic Beam. VariableMoment of Inertia, I. If the J is vari- 
able, (e.g., if the beam tapers) it must be retained on the right of the integral 
sign in the expressions for cj) and Jy and then expressed as a function of x be- 
fore the integration can be proceeded with. In some cases I may be constant 
within the limits of definite portions of the beam and then the procedure is 
simple. For instance, if the beam in Fig. 444^ has a constant value, = I^, for 

3 
the portions OB and FC, and a larger (but constant) value, of J,) = h ^i» 

for all the sections from B to F, the following takes the place of eq. (1) above : 
. P fa Pa /'2a 4 Pa^ 

397. Properties of Moment Diagrams (Moment-Areas and Cen- 
ters of Gravity). Prismatic Beams in Horizontal Position. Vertical 
Loads and Reactions. In Fig. 445 let AI) be the bent condition 
(i.e., elastic curve) of the axis of a straight prismatic homo- 
geneous beam supported on supports at, or nearly at, the 
same level (so that all tangent lines to the elastic curve deviate 
but slightly from the horizontal. That is, the bending is slight). 
Also, let A"D"B"' 0'" be the corresponding moment diagram (as 
defined and illustrated on pp. 265 to 309). For instance, for 
any point m of the elastic curve the moment of stress-couple (or 
"bending moment ") in that section of the bent beam is repre- 



FLEXURE OF BEAMS; GEOMETRICAL TREATMENT, 491 



sented (to scale) by the ordinate m"m"' or M, in the same 
vertical as m. 

If now a small horiz. distance dx^ or m'V, be laid off 
from m" and a vertical r . .8 
be drawn through r, the pro- 
duct M . dx would be proj)or- 
tional to, and may be repre- 
sented by, the area of the 
vertical strip m"rsm"'. Now 
dx being inches (say) and M 
being inch-lbs., this product 
might be called so many "sq. 
inch-lbs." of moment-area (as 
it will be called). But the 
angle <^ between the tangent- 




i^. 



,^ ^ Mom. Diagraml 

Fig. 445. 

lines drawn at an}- two points On and B of the elastic curve is 

1 r^ 

equal to — - / Mdx ; and hence we may write 



<!> 



[total " moment-area 
between and B 



]' 



EI 



(la] 



or, for brevity, ((> ={A^^ ) ^ EI . . . . . . . . . . (1) 

This " moment-area," then, between and B is the pro- 
duct of the base O'^B" (inches) by the average moment be- 
tween and B regarded as the average altitude of the figure, 
0"B"B"V", this altitude being inch-lbs. 

Again, if the elementary " moment-area " Mdx be multiplied 
by x, its horizontal distance from 0" (i.e. from and 0„), and 
these products summed up for all the dx'^ between and 5, 

there results the expression I (M , dx) . x which may be 

written {A^^.x, where x denotes the horiz. distance of the 
center of gravity of the moment-area O^'B'" from 0''0"' (since, 
from the theory of the center of gravity, the sum of the pro- 
ducts of each ^tri]} of an area by its x co-ordinate is equal to 
the product of the whole area by the distance of its center of 
gravity from the same axis). In the figure the center of grav- 
ity of the moment-area 0"B"' is shown at C" •, and the cor- 
responding X is marked. 



492 



MECHANICS OF ENGINEERING. 



But we have 
^y, or OOn, - 



\J Mxdx 1 -5- E/ =r r Mdx ,x'\^ EI\ 



Sv — - 



and hence we may write 

00,,, or AV, = [(A^) ,i] -^ E7 . . . . (2) 
which furnishes us with a simple means of determining the dis- 
placement of any point 0^ in the elastic curve of the bent beam 
from the tangent-line at any other point B in that elastic curve. 
Evidently, from equations (1) and (2) we have A?/, = 00^^ 
= j>x ; and can therefore state that the intersection of the two 
tangent-lines, one drawn at 0, the other at B, lies in the same 
vertical as the center of gravity of the intervening moment-area- 

(N.B. Instead of the product {Ao)'X, we may, of course, 
use the algebraic sum of similar products for any component 
parts into which it may be convenient to subdivide the total 
moment-area.) 

398. Examples of TJse of Eqs. (1) and (2) of Preceding Paragraph. 
Example I. Simple Cantilever. Concentrated load at free end. Fig. 444i. 
Constant E and I. {Prism.) Here the moment-diagram for whole length is a 

triangle (§ 249) whose base is I inches and whose 
altitude is PI inch-lbs. 

• Hence, with and B taken as in Fig. 
445,, we note that 




AB-. 



'?=('•?). 
...,= [.?]. 



2 
and that x = ^ Z. 

o 



E.I.== 



at On', while OOn = 



T^ 1 



Fig. 445i. 



I.e., OOn^^ 



EI 



['•"] 



PP ^ 

2 EI' 

1_ 

EI 

pn 2 

3 



for the slope 



i^l)' ^> 



Z = 



PP 



3. EI 

Example II. Prismatic Beam on Two End-Supports, Load Uniformly dis- 
tributed over the whole span or length, I', W = wl. From p. 268 we know that 
the moment-diagram (Fig. 4462) is a symmetrical segment of a parabola with 
axis vertical, and that the moment at the middle section is Wl -^ 8. Also, 
from p. 12 of Notes, etc., in Mechanics, we find the x of the left-hand half of this 
moment-figure, measured from the left-hand extremity On, is ^ Z — | of 1 1; i.e., 
i = f of 1 1. 

The area of this semi-pardbolic-segment is two-thirds that of the circum- 
scribing rectangle. From symmetry, the tangent-line drawn at B, the middle 
point of the elastic curve, is parallel to On D, so that the displacement OnO of 
from that tangent is equal to the deflection of B from OnD. Hence 

-2 IWl 5 I- 
3 •2* 8 



OnO, 



(Ao).x 
EI ' 



a^ 



384 EI 



ELBXUEE OF BEAMS: GEOMETRICAL TREATMENT. 493 



Example III. Prismatic Beam. Ends Supported. Two Concentrated Loads 
Equidistant from Supports. Fig. 4453. 

Here, as before, from symmetry the tangent at B is horizontal, parallel to 



V yvYvVVvVYYVVYVVVYVV 

On C • D. 






-^ 



j^- 




FiG. 4452. 



Fig. 4453. 



OnD; so that On equals the deflection of B from C (its position before load- 
ing of beam). Each load P is in middle of a half -span. Required OnO ;, i.e., 
CB=? 

In this case the moment-diagram is easily shown to consist of a triangle at 
each end with a central rectangle of altitude = Pa (inch-lbs.). To find OnO 
we need the product (A^).x. But this A^ consists of the triangle 0"A"N 
with its center of gravity distant f of a from and of the rectangle A"B"KN 
whose center of gravity is at a distance of | of a from 0. Utilizing, therefore 
the principle stated in the N.B. of § 397, we write 



Q-Q _ (^)O- 



1 r Pa 



2a -r. 3 a 

--l-a.Pa-^ 



n Pa^ 
~(S-Er' 



Example IV. Prismatic Beam on End Supports. Single Eccentric Load, P. 
Fig. 445^. Here a tangent drawn to the elastic curve at the load-point B, not 
being horizontal, is not par- 
allel to OnCn, and hence OnO 
does not = the deflection, 5, 
otB. 

However, the displace- 
ments ( = (^1 andd.,), of Ofrom 
5's tangent and of C from 
B's tangent, are easily found, 
the moment-diagram 0"NC" 
having been drawn, in which 
'B"N = {Pa,a, -^ I) inch-lbs. 
(§260). Call" the "moment- Fig. 445,. 

area" of triangle 0"B"N, A' ■ and that on right of load, viz. of C"B"N, A". 

Then, from eq. (2) of § 397, we may write 

Eld^ = A'z,; and Eld., = A"x.,. 
If now we draw a horizontal line, HI)., through the point B of the elastic 
curve, we note, from the similar triangles thus formed, the proportion 

-J — -r = — . From these three equations d^ and d., may be eliminated and d 




494 



MECHANICS OF ENGlISEEillNG. 



obtained; (since A 
and Xj = I a, ■) 



Pa{a^ ~ I, and A" = ^ Pa^^ -i- I; while Xj = ja^, 



We thus obtain 5 = (J Pa^^a.^-) -^ {EI, I) 



(3) 



This is for the load-point. For the maximwn deflection see next example. 

Example V. Maximum Deflection of Prismatic Beam. End Supports. 
Single Eccentric Load. Fig. 4i5g. To locate the lowest point D of elastic 

curve and determine its deflection, 
d, below the horizontal 0„B. 

Draw a tangent at D, also at 
B whose distance n from D is, as 
yet, unknown. Note that the tan- 
gent at D is horizontal. 

The moment-diagram is a tri- 
angle of altitude ik' ; (M' = Pah^l); 
denote the moment at B by m'. 
We have m' =.(n -h 6) .M'. Now 
the angle (p =d' -h I, and 

d' = (AI) .x-^EI = 
Fig. 4455. ^ M'l (a+| [J (a + &) - a])--E'I. 

. •. 6 EI4> = M' (2 a -j- 6). But0, = (^^) -- ^I, = n . ?n' ^ 2EI, and ^ = 0^ ; 
.-., finally, we have n = \^\b{2a+ b), which locates the point R. 

Now note that the intersection C lies in the vertical through the center of 
gravity of the shaded triangle (§ 397). Hence CB = In and therefore from 
similar triangles B8 = \ns. But RB, =d, = BS- BS, and BS = <p^.CB = 
<i>.\n. Hence d = |0n and finally by substitution, and with M' placed = 
Pah -H I, we have (with h> a) . 




Pah 



^-\ EI 



[2a + 6] Vifi (2a + h) 



same as 
on page 258 



399. The "Normal Moment Diagram.'' If a portion, OB, of a 
horizontal beam carrying loads, be conceived separated from 
the remainder of beam and placed on two supports at its 
extremities and B, while carrying the loads [say P^ and PJ 
originally lying between and B, the corresponding moment- 
diagram, 0'"TB"' of Fig. 446 may be called the "normal-moment 
diagram" for portion* OB (of original beam) and its load. If 
Vq is the pier reaction at left, we have for any section t (say 
between P^ and P^) x ft. from 0, the moment of stress-couple 
[call it Mn or " normal moment "] 



M„ 



Vf^x — P,(x — a) 



(1) 



Now consider OB in its original condition (see lower part 
of Fig. 446) when forming part of a much longer beam sup- 



FLEXURE OF BEAMS: GEOMETKICAL TREATMENT. 495 



ported in any manner. If we consider OB, now, as a " free 
body," M^e must put in, besides the loads P^ and P^, a shear Jq 
and a stress-couple of moment Mq in section at 0, and Jq and 
couple of moment Mb at B. The moment in any section t of 
OB is now M = Mo + /o^ — Pi (x — a). Let 7 = difference 
between J^ and Vq, 



I.e., 
then 



M = Mo + 7x + [7o:c-P,(a;-a)] . (2) 



i.e. [see (1)], 



M = M^ + Vx + M„ 



(3) 




N, ~T^° :- 



Hence the moment M {=kwin. Fig. 446) of any section of OB 
is made up of a constant 
part Mw a part proportional 
to X, and a third part equal 
to the " normal moment " of 
th&,t section. Therefore, if, 
in the moment-diagram 
0'B'B"wO" for OB we join 
0" and B" by a straight 
line, and also draw a hori- 
zontal through 0'^ the ver- 
tical intercepts [such as 
uvo] between the line 0"B" 
[or "chord"] and the broken 
line 0"wB" are the normal 
moments for OB and its 
load, and the area (mom.- 
area) of the figure formed by 
these intercepts is equal to 
that of the normal moment 
diagram. 

It is also evident that the center of gravity of the figure 
0"B"w lies in the same vertical as that of the normal moment 
diagram. 

(In the next paragraph the trapezoid 0'B'B"0" will be 
divided into two triangles, instead of into a triangle and a 
rectangle. ) 




Fig. 446. 



496 



MECHANICS OF ENGINEERING. 



400. The Theorem of Three Moments. Let 0, B, and C, 
Fig. 446a, be any three points in the elastic curve of a 

homogeneous, continuousy 
and prismatic beam, origi- 
nally straight and hori- 
zontal but now slightly 
bent under vertical forces 
(some of which are reac- 
tions of supports; no loads 
or forces are shown in the 
figure). 

Let Mq, Mj, and M^ be 
the moments of the couples 
in sections 0, B, and C. 
The moment-diagram for 
portion OBC is 0'C'C"T^B"TP". Join 0"B' and C"B' ; also 
0"B" and C"B" . At the point B of elastic curve draw a tan- 
gent mjn^ and join OC. Then Omo, or d^, is the displacement 
of point from 5's tangent, and d^ = Cm^, is the displacement 
of Cfrom the same tangent; while S is the deflection of Bfrcm 
the straight line joining and C. ■ 

Now the vertical displacement d^ = [mom .-area O'B'T^ X 
distance of its cent. grav. from 00"'\ -^ El. But the moment- 
figure O'B'T^, under OB, is composed of the two triangles shown 
and the '■'■normal moment-diagram ''for OB, viz. : 0"B"T^, whose 
mom.-area may be called A^ and whose center of gravity is x^ ft. 
from 00", while the corresponding distances for the triangles 
are i a and | a . 




Hence, from eq. (2), § 397, we have: 



(1) 



and similarly, with corresponding notation, for the right-hand 
portion, or- segment, BC-, of OBC (denoting the " normal mom.- 
area " C"B"T^ by A^ and reckoning x^, etc., from CC"), 



Eld. = i M^a^ 



¥ ^2 + 



i M,a, . 7 a, + A.x^. 



(2) 



If now a straight line be conceived to be drawn 'through B 
parallel to OC, we have, from the similar triangles so formed. 



FLEXURE OF BEAMS ; GEOMETRICAL TREATMENT. 497 



(as ill Fig. 4454), {d, -h) ^ a, = {h- d,) -^ a, 
this with eqs. (1) and (2) we have finally 



Combining 



Mpg, M^(a, + aS) M^a^ A^ A^x 



+ 



+ 



+ 



+ 



EI8 



(4) 



6 ' 3 '6 

1 

which is the "Theorem of Three Moments." 

E is the modalus of elasticity of material of beam, I the 
" moment of inertia " of its cross-section ; M^, M^, M^, the 
moments of stress-couples (" bending-moments ") at 0, B, and 
C respectively. Distances a^ and a^ are shown in Fig. 446a, 
while A^, A^, x^, and x^ are as above ; 8 being the deflection of 
point B from the straight line joining and C. 

U.B. It should be carefully noted that eq. (4) does not 
apply unless the part of beam from to C is continuous and pris- 
matic ; also that in its derivation, the elastic curve is considered 
concave upward throughout ; hence if a negative number is ob- 
tained for Mq, M,, or M^, in any example by the use of eq. 
(4), it implies that at that section the beam is convex upward, 
instead of concave ; in other words that the upper fibers are in 
tension and the lower in compression (instead of the 'reverse, 
as in Fig. 446a). 

401. Values of ^,1-, and A2X2 iii Special Cases. The Theorem of Three 
Moineuts involves the use of the (imaginary) normal mom.-area of each of the 
two portions (left and right "spans", or "panels"), OB and BC ; i.e. of the 
products ^jXi and A^x.^, where Xj is measured from the left end of the left panel, 
and x, from the right end of the right panel. "We are now to determine values 
of ^jX, and A^x., for several ordinary cases of loading. 



I. Single Cen- 
tral Concentrated Load, 
P, Fig. 446j. Here, for 
a left-hand panel, 
PI I I 



4 2 2' 



A^x,= 

— ; and for a right- 



hand panel, 



AoX., = 



PP 
16"' 




1^ X 

Fig. 446i. 

Case II. Single Non-central Concentrated Load, P. 
case as a left-hand panel, 

Pbc (I + c) 



A,x^= [- 



Fig. 4462. 

Fig. 4463. For this 

while, as a right-hand 



panel, -42X2= 



498 



MECHANICS OF ENGIKEEKING. 



Case III. Two (or more) Concentrated Loads. Fig. 4463. 

A^, = I [P'b'C (I + b') + P"b"c" {I + &")] ; and for each load more than 

two add a proper term in the bracket. 

For A.^2 interchange b' and c', b" and c", etc. 




' B o 



-hi 




^%- 



Fig 446,. 



Fig. 4464. 



Case IV. Any Continuous Load over a Part or the Whole of the Span; of w 
ibs. per linear foot, w being variable or constant. Fig. 446^. The load on a 
length dx (of loaded part) is wdx lbs. ; comparing which with the P of Case II, 
(or one of the P's of Case III), we note that x corresponds to b, and l—x to c ; 

— 1 /^X=Ci 1 /»Cl 

hence AjX^ = ^ t wdx {I — x)x{l + x) =g- t wx {P — x') dx. 

If w is variable it must first be expressed in terms of x. (For A^^ we 
measure x from the right-hand end, B.) 

Case V. Uniformly Distributed Load over Whole Span ; (i. e. , w is constant). 
Let W, = wl, = whole load, lbs. Fig. 4465. 



o liilUIIIUiUlB oliiiUUU 



A^x, = A.x^ 
-2 I Wl I 

^WP 
~ 24* 



Case VI. Uniformly 
Distributed Load Ad- 
joining one End of 
Span; (left end for 
example). Fig. 446g. 
Total load =W= wb. Applying method of Case IV, withCj = &, and b^ =0, 
we have A^^ = J^ Wbd^ — \ b^). Also from Case IV, now measuring x from 
B,A.j^, = ^\W{l'-c^) (l + c). 




Parabola 
Fig. 446s. 



FLEXUJRE OF BEAMS ; GEOMETRICAL TREATMENT. 499 



Case VII. Uniformly Distributed Load Not Adjoining either End of the 
Span. Fig^. 446. Whole load = W= w (e- 6). By Case IV, we tind 



A,x, — 



W{e + b) 



-[''- 



e2 + 62 ■ 



■A^2 ~~ 



12 , L ' 2 J ' 



12 



■D'-^l 




It is now seen how A^x^^ and ^2*2 ™^y 
be obtained for any loading. 

402. Continuous Girders Treated 
by the Theorem of Three Moments. 

This theorem is of special advan- 
tage in solving continuous beams 
(p. 271) ; and examples will now 
be given. 

Example I. Fig. 447^. A straight, homogeneous, prismatic 
beam or girder, 35 feet long, is placed upon three supports at 
the same level, forming two spans of 15' and 20'; two concen- 
trated loads in the left span, a uniformly distributed load on part 
of right span. Required the maximum moment, and maximum 
shear. (Neglect weight of beam.) 

Take 0, B, and C, as the three sections where the three 
moments Mo, M^, and M^ are situated [respectively] used in the 
theorem of § 400. But both Mo and M^ are zero in this case, 
and 8 (deflection of point B from line joining and C) is also 
zero (since the supports are on the same level). Hence M^ (i.e., 
at B) is the only unknown quantity in applying the theorem of 
§ 400 (eq. (4) ) to this problem. 

Taking the A^x^ from Case III, and A^x^ from Case VII (with 
/= 0), of § 398, we obtain (using the foot and ion as units), 

Mr^^ + ^O) _^ ^ _^_ _1 1-6x4x11x19+8x10x5x251 

3 6 X 15 •- -" 



+ 



16x16 

12x20 



202- 



16^ 



1=0; and .'. M, = - 39.2 ft.-tons. 



The negative sign shows that at section B the elastic curve 
is convex on its upper side (see N. B. in § 400). To follow up 
the solution from this point, let us draw the actual moment- 
diagram somewhat differently from that in Fig. 446a, (which 
see), where the actual moment for any section is measured from 



500 



MECHANICS OF ENGINEERING. 



a continuous horizontal line, O'C, as an axis. Let the 
« chords " 0"B" and B"C" of the two normal moment fig- 



AVo 



>'E 



-f.^>-^6'- 



>'F 



Vb 



10 urns 
Hy i y V I i l i i I i i I i 



AVo 






k 



-^—i5'-j- — M-t-- 



—16'— >-l: 




Fig. 447i. 



ures, 0"B"T^ 
and C"B"T^, be 
made a contin- 
uous horizontal 
line by an up- 
ward shifting 
{each in its own 
vertical) of the 
intercepts of 
those figures. 
The intercepts 
0"0', B"B', 
and C"C', and 
the four tri- 
angles involved 
with them, now 
extend upward 
from that hori- 



zontal line. But in our present problem both M^^ and M^ are 
zero; hence the two upper triangles disappear and the two inner 
triangles project above the horizontal line, with M^ as a com- 
mon base. The actual moments of the points of the elastic 
curve are now measured (in general) by the vertical intercepts 
between the lower boundary of the normal moment figures and 
the upper edges of the two triangles ; but since in the present 
case ilf J is negative, M^ must be laid off below the (new) hori- 
zontal line so that the lines 0"B" and C"B" will cross the 
lower boundaries of the normal figures ; the actual moments being 
now measured by the vertical intercepts between these two oblique 
lines and the curved (or broken^ boundaries of the normal figures. 

This re-arrangement has been observed in the moment-diagram 
0"B" G" of Fig. 447j, where line-shaded areas correspond to the 
parts of the elastic curve which are concave upward; and the dot- 
shaded areas, to parts convex upward (or upper fibers in tension). 

Before determining the shears, J, along the beam, we must 
first determine the reactions at the support, viz. Vq, Fgj and V^ 
Consider the portion OB as a " free body", cutting just on left 



FLEXURE OF BEAMS: GEOMETRICAL TREATMENT. 501 



of support B, and put ^ (moments) = about the neutral 

axis of section at B -, deriving 

6x11 + 8x5- 39.2 _ 7^ x 15 = 0; and .-. Vo= 4.5 tons. 

Similarly, with EC as free body, taking moments about B, 
16 x 12 - 39.2 - y^, X 20 = 0; and .-. Vo = 7.6 tons; and 
hence, since V o + Vb + Vc = ^^ tons, Vg = l'^-^ tons. The 
shear-diagram is now easily formed (see Fig. 417^) ; the Jiiax- 
imum shear being evidently 9.55 tons, occurring in the section 
just on the left of support B. 

We note that the shear changes sign three times, corre- 
sponding to the three (local) maximum moments (at E, B, and 
■K). To locate, and determine, M^, note that the change of sign 
of the shear at I) is gradual and that hence the shear is exactly 
zero at K; which requires that "the load between K and the 
support be equal to the reaction at C, (from the free body 
concerned). Since w along HC is one ton per foot, the dis- 
tance ^Cmust be 7.64 tons -v- 1.00, = 7.6 ft. From this free 
body, KC, we now find, by moments, that ilf^ = 29.2 ft.-tons. 

As to the other maximum moment, i.e., at E, we note that 
the moment at E in the normal moment^figure would be E^'' E" 
= 28.2; from which by deducting t*^ of Mb (i.e., of 39.2) we 
obtain M^, = 17.7 ft.-tons. Hence, the greatest moment to be 
found in the beam is that at B, viz. 39.2 ft.-tons, and upon 
this depends the choice of a safe and economical beam. . 



Example II. Fig. 4472, Con- 
tinuous prismatic beam OG. 
Three supports at same level. 
Find maximum moment, etc., 
under given loading, the 12 tons 
being uniformly distributed over 
whole of right-hand span. Neg- 
lect weight of beam. 

i/ max. =ilfi,= 16.6 ft.-tons. Ans. 



rj ions 



-^5 ■ 




"■B- 



us 20- 




-40- 



''10 if 10 tons 

Fig. 4473. 



■^—16'— 4^ 



> ' 16 tons 






Fig. 4472. 

Example III. Continuous 
prismatic beam on three supports 
O, B, C, at same level. Three 
concentrated loads. Neglect 
weight of beam. Find Mb and 
maximum moment, etc. 

Mb= - 92.6 ft.-tons. Max. 
if = 116 ft.-tons, at D. Ans. 



502 



MECHANICS OF ENGINEERING. 



Example IV. Continuous prismatic beam, 40 ft. long and extending ovei 
four supports at the same level. The loading is symmetrical, as shown (Fig. 



lUlUiUiUl 



UllilUllUio 




4474). Here we note that from symmetry Jf^ must = Mc; also Mo and Mj) 
each = 0. Applying the Three-Moment Theorem to O, B, and C, (with 5=0) 
we find 

+ 1 Jf5X26+lilfBXl2+ l.?|xl7+ J-^.i X 12^ =0; 

and . •. Ms = — 23.7 ft. -tons, ( = Mc, also). 

Completing the mom. -diagram we find that Mb(= Mc), or 23.7, is greater 
than any other moment along the beam; .-. M max. =23.7 ft.-tons. The 
reactions of the supports are found to be : Fq (and Vd) =8.3 tons ; and Vb 
25 tons (and Vc) = 16.6 tons. Evi- 

illliJ,!! dently the elastic curve is con- 

vex up, over both B and C. 



' 10 tons 



-<r—:is!- >\ 



^-6': 




\<—7'- 



S! ^ 



Fig. 4476- 



The maximum shear is 11.8 
tons, close on the left of sup- 
port B, (or close on right of C) . 
Example V. Continuous 
prismatic beam, 38 ft. long, on 
three supports at the same level. 

Fig. 447^. Uniformly distributed loads over portions of the length. Find the 

maximum moment and maximum shear, (J). 

Max. if = - 39.6 ft.-tons (at B) ; 
Max. J = 10.2 tons (close at right of B) 

Example VI. Fig. 447„. Continuous prismatic beam, 50 ft. long, on /our 
supports at same level ; but the arrangement of loading and span-lengths is non- 
symmetrical. Fig. 4476. Find the max. M and max. J. In this case Mo and 
Md are each = 0, but Mb is not = Mc- We are therefore compelled to apply 
the Theorem of Three Moments twice, viz. : first to the three points 0, B, and 
C; and then to the three points B, C, and B ; whence we have 

- Mb (14 + 20) McX 20 16 X 6 X 8 (14 + 6) 20 x 2D» 



Ans. 



6 X 14 



24 X20 



Mb X 20 Mc (20 + 16) 



+ + 



20 X 203 16 X 7 X 9 qe + 7) _ 



24 X 20 



6X16 



FLEXURE OF BEAMS; GEOMETRICAL TREATMENT. 503 



(1) 

(2) 



or, 68Jlf5 + 20 if c+ 1097.1 + 2000 = 0; .... 

and 20 Mb + 72 if c + 2000 + 1358.3 = ; .... 

two simultaneous equations, foi- determining ifs and Mc- 

Solving, we find if 5 = - 34.6, and Mo = - 37.0, ft.-tons. 

The three "normal mom. -diagrams " having been drawn to the common 

base 0"B"C"D" in the figure, we lay off B"B' downward from B" and 



aiiiuiiiiuiiiic 



■''-\-s'- 



20'. 



9^ 



—7^ 



Iq" Ic" Id" 



1^— -:J4 



16'-\- 




Fig. 4476. 



= 34.6 ft.-tons; and C"C', also downward, = 37.0; and draw the straight 
lines 0"B\ B'C, and C'B" ; thus completing the mom. -diagram, in which, 
as before, the differently shaded portions show whether the elastic curve is con- 
cave up (line-shading), or convex up (dot-shading), in the corresponding part of 
beam. 

The four reactions of supports are then found, viz. : To, F^, Fc, and Vd, 
= 6.7, 19.2, 19.0, and 6.1 tons, respectively. Shears are now easily found and 
are shown in the shear diagram, the max. J being 9.8 tons, occurring just on 
the right of support B. The max. if is found to occur at E, and to have a value 
of 42.7 ft.-tons. 

402a. Continuous Beam with One Span Unloaded. In Fig. 448 
we have a continuous prismatic beam supported at three points 
0, S, and C at the same level ; but the support at is above 
the beam, instead of below; since that end tends to rise, there 
being no load in the left-hand span. (Case of a drawbridge with 
the left-end " latched down "). Neglect weight of beam. By 
the Theorem of Three Moments applied to 0, B, and C, with 
Mq and Mc = 0, and Mb unknown, we find Mb = — 4.9 ft.-tons. 
In forming the moment-diagram here, we note that since there is 
no load from to B the lower edge of the "normal mom.-diagram" 



604 



MECHANICS OF ENGINEERING. 




J), tons 



■ Fig. 448. 



for OB coincides with its upper edge, i.e., with the axis itself, 
viz. 0"B". The "normal mom.-diagram " for BG is, a. triangle, 
with. B"Q" as base. Laying off B"B' = 4.9 downward from 

B', and joining 0" B' and 
B'C'^-, we complete the 
actual (shaded) mom.- 
diagram, as shown. 
Max. moment is found 
under the load and = + 
9.55 ft. -tons. 

To find the reaction 
at 0, take OB as " free 
body", cutting close on 
left of ^. (See (i^) in 
the figure. Note the 
position of stress-couple 
at right-hand end of this 
body.) By moments about B we have V a X 10 — 4.9 = 0, 
whence V q = (say) 0.5 tons. The other reactions and the shear 
diagram are now easily d'etermined. 

403. Supports out of Level. In the foregoing examples the 
quantity S has been zero in each instance of the application of 
the Theorem of Three Moments ; but when such is not the 
case the quantities E and I are brought into play. In this con- 
nection it must be remembered that any unequal settling of the 
supports (originally at same level) after the beam has been put 
in place, may cause considerable changes in the values of the 
various moments and shears, and consequent stresses in the 
material. (See lower half of p. 323.) 

404. Continuous Beam with " Built-in" Ends. Fig. 449. As 
a case for illustration take the prismatic beam in Fig. 449, 
" huilt-in " or clamped, horizontally, at B and at C ; at the 
same level. A load P is placed as 
shown. On account of the mode of 
support the tangents to the elastic 
curve at B and will be horizontal 
and are coincident ; so that portions 
of the curve near the ends are convex 
up 




a-> 



Fig. 449. 

Now conceive the beam to be sustained at i? by a simple 



FLEXURE OF BEAMS: GEOMETRICAL TREATMENT. 505 



support underneath and to extend toward the left a length Qq 
at the end of which a support, 0, is placed above the beam, and 
at same level as B and C (allowing for thickness of beam). 
This makes an additional span (with M q = zero) ; and the tan- 
gent to elastic curve at B will no longer be horizontal. But it 
may he made as nearly horizontal as we please, by taking ttg small 
•enough (supposing no limit to strength of beam). When a^ = 
zero the tangent at B will be in its actual position (horizontal). 
We may therefore apply the Theorem of Three Moments (§400) 
to 0, B, and C, [noting that there is no load on 5], if we 
write both Mq, and a^, = 0, whence (see also Case II of § 401), 

a(Sa + 2a) 



^^ Ms (0 + 3a) ^ Mc^ ^ Q ^ P^a 



0. 



3 ' 6 ' ' 6 x,3a 

Similarly, by conceiving the beam extended to tlie right, a dis- 
tance a' to a point D, for another support, etc., we may apply 
the theorem to the three points, B, C, and D in like fashion, 
with Mo and a' = 0, obtaining 



Ms^a , M^(3a + 0) 



6 



3 



+ + 



P2a. a (3a + a) 
6 X 3a 



+ = 0. 



Elimination gives Mb = — t: Pa-, and Mc = 

V 



-Pa, ft. -tons. 

y 

405. Deflections found by the Theorem of Three Moments for 
Prismatic Beams. Since this theorem contains h (see Fig. 446a) 
the deflection of the point B 
of the elastic curve of a con- 
tinuous prismatic beam from 
the line joining the two 
others, and C, we may 
use the theorem in many 
cases for determining deflec- 
tions when the three mcments are known. 

Example I. Fig. 450. Case of two end-supports and a 
single non-central load, P ; (with weight of beam neglected). 
Taking 0, B, C, as the three points (i.e., OB is the left-hand, 
and BO the right-hand, span: icith no lead on either span) we 
have, with Mq and Mc = 0, and Mg = Pa^a, -i- I, 




Fig. 450. 



+ ~ i^ - + -h 

I 3 



= ElS 



a J 



d = 



Pa,%^ 
SEI ' 



506 MECHANICS OF ENGINEERING. 

Example n. If n is any point between and 5, at x ft. 
from 0, and 0, n, and C are taken as the three points for the 
theorem, we may find 8„, the deflection of n below OC^ That 
is, with Mo and Mc = and Mn = P{a^ ^ I) x, 

+^^L£-ii + 4- + -P^2(c^t -^)i{l -^) +^ 2] 
3Z ' Q{1 — x) 



= EI8„ 



X I — xj 



or, after reduction, 

^"^eif^ [z^-<-a:^> (4) 

Now if ttj > a^, we may find the distance x\ from 0, of the 

point of maximum deflection, by putting " ■ = : whence is 

dx 



obtained x' = V ^ {l^ - a,') C^) 

and this substituted in (4) gives 

Pa 
)EI 
(Compare with pp. 258 and 494.) 



max. deflection, = -^^ (f - a}) V\ {I' - a/) ... (6) 



(The following example is the one referred to at the loot of page 514.) 

Example. — A hollow sphere of mild steel, of thickness 2 in. and internal 
radius of ?'o = 4 in., contains fluid at a pressure of 2 tons/in.^ Find max. 
stress and max. strain; with £'=15,000 tons/in.^ and A; = 0.30. Here 
»i = 1.5; and by substitution in eq. (30) we obtain max. hoop stress to be 
§■0= — 2.26 tons/in.^ (tension), while from eqs. (22) and (23) the tangential, 
or hoop strain, at inner surface is found to be £3=" 0-000145 (elongation), 
and the radial strain to be £1= +0.000224 (shortening). 

The latter strain, £■^, is seen to be the greater and the ideal ^'equivalent 
simple stress" (see § 4056) is ££1=4-3.35 tons/in.^, compression, i.e., much 
larger than the actual max. stress (2.26) in this case. On the "elongation 
theory" (see § 4056), the 3.36, and not the 2.26, tons/in.^, is the figure that, 
for safety, should not pass a prescribed unit stress ais Liferred from com- 
pressive tests with an ordinary testing machine. 



THICK HOLLOW CYLINDERS AND SPHERES. 



507 



A' d\ 



CHAPTER XIII. 

THICK HOLLOW CYLINDERS AND SPHERES. 

405a. General Relations between Stress and Strain. — (Elas- 
tic limit not passed.) If a small cube of homogeneous and 
isotropic material, dx inches long on each edge, is subjected 
to a compressive stress of pi Ibs./in.^ on two opposite faces, 
not only is its length in direction of the stress diminished, 
and by an amount dX, but its lateral dimensions are increased 
by an amount d/' which is a certain fraction (from 0.20 to 
0.35 for metals) of dL This ratio, or fraction, is called Poisson's 
Ratio, and will be denoted by k. Thus, in Fig. 450a we have 
such a cube, AD being its unstrained 
form. Axes 1 and 2 are in the plane of 
the paper while axis 3 is 1 to the paper. 
On the left and right faces is shown acting 
the compressive unit-stress pi Ibs./in.^, 
A'D being the form of the cube under 
this stress. If now E represent the 
modulus of elasticity (Young's) of the 
material, we have (see p. 203) denoting 
the ratio dX-^dx, or relative decrease in length, by si, ei = px -r-E; 
so that if dX" is the increase in length of the vertical edges we 
havedX" -i-dx (call this ratio £2=^—kpi^E; while the relative 
increase of length in the horizontal edges 1 to paper will 
be an equal amount, viz., e^^—kpi-^E. These ratios £i, £2, 
and £3 are called the strains along the three axes 1, 2, and 
3, respectively, and are abstract numbers. Hence the three 
strains produced by the stress pi acting alone are 

P\ kpi kpx . 

£i = ^; ^2=-~-^; and £3=--^. . . (1) 

Now if while pi is still in action a compressive stress of 
P2 lbs./in.2 acts on the two horizontal faces, and also a com- 
pressive stress of ps on the two vertical faces which are parallel 
to the paper, the total strain in the direction of axis 1 (that is, 
the relative shortening of the cube in that direction) will be, 

by superposition, £r 




1- dx 

Fig. 450a. 



Pi k{p2 + Pz) J • M 1 • +U 

-^ ^ — ; and similarly, m the 



directions of the other two axes, we have 



£0 — ^ — 



P2 kipi + ps) 



E 



E 



and £3 = Ts — 



Ps k(pi + p2) 
E E 



(2) 



508 MECHANICS OF ENGINEERING. 

(This form of stress-strain relation is due to Grashof.) 
Note that if either p^, p2, or ps is a tensile stress, a negative 
number must be substituted for it; and that if a negative num- 
ber is obtained for si, £2, or £3, in any problem, it indicates a 
lengthening instead of a shortening. Similarly, if the con- 
dition is imposed that the strain £2 (say) shall be a relative 
elongation of 0.00020,-0.00020 must be substituted for it in 
above relation. 

405b. "Elongation Theory" of Safety. — In all preceding 
chapters the criterion of safety has been that the unit-stress 
in the element of the elastic body where the stress is highest, 
regardless of stress on the side faces, should not exceed a cer- 
tain value, or working stress, =R' Ibs./in.^, as determined 
upon by a consideration of the stress at "elastic limit; " this 
"elastic limit " being itself determined by the ordinary ex- 
periments on "simple " tension or compression of rods of the 
material in question, there being no stress on the sides of the 
rod. In such experiments, however, an element with four 
faces parallel to the axis is subjected to stress, say p, on two 
(end) faces only; and the question naturally arises whether 
the elastic limit would be reached for the same value of p as 
before, in case there were also present tensile or compressive 
stress acting on the side faces of the element. Experiments 
which would throw much light on this point are unfortunately 
wanting, and some authorities,, notably on the continent of 
Europe, contend that the extreme limit of safety, as regards 
state of stress in isotropic materials, is when the greatest rela- 
tive strain (elongation or shortening), say £1, is as great as 
would be produced at the elastic limit in an experiment involving 
only "simple " tension, or compression (as above described), 
in an ordinary testing machine. This view would make the 
greatest "strain," or deformation (change of form), the cri- 
terion of safety instead of the greatest stress. Now if a stress 
of "simple " tension, =p', (no side stresses) acts on an element, 
the highest strain produced is in thQ direction of this stress 
and has a value e' = p' ^E, since Young's modulus, E, is deter- 
mined by experiments of this very nature; that is, p' = Ee'. 
Hence if the greatest strain in an element in some compound 
state of stress, as in § 405a, is £1, and it is desired to place it 
equal to | (say) of the £1 in simple tension at elastic limit, 
we may write £i = f £' = |(p'-^£'); or E£i = lp'. If now we 



THICK HOLLOW CYLINDEES AND SPHERES. 



509 



denote |p' by p" we may write Eei==p" and describe j/' , or Eei, 
as the ideal tensile stress which would produce a strain, or relative 
elongation, equal to £i in case there were no side stresses; Cotterill 
calls this ideal stress (Esi) the ^'equivalent simple stress." 

For instance, if on an element of the shell of a cylindrical 
steam-boiler of soft steel the "hoop stress " (p. 537) is pi on 
two end faces and the stress on two of the other faces is p2, 
= -|pi, (the stress on the remaining two opposite faces being 
ps = practically zero in this connection) we have for the strain 
in direction 1, by eq. (2), Eei = pi-k{hpi + 0). . . . (2') 

Let p", =—15,000 lbs./in.2, tension, be the safe working 
stress for the metal in simple tension; with £' = 30,000,000 
lbs./in.2, and Poisson's ratio = A; = 0.30. Then according to 
the view of preceding chapters the greatest safe value for the 
stress pi would be —15,000 Ibs./in.^ But according to the 
new view now being illustrated the safe value of pi must be 
determined by limiting the strain si to a value which would 
be produced by 15,000 Ibs./in.^ in simple tension, i.e., —0.00050, 
(which =^p" -^E); which amounts to the same thing as re- 
quiring that the ''equivalent simple stress" shall =15,000 
lbs./in.2 Hence, substituting in eq. (2'), we have 

- 15,000 = pi(l- J. (0.30)); i.e., pi = -17,600 lbs./ in.2 
tension; which is considerably greater than the 15,000 allowed 
by the older theory. The relation thus brought out in this 
case that the tenacity of a material is increased by the presence 
of lateral tension "can hardly be considered as intrinsically 
probable, and such direct experimental evidence as exists is 
against the supposition " (Cotterill). 

But in many cases the results of this "elongation theory " 
are more probable than those based on the older theory ; hence 
the former is much favored by continental writers. 

405c. Thick Hollow Cylinder. Stresses and Strains. — Fig. 
4506 shows a longitudinal section of a thick hollow cyUn- 
der of homoge- 



neous and isotropic 

material (say steel 

or iron) provided 

with end stoppers 

(^no iriciion nor vj/,,,,,,,^,,,,,,. „y ,,^yy„,,^//,/,,///////,/// 

leakage); and Fig. Fig. 4506. 

450c a transverse section, giving dimensions. 




Fig. 450c. 

ro is the inner 



510 MECHANICS OF ENGINEERING. 

radius and nro the outer radius {n is a ratio). Fig. 450c also 
shows (dotted Knes) an elementary hoop, or shell, of inner 
radius r and outer radius r + dr. The interior of the cylinder 
is filled with fluid under a high pressure, po Ibs./in.^; and it 
is required to determine the stresses and strains in a cubic 
element in any elementary hoop or sign such as ABC, Fig. 450c. 
Let the half hoop ABC of Fig. 450c be considered as a "free 
body " in Fig. 450d, showing also at 3 a small cubic element, 
as mentioned above. The compressive stress 
P+dpk \ A-j-,'^ on the inner surf ace of the hoop isp (radial), 
\l jK^^'^^jT exerted on it by the adjacent inner hoop; 
while on the outer surface of the elementary 
\ I hoop, and exerted on it by the adjacent 
r___i-Ji— a'- outer hoop, is the compressive stress p+dp. 
The thickness of the hoop is dr. The stress 
on the edges, A and C, of this free body 
(half hoop), will be taken as compressive 

/\ ^h-y^l" at first, of intensity q Ibs./in.^ Let the 

21 / / f s ^ hoop or thin shell have a length =Z, 1 to 
Fig. 450d. ^^^^^ -^ ^ ^ longitudinally (see Fig. 4506). 

Now for this free body put IX = and we have (see pp. 525 
and 526) {p + dp)i2r + 2dr)l-2ql . dr~p{2rl)=0 . .(3) 

i.e., pr + r . dp + p . dr + dp . dr — q . dr—pr = . . (4) 

and hence, omitting the term dp . dr of the second order, 

r . dp + p . dr^q . dr (5) 

which is a differential equation of stress. Next consider the 
relations of stress and strain to be found in the small cube at 
3, Fig. 450d. It is subjected to a compressive stress p along 
the radial axis 1 ; to a compressive stress q a long the tangential 
axis 2; while on the front and back faces the stress is p3 = zero, 
parallel to axis 3 (t to paper). Let now oi denote the radial 
strain, £2 the tangential, and £3 the axial strain, this latter 
being parallel to the axis of the cylinder. All of these strains 
are supposed to be shortenings for the present; and from the 
circumstances of the case the third strain £3 (axial) is con- 
sidered constant (i.e., the same for all values of the variable 
r), since the cylinder is under no constraint as to longitudinal 
change of form. 

We may therefore write the relations (see § 405a) 
Ee, = p-k.q, (6); Ee2 = q-k.p, (7); Ee3 = 0-k(p + q), (g) 



THICK HOLLOW CYLINDERS AND SPHERES. 511 

From (8) we have q=—'p—{Eez-^k), which in (5) gives 

r .dp + 2pr .dr=-{Ee3^k)dr .... (9) 

Now multiply by r (integrating factor) and denote Eos-^k 

by A, an unknown constant, (unknown since it contains the 

strain £3) and we have 

r2. dp + 2pr . dr=—Ar . dr; . . . (10) 

that is to say, d[r^p]= —Ar . dr; . .' . (11) 

which may be integrated; giving, r^p=—^Ar^-\-C; . . (12) 

where C is a constant of integration. The two constants A 

and C may now be determined by substituting in (11) the 

values To and po which the two variables r and p assume at 

the inside surface of the cylinder. Similarly, at the outside 

surface r and p have the values nro and (atmospheric pressure 

relatively small and hence neglected); which being placed in 

(11) give rise to a second equation, which like the first contains 

constants only. From these two equations we easily find 

. 2po , „ n^ro^po 

A = -^ — r; and C = ^ — r-; 
n^ — r n^ — 1 

and hence finally, from equations (12) and (8), 
P=^[^-'l (13);and,= -J^j[5>Vl]. (14) 

From (13) and (14) we may find the stresses p and q for 
any value of the variable distance, r, from the axis. The 
negative sign for q shows that it is in reality a tensile stress, 
the reverse of the character assigned to it at first; i.e., it is 
a "negative compressive " stress for this case of fluid pressure 
acting inside the cylinder. Both p and q have maximum values 
at the inner surface and diminish toward the outside. 

Example. — With inner radius rQ = 4 in.; and outer, =5 in. (hence w = 5/4 
or 1.25); and j9o = 800 lbs. /in. ^; we find q max. (or q^), at inner surface, 
to be 3644 Ibs./in.^; while at outer surface g= — 2844 lbs. /in. ^, tension. 
If the metal is cast iron we may put ^ = 0.23 (see p. 230) and £' = 15,000,000 
lbs. /in. ^, and thus obtain for the radial strain at the inner surface, e,= 
+ 0.000109, indicating a shortening; and for the "hoop" strain (or tan- 
gential strain) at the same place, £2= ~ 0-000255, i.e., a relative elongation 
of about 2^ parts in 10,000. (The student should verify all the details of 
this example, carefully noting the proper signs to be used). 

405(1. Thick Hollow Cylinder under External Fluid Pressure. — If the cylinder is 
siirrounded by fluid under high pressure, pn lbs. /in. ^, the internal pressure 
Po being practically zero (atmosphere) in comparison, we must determine 
the constants A and C of eq. (12) on the basis that p = for r = ro and 
that p = pn for r = nr^ ; whence, finally, we obtain 

^=5^('-'^)^ ■ ■ ■ ('« "-^^ii^^')) a« 

for the stresses at any distance r from the axis. Here both p and q are com- 



512 MECHANICS OF ENGINEERING. 

pressive stresses; the latter increasing, and the former diminishing, toward 
the center. Evidently if the cylinder were not hollow, but solid, r^ would = 0, 
and n = cc ; and both p and q would be constant, =pn, at all points. 

405e. Approximate equalization of the tensile hoop stress in 
successive rings of a thick hollow cylinder under internal 
fluid pressure may be brought about by using two or more 
separate, cylinders of which fhe outer ones are successively 
"shrunk on" before the fluid is introduced. For instance, 
with only two cylinders, the outer one is first heated to such 
an extent that it barely fits over the inner one, the latter being 
cold. When the compound cylinder has cooled the outer one 
has shrunk and is in a state of hoop tension, while the inner 
one is in a state of hoop compression. The radial pressure 
between them, at their siirface of contact, is an internal or 
bursting pressure for the outside cylinder, and an external 
or collapsing pressure for the inside cylinder. The original 
and final temperatures being known, we are able to make 
use of the foregoing equations [(6) to (16)] to compute all 
stresses and strains thus induced before the fluid is intro- 
duced into the interior of the smaller cylinder. When the 
internal pressure is eventually produced, the hoop stresses 
in the smaller cylinder, initially compressive, will be con- 
verted into moderate tensions and the tensile stresses in the 
external cylinder will be increased; but the maximum ten- 
sion is not so great as if the complete cyhnder had been origi- 
nally continuous.. (See Prof. Ewing's Strength of Materials). 
In the case of thick hollow cylinders subjected to the severe 
internal pressures needed in the manufacture of lead pipe, to 
produce ''flow" of the metal, it is well known (Cotterill) 
that a permanent increase in the internal diameter takes place, 
showing that in the inner layers of the cylinder the elastic 
limit has been passed. In this way an approach is made to 
equalization in the hoop stresses of all the layers and the above 
formuliE no. longer hold; but the cylinder as a whole is not 
injured, having simply adapted itself better to its function. 
Cast-iron hydraulic press cylinders are sometimes subjected 
to the high internal pressure of 3 tons/in.^ If the cylinder 
is short, its resistance is doubtless much increased by connec- 
tion with the end plates, or ''domes." 

405f. Equation of Continuity for Thick Hollow Cylinder under Stress. — (Cotterill.) 
In Fig. 450e we have mABCD the form and position of a " cubical" element (of 
an elementary hoop) between the two radial planes ABO and Z)CO, in its con- 



THICK HOLLOW CYLINDERS AND SPHERES. 



513 



dition of no stress. After it is subjected to stress it will still be iound*between 
the same radial planes and will occupy some position A'B'C'D'. The radial' 
thickness AB = dr, or t, f i 

will have changed to t', / /L ? ' 

BC has changed to B'C, 
etc. With axes 1 and 2 
as shown, 1 being radial, 
and 2 tangential (or cir- I" 
cumferential), we note 
that the whole circumfer- 
ence of which BC is a part 
has shortened from a 
length 2Tzr to 27rs, so that 27rs=2ffr(l— Sj), £2 being the value of the tan- 
gential strain, or "hoop" strain, at distance r from center 0; and similarly 
2ns' = 27zr'{l— £'2), where £3' is the hoop strain at distance r' (i.e., r+dr) 

from 0. But £2 varies with r, so £2'= ^2 + t ■ -j^- Hence, subtracting. 




FiG. 450e. 



t' = t-t£,-r't 



s=(r'- 

ds2 
dr 



■r){l~e2)- 



dr' 

-r't-r. 
dr 



or. 



whence 



£1 — e. = r 



d£. 



-r't'^- 
dr ' 



dr 



(17) 
(18> 
.(19) 



which is the "equation of continuity of substance," of the cylinder. 

Since in the foregoing cases of thick hollow cylinder, under bursting or' 
collapsing fluid pressure, both s^ and $2 may be expressed as functions of^ 
r, it is a simple matter for the student to show that (19) is verified in those 
cases, and that hence the solutions given are adequate. 

Evidently eq. (19) also holds in the case of the thick hollow sphere, where 
there is a hoop strain £3, q to paper in Fig. 450, equal to that, £2? along axis 2. 

lOcg. Thick Hollow Sphere under Internal Fluid Pressure. — (For thin- 
walled spheres see p. 536). As thick hollow spheres are sometimes used' 
to hold fluids under high pressure, and the halves of such spheres fre- 
quently form the ends of thick hollow cylinders, the following treatment 
will be of practical value. In Fig. 450/ we have a cross-section of the sphere 
through the center. The inner radius is Tq; the 
outer, nro (where n is a ratio) ; while the (variable) 
r is the inner radius of 
any thin spherical shell, 




Fig. 450/. Fig. 450?. 

of thickness dr, an infinite number of which make up the complete sphere. 
Though each of these shells is under a "hoop tension," when the internal 
fluid pressure is in action, we shall at first deal with this stress as if 
compressive, for uniformity in applying the principles of § 450a. 



514 MECHANICS OF ENGINEERING. 

Consider as a free body any hemispherical shell as shown in Fig. 450g. 
The pressure (radial) on the inside, from the adjacent shell, is p lbs. /in.*; 
and that from the adjacent shell on the outside is p + dp, or p'. The radius 
of the outside will be called r', { = r + dr). The "hoop compression" on the 
thin edge of the shell is q Ibs./in.^ These three quantities, p, r, and q, 
are variables, i.e., refer to any infinitesimal shell of the sphere. The strains 
affecting any small "cubical" block in any shell are p, radial; q, tangential; 
and 53, =q, along a tangent T to the first mentioned. 

Taking an axis X through the center and T to sectional plane AB, and 
putting 2 (X-components) = 0, for equilibrium, we have 

7ir'^p' — 7rr^p—27:r.dr.q = 0; i.e., r'^p' — r^p = 2qr . dr. 

But the difference, r'^p' — r^p, is nothing more than the increment accruing 
to the product r^p when r takes the increment dr, and is therefore the 
differential of the quantity or product r^p ; hence we may write 

d{r^p)=^2qr . dr; or, r^ . dp+2pr . dr = 2qr . dr . . . (20) 

We next consider the relations of stress and strain in the small cubical 
element shown in Fig. 450/i, having four radial and two tangential faces. 
The unit stresses on the faces are as there shown; p on the two tan- 
gential, and q on the four radial, faces. Radial strain = e^ and hoop strain 
(tangential) = £2 = same for any tangent. It has already been proved in 

§405/ that e,-e, = r.^; (21) 

and we also have, from § 405a, Eei = p—2kq; (22) 

and Ee2 = q-k{p + q) (23) 

From the four equations, (20), (21), (22) and (23), containing the five 
variables p, q, £1, £2) and r, we may by elimination and integration finally 
determine p and q, each as a function of r; as follows: (C, C", Cj, C^, etc., 
will denote constants of integration, or, constant products involving E and k. 

From (22) and (23) we obtain a value for £1— £n which in (21) gives rise 
to an expression for p—q. Another expression for p—q is obtained from 
(20) . Equating these two expressions we derive —dp = Ci . dsi', that is, 
by integration, — p = Ci£2 + C2 (24) 

By elimination of £2 between (24) and (23) we obtain q in terms of p which 
substituted in (20) produces r^ . dp + 2>pr . dr = C^r . dr .... (25) 

Eq. (25) is made integrable by multiplying by r (integrating factor); 
whence r^ . d'p + 3pr^ . dr = C^r'^ . dr 

The left-hand member is evidently d[r^p\ Therefore (^[r^^?] = C^r^ . dr ; 

C" 
or, integrating, Hp = ^Cor^ + C^ ; that is, p =C'-\ — 3-, . . (26) 

dp 3C" C" 

whence, also, -f^= ^; which in (20) gives rise to q = C' — -^. . (27) 

We may now determine the two constants C and C" substituting in eq. (26), 
first p = P(, and r = rg ; and then p = with r = nr^. The values of C" and C" so 
obtained are placed in (26) and (27), resulting finally in the relations 

P=„-^5?^-l)^ • (28) and ,_--.?i-.("^+l). . (29) 

The negative sign of q shows that it is a tensile stress, or "hoop tension." 
It is evidently a maximum for r=rf), this maximum value being 

Pi 



^(^+0 <^« 

(For a numerical example see foot of p. 506). 



go, =gmax.,= . ^^ 



PART IV. 

HYDRAULICS. 



CHAPTEE I. 

DEFINITIONS— FLUID PRESSURE-HYDROSTATICS BEGUN. 

406. A Perfect Fluid is a substance the particles of which 
are capable of moving upon each other with the greatest free 
dom, absolutely without friction, and are destitute of mutual 
attraction. In other words, the stress between any two con- 
tiguous portions of a perfect fluid is always one of comjpression 
and normal to the dividing surface at every point ; i.e., no 
shear or tangential action can exist on any imaginary cutting 
plane. 

Hence if a perfect fluid is contained in a vessel of rigid ma- 
terial the pressure experienced by the walls of the vessel is 
normal to the surface of contact at all points. 

For the practical purposes of Engineering, water, alcohol, 
mercury, air, steam, and all gases may be treated as perfect 
fluids within certain limits of temperature. 

407. Liq[uids and Gases. — A fluid a definite mass of which 
occupies a definite volume at a given temperature, and is in- 
capable both of expanding into a larger volume and of being 
compressed into a smaller volume at that temperature, is called 
a Liquid, of which water, mercury, etc., are common examples; 
whereas a Gas is a fluid a mass of which is capable of almost 
indefinite expansion or compression, according as the space 
within the confining vessel is made larger or smaller, and al- 
ways tends to fill the vessel, which must therefore be closed in. 
every direction to prevent its escape. 

515 



616 MECHANICS OF ENGINEERING. 

Liquids are sometimes called inelastic fluids, and gases 

elastic fluids. 

408. Eemarks. — Though practically we may treat all liquids 
as incompressible, experiment shows them to be compressible 
to a slight extent. Thus, a cubic inch of water under a pres- 
sure of 15 lbs. on each of its six faces loses only fifty millionths 
(0.000050) of its original volume, while remaining at the same 
temperature ; if the temperature be sufficiently raised, how- 
ever, its bulk will remain unchanged (provided the initial tem- 
perature is over 40° Fahr.). Conversely, by heating a liquid in 
a rigid vessel completely filled by it, a great bursting pressure 
may be produced. The slight cohesion existing between the 
particles of most liquids is too insignificant to be considered in 
the present connection.* 

The property of indefinite expansion, on the part of gases, 
by which a confined mass of gas can continue to fill a confined 
space which is progressively enlarging, and exert pressure 
against its walls, is satisfactorily explained by the " Kinetic 
Theory of Gases," according to which the gaseous particles are 
perfectly elastic and in continual motion, impinging against 
each other and the confining walls. Nevertheless, for prac- 
tical purposes, we may consider a gas as a continuous sub- 
stance. 

Although by the abstraction of heat, or the application of 
great pressure, or both, all known gases may be reduced to 
liquids (some being even solidified); and although by con- 
verse processes (imparting heat and diminishing the pressure) 
liquids may be transformed into gases, the range of tempera- 
ture and pressure in all problems to be considered iu this work 
is supposed kept within such limits that no extreme changes of 
state, of this character, take place. A gas approaching the 
point of liquefaction is called a Vapor. 

Between the solid and the liquid state we find all grades of 
intermediate conditions of matter. For example, some sub- 
stances are described as soft and plastic solids, as soft putty, 
moist earth, pitch, frosh mortar, etc.; and others as viscous and 
sluggish liquids, as molasses and glycerine. In sufficient bulk, 

* Water has recently been subjected to a pressure of 65,000 Ibs./in.^; 
resulting in a reduction of 10 per cent in the volume. See Engineering 
News, Oct. 1900, p. 236. 



DEFINITIOlSrS — FLUID PRESSURE — HYDROSTATICS. 517 

however, the latter may still be considered as perfect fluids. 
Even water is slightly viscous. 

409. Heaviness of Fluids. — The weight of a cubic unit of a 
homogeneous fluid will be called its heaviness,* or rate of 
weight (see § 7), and is a measure of its density. Denoting it 
hy y, and the volume of a definite portion of the fluid by Vy 
Ive have, for the weight of that portion, 



6^= Fr. 



a) 



This, like the great majority of equations used or derived in 
this work, is of homogeneous form (§ 6), i.e., admits of any sys- 
tem of units. E.g., in the metre-kilogram-second system, if y 
is given in kilos, per cubic metre, Y must be expressed in 
cubic metres, and G will be obtained in kilos.; and similarly 
in any other system. The quality of 7, = 6^ -f- "F, is evidently 
one dimension of force divided by three dimensions of length. 

In the following table, in the case of gases, the temperature 
and pressure are mentioned at which they have the given 
heaviness, since under other conditions the heaviness would be 
different ; in the case of liquids, however, for ordinary pur- 
poses the effect of a change of temperature may be neglected 
(within certain limits). 



HEAVINESS OF VARIOUS FLUIDS.* 
[In ft. lb. sec. system; y = weight in lbs. of a cubic foot.] 



Liquids. 



Freshwater, y =^ 63.5 

Sea water 64.0 

Mercury 848.7 

Alcohol 49.3 

Crude Petroleum, about 55.0 

(N.B. — A cubic inch of water 
weighs 0.0361 lbs.; and a cubic 
foot 1000 av. oz.) 



p(„„„„ J At temp, of melting ice; and 14.7 
^^tdbet, j jjjg pgj, gq jj^_ tension. 



Atmospheric Air 0807C 

Oxygen 0.0893 

Nitrogen ...0.0786 

Hydrogen 0.0056 

Illuminating ) from 0. 0300 

Gas, )to 0.040(^ 

Natural Gas, about 0.0500 



* Sometimes called "specific weight;" while its reciprocal, or \^x 
may be styled the "specific volume" of the substance, i.e., the volume 
of a unit of weight. 



518 



MECHANICS OF ENGINEEEING. 



For use in problems where needed, values for the heaviness 
of pure fresh water are given in the following table (from 
Rossetti) for temperatures ranging from freezing to boiling ; 
as also the relative density, that at the temperature of maxi- 
mum density, 39°. 3 Fahr. being taken as unity. The temper- 
atures are Fahr., and y is in lbs. per cubic foot. 



Temp. 


Rel. 
Dens. 


7- 


Temp. 


Rel. 
Dens. 


V- 


Temp. 


Rel. 
Dens. 


V- 


32° 


.99987 


62.416 


60° 


.99907 


62.366 


140° 


.98338 


61.886 


35° 


.99996 


62.421 


70° 


.99802 


62.300 


150° 


.98043 


61.203 


39°. 3 


1.00000 


62.424 


80° 


.99669 


62.217 


160° 


.97729 


61.006 


40° 


.99999 


62.423 


90° 


.99510 


62.118 


170° 


.97397 


60.799 


43° 


.99997 


62.422 


100° 


.99318 


61.998 


180° 


.97056 


60.586 


45° 


.99992 


62.419 


110° 


.99105 


61.865 


190° 


.96701 


60.365 


50° 


.99975 


62.408 


120° 


.98870 


61.719 


200° 


.96333 


60.135 


55° 


.99946 


62.390 


130° 


.98608 


61.555 


212° 


.95865 


59.843 



From D. K. Clark's] for temp. = 
" Manual." \ y =^ 



230° 
59 4 



250° 270° 
58.7 58.2 



290° 
57.6 



298° 
57.3 



338° 
56.1 



366° 
55.3 



390° 
54.5 



Example 1. What is the heaviness of a gas, 432 cub. in. of 
which weigh 0.368 ounces? Use ft.-lb.-sec. system. 

432 cub. in. = \ cub. ft. and 0.368 oz, = 0.023 lbs. 

.*. y =. -z=: -'— — = 0.092 lbs. per cub. foot. 

y i 



Example 2. Required the weight of a right prism of mer- 
cury of 1 sq. inch section and 30 inches altitude. 

30 



F=30 X 1 = 30 cub. in 
table, y for mercury = 848.Y lbs. per cub. ft. 

30 



^ ^^ - cub. feet : while from the 

1Y28. ' 



its weight = ^ = Yy 



17! 



X 848 r ■= 14.73 lbs. 



410. Definitions. — By Hydraulics we understand the me- 
<'iianics of fluids as utilized in Engineering. It may be divided 
i7Jto 

Hydrostatics^ treating of fluids at rest ; and 

Hydrokinetics^ which deals with fluids in motion. (The 
name Pneumatics is sometimes used to cover both the statics, 
and kinetics of gaseous fluids.) 



DEFINITIONS — FLUID PRESSURE — HYDROSTATICS. 519 

411. Pressure per Unit Area, or Intensity of Pressure. — As in 

§ 180 in dealing with solids, so here with fluids we indicate the 
pressure per unit area between two contiguons portions of 
fluid, or between a fluid and the wall of the containing vessel, 
by J?, so that if dP is the total pressure on a small area dF^ 
we have 

dP ... 

^^dF (^) 

as the pressure per unit area, or intensity of pressure (often, 
though ambiguously, called the tension in speaking of a gas) 
on the small surface dF. If pressure of the same intensity 
exists over a finite plane surface of area = F, the total pres- 
sure on that surface is 

P = fpdF=pfdF= Fp, 1 

P ^ .... (2) 

or i? = p. J 

(N.B. — For brevity the single word " pressure" will some- 
times be used, instead of intensity of pressure, where no am- 
biguity can arise.) Thus, it is found that, under ordinary con- 
ditions at the sea level, the atmosphere exerts a normal pressure 
(normal, because fluid pressure) on all surfaces, of an intensity 
of about p = 14:7 lbs. per sq. inch (= 2116. lbs. per sq. ft.). 
This intensity of pressure is called pne atmosjphere. For ex- 
ample, the total atmospheric pressure on a surface of 100 sq. 
in. is [inch, lb., sec] 

P=i^p=. 100X14.7 = 1470 lbs. ( = 0.735 tons.) 

The quahty of p is evidently one dimension of force 
divided by two dimensions of length. 

By one ^' atmosphere,^' then (or "standard atmosphere;" 
an arbitrary unit), is to be understood a unit-pressure of 
14.70 Ibs./sq. in., or 2116.8 Ibs./sq. ft. This would be the 
weight of a prismatic column of w^ater one sq. in. in section 
and 33.9 ft. high (commonly considered 34 ft. for ordinary 
computations); or of a prismatic column of mercury 30 in. 
high and one sq. in. section. These numbers, 14.70, 34, 



520 



MECHANICS OF ENGINEERING. 



and 30, with their meanings as above, should be memorized 
by the student; as they are to be of very frequent service 
in this study. 

At high altitudes the actual pressure of the air is much 
smaller than the conventional "atmosphere." E.g. (see 
p. 621) at 7000 ft. above the sea the height of a mercury 
column measuring the air pressure- is only about 24 in., 
instead of the 30 in. above cited; varying somewhat, of 
course, with the weather. 

412. Hydrostatic Pressure; per Unit Area, in the Interior of a 
Fluid at Rest. — In a body of fluid of uniforin heaviness, at 
rest, it is required to find the mutual pressure per unit area be- 
tween the portions of fluid on opposite sides of any imaginary 
cutting ])]ane. As customary, we shall consider portions of 
the fluid as free bodies, by supplying the forces exerted on 
them by all contiguous portions (of fluid or vessel wall), also 
those of the earth (their weights), and then apply the condi- 
tions of equilibrium. 

First, cutting plane horizontal. — Fig. 451 shows a body of 
homogeneous fluid confined in a rigid 
vessel closed at the top with a small air- 
tight but frictionless piston (a horizontal 
disk) of weight = G and exposed to at- 
mospheric pressure {=Pa per ^ii^it area) 
on its upper face. Let the area of piston- 
face be = ^. Then for the equilibrium 
of the piston the total pressure between 
its under surface and the fluid at must 
be 



WSM 




Fm. 451. 



F=G + Fpa, 
and hence the intensity of this pressure is 

' G 



i^o 



F 



-\-Pa' 



(1) 



It is now required to find the intensity, p, of fluid pressure 
between the portions of fluid contiguous to the horizontal cut- 
ting plane BCa.t a vertical distance = A vertically below the pis- 
ton 0. In Fig. 452 we have as a free body the right parallel©- 



FLUID PEESSUEE. 



521 



piped OBC oi Fig. 451 with vertical sides (two || to paper and 

four ~j to it). The pressures acting on its six faces are normal 

to them respectively, and the weight of the prism is = vol. 

X;k = FhVi supposing y to have the same value at all parts of 

the column (which is practically true for any height of liquid 

and for a small height of gas). Since the 

prism \2 in equilibrium under the forces 

shown in the figure, and would still be so 

were it to become rigid, we may put (§ 36) >i 

2 (vert, compons.) = and hence obtain *" j 

Fp-F:p,- Fhy = 0. . . (2) r ! ^^^ 

>JT-i-T— rx" 

(In the figure the pressures on the ver- ^ 

tical faces i| to paper have no vertical com- ^P 

ponents, and hence are not drawn.) From fig. 452, 

(2) we have 



JP =P. + hy. 



(3) 



{hy, being the weight of a column of homogeneous fluid of unity 
cross-section and height A, would be the total pressure on the 
base of such a column, if at rest and with no pressure on the 
upper base,, and hence might be called intensity due to weigJd.) 
Secondly, cutting plane oblique. — Fig. 453. Consider free 
an infinitely small right triangular prism bed, whose bases are 

li to the paper, while the three side 
faces (rectangles), having areas = dF, 
dF^ , and dF^ , are respectively hori- 
zontal, vertical, and oblique ; let angle 
cbd = a. The surface he is a portion 
^_ V^h of the plane BC oi Fig. 452. Given 
H — V j? (= intensity of pressure on dF) and 
" ) Of, required ^2? the intensity of pressure 
on the oblique face hd, of area dF^. 
SJS,. B. — The prism is taken very small 
in order that the intensity of pressure may be considered con- 
stant over any one face ; and also that the weight of the prism 
may be neglected, since it involves the volume (three dimen- 




Fm. 453. 



522 MECHANICS OF ENGINEEEIFG. 

sions) of the prism, while the total face pressures involve onlj 
two, and is hence a differential of a higher order.] 
From ^ (vert, compons.) = we shall have 

p^dF^ cos a —j>dF= ; but dF-^ dF^ = cos or ; 

I>.=P, (4) 

which is independent of the angle a. 

Hence, the mtensity of fluid 'pressure at a given point is 
the same on all imaginary cutting planes containing the 
point. This is the most important property of a fluid, and is 
true whether the liquid is at rest or has any kind of motion ; 
for, in case of rectilinear accelerated motion, e.g., although the 
sum of the force-components in the direction of the accelera- 
tion does not in general = 0, but = mass X ace, still, the 
mass of the bodj in question is = weight -i- g, and therefore 
the term mass X ace. is a dijfferential of a higher order than 
the other terms of the equation, and hence the same result 
follows as when there is no motion (or uniform rectilinear 
motion). 

413. The Intensity of Pressure is Equal at all Points of any 
Horizontal Plane in a body of homogeneous fluid at rest. If 
we consider a right prism of the fluid in Fig. 451, of small 
vertical thickness, its axis lying in any horizontal plane £0^ 
its bases will be vertical and of equal area dF. The pressures 
on its sides, being normal to them, and hence to the axis, have 
no components |1 to the axis. The weight of the prism also 
has no horizontal component. Hence from 2 (hor. comps. 
II to axis) = 0, we have,^i smdp^ being the pressure-intensi- 
ties at the two bases, 

p,dF-p,dF=0; .:p=p,, .... (1) 

which proves the statement at the head of this article. 

It is now plain, from this and the preceding article, that 
the pressure-intensity p at any point in a homogeneous fluid 
at rest is eqiial to that at any higher point, plus the weight 



FLUID PRESSURE. 



523 



{hy) of a column of the jiuid of section unity and of altitude 
\fi) = vertical distance between the joints. 



^.(^., 



p =p. + hy, 



(2) 




whether they are in the same vertical or not^ and whatever he 
the shajpe of the containing ^ 
vessel {or pipes), provided the 
fluid is continuous hetween 
the two points I for, Fig, 454, 
bj considering a series of 
small prisms, alternately ver- 
tical and horizontal, ohcde, we 
know that 

Pd=Pc — Ky ; and Pc—Pd'i 
hence, finally, by addition we have 

(in which A = Aj — h^. 

If, therefore, upon a small piston at <?, of area = ^o, a force 
jP„ be exerted, and an inelastic fluid (liquid) completely fills the 
vessel, then, for equilibrium, the force to be exerted upon the pis- 
ton at 6, viz., Pg , is thus computed : For equilibrium of fluid 
p^ =.p^ -\- hy ; and for equil. of piston o, j?„ = P„ -^ F^ ; also, 



P, = ^^P,-\-FM. 



(3) 



From (3) we learn that if the pistons are at the same level 
{h, = 0) the total pressures on their inner faces are directly 
proportional to their areas. 

If thie fluid is gaseous (2) and (3) are practically correct if 
h is not > 100 feet (for, gas being compressible, the lower 
i^^trata are generally more dense than the upper), but in (3) the 
pistons must be fixed, and P^ and P„ refer solely to the in- 
terior pressures. 



524 MECHANICS OF EISTGIlSrEERIlS'G. 

Again, if A is small or jp^ very great, the term hy may be 
omitted altogether in eqs. (2) and (3) (especially with gases, 
since for them y (heaviness) is usually small), and we then 
have, from (2), 

i^=i?o; (4) 

being the algebraic form of the statement: A l)ody of Jkiid 
at 7'est transmits pressure with equal intensity in every direc- 
tion and to all of its parts. [Principle of "Equal Transmis. 
sion of Pressure."] 

414. Moving Pistons. — If the fluid in Fig. 454 is inelastic 
and the vessel walls rigid, the motion of one piston (c) through 
a distance s^ causes the other to move through a distance s^ de- 
termined by the relation F^s^ = F^s^, (since the volumes de- 
scribed by them must be equal, as liquids are incompressible) ; 
but on account of the inertia of the liquid, and friction on the 
vessel walls, equations (2) and (3) no longer hold exactly, still 
are approximately true if the motion is very slow and the 
vessel short, as with the cylinder of a water-pressure engine. 

But if the fluid is compressible and elastic (gases and vapors ; 
steam, or air) and hence of small density, the effect of inertia 
and friction is not appreciable in short wide vessels like the 
cylinders of steam- and air-engines, and those of air-compres- 
sors ; and eqs. (2) and (3) still hold, practically, even with high 
piston-speeds. For example, in the space ABy 
Fig. 455, between the piston and cylinder-head 
of a steam-engine (piston moving toward the 
right) the intensity of pressure, ^, of the 
steam against the moving piston B is prac- 
FiG. 455. tically equal to that against the cylinder-head 

A at the same instant. 

415. An Important Distinction between gases and liquids 
(i.e., between elastic and inelastic fluids) consists in this: 

A liquid can exert pressure against the walls of the contain- 
ing vessel only by its weight, or (when confined on all sides) 
by transmitted pressure coming from without (due to piston 
pressure, atmospheric pressure, etc.); whereas — 




FLUID PRESSUEE. 625 

A gas, confined, as it must be, on all sides to prevent dif- 
fusion, exerts pressure on the vessel not onlj by its v^eiglit, 
but by its elasticity or tendency to expand. If pressure from 
without is also applied, the gas is compressed and exerts a still 
greater pressure on the vessel walls. 

416. Component, of Pressure, in a Given Direction. — Let 
A BCD, whose area = c.Zi^ be a small element of a surface, 
plane or curved, and^ the intensity of A 

fluid pressure upon this element, then ^ap\ /i\ 
the total pressure upon it is pdJF, and is \/^ I \q 

of course normal to it. Let A'B' CD be / ---'''^^ > 

the projection of the element dJc upon cc X \|b'^,--' 

a plane CDM making an :?.ngle a with y" ^X,-^-'"'' 

the element, and let it be required to j ' 

find the value of the component oi jpdF ^^^' '*°^" 

in a direction normal to this last plane (the other component 
being understood to be 1| to the same plane). We shall have 

Compon. ofpdF ~\ to CDM = pdF cos a =j>{dF. Goa a). (1) 

But dF . cos a = area A'B' CD ^ the projection of <i^upon 
the plane CDM, 

.', Compon. 1 tojplane CDM ■=p X {project, of dF on CDM)\ 

i.e., the component of fluid pressure (on an element of a sur- 
face) in a given direction (the other component being ~1 to 
the first) is found hy midtiplying the intensity of the pressure 
hy the area of the projection of the element xpon a plane 1 to 
the given direction. 

It is seen, as an example of this, that if the fluid pressures 
on the elements of the inner surface of one hemisphere of a 
hollow sphere containing a gas are resolved into components ~| 
and II to the plane of the circular base of the hemisphere, the 
sum of the former components simply = n'r^p, where r is the 
radius of the sphere, and^ the intensity of the fluid pressure ; 
for, from the foregoing, the sum of these components is just 
the same as the total pressure would be, having an intensity p., 



526 



MECHANICS OF ENGINEERING, 



Oil a great circle of the sphere, the area, Trr^, of this circle being 
the sun) of the areas of the projections, upon this circle as a 
base, of all the elements of the hemispherical surface. (Weight 
of fluid neglected.) 

A similar statement may be made as to the pressures on 
the inner curved surface of a right cylinder. 

417. Non-planar Pistons. — From the foregoing it follows that 
the sum of the components || to the piston-rod, of the fluid 
pressures upon the piston at A, Fig. 457, is just the same as at 
_5, if the cylinders are of equal size and the steam, or air, is at 
the same tension. For the sum of the projections of all the 
elements of the curved surface of A upon a plane ~\ to the 
piston-rod is always = Ttr'^ = area of section of cylinder-bore. 




Fig. 457. 

If the surface of A is symmetrical about the axis of the cylin- 
der the other components (i.e., those ~] to the piston-rod) will 
neutralize each other. If the line of intersection of that sur- 
face with the surface of the cylinder is not symmetrical about 
the axis of the cylinder, the piston may be pressed laterally 
against the cylinder-wall, but the thrust along the rod or 
" working force'' (§ 128) is the same (except for friction in- 
duced by the lateral pressure), in all instances, as if the surface 
were plane and 1 to piston-rod. 

418. Bramah, or Hydraulic, Press. — This is a familiar instance 
of the principle of transmission of fluid pressure. Fig. 458. 
Let the small piston at O have a diameter <^ = 1 inch = -^ ft., 
while the plunger E, or large piston, has a diameter d' = AB 
= CD=lh in. = I ft. The lever MJ^ weighs <?, = 3 lbs., 
and a weight G — 4S) lbs. is hung at M. The lever-arms of 
these forces about the fulcrum N are given in the figure. 
The apparatus being full of water (oil is often used), the fluid 
pressure P„ against the small piston is found by putting 



FLUID PRESSUEE. 



527 



5(moms. about JV) == for the equilibrium of the leverj 
whence [ft., lb , sec] 

P„ X 1 - 40 X 3 - 3 X 2 = 0. /, P„ = 126 IbSo 




Fig. 458. 



But, denoting atmospheric pressure by ^„, and that of the 
water against the piston by p^ (per unit area), we may also 
write 

Solving for p^ , we have, putting p^, = 14.7 X 144 lbs, per 
aq. ft., 

p, = [126 -^ ~ {-^yl + 14.Y X 144 = 25286 lbs. per sq. ft. 

Hence at e the press, per unit area, from § 409, and (2), § 4185 m 
p^ = j>„ 4- A;j/ = 25236 + 3 X 62.5 = 25423 lbs. per sq. ft. 

= 175.6 lbs. per sq. inch or 11.9 atmospheres, and the total 
upward pressure at e on base of plunger is 

P = FePe =7t'±-p, = i 7r{iy X 25423 = 81194 lbs., 

or almost 16 tons (of 2000 lbs. each). The compressive force 
upon the block or bale, C, = P less the weight of the plunger 
and total atmos. pressure on a circle of 15 in. diameter. 



528 



MEGHAlSriCS OF ENGINEERING. 



419, The Dividing Surface of Two Fluids (which do not mix) in 
Contact, and at Sest, is a Horizontal Plane, — For, Fig. 459, sup- 
posing any two points e and O of this, sur- 
face to be at different levels (the pressure 
at being ^o, that at ejCg, and the teavi- 
nesses of the two fluids 7, and y^ respec- 
tively), we would have, from a considera- 
tion of the two elementary prisms eb an 
to (vertical and horizontal;, the relation 




Fig. 459. 



while from the prisms eo and gO^ the relation 

These equations are conflicting, hence the aoove supposition 
is absurd. Therefore the proposition is true. 

For stable equilibrium, evidently, the heavier fluid must oc- 
cupy the lowest position in the vessel, and if there are several 
fluids (which do not mix), they will arrange 
themselves vertically, in the order of their den- 
sities, the heaviest at the bottom, Fig. 460. On 
account of the property called diffusion the par- 
ticles of two gases placed in contact soon inter- 
mingle and form a uniform mixture. This fact 
gives strong support to the " Kinetic Theory of 
Gases" (§ 408). 



Fig. 460, 



420. Free Surface of a Liquid at Rest. — The surface (of a 
liquid) not in contact with the walls of the containing vessel 
,...,._.,...,,.., is called a free surface^ and is necessarily 
^fi^'^^^^^^^^m^:^'^ T' horizontal (from § 419) when the liquid is at 
:..;,(.:,.- w-i;: .,•' yq%\.. Fig. 461. (A gas, from its tendency 
to indefinite expansion, is incapable of hav- 
ing a free surface.) This is true even if the 
space above the liquid is vacuous, for if the 
surface were inclined or curved, points in the 
body of the liquid and in the same horizon- 
tal plane would have different heights (or " heads") of liquid 



Fig. 461. 



TWO LIQUIDS IIS'^ BEISTT TUBE. 



529 



between ttein and the surface, producing different intensities 
of pressure in the plane, which is contrary to § 413. 

When large bodies of liquid like the ocean are considered, 
grayitj can no longer be regarded- as acting in parallel lines ; 
consequently the free surface of the liquid is curved, being ~\ 
to the direction of (apparent) gravity at all points. For ordi- 
nary engineering purposes (except in Geodesy) the free surface 
of water at rest is a horizontal plane. 

421. Two Liquids (whicli do not mix) at Rest in a Bent Tube 
open at Both Ends to the Air, Fig. 460 ; water and mercury, for 
instance. Let their heavinesses be y^ 
and y^ respectively. The pressure at e 
may be written (§ 413) either 



or 






according as we refer it to the water 
column or the mercury column and 
their respective free surfaces where the 
pressure j?Oj =i?Og = Pa = atmos. press. 
€ is the surface of contact of the two liquids. 




_.-Xv_>« 



Hence we have 



l>a + Kr,=Pa + Kn; i.e., ^, : K-'-n- r^- • (3) 

le., the heights of the free surfaces of the two liquids above the 
surface of contact are inversely proportional to their respec- 
tive heamnesses. 

ExamJ'le. — If the pressure at ^ = 2 atmospheres (§ 896) we 
shall have from (2) (inch-lb.-sec. system of units) 

KYx = JPe — /?a = 2 X 14.7 — 14.7 = 14.7 lbs. per sq. inch. 
.-. \, must = 14.7 -j- [848.7 -H 1728] = 30 inches 

(since, foi mercury, y^ = 848.7 lbs. per cub. ft.). Hence, 
from (3), . 

, h,y, 30 X [848.7 -- 1728] , „g . , „ . ._^ 

^' = 7r" 6275-^1728 = ^^^ ^^'^"' = ^^ ^- 



530 



MECHANICS OF ENGINEEKING. 



i.e., for equilibrium, and that j?e may = 2 atmospheres, k^ aud 
Aj (of mercarj and water) must be 30 in. and 34 feet respec- 
tively. 

422. City Water-pipes. — If h = vertical distance of a point 
^ of a water-pipe below the free surface of reservoir, and tlie 
water be at rest, the pressure on the inner surface of the pipe 
at B (per unit of area) is 

p =p^-\- hy ; and here j!?o =^„ = atmos. press. 

Example. — If h = 200 ft. (using the inch, lb., and second) 

P = 14.7 + [200 X 12][62.5 -=- 1T28] = 101.5 lbs. per sq. in. 

The term hy, alone, = 86.8 lbs. per sq. inch, is spoken of as the 
'hydrostattG press-ure due to 200 feet height, or "Head," of 
water. (See Trautwine's Pocket Book for a table of hydro- 
static pressures for various depths.) 

If, however, the water {^flowing through the pipe, the pres- 
sure against the interior wall becomes less (a problem of Hy- 
drokinetics to be treated subsequently), while if that motion 
is suddenly checked, the pressure becomes momentarily much 
greater than the hydrostatic. This shock is called '■ water- 
ram" and " water-hammer," and may be as great as 200 to 300 
lbs. per sq. inch.* 

423. Barometers and Manometers for Fluid Pressure. — If a 
tube, closed at one end, is filled with water, and ihe other ex- 
tremity is temporarily stopped and afterwards 
opened under water, the closed end being then 
a (vertical) height = h above the surface of 
the water, it is required to find the intensity, 
jp^ , of fluid pressure at the top of the tube, sup- 
posing it to remain filled with water. Fig. 
463. At E inside . the tube the pressure is 
14.7 lbs. per sq. inch, the same as tljat outside 
at the same level (§ 413) ; hence, from Pe=^ P<s 




Fig. 463. 



H-Vj 



P.=I>E-hy. 



w 



* See pp. 203-214 of the author's "Hydraulic Motors." 



BAROMETERS. 531 

Example. — Let A = 10 feet (with inch-lb.-sec. system) ; then 
f^ = 14.Y - 120 X [62.5 -^ 1T28] = 10.4 lbs. per sq. inch, 

or about f of an atmosphere. If now we inquire the value 
of A to make 'p^ = zero, we put j?^ — hy = and obtain h =z 
408 inches, = 34 ft., which is called the height of the water- 
haroineter. Hence, Fig. 463^^, ordinary atmospheric pressure 
will not sustain a column of water higher than 34 feet. If 
mercury is used instead of water the height supported by one 
atmosphere is 

I = 14.Y -^ [.848.7 -=- 1723] = 30 inches, 



Fig. 463a. 



= 76 cefatims. (about), and the tube is of more manageable 
proportions than with water, aside from the ad- 
vantage that no vapor of mercury forms above 
the liquid at ordinary temperatures. [In fact, the 
water-barometer height 5 = 34 feet has only a 
theoretical existence since at ordinary tempera- 
tures (40° to 80° Fahr.) vapor of water would 
form above the column and depress it by from 
0.30 to 1.09 ft.] Such an apparatus is called a 
Barometer^ and is used not only for measuring 
the varying tension of the atmosphere (from 14.5 
to 15 lbs. per sq. inch, according to the weather and height 
above sea-level), but also that of any body of gas. Thus, Fig. 
464, the gas in D is put in communication with 
the space above the mercury in the cistern at 
(7; and we have jp = hy^ where y = heav. of 
mercury, and p is the pressure on the liquid in 
the cistern. For delicate measurements an at- 
tached thermometer is also used, as the heavi- 
ness y varies slightly with the temperature. 

If the vertical distance CD is small, the ten- 
sion in Cis considered the same as in D. 

For gas-tensions greater than one atmosphere, 
the tube may be left open at the top, forming an open ma- 



n. 



Fig. 464. 



532 



MECHANICS OF ENGINEERITiG. 



nometevy Fig. 4C5. In this case, tlie tension of the gas above 
the mercury in lh% cistern is 



c 


-.^ 


AjR-:-':,-' 
h 


m 


■;;;; 


— 




■ ■■'■ Y ■ 


i 








D 


1 


'■'■'■'■-CS 





Fig. 465. 



J? = (A -f h)r 



(1) 



in which 5 is tlie height of mercury (abf>ut 30 
in,) to which the tension of the atmosphere above 
the mercury column is equivalent. 

Example. — If A = 51 inches, Fig. 465, we 
have (ft., lb., sec.) 



p = [4.25 ft. + 2.5 ft.] 848.Y = 5728 lbs. per sq. foot 
= 39.7 lbs. per sq. incli = 2.7 atmospheres. 



Another form of the open manometer consists of a U tube, 
Fig. 464, the atmosphere having access to one branch, the gas 
to be examined, to the other, while the 
mercury lies in the curve. As before, we 

•^ ' AIR 

nave -^ ^ -" 



jp = qi-^l)y = hy -\-2>^ 



(2) 



r^:^ 



where j^ei = atmos. tension, and h as above. 
The tension of a gas is sometimes spoken , ^j 
of as measured by so many inches of 7ner- •"- ^^^^ 

G^iry. For example, a tension of 22.05 fig. 466. 

lbs. per sq. inch {1^ atmos.) is measured by 45 inches of mer- 
cury in a vacuum manometer (i.e., a common barometer), 
Fig. 464. With the open manometer this tension (1-|- atmos.) 
would be indicated by 15 incbes of actual mercury, Figs. 465 
and 466. An ordinary steam-gauge indicates the eaaoess of 
tension over one atmosphere ; thus " 40 lbs. of steam" implies 
a tension of 40 + 14:.7 = 54.7 lbs. per sq. in. 

The Bourdon steam-gauge in common use consists of a 
curved elastic metal tube of flattened or elliptical section 
^^with the long axis ~] to the plane of the tube), and has one 
end fixed. The movement of the other end, which is free and 



DIFFERENTIAL MANOMETER. 



533 



closed, bj proper mechanical connection gives motion to the 
pointer of a dial. This movement is caused by any change of 
tension in the steam or gas admitted, through the fixed end, to 
the interior of the tube. As the tension increases the ellip- 
tical section becomes less flat, i.e., more nearly circular, caus- 
ing the tv70 ends of the tube to separate more widely, i.e., the 
free end moves away from the fixed end ; and vice versa. 

Such gauges, however, are not always reliable. They are 
graduated by comparison with mercury manometers ; and 
should be tested from time to time in the same way.* 

424. The Differential Manometer.— In Fig. 467 OO'NK is a 
portion of a pipe with the upper wall 00' horizontal. In 
this pipe water is flowing from left to right in so called 
"steady flow; " that is, there is no change, as time goes on, 
in the velocity or internal pressure of the water at a given 
section, as at or 0'. At 0' the velocity is greater than at 
0, since the sectional area is smaller and the pressure po is 
smaller than that, po, at 0; (as explained later) (p. 654). 

The U-tube dmm contains mercury weighing 7-^ lbs. /eft. 
in its lower part and is connected by the tubes aO and hO', 
as shown, with holes in the upper wall of the pipe at and 
0'. Air previously contained in these 
tubes has been expelled through the 
cocks a and 6, which are now closed. 
The water columns Oac and dhO' and 
the mercury in cm have adjusted them- 
selves to a state of rest and are therefore 
in a hydrostatic condition. The water 
in the pipe exerts an upward pressure, yo, 
as it flows by, against the base of the 
stationary Uquid in tube Oa; and at 0' a 
smaller upward pressure, p^ , against 
the base 0' of the stationary water 
column O'h. If the height h (between summits of the 
mercury columns) be read from a scale, we are enabled to 
compute the value of the difference, po — p^ , of the pressures 
at and 0'; as may thus be shown (with p^ and p^ denoting 
the pressures at c and d, respectively) : — 

Since between and c we find stationary and continuous 
water (heaviness = 7-) , we have po = Pc+^r- • • • (1) 

* Of late years gauges have come into use constrncted of boxes with cor- 
rugated sides of thin metal like the aneroid barometer. Motion of the sides, 
under varying internal fluid pressure, causes movement of a pointer on a dial. 








534 



MECHANICS OF ENGINEERING. 



Similarly, between 0' and d, po' -=pd+(hi-x)y; 
while between d and c we have, for the mercury, 

By elimination there easily results 



(2) 

(3) 



po- 



^-^ = /if^-l 



^0-* 



•■>f«.-v.-?-V-V..:V.> 



■Po'=h{rm-r); or ^-^^=h[^j-ij. . (4) 

Evidently, if in place of the mercury we use a Hquid only 
slightly heavier than water and that 
does not mix with the latter, h would 
be quite large for a small value of 
po — po'', i-G-, the instrument would 
be more sensitive. If kerosene oil, 
which is a little lighter than water, 
were used instead of mercury, an 
arrangement of tubes like that in 
Fig. 467a would be necessary, and 
similar analysis, (if yk denote the 
heaviness of kerosene) gives rise to 
the formula. 




Po 



-po'^Hr-n); or 2»-^'=a(i-Q), 



(5) 



■c— ^ 



:is 



■E 4~yi 



^■•■- -r I-'- ■ 

Fig. 468. 




425. Safety-valves. — Fig. 468. Eequired the proper weight 
G to be huDg at the extremity of the horizontal lever AB, 

with fulcrum at B, that the flat 
disk- valve ^sliall not be forced 
upward by the steam pressure,^', 
until tlie latter reaches a given 
value =p. Let tlie weight of 
the arm be G^ , its centre of grav- 
ity being at 0, a distance = o 
from JB ; the other horizontal distances are marked in the 
figure. 

' Suppose the valve on the point of rising; then the forces 
acting on the lever are the fnlcrum-reaction at B, the weights 
G and G^ , and the two fluid-pressures on the disk, viz. : JPp^ 
(atmospheric) downward, and I^p (steam) upward. Hence, 
from ^(moms. ^) =0, 

Gb + G,G + F;p^a - Fpa = 0. ... (1) 



BUKSTING OF PIPES. 



635 



Solving, we have 



(^ = IF{P-I>a)-G,t. 



(2) 



Example. — "With a = 2 inches, 5 = 2 feet, c — 1 foot 
G^=: 4c Ibs.,^ = 6 atmos., and diam. of disk = 1 inch; with 
the foot and pound, 

G = I. . ^'fAVce X 14.7 X IM - 1 X 14.7 X 144] - 4 X^. 



24*4 \12 



.-. G = 2.81 lbs. 



[Kotice the cancelling of the 144; for J^{p —j)a) h pounds, 
being one dimension of force, if the pound is selected as the 
unit of force, whether the inch or foot is used in both fac- 
tors.] Hence when the steam pressure has risen to 6 atmos. 
(= 88.2 lbs. per square inch) (corresponding to 73.5 lbs. persq. 
in. by steam gauge) the valve will open if 6^^ = 2.81 lbs., or be 
on the point of opening. 



426. Proper Thickness of Thin Hollow Cylinders (i.e., Pipes 
and Tubes) to Resist Bursting by Fluid Pressure. 

Case I. Stresses in the G?vss-section due to End Pressure j 
Fig. 469. — Let AB be the circular cap clos- 
ing the end of a cylindrical tube containing 
fluid at a tension =:^. Let i>^ = internal 
radius of the tube or pipe. Then considering 
the cap free, neglecting its weight, we 
have three sets of || forces in equilibrium 
in the figure, viz. : the internal fluid pres- 
sure =:: nr'^jp ; the external fluid pressure 
= nr^'Pa ; while the total stress (tensile) on 
the small ring, whose area now exposed is 
^Tcrt (nearly), is = '^■rcrtj^^ , where t is the thickness of the pipe, 
aTid^?j the tensile stress per unit area induced by the end-pres- 
sures (fluid). 




Fig. 469. 



536 



MECHANICS OF ENGINEERING. 



For equilibrium, therefore, we may put ^(hor. comps.) = ; 



1.6., 



KP -Pa) 



.i>i = 



^t 



(1) 



(Strictly, the two circular areas sustaining the fluid pressures 
are different in area, but to consider them equal occasions but 
a small error.) 

Eq. (1) also gives the tension in the central section of a thin 
hollow sphere, under bursting pressure. 

Case II. Stresses in the longitudinal section of pipe, due to 
radial 'fluid pressures.^ — Consider free the half (semi-circular) 

of any length I of the pipe, be- 
tween two cross-sections. Take an 
axis X (as in Fig. 470) ~\ to the 
longitudinal section which has been 
made. Let p^ denote the tensile 
stress (per unit area) produced in 
the narrow rectangles exposed at A 
and B (those in the half-ring edges, 
having no X components, are not 
drawn in the figure). On the in- 
ternal curved surface the fluid pres- 
sure is considered of equal intensity 
=.p at all points (practically true even with liquids, if 2r io 
small compared with the head of water producing p). The 
fluid pressure on any dF or elementary area of the internal 
curved surface is =. pdF. Its X component (see § 416) is 
obtained by multiplying j? by the projection of dF on the ver- 
tical plane ABO, and since p is the same for all the dF^% of 
the curved surface, the sum of all the JT components of the in- 
ternal fluid pressures must = 2^ multiplied by the area of rect- 
angle ABCD, = 'iirlp I and similarly the X components of the 




Fig. 470. 



* Analytically this problem is identical with that of the smooth cord on 
a smooth cylinder, § 169, and is seen to give the same result. 



BUESTING OF PIPES. 637 

external atraos. pressures = '^rljp^ (nearly). The tensile stresses 
( II to X) are equal to 'ilt])^ ; hence for equilibrium, '2X = 
gives 

mp^ - 2rZ^ + 2r^j9a = ; 

"Pi — ^ \^) 

This tensile stress, called hoop tension, p^, opposing rupture by 
longitudinal tearing, is seen to be double the tensile stress ^:>i 
induced, under the same circumstances, on the annular cross' 
section in Case I. Hence eq. (2), and not eq. (1), should be 
used to determine a safe value for the thickness of metal, t, or 
any other one unknown quantity involved in the equation. 

For safety against rwpture, we must put p^ = T', a safe 
tensile stress per unit area for the material of the pipe or tube 
(see §§ 195 and 203) ; 

,,t = TiPp^ (8) 

(For a thin hollow sphere, t may be computed from eq. (1) ; 
that is, need be only half as great as with the cylinder, other 
things being equal.) 

Example. — A pipe of twenty inches internal diameter is to 
contain water at rest under a head of 340 feet ; required the 
proper thickness, if of cast-iron. 

340 feet of water measures 10 atmospheres, so that the in 
ternal fluid pressure is 11 atmospheres ; but the external pres 
sure Pa being one atmos., we must write (inch, lb., sec.) 

{p —pa) = 10 X 14. Y = 147.0 lbs. per sq. in., and r = 10 in., 

while (§ 203) we may put T' =^oi 9000 = 4500 lbs. per sq. 
in. ; whence 

. 10 X 147 ^ Qo« • 1 
^ = — Tir^ — == 0.326 mches. 
4500 



638 MECHANICS OF ENGINEERING. 

But to insure safety in handling pipes and iraperviousnesB to 
the water, a somewhat greater thickness is adopted in practice 
than given by the above theory.* 

Thus, Weisbach recommends (as proved experimentally also) 

for 

Pipes of sheet iron, t = [0.00172 rA + 0.12] inches; 
" " cast " t = [0.00476 rA + 0.34] " 
" " copper t = [0.00296 rA 4- 0.16] " 
" " lead t = [0.01014 rA + 0.21] " 

" " zinc t = [0.00484 rA + 0.16] " 

in which t =■ thickness in inches, r = radius in inches, and A 
= excess of internal over external fluid pressure (i.e., p — Pa) 
expressed in atrnmjpheres. 

For instance, for the example just given, we should have 
(cast-iron), ^-.00476x10X10 + 0.34 = 0.816 in. 

With riveted steel pipes, if the longitudinal seams are pro- 
vided with two rows of rivets, a value of 10,000 Ibs./ini^ 
may be used for the T' of eq. (3). This makes a fair allow- 
ance for the weakening of the steel plates by the rivet holes. 
The East Jersey Water Co. uses such pipes 2r — 4 ft. in 
diameter, with a thickness of i = f in., under a head of 340 ft. 
At the Mannesmann Works in Hungary, special steel tubes 
4 in. in diameter and | in. thick have been made, safely 
withstanding an internal fluid pressure of 2000 Ibs./in.^ 

Water Ram. — When water flowing in pipes is subject to sudden arrest 
of motion, a high bursting pressure, called '"water ram,", or "water 
hammer," may be produced. See pp.- 204-211 of the writer's Hydraulic 
Motors. 

In thick hollow cylinders, on account of the thickness of the walls, 
the stress in the nietal is not uniformly distributed. See pp. 507, etc., 
of this book. 

427. Collapsing of Tubes under Fluid Pressure. (Cylindrical 
boiler-flues, for example.) — If the external exceeds the internal 
fluid pressure, and the thickness of metal is small compared 
with the diameter, the slightest deformation of the tube . or 
pipe gives the external pressure greater capability to produce 
a further change of form, and hence possibly a final collapse; 
just as with long columns (§ 303) a slight bending gives great 
advantage to the terminal forces. Hence the theory of § 426 



COLLAPSE OF TUBES. 539 

is inapplicable. According to Sir Wm. Fairbairu's experi- 
ments (1858) a thin wrouglit-iron cylindrical (circular) tube 
will not collapse until the excess of external over internal 
pressure is 

^(in lbs. per sq. in.) = 9672000^. . . (1) . . (not homog.) 

(f, I, and d must all be expressed in the same linear unit.) 
Here t = thickness of the wall of the tube, d its diameter, and 
I its length ; the ends being understood to be so supported aa 
to preclude a local collapse. 

Example. — ^With 1 = 10 ft. = 120 inches, d = 4: in., and t = 
^ inch, we have 



p = 9672000 



J 



-^ H- (120 X 4) = 201.5 lbs. per sq. inch. 



For safety, ^ of this, viz. 40 lbs. per sq. inch, should not be 
exceeded ; e.g., with 14.7 lbs. internal and 54.7 lbs. external. 

[Note. — For simplicity the power of the thickness used in eq. (1) above 
has been given as 2.00. In the original formula it is 2.19, and then all 
dimensions must be expressed in incJies. A discussion of the experiments 
of Mr. Fairbairnwill be found in a paper read by Prof. Unwin before the 
Institute of Civ. Engineers (Proceedings, vol xlvi.). See also Prof. Unwin's 
" Machine Design," p. 66. It is contended by some that in the actual con- 
ditions of service, boiler-flues are subjected to such serious straining 
actions due 1o unequal expansion of the connecting parts as to render the 
above formula quite unreliable, thus requiring a large allowance in ita 
application.] 

437a. Collapsing Pressure of Steel Tubes. — Recent experiments by Prof. 
R. T. Stewart (see Engineering News of May 10, 1906, p. 528) on Bes- 
semer steel lap-welded tubes of 8§- in. in diameter and all commercial 
thicknesses of wall and in lengths of 2]-, 5, 10, 15, and 20 ft.; and also on 
single lengths of 20 ft. (between end connections) in seven sizes from 3 to 
10 inches outside diameter and in all commercial thicknesses obtainable; 
have shown that length has practically no influence on the strength, if the 
length is greater than six times the diameter. From these experiments 
Prof. Stewart has deduced the following formulae in which p is the col- 
lapsing unit-pressure in lbs. per sq. inch, d the outside diameter of the 
tube in inches, and t the thickness of wall of tube, also in inches : — 



p = 1000(l-/j/l-1600|^), (4) 

p = 86670 J— 1386.0 (5) 

Eq. (4) is for use where f-j-d is less than 0.028, and eq. (5) for larger 
T^alues of that ratio. 



540 MECHANICS OF ENGINEEKIKCh. 



CHAPTEE II. 

HYDROSTATICS (Cow^mwecZ)— PRESSURE OF LIQUIDS IN TANKS. 
AND RESERVOIRS. 

428. Body of Liquid in Motion, but in Relative Equilibrium.— 

By relative equilibrium it is meant that the particles are not 
changing their relative positions, i.e., are not moving among 
each other. On account of this relative equilibrium the fol- 
lowing problems are placed in the present chapter, instead of 
under the head of Hydrodynamics, where they strictly belong. 
As relative equilihriwm is an essential property of rigid bodies, 
we may apply the equations of motion of rigid bodies to bodies 
of liquid in relative equilibrium. 

Case L All the particles moving in parallel right lines 
with equal velocities ^ at any given instant (i.e., a motion of 
translation.) — If the common velocity is constant we have a 
uniform translation, and all the forces acting on any one par- 
ticle are balanced, as if it were not moving at all (according to 
iNewton's Laws, § 54); hence the relations of internal pressure^ 
free surface, etc., are the same as if the liquid were at rest. 
Thus, Fig. 471, if the liquid in the moving tank is at rest rel- 
^ atively to the tank at a given instant, with 
" ?____ _n its free surface horizontal, and the motion 

^^^^^^^ of the tank be one of translation with a uni- 

y^J \ y^ ' '" form velocity, the liquid will remain in this 
mdmm^^ condition of relative rest, as the motion 

Fig. 471. , 

proceeds. 
But if the velocity of the tank is accelerated with a consta/nt 
acceleration ^=p (this symbol must not be confused with p 
for pressure), the free surface will begin to oscillate, and finally 
come to relative equilibrium at some angle oc with the horizon- 
tal, which 18 thus found, when the motion is horizontal. See 
Fig. 472, in which the position and value of a are the same, 
whether the motion is uniformly accelerated from left to rigbt 



EELATIVE EQUILIBRIUM OF LIQUIDS. 



541 



Let be the lowest 



-->p 




Fig. 473. 



or uniformly retarded from right to left, 
point of the free surface, and Oh a 
small prism of the lig^uid with its 
axis horizontal, and of length = x ; 
rib is a vertical prism of length = 
0, and extending from the extremity 
of Oh to the free surface. The 
pressure at both and n \% jp^i=. 
atmos. pres. Let the area of cross- 
section of both prisms be = dF. 

l!^ow since Oh is being accelerated in direction ^(horizont.), 
the difference between the forces on its two ends, i.e., its ^Xy 
must = its mass X accel. (§ 109). 

.'.jp^dF-jpadF^ixdF.y — g']]). . . . (1) 

{y = heaviness of liquid ; pi, = press, at h) ; and since the ver- 
tical prism nh has no vertical acceleration, the -2(vert. com- 
pons.) for it must = 0. 

.\pSF-padF-zdF.y=^, ..... (2) 

From (1) and (2), 



xv — 
-^.p = zy\ 



X g° 



(3) 



... (4) 



Hence On is a right line, and therefore 

z V 

tan a, or — , =-^. . . 

X g 

[Another, and perhaps more direct, method of deriving this 
result is to consider free a small particle of the liquid lying in 
the surface. The forces acting on this particle are two : the 
first its weight = dG ^ and the second the resultant action of 
its immediate neighbor-particles. ]^ow this latter force (point- 
ing obliquely upward) must be normal to the free surface of 
the liquid, and therefore must make the unknown angle a with 
the vertical. Since the particle has at this instant a rectilinear 
accelerated motion in a horizontal direction, the resultant of the 
two forces mentioned must be horizontal and have a value = 
mass X acceleration. That is, the diagonal formed on the two 



542 



MECHANICS OF ENGINEERING, 



forces must be horizontal and have the value mentioned, = 
{dG -V- g)jp ; while from the nature of the figure (let the stu- 
dent make the diagram for himself) it must also = dG tan a. 



dG 



P 



Q. E. D. 



.*. dG tan a = — .p ; or, tan a = 
9 9 

If the translation were vertical, and the acceleration xijpward 
[i.e., if the vessel had a uniformly accelerated upward motion 
or a uniformly retarded downward motion], the free surface 
would be horizontal, but the pressure at a depth = h below the 
surface instead oi jp =^Pa + ^7 would be obtained as follows: 
Considering free a small vertical prism of height = Ji with 
upper base in the free surface, and putting 2(vert. compons.) 
= mass X acceleration, we have 

hdF. y 



dF . p — dF . jpa — hdF . y 



9 



p 



'P=I>a-[-hy 



' 9+P 
L g .. 



(5) 



If the acceleration is downward (not the velocity necessarily) 
we make J? negative in (5). If the vessel falls freely, j} =— g 
and .'.p =pai in all parts of the liquid. 
Query : Suppose jp downward and > g. 
Case II. Uniform Rotation about a YerticalAxis. — If the 
narrow vessel in Fig. 473, open at top and containing a liquid, 
^c be kept rotating at a uniform angu- 

lar velocity cl> (see § 110) about a 
vertical axis Z, the liquid after some 
oscillations will be brought (by fric- 
tion) to relative equilibrium (rotat- 
ing about Z, as if rigid). Required 
the foi-m of the free surface (evi- 
dently a surface of revolution) at 
each point of which we know 

JP=Pa- 

Let be the intersection of the 
axis Z with the surface, and n any point in the surface ; J being 



V 



1 


71 

/ 


1 


/ 






\ 1 / 




-'k. n / 




?\:~~-^^f 




3S — 


/ 



Fig. 473. 



TJlSriFOEM EOTATION- OF LIQUID IIST VESSEL. ^43 

a point vertically under n and in same horizontal plane as 0. 
Every point of the small right prism nh (of altitude = s and 
sectional area 6^^) is describing a horizontal circle ahont Z. nnd 
has therefore no vertical acceleration. Hence for this prism, 
free, we have 2Z = 0; i.e., 

dF.j?t, - dF.pa - zdF. y = (1) 

!Now the horizontal right prism Oh (call the direction ...h^ 
^) is rotating uniformly about a vertical axis through one ex- 
tremity, as if it were a rigid body. Hence the forces acting 
on it must be equivalent to a single horizontal force, — ca/Mp. 
(§122«,) coinciding in direction with JT. IM= mass of prism 
= its weight -^ g, and p = distance of its centre of gravity 
from ; here p = ^x= ^ length of prism]. Hence the 2X 

xdF 
of the forces acting on the prism Oh must = — ta^ y^x. 

But the forces acting on the two ends of this prism are their 
own ^components, while the lateral pressures and the weights 
of its particles have no Xcompons. ; 

JTT JTT —Gifx^dF.y ,^. 

,'.dF.jc>a — dF.pt — o ~' • • (2) 

From (1) and (2) we have 

^--27 -2^' ...... (3) 

where v = axe = linear velocity of the point n in its circular 
path. 

[As in Case T, we may obtain the same result by considering 
a single surface particle free, and would derive for the resultant 
force acting upon it the value dG tan n' in a horizontal direc- 
tion and intersecting the axis of rotation. But here a is dif- 
ferent for particles at different distances from the axis, tan a 

being the -j- of the curve On. As the particle is moving uni- 
formly in a circle the resultant force must point toward the 



544 MECHANICS OF ENGIKEERINa. 

eectre of the circle, i.e., horizontally, and have a value — . — , 

g 10 

where x is the radius of the circle [§ 74, eq. (5)] ; 

jj^ . dG (goxY ^ dz Go^x 

.*. aijr tan a = -^ — -^ ; or tan «,=—-,= — ; 

g X ax g 

.*. / dz — -~ I xdxi or, z = — . -. . Q. E. D. 

Hence any vertical section of the free surface through the 
axics of rotation Z is a, parabola, with its axis vertical and vertex 
at 0; i.e., the free surface is 2i paraboloid of revolution, with 
Z as its axis. Since cox is the linear velocity v of the point 
h in its circular path, ^ = " height due to velocity" v [§ 52], 

Example. — If the vessel in Fig, 4Y3 makes 100 revol. per 
minute, required the ordinate s at a horizontal distance of a? = 
4 inches from the axis (ft.-lb.-sec. system). The angular veloc- 
ity OS =. [^Tt 100 -V- 60] radians per sec. [K. B. — A radian = 
the angular space of which 3.1415926 . . . make a half-re vol., 
or angle of 180°]. With a? = i f t. and g — 32.2, 

and the pressure ht b (Fig. 473) is (now use inch, lb., sec.) 

62 5 
P& —Pa + ^r— 14:.7 +^X j;^ = 14.781 lbs. per sq. in. 

Picf. Mendelejeff of Russia has recently utilized the fact an- 
nounced as the result of this problem, for forming perfectly 
true paraboloidal surfaces of plaster of Paris, to receive by 
galvanic process a deposit of metal, and thus produce specula 
of exact figure for reflecting telescopes. The vessel contain- 
ing the liquid plaster is kept rotating about a vertical axis 
at the proper uniform speed, and the plaster assumes the de- 
sired shape before solidifying. A fusible alloy, melted, may 
also be placed in the vessel, instead of liquid plaster. 



EELATIYE EQUILIBKIUM. 



545 



top, ex- 



Remaek. — If the vessel is quite full and closed on 
except at 0' where it communicates 
bj a stationary pipe with a reser- 
voir, Fig. 474, the free surface 
cannot be formed, but the pres- 
sure at any point in the water is 
just the same during uniform rota- 
tion, as if a free surface were formed 
with vertex at ; 

i.e., p-„=p^ + (A„ 4- z)y. . (4) 

See figure for h^ and z. (In subse- 
quent paragraphs of this chapter 
the liquid will be at rest.) 

Fig. 474. 

428a. Pressure on the Bottom of a Vessel containing Liquid at 
Rest. — If the bottom of the vessel is plane and horizontal, the 
intensity of pressure upon it is the same at all points, being 







^ISi^ 



Fig. 475. 




p—ip^J^hy (Figs. 475 and 476), and the pressures on the ele- 
ments of the surface form a set of parallel (vertical) forceSo 
This is true even if the side of the vessel overhangs, Fig. 476, 
the resultant fluid pressure on the bottom in both cases being 



P = Fp-F2?a = Fhy. 



(1) 



(Atmospheric pressure is supposed to act under the bottom.) 
It is further evident that if the bottom is a rigid homogeneous 
plate and has no support at its edges, it may be supported at a 



-U^='- 




^6 MECHANICS OF ENGINEEEllSTG. 

single point (Fig. 477), which in this case (horizontal plate) 
is its centre of gravity. This point is calle*] 
the Centre of Pressure, or the point of apjili- 
cation of the resuliant of all tlie fluid pressures 
acting on the plate. The present case is such 
that tliese pressures reduce to a single result- 
ant, but this is not always practicable. 

Example. — In Fig. 476 (cylindrical vessel 
containing water), given A = 20 ft., h^ = 
15 ft., r^ = 2 ft., )\ = 4 ft., required the pressure on the bot- 
tom, the vertical tension in the cylindrical wall CA^ and the 
hoop tension (^§ 426) at G. (Ft., lb., sec.) Press, on bottom = 
Fhy= Ttr^hy = vtlQ X 20 X 62.5 = 62857 lbs.; while the 
upward pull on CA = 

{m;' - 7rr;)h,y = ;r(16 - 4)15 X 62.5 = 35357 lbs. 

If the vertical wall is t = -^-q inch thick at C'this tension will 
be borne by a ring-shaped cross section of area = 27T7\t (nearly) 
= 27r48 X tV = 30.17 sq. inches, giving (35357 -^ 30.17) = 
about 1200 lbs. per sq. inch tensile stress (vertical). 
The hoop tension at C is horizontal and is 

p" = r^P —Pa) -^ i (see § 426), where ^ =p^ -f h,y ; 

„ 48 X 15 X 12 X (62.5 ^ 1728) __„ ,, 
.\j}" z= ^ i = 3125 lbs. per sq. in., 

(using the inch and pound). 

429. Centre of Pressure. — In subsequent work in this chapter, 
since the atmosphere has access both to the free surface of 
liquid and to the outside of the vessel walls, and p^ would can^ 
eel out in finding the resultant fluid pressure on any elemen, 
tary area d.F of those walls, w^e shall write : 
' The res^dtant fluid pressure on any dF' of the vessel wall is 
normal to its surface and is dP z= pdF "= zydF. in which s 
is the vertical distance of the element below the free surface 
of the liquid (i.e., s = the '■'head of water"). If the surface 
pressed on is plane, these elementary pressures form a system 
of parallel forces, and may be replaced by a single resultant 



CENTRE OF PEESSURE. 



547 



(if the plate is rigid) which will equal their sum, and whose 
point of application, called the Centre of Pressure, may be 
located by the equations of § 22, put into calculus form. 

If the surface is curved the elementary pressures form a sys 
tem of forces in space, and hence (§ 38) cannot in general he 
reduced to a single resultant, but to two, the point of applica- 
tion of one of which is arbitrary (viz., the arbitrary origin, 
§38). 

Of course, the object of replacing a set of fluid pressures by 
a single resultant is for convenience in examining the equi- 
librium, or stability, of a rigid body the forces acting on which 
include these fluid pressures. As to their effect in distorting 
the rigid body, the fluid pressures must be considered in their 
true positions (see example in § 264), and cannot be replaced 
by a resultant. 



430. Resultant Liquid Pressure on a Plane Surface forming 
Part of a Vessel "Wall. Co-ordinates of the Centre of Pressure. — 

Fig. 478. Let JL5 be a portion (of any shape) of a plane 

surface at any angle with the 

horizontal, sustaining liquid 

pressure. Prolong the plane 

of AB till it intersects the free 

surface of the liquid. Take 

this intersection as an axis Y, 

being any point on Y. The 

axis X, ~1 to Y, lies in the 

given plane. Let a = angle 

between the plane and the free 

surface. Then x and y are the 

co-ordinates of any elementary 

area dF oi the surface, referred to Xand Y. z = the "head 

of water," below the free surface, of any dF. The pressures 

are parallel. 

The norinal pressure on any dF =■ zydF", hence the swn of 
these, = their resultant. 




Fig. 478. 



=.P, = yfsdF=Fzy', 



(1) 



648 MECHANICS OF ENGINEEEIJSTG, 

in which z = the " mean s," i.e., the b of the centre of gravity 
G of the plane figure AB, and ^= total area of AB \_F z — 
fzdF^ from eq. (4), § 23]. y = heaviness of liquid (see § 409). 
That is, the total liquid py^essure on a plane figure is equal 
to the weight of an imaginary jprism of the liquid hamng a. 
'base = area of the given figure and an altitude = vertical 
depth of the centre of gravity of the figure helow the surface 
of the liquid. For example, if the figure is a rectangle with 
one base (length = V) in the surface, and lying in a vertical 
plane, 

P = lh. \hy = ^hhy. 

Evidently, if the altitude be increased, P varies as its square. 

From (1) it is evident that the total pressure does not de- 
fend on the horizontal extent of the water in the reservoir. 

Now let a?c and y^ denote the co-ordinates, in plane YOX^ 
of the centre of pressure., G, oy point of application of the re- 
sultant pressure P, and apply the principle that the sum of 
the moments of each of several parallel forces, about an axis "1 
to them, is equal to the moment of their resultant about the 
same axis [§ 22]. First taking OY 2iS> an axis of moments, 
and then OX, we have 

Px^ = J* {zydF)x, and Py^ = f {zydF)y. . (2) 

But P = Fzy = Fx{sm a)y, and the z of any dF= X sin a. 
Hence eqs. (2) become (after cancelling the constant, y sin a) 

F X Fx Fx 

in which Iy = the " tnojn. of inertia''^ of the plane figure re- 
ferred to Y (see § 85). [_'N. B. — The centre of pressure as 
thus found is identical with the centre of oscillation (§ IIY) 
and the centre of percussion [§ 113] of a thin homogeneous 
plate, referred to axes X and Y, Y being the axis of suspen- 
sion.] 

Evidently, if the plane figure is vertical a = 90°, a? = ^ for 



CENTRE OF PRESSUEE. 



549 



all 6?^'s, and a? = s. It is also noteworthy that the position 
of the centre of pressure is independent of a. 

JSToTE. — Since the pressures on the equal cZ^'s lying in any 
horizontal strip of the plane figure form a set of equal parallel 
forces equally spaced along the strip, and are therefore equiva- 
lent to their sum applied in the middle of the strip, it follows 
that for rectangles and triangles with horizontal bases, the 
centre of pressure must lie on the straight line on which the 
middles of all horizontal strips are situated. 

431. Centre of Pressure of Rectangles and Triangles with Bases 
Horizontal. — Since all the dF''% of one horizontal strip have 
the same a?, we may take the area of the strip ^ 

for <i^ in the summation /b'^^^ Hence for T 
the rectangle AB, Fig 4Y9, we have from eq. \ 
(3), § 430, ynt\\dF= hdx, I 



x, = KG = 






tJu CL'to 



KK-K) 



h,-\-h, 






dx 



— -5--— ->B 
Fia. 479. 



while (see note, § 430) y^ = i^- 

Whe7i the tipjyer hase lies in the surface, h^ = 0, and x^ = 
1^2 =: ^ of the altitude. 

For a triangle loith its hase horizontal and vertex tip. Fig. 
480, the length %i of a horizontal strip is variable and dF=- 

udx. From similar triano;les it = -{x — h^ \ therefore 




\ — A, 



K 



I x^{x — h}jdx 



lh{K 



K)Uh-^Wh-K)] 



x'ix — h^dx = 



U,V4 



'2 /a?' 7 X 



-6--B > 

Fig. 480. 



J, 3A/ + 2AA + A; 



(2) 



650 MECHANICS OF ENGINEEEING. 

Also, since the centre of pressure must lie on the line AB join- 
ing the vertex to the middle of base (see note, § 430), we easilji 
determine its position. 
Evidently for A, = 0, i.e., when the vertex is in the surface, 



o 



^ jY Xc = f Aj. Similarly, /br a triangle with 

] y*i hase horizontal and vertex doion^ Fig. 481, 

mV 7~i /D we find that 

\l....±..^ If tli6 base is in the surface, A^ = and 

Fig. 481. (3) rcduccs to Xc = ^h^. 

It is to be noticed that in the case of the triangle the value 
of Xc is the same whatever be its shape, so long as h^ and k^ 
remain unchanged and the base is horizontal. If the base is 
not horizontal, we may easily, by one horizontal line, divide 
the triangle into two triangles whose bases are liorizontal and 
whose combined areas make up the area of the first. The re- 
sultant pressure on each of the component triangles is easily 
found by ths foregoing principles, as also its point of applica- 
tion. The resultant of the two parallel forces so determined 
will act at some point on the line joining the centres of pres- 
sures of the component triangles, this point being easily found 
by the method of moments, while the amount of this final re- 
sultant pressure is the sum of its two components, since the 
latter are parallel. An instance of this procedure will be 
given in Example 3 of § 483. Similarly, the rectangle of Fig. 
479 may be distorted into an oblique parallelogram with hori- 
zontal bases without affecting the value of x^ , nor the amount 
of resultant pressure, so long as h^ and h^ remain unchanged. 

432. Centre of Pressure of Circle. — Fig. 482. It will lie on 
the vertical diameter. Let r = radius. From eq. (3), § 430, 

° 1 i ] ^ /^ L^Fla" iTtr' -^ Ttr'a" 

■ ^4-1 — ^ -T a? jT a? Tvrx 

M 1^l\ C^®^ ®<i- (^)' § ^^' ^^^ ^^^^ § ^^0 

\ _.i__Jc J 12 

X___y .'.x, = x-\-j.-. ... (4) 

l<a. 482. ^ X 



CENTRE