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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
thumbscrews 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 vernierarm 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
r
^^'o^ .<t
\
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 oilengines, the Cippolletti weir, losses of heads in
pipes and bends, the hydraulic gradeline, the Venturi meter, currentmeters,
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 halftone cut of their currentmeter;
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 sealevel,
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 onethird 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 sealevel), 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 = ab, 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 10inch 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 =
10.3010301.000000 (or, 19.30103010). 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., =0.92741.0000.
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{abx^).
\abx^ SVab'"* Vabbx' jabx^
Numerical Constants. — The acceleration of gravity, g, (for the English
foot and second) is 32.16 for the latitude of Philadelphia at sealevel, and
for any latitude ji, and elevation h above sealevel, is
32.17230.0833 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 1619. Composition and Equilibrium of Concurrent Forces. ... 8
Chapter II. Parallel Forces and the Centre op Gravity.
§§ 2022. Parallel Forces 1^
§§23276. Centre of Gravity. Problems. Centrobaric Method. . . 18=
Chapter III. Statics of a Rigid Body.
g§ 2834. Couples 27
^§3539. Composition and Equilibrium of Nonconcurrent Forces. 31
Chapter IV. Statics of Flexible Cords.
§§ 4048. Postulates. Suspended Weights. Parabolic Cord. Cat
enary 4r
PART II. KINETICS.
Chapter I. Rectiijnear AIotion of a Material Poini
^§ 4955. 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 Q16&, Definitions and Propositions. Connectingrod. Prob
lems 67
yiil CONTENTS.,
•^ChAPTBR III. CUKVILINEAR MOTION OP A MATERIAL POINT.
PAOS
§§ 7074. Composition of Motions, of Velocities, etc. General
Equations 72
§§ 7584 Normal Acceleration. Centripetal and Centrifugal
Forces. Simple Pendulums. Projectiles. Rela
tive Motion 77
Chapter IV. Moment of Inertia.
%% 8594. Plane Figures. Rigid Bodies. Reduction Formulae.
The Rectangle, Triangle, etc. Compound Plane
Figures. Polar Moment of Inertia 9t
§S 95104. Rod. Tliin Plates. Geometric Solids 98
■§§105107. Numerical Substitution. Ellipsoid of Inertia ICo
Chapter V. Kinetics of a Rigid Body.
§§108115. Translation. Rotation about a Fixed Axis. Centre of
Percussion 105
§ 117121. Compound Pendulum. The Flywheel 118
§§ 122123. Uniform Rotation. " Centrifugal Action," Free Axes. 125
§ 124136. Rolling Motions. Parallel Rod of Locomotive 130
^ Chapter VI. Work, Energy, and Power.
§§ 127134. Work. Power, Horsepower. Kit etic Energy...... 133
§§ 135138. Steamhammer. Piledriving. Inelastic Impact . 188
§§ 139141. Rotary Motion. Equivalent Systems of Forces. Any
Motion of a Rigid Body .„ 143
§§ 142146. Work and Kinetic Energy in a Moving Machine of
Rigid Parts 147
§§ 147155. ^ Power of Motors. Potential Energy. Heat, etc. Dy
namometers. Boatrowing. Examples 153
Chapter VII. Friction.
§§ 156164. Sliding Friction. Its Laws. Bent Lever 164
§§ 165171. Axlefriction. Friction Wheels. Pivots. Belting.
Transmission of Power by Belting 175
§ 172177. Rolling Friction. Brakes. Friction of Car Journals;
and of Welllubricated Journals. Rigidity of
Cordage. Examples 186
CONTENTS.
IX
PART III. STRENGTH OF MATERIALS.
(or mechanics of matekials.)
OHAPTER I. ELEMENTARY STRESSES AND STRAINa
^§ 178183. Stress and Strain i of Two Bands. Oblique Section
of Rod in Tension . 195
§§ 183a190o 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
§§ 191195.
i§ 196199b
§§ 30030Sp
§§ 303306. 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,
§§ 314330. Angle and Moment of Torsion. Torsional Strengtila,
Stiffness, and Resilience. NonCircular Shafts 383
§§ 331333. Transmission of Power. Autographic Testing Ma
chine. Torsion Clinometers. Examples 238
OHAPTER in. FLEXURE OF HOMOGENEOUS PRISMS UNDER
PERPENDICULAR FORCES IN ONE PLANE.
§§ 334r333ao The Common Theory. Elastic Forces. Neutral
Axis. The "Shear" and "Moment." Flex
lual Strength and Stiffness. Radius of Curva
Uire. Resilience c ... c . ^ ...., ^  .. . 344
ELASTIC CURVES.
^§ 333338. Single Central Load ; at Rest, and Applied Suddenly.
Eccentric Load. Uniform Load. Cantilever. . 353
X co:n'tents.
SAFE LOADS.
§§ 239246. Maximum Mom^ent. Shear = xDerivative of the
Moment. Simple Beams With Various Loads.
Comparative Strength and Stiffness of Rectan
gular Beams 262
§§ 247252. Moments of Inertia. Rolled Steel IBeams, etc.
Cantilevers. Tables. Numerical Examples . . . 273^
SHEARING STRESSES IN FLEXURE.
§g 253257. Shearing Stress Parallel to Neutral Surface ; and in
Cross Section. Web of IBeam. Riveting of
Built Beams . , 284
SPECIAL PROBLEMS IN FLEXURE.
§§ 258265. Designing Sections of Built Beams. Moving Loads.
Special Cases of Quiescent Loads. Hydrostatic
Load 395
§§ 266270. Strength of Flat Plates. Weight Falling on Beam.
Crank Shaft. Other Shafts. Web of IBeam . 310^
§§ 271276.
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 NONPRISMATIC BEAMS.
§§ 277279. Double Truncated Wedge, Pyramid, and Cone 332
NONPRISMATIC BEAMS OF UNIFORM STRENGTH.
§§ 280289. Parabolic and WedgeShaped Beams. Elliptical
Beams. (See Appendix, p. 841, for Cantilevers). 335
CHAPTER V. FLEXURE OF REINFORCED CONCRETE
BEAMS.
§§ 284287. Concrete, and "ConcreteSteel" Beams. Beams of
Rectanaiular Section. Horizontal Shearing
Stresses in Latter. Examples 33&
§§ 288294. ConcreteSteel Beams of TForm Section. Deflection
of ConcreteSteel Beams. Practical Fotmulse
and Diagrams 346
CONTENTS. XI
CHAPTER VI. FLEXURE. COLUMNS AND HOOKS. OBLIQUE
LOADS.
§§ 294300. Oblique Cantilever. Moment, Thrust and Shear.
Experimental Proof. Common Theory of
CraneHooks. WinklerBach Theory 352
§ 301314. Long Columns. EndConditions. Euler's and Ran
kine's Formulae. Examples. Radii of Gyra
tion. Built Columns. Other Formul:e: Merri
manRitter ; ' ' StraightLine ; ' ' 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.)
§§ 315823. Inverted Catenary. Parabolic Arch. Circular Arch.
Transformed Catenary as Arch 886
CHAPTER VIII. ELEMENTS OF GRAPHICAL STATICS
§§ 831326. Force Polygons. Concurrent and NonConcurrent
Forces in a Plane. Force Diagrams. Equili
brium Polygons 897
§§ 327832. Constructions for Resultant, PierReactions, and
Stresses in Roof Truss. Bow's Notation. The
Special Equilibrium Polygon 403
CHAPTER IX. GRAPHICAL STATICS OF VERTICAL FORCES.
§g 333336. Jointed Rods. Centre of Gravity 413
§§ 337848. Useful Relations Between Force Diagrams and Their
Equilibrium Polygons 415
CHAPTER X. RIGHT ARCHES OF MASONRY.
§§ 344353. Definitions. Mortar and Friction. Pressure in Joints.
Conditions of Safe Equilibrium. True Linear
Arch 421
§§ 353357. Arrangement of Data for Graphic Treatment 428
§§ 358363. Graphical Treatment of Arch. Symmetrical and
Unsymmetrical Cases 431
CHAPTER XI. ARCHRIBS.
§§ 334374, Mode of Support. Special Equilibrium Polygon and
its Force Diagram. Change in Angle Between
Rib Tangents. Displacement of Any Point on
Rib 438
§§ 374a378a. Graphical Arithmetic. Summation of Products. .
Moment of Inertia by Graphics. Classification
of ArchRibs 450
XU CONTENTS.
§§ 379388. Prof. Eddy's Graphical Method for ArchRibs of
Hinged Ends ; and of Fixed Ends. Stress
Diagrams. Temperature Stresses. Braced
Arches 461
§§ 389391. Cucular Ribs and Hoops 479
CHAPTER XII. FLEXURE OF BEAMS, SIMPLE AND CONTIN
UOUS, GEOMETRICAL TREATMENT.
§§392398. 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
§§ 399405. 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 "Builtin" Ends.
Deflections Found by Theorem of Three Mo
ments 494
CHAPTER XIII. THICK HOLLOW CYLINDERS AND SPHERES.
§§ 405a405sr. 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." Nonplanar Pistons .. . 515
§§ 418427. Hydraulic Press. Free Surface of Liquid. Barometers
and Manometers. The Differential Manometer.
Safetyvalves. Strength of Thin Hollow Cylinders
against Bursting and Collapse 526
CHAPTER II.— HYDROSTATICS CONTINUED. PRESSURE OF
LIQUIDS IN TANKS AND RESERVOIRS.
§§428434. Liquid in Motion, but in "Relative Equilibrium."
Pressure on Bottom and Sides of Vessels. Centre
of Pressure of Rectangles, Triangles, etc 540'
§§ 435444. 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.
§§ 445455. 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.
§§456460. Buoyant Effort. Examples of Immersion. Specific
Gravity. Equilibrium of Flotation. Hydrometer.. 586
§§ 461465. Depth of Flotation. Draught and Angular Stability of
Ships. The Metacentre 593.
xiii
XIV CONTENTS.
CHAPTER v.— HYDROSTATICS CONTINUED. GASEOUS
FLUIDS.
§§466478. Thermometers. Absolute Temperature. Gases and Va
pors. Critical Temperature. Law of Charles.
Closed Airmanometer. Mariotte's Law. Mixture of
Gases. Barometric Levelling. Adiabatic Change.. G04
§§479489. Work Done in Steamengine Cylinders. Expanding
Steam. Graphic Representation of Change of Stat?
of Gas. Compressedair Engine. Aircompressor.
Hotair Engines. Gasengines. Heat efficiency.
Duty of Pumpingengines. Buoyant Effort of the
Atmosphere 624
CHAPTER VL— HYDROKINETICS BEGUN. STEADY FLOW OF
LIQUIDS THROUGH PIPES AND ORIFICES.
§§ 489a495. Phenomena of a " Steady Flow." Bernoulli's Theorem
for Steady Flow without Friction, and Applications,
Orifice in " Thin Plate" 646
§§ 496500. Rounded Orifice. Various Problems involving Flow
through Orifices. Jet from Forcepump. Velocity
and Density; Relation. Efflux under Water. Efflux
from Vessel in Motion. Barker's Mill 663
§§ 501508. 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
§§ 509513. 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
§§519526. Chezy's Formula. The "Hydraulic GradeLine."
Pressureenergy. Losses of Head due to Sudden
Enlargement of Section; Borda's Formula. Dia
phragm in Pipe. Venturi Watermeter 71'!
.?§ 537536. Sudden Diminution of Section. Losses of Head due to
Elbows, Bends, Valvegates, and Throttlevalves.
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
§§ 538542a. Nomenclature. Velocity Measurements and Instru
ments for the same. RitchieHaskell Direction
Currentmeter. Change of Velocity with Depth.
Pitot's Tube and W. M. White's Experiments.
Currentmeters. Gauging Streams. Chezy's For
mula for Uniform Motion in Open Channel.
Experiments 749
§§ 54Ji6647. 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.
§§ 548556. Theorem for Steady Flow of Gases without Friction.
Flow through Orifices by Waterformula; 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
§^ 567561a. 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
§§ 562569. Reaction of a Jet of Liquid. Impulse of Jet on Curved
Vanes, Fixed and in Motion. Pitot's Tube. The
California "Hurdygurdy." Impulse on Plates.
Plates Moving in a Fluid. Plates in Currents of
Fluid .... 798
B§ 57C'575. Windpressure. Smitlisoniau Scale. Mechanic; of the
Sailboat. 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. Firestreams, Hose friction, etc  818
Appendix. Miscellaneous Addenda and Tables 835854
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. Pressuie, attraction, repulsion, and
traction are instances in point. Muscular sensation conveys
the idea of force, while a springbalance gives an absolute
measure of it, a beambalance 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 mejely 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
Brickwork, 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, Mliose 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
SLga
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 numbeis,
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 nonconcurrent 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 nonconcurrent 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 Antiresultant.
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 = Vf" + §' + 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.
Siniilaily 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.
similaily 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
•deteimine 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 EreeBody Method. — The foregoing rather labored so
lutions of very simple problems have been made such to illus
trate what may be called the "freebody 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 "forceahle" 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 castiron 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, 0.7660P0.9396Q = 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 4ton weight is suspended on the bolt C,
which passes through the ends of boom OC^ and tierod DC. Bolt C
is also subjected to a horizontal pull tov/ard the left, due to the 2ton
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 tierod 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°(?20 = 0; (5)
iF = Q sin41°Psin 15°Gi0 = 0; (6)
or, numerically, 0.7547Q0.9659P2 = 0, (7)
and 0.6560Q0.2588P4 = (8)
From (7), Q = 2.6514 1.279P, which in (8) gives
0.6560(2.651 + 1.279P)  0.2588P = 4 ;
that is, 1.740H0.8390P0.2588P = 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 nonconcurrent. In
most problems in statics, however, the forces acting are in one
plane: we accordingly begin by considering nonconcurrent
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 sL3/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;
■XH
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^cj 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 leverarm.
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^ — ai
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 IP2&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 coordinates of their points of application
as referred to an arbitrary set of three coordinate 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 coordinates 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 coordinate z'\ and we obtain
P"z" = P'z' + P3S3, ie., 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,
PP. + i'.+ i^s + etc.;
and for the vertical coordinate 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 coordinates.
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^ Lil 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 + 66 + 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 footpounds.) 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 fieo 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 \ ' ^ ^
?^\ — gis 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 socalled attraction
of the latter (i.e., gravitation), as shown by a springbalance,
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 coordinates of the centre of gravity
of a body referred to three coordinate 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 coordinates 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 coordinate 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 coordinates
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. zMl. ,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 coordinates 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 crosssection, 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
,,. rg^ 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 nonparallel
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 coordinates, 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 (;rmeasure) ; e.g., for a quad
rant, yS = I = 1.5707*+ ; for 30°, /? = ^ ; or, in general, if fi
m degrees := , then p in ;rmeasure = — .
° 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. 10
 _ 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
halfheight 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 halfchord, 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 gravityline 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 gravityline 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 — »«5i of the crosssection 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.3f2[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 jbotpounds; or 0.180 foot
tons; or 4320 inchpounds; or 2.16 inchtons.
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 throuoh 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 = PfxSg,
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
hkff" 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 Nonconcurrent 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.
coordinates 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 directioncosine 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 ^{YxXy) = 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 nonconcurrent 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 nonconcurrent
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 "momentequation," 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 " leverarmj^'' 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 Nonooncurrent 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^^^epressure 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 nonconcurrent 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 hingepressure 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 bridgetruss rests on the horizontal
smooth abutmentsurfaces 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. " Twoforce 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+AhVi2a+Pi ■ fa+P2 • ia = 0;
and hence A = {ia^h)[Wi3Pi^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/i7ia+Pia=0; /. C=(iaJ/i)[37i2PJ,
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 + 7iPiP2 = 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,FJcos9>
^^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 hingepin. Fig. 41.
Problem 5. — Rooftruss; pinconnection; all loads at joints ;
windpressures 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 Nonconcur
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 coordinates
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, (ftlbs., 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
coordinate 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 directionangles
^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 nonconcur
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 nonconcurrent forces in space is in
equilibrium, the plane systems formed hy projecting the given
systein upon each of three arbitrary coordinate 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 nonconcurrent 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 = GiP, 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 nonconcurrent system of forces in space; in equi
librium. Since there are no Ycomponents 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 "endview" 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.)
+ PX1.5800XH0 = 0; .. P= 177.77 lbs. . . . (1)
Next projecting on the vertical plane ZY, containing axis of shaft
(i.e., taking a "sideview" 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+800X2ZiX3PX0.50X4 + 0=0 ... (2)
while moms, about Oigives+ZoX 3 Px0.5Xl200Xli 800 1 = (3)
Finally, projecting on the horizontal plane XZ ("topview"), the
forces in this projection are P cos 30°, X^, and X^; so taking moms.
about point Oj,
+ZoX3Px0.8660Xl = 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 XqhZq, (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 balland
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 cordtension 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 "sprocketteeth" 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)rri. 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, tierod 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 tierod 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 1400X9400X7 = (1)
i.e., TX 9 X 0.8391 = 6000 ft.lbs.; .. 2" = 794.3 lbs., (tension in tierod.)
IX = Xi, .•.Xo400cos40°r = 0; .. Xo = 794.3H400X .766= 1100.7 lbs.
^F=0, .. Yo^OO sin40°400400 = 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., Tj =
. (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 tangentline 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 suspensionbridge m
which the suspensionrods 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
qrx
Fis. 53.
M, = U,\ .. the tension at m = \^ H^ + Y^= ^{_qb V4:a'\ b'].
Act
The semilength, 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. Requiied 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 2J^ = ^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 40foot 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 halfspan. 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, ?/lc = 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/Pd^ = ch2ce 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
=ccoshl —
(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^— gw] is called the "hyperbolic sine" of the number
u, or sinh (u); and ^[e^+eu] 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.— ■\/Pd^ = \/9100 = 95A ft., the lefthand member of (7).
Assuming c = 20 ft. as a first trial, we find fcr2c = 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;r2c)
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
L0
* 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{GiV)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 startingpoint
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=40x3i(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 = ctiipit^; , . . (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 = 48 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 coordinates 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 = riSHSi.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 coriesponding 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 crankpin of a reciprocating engine moved uniformly
in its circular path, the piston would have a harmonic motion
if the connectingrod 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
crankpin, and M = mass of the piston and rod A£. Then
the velocity of M at midstroke must = c, at the deadpoints,
zero; its acceleration at midstroke zero; at the deadpoints
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 deadpoint a pressure will
be exerted on the crankpin, equal to mass X ace. = M& ^ r
at those points, independently of the force transmitted due to
steampressure on the pistonliead, 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 freebody method,
considering AB free immediately after passing the deadpoint
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 pistonhead ; and P', the unknown pressure of
crankpin 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 deadpoint we would obtain P' =^ P {
Mc^ ^ r. In highspeed engines with heavy pistons, etc.,
Mg^ ^ r is no small item. [The upper halfrevol., 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 steampressure 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 backpressure.
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 crosssection 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= [_SAG^~M. . . . . (1)
Hence (eq. III.) Mvdv = [S — A — G'jds, and .*. for the
whole stroke
J/j^ vdv=J^ [SA G]ds; i.e.,
= Sj^ & + <Sa/ ^A^deoX 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 cutoff) 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 backpressure 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 masscentres 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 eMlcc: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 righthand 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 nonconcurrent 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 nonconcurrent forces have not yet been investigated,
we mav ima^ne any slioht 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 radiusvector of the point of application of P^. and
«j its leverarm 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 nonconcurrent 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. ConnectingRod by Virtual Velocities. — Let the effective
steampressure P be the means, througli the connectingrod
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 (steampressure), N
(vertical reaction of guide), and N' and :7^(the tangential and
normal components of the crankpin 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' XOTds = 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 steampressure (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 \^^/\pa \^
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 copyingpress 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
screwshaft. See Fig. 75a.
Solution. — Considering free the rigid body consisting of the wheel and
screwshaft, 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 . rddR . 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 coordinate axes X. and 7^ as may be ex
plained thus : Fig. 76. Consider . the
drawingboard CD as fixed, and let the
head of a Tsquare 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 startingpoint,
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; dXf
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 daidt. 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^ir,
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 railpressure r^
against the wheels. So long as the car $^_ t — RH;;
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 springbalancej
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 springbalaTice and MA is its ..^G
line of action (plumbline) 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 footsecond units,*
^ = 32.lY230.083315cos2,50.000003/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 sealevel 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* jri,
* Compare with lower part of p. 83.
CURVILINEAR MOTION OF A MATElilAL POINT.
81
whence P^ the cordtension 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 halfoscillation,
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 fuither 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
dydii
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 pressuie 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^ir; 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 cannonball,
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 masthead 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 waterwheel channel, which the water is to enter
without sudden deviation, or impact, the channelpartition
■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 sailboat,"^ 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 windpressure, having a small foreand
aft component, which in the case of an iceboat may exceed
the small foreandaft resistance of such a craft, and thus v^ will
be still further increased ; i.e., an iceboat 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 platformcar on a straight Ievel 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 Xmotion (horizontal) of the ring is that of the car and
has a constant velocity Cx = 20X5280^3600 = 29.34ft./sec. Its Fmotion
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 10ft. fall = i/2X10h32.2 = 0.788
sec. and the Fvelocity 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 velocityunits (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 Fmotion. 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 = 6h (25^5)0.788 = 9.94 ft./sec, which is the velocity of the
JCmotion 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 cordlength
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 = 7rj/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 hrl=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 10lb. body at M is c', =(r'Hr)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 = 54 [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 airparticles
(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 figuie (§ 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 (§§ 9093). 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 = Vlr 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., fourdimension 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^'^^', ie, = {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\{yhf
= (£»' + 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. gravityaxis
(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
ityaxis 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 semiaxes ; 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 semidiameters of this ellipse ; called
therefore the Ellijpse of Inertia. It follows therefore that the
moments of inertia about all gravityaxes 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 coordinate, 0, we may take
dz the area of such a strip for dF ■=^ hds\
.. Ib= z'dF= I / z'dz
Lo
Secondly, about a gravityaxis 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 reductionformulae 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
coordinates, Ig =^ fz^dF. Here we take dF=^ area of an ele
mentary rectangle = pdq) . dp, while z=^ p sin cp.
■» h — •
1
fi^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)(00)_.
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 gravityaxis, 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, wuiting areas instead of volumes, or, experi
mentally, having cut the given shape out of sheetmetal or
cardboard, we may balance ition a knifeedge. 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
ityaxis (parallel to base) of the
compound figure is also a grav
FiG. 106a. ity axis (parallel to base) of each
of the two 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 Tform.) Similarly for the ring,
Fig. 107.
Fig. 108.
Fig. 107, or space between two concentric circumferences, we
have, about any diameter or ^ (§ 91),
Io = \«r:) (4)
MOMENT OF INERTIA. 97
The rhorribus about a gravityaxis, 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 handbook. )
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 crosssection (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 writingJ!/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. Gravityaxis 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
gravityaxis peipendicular 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) crosssection.
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 gravityaxis (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 gravityaxis 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 gravitjaxis 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 gravityaxis parallel to the hase. — From
eq. (3), § 88, with d = ^h (see Prob. 7, § 26), and the result
just obtained, we have /= J^io[_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, semiaxes 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'
PocketBook may be consulted ; also Trautwine's Civil Engi
neers' PocketBook.
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 fourdimension^
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)[Kr2r)2+rr2] .... (4)
as an approximation (obtained by reduction from the axis at
the extremity of an arm to a parallel gravityaxis, then to the
required axis, then multiplying by four). In most flywheels,
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 footpoundsecond system of units we then
have for the moment of inertia of the huh (120^gf)^[^]^= 6.66 ig';
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 (500fgr)—
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 ig (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 midpoint
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 coordinate
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 coordinate 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 reestablished, 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
antiresultant (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 (Gh^)
= 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 = 42.5=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 refeience, 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
timeintervals, 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. 421(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 2mm= 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, F20080=5/dM:cw2/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
V2mSQ = eMxw'^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 720080=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=efdMyw^fdMx,= eMyw^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 stiaight,
homogeneous, and of constant
(small) crosssection. 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
PYo =/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={6MyGo'Mx; . . . . (IX.)
:SY :=6Mxco'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 (tangential) dfj of
the mass dM, is due to the socalled 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,{lea^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, .0,0Y^ .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 wroughtiron ; 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 + (48H32.2)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(l15.36Xi^?) =48(10.24) = 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 castiron
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.) ) :
= ls =
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 knifeedges. 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 knifeedges is the length, I, of the simple pendulum
vibrating in the computed time (if the knifeedges 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 = 3.259740.008441 cos 2/30.0000003/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 gramtyplane, 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'rr4:7r') — kJ. . . (1)
Hence for a given f, there are two positions for the axis O
parallel to any axis through C, in any gravityplane, 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 inchpoundsecond system
of units, first consulting § 51.)
120. The FlyWheel in Fig. 133 at any instant experiences
a pressure P' against its crankpin from the connectingrod
and a resisting pressure P" from the teeth of a spurwheel 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 steampressure on the piston at any
instant, on the obliquity of the connectingrod, and in high
speed engines on the masses and motions of the piston and con
nectingrod. Let r. = radius of crankpin 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 pistonrod due
to steam pressure on piston.
122 MECHANICS OF ENGHSTEEKING.
as the angular acceleiatioii 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 crankpin
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 deadpoint 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 deadpoint 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^^ TrdaP"J^^ ada. . . (3>
But rda = 6/s = an element of the path of the crankpin, and
also the " virtual velocity" of the force T, and ada = ds", an.
element of the path of a point in the pitchcircle 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 steampressnre, the varying obliqnit}^ of the connectingrod,
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 steampressure on the piston in the cylinder at any in
stant, dx the small distance described by the piston while the
crankpin describes any ds^ and n' and m' the positions of the
piston (or of crosshead, as in Fig. 133) when the crankpin 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 connectingrod 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. FlyWheel.— (See Fig. 133 and
the equations of § 120.) Suppose the engine is noncondensing
and nonexpansive (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 lefthand member ; fPdx = P X 2 f t. = 5500 lbs. X 2 ft.,
or 11,000 footpounds (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
crankpin is describing its upper semicircle) the max. angular
velocity g?^ shall exceed the minimum go^ by only ^Lg?', 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 crankpin are found that
T, the tangential component of the thrust P' produced in the
connectingrod by the steampressure P (which may be resolved
into two components, along the connectingrod 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 crankpin, 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 flywheel 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 flywheel 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 (7rmeasure
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^ityline, 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 righthand 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 pressuie
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 flywheel 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
tonsecond 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 ilywheel depends solely on the tension in the rim,
required its amount. The figure shows the half
rim fiee, 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 coordinate, 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 gravityline 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 aie 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
ityline 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 cordconnection with a fixed p
point in this axis. Fig. 140. Given oo, (p, I,
(p =: \l cos cp), and F the crosssection 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 cordtension 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 gravityaxis.
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^ gravityaxis. 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 gravityaxis 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 gravityaxis ; 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. ParallelRod of a Locomotive. — "When the locomotive
moves uniformly, each dJf of the rod between the two (or
three) drivingwheels 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 onethird 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 pincentre, 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 aj 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
footpounds, inchpounds, kilogrammeters, according to the
system of units employed. The ordinary English unit is the
footpound, 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 watersupply, 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 footpound 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. HorsePower. — 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 horsepower, or chevalvapeur^ is slightly less than the
English, being 75 kilogrammeters per second, or 32,550 ft.lbs.
per minute.) This value for the horsepower 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 horsepower. A
man can work at a rate equal to about J^ of a liorsepowe» ,
with proper intervals for eating and sleeping,
133. Kinetic Energy. Retarded Translation. — In a retarded
translation of a rigid body whose mass = Jf, suppose theie
are no workingforces, 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~"); ,ie., 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 agravityline* 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 flywheel 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 SteamHammer. — Let us apply eq. {e) to
determine the velocity v' attained by a steamhammer at the
lower end of its stroke (the initial velocity being = 0), just
before delivering its blow upon a forging, supposing that
the steampressure i*^ ^^ "^ stages of the downward stroke is
given by an indicator. Fig. 148. Weight of moving mass
is 322 lbs.; .. J!/' =10 (footpoundsecond 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 exhaustpressure (16
lbs. per sq. inch X 20 sq. inches pistonarea) =
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 spaceaverage, to dis
tinguish it from a timeaverage, 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. PileDriving.* — 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 stiiking 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 bridgepier
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 foottonsecond 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
flangepressure 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 flangepressure 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 workingforces G, and
G^ cannot be neglected, and would appear in the term(^i
f G^s' in the righthand 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 stietch
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 flywheel, then, is a reservoir of kinetic energy.
Example. — The 668lb. 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;^ = (66873) 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(668h32.2) 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 + [080,810]; i.e., n643 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 duiing 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 workingforces 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 workingforce 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 paits
forming a machine, or any part of a machine (a similar proof
will apply to thiee 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 gravityaxis 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, =107 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 ; connectingrod ; flywheel, 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 crankpin is at its deadpoints o and n\ i.e., we
deal with the progress of the pieces made while the crankpin
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
cranlvpin. 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 connectingrod 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 lefthand 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 crankpin 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 crankpin 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 crankpin)
instead of Y^. Hence the righthand 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^~ +My~']. . . (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. — (Footpoundsecond
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 crankpin).
Putting (1) = (2) we have, solving for the unknown Vn,
K =
2Tf
iv:.
(3)
For above values,
W = 12,000  400  125.T  Y590.0  141 Y.3
= 2467 footpounds ;
while ^ = 0.5 + 40.3 f 225.0 = 265.8 massunits;
whence F„ = 4/18.56 + 16 = VsU^ = 5.88 ft. per second.
As to whether the crankpin actually reaches the deadpoint
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 deadpoint the (K. E.),^ of M' is not
zero, and must be expressed in terms of F^, ;uid that the
(K. E.)to of the connectingrod J/'''^raust be obtained from § 143.
146. Eegulation of Machines. — As already illustrated in
several exauiples (§ 121), a flywheel 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 steampressure on a piston, or waterimpulse 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 tluough 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' worhingforces = 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 steamengine, or turbine) woi'ks at a constant rate, i.e., de
velops a definite horsepower (H.P.), which is thus found in
the case of reciprocating steamengines (doubieactingj,
II.P. = total mean effective \ l distance in feet ]
steampressure 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 backpressure^
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 {footjpound 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 steampressure 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) steamengine
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
crankpin is 10 in. (so that the length of stroke is 1 = 20 inches). The^
flywheel turns at rate of 330 revs./min. Find the horsepower 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 ieckoning
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 drivingwheel, 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(644f32.2)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 steamengine 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 wellknown
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 mainshaft, A^ Fig. 161, being the drivingpulley
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 = (PPOv, 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 scalepan 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 scalepan 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 actionlines 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 transmissiondynamometers 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
mements 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 {PP')a = Gb, PP' = 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,
= {PP')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 steamengine 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 pulleyrim may be
made constant at any desired value (within certain limits) by
the consequent friction, y is measnied 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 bellcrank must be inter
posed between the arm and the scalepan.)
WORK, ENERGY, AND POWER.
159
Example. — A vertical pulley of a = 2 ft. radius and run by a turbine
waterwheel, 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 pulleyrim
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 pulleyrim 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 footnote 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 "selfadjusting."
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 backpres
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 flywheel 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 " brakepower.^^
Example. — If from indicatorcards 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. BoatRowing. —x^'ig. 166. During the stroke proper,
let /* = mean pressure on one oarhandle ; hence the pressures
on the footrest 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 oarblade 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
oarliandle tlie space s„ = AB, wliile tlie oarblade 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 oarhandle 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 liquidresistance, 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 machinefriction, 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 ?
{Foottonsecond 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 drawbar
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 upgrade, 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
soibed 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 crankshaft makes 60
"WORK, ENERGY, AND POWEE. 163
revolutions per minute, the crankpin describing a circle of 15
incbes radius, required the average* value of the tangential
component of the thrust (or pull) of the connectingrod against
the crankpin. Ans. 26890 lbs.
Example 7. — A solid sphere of castiron 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 crankpin 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
; tangentplane ; 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 frictiongearing
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 brakeshoe and the rim of the carwheel 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 .*.
8fG^^{G,^g):p', (1)
while for the otlier free body in Fig. 61 we have, as before,
G8={G^g):p (2).
From (1) and (2), 8 the cordtension 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 "Pocketbooks" 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 GaltonWestinghouse experiments
with railroad brakes (castiron brakeshoes on steeltired 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 brakeshoe 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
aougli 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 guesswork 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,
iZ^O; i.e.,/iVi+mGiVP = 0; . . . (1)
27 = 0; i.e., i\ri + /ArmG = 0; .... (2)
i'(moms.)c=0; i.e., iVaA^'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^ = t71
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 halflength. 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 midpoint of CK, since B is the midpoint 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
frictionangle, 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
'QJSr,+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 57.6153.6=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 pulleyrim;
is 9 ft. per second ?
The limitvalue 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 screwpropeller 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 BentLever with Friction. Worn Bearing. — Fig.
176. Neglect the weight of the lever, and suppose the plumb
erblock 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 gearwheel) 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 toothpressures upon two spurwheels,
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 tangentpoint of the friction
circle.
Another way of stating this is as follows : So long as the
resultant of I^ and P falls within the " deadangle" 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. BentLever 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 toothpressures on two
FRICTION.
175
spurwheels 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. AxleFriction. — 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 axlefriction may be
written
in which f is a coejficieni of axlefriction 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 axlefriction 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
" deadangle."
166. Experiments with AxleFriction. — Prominent among
recent experiments liave been those
of Professor Thurston (187273),
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 axlefriction 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 wellsustained
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 marineengine 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. FrictionWheels — (Or, rather, antifriction wheels).
In Fig. 179, M and M (and two more behind) are the "fric
tionwheels," with axles Ci and C\ in fixed bearings.
G is the weight of a heavy wheel, Pi is a known vertical
resistance (tootlipressnre), and P an
unknown vertical working force,
whose value is to be determined to
maintain a uniform rotation. The
utility of the frictionwheels 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
frictionwheels. 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 frictionwheels 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
frictionwheels,
Whereas, if C revolved in a fixed bearing,
Work lost per revol. )
without V =f'P'i7tr.
frictionwheels, )
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 frictionwheels 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 frictionwlieels (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 frictionwheels 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 frictionwheels, 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
frictionwheel 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 waterwheel 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 frictionwheels (adopting the footpoundsec
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 frictionwheels, 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 flatended 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) {footlb. 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 axlefriction also occurs its effect must be added.
If the flatended 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 leverarm 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 leverarm = f r, about tlie axis A, and a value
> than for a flatended pivot ; for, on account
of the wedgelike 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 flatended pivot of the same diameter = 2r.
In Schiele's " antifriction" 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.,yr^^ — 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 leverariTi^. 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 tiansfonning,
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
TTmeasure.
Since 8n evidently increases very rapidly as oc becomes
larger, 8^ remaining the same, we have the explanation of the
wellknown fact that a comparatively small tension, 8^, exerted
by a man, is able to prevent the slipping of a rope around a
pilehead, 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^ = lBQSI, ;
= "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 drivingside) 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 axlefriction 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 gravityaxis, 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 workingforce, ^ 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 tensionweight 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 pulleyaxles 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 fiiction"
(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 "frictionrollers,'^
"ballbearings," and "rollerbearings."
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 axlefriction, 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 axlefriction
(total) through the (relative) distance — sin the journal boxes j
whence, the change in kinetic energy being zero,
1 ^ b ^ a
Gs
^Ga^Gsfas = 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 brakepressure 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 wheeltread, 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 twentyton 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 brakeshoe on wheelrim 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 Carjournals in Brass Bearings. — :(Prof. J.
E. Denton, in Vol. xii Transac. Am. See, Mecli. Engs.,
p. 405; also Kent's PocketBook.) 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 axlefriction. 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 leverarm, a, of P is evidently < h
that of Q. If axlefriction 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 axlefriction. When a hemprope 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 axlefriction. (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 (eyebar 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 straightline 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=14400j'
^=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 crosssection 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 dovetail 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=PhF, (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 fibreplane 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 coefficient
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
subscript " 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 EndFaces.
Ellipse of Stiess.— In Fig. 193, p. 198, the parallelopiped RKNS is sub
jected to stress on the two endfaces only. Let us now consider a small
squarecornered element of material subjected to a normal stress p^
(tension) on the two vertical endfaces, 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 righthand
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 pjid lbs. Since the total stress on AC
is the antiresultant 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 semiprincipal axes are OH^
and OH 2, i.e., p^ and pj. This ellipse is called the Ellipse of Stress;
Om being a semidiameter, 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 semidiameter 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 " pinconnections " 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=PrF. 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 subdivision
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. StressStrain Diagrams. — If the relative elongations
or "strains " (s) corresponding to a series of values of the
tensile unitstresses (p) (Ibs./in.^) to which a rod of metal
has been subjected in a testing machine, are plotted as
abscissae, and the unitstresses 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
"unitelongation " is one hundred times as great as that
used in the righthand 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 ry^ — ^
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 testpiece of iron or steel, showing that on reapplying
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 + 00; 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=dAhdx [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 crosssection
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 dXidx 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 coefficient of expansion) a length =1 would be con
tracted an amount X=fjl(tt') 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)
vF _
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 archring
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 yieldpoint 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 yieldpoint, then relieved and heated for a few minutes
to 100° Cent., then stretched just to its new yieldpoint,
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
Ciovernment, 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 lapjoint) 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 coverplates, 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 steamtight, 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
halfdiagonal 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 tierod, 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 lapjoint, 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 buttjoint ; (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 sidecrushing 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»
testingmachine 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 cementmortar, 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 rightangled, 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')t1; 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^rUs; 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 helixangles.
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 bfthand 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 righthand 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 Ip=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 righthand 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 crosssection
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. IfonCircular Shafts. — If the crosssection 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 crosssection. The
warping of the crosssection in such a case is easily veri '
fied by the student by twisting a bar of indiarubber 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 cogwheel,
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
unittime, the work transmitted {transmitted ; not expend^
TOESION. 239
ed in twisting the shaft whose angle of torsion remains
constant, corresponding to Fa) per unittime, 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 crosssection is
12' = 27rroi5;' = 27rlOXiX 5000 = 78,540 lbs., whHe the velocity of the
midthickness 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 testpiece is of a standard shape and size, its
central cylinder being subjected to torsion. A jaw, carry
ing a handle (or gearwheel turned by a worm) and a drum
on which paper is wrapped, takes a firm hold of one end
of the testpiece, 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
testpiece, which is thus subjected to an increasing tor
sional moment. The axis of the testpiece 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 testpiece) 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 " elasticityline," 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 testpiece 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
horsepower. Then H. V.^Pu^na^N, which for the foot»
poundminute system of units gives
H. P.=2,000x50x27rx333,000=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. ;7measure), 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 onethird 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 forceplane ; the forceplane contains the
axis of the piece and cuts each crosssection symmetri
cally ;
(2.) The crosssections 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
forceplane, 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 crosssections, 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 crosssection 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 crosssection
at m / this crosssection, 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 crosssection 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 forceplane (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 crosssection 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
CrossSection. — Fig. 222. Let e— the distance of the outer
most elem.ent of the crosssection 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 crosssection 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 crosssection 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, =3X204X5 = 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^yrdx^ 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 springingback), 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 " stresscouple." 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^yi
'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?antiderivative 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/ic?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"i2'^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 cpantiderivative 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 crosssection 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 righthand pier (or left).
From the free body in Fig. 225 we have, using the form Elip for
the moment of the stresscouple 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 cinve 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 stresscouple 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 = PP716£'/, 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 stresscouple)
in place of the present lefthand 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 halflength; 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 + 0Q (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 =PZhZi, 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 coordinates 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{yx)—p,{k—x)', , . (1)
(X OC
whencO;. by taking the x antiderivatives 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 antiderivatives, 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 footnote 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
coordinate axes.)
From (4) we have in the ordinary way (ajantideriv.)
EI"^ =Pil,x J^)+C" . . (5X
ax 2
To determine C", consider that the curves CB and OG
have the same slope (dyrdx) at G where x=l; hence put
x~l in the righthand 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
dyhdx of that curve, as expressed in equation (2), =0.
Also for O.Q write its value from (7), having put Pi=P?rZij
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?TZi, 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 ccantiderivative of both sides of (1),
^^Tx =yM}{l^'^—}i^')+{G=0) (2)
as the equation of slope. (The constant is =0 since at
both dyidx 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'^yrdx^
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 stresscouple 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 inchlbs. 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?
tpdF ^
240, The Shear is the First xDerivative 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 inchpounds, or footpounds,
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 ifiZixiM^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= jilx^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 dovetail 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=PoPi
F " H,J^P^P^P^
H " B, J=PoP\P2Ps
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
?^PolP.{kk) (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 crosssection, 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
inchtonsecond 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 inchtons, 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 crosssection 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 socalled " Ibeam " 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 Ibeams, Boxgirders, Etc. — In
common with other large companies, the Cambria Steel
* This function, /re, of the plane figure formed by the crosssection
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 "sectionmodulus." 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 Ibeams, deckbeams (or " bulb beams "), rails, angles,
Tbars, channels, Zbars, etc., according to the form of their
sections. See Fig. 239 for some of these forms. The company
d
^
T
CHANNEL. DECKBEAM. RAIL.
Fie. 239.
publishes a pocketbook 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 footnote, p. 274.)
For beams of still greater stiffness and strength com
binations of plates, channels, angles, etc., are riveted to
gether, forming " builtbeams,"' 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
pocketbook 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 rivethole).
For example, a boxgirder 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 rivetholes  in, wide and If in.
long. That is, 510; i = l; d=8; f = f; and d' = l in.
Also from the handbook we find that for the channel in
question C==320 in,* (i.e., biquad. in.). Hence, eq. (1),
7^=640 + 2X10X1X644x1. 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 boxgirders in buildings, we have
" R'l 15000X1625 . . .,
— = ^^ = 2, 867, 500 inchlbs.
e 8.5 ' '
That is, the max, safe ^'moment of resistances^ of the box
girder is M^ = 2,867,00O inchlbs. = 1433.7 inchtons; this
quantity having to do with normal stresses in the section. The
greatest " hendingmomenf^ 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
webthickness for /beams and " builtgirders." 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, crosssections 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 Ibeams, 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, straightgrained, 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 handbook 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 Ibeams 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 archfilling of a fireproof 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=eUTp, 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 forceplane (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 castiron 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;^ ^(256150)r[144x4] .. P=3,470 lbs.
The weight of the beam itself is (r= Vy, (§ 7), i.e.,
Q^^^r,'^ri)lr= f(1612i^)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 Ibeam, 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 crosssection. 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 "sectionmodulus," 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
7beam in whose outer fibres no
n
greater normal stress than 8,000
*£ lbs. per sq. inch is allowable. If
Pio. 245. a number of Ibeams 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
crosssections are Similar Figures, the moment of inertia of
the 15inch beam being 800 biquad. inches ?
The 12inch 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 " freebody " 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 coordinates 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. 234237 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 crosssection
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(rin)
J{r{'—r.i')
3 ;r(ri*— r2*)(ri— rg)
Applying 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[6442.8]
3x22[256150](43.5)
= 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,V5,V ]
' %ihji,'hjh%hh,) 2* [6A^M/][6;&,]
The same equation holds good for Fig. 257 (Ibeam with
square corners) but then &2 denotes the sum of the widths
of the hollow spaces.
256. Shearing Stress in the Web of an IBeam. — It is usual to
consider that, with Ibeams (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 (Jrhoh^.
Similarly, the shearing stress per unit area at n, the
upper edge of the web, is also nearly equal to e7f 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 Ibeam 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 Ibeam 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 15incli Ibeam 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 Ibeam 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 Ibeam* is formed by riveting together, in the
manner shown in the figure, a " vertical stem plate " or
web, four "anglebars," and two "flangeplates," each of
Fig. 259.
Fig. 260.
these seven pieces being continuous through the whole
length of the beam. Fig 260 shows a boxgirder. 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 Ibeam is usually designated a " plategirder. '\ 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 handbook, 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 (boxbeam) 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 Ibeam 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 20inch 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
halfspan, (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 plategirder, 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,^ inchlbs. (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 handbook of the Cambria Steel Co. gives tables
of safe loads of beam boxgirders and plategirders 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 handbook. The
value of 15,000 Ibs./sq. in. for R' has been used in computing
these tables.)
Example. — (Units, inch and pound) . A plategirder 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 flangeplates 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 handbook 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* i6.,
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(xa'y]—P^hPih\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 dMihdx=0, i. e.
m
j{l2x+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 equidistant 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 onethird to onehalf
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 inchlbs. 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 boxgirder.
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 dovetail 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 12 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—BiX — 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
dMr 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
PxP {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 MTdx, 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 Ibeam, 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 ~ \ %lil,^] 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 onethird 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 inchpoundsecond
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 halfplate, 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 stresscouple, 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., cylinderhead 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 crosssections (similarly placed),
is seen to be proportional to fheii volumes; while if further
more their crosssections 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
connectingrod against the
crankpin 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 crosssection 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, =j9sin2^42^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 *
gmax =;^[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 crankpin 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 cogpressure) is
B expended in overcoming the
resistance (another vertical
cogpressure) 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 IBeam. 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 Ibeam 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 xantiderivative 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
Ixt^+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?xPA^^—^'\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,pfy,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'^^ lFxlx'+^]F,[lx ^]+[Coiist=0] (2)
[Const. =0 for at both dyidx 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 . ty^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
"stresscouple " 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 dMidx—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 xderivative 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 — — — ViT^ 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 stresscouple) 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 " stresscouples " 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 lx) + 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:antiderivative 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 tangentline 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)]
y2PxM, + 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 inchpounds
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 halfspans, 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 stresscouple 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+i4WxMo
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 NONPEISMATIC BEAMS.
333
of the stresscouple, as successive sections along the beam
were examined.
But for a nonprismatic 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 nonprismatic 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_ op & {x\cy — 2a;(a;tc)
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 NONPRISMATIC 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 midspan)
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 midspan section) ;
^^ ^R' (l^o)' ^%Fc,ioxy2l> %c (13)
(where r^ = radius of extremity.)
IS^ONPMSMATIC 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 fibrestress 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. IfONPIlISMATIC 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, halfway 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 wedgeshaped ; 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'^ = xiuv^; 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](lxx') (5)
This relation between the halfheight ^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 "ConcreteSteel" 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 concretesteel 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 concretesteel beam
flange
teeb, or stem
of Tsection, 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. ConcreteSteel Beam of Rectangular Section. Flexure
Stresses. — As in the common theory of flexure of homogeneous
beams, it will be assumed that crosssections 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 crosssection, 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 lefthand 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 righthand, an endview 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 " cinderconcrete ; " 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 onethird
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 unitstress 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 unitstress (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 lefthand end of
the steel; the tension on the righthand end of these rods
[being something greater (say) and expressed by F' (^p' \ dp'^
in which dp is the difference between the unittensions at the
two ends of the steel " reinforcement "] ; 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 lefthand extremity of the same,
while p \ dp expresses the unit compressive stress in the same
outer fiber at the righthand 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(aie) •
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 ConcreteSteel 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, . .
^iOQT(V .64X15 +ll)=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. ConcreteSteel Beam of TForm 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 " reinforcement." Fig.
297 shows a side view and also an endview of a portion of the
beam in Fig. 291 extending from the lefthand 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 unitstress 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 unitstress 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 (^^ stresscouple '^ of the section; with an arm = 6r6r',
= a"), and we have
pi2ed)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 " stresscouple ") 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
Fn.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 stresscouple, 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 + ae)~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 TForm ConcreteSteel 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 Tform
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 + ae) (^^'
290. Deflections of ConcreteSteel Beams. The deflection of a
loaded prismatic concretesteel beam resting on two supports at
its extremities, may be obtained for the cases dealt with in
§§ 233236 inclusive, in connection with homogeneous beams ;
provided the product EI occurring in the expressions for these
deflections be replaced by ■ [ a— ^), for concretesteel beams
of rectangular section; and by E' F' {a — e) (c + a — e), for
those of Tform section.
291. Practical Formulae and Diagrams for use with ConcreteSteel 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.
it.. 60
s =
/«
»
:;r:?:
 T ^^
~ «>«./'„ . „ \ '
 _l _
:.^y.
in which ^c
::z5:::t:
,?i i
I'.Tl'i
 \^
■ J.
y
y
..2f_
■^• 50
—
— 1
/
— 1 — ^.
1 ^1

 L V
^ y J
J& 1
.._,.(«
y
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.,2^
y
■'y r
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\ d ah 1
xc
/
: _Ly
1 "[
\ J
^5 ^ ^ n Q
'"i""
'/
/

y
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1
' y
 k. An
y
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CO
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in
<•
o / O "^
/I
y
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1
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i! ' .. oc
'
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' Z y
T ■ 35
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y ?^
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y
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 4 
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. .'. . . L.
1 1
1
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 , ,
UIMbKMm li.
\
sj
1
1
3r t2^
^
V
t
\
s
V
1
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
■4^
, 1 s
' m •
\
\
\
1 »
1
. 1
X.i.i.
T y "p uj
+ ..,x.._l 1
\
\
s.
^s
Pv
. ' ;,i.:4..
I.. ',.J m
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s
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\ ^^ i ••
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t
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r'ii
IS
\
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: 'K,\ ^
\
Nj
\
\
1
^ 1
i.
^\ 'K i
i. :i..J..i.
K
N,
\
1
t ^ T ^

!
\
\
1
1
1 s
1 ^^
'v 1 I **;;..
1
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kH 's
1
1
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1
1
1
_ J_ 
' ..jitJ... D
'



1
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 = nh[2r{r+n)] (25)
3 J'42 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
crosssection 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 inchlbs. 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 inchlbs., 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 inchlbs. (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 " stresscouple^^ due to the moment of P (i.e., Pu) about ^, the
center of gravity of the crosssection m'm.
More in detail, consider in Fig. 300 a portion AB of the
prism, being the part lying above a crosssection 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
crosssection.
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 crosssection, 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 crosssection, 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 sideview, and that they are all compressive, they are
equivalent to a rectangle, m't'tm, of stress of uniform compressive
unitstress 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 momentsum 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 crosssection
* 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 stresscouple,
' '" C, whose moment is ^ in.
FiG. 301. e
lbs., in which pj = Pm — P^ ^^^ I is the " moment of inertia "
of the crosssection (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 crosssection 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 straightgrained pine
wood, 12 inches in length and of square
crosssection (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 righthand 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 unitstress 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 totalthrust (p,F^ of 4500 lbs.;
a unitthrust (w^) of 4500 lbs. per sq. in. ; and a stresscouple
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. CraneHooks. First (Imperfect) Theory. Fig. 303 shows
a common cranehook 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 stresscouple 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. CraneHooks. 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 crosssection 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 crosssection 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 unitstress 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 crosssection 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 unitstresses 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 fiberlengths 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 unitstresses, 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 unitstress p^ =^ P i F and
lay it off, == is, upward from hi, and draw the horizontal 6s, we
thereby rearrange the stresses into a uniformly distributed pull
(or " negative thrust ") p^F ^hs., represented by the rectangle
hsih, and a stresscouple 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. CraneHooks. WinklerBach 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 WinklerBach theory it results that the unitstress 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 crosssection 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
unitstress, 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 bridgetrusses and roofs, the " bents " of a
trestle work, and the pistonrods and connectingrods of
steamengines, are the principal practical examples of long
columns. That they should be weaker than short blocks
of the same material and crosssection 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 ballandsocketjoints ; 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 roundended as regards flex
ure in the plane 1 to the pins, but as squareended 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 " forceplane " spoken of in previous chap
ters), or plane of the axis of column when bent.
303. Euler's Formula. — Taking the case of a roundended
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 stresscouple 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
ay.
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{ary)
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!=±yp (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 Tsquai eblade 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 flatended 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 roundended column. Also, the tangents
to the elastic curves at and G being pre
served vertical by the f rictionless guideblocks
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 flatended column. In
other words, the breaking load of a round
ended column of length =^Z, is the same as
that of a flatended column of length —I.
Hence for the I oi eq. (8) write %l and we
have as the breaking load of a column with
flatends and of length =1.
}il
f/MWM
Fig. 315.
TLEXURE. LONG COLUMNS. 365
r.^4.m^ .... (9)
Similar reasoning, applied to tlie " pinandsquare "
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 ^'pinandsquare "
column, of length = I, is
P^=l ^/^ . . . (10)
Comparing eqs. (8), (9), and (10), and calling the value of
Pi (flatends) unity, we derive the following statement :
The breaking loads of a given column are as the numbers
1
flatends
9/16
pinandsquare
y^ j according to the
roundends \ 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 leverarm of the load about the wallsection, 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 unitstress 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 unitstress (compressive) on the
crosssection and l^k is a ratio measuring the slenderness of
the column. (Of course, when the column actually gives way
by buckhng, the unitstress 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 lik ranges from 20 to 480.
As an example consider a 3"x3"Xi" anglebar (or "angle")
of wrought iron, with Z = 15 ft., to be
used as a column. Fig. 315a shows
the crosssection 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 ballbearing 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
nessratio;' 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 unitstress 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 PoHi^ and Zh 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 coefficients requiring experimental de
termination. It may be developed as follows :
Since in the flatended 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 stresscouple, 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 stresscouple, 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
stresscouple 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 crosssection 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 roundended
column we substitute 21 for I; for a column with one end round
and the other '^fiat " or ''fixed " (i.e., for a " pinandsquare "
column), ^l for I; and obtain
Breaking load for a round 1 FC ^
ended column \^^^TTW(hW' • • • (^)
Breaking load for a ''pin 1 FC
andsquare " 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 endconditions 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 unitstress), 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;) = "slendernessratio " = 96j0.6 = 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 castiron 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 biquadratic 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 (connectingrod 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 ''roundends" if the
axis of the column were to bend in a plane T to the axes of
the pins (as in Fig. 312), but ''flatends" [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 roundended 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 flatended 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 gravityaxis. By giving the least value oi
h^ it is implied that the plane of flexure is not determined
by the endconditions 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 pinJoint the column may need to
be treated as having a flatend 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
roundended 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
AngleIron of Equal j^ig. 317^^) A;2=^62
ribs
F:
62^*
A^ngleIion of unequal , ^g g^g^^^^ ^= 12(F+3?)
Cross of equal arms. Fig. 318 (6). ^=4^'
IBeam 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, Zbars, 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 zigzag, 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 crosssection 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 Ibeams
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 Ibeams by lattice
work, Fig. 321, (a). (While the latticework 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 Ibeams.
Let F' = the sectional area of one of the Ibeams, Fx
Tsee Fig. 321(a) its moment of inertia about its webaxis,
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 handbook of the iron company, for each size
of rolled beam).
Then the
total 7x = 2rx ; and total I^ = 2ri'v f WYl
(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 Ibeam, 105 lbs. per yard, of the N. J. Steel and Iron
Co., in whose handbook 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± (185.69.43) Vera = 8.21 inchea.
V 1 0.44.
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 Ibeams united
by plates instead of lattice. Let the thickness of the plate
■= t, Fig. 321, (&). Neglect the rivetholes. The distance
a is known from the handbook. 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 castiron
flatended 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 [828]
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 MerrimanRitterFormula 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 unitstress 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, pinandsquare,
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 "StraightLine 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 lik 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 crosssection 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 shortleaf 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 coordinate x and a horizontal coordinate 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)[lcos (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 crosssection 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 (1.1901)
= 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 + ay) + {i)w{P4x^) .... (5)
Since (d^yrdx^) is a variable, let us denote i—EIiP)(d^yTdx^) by u,
as an auxiliary variable; and eq. (5) wUl now read
yu=c+a+[{i)w{P4: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 PiEI 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 = dMf (VC + Zim— 6V), which integrates into
a;=A.smM , — +C^ (8)
where C is a constant. Transfo rmation of (8) gives
(Vh' + ACb') sin [b(xC')]+h = 2b^u (9)
Eliminating u by aid of eqs. (6) and (9) we have
2b'y==\/h' + 4Cb'sin [b(xC')] + h + 2b\c+a) + (i)b^h{P4x^) (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 \ricos (6a: "1 ,, , ,^„^
^+26^jL^os(i60J"*^'^' ^^'^
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 crosssection 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,
Vl^^f (15)
M„=P
FLEXURE. LONG COLUMNS.
383
Since 6 and h denote VPiEI 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 WebPlates in Built Girders. — In §257 men
tion was made of the fact that very high web plates in
built beams, such as /beams and boxgirders, 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 Ibeam,
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
'ir/ 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 tangentline 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 semicircle, 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 coefficients, 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 lefthand members of these equations
= to that of the righthand 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 coordinates 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 archring
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
lefthand 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 antiresultant), 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. NonConcurrent Forces in a Plane. — Given a system of
nonconcurrent 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 (halfparallelograms) 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 forcedia
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 nonconcurrent 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 forcediagram 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 spacediagram, 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 Fj, 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;ticonciirrent
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 archribs, 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 spacediagram (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 spacediagram 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 coincide
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 antiresultant 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 antiresultant 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 antiresultant 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 " loadline " AD
GRAPHICAL STATICS.
405
(see (h.) Fig. 854) as a beginning for a forcediagram ; 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 forcediagram. This
last line will cut ths " loadline " 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 " closingline," 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 framework, we proceed
tc find the pulls and thrusts in the individual pieces, on
the following plan. The truss being pinconnected, 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 trusspieces. In this way we can speak of
the force Wi as the force BC, of W2 as the force CE, 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 spacedia~
gram, small letters will be used at the vertices of the closed
forcepolygons (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 righthanded
(or clockwise) 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 stressdiagram may be formed by super
posing the polygons already found in such a way as to
make equal sides coincide ; 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 framework (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
bracedarch (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 archtruss. 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 crosssection of the portion O'A of
the beam, are due to P^ alone ; those of any crosssection
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 " antistressresuU
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 HonConcurrent Forces m a plane, and the special equi
librium polygon has been draivn, then each ray of tlie force
diagram is the antistressresultant 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
" antistressresultant " 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 archrib,
or the blockwork of an archring, 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 suspensionbridge 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 "loadline," 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 loadline 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 abutmentvertical is struck in some
point B. AB is the " abutmentline " 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 archrib or stonearch, 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 loadline.
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 endcurves 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
loadline 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 abutmentline, and draw a
parallel to it through 0, thus fixing the point n' on the
416
MECHANICS or ENGINEERIKG.
Pi
P. t
l'
Fig.
loadline. This point %' , as above determined, is indepen'
dent of the location of the pole, 0, (proved in § 329) and
divides the loadline 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 loadline, thus dividing the loadline
into two portions which are V^ (upper portion) and F^.
However, if A and B be connected by a tierod, 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, loadverticals, and
abutment verticals, are inversely proportional to their H^s {or
"pole distances "). We here regard an equil. polygon and
its abutmentline as a closed figure. Thus, in Fig. 369,
we have two forcediagrams (with a common loadline, 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 forcediagram 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 abutmentline
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 poledistance].
§ 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 loadverticals, 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 loadline, and
with any convenient pole, Oj, construct a forcediagram,
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 loadline (on the left of the loadline 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 loadline, 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 loadline 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" storcearcli (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 threecentred 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.
" springingline."
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 voussoirarches 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 onethird the joint
length from A. Similarly in Fig. 377, it acts onethird
the jointlength 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 sandstone, 2inch cube,     8955
4 " "     11720
Limestone, Sebastopol, 2inch cube {chalk)^   1075
Limestone from Caen, France,   . . 3650
Limestone from Kingston, ]^. Y.,   .  13900
Marble, Vermont, 2inch cube,   8000 to 13000
Granite, New Hampshire, 2inch 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 LinearArch, 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 keystone 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 archring (the archring 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 archstones 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 archring, 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 archring as an elastic curved beam
inserted in firm piers (i.e., the blocks at the springingline
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 archring 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 archring 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 archring 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 springingline, 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 archring.
II. If (with firm piers, as before) no equilibrium poly
rgon can be drawn Avithin the middle third, and only one
within the archring 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 archring, 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. Cliaracter of Load. — In most large stone arch bridges
the load (permanent load) does not consist exclusively of
masonry up to tlie roadway 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 springingline 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 springingline, 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 LoadContour. — 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 archring as a standard, we must find for
each point "p of the extrados an imaginary height z of the
archring 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 LoadContour," i.e.,
the imaginary line, AM, the area between which and the
extrados of the archring 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 LoadContour. — Fig, 388. Given
an archring 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 LoadContour 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 rightline 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
archring 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 LoadContour.
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 semiarches, plus the load im
* If the earthfilling 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 abutmentpier.
Fig. 39;^.
GRAPHICAL. TREATMENT OF ARCH.
358. — Having found the " reduced loadcontour," 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 archring 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 semiarch 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 archring). 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
loadline, 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 halfarch. Draw joints in the archring 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 archring,
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 coinciding 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.
abutmentjoint at the end of the span vihich is the more
heavily loaded ; in the other abutmentjoint take B at tht
upper limit of the middle third ; and take p in the middle
of the crownjoint. 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 archring and loading is not
to be considered satisfactory until it is ascertained that for
the dead load and any possible combination of liveload
'(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. Archring under Uon vertical Forces. — An example of
this occurs when a vertical archring is to support the pressure
of a liquid on its extrados. Since waterpressures are always
at right angles to the surface pressed on, these pressures on the
extradosal surface of the archring form a system of nonparal~
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 archring, these resultants form a series of nonvertical
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 crownjoint 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 archring 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 circularGentreline of the
wrchring is the " true linear archP
4:38 MECHAIiflCS OF ENGLNiiJEJiLNG.
CHAPTER XI.
ARCHRIBS.
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 arclirib (or elastic
arch, as distinguished from a blockwork 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 crosssection
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 endtangent
line to turn in the vertical plane of the rib when a load is
applied); or may 'hef,xed, i.e., so builtin, or bolted rigid
ly to the pier, that the en^tangentline 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
suspensioncable.
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 antistress 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 crosssection of the rib as at m. Let the
ARCH RIBS.
Ml
FiG. 404.
of the forcediagram on the right be 200 lbs. to the
inch, say, and that of the spacediagram (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 antistressresultant is a force R2 acting
in the line 6c, and of an amount given in the forcediagram.
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 stresscouple, 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 stresscouple 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 archrib, of given mode of support,
p.nd under a given loading, then in any crosssection 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
/ 111 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 poledistance) 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 stresscouple 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 crosssections 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 archribs, 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 ribaxis) was originally =ds
(nearly) and the two consecutive tangentlines, 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 pi 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 tangentlines 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 stresscouple
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 archrib 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 endtangents 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 tangentline 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 tangentline 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 stresscouple) 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 coordinates 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 righttriangles,
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 crosssection, 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 ^
tangentlines and ^ [ ~ Jo ^ ' o • • W
The XProjection of O's Displacement "]
Relatively to B and B's tangent I A^Myds /tt ^
line ; {the origin being at 0) I — / — ^fjr • • • (JJJ
and the axes X and F 1 to [ ^o i^i
each other) •
450
MECHAKICS OF EKGINEEKING.
The YPrqjection of O's Displacement,  _
etc., as above.
. (m.)
Hviie X anv.. y are the coordinates of points in the rib
curve, ds an element of that curve, M the moment of the
stresscouple 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 ribaxis 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.
AECHHIBS. 451
rium polygons may be drawn for tlie given loads and
abntmentverticals, 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 loadline 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 loadline, 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 archribs, 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 loadline
taking care to
"^ lay off % and
.. % upuard in
^« their turn.
Take any con
FiG. 416. venient jola
; draw the rays ... 1, ... 2, etc.; then, having pre
viously drawn vertical lines whose horizontal distances
from an extreme lefthand 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.
AECHKIBS.
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 beingapplied 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.pprox.=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.
ARCHEIBS.
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 vim.,), 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^.
AECHEIBS.
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 equidistant from said vertical line.
378, Classification of ArchEibs, 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 archribs 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
gentlines 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 windpress
ure on which is to be considered as well as the weights of
the materials composing the roofcovering, 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.
ARCHEIBS. 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 interr
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. ArchRib 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 ribcurve 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 coordinates 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
ribcurve 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 ribcurve into an even number of equal As's will be ob
served in all problems henceforth.)
379a. Detail of the Construction. Given the archrib 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 loadline 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
mentline " 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 midpoints 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)
ABCnKIBS.
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 coordinate 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^^Zlf}.'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 referenceline,
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
AECHEIBS.
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 / onehalf 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 loadline 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 loadline 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. :
AUCHEIBS.
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 distance 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 loadline. 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 crosssection 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 ribshortening, will be con
sidered in a subsequent paragraph. They have no exist
ence for an archrib 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 ribcurve, 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
AECnKIBS. 471
aggerated ribcurve, 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 stresselements
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 stresscouples 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 tangentline
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 crosssection) may be computed. Hence the
greatest compression per sq. inch (Pi+p^) may be found in
each section. A separate stressdiagram 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?^ Ishaped,
or boxshaped. 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 ArchRib 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 {tQ/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{tQrj=—Jfds;
hence for graphics, and
equal Js's, we have
Ell{tt,)y^=HJs I^y' . . . . (1)
From eq. (1) we determine H, having divided the ribcurve
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 stresscouple = Hy,
AKCHKIBS.
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 ArchRibs 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., WOVBW, i.e., \tt:)r^,=^J{y^yds . (2)
or, graphically, with equal As's
Fig. 435.
EIl{t
Qr
HAs
IfdSly
(3)
Also, since there has been no change in the angle betweeij
endtangents, we must have, from § 374,
^_rMds=0; i.e., — / 2c?s=0;i.e., nyd)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 stresscouple
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 RibShortening — 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.
AECHRIBS. 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 stresscouple 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 archrib 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 ribshortening 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 crosssection, 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 ArchTruss, or braced arch. An openwork
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
archrib.* 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 archrib, 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,
ARCHEIBS. 479
pressionB (or tensions) in which are thus found : For the
point m, of ribaxis, 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 ribaxis 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 tangentline at 0„ makes with its former fhorizon
tal) position OX. The beam is homogeneous and of constant
crosssection ; 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 coordinates x and y of any point, m, of
the axis of the beam are as shown in the figure. Taking now
polar coordinates, 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 crosssection 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. SemiCircular ArchRib. Hinged at the Two Piers or Sup
ports, and Continuous Between. Fig. 440. The supports are at the
same level. The archrib, 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 tangentline 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 smdcosdddH 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 + li + 0J//^00(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 tangentline 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 pipewall, i, and length I, (so that
the moment of inertia of the crosssection 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 pipewall, 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 stresscouple ; 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 lefthand 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 stresscouple,
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 stresscouple 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 stresscouple at B). At B,
therefore, we find only a stresscouple, 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 tangentlines 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
couplemoment 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 fjee 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 unitstress 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 tangentline.
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 unitstress 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 pipelength.
Although the load is called "concentrated" as regards the endview 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 EndTangents 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 crosssections, 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 stresscouple 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 7rmeasure ;
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 endtangents 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 crosssection 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 stresscouple 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 lefthand 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 tangentline 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 stresscouple 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 crosssections, 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 tangentlines 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 tangentline OB, the length 00,^ is evi
dently O's displacement relatively to 5's tangentline, 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 stresscouple in the crosssection 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 stresscouple 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 endsupports. 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 endsupports at same level, the load
being uniformly distributed over the whole span, I. Fig. 4443. That is, W = wl,
1 rB 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 tangentline 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 pBrWX'^ WX3  1 r WX^ WXn 5
°^« = MJo I ^^" 2^^J= ^[6 Xjo =
5
584
EI
490 MECHANICS OF ENGINEEEIKG.
Example IV. Prismatic beam on endsupports, 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 stresscouple 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 tangentline at the middle point B is horizontal, and
parallel to OnC. Otherwise OnO would not he equal to the deflection of B.)
396. Nonprismatic 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 (MomentAreas 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 stresscouple (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 inchlbs., this product
might be called so many "sq.
inchlbs." of momentarea (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 " momentarea
between and B
]'
EI
(la]
or, for brevity, ((> ={A^^ ) ^ EI . . . . . . . . . . (1)
This " momentarea," 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 inchlbs.
Again, if the elementary " momentarea " 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 momentarea 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 coordinate 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 momentarea 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 tangentline 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
tangentlines, one drawn at 0, the other at B, lies in the same
vertical as the center of gravity of the intervening momentarea
(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
momentarea.)
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 momentdiagram for whole length is a
triangle (§ 249) whose base is I inches and whose
altitude is PI inchlbs.
• 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 EndSupports, Load Uniformly dis
tributed over the whole span or length, I', W = wl. From p. 268 we know that
the momentdiagram (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 lefthand half of this
momentfigure, measured from the lefthand extremity On, is ^ Z —  of 1 1; i.e.,
i = f of 1 1.
The area of this semipardbolicsegment is twothirds that of the circum
scribing rectangle. From symmetry, the tangentline 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 momentdiagram is easily shown to consist of a triangle at
each end with a central rectangle of altitude = Pa (inchlbs.). 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
QQ _ (^)O
1 r Pa
2a r. 3 a
la.Pa^
n Pa^
~(SEr'
Example IV. Prismatic Beam on End Supports. Single Eccentric Load, P.
Fig. 445^. Here a tangent drawn to the elastic curve at the loadpoint 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 momentdiagram 0"NC"
having been drawn, in which
'B"N = {Pa,a, ^ I) inchlbs.
(§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 loadpoint. 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 momentdiagram 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 "normalmoment
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 stresscouple
[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 stresscouple 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:cP,(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 momentdiagram
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 momentdiagram 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 momentdiagram ''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 righthand
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 crosssection ; M^, M^, M^, the
moments of stresscouples (" bendingmoments ") 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 lefthand 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 Noncentral Concentrated Load, P.
case as a lefthand panel,
Pbc (I + c)
A,x^= [
Fig. 4462.
Fig. 4463. For this
while, as a righthand
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 righthand 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 16x4x11x19+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 — Mt
—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 rearrangement has been observed in the momentdiagram
0"B" G" of Fig. 447j, where lineshaded 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
sheardiagram 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 righthand 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 ThreeMoment 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 spanlengths 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 (lineshading), or convex up (dotshading), 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 lefthand span. (Case of a drawbridge with
the leftend " 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 momentdiagram 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 stresscouple
at righthand 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 " Builtin" Ends. Fig. 449. As
a case for illustration take the prismatic beam in Fig. 449,
" huiltin " 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 endsupports and a
single noncentral load, P ; (with weight of beam neglected).
Taking 0, B, C, as the three points (i.e., OB is the lefthand,
and BO the righthand, 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=" 0000145 (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=43.35 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 unitstress 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 rE;
so that if dX" is the increase in length of the vertical edges we
havedX" idx (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 stressstrain 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 unitstress
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
steamboiler 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 = pik{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 \ Aj,'^ 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___iJi— 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
/\ ^hy^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)l2ql . 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, = pk.q, (6); Ee2 = qk.p, (7); Ee3 = 0k(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= ~ 0000255, 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.
Castiron 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' = tt£,r't
s=(r'
ds2
dr
■r){l~e2)
dr'
r'tr.
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 crosssection 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 (Xcomponents) = 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 = qk{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 lefthand 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 PRESSUREHYDROSTATICS 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 metrekilogramsecond 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 unitpressure 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
FpF:p, Fhy = 0. . . (2) r ! ^^^
>JTiT— 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
crosssection 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 forcecomponents 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 pressureintensi
ties at the two bases,
p,dFp,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 pressureintensity 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 waterpressure 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 airengines, and those of aircompres
sors ; and eqs. (2) and (3) still hold, practically, even with high
pistonspeeds. For example, in the space ABy
Fig. 455, between the piston and cylinderhead
of a steamengine (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 cylinderhead
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. Nonplanar Pistons. — From the foregoing it follows that
the sum of the components  to the pistonrod, 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
pistonrod is always = Ttr'^ = area of section of cylinderbore.
Fig. 457.
If the surface of A is symmetrical about the axis of the cylin
der the other components (i.e., those ~] to the pistonrod) 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 cylinderwall, 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 pistonrod.
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 leverarms 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
:..;,(.:,. wi;: .,•' 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) (inchlb.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 Waterpipes. — If h = vertical distance of a point
^ of a waterpipe 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 pressure 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 " waterhammer," 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.
HVj
P.=I>Ehy.
w
* See pp. 203214 of the author's "Hydraulic Motors."
BAROMETERS. 531
Example. — Let A = 10 feet (with inchlb.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
waterbarometer 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 sealevel), 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 gastensions 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 steamgauge 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 steamgauge 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 Utube 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+(hix)y;
while between d and c we have, for the mercury,
By elimination there easily results
(2)
(3)
po
^^ = /if^l
^0*
•■>f«.v.?VV..:V.>
■Po'=h{rmr); or ^^^=h[^jij. . (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'', iG, 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'^Hrn); or 2»^'=a(iQ),
(5)
■c— ^
:is
■E 4~yi
^■•■ r I' ■
Fig. 468.
425. Safetyvalves. — 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 fnlcrumreaction at B, the weights
G and G^ , and the two fluidpressures 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{PI>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?vsssection 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 endpres
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 (semicircular)
of any length I of the pipe, be
tween two crosssections. 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 halfring 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 castiron.
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
(castiron), ^.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. 204211 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
boilerflues, 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 wrouglitiron 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, boilerflues 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 lapwelded 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 unitpressure 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/l1600^), (4)
p = 86670 J— 1386.0 (5)
Eq. (4) is for use where fjd 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^dFjpadF^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.
.\pSFpadFzdF.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 neighborparticles. ]^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 foim 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 halfre 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 = FpF2?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 ringshaped 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. Coordinates 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
coordinates 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 coordinates, 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
KKK)
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^WhK)]
x'ix — h^dx =
U,V4
'2 /a?' 7 X
6B >
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"
■ ^41 — ^ 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 OF PKESSUEE.
551
433. Examples. — It will be noticed that although the total
pressure on the plane figure depends for its value upon the
head, s, of the centre of gravity, its point of application is al
ways lower than the centre of gravity.
Example 1. — If 6 ft. of a vertical sluicegate, 4 ft. wide,
Fig. 483, is below the watersurface, the total
water pressure against it is (ft., lb., sec. ; eq.
(1), § 430)
P  Fzy = 6 X 4 X 3 X 62.5 = "4500 lbs.,
and (so far as the pressures on the vertical
posts on which the gate slides are concerned)
is equivalent to a single horizontal force of
that value applied at a distance a?c = f of
6 = 4 ft. below the surface (§ 431).
Example 2. — To (begin to) lift the gate in Fig. 483, the
gate itself weighing 200 lbs., and the coefficient of friction
between the gate and posts being/" = 0.40 (abstract numb.) (see
§ 156), we must employ an upward vertical force at least
= P' = 200 + 0.40 X 4500 = 2000 lbs.
Fig. 483.
Example 3. — It is required to find the resultant hydro
static pressure on the trapezoid in Fig. 483a with the dimen
sions there given and its bases horizontal ; also its point of ap
plication, i.e., the centre of pressure of
the plane figure in the position there
shown. From symmetry the C. of P. will
be in the middle vertical of the figure,
as also that of the rectangle B CFE^ and
that of the two triangles ABE and
CDF i2tkQ,r\ together (conceived to be
shifted horizontally so that CF and
BE coincide on the middle vertical,
thus forming a single triangle of 5 ft. ba.se, and having the
same total pressure and C. of P. as the two actual triangles
taken together). Let P^ = the total pressure, and xj refer to
the C, of P., for the rectangle ; P^ and x/, for the 5 ft. tri
EF= 5
Fig. 4S3a.
552
MECHANICS OF ENGHSTEEEING.
angle; Aj = 4 ft. and h^ = 10 ft. being the same for both.
Then from eq. (1), § 430, we have (with the ft., lb., and sec.)
P^ = 30 X 7r = 210x ; and P, = ^ x 6 X 5 X 6;k = 90^;
while from eqs. (1) and (3) of § 431 we have also (respectively)
, 2 1000  64 2 936
3 ■ 100  16 3
,_1 48 + 80 + 100
84
= 7.438 feet;
228
8+10
2X18
= 6.333 feet.
The total pressure on the trapezoid, being the resultant of
jPj and Pj , has an amount ^ P^\ P^ (since thej are parallel),
and has a leverarm x^. about the axis OY io be found by the
principle of moments, as follows :
^ _ P,x: + P,x: _ (210 X 7.438 + 90 X 6.33)^ _ » naft
The total hydrostatic pressure on the trapezoid is (for fresh
water)
P = P,^P,^ [210 + 90] 62.5 = 18750 lbs.
Example 4. — Required the horizontal force P\ Fig. 484, to
be applied at N (with a leverage of a' = 30 inches about the
N fulcrum M) necessary to (begin
to) lift the circular disk ^^ of
radius r = 10 in., coveri