cos sin 9? = y.fdMy = 3fy ~ ; etc.) :
:2X = fdP - fdT sin cp - fdN Qo&o^j free., the acting forces
are G (known) and N and P., the un- ^ ...-
known normal and tangential compo- ''V
nents of the action of the plane on the
roller. If slipping occurs, then P is the ^'*^- ^^^
gliding friction due to the pressure Ni^ 156); here, however, it is
less by hypothesis (perfect rolling). At any instant the four
unknowns are found by the equations
KINETICS OF A RIGID BODY. 131
JSX, i.e., G sin fi - P, = {G-^ g)p ; . (1)
:SY, i.e., (? cos ^ - if, = ; . . . . (2)
^ moms.^, i.e., Pa, = OMkz ; • • (^)
while on account of the perfect rolKng, da = p . . • (4)
Solving, we have, for the acceleration of translation,
_2? = ^ sin /i - [1 + (/.■/ - a')l
(If the body slid wirhont friction, j9 would = p'sin /?.) Hence
for a cylinder (§ 97), kz" beiiio- = -^a", we have^ = f^ sin /3 ;
and for a sphere (§ 103) j> = ^g sin /3.
(If the plane is so steep or so smooth that both rolling and
slipping occur, then da no longer = _p, but the ratio of i-* to iV
is known from experiments on sliding friction ; hence there are
still four equations.)
The motion of translation being thus found to be uniformly
accelerated, we may use the equations of § 56.
Numerically, if a homogeneous solid sphere took 1.20 sec. to descend
(from rest() 10 ft. along a rough inclined plane, with /? = 30°, did any-
slipping occur, or was the motion perfect rolling? From p. 54 we have
s = -|-p/^ that is, 10 = -|- . ^ .^ sin 30° . t^, for perfect rolling; from which
we obtain i=1.32 seconds, which is >1.20 sec. Hence some slipping
must have occurred. (The time of descent would have been only
i = "|/2s-=-^g = 1.114 sec, if the surfaces had been perfectly smooth; and
the sphere would have had simply a motion of translation, the force P
being zero).
N.B. — A hollow sphere would occupy a longer time than a solid one in
descending the plane (if rough) ; since the ratio kz^a is greater for the
former.
125. Parallel-Rod of a Locomotive. — "When the locomotive
moves uniformly, each dJf of the rod between the two (or
three) driving-wheels rotates with j \ ;
uniform velocity about a centre of its
own on the line j5i>, Fig. 143, and with
a velocity -y* and radius r common ^._.^
to all, and likewise has a horizontal ( * js : , ; m t )
■M/l'^/or■7;^ motion ot transhition. Hence (ii.)
if we inquire what are the reactions P ^^^- ^^3.
* This velocity is that which the dM ?ias relatively to the frame of the
locomotive, in a circular path. E.g., if the locomotive (frame; has a velocity
of GO miles per hour and the radius r is one-third of the radius of the driver,
then V is 20 miles per hour.
132 MECHANICS OF ENGI]S'EEEING.
of its supports, as induced solely T)y its weight and motion^
w'.ien in its lowest position (independently of any thrust along
the rod), we put JSJT of (I.) = 2Y of (II.) (II. shows the
imaginary equivalent system), and obtain
2P - G =fdN =fdMo' -^r^iv'-^ r)fdM = Mv' :- r.
Example. — Let the velocity of translation = 50 miles per
hour, the radius* of the pins be 18 in. = f ft., and = half that
of the driving -wheels, while the weight of the rod is 200 lbs.
With g = 32.2, we must use the foot and second, and obtain
V = i[60 X 5280 ^ 3600] ft. per second = 36.6;
while Jf = 200 ^ 32.2 = 200 X .0310 = 6.20 ;
and finally P = i[200 + 6.2(36.6)^-=- f] = 2868.3 lbs.,
or nearly If tons, about thirty times that due to the weight
alone.
126. So far in this chapter the motion has been prescribed,
and the necessary conditions determined, to be fulfilled by the
acting forces at any instant. Problems of a converse nature,
i.e., where the initial state of the body and the acting forces
are given while the resulting motion is required, are of much
greater complexity, but of rare occurrence in practice.
For further study in this direction the reader is referred to Routh's
"Rigid Dynamics," Rankine's "Applied Mechanics," Sehell's " Theorie
der Bewegung und der Kraefie," and Worthington's "Dynamics of Rota-
tion" (this last being a small but clearly written and practical book).
In Wood's "Analytical Mechanics" will be found the proof of "Euler's
Equations," which are the basis of the treatment of the gyroscope in the
book of that name by Gen. J. G. Barnard (Van Nostrand's Science Series,
No. 90). The article on the gyroscope in Johnson's Cyclopaedia is by
Gen. Barnard. Perry's "Spinning Tops" is an interesting popular book.
The Brennan "Monorail Car" (model) is described in the Engineering
Record for Aug. 31, 1907, p. 226, and depends for its stability (there
being but one rail under the car) upon two gyroscope wheels revolving
at 7000 revs, per min. in a vertical plane parallel to the rail. See also
McClure's Magazine for Dec, 1907, p. 163.
* Or, rather, the radius of the circular path of the pin-centre, whose
velocity in this path is 25 miles per hour.
WORK, ENEKGY, AND POWER. 18b
CHAPTER YI.
WORK, ENERGY, AND POWER.
127. Remark. — These quantities as defined and developed
in this cliapter, though compounded of the fundamental ideas
of matter, force, space, and time, enter into theorems of such
wide application and practical use as to more than justify their
consideration as separate kinds of quantity,
128, Work in a XJniform Translation. Definition of Work. —
Let Fig. 144 represent a rigid body having a motion of trans-
lation parallel to X, acted on by a yrp^
system of forces P^, P^, H^, and Ii^, ^f — ''~'&, of the body; a final, n
141. Relation of Work and Kinetic Energy for any Extended
Motion of a Rigid Body Parallel to a Plane. — (If at any instant
any of the forces acting are not
parallel to the plane mentioned,
their components lying in or
parallel to that plane, will be used
instead, since the other compo-
nents obviously would be neither
working forces nor resistances.)
Fig. 153 shows an initial position,
and anj' intermediate, as q. The
forces of the system acting may vary in any manner during
the motion.
In this motion each dM describes a curve of its own with
varying velocity v, tangential acceleration j^t-, ^^^d radius of
curvature r ; hence in any position ^, an imaginary system JB
(see Fig. 154), equivalent to the actual system A (at q in Fig.
153), would be formed by applying to each dM a
tangential force dT =^ dMpt, and a normal force
dN' = dMv'^ -V- r. By an infinite number of con-
secutive small displacements, the body passes from
o to n. In the small displacement of which q is the
initial position, each 6? J/^ describes a space ds^ and
dT does the work dTds = dMvdv, while dJV does the work-
dJV X = 0. Hence the total work done by £ in the small
displacement at q would be
dN
dT
dM'v'dv' + dM"n}"dv" -f etc.,
(1)
including all the dM^& of the body and their respective veloci-
ties at this instant.
But the work at q in Fig. 153 by the actual forces (i.e., of
system A) during the same small displacement must (by § 140)
be equal to that done by B. hence
P,du, -f P,du^ + etc. = dM'v'dv' + dM"v"dv" -f etc. (^)
Now conceive an equation like {g) written out for each of
WORK, ENERGY, AND POWER. 147
the small consecutive displacements between positions o and
yi and corresponding terms to be added ; this will give
P/hc^ -\- 1 P^du^ -\- etc.
= dW / v'dii' + dM" / v"dv" + etc.
= \dM'{v^^ - ^;=) + i^^Jf'X V - <'') + etc.
The second member may be rewritten so as to give, finally,
/ P,dit,+ P,du,-^etG.=:S{idMv,')-:S{^dMv,'),{XY.)
or, in words, the worTi, done hy the acting forces {treating a re-
sistance as a negative worhing force) between any two posi-
tions is equal to the gain {or loss) in the aggregate Icinetic
energy of the particles of the hody hetwee7i the tioo positions.
To avoid confusion, 2 has been used instead of the sign y in
one member of (XY.), in which v^ is the final velocity of any
dM {not the same for all necessarily) and v^ the initial.
(The same method of proof can be extended to three dimen-
sions.)
Since kinetic energy is always essentially positive, if an ex-
pression for it comes out negative as the solution of a problem,
some impossible conditions have been imposed.
142. Work and Kinetic Energy in a Moving Machine. —
Defining a mechanism or machine as a series of rigid bodies
jointed or connected together, so that working-forces applied
to one or more may be the means of overcoming resistances
occurring anywhere in the system, and also of changing the
amount of kinetic energy of the moving masses, let us for
simplicity consider a machine the motions of whose parts are
all parallel to a plane, and let all the forces acting on any one
piece, considered free, at auy instant be parallel to the same
plane.
Now consider each piece of the machine, or of any series of
its pieces, as a free body, and write out eq. (XY.) for it be-
tween any two positions (whatever initial and final positions are
selected for the first piece, those of the others must be corre-
sponding initial and corresponding final positions), and it will
348
MEOIIA.MCS OF EXGINEEIilNG.
be found, on adding np corresponding members of these equa-
tions, that the terms involving tliose components of the mutual
pressures (between the pieces considered) which are normal
to the rubbing surfaces at any instant will cancel out, while
their components tangential to the rubbing surfaces {i.e., fric-
tion, since if the surfaces are perfectly smooth there can be
no tangential action) will appear in the algebraic addition as
resistances multiplied by the distances rubbed through, meas-
ured on the rubbing surfaces. For example. Fig. 155, where
one I'otating piece both presses and rubs on another. Let the
normal pressure between them at A be R^ = P^ ; it is a work-
ing force for the body of mass M" , but a resistance for M' ,
hence the separate symbols for the numerically equal forces
(action and reaction).
Similarly, the f liction at ^ is i?3 = ^Pg ; a resistance for M' ,
a working-force for M" . (In some cases, of course, friction
may be a resistance for both bodies.) For a small motion, A
describes tlie small arc AA' abont 0' in dealing with M\ but
for M" it describes the arc AA" about 0" . A' A" being
parallel to the surface of contact AD, while AB is perpen-
FiG. 156.
Fig. 157.
Fig. 155.
dicular to A' A" . In Figs. 156 and 15Y we see M' and M"
free, and their corresponding small rotations indicated. During
these motions the kinetic energy (K. E.) of each mass has
clianged by amounts ?' ^
a force jP in an axial direction ' ' ' / j
(they are equal, supposing the Fig. leo.
mass of the spring small). As the machine operates of which
it is a member, it moves to a new consecutive position J^,
suffering a further elongation dX in its length (if F is increas-
ing). P on the right, a woi'king force, does the work Pdx'',
how is this expended ? ^ on the left has the work Pdx done
upon it, and the mass is too small to absorb kinetic energy or
to bring its weight into consideration. The remainder, Pd'x'
— Pdx = Pdx, is expended in stretching the spring an addi-
tional amount dX, and is capable of restoration if the spring
retains its elasticity. Hence the work done in changing the
form of bodies if they are elastic is said to be stored in the
form of potential energy. Tliat is, in the operation of ma-
chines, the name potential energy is also given to the energy
stored and restored periodically in the changing and regaining
of form of elastic bodies.
156 MECHANICS OF ENGINEERING.
EKatiif)le. — A given helical spring, when held stretched s'=J ft. beyond
its "natural" (or unstrained) length, exerts a pull of i2' = 1200 lbs.
at its two ends; and the "potential energy" residing in it is — mean forceX
distance, =^R's\ = (i) 1200 (^), =300 ft.-lbs. If such a stretched
spring be placed on a car of 644 lbs. weight on a level track and properly
connected with a driving-wheel, which does not slip on the track, its
recovery of natural length may be made the means of starting the car
into motion and causing it to attain a final velocity of v = 5A7 ft. /sec.
(if no friction is met with); from i(644-f-32.2)i;2 = 300.
149. Other Forms of Energy. — Numerous experiments witli
various kinds of apparatus have proved that for every 7Y8
(about) ft.-lbs. of work spent in overcoming friction, one British
unit of heat is produced (viz., tlie quantity of lieat necessary to
raise tlie temperature of one pound of water from 32° to 33°
Fahrenheit); while from converse experiments, in which the
amount of heat used in operating a steam-engine was all carefully
estimated, the disappearance of a certain portion of it could only
be accounted for by assuming that it liad been converted into
work at the same rate of (about) 778 ft.-lbs. of vs^ork to each
unit of heat (or 427 kilogrammetres to each French unit of
lieat). This number 778 or 427, according to the system of
units employed, is called the Mechanical Equivalent of Htot
Heat then is energy, and is supposed to be of the kinetic
form due to tJie rapid motion or vibration of the molecules of
a substance. A similar agitation among the molecules of the
(hypothetical) ether diffused through space is supposed to pro-
duce the phenomena of light, electricity, and magnetism.
Chemical action being also considered a method of transform-
ing energy (its possible future occurrence as in the case of coal
and oxygen being called potential energy), the well-known
doctrine of the Conser nation of Energy^ in accordance with
which energy is indestructible, and the doing of work is simply
the conversion of one or more kinds of energy into equivalent
amounts of others, is now an accepted hypothesis of physics.
Work consumed in friction, though practically lost, still re-
mains in the universe as heat, electricity, or some other subtile
form of energy.
150. Power Required for Individual Machines. Dynamome-
ters of Transmission. — If a machine is driven by an endless
belt from the main-shaft, A^ Fig. 161, being the driving-pulley
WORK, ENERGY, AND POWER.
167
Fig. lUl.
on the machine, the working force which drives the machine,,
in other words tlie " grip" with which the
belt takes hold of the pulley tangentially^
= /^ — P' ^ P and P' being the tensions
in the "driving" and ''following" sides of
the belt respectively. The belt is supposed
not to slip on the pulley. If v is the ve-
locity of the pulley -circumference, the
work expended on the machine per second, i.e., \\\q lyower, is
L = (P-POv, ft.-lbs. per sec. .... (1)
To measure the force {P — P')^ an apparatus called a Dy-
nainometer of Transmission may be placed between the main
shaft and the machine, and the belt made to pass through it in
such a way as to measure the tensions P and P' ^ or princi-
pally their difference, without meeting any resistance in so do-
ing ; that is, the power is transmitted^ not absorbed, by the
apparatus. One invention for this purpose (mentioned in the
Journal of the FranMin Institute some years ago) is shown
{in principle) in Fig. 162. A ver-
tical plate carrying four pulleys and
a scale-pan is first balanced on the
pivot C. The belt being then ad-
justed,- as shown, and the power
turned on, a sufficient weight G is
placed in the scale-pan to balance
Fig. 163. tlie plate again, for whose equilib-
rium we must have Gh = Pa — P'a, since the P and P' on
the right have action-lines passing through C. The velocity of
belt, V, is obtained by a simple counting device. Hence
(P —P') and V become known, and .'. L from (1).
Many other forms of transmission-dynamometers are in use,
some applicable whether the machine is driven by belting or
gearing from the main shaft. Emerson's Hydrodynamics de-
scribes his own invention* on p. 283, and gives results of meas-
m-ements with it ; e.g., at Lowell, Mass., the power required
to drive 112 looms, weaving 3 6 -inch sheetings, No. 20 yarn,
60 threads to the inch, speed 130 picks to the minute, was
found to be 16 H.P., i.e., \- H.P., to each loom (p. 335).
* Prof . Flather's '^Dynamometers" is a standard book (1907).
158
MECHANICS OF ENGINEERING.
Example. — The endless belt connecting the pulley (running at n=180
revs./min., with a radius of r = 2 ft.) of an engine shaft with that of a
planing machine is led over the idle pulleys of the apparatus in Fig. 162,
as there shown (engine pulley on left, and that of machine on right;
but neither shown in figure). To balance the plate in position shown
(with a = 2 ft. and 6 = 4 ft.) is found to require a weight G = 210 lbs.
We have, therefore, from {P-P')a = Gb, P-P' = 210X4 --2 = 420 lbs.
as the net working force operating the machine; while the velocity of
the belt is v, =n2;rr, =(180-^60) 2;r2= 18.85 ft. /sec. Hence the
power transmitted through the dynamometer of transmission is L,
= {P-P')v, =420X18.85 = 79,170 ft.-lbs. per sec, or 14.4 H.P
151. Dynamometers of Absorption. — These are so named
.since they furnish in themselves the resistance (friction or a
weight) in tlie overcoming (or raising) of which the power is
expended or absorbed. Of these the Prony Friction Brake
is the most common, and is nsed for measuring the power
developed by a given motor (e.g., a steam-engine or turbine)
mot absorbed in the friction of the motor itself. Fig:. 163
«hows one fitted to a vertical pulley driven by the motor. By
tightening the bolt B^ the velocity i) of pulley-rim may be
made constant at any desired value (within certain limits) by
the consequent friction, -y is measni-ed by a counting appara-
tus, while the friction (or tangential components of action be-
tween pulley and brake), = I\ becomes known by noting the
weight G which must be placed in the scale pan to balance the
arm between the checks; then with G^''= weight of brake and
h' =tlie horizontal distance of its center of gravity from the
center of pulley, we have, for the equilibrium of the brake
(moments about pulley center),
Fa=Gb + GV', (1)
and the work done on F per unit of time, or power, is
L=i^y, ft.-lbs. per sec (2)
(In case the pulley is horizontal, a bell-crank must be inter-
posed between the arm and the scale-pan.)
WORK, ENERGY, AND POWER.
159
Example. — A vertical pulley of a = 2 ft. radius and run by a turbine
water-wheel, is gripped by a Prony brake, as in Fig. 163, with arm
fe = 4 ft. 9 in. A load of G = lQO lbs. is placed in the scale pan, the water
turned on, and the bolt B screwed up until the friction F of pulley-rim
on brake is just sufficient to lift the weight and hold the brake in equi-
librium Weight of brake is (r' = 40 lbs., with centre of gravity 6' = 1.5 ft.
on right of pulley centre. The speed to which the pulley now adjusts
itself is at rate of 210 revs./min. The friction is F ={Gb + Gb')^a =
(160X4.75+40X1.5)-h2 = 410 lbs.; the velocity of pulley-rim is v =
(210^60) 27rX2 = 44 ft. /sec; hence the power developed is F'?; = 410X44
= 18,040 ft.-lbs. per sec. ; or 32.8 H.P.
Note. — For an account of various modern designs of absorption and
transmission dynamometers, the reader is referred to Prof. Flather's
book, already mentioned in the foot-note on p. 157. This is a recent
and a standard work. In the "Notes and Examples in Mechanics" by
the present writer, brief descriptions are given (pp. 96 and 97) of the
Appold and the Carpentier dynamometers of absorption, with the theory
of the same; both of these are "automatic" or "self-adjusting."
It must be carefully noted by the student that in the absorption dyna-
mometer, which for purposes of test temporarily takes the place of useful
machines, the power is absorbed and converted into heat, necessitating
cooling devices if the parts are to run smoothly and lubricants are to
remain unaffected; whereas in the dynamometer of transmission the
power simply passes through without heating effect.
152, The Indicator, used with steam and other fluid engines,
is a special kind of dynamometer in which the automatic mo-
tion of a pencil describes a curve
on paper whose ordinates are
proportional to the fluid pres-
sures exerted in the cylinder at
successive points of the stroke.
Thus, Fig. 164, the back-pres-
sure being constant and = P^, fig. lu.
the ordinates P^, P^, etc.. represent the effective pressures at
equally spaced points of division. The mean effective pressure
P' (see § 147) is, for this figure, by Simpson's Rule (six equal
spaces),
P' = tV[^o + 4(P, + P3 + P.) + 2(P. + P.) + Pe].
This gives a near approximation. The power is now found by
§147^
153. Mechanical Efficiency of a Steam or Vapor Engine (gas,
petroleum, gasoline, or alcohol vapor, etc.). By the term
*' mechanical efficiency " is meant the ratio of the power obtained
< --
l
!
^\/
fk<
P1
P2
P3
P4
Pa
T 1
Pe
l^b
2ERC
3 LINE
X
160 MECHANICS OF ENGINEERING.
at tlie rim of tlie pnlley or fly-wheel on the main shaft of the
engine (where it would be connected with machinery to be
operated or where in a test the resistance of brake -friction
wonld be overcome) to the power exerted directly on the piston
of the engine by the pressure of the fluid concerned. This
latter item becomes known through the use of the ' ' indicator ' '
(see preceding paragraph) and is hence often called the 'indi-
cated 'power ; " the power spent on friction provided by a Prony
brake, for testing purposes, being called the " brake-power.^^
Example. — If from indicator-cards the value of P', or total mean
effective pressure on the piston of an engine, is found to be 12,000 lbs.,
the piston speed being at the (mean) rate of 6 ft. per sec, the "indi-
cated power" of the engine is = 72,000 ft.-lbs./sec. Now, when the
engine is running under these same conditions of pressure and speed,
if it is found by the use of a Prony friction brake that the power spent
on brake friction consists of overcoming a friction of 6000 lbs. through
10 ft. each second, and that therefore the power obtained at the brake,
or "brake power," is equal only to 60,000 ft.-lbs./sec, the mechanical
efficiency of the engine (in this test) is 60,000-^72,000, =0.833 or 83 J per
cent. In other words, 16t per cent of the power exerted by the fluid
pressure on the piston, or "indicated power," is lost in the overcoming
of the friction of the engine itself, i.e., among the moving parts situated
between the piston and the rim of the test pulley.
153a. Efficiency of Power Transmission. — In transmitting
power through a long line of shaftmg, or by ropes or belts, or
water in pipes, or by electric current, the efficiency is the ratio
of the power put in at the sending station to that obtained at
at the receiving station. For example,
Example. — An engine exerts power at the rateof (say) 600,000 ft.-lbs./sec,
in running a "dynamo" at the sending or power station. The electric
current so generated is conducted 60 miles through wires to a receiving
station, where by operating an electric motor it enables a pulley to be
run within a Prony brake from whose indications it is found that a power
of 360,000 ft.-lbs./sec. is there obtainable. Hence the efficiency of
transmission is 360,000-^600,000, =60 per cent.
154. Boat-Rowing. —x^'ig. 166. During the stroke proper,
let /* = mean pressure on one oar-handle ; hence the pressures
on the foot-rest are 2P, resistances. Let J!f"= mass of boat
and load, v^ and Vn its velocities at beginning and end of stroke.
Pj = pressures between oar-blade and water. Ji = mean re-
sistance of water to the boat's passage at this (mean) speed.
These are the only (horizontal) forces to be considered as act-
ing on the boat and two oars, considered free collectively.
During the stroke the boat describes the space s^ = CD, the
oar-liandle tlie space s„ = AB, wliile tlie oar-blade slips back-
WORK, ENERGY, AND POWER.
161
ward through the small space (the " slip") = s^ (average).
Hence by eq. (XYI.), § 142,
i.e., 2P{s,-s,)=2FxAJi=2Fs =Bs,+2P,s,-\- iMiv^'-v,');
or, in words, the product of the oar-handle pressures into the
distance described by them measured on the hoat, i.e., the work-
done by these pressures relatively to the hoat, is entirely ac-
counted for in the work of slip and of liquid-resistance, and in^
■■..si>^
Fig. 166.
creasing the kinetic energy of the mass. (The useless work
due to slip is inevitable in all paddle or screw propulsion, as
well as a certain amount lost in machine-friction, not considered
in the present problem.) During the '' recover" the velocity
decreases again to v^. (See example on p. 9T, Notes, etc.)
155. Examples. — 1. What work is done* on a level traci^, in
bringing up the velocity of a train weighing 200 tons, from
zero to 30 miles per hour, if the total frictional resistance (at
any velocity, say) is 10 lbs. per ton, and if the change of speed
is accomplished in a length of 3000 feet ?
{Foot-ton-second system.) 30 miles per hour = 44 ft. per
sec. The mass
- 200 -^ 32.2 = 6.2 ;
.'. the change in kinetic energy,
(= Wv-' -iM X 0%
= i(6.2) X 44* = 6001.6 ft.-tons.
* That is, what work is done by the pull, or tension, P, in the draw-bar
between the locomotive and the "tender."
162 MECHANICS OF ENGlNEEKliMG.
The work done in overcoming friction = Fs, i.e.,
= 200 X 10 X 3000 = 6,000,000 ft.-lbs. = 3000 ft.-tons ;
.-. total work = 6001.6 + 3000 = 9001.6 ft.-tons.
(If the track were an up-grade, 1 in 100 say, the item of
200 X 30 = 6000 ft.-tons would be added.)
Exmnjple 2. — Required the rate of work, or power, in Ex-
ample 1. The power is variable, depending on the velocity of
the train at any instant. Assume the motion to be uniformly
accelerated, then the working force is constant ; call it P.
The acceleration (§ 56) will be ^='y'-=- 2^=1936-^6000 = 0.322
ft. per sq. sec; and since P — J^= Mjp^ we have
P = 1 ton + (200 -=- 32.2) X 0.322 = 3 tons,
which is 6000 -^ 200 = 30 lbs. per ton of train, of which 20 is
due to its inertia, since when the speed becomes uniform the
work of the engine is expended on friction alone.
Hence when the velocity is 44 ft. per sec, the engine is
working at the rate of P'o = 264,000 ft.-lbs. per sec, i.e., at the
rate of 480 H. P.;
At i of 3000 ft. from the start, at the rate of 240 H. P., half
as much ;
At a uniform speed of 30 miles an hour the power would be
simply 1 X 44 = 44 ft. -tons per sec. = 160 H. P.
Example 3. — The resistance offered by still water to the
passage of a certain steamer at 10 knots an hour is 15,000 lbs.
What power must be developed by its engines, at this uniform
speed, considering no loss in " slip" nor in friction of ma-
chinery ? A71S. 461 H. P.
Example 4, — Same as 3, except that the speed is to be 15
knots (i.e., nautical miles ; each = 6086 feet) an hour, assum-
ing that the resistances are as the square of the speed (approxi-
mately true). Ans. 1556 H. P.
Example 5. — Same as 3, except that 12^ of the power is ab-
soi-bed in the " slip" (i.e., in pushing aside and backwards the
water acted on by the screw or paddle), and 8^ in friction of
machinery. Ans. 576 H. P.
Example 6. — In Example 3, if the crank-shaft makes 60
"WORK, ENERGY, AND POWEE. 163
revolutions per minute, the crank-pin describing a circle of 15
incbes radius, required the average* value of the tangential
component of the thrust (or pull) of the connecting-rod against
the crank-pin. Ans. 26890 lbs.
Example 7. — A solid sphere of cast-iron is rolling up an in-
cline of 30°, and at a certain instant its centre has a velocity of
36 inches per second. Neglecting friction of all kinds, how
much further will the ball mount the incline (see § 143) %
Ans. 0.390 ft. \
Example 8.— In Fig. 163, with J = 4 f t. and a = 16 inches,
it is found in. one experiment that the friction which keeps the
speed of the pulley at 120 revolutions per minute is balanced
by a weight G — 160 lbs. Eequired the power thus measured..
Ans. 14.6 H. R
Although in Examples 1 to 6 the steam cylinder is itself in
motion, the work per stroke is still ■=■ mean effective steam-
pressure on piston X length of stroke, for this is the final form
to which the separate amounts of work done by, or upon, the
two cylinder heads and the two sides of the piston will re-
duce, when added algebraically. See § 154.
* By " average value" is meant such a value, Tm, as multiplied into the
length of path described by the crank-pin per unit of time shall give the
power exerted.
164 MECHANICS OF EWGINEEKING.
CHAPTEK YIL
FRICTION.
156. Sliding Friction. — When the surfaces of contact of two
bodies are perfectly smooth, the direction of the pressure or pair
of forces between them is normal to these surfaces, i.e., to their
; tangent-plane ; but when thej are rough, and
'Y"'f \ moving one on the other, the forces or ac-
pV 4N tions between them incline awaj from the
i \ I ij -, P- normal, each on the side opposite to the di-
WmJ^ S^ W//Jw///M ^^ction of the (relative) motion of the body
/-. ^q/ ///M ^^ which it acts. Thus, Fig. 167, a block
Fig. 167. whose weight is G, is drawn on a rough
horizontal table by a horizontal cord, the tension in which is
P. On account of the roughness of one or both bodies the ac-
tion of the table upon the block is a force P^^ inclined to the
normal (which is vertical in this case) at an angle = (p away
from the direction of the relative velocity -y. This angle q) is
called the angle of friction^ while the tangential component of
P^ is called the friction = F. The normal component N^
which in this case is equal and opposite to G the weight of tlie
body, is called the normal pressure.
Obviously i^= iV^tan
If G, Fig. 59, is made just sufficient to start the block, or
sledge, G^, we have for the friction of rest
. /=|.. ....... ii)
160. Results of Experiments on Sliding Friction. — For accounts of recent
experiments (and others) and deductions therefrom, the reader may consult
the Engineer's "Pocket-books" of Kent and Trautwine; also Thurston's
"Friction and Lost Work," Barr and Kimball's "Machine Design," and
" Lubrication and Lubricants, ' ' by Archbutt and Deely. The following
table gives a few values for the coefficient of friction, /, for slow motion,
taken from the results obtained by Morin and others. Small changes in
the condition of the surfaces may produce considerable variation in the
value of /. Our knowledge is still quite incomplete in this respect.
168 MECHANICS OF ENGINEEKING.
TABLE FOR FRICTION OF SLOW MOTION.
No.
Surfaces.
Unguent.
Angle cp, this tangential resistance being oiiij fN
and < P sin /?, A will begin to slip, with an acceleration.
162. Problems in Sliding Friction. ^ — In the following prob-
lemsy is supposed known at points where rubbing occurs, or
is impending. As to the- pressure iV^ to which the friction is
due, it is generally to be considered unknown until determined
by the conditions of the problem. Sometimes it may be an
advantage to deal with the single unknown force P\ (resultant
of N 'Aw^fN^ acting in a line making the known angle 9? with
the normal (on the side away from the motion).
Problem 1. — Keqnired the value of the weight P^ Fig. lYl,
the slightest addition to which will cause motion of the hori-
zontal rod OB, resting on rough planes at 45°. The weight
G of the rod may be applied at the ys^^y
middle. Consider the rod free ; at
€ach point of contact there is an un-
known JSf and a friction due to it
fN\ the tension in the cord will be
= P, since there is no acceleration
and no friction at pulley. Notice fig. 171.
the direction of the frictions, both opposing the impending
motion. Take axes OX and OF as shown, and note the inter-
sections, A and C, of the line of G with axes OX. and OY .
The student should not rush to the conclusion that, if G were
transferred to A and there resolved into components along
AO and AB, the value of 'N (and A^i) would be equal to that of
one of these components, viz., mG (where m denotes sin 45°).
Few problems in mechanics are so simple as to admit of an
immediate mental solution, and guess-work should be care-
fully avoided.
It will be found of advantage to take C as a center of
moments. The line of P at is considered as passing through
point C , as also the lines of /AT and /iVi- For equilibrium
(impending slipping) we have, therefore,
i-Z^O; i.e.,/iVi+mG-iV-P = 0; . . . (1)
27 = 0; i.e., i\ri + /Ar-mG = 0; .... (2)
i'(moms.)c=0; i.e., iVa-A^'ia = 0. ... (3)
170
MECHANICS OF ENGINEERING.
The three unknowns P, N, and Ni, can now be founds
From (3) we have N = Ni, which in (2) gives A^ = t-71-
Also from (1) we now find P=^^jN; and hence finally
P
1 +/"!+/■
G.
Peoblem 2. — Fig. 1Y2. A rod, centre of gravity at middle,.,
leans against a rough wall, and rests on an equally rough floor;
how small may the angle a become before it
slips ? Let a = the half-length. The figure-
p shows the rod free, and following the sugges-
P tion of § 162, a single unknown force ^,
p making a known angle (p (whose tan =/")
P with the normal D^, is put in at D, leaning^
away from the direction of the impending
motion, instead of an JV and /"iV; similarly
7^2 acts at C. The present system consisting^
of but three forces, the most direct method of finding or, with-
out introducing the other two unknowns jP^ and P*^ at all, is
to use the principle that if three forces balance, their lines
of action must intersect in a point. That is, P*^ must inter-
sect the vertical containing G, the weight, in the same point
as Pi, viz., A.
ISTow, since CF is 1 to FD and the two angles are equal,
CA is T to DA and therefore DAC is a right triangle. "We
also note that if CA be prolonged to meet DF in some point K,
A must be the mid-point of CK, since B is the mid-point of
CF ; and therefore it follows that in triangle DCK not only
does DA bisect the side KC but is 1 to it. In other words,
KDC is an isosceles triangle, with DK and DC as the two
equal sides, and therefore the line DA bisects the angle KDC.
Hence the angle KDC = '2(p. That is to say, the angle a,
which was to be determined, is the complement of double the
friction-angle, or
a: = 90°- 20.
PKICTION.
171
Pkoblem 3. — Fig. 173. Given the resistance Q, acting
parallel to tlie fixed guide C, the angle a, and tlie (equal) co-
efficients of friction at the rubbing surfaces, required the
Fig. 173.
amount of the horizontal force P, at the head of the block A
(or wedge), to overcome Q and the frictions. D is fixed, and
ah is perpendicular to cd. Here we have four unknowns, viz.,
JP, and the three pressures iV^, JV^, and JV^, between the blocks.
Consider ^ and 5 as free bodies, separately (see Fig. 174), re-
membering l^ewton's law of action and reaction. The full
values {e.g.,f]V) of the frictions are put in, since we suppose
.a slow uniform motion taking place.
For A, 2X= and :S r = give
i^i — iVcos a -|-y^sin a — J^ sin a = ; .
f¥^ + N sin a -[-/iTcos a — P cos a = 0. .
'FoyB, :SXand ^Zgive
'Q-JSr,+fJV, = 0;....iS) and W,-fJ^, = 0.
Solve (4) for JV^ and substitute in (3), whence
j^.o^-r) = Q
(1)
(2)
m
(5)
Solve (2) for JV, substitute the result m (1), as also the value
of i\r^ from (5), and the resulting equation contains but one un-
known, P. Solving for P, putting for brevity
ycos a-\- sin a
■we have P =
m and cos a — /"sin a = n,
{w,-\-fn)Q
or
{n . cos a -\- m . sin a)(l — yy
{T}
172 MECHANICS OF ENGINEERING.
Numerical Examjple of Problem 3. — If Q = 120 lbs., /"
= 0.20 {an abstract number, and .*. the same in any system of
units), while a = 14°, whose sine = 0.240 and cosine = .970,,
then
m = 0.2 X. 97 + 0.24 = 0.43 and t^ = .97 — .2X.24 = 0.92,.
whence P = OMQ = 76.8 lbs.
While the wedge moves 2 inches P does the work (or exerts
an energy) of 2 X 76.80 = 153.6 in.-lbs. = 12.8 ft.-lbs.
For a distance of 2 inches described by the wedge horizon-
tally, the block P (and .•. the resistance Q) has been moved
through a distance = 2 X sin 14° = 0.48 in. along the guide-
G, and hence the work of 120 X 0.48 = 57.6 in.-lbs. has been
done upon Q. Therefore for the supposed portion of the
motion 153.6 — 57.6 = 96.0 in.-lbs. of work has been lost in
friction (converted into heat).
For the "mechanical efficiency" of this machine (see § 153)
we have 57.6-153.6=0.375. Also note that for / = 1.0(>
P = '^ ; andfor a: = 90°, P = Q.
Problem 4. Numerical. — With what minimum pressure
P should the pulley A be held against P, which it drives by
■^ n x riA " frictioiial gearing," to transmit 2 H. p.;.
^ - p -(--- ^ > if a = 45°, f for impending (relative)
«y 7>! ^ motion, i.e., for impending slipping =
Fig. 175. 0.40, and the velocity of the pulley-rim;
is 9 ft. per second ?
The limit-value of the tangential " grip"
T = 2/i\^= 2 X 0.40 X P sin 45°,
2 H. P. = 2 X 550 = 1100 ft.-lbs. per second.
Putting T X 9 ft. = 1100, we have*
2 X 0.40 X 4/5 X P X 9 = 1100 ; .: P = 215 lbs..
Problem 6. — A block of weight G lies on a rough plane,
inclined an angle ^ from the horizontal ; find the pull P, mak-
ing an angle a with the first plane, which will maintain a uni-
form motion up the plane.
* In this problem the student should note that, in general, when a is not 45°,
we have N = iP -v- cos a (since in such a case the parallelogram of. forces i&
not a square).
FRICTION. 173
Pkoblem 7. — Same as 6, except that the pull P is to permit
a uniform motion down the plane.
Pkoblem 8. — The thrust of a screw-propeller is 15 tons.
Tlie ring against which it is exerted has a mean radius of 8
inches, the shaft makes ©ne revolution per second, andy = 0.06.
Required the H. P. lost in friction from this cause.
Ans. 13.7 H. P.
163. The Bent-Lever with Friction. Worn Bearing. — Fig.
176. Neglect the weight of the lever, and suppose the plumb-
er-block so worn that there is
contact along one element only of
the shaft. Given the amount and
line of action of the resistance R^
and the line of action of jP, re-
quired the amount of the latter for
impending slipping in the direction
of the dotted arrow. As P grad-
ually increases, the shaft of the
lever (or gear-wheel) rolls on its fig. 176.
bearing until the line of contact has reached some position Ay
when rolling ceases and slipping begins. To find A^ and the
value of P^ note that the total action of the bearing upon the
lever is some force ^,, applied at A and making a known
angle q) (^f =: tan q)) with the normal A C. P^ must be eqnal
and opposite to the resultant of the known P and the unknown
P, and hence graphically (a graphic is much simpler here than
an analytical solution) if we describe about C 2i circle of radius
= r sin 9?, r being the radius of shaft (or gudgeon), and draw
a tangent to it from P, we determine PA as the line of action
of P^. If PG is made = P, to scale, and ^^ drawn parallel
to P . . . P, P is determined, being = PP, while P^ = PK
If the known force P is capable of acting as a working force,
by drawing the other tangent PP from P to the " friction-
circle," we have P = PP, and P^ = PK, for impending
rotation in an opposite direction.
If P and P are the tooth-pressures upon two spur-wheels,
keyed upon the same shaft and nearly in the same plane, the
174 MECHANICS OF ENGINEERING.
y = [Pj sin (p]27rr.
same constructions hold good, and for a continuous uniform
motion, since the friction = P, sin cp,
the work lost in friction
per revolution,
It is to be remarked, that without friction Pj would pass
through 0, and that the moments of i? and I^ would balance
about C (for rest or uniform rotation) ; whereas with friction
thej balance about the proper tangent-point of the friction-
circle.
Another way of stating this is as follows : So long as the
resultant of I^ and P falls within the " dead-angle" BDA,
motion is impossible in either direction.
If the weight of the lever is considered, the resultant of it
and the force M can be substituted for the latter in the fore-
going.
164. Bent-Lever with Friction. Triangular Bearing. — Like
the preceding, the gudgeon is much exaggerated in the figure
(1Y7). For impending rotation in
direction of the force P, the total
actions at A^ and A^ must lie in
known directions, making angles = cp
with the respective normals, and in-
clined away from tlie shpping. Join
the intersections D and L. Since
the resultant of P and R i^i D must
act along PL to balance that of P^
and P^^ having given one force, say
Fig. 177. B, wc easily iind P = PE, wliile
P^ and P^ = ZJf^and ZiV respectively, LO having been made
= PP, and the parallelogram completed.
(If the direction of impending rotation is reversed, the change
in the construction is obvious.) If P^ = 0, the case reduces
to that in Fig. 176 ; if the construction gives P^ negative, the
supposed contact at A^ is not realized, and the angle A^OA^
should be increased, or shifted, until P^ is positive.
As before, P and P may be the tooth-pressures on two
FRICTION.
175
spur-wheels nearly in tiie same plane and on the same shaft ;
if so, then, for a uniform rotation.
"Work lost in fric. per revol. = [P^ sin cp -\- P^ sin cp\^7tr.
165. Axle-Friction. — Tiie two foregoing articles are intro-
ductory to the subject of axle -friction. When the bearing is
new, or nearly so, the elements of the axle which are in contact
with the bearing are iniinite in number, thus giving an infinite
number of unknown forces similar to P^ and P^ of the last
paragraph, each making an angle cp with its normal. Refined
theories as to the law ox distribution of these pressures are of
little use, considering tne uncertainties as to the value of
y ( ^ tan , ....... (1)
in which T is the tangential action, or "grip," between one
pair of friction-wheels and the axle C which rolls upon them.
^ would noL equal yiV unless slipping took place or were im-
])ending at E^ but is known bj considering a pair of friction-
wheels free, when ^ (-P«) about C^ gives
2 ' cos «'
which in (1) gives finally
b T T
P = iP,^ ^-f'R. (2)
' ' »! cos a*^ ^ '
Without friction-wheels, we would have
P^\p,^fR\ : . (3)
The last term in (2) is seen to be less than that in (3) (unless
a is too large), in the same ratio as already found for the saving
of work, supposing the jS's equal.
If P^ were on the same side of C as P^ it would be of an
opposite direction, and the pressure i? would be diminished.
Again, if P were horizontal, R would not be vertical, and the
friction-wheel axles would not bear equal pressures. Since P
depends on Pj, G^ and the frictions^ while the friction depends
on R^ and R on P^^ G, and P, an exact analysis is quite
complex, and is not warranted by its practical utility.
Example. — If an empty vertical water-wheel weighs 25,000
lbs., required the force P to be applied at its circumference to
maintain a uniform motion, with « = 15 ft., and r = 5 inches.
Here P^ = 0, and R = G (nearly ; neglecting the influence of
P on R), i.e., R = 25,000 lbs.
Eirst, iDiihout friction-wheels (adopting the foot-pound-sec-
ond system of units), withy =: .07 (abstract number). Frona
eq. (3) we have
P =: + 0.07 X 25,000 X (tV "^ 1^) = 48.6 lbs.
FRICTION, 179
The work lost in friction per revolution is
f'B^Tir = O.or X 25,000 X 2 X 3.14 X A = ^580 ft.-lbs.
Secondly, with friction-wheels, in which r^ '. a^ =: ^ and
cos a = 0.80 (i.e., a = 36°). From eq. (2)
J> = 0-^^.\^X 48.6 = only 12.15 lbs.,
while the work lost per revolution
= 1. . jMi X 4580 = 1145 ft.-lbs.
Of course with friction- wheels the wheel is not so steady as
without.
In this example the force J* has been simply enough to
overcome friction. In case the wheel is in actual use, JP is the
weight of water actually in the buckets at any instant, and does
the work of overcoming I^^, the resistance of the mill machinery,
and also the friction. By phicing J*^ pointing upward on the
same side of C as P, and making h^ nearly ■=!), H will = G
nearly, just as when the Avheel is running empty; and the
foregoing numerical results will still hold good for practical
purposes.
168. Friction of Pivots. — In the case of a vertical shaft or
axle, and sometimes in other cases, the extremity requires sup-
port against a thrust along tlie axis of the axle or pivot. If
the end of the pivot \%flat and also the surface
against which it rubs, we may consider the
pressure, and therefore the friction, as uniform
over the surface. With a flat circular pivot,
then. Fig, 180, the frictions on a small sector
of the circle form a system of parallel foices
whose resultant is equal to their sum, and is ^^'^' "^^°"
applied a distance of -|r from the centre. Hence the sum of
the moments of all the frictions about the centre =y!^|r, in
which .^ is the axial pressure. Therefore a force P necessary
to overcome the friction with uniform rotation must have a
moment
Pa =fR\r,
180 MECHAl^ICS OF ENGINEERING,
and the work lost in friction per revolution is
^fR^Tt .\T = ^,7tfRr. . . . . (1)
As the pivot and step become worn, the resultant frictioii*
in the small sectors probably approach the centre; for the
greatest wear occurs first near tlie outer edge, since there the
product ^J)^> is greatest (see § 166). Hence for \r we may more
reasonably put ^\
Exam]jle. — A vertical flat-ended pivot presses its step with
a force of 12 tons, is 6 inches in diameter, and makes 40 revolu-
tions per minute. Required the H. P. absorbed by the friction.
Supposing the pivot and step new, and /"for good lubrication
= 0.07, we have, from eq. (1) {foot-lb. -second),
"Work lost per revolution
= .07 X 24,000 X 6.28 X I • i = 1758.4 ft.-lbs.,
and .*. work per second
= 1758.4 X |-t = 1172.2 ft.-lbs.,
which -i- 550 gives 2.13 H. P. absorbed in friction. If ordi-
nary axle-friction also occurs its effect must be added.
If the flat-ended pivot is hollow, with radii r^ and r^, we may
put ^{i\-\-')\) instead of the fr of the preceding.
It is obvious that the smaller the lever-arm given to the
resultant friction in each sector of the rubbing surface the
smaller the power lost in friction. Hence pivots should be
made as small as possible, consistently with strength.
For a conical pivot and step, Fig. 181, the resultant friction
in each sector of the conical bearing surface has
a lever-arm = f r, about tlie axis A, and a value
>- than for a flat-ended pivot ; for, on account
of the wedge-like action of the bodies, the
pressure causing friction is greater. The sum of
the moments of these resultant frictions about
A is the same as if only two elements of the
cone received pressure (each = iV= ^R -f- sin or). Hence the
FRICTION. 181
moment of friction of the pivot, i.e., the moment of the force
necessary to maintain uniform rotation, is
'^ 3 ' "^ sm or 3 "
4 B
and work lost per revolution = o'^f~ '^v
o sin oi
By making r^ small enough, these values may be made less
than those for a flat-ended pivot of the same diameter = 2r.
In Schiele's " anti-friction" pivots the outline is designed
according to the following theory for securing uniform vertical
wear. Let j!? = the pressure per r— — ^^ — _ —
horizontal unit of area (i.e., Ip jA ip
= -^ -r- horizontal projection of .^ ! ■ L lowing will determine. Consider a semi-
circle of the 2'2vd tree, neglecting its weight.
Fig. 183. The forces holding it in equilib-
rium are tlie tensions ar the two ends (these
are equal, manifestly, the cylinder being
„ smooth ; for tnev are the only two forces
* / 7/1 having moments about c/, and each has the
Fig. 183. smus lever-ariTi^. and the normal pressures,
which are infinite in number, but nave an intensity, p^ per
linear unit, which must be constant along the curve since 8 is
the same at all points. The normal pressure on a single ele-
ment, ds, of the cord is = //Jsr. aiid its JT component =
pds cos 6 — prdd cos ^. Hence S.X = gives
cos BdQ — 2/S' = 0. i.e., rjp\ sin d = 28;
.-in
.'. rp[l — (— 1)] = 26' or z> =
S
(1)
170. Belt on Eough Cylinder. ImDending Slipping. — If fric-
tion is possible between the two bodies, the tension may vary
»k)ug the arc of contact, so that ^also varies, and consequently
Fig. 184.
Fig. 185.
the friction on lui element (^s being =^ds =f{8-^ r)ds, also
varies. If slipping is impendina. the law of variation of the
tension S may be found, as follows ° Fig. 184, in which the
ERICTIWN. 183
impending slipping is toward the left, shows the cord free.
For any element, ds, of the cord, we have, putting 2 (moms,
about 0) = Q (Fig. 185),
{8+ dSy = Sr + dFr ; i.e., dF= dS,
or (see above) dS =f{S -^ r)ds.
But ds = rdO ; hence, after ti-ansfonning,
fde = §. (1)
In (1) the two variables and S are separated ; (1) is there-
fore ready for integration.
fa = loge 8n — l0g« 8, = l0ge[_^J. (2)
Or, by inversion, 8^ef"- — 8n, (3)
the above value in (T) ; corresponding values of the
two tensions may then be found from (5), and from the rela-
tion (see § 150)
{8^-8y = L (64
These new values of the tensions will be found to satisfy the-
condition of no slip, viz.,
(^,:xS'„)<^-(§170).
For leather on iron, ef"" = 2.2 (see example in § lYO), as a.
low value. The belt should be made strong enough to with-
stand 8n safely.
As the belt is more tightly stretched, and hence elongated,,
on the driving than on the following side, it ^' creeps'^ back-
ward on the driving and forward on the driven pulley, so that
the former moves slightly faster than the latter. The loss of
work due to this cause does not exceed 2 per cent with ordinary
belting (Cotterill).
In the foregoing it is evident that the sum of the tensions in
the two sides = G„, i.e., is the same, whether the power is^
being transmitted or not ; and this is found to be true, both in
theory and by experiment, when a tension-weight is not used,
viz., when an initial tension S is produced in the whole belt
before transmitting the power, then after turning on the latter
the sum of the two tensions (driving and following) always
= ^S, since one side elongates as much as the other contracts ;
it being understood that the pulley-axles preserve a constant
distance apart.
172. Rolling Friction. — The few experiments which have-
been made to determine the resistance offered by a level road-
FRICTION. 187
"Way to the uniform motion of a roller or wheel rolling upon it
corroborate approximately the following theory. The word
friction is hardly appropriate in this connection (except when
the roadway is perfectly elastic, as will be seen), but is sanctioned
by usage.
Firsts let the roadway or track be compressible, but inelastic,
O the weight of the roller and its load, and P the horizontal
force necessary to preserve a uniform motion ^ — ~\ — >
(both of translation and rotation). The track / |g \
(or roller itself) being compressed just in h"^ j'"*'^
front, and not reacting symmetrically from \ q[\ ,_^^
behind, its resultant pressure against the //////////mmW^y^-'^'^^^-
roller is not at vertically under the centre, ^^**" ^^^'
but some small distance, OD = h, in front. (The successive
crushing of small projecting particles has the same effect.)
Since for this case of motion the forces have the same relations
as if balanced (see § 124), we may put 2 moms, about D = 0,
.'.Fr=Gh; or, P = ~G (1)
According to Professor Goodman we have the following
values of b, approximately :
Inches.
Iron or steel wheels on iron or steel rails. . 6 = 0.007 to 0.015
" " " " " wood 0.06 " 0.10
" " " " " macadam 0.05 " 0.20
" " " " " soft ground 3.0 "5.0
Pneumatic tires on good road, or asphalt.. 0.02 " 0.022
'' " heavy mud 0.04 " 0.06
Solid rubber tires on good road, or asphalt 0.04
" " heavy mud 0.09 " 0.11
According to the foregoing theory, P, the " rolling fi-iction"
(see eq. (1)), is directly proportional to G, and inversely to the
radius, if h is constant. The experiments of General Morin and
others confirm this, while those of Dupuit, Poiree. and Sauvage
indicate it to be proportional directly to G, and inversely to the
square root of the radius.
188 MECHAlSriCS OF ENGINEEEING.
Although J is a distance to be expressed in linear units, and
not an abstract number like the /"and f for sliding and axle-
friction, it is sometimes called a " coefficient of rolling fric-
tion." In eq. (1), h and r should be expressed in the same
unit.
Of course if P is applied at the top of the roller its lever-
arm about D is 2r instead of r^ with a corresponding change
in eq. (1).
With ordinary railroad cars the resistance due to axle and
rolling frictions combined is about 8 lbs. per ton of weight on
a level track. For wagons on macadamized roads & = |- inch,
but on soft ground from 2 to 3 inches.
Secondly^ when the roadway is jperfectly elastic. This is
chiefly of theoretic interest, since at first sight no force would
be considered necessary to maintain a uniform rolling motion.
But, as the material of the roadway is compressed under the
roller its surface is first elongated and then recovers its former
state ; hence some rubbing and consequent sliding friction must
occur. Fig. 189 gives an exaggerated view of the circum-
stances, P being the horizontal force applied at the centre
necessary to maintain a uniform motion. The roadway (rub-
ber for instance) is heaped up both in front and behind the
roller, being the })oint of greatest pressure and elongation
of the surface. The forces acting are ^, P^ the normal
pressures, and the frictions due to them, and must form a
balanced system. Hence, since G and P^ and also the normal
pressures, pass through C^ the resultant of the frictions must
also pass through G\ therefore the frictions, or tangential
actions, on the roller must be some forward and some backward
(and not all in one direction, as seems to be asserted on p. 260
of Cotterill's Applied Mechanics, where Professor Reynolds'
FRICTION. 189
explanation is cited). The resultant action of the roadwaj
upon the roller acts, then, through some point J9, a distance
OD = h ahead of (9, and in the direction DC, and we have as
before, with 2? as a centre of moments,
Pr=Gh, or P=-G.
If rolling friction is encountered above as
well as helow the rollers, Fig. 190, the
student may easily prove, by considering
three separate free bodies, that for uniform
motion
p = '-^<^'
Fia. 190.
where h and h^ are the respective " coefficients of rolling fric-
tion ' ' for the upper and lower contacts. (See Kent's " Pocket-
Book for Mechanical Engineers'' for "friction-rollers,'^
"ball-bearings," and "roller-bearings."
Exa/mjple 1. — If it is found that a train of cars will move
uniformly down an incline of 1 in 200, gravity being the only
working force, and friction (both rolling and axle) the only
resistance, required the coefficient, f\ of axle-friction, the
diameter of all the wheels being 2f = 30 inclies, that of the
journals 2a = S inches, taking h = 0.02 inch for the rolling
friction. Let us use equation (XYI.) (§ 142), noting that while
the train moves a distance s measured on the incline, its weight
1 A
G does the work G ^z s, the rolling friction — G (at* the axles)
has been overcome through the distance s, and the axle-friction
(total) through the (relative) distance — sin the journal boxes j
whence, the change in kinetic energy being zero,
1 ^ b ^ a
Gs
-^Ga-^Gs--fas = 0.
Gs cancels out, the ratios h : r and a : r are = tAtt ^"<5 iV*
respectively (being ratios or abstract numbers they have the
* That is, the ideal resistance, at centre of axles and || to the incline, equiV'
alent to actual rolling resistance.
190 MECHANICS OF ENGINEERING.
same numerical values, whether the inch or foot is used), an (J
solving, we have
/ = 0.05 - 0.0133 = 0.036.
Examjple 2. — How many pounds of tractive effort per ton
of load would the train in Example 1 require foi- uniform mo-
tion on a level track ? Ans. 10 lbs.
173. Eailroad Brakes.* — During the uniform motion of a
railroad car the tangential action between the track and each
wheel is small. Thus, in Example 1, just cited, if ten cars of
eight wheels each make up the train, each car weighing 20 tons,
the backward tangential action of the rails upon each wheel is
only 25 lbs. When the brakes are applied to stop the train
this action is much increased, and is the only agency by which
the rails can retard the train, directly or indirectly : directly^
when the pressure of the brakes is so great as to prevent the
wheels from turning, thereby causing them to "skid" (i.e.,
slide) on the rails ; indirectly^ when the brake-pressure is of
such a value as still to permit perfect rolling of the wheel, in
which case the rubbing (and heating) occurs between the brake
and wheel, and the tangential action of the rail has a value
equal to or less than the friction of rest. In the first case,
then (skidding), the retarding influence of the rails is the/r^c-
iion of motion between rail and wheel; in the second, a force
which may be made as great as the friction of rest between rail
and wheel. Hence, aside from the fact that skidding produces
objectionable flat places on the wheel-tread, the brakes are
more effective if so applied that skidding is impending, but
not actually produced ; for the friction of rest is usually greater
than that of actual slipping (§160). This has been proved
experimentally in England. The retarding effect of axle and
rolling friction has been neglected in the above theory.
Example 1. — A twenty-ton car with an initial velocity of 80
feet per second (nearly a mile a minute) is to be stopped on a
level within 1000 feet ; required the necessary friction on each
of the eight wheels.
Supposing the wheels not to skid, the friction will occur
* See statement on p. 168, as to diminution of the coefficient / with
; speed.
iJ'RICTION. 191
between the brakes ana wheels, and is overcome through the
(relative) distance 1000 feet. Eq. (XYI.), § 14:2, gives (foot-
Ib.-second system)
1 40000
- 8i^'X 1000 = - 1 ^^(80)^
from which F { = friccion at circumference of each wheel)
= 496 lbs.
Note. — This result of 496 lbs. must be looked upon as only an average
value. For a given pressure, A'^, of brake-shoe on wheel-rim on account
of the variation of the coefficient /' with changing speed (see p. 168)
the friction will be small at first and gradually increase. This same
remark applies to Examples 3 and 4, also.
1/
Examjple 2.— Supoose sisiddins^ to be impending in the fore-
going, and tlie coefhcient of friction of rest (i.e., impending
slipping) between ran ana wneel to be/'=0,20o In what
-distance will the. car oe stopped? Ans. 496 ft
Example 3. — Supoose tne car in Example 1 to be on an up»
grade of 60 feet to tne mile. Qn applying eq. (XVI.) here,
the weight 20 tons win enter as a resistance.) Ans. 439 lbs.
Example 4. — In Ji,xample 3. consider all four resistances,
viz., gravitj^, rolling triction. and brake and axle frictions, the
distance being 1000 ft., and \F the unknown quantity.
(Take the wheel dimensions of p. 189.) Ans. 414 lbs.
174. Friction of Car-journals in Brass Bearings. — :(Prof. J.
E. Denton, in Vol. xii Transac. Am. See, Mecli. Engs.,
p. 405; also Kent's Pocket-Book.) A new brass dressed
with an emery wheel, loaded with 5000 lbs., may have an
actual bearing surface on the journal, as shown by the polish
of a portion of the surface, of only one sq. inch. "With this
pressure of 5000 Ibs./sq.in. the coefficient of friction may be
0.060 and the brass may be overheated, scarred, and cut ; or,
on the contrary, it may wear down evenly to a smooth bearing,
giving a highly polished area of contact of 3 sq. in., or more,
inside of two hours of running, gradually decreasing the
pressure per sq. in. of contact, and showing a coefficient of
friction of less than 0.005. A reciprocating motion in the
direction of the axis is of importance in reducing the friction.
With such polished surfaces any oil will lubricate and the
192 MECHANICS OF ENGINEERING.
coefficient of friction then depends on the viscosity of the oil.
With a pressure of 1000 lbs, per sq. in. , revolutions from 170
to 320 per min., and temperature of 75° to 113° Fahr., with
both sperm and parraffine oils, a coefficient as low as 0.0011
has been obtained, the oil being fed continuously by a pad.
175. Well Lubricated Journals. Laws of Friction. — In the
Proc. Inst. Civ. Engs. for 1886 (see also Engineering News
for Mar. 31, April 7 and 14, 1888) Prof. Goodman presents
the conclusions arrived at by him as to the laws of friction of
well lubricated journals as based on the experiments made by
Thurston, Beauchamp Tower, and Stroudley. They are as
follows :
1 . The coefficient friction with the surfaces efficiently lubri-
cated is from ^ to ^ that for dry or scantily lubricated surfaces.
2. The coefficient of friction for moderate pressures and
speeds varies approximately inversely as the normal pressure ;
the frictional resistance varies as the area- in contact, the
normal pressure remaining constant.
3. At very low journal speeds the coefficient of friction is
abnormally high, but as the speed of sliding increases from
about 10 to 100 ft. per min. the friction diminishes; and
again rises when that speed is exceeded, varying approximately
as the square root of the speed.
4. The coefficient of friction varies approximately inversely
as the temperature, within certain limits, viz., just before
abrasion takes place.
In one of Mr. Tower's experiments it was found that when
the lubrication was made by a pad under the journal (which
received pressure on its upper surface) the coefficient was
some seven times as large as when an ' ' oil bath, ' ' or copious
supply of oil, was provided; (0.0090 as against 0.0014).
176. Rigidity of Ropes. — If a rope or wire cable passes over
a pulley or sheave, a force J-* is required on one side greater
than the resistance Q on the other for uniform motion, aside
from axle-friction. Since in a given time botli I^ and Q
describe the same space s, if ^ is > Q, then I^sis > Qs, i.e.,
the work done by i^ is > than that expended upon Q. This
is because some of the work J*s has been expended in bending
the stiff rope or cable, and in overcoming friction between the
strands, both where the rope passes upon and where it leaves
FRICTION.
198
the pulley. With hemp ropes, Fig. 191, the material being
nearly inelastic, the energy spent in bending it on at D is
nearly all lost, and energy must also be spent in straightening
Fig. 191.
it at E\ but with a wire rope or cable some of this energy is
restored by the elasticity of the material. The energy spent
in friction or rubbing of strands, however, is lost in both cases.
The iigure shows geometrically why P must be > ^ for a
uniform motion, for the lever-arm, a, of P is evidently < h
that of Q. If axle-friction is also considered, we must have
Pa=qb^f{P-^Q)r,
r being the radius of the journal.
Experiments with cordage have been made by Prony, Cou-
lomb, Eytelwein, and Weisbach, with considerable variation in
the results and forraulse proposed. (See Coxe's translation of
vol. i., "Weisbach's Mechanics.)
With pulleys of large diameter the effect of rigidity is very
slight. For instance, Weisbach gives an example of a pulley
five feet in diameter, with which, Q being = 1200 lbs., P
= 1219. A wire rope f in. in diameter was used. Of this
difference, 19 lbs., only 5 lbs. was due to rigidity, the remainder,
14 lbs., being caused by axle-friction. When a hemp-rope 1.6
inches in diameter was substituted for the wire one, P — ^=27
lbs., of which 12 lbs. was due to the rigidity. Hence in one
case the loss of work was less than \ of \%. in the other about
1^, caused by the rigidity. For very small sheaves and thick
ropes the loss is probably much greater.
13
194 MECHANICS OF ENGINEERING.
/Vl'^, Miscellaneous Examples. — Example 1. Tiie end of a
\ /shaft 12 inches ill diameter and making 50 revolutions per min-
1/ ute exerts against its bearing an axial pressure of 10 tons and
/ a lateral pressure of 40 tons. With /" ^y = 0.05, required
the H. P. lost in friction. Ans. 22.2 H. P.
Example 2. — A leather belt passes over a vertical pulley,
covering half its circumference. One end is held by a spring
balance, which reads 10 lbs. vi'hile the other end sustains a
vreight of 20 lbs., the pulley making 100 revolutions per min-
ute. Required the coefficient of friction, and the H. P. spent
in overcoming the friction. Also suppose the pulley turned
in the other direction, the weight remaining the same. The
diameter of the pulley is 18 inches. . {f = 0.22 ;
^^' I 0.142 and .284 H. P.
Example 3. — A grindstone with a radius of gyration = 12
inches has been revolving at 120 revolutions per minute, and
at a given instant is left to the influence of gravity and axle
friction. The axles are 1|- inches in diameter, and the wheel
makes 160 revolutions in coming to rest. Required the coeffi-
(jient of axle-friction. (Average.) Ans. / = 0.039.
Example 4. — A board A, weight 2 lbs., rests horizontally on
another B:, coefficient of friction of rest between them being
f = 0.30. B is now moved horizontally with a uniformly
accelerated motion, the acceleration being = 15 f set per " square
seco'id ;" will A keep company with it, or not ? Ans, " if o."
y
V
STRENGTH OF MATERTALSo
[Or Mechanics of Materials] =
CHAPTEE I.
ELEMENTARY STRESSES AND STRAINS.
178. Beformation of Solid Bodies. — In tlie preceding por-
tions of this work, what was called technically a " rigid
body," was supposed incapable of changing its form, i.e.,
the positions of its particles relatively to each other, under
the action of any forces to be brought upon it. This sup-
position was made because the change of form which must
actually occur does not appreciably alter the distances,
angles, etc., measured in any one body, among most of
the pieces of a properly designed structure or machine.
To show how the individual pieces of such constructions
should be designed to avoid undesirable deformation or
injury is the object of this division of Mechanics of En-
gineering, viz., the Strength of Materials,
,^6
'€5.
D
Fis. 193. § 178.
AS perhaps tne simplest instance of the deformation or
distortion of a solid, let us consider the case of a prismatic
rod in a state of tension, Eig. 192 (eye-bar of a bridge
195
196 MECHANICS OF EI>J GINBERmG.
truss, e.g.). The pull at each end is P, and the body is
said to be under a tension of P (lbs., tons, or other unit),
not 2P. Let ABGD be the end view of an elementary
parallelopiped, originally of square section and with faces
at 45° with the axis of the prism. It is now deformed, the
four faces perpendicular to the paper being longer"^ than
before, while the angles BAD and BCD, originally right
angles, are now smaller by a certain amount d, ABC and
ADG larger by an equal amount d. The element is said
to be in a state of strain, viz.: the elongation of its edges
(parallel to paper) is called a tensile strain, while the alter-
ation in the angles between its faces is called a shearing
strain, or angular distortion (sometimes also called a slid-
ing, or tangential, strain, since BG has been made to slide,
relatively to AD, and thereby caused the change of angle).
[This use of the word strain, to signify change of form and
not the force producing it, is of recent adoption among
many, though not all, technical writers.]
179. Strains. Two Kinds Only. — Just as a curved line may
be considered to be made up of small straight-line ele-
ments, so the substance of any solid body may be consid-
ered to be made up of small contiguous parallelopipeds,
whose angles are each 90° before the body is subjected to
the action of forces, but which are not necessarily cubes.
A line of such elements forming an elementary prism is
sometimes called a> fibre, but this does not necessarily imply
a fibrous nature in the material in question. The system
of imaginary cutting surfaces by which the body is thus
subdivided need not consist entirely of planes ; in the sub-
ject of Torsion, for instance, the parallelopipedical ele-
ments considered lie in concentric cylindrical shells, cut
both by transverse and radial planes.
Since these elements are taken so small that the only
possible changes of form in any one of them, as induced
by a system of external forces acting on the body, are
* When a is nearly 0° (or 90°) BG and AD (or AB and DG) are shorter
than before, on account of lateral contraction; see § 193.
ELEMENTARY STRESSES, ETC, 197
elongations or contractions of its edges, and alteration of
its angles, there are but two kinds of strain, elongation
(contraction, if negative) and shearing.
180. Distributed Forces or Stresses. — In the matter preced-
ing this chapter it has sufficed for practical purposes to
consider a force as applied at a point of a body, but in
reality it must be distributed over a definite area ; for
otherwise the material would be subjected to an infinite
force per unit of area. (Forces like gravity, magnetic at-
traction, etc., we have already treated as distributed over
the mass of a body, but reference is now had particularly
to the pressure of one body against another, or the action
•of one portion of the body on the remainder.) For in-
stance, sufficient surface must be provided between the
end of a loaded beam and the pier on which it rests to
avoid injury to either. Again, too small a wire must not
be used to sustain a given load, or the tension per unit
of area of its cross section becomes sufficient to rupture
it.
Stress is distributed force, and its intensity at any point
of the area is
• o e (1)
"where dF is a small area containing the point and dP the
force coming upon that area. If equal dP^s (all parallel)
act on equal dF'soi a plane surface, the stress is said to
be of uniform intensity, which is then
i>=p . . . . (2)
where P=-= total force and ^the total area over which it
acts. The steam pressure on a piston is an example of
stress of uniform intensity.
X98 MECHANICS OF ENGINEEKING.
For example, if a force P= 28800 lbs, is uniformly dis-
tributed over a plane area of ^=72 sq. inches, or ^ of a
sq. foot, the intensity of the stress is
28800 ,^^,, . ,
p= =400 lbs. Der sq. inch,
(or jp = 28800^ >^ =57600 lbs. per sq. foot, or p=14400-j'
^=28.8 tons per sq. ft,, etc...
181. Stresses on an Element : of Two Kinds Only. — When a
solid body of any material is in eauiiibrium under a sys-
tem of forces which do not rupture it. not only is its shape
altered (i.e. its elements are strained), and stresses pro-
duced on those planes on which the forces act, but other
stresses also are induced on some or all internal surfaces
which separate element from element, f over and above the
forces with which the elements mav have acted on each
other before the application of the external stresses or
" applied forces "). So long as the whole solid is the "free
body " under consideration, these internal stresses, being
the forces with which the portion on one side of an imag-
inary cutting plane acts on the portion on the other side,
do not appear in any equation of eauiiibrium (for if intro-
duced they would cancel out); but if we consider free a
portion only, some or all of whose bounding surfaces are
cutting planes of the original bodv. the stresses existing
on these planes are brought into the eauations of equilib-
rium.
Similarly, if a single element of the body is treated by
itself, the stresses on all six of its faces, together with its
weight, form a balanced system of forces, the body being
supposed at rest.
FiS. 138.
ELEMENTAE.T STRESSES, ETC.
199
As an example of internal stress, consider again the case
of a bar in tension ; Fig. 193 shows tlie whole bar (or eye-
bar) free, the forces P being the pressures of the pins in.
the eyes, and causing external stress (compression here)
on the surfaces of contact. Conceive a right section made
through BS, far enough from the eye, (7, that we may con-
sider the internal stress to be uniform * in this section, and
consider the portion BSG as a free body, in Fig. 194. The
stresses on R8, now one of the bounding surfaces of the
free body, must be parallel to P, i.e., normal to B8;
(otherwise they would have components perpendicular to
P, which is precluded by the necessity oi lY being = 0,
and the supposition of uniformity.) Let .^ = the sec-
FlG. 194,
Fig. 195.
tional area RS, and p = the stress per unit of area ; then.
P
IX-= gives P= Fp, i.e., p=
F
.(2).
The state of internal stress, then, is such that on planes
perpendicular to the axis of the bar the stress is tensile and
normal (to those planes). Since if a section were made
oblique to the axis of the bar, the stress would still be
parallel to the axis for reasons as above, it is evident that
on an oblique section, the stress has components both nor-
mal and tangential to the section, the normal component
being a tension.
* As will be shown later (§ 295) the line of the two P's in Fig. 193 must
pass through the centre of gravity of the cross-section RS (plane figure) of
the bar, for the stress to be uniform over the section.
200
MECHANICS OF EISTGINEERLNG.
The presence of the tangential or shearing stress in ob-
lique Sections is rendered evident by considering that if an
oblique dove-tail joint were cut in the rod, Fig. 195, the
shearing stress on its surfaces may be sufficient to over-
come friction and cause sliding along the oblique plane.
If a short prismatic block is under the compressive ac-
tion of two forces, each =P and applied centrally in one
base, we may show that the state of internal stress is the
same as that of the rod under tension, except that the nor-
mal stresses are of contrary sign, i.e., compressive instead
of tensile, and that the shearing stresses (or tendency to
slide) on oblique planes are opposite in direction to those
in the rod.
Since the resultant stress on a given internal plane of a
body is fully represented by its normal and tangential
components, we are therefore justified in considering but
iwo kinds of internal stress, normal or direct, and tangen-
tial or shearing.
182. Stress on Obliq[ue Section of Rod in Tension, — Consider
free a small cubic element whose
edge =a in lengthy it has two
faces parallel to the paper, being
taken near the middle of the rod
in Fig. 192. Let the angle which
the face AB, Fig. 196, makes with
the axis of the rod be = a. This
angle, for our present purpose, is
considered to remain the same
while the two forces P are acting,
as before their action. The re-
sultant stress on the face AB hav-
ing an intensity p=P-h-F, (see eq.
2) per unit of transverse section
of rod, is = jp (a sin a) a. Hence
its component normal to AB is
■pa^ sin^ a ; and the tangential or shearing component along
AB '=*pa^ sin a cos a. Dividing by the area, a^, we have
the following :
For a rod in simple tension we have, on a plane making
an angle, a, with the axis :
ELEMENTARY STRESSES, ETC. 201
a Normal Stress =p Bin? a per unit of area . . (1)
and a Shearing Stress =p sin a cos a per unit of area . (2)
" Unit of area " here refers to the oblique plane in ques-
tion, while p denotes the normal stress per unit of area of
a transverse section, i.e., when a=90°. Fig. 194.
The stresses on CD are the same in value as on AB,
while for BG and AD wq substitute 90° — a for a. Fig.
197 shows these normal and shearing stresses, and also,
much exaggerated, the strains or change of form of the
element (see Fig. 192).
182a, Eelation between Stress and Strain. — Experiment
shows that so long as the stresses are of such moderate
value that the piece recovers its original form completely
when the external forces which induce the stresses are re-
moved, the following is true and is known as Hoohe's Law
(stress proportional to strain). As the forces P in Fig.
193 (rod in tension) are gradually increased, the elonga-
tion, or additional length, of BK increases in the same
ratio as the normal stress, p, on the sections BS and KI^^
per unit of area [§ 191].
As for the distorting effect of shearing stresses, considei
in Fig. 197 that since
p sin a cos a = p cos (90° — a) sm (90° — a)
the shearing stress per unit of area is of equal value on all
four of the faces (perpendicular to paper) in the elementary
block, and is evidently accountable for the shearing strain,
i.e., for the angular distortion, or difference, d, between
90° and the present value of each of the four angles. Ac-
cording to Hooke's Law then, as P increases within tlvg
limit mentioned above, d varies proportionally to
p sin a cos a, i.e. to the stress.
182b. Example. — Supposing the rod in question were of
a kind of wood in which a shearing stress of 200 lbs. per
sq. inch along the grain, or a normal stress of 400 lbs. per
8q. inch, perpendicular to a fibre-plane will produce rup-
ture, required the value of a the angle which the grain
must make with the axis that, as P increases, the danger
of rupture from each source may be the same. This re-
202 MECHANICS OE ENGIl>rEBEI2fG.
quires that 200:400::p sin a cos a :p sin^a, i.e. tan. a must
= 2.000.-.a=63i^°. If the cross section of the rod is 2 sq.
inches, the force P at each end necessary to produce rup-
ture of either kind, when a=63^°, is found by putting
p sin a cos ^=^00. '.^=500.0 lbs. per sq. inch. "Whence, since
p=P-^F, P=1000 lbs. (Units, inch and pound.)
183. Elasticity is the name given to the property which
most materials have, to a certain extent, of regaining their
original form when the external forces are removed. If
the state of stress exceeds a certain stage, called the Elastic
Limit, the recovery of original form on the part of the ele-
ments is only partial, the permanent deformation being
called the Set.
Although theoretically the elastic limit is a perfectly defi-
nite stage of stress, experimentally it is somewhat indefi-
nite, and is generally considered to be reached when the
permanent set becomes well marked as the stresses are in-
creased and the test piece is given ample time for recovery
in the intervals of rest.
The Safe Limit of stress, taken well within the elastic
limit, determines the working strength or safe load of the
piece under consideration. E.g., the tables of safe loads
of the structural steel beams for floors, made by the Cambria
Steel Co. , at Johnstown, Pa, , are computed on the basis that
the greatest normal stress (tension -or compression) occurring
on any internal plane shall not exceed 16,000 lbs. per sq. inch;
and, again, by the building laws of Philadelphia, the greatest
shearing stress to be permitted in ' ' web plates " of " mild
steel" is 8750 lbs. /in. 2
The tJltimate Limit is reached when rupture occurs.
184. The Modulus of Elasticity (sometimes called co-efficient
of elasticity) is the number obtained by dividing the stress
per unit of area by the corresponding relative strain.
Thus, a rod of wrought iron ^ sq. inch sectional area
being subjected to a tension of 2^ tons =5,000 lbs., it is
ELEMENTARY STRESSES, ETC. 203
iound that a length whicli was six feet before tension is
»= 6.002 ft. during tension. The relative longitudinal strain
or elongation is then= (0.002)-^6= 1 : 3,000 and the corres-
ponding stress (being the normal stress on a transverse
plane) has an intensity of
i?t=P^i^= 5,000-^ 1^=10,000 lbs., per sq. inch.
Hence by definition the modulus of elasticity is (for ten-
sion), if we denote the relative elongation by s,
Bt-=Pt'^ £=10,000^ g-^ =30,000,000 lbs. per sq. inch, (the
sub-script " t " refers to tension).
It will be noticed that since £ is an abstract number, Et
is of the same quality as p^, i.e., lbs. per sq. inch, or one di-
mension of force divided by two dimensions of length.
(In the subject of strength of materials the inch is the
most convenient English linear unit, when the pound is
the unit of force ; sometimes the foot and ton are used to-
gether.)
The foregoing would be called the modulus of elasticity
of lorought iron in tension in the direction of the fibre, as
given by the experiment quoted. • But by Hooke's Law p
and £ vary together, for a given direction in a given ma-
terial, hence ivithin the elastic limit E is constant for a given
direction in a given material. Experiment confirms this
approximately.
Similarly, the modulus of elasticity for compression E^
in a given direction in a given material may be determined
by experiments on short blocks, or on rods confined lat-
erally to prevent flexure.
As to the modulus of elasticity for shearing, E^, we
divide the shearing stress per unit of area in the given
direction by (? (in radians) the corresponding angular strain
or distortion; e.g., for an angular distortion of 0.10° or
,^ = .001T4, and a shearing stress of 15,660 lbs. per sq. inch,
we have £;=^^ = 9,000,000 lbs. per sq. inch.
204 MECHANICS OF ENGINEERING.
184a. Young's Modulus is a name frequently given toEf and
Ec, it being understood that in the experiments to determine
these moduli the elastic limit is not passed, and also that the
rod or prism tested is not subjected to any stress on the sides.
See p. 507.
185. Isotropes. — This name is given to materials which
are homogenous as regards their elastic properties. In
such a material the moduli of elasticity are individually
the same for all directions. E.g., a rod of rubber cut out
of a large mass will exhibit the same elastic behavior when
subjected to tension, whatever its original position in the
mass. Fibrous materials like wood and wrought iron are
not isotropic ; the direction of grain in the former must
always be considered. The " piling " and welding of nu-
merous small pieces of iron prevent the resultant forging
from being isotropic.
186. Resilience refers to the potential energy stored in a
body held under external forces in a state of stress which
does not pass the elastic limit. On its release from con-
straint, by virtue of its elasticity it can perform a certain
amount of work called the resilience, depending in amount
upon the circumstances of each case and the nature of the
material. See § 148.
187. General Properties of Materials. — In viev/ of some defi-
nitions already made we may say that a material is ductile
when the ultimate limit is far removed from the elastic
limit ; that it is brittle like glass and cast iron, when those
limits are near together. A small modulus of elasticity
means that a comparatively small force is necessary to
produce a given change of form, and vice versa, but implies
little or nothing concerning the stress or strain at the
elastic limit ; thus Weisbach gives E^, lbs. per sq. inch for
wrought iron = 28,000,000= double the E^ for cast iron
while the compressive stresses at the elastic limit are the
same for both materials (nearly).
ELEMENTARY STRESSES, ETC.
205
188. Element with Normal Stress on Sides as well as on End-Faces.
Ellipse of Stiess.— In Fig. 193, p. 198, the parallelopiped RKNS is sub-
jected to stress on the two end-faces only. Let us now consider a small
square-cornered element of material subjected to a normal stress p^
(tension) on the two vertical end-faces, while on the horizontal side faces
there acts a normal (also tensile) stress of pj Ibs./in.^; (but no stress
on the vertical side ^- ^n'---^ /
faces). In Fig. 197a '''' ' ^""- '^
is shown, as a free
body in equilibrium,
a triangular prism
ABC, which is the
upper right-hand
half of such an ele-
ment; obtained by
passing the cutting
plane AC along a
diagonal of the side
plane (plane of
paper) on which
there is no stress,
and 1 to it. The
angle 6 may have
any value and it
is desired to deter-
mine the unit stress "iG- 197a.
5o induced on the oblique plane AC by the normal stresses Pi and Pz
acting respectively on the end face BC and on the side face AB. The
unit stress q^ on the face AC is not 1 to that face but makes with it
some angle ^. Let AB = h inches, BC = n in., and AC = c in.; each
of the rectangular areas having a common dimension, =d in., T to
the paper. Then the total (oblique) stress on face AC is q^cd lbs., that
on AB is P'pd, and that on BC is p-jid lbs. Since the total stress on AC
is the anti-resultant of the other two, and these are T to each other,
we have
{,q^cdy={,PTndy+{p^hdy; i.e., ?o^ = fpiyj + (pj-
But, since ?i^c = sin d, and fe-^c = cos 6, this may be written
q-'={p,smey+{p,coBdy (i)
Eq. (1) gives the magnitude of q^ for any value of angle d; but both
position and magnitude are best shown by a geometrical construction.
being any point on AC, draw a circle with center at and radius,
OH^, equal by scale to the unit stress p^. Similarly, with radius OH 2,
equal (on same sea'"") to the unit stress P2, draw the circle H2E2. Through
draw EfiN normal to the face AC on which the stress 50 is to be deter-
mined, and note the intersections E^ and £'2 (both on left of 0) with
the two circles respectively. A vertical line through E^ and a hori-
zontal through E2 intersect at some point m. Cm is the magnitude
and position of the stress q^; since mD2 = EiDi = pi sin 0, and ODj =
P2COSO; hence from eq. (1) Om=5o.
206 MECHANICS OF ENGINEERING.
T he po int m is a point in an ellipse whose semi-principal axes are OH^
and OH 2, i.e., p^ and pj. This ellipse is called the Ellipse of Stress;
Om being a semi-diameter, determined in the way indicated. (Similarly,
if the elementary right parallelepiped is subjected to the action of three
normal stresses, Pi, p2, and p^, on all three pairs of faces, respectively,
the unit stress on any oblique plane is a semi-diameter of an Ellipsoid
of Stress).
The unit shearing stress on the oblique face AC is qs=^qo cos ,u; and
the unit normal stress is q=qo sin /;.
In case the normal stress P2 on the face AB were compressive, p^ being
tensile, a horizontal would be drawn through E'2 on the circle of radius
OH2, instead of through E2, to meet the vertical through E^, and would
thus determine Om', instead of Om, as the stress on AC. If, in such
a case, P2 were numerically equal to p^, and d were 45°, go would = Pi = pj,
and would lie in the surface AC (pure shear; compare with Exam. 5,
p. 242). With Pi = P2, and both tensUe, or both compressive, 50 would
be equal to Pi, =P2, for all values of d.
189. Classification of Cases. — Althougli in almost any case
whatever of the deformation of a solid body by a balanced
system of forces acting on it, normal and shearing stresses
are both, developed in every element which is affected at
all (according to the plane section considered,) still, cases
where the body is prismatic, and the external system con-
sists of two equal and opposite forces, one at each end of
the piece a,nd directed away from each other, are commonly
called cases of Tension; (Fig, 192); if the piece is a short
prism with the same two terminal forces directed toward
each other, the case is said to be one of Compression ; a case
similar to the last, but where the prism is quite long
(" long column "), is a case of Flexure or bending, as are also
most cases where the " applied forces " (i.e., the external
forces), are not directed along the axis of the piece. Rivet-
ed joints and " pin-connections " present cases of Shearing;
a twisted shaft one of Torsion. When the gravity forces
due to the weights of the elements are also considered, a
combination of two or more of the foregoing general cases
may occur.
In each case, as treated, the principal objects aimed at
are, so to design the piece or its loading that the greatest
stress,* in whatever element it may occur, shall not exceed
a safe value ; and sometimes, furthermore, to prevent too
great deformation on the part of the piece. The first ob-
ject is to provide sufficient strength; the second sufficient
stiffness.
* See § 405b for mention of the "elongation theory" of safety. This
is based on considerations of strain, or deformation, instead of stress.
TMNSION.
207
te:nsion.
191. Hooke's Law by Experiment. — As a typical experiment
in the tension of a long rod of ductile metal sncli as
wrought iron and the mild steels, the following table is quot-
ed from Prof. Cotterill's " Applied Mechanics." The experi-
ment is old, made by Hodgkinson for' an English Railway
Commission, but well adapted to the purpose. From the
great length of the rod, which was of wrought iron and
0.517 in. in diameter, the portion whose elongation was
observed being 49 ft. 2 in. long, the small increase in length
below the elastic limit was readily measured. The succes-
sive loads were of such a value that the tensile stress
p=P^F, or normal stress per sq. in. in the transverse
section, was made to increase by equal increments of 2657.5
lbs. per sq. in., its initial value. After each application of
load the elongation was measured, and after the removal
of the load, the permanent set, if any.
Table of Elongations of a Wrought Iron Rod, of a
Length = 49 Ft. 2 In.
p
X
JA
e^X^l
X'
Load (lbs
square ii
. per Elongation,
ich.) (inches.)
Increment
of
Elongation.
s, the relative
elongation, (ab-
stract number.)
Permanent
Set.
(inches.)
1X266
7.5 .0485
.0485
0.000082
2X '
. 1095
.061
.000186
3X '
. 1675
.058
.000283
0.0015
4X '
.224
.0565
.000379
.002
5X '
.2805
.0565
.000475
.0027
6X '
.337
.0565
.000570
.003
7X '
.393
.056
.004
8X '
.452
.059
.000766
.0075
9X '
.5155
.0635
.0195
lOX '
.598
.0825
.049
IIX '
.760
.162
.1545
12X '
1.310
.550
.667
etc.
208
MECHANICS OF E]SrGIl!fEEEIl!fG.
Referring now to Fig. 198, the notation is evident. P
is the total load in any experiment, F the cross section of
the rod ; hence the normal stress on the transverse section
is p=P-r-F. When the loads are increased by equal in-
crements, the corresponding increments of the elongation
a should also be equal if Hooke's law is true. It will be
noticed in the table that this is very nearly true up to the
8th loading, i.e., that JX, the difference between two con-
secutive values of }., is nearly constant. In other words the
proposition holds good ;
if P and Pi are any two loads below the 8th, and X and ki
the corresponding elongations.
The permanent set is just perceptible at the 3d load, and
increases rapidly after the 8th, as also the increment of
elongation. Hence at the 8th load, which produces a ten-
sile stress on the cross section of j9= 8x2667.5= 21340.0
lbs. per sq. inch, the elastic limit is reached.
As to the state of stress of the individual elements, if
we conceive such sub-division
of the rod that four edges of
each element are parallel to the
axis of the rod, we find that it
is in equilibrium between two
normal stresses on its end faces
''^^ (Fig. 199) of a value ^pdF==
{P^F)dF where dF is the hor-
izontal section of the element.
If dx was the original length,
and dX the elongation produced by pdF, we shall have,
since all the dx's of the length are equally elongated at the
dX X
same time, w" ^ T
where Z = total (original) length. But dX^dx is the relative
elongation e, and by definition (§ 184) the Modulus of Elas-
ticity for Tension, Ei, = jp^e, {Young's Modulus, § 184a).
TENSION. 209
.'.E.=-4rr or E,=^^ .... (1)
Jblq. (1) enables us to solve problems involving the elonga-
tion of a prism under tension, so long as the elastic limit
is not surpassed.
The values of E^ computed from experiments like those
just cited should be the same for any load under the elas-
tic limit, if Hooke's law were accurately obeyed, but in
reality they differ somewhat, especially if the material
lacks homogeneity. In the present instance (see Table)
we have from the
2d Exper. ^=^-^£=28,680,000 lbs. per sq. in.
5th " Ec= " =28,009,000
8th " ^t= " =27,848,000
192. Stress-Strain Diagrams. — If the relative elongations
or "strains " (s) corresponding to a series of values of the
tensile unit-stresses (p) (Ibs./in.^) to which a rod of metal
has been subjected in a testing machine, are plotted as
abscissae, and the unit-stresses themselves (p) as ordinates,
we have in the curve joining these points a useful graphic,
representation of the results of experiment.
Fig. 200 shows some of these curves, giving average re-
sults for the principal "ferrous " metals. On the left, in
(I), the scale adopted (horizontal) for the "strain " (e) or
"unit-elongation " is one hundred times as great as that
used in the right-hand diagram, (II) ; while the vertical
scale for stress (p) in (I) is only twice as great as that in
' (II) . The change of form within the elastic limit is so
small compared with that beyond, that this difference in
scale is quite necessary in order that diagram (I) may show
what occurs within the elastic limit and a Httle beyond.
Diagram (II) shows the remainder of the curves of wrought
iron and soft steel, up to the point of rupture.
We have here the means of comparing the properties of the
four typical metals represented, as to elasticity and tenacity.
Up to the respective elastic hmits, B, B' , B" , and B'" , stress
is fairly proportional to strain, and a straight line is the result;
210
MECHANICS OF ENGINEERING.
the "true elastic limit " being regarded as the point where
such proportionahty ceases. In the case of wrought iron and
soft steel there is a point Y, called the "yield point," a little
above the true elastic limit, and sometimes called the "apparent
elastic limit/' or "commercial elastic limit/' immediately be-
yond which further slight increments of stress produce rela-
tively great increments of strain, permanent set becoming
very marked; i.e., the part YD of the curve is almost hori-
zontal. Beyond D the curve rises again, more steeply, but
just before rupture [see (II)] may descend somewhat; since,
50,000 r r--y^ — ^
40,000
10,000
Ibs./iv}
Fig. 200.
on account of the lateral contraction mentioned in the next
paragraph, here plotted, the stress being computed by dividing
the total pull by the original sectional area, is less toward
rupture than at stages closely preceding.
If at any point beyond the elastic limit, as at C (see curve
for wrought iron) in (I), the stress be gradually removed, the
relations of stress and strain during this gradual diminution
of stress, are shown by the straight line CC. The position of
the point C indicates that there is in the rod (now under no
stress) a permanent set, or relative elongation, of £ = 0.0015,
or 15 parts in 10,000, an elastic recovery having occurred from
0.0028 to 0.0015 (see horizontal scale).
Since by definition the modulus of elasticity E = p^s, the values
of the respective moduli for the metals in diagram (I) are propor-
TENSION.
211
tional to the tangent of the angle which the corresponding
straight portions OB, OB' , etc., make with the horizontal axis.
From the various ordinates and abscissae for the points B, B',
etc., we find E for cast iron to be 14,000,000 Ibs./in.^ and for
the other three metals 28,000,000, 30,000,000, and 40,000,000,
respectively. The curve for the "harder steel" is not
shown in (II), being beyond the limits of the diagram, as
to stress; and the complete curve for cast iron is contained
within the limits of diagram (I), since the elongation at
rupture is very small in the case of this metal, only about
3/10 of one per cent, or 3 parts in 1000; whereas that for
wrought iron or soft steel is 300 parts in 1000 (or 30 per
cent). In the case of cast iron the elastic limit is very ill-
defined and the proportion of carbon and the mode of manu-
facture have much influence on its behavior under test.
"Soft steel" is another name for "structural steel," used
in construction on a large scale, as in buildings and bridge
trusses; "medium steel " being a somewhat harder grade
of the same. Many grades of steel are made which are
much stronger and harder than these, such as tool steel,
nickel steel, and piano wire (whose rupturing stress may
be as high as 300,000 Ibs./in.^). Wrought iron in the form
of wire is much stronger than in bars.
Note. — Such a line as CC, showing the relation of stress and strain
as the stress is gradually removed, will be called an "elasticity line"
on p. 241. In § 206 some mention will be made of the phenomena of
"overstraining" a test-piece of iron or steel, showing that on re-applying
stress after a certain period of rest the plotted results of stress and
strain relations show that the line C'C is retraced to C and continues
in the same straight line prolonged, to a new elastic limit higher than C,
before curving off to the right.
193. Lateral Contraction. — In the stretching of prisms of
nearly all kinds of material, accompanying the elongation
of length is found also a diminution of width whose rela-
tive amount in the case of the three metals just treated is
about ^ or i^ of the relative elongation (within elastic
limit). Thus, in the third experiment in the table of § 191,
this relative lateral contraction or decrease of diameter
~ H ^^ /i ^^ ^> ^■^•> about 0.00008. In the case of cast
iron and hard steels contraction is not noticeable ex-
212 MECHANICS OF ENGINEEBING
csept by very delicate measurements, both within and with-
out the elastic limit ; but the more ductile metals, as
wrought iron and the soft steels, when stretched beyond
the elastic limit show this feature of their deformation
in a very marked degree. Fig. 201 shows by dotted lines
the original contour of a wrought iron rod, while the con-
tinuous lines indicate that at rupture. At the cross section
of rupture, whose position is determined by some
local weakness, the drawing out is peculiarly
pronounced.
The contraction of area thus produced is some-
times as great as 50 or 60% at the fracture.
194. "Flow of Solids." — When the change in re-
lative position of the elements of a solid is ex-
treme, as occurs in the making of lead pipe,
I drawing of wire, the stretching of a rod of duc-
j tile metal as in the preceding article, we have
Fig. 201. instances of what is called the Flow of Solids, in-
teresting experiments on which have been made by
Tresca.
195. Moduli of Tenacity. — The tensile stress per square
inch (of original sectional area) required to rupture a
prism of a given material will be denoted by T and called
the modulus of ultimate tenacity ; similarly, the modulus oj
safe tenacity, or greatest safe tensile stress on an element,
by T' ', while the tensile stress at elastic limit may be
called T". The ratio of T' to T" is not fixed in practice
but depends upon circumstances (from j4, to ^).
Hence, if a prism of any material sustains a total pull
or load P, and has a sectional area=jP, we have
P= FT for the ultimate or breaking load. \
P'=FT' " " safe load. f ' ' (^^
P"=FT" " " load at elastic limit. )
Of course T' should always be less than T". (The hand-,
book of the Cambria Steel Co. , in quoting from the building
laws of various cities of the U. S., gives allowable unit-
stresses for ordinary materials, both in tension and com-
pression.)
TENSION. 213
196. Resilience of a Stretched Prism. — ^Fig. 202. In the
gradual stretcliing of a prism, fixed at one extremity ^ the
value of the tensile force P at the other necessarily de-
pends on the elongation A at each stage of the lengthening,
according to the relation [eq. (1) of § 191.]
' ^^ (8)
FE,
within the elastic limit. (If we place a weight G on the
^^ flanges of the unstretched prism and then leave
^ it to the action of gravity and the elastic action
of the prism, the weight begins to sink, meeting
an increasing pressure P, proportional to l, from
the flanges). Suppose the stretching to continue
until P reaches some value P" (at elastic limit
\\ say), and I a value X'. Then the work done so
N^ far is (see p. 155)
Fig 802 ^7= mean force X space = ^ P" /I" . . (4)
But from (2) F'=FT", and (see §§ 184 and 191)
K"=e"l
.-. (4) becomes XJ=y2 T e". Fl=}^ T e" V . . (6)
where Fis the volume of the prism. The quantity }4T"£",
or work done in stretching to the elastic limit a cubic
inch (or other unit of volume) of the given material, may
be called the Modvlus of Resilience for tension. From (5)
it appears that the amounts of work done in stretching to
the elastic limit prisms of the same material but of differ-
ent dimensions are proportional to their volumes simply.
The quantity }4T"e" is graphically represented by the
area of one of the triangles such as OA'B, OA'B" in Fig.
200 ; for (in the curve for wrought iron for instance) the
modulus of tenacity at elastic limit is represented by ^'P,
and e" (i.e., e for elastic limit) by OA'. The remainder of
214 MEOHAE^ICS OF ENGINBERI^a.
the area OBG included between the curve and the hori-
zontal axis, i.e., from B to G, represents the work done in
stretching a cubic unit from the elastic limit to the point
of rupture, for each vertical strip having an altitude =p
and a width =de, has an area ^pde, i.e., the work done by
the stress p on one face of a cubic unit through the dis-
tance de, or increment of elongation.
If a weight or load = (r be " suddenly "applied to stretch
the prism, i.e., placed on the flanges, barely touching
them, and then allowed to fall, when it comes to rest again
it has fallen through a height X^, and experiences at this
instant some pressure P\ from the flanges; Pi=:?. Apply-
ing to this body the "Work and Energy" method (p. 138),
noting that its initial and final kinetic energy are each zero
and that the force G is constant, while the upward force P
(from the flanges) is variable, with an average value of JPi,
we have
GAi = iPiAi + 0-0; whence Pi = 2(?.
Since Pi = 2G, i.e., is >(t, the body does not remain in
this position but is pulled upward by the elasticity of the
prism. In fact, the motion is harmonic (see §§ 59 and
138). Theoretically, the elastic limit not being passed, the
oscillations should continue indefinitely.
Hence a load O " suddenly applied " occasions double the
tension it would if compelled to sink gradually by a sup-
port underneath, which is not removed until the tension is
just = Q, oscillation being thus prevented.
If the weight G sinks through a height —h before strik-
ing the flanges. Fig. 202, we shall have similarly, within
elastic limit, if ^i= greatest elongation, (the mass of rod
being small compared with that of G).
G{h^K)=%P,K .... (6)
If the elastic limit is to be just reached we have from eqs.
(5) and (6), neglecting ^ compared with h,
Gh=%T"B"V . . . (7>
TBNSIOX.
215
nn equation of condition that tlie prism shall not be in-
jured.
Example. — If a steel prism have a sectional area of i/
eq. inch and a length ^=10 ft. =120 inches, what is the
greatest allowable height of fall of a weight of 200 lbs.,
that the final tensile stress induced may not exceed T" =
30,000 lbs. per sq. inch, if e" z=.002 ? From (7), using tha
inch and pound, we have
h=
T"e"V
30,000 X. 002x1^x120.
2^
2x200
:4.5 inches.
197. Stretching of a Prism by Its Own Weight. — In the case
of a very long prism such as a mining-
pump rod, its weight must be taken into
account as well as that of the terminal
load P , see Fig. 203. At (a.) the prism
is shown in its unstrained condition ; at
(&) strained by the load P^ and its own
weight. Let the cross section be =jP, the
heaviness of the prism =y. Then the rela-
tive extension of any element at a distance
Fig. 203.
jy from o is "^
^_dX {P,+rFx)
dx'
FE,
(1)
{See eq. (1) § 191) ; since P^-^-Fjx is the load hanging upon
the cross section at that locality. Equal c?a?'s, therefore,
are unequally elongated, x varying from to I. The total
elongation is
=/*=jk/f^''"+''^"'"^=/i;
'/2GI
FE,
Le., k= the amount due to Pi, plus an extension which
half the weight of the prism would produce, hung at the
lower extremity.
PI
* In A. = put dX for A, dx for I, and (P, + j^^.r) for P,
FEt
216 MECHANICS OF ENGINEERING.
The foregoing relates to tlie deformation of the piece,
and is therefore a problem of stiffness. As to the strength
of the prism, the relative elongation e=dA-h-dx [see eq. (1)],
which is variable, must nowhere exceed a safe value e'=
T'^E, (from eq. (1) § 191, putting P=FT', and X=X),
Now the greatest value of the ratio dX : dx, by inspecting
eq. (1), is seen to be at the upper end where x=l. The
proper cross section F, for a given load Pj, is thus found.
Putting ^]^-~^ ^e have F =^^ . (2)
198. Solid of Uniform Strength in Tension, or hanging body
of minimum material supporting its own
weight and a terminal load Pj. Let it be a
solid of revolution. If every cross-section
P at a distance =x from the lower extrem-
ity, bears its safe load FT', every element
of the body is doing full duty, and its form
is the most economical of material.
The lowest section must have an area
Fi(j.204. Fo=P^-^T', since Pi is its safe load. Fig.
204. Consider any horizontal lamina ; its weight is yFdx,
(j= heaviness of the material, supposed homogenous), and
its lower base Pmust have Pi-\-G for its safe load, i.e.
G+F,=FT' ... a)
in which G denotes the weight of the portion of the solid
below F. Similarly for the upper base F-\-dF, we have
G+P,+r^dx={F+dF)T' . , (2)
By subtraction we obtain
rFdsc=T'dFi ie. -l.dx^ ^
T F
TENSION. 217
in whicli the two variables x and F are separated. By in-
tegration we now have
;or^,=log.e^ . . (3)
. \x p yx
1.6., F=Foer =-1, e~ (4)
from which i^may be computed for any value of x.
The weight of the portion below any F is found from (1)
and (4); i.e.
while the total extension ^ will be
^=^"^1 (6)
"the relative elongation dX-i-dx being the same for every dx
and bearing the same ratio to e" (at elastic limit), as T'
does to T".
199. Tensile Stresses Induced by Temperature. — If the two
ends of a prism are immovably fixed, when under no strain
and at a temperature t, and the temperature is then low-
ered to a value t', the body suffers a tension proportional
to the fall in temperature (within elastic limit). If for a
rise or fall of 1° Fahr. (or Cent.) a unit of length of the
material would change in length by an amount t^ (called
the co-efficient of expansion) a length =1 would be con-
tracted an amount X=-fjl(t-t') during the given fall of tem-
perature if one end were free. Hence, if this contraction
is prevented by fixing both ends, the rod must be under a
:tension P, equal in value to the force which would be
218 MECHANICS OF exgi]s:eeeing.
necessary to produce the elongation X, just stated, under
ordinary circumstances at the lower temperature.
!From eq. (1) §191, therefore, we have for this tension
dtite to fall of temperature
For 1° Cent, we may write
For Cast iron -f] = .0000111 ;
« Wrought iron = .0000120 ;
« Steel = .0000108 to .0000114 J
« Copper yj = .0000172 ;
« Zinc 7^ = .0000300,
COMPRESSION OF SHORT BLOCKS.
200, Short and Long Columns. — In a prism in tension, its-
own weight being neglected, all the elements between thl
jocaiities of application of the pair of external forces pro-
ducing; the stretching are in the same state of stress, if the
external forces act axially (excepting the few elements in the
immediate neighborhood of the forces ; these suffering
local stresses dependent on the manner of application of
the external forces), and the prism may be of any length
without vitiating this statement. But if the two external
forces are directed toivard each other the intervening ele-
ments will not all be in the same state of compressive
stress unless the prism is comparatively short (or unless
numerous points of lateral support are provided). A long
prism will buckle out sideways, thus even inducing tensile
stress, in some cases, in the elements on the convex side.
Hence the distinction between sTiort UocTcs and long
columns. Under compression the former yield by crush-
ing or splitting, while the latter give way by flexure (i.e.
bending). Long columns, then will be treated separately
COMPRESSION OF SHORT BLOCKS. 219
In a subsequent chapter. In the present section tlie blocks
treated being about tliree or four times as long as wide,
^11 tlie elements will be considered as being under -equal
compressive stresses at tbe same time.
201. Notation for Compression. — By using a subscript c,
we may write
E^= Modulus of Elasticity;* i.e. tlie quotient of the
compressive stress per unit of area divided by the relative
shortening. (Young's Modulus; no stress on sides);
C= Modulus of crushing ; i.e. the force per unit of sec-
tional area necessary to rupture the block by crushing ;
G'= Modulus of safe compression, a safe compressive
stress per unit of area ; and
G"= Modulus of compression at elastic limit.
For the absolute and relative shortening in length we
may still use X and e, respectively, and within the elastic
limit may write equations similar to those for tension, F
being the sectional area of the block and F one of the ter-
minal forces, while p = compressive stress per unit of area
of Ff viz.:
. (1)
v-F _
-dx
X-
rF_
■A ~
.PI
~FX
ithin the elastic limit.
Also for
a short block
Crushing force =FG
Compressive force at elastic limit =iFG" }• . (2]f
Safe compressive force =FG'
limit r=FG" \ .
7' )
202. Remarks on Crushing. — As in § 182 for a tensile
stress, so for a compressive stress we may prove th
>r*
T
Fig. 205.
Fig, 206.
In Fig. 205 (a lap-joint) the rivet is said to be in single
shear ; in Fig. 206 in double shear. If P is just great
enough to shear off the rivet, the modulus of ultimate shear-
ing, which may be called S, (being the shearing force per
unit of section when rupture occurs) is
F iTid^
(1)
in which i^==the cross section of the rivet, its diameter
being =d. For safety a value S'= }{ to ^ of >S' should
be taken for metal, in order to be within the elastic limit.
As the width of the plate is diminished by the rivet
hole the remaining sectional area of the plate should be
ample to sustain the tension P, or 2P, (according to the
plate considered, see Fig. 206), P being the safe shearing
force for the rivet. Also the thickness t of the plate
should be such that the side of the hole shall be secure
against crushing ; P must not be > C'td, Fig. 205.
Again, the distance a, Fig. 205, should be such as to
prevent the tearing or shearing out of the part of the
plate between the rivet and edge of the plate.
226
MECHANICS OF E:NGINEEKING.
For economy of material tlie seam or joint sliould be
no more liable to rupture by one tban by another, of the
o o ^ o
■.t
Fig. 307.
four modes just mentioned. The relations which must
then subsist will be illustrated in the case of the " butt=
joint " with two cover-plates, Fig. 207. Let the dimen-
sions be denoted as in the figure and the total tensile force
on the joint be = Q. Each rivet (see also Fig. 206) is ex-
posed in each of two of its sections to a shear of I2 Q}
hence for safety against shearing of rivets we put
12 Q-
-% Tides'
(1)
Along one row of rivets in the main plate the sectional
area for resisting tension is reduced to {b — ^d)t,, hence for
safety against rupture of that plate by the tension Q, we
put
Q=(h—3d)t,T' , (2)
Equations (1) and (2) suffice to determine d for the rivets
and ^1 for the main plates, Q and b being given ; but the
values thus obtained should also be examined with refer-
ence to the compression in the side of the rivet hole, i.e.,
J^ Q must not be > C't.d. [The distance a, Fig. 205, to the
edge of the plate is recommended by different authorities
to be from d to 3d.]
Similarly, for the cover -plate we must have
and 12^ not > G'td,
^4QoT(b—dd)tT'
<
(8)
SHEARi:srG. 227
If the rivets do not fit their holes closely, a large margin
should be allowed in practice. Again, in boiler work, the
pitch, or distance between centers of two consecutive rivets
may need to be smaller, to make the joint steam-tight, than
would be required for strength alone.
208, Shearing Distortion. — The change of form in an ele-
ment due to shearing is an angular deformation and will
be measured in tt -measure. This angular change or dif-
ference between the value of the corner angle during strain
and ^4i'^, its value before strain, will be called d, and is
proportional (within elastic limit) to the shearing stress
per unit of area, p^, existing on all the four faces whose
angles with each other have been changed.
Fig. 208. (See § 181). By § 184 the Modulus of Shearing
Elasticity is the quotient obtained by dividing p^hj d \ i.e.
{elastic limit not passed)^
^s=^ . . . . (1)
or inversely, d=p^-^E^ (1)'
The value of E^ for different substances is most easily
I determined by experiments on torsion
in which shearing is the most promi-
/ nent stress.* (This prominence depends
y\ on the position of the bounding planes
j^ of the element considered ; e.g., in Fig.
_ __.L 208, if another element were considered
/ '---cix->, within the one there shown and with
Fig. 208. its plaues at 45° with those of the first,
we should find tension alone on one pair of opposite faces,
compression alone on the other pair.) It will be noticed
that shearing stress cannot be present on two opposite
faces only, but exists also on another pair of faces (those
perpendicular to the stress on the first), forming a couple
of equal and opposite moment to the first, this being
necessary for the equilibrium of the element, even when
* For instance, see numerical example on p. 237, giving a value of
Es as resulting from a torsion test made by students in the Civil Engi-
neering Laboratory at Cornell University, April, 1904.
228
MECHAIflCS OF ENGINBERIKG.
tensile or compressive stresses are also present on the
faces considered.
209. Shearing Stress is Always of the Same Intensity on the
Four Faces of an Element. — (By intensity is meant per unit
of area ; and the four faces referred to are those perpen-
dicular to the paper in Fig. 208, the shearing stress being
parallel to the paper.)
Let dx and dz be the width and height of the element
in Fig. 208, while dy is its thickness perpendicular to the
paper. Let the intensity of the shear on the right hand
face be =q^, that on the top face =Ps. Then for the ele-
ment aw a free body, taking moments about the axis per-
pendicular to paper, we have
q^ dz dy X dx — ^g dx dy x dz =0 .•. qs =p^
{dx and dz being the respective lever arms of the forces
q^ dz dy and p^ dx dy.)
Even if there were also tensions (or compressions) on
one or both pairs of faces their moments about would
balance (or fail to do so by a differential of a higher order)
independently of the shears, and the above result would
still hold.
210. Table of Moduli for Shearing.
d"
^s
8"
s
Material.
i.e. 6 at elastic
limit.
Mod. of Elasticity
for Shearing.
(Elastic limit.)
(Rupture.)
arc in radians.
lbs. per sq. in.
lbs. per sq. in
lbs. per sq. in.
Soft Steel,
9,000,000
30,000
70,000
Hard Steel,
0.0033
14,000,000
45,000
90,00C
Cast Iron,
0.0021
7,000,000
15,000
30,00C
Wrought Iron,
0.0022
9,000,000
20,000
50,000
Brass,
5,000,000
Glass,
Wood, across (
fibre, 1
1,500
to
8,000
Wood, along (
fibre, (
500
to
3,200
SHEAKIISG.
229
As in tlie tables for tension and compression, tlie above
values are averages. The true values may differ from
these as mucli as 30 per cent, in particular cases, accord-*
ing to the quality of the specimen.
211. Punching rivet holes in plates of metal requires the
overcoming of the shearing resistance along the convex
surface of the cylinder punched out. Hence ii d = diam-
eter of hole, and t= the thickness of the plate, the neces-
sary force for the punching, the surface sheared being
F= tjvd^ is
P=8t^d
(2)-
Another example of shearing action is the " stripping "
of the threads of a screw, when the nut is forced off lon-
gitudinally without turning, and resembles punching in
its nature.
212. EandEgj Theoretical Relation. — In case a rod is in
iension within the elastic limit, the relative (linear) lateral
contraction (let this =m) is so connected with E^ and E^
ihat if two of the three are known the third can be de-
duced theoretically. This relation is proved as follows,
by Prof. Burr. Taking an elemental cube with four of its
faces at 45° with the axis of the piece, Fig. 209, the axial
half-diagonal AD becomes of a length AD'=AD-\-s.AD
under stress, while the transverse half diagonal contracts
to a length B'D'=AD — m.AD, The angular distortion d
.-\U.S .
•
X<'>^'
/^<\^^^
A<
^ D D/^ ^^'
,.
\>6-^
Fig. 209. § 212.
Fig. 210.
230 MECHAIflCS OF EI^GINEEKING.
is supposed very small compared with 90° and is due to
the shear j9g per unit of area on the face BG (or BA\
From the figure we have
tan(45°— -) = __^,=__=1— m— s, approx.
[But, Fig. 210, tan(45° — x)=l — 2x nearly, where a; is a
small angle, for, taking CA=unitj=AE, ian AD=AF=
AE—EF. Now approximately EF= EG, y2 and EG=
BDa^^=x^^ .'. AF= 1 — 2a7 nearly.] Hence
1 — d= 1 — m — £ ; or d= m+e . . (2)
Eq. (2) holds good whatever the stresses producing the
deformation, but in the present case of a rod in tension,
if it is an isotrope, and if ^ = tension per unit of area on
its transverse section, (see § 182, putting «=45°), we have
^t==p-=-£ and E^={psOJx BG')-^d=}^p-^d. Putting also
(m : £)= h, whence m=k£, eq. (2) may finally be written*
>4-==(^ + l)4-; i.e., ^s=-^^ . . (3)
Prof, Bauschinger, experimenting with cast iron rods,,
found that in tension the ratio m: £was =m} as an average,
which in eq. (3) gives
^=12^^,= !^, nearly. , , . (4)
246 5 ^ ^ ^
His experiments on the torsion of cast iron rods gave
^,= 6,000,000 to 7,000,000 lbs. per sq. inch. By (4), then,
E, should be 15,000,000 to 17,500,000 which is approxi-
mately true (§ 203).
Corresponding results may be obtained for short blocks
in compression, the lateral change being a dilatation in-
stead of a contraction.
* This ratio, m-^e, denoted by k, is called Poisson's Ratio. For metals
its value lies approximately between 0.20 and 0.35. See also p. 507.
313. Examples in Shearing. — Example 1. — Kequired the^
proper length, a, Fig. 211, to
guard against the shearing off,
along the grain, of the portion
ah, of a wooden tie-rod, the force
P being = 2 tons, and the width
of the tie = 4 inches. Using a
value of S' = 100 lbs. per sq. irL.s
we put 6a/S''= 4,000 cos 45° ; i.e.
■Fi^m. a= (4,000x0. 707) --(4x100)= 7.07
inches.
Example 2.— A ^ in. rivet of wrought iron, in single
shear (see Eig. 205) has an ultimate shearing strength
P= FS=}(7T(PS= %7t{ Yq y X 50,000= 30,050 lbs. For safety,
putting aS"= 8,000 instead of aS',P'=4,800 lbs. is its safe
shearing strength in single shear.
The wrought iron plate, to be secure against the side-
crushing in the hole, should have a thickness t, computed
thus I
P'=tdC' ; or 4,800=^.^ x 12,000 ,-. ^=0.46 in.
If the plate were only 0.23 in. thick the safe value of P
would be only ^ of 4,800.
Example 3. — Conversely, given a lap-joint, Fig. 205, in
which the plates are ^ in. thick and the tensile force on
the joint = 600 lbs. per linear inch of seam, how closely
must ^ inch rivets be spaced in one row, putting jS"=8,000
and 6" =12,000 lbs. per sq. in. ? Let the distance between
centres of rivets be =x (in inches), then the force upon
each rivet =600a7, while its section P=0.44 sq. in. Having
regard to the shearing strength of the rivet we put 600cc=
0.44x8,000 and obtain a?=5.86 in.; but considering that the
safe crushing resistance of the hole is =1^-^.12,000=
2,250 lbs., 600aj=2,250 gives a;=3.75 inches, which is the
pitch to be adopted. What is the tensile strength of the.
reduced sectional area of the plate, with this pitch '?
232 MECHAXIC3 OF EKGIISTEBIilNG.
Example 4 — Double butt-joint ; (see Fig. 207) ; ^s iiich
plate; ^ in. rivets; F =C' =12,000 ; S' =8,333; width of
plates=14 inches. Will one row of rivets be sufficient at
each, side of joint, if ^=30,000 lbs.? The number of rivets
^^ ? Here each rivet is in double shear and has therefore
a double strength as regards shear. In double shear the
safe strength of each rivet =2i^>S"= 7,333 lbs. Now 30,000^
7,333=40 (saj). With the four rivets in one row the re-
duced sectional area of the main plate is =[14 — 4x ^] X^s
=4,12 sq. in., whose safe tensile strength is =i^J"=4„12x
12,000=49,440 lbs.; which is > 30,000 lbs. .% main plate is
safe in this respect. But as to side-crushing in holes
in main plate we find that G't^d (i,e, 12,000 X Vs >^ M^^'^'^^
lbs.) is <.%Q i,e. <7,500 lbs., the actual force on side of
hole. Hence four rivets in one row are too few unless
thickness of maiiL plate be doubled. Will eight in one
row be safe ?
213a, (Addendum to § 206.) Elasticity of Stone and Cements.
— Experiments by Gen. Gill more with the large Watertowi»
testing-machine in 1883 resulted as follows (see p. 221 for
notation) :
With cubes of Haverstraw Freestone (a homogeneous brown-
stone) from 1 in. to 12 in. on the edge, E^ was found to be
from 900,000 to 1,000,000 lbs. per sq. in. approximately ; and
C about 4,000 or 5,000 lbs. per sq. in. Cubes of the same
range of sizes of Djckerman's Portland cement gave E^ from
1,350,000 to 1,630,000, and G from 4,000 to 7,000, lbs. per sq.
in. Cubes of concrete of the above sizes, made with the
Newark Cc.'s Rosendale cement, gave E^ about 538,000^ while
cubes of cement-mortar, and some of concrete, both made with
National Portland cement, showed E^ from 800,000 to 2,000,-
OOO lbs. per sq. in.
The compressibility of hrick jpiers 12 in. square in section
and 16 in. high was also tested. They were made of common
North River brick with mortar joints f in. thick, and showed
a value for E„ of about 300,000 or 400,000, while at elastic
limit C" was on the average 1,000, lbs. per sq. in.
TORSIOM.
233
CHAPTER IL
TOKSION.
S14. Angle of Torsion and of Helix. "When a cylindrical
beam or shaft is subjected to a twisting or torsional action,
I. e. when it is the means of holding in equilibrium two
couples in parallel planes and of equal and opposite mo-
ments, the longitudinal axis of symmetry remains straight
and the elements along it exper-
lience no stress (whence it may be
I called the "line of no twist"),
while the lines originally parallel to
Fig. 212. i^ assume the form of helices, each
element of which is distorted in its angles (originally
right angles), the amount of distortion being assumed pro-
portional to the radius of the helix. The directions of the
faces of any element were originally as follows : two radial,
two in consecutive transverse sections, and the other two
tangent to two consecutive circular cylinders whose com-
mon axis is that of the shaft. E.g. in Fig. 212 we have
an unstrained shaft, while in Fig. 213 it holds the two
234
MECHANICS OF ENGINEEHmG.
couples (of equal moment Pa = Qh) in equilibrium. These
couples act in parallel planes perpendicular to the axis of
the prism and a distance, ?, apart. Assuming that the
transverse sections remain plane and parallel during tor-
sion, any surface element, m, which in Fig. 212 was entire-
ly right-angled, is now distorted. Two of its angles have
been increased, two diminished, by an amount d, the angle
between the helix and a line parallel to the axis. Suppos-
ing m to be the most distant of any element from ihe axis,
this distance being e, any other element at a distance s
from the axis experiences an angular distortion =- <§„
If now we draw B' parallel to 0' A the angle B B',
=a, is called the Angle of Torsion, while d may be called the
helix angle', the former lies in a transverse plane, the latter
in a plane tangent to the cylinder. Now
tan d = (linear arc B B')-t-1; but lin. arc B B' =' ea; hence,
putting d for tan d, (3 being small)
(1)
(d and « both in radians).
215. Shearing Stress on the Elements. The angular distor-
tion, or shearing strain, d, of any element (bounded as al-
ready described) is due to the shearing stresses exerted on
it by its neighbors on the four faces perpendicular to the
tangent plane of the cylindri-
cal shell in which the element
is situated. Consider these
neighboring elements of an
outside element removed, and
the stresses put in ; the latter
are accountable for the dis-
^®- ^*^ tortion of the element and
-pdF
TOKSiou". 235
hold it in equilibrium. Fig. 214 shows this element
"free." Within the elastic limit ^ is known to be propor-
tional to jOg, the shearing stress per unit of area on the
faces whose relative angular positions have been changed.
That is, from eq. (1), § 208, S -^^p^-r-Us; whence, see (1) of
§ 214,
In (2) Ps, and e both refer to a surface element, e being
the radius of the cylinder, and p^ the greatest intensity of
shearing stress existing in the shaft. Elements lying nearer
the axis suffer shearing stresses of less intensity in pro-
portion to their radial distances, i.e., to their helix-angles.
That is, the shearing stress on that face of the element
which forms a part of a transverse section and whose dis-
tance from the axis is z, is p, =— p^, per unit of area, and
the total shear on the face is pdF, c?^ being the area of the
face.
216. Torsional Strength. — ^We are now ready to expose tlia
full transverse section of a shaft under torsion, to deduce
formulae of practical utility. Making a right section of
the shaft of Fig. 213 anywhere between the two couples
and considering the left hand portion as a free body, the
forces holding it in equilibrium are the two forces P of
the bft-hand couple and an infinite number of shearing
forces, each tangent to its circle of radius s, on the cross
section exposed by the removal of the right-hand portion.
The cross section is assumed to remain plane during tor-
sion, and is composed of an infinite number of dF's, each
being the area of an exposed face of an element | see !Fig.
236
elementary shearing force = S p^dF, and s is its
lever arm about the axis Oo . For equilibrium, S (mom.),
about the axis Oo must =0 ; i.e. in detail
_p^o^p^a+ f ( -£ p,dF)%^ii
©r, redncing.
h rz^dF=Pa\ or, A.^Pa
eJ e
(3)
Eq, (3) relates to torsional strength, since it contains ^s, tha
greatest shearing stress induced by the torsional couple,
whose moment Pa is called the Moment of Torsion, the
stresses in the cross section forming a couple of equal and
opposite moment. Pa is also called the "torque."
Ip is recognized as the Polar Moment of Inertia of the cross
section, discussed in § 94 ; e is the radial distance of the
outermost element, and = the radius for a circular shafto
217. Torsional Stiffness. — In problems involving the angle
of torsion, or deformation of the shaft, we need an equa-
tion connecting Pa and a, which is obtained by substitut-
ing in eq. (3) the value of p^ in eq. (2), whence
I
=Pa.
(4)
From this is appears that the angle of torsion, a, is pro-
portional to the moment of torsion, or " torque," Pa inch -lbs.,
within the elastic limit; a must be expressed in radians.
TORSION. 237
Example. — A portion 3.4 ft. long, of a solid cylindrical shaft of soft
steel, of diam. = 1.5 in., is found by the use of "Torsion Clinometers"
(see frontispiece) to be held at an angle of torsion of a = 5.41°, =0.0944
radians, just before the elastic limit is reached, by a "torque," =Pa, of
10,200 in. -lbs. Compute the Modulus of Elasticity for Shearing.
Substituting in eq. (4), with I-p=Tzr^l2, (§94), =7r(0.75)^^2, =0.497
in.*, and Z = 3.4X 12 = 40.8 in., we have
10,200X40 .8 Qg^nnnniv.
' ^ 0.0944 X 0.4 97 ^ 8,870,000 lbs. per sq. m.
218. Torsional Resilience is the work done in twisting a
shaft from an unstrained state until the elastic limit is
reached in the outermost elements. If in Fig. 213 we
imagine the right-hand extremity to be fixed, while the
other end is gradually twisted through an angle each
force P of the couple must be made to increase grafdually
from a zero value up to the value Pj, corresponding to ai.
In this motion each end of the arm a describes a space
= ^aai, and the mean value of the force = }4Pi (compare
§ 196). Hence the work done in twisting is
Ui=}4FiX}4aaiX2=}4Piaaj^ . . (5)
By the aid of preceding equations, (5) can be written
If for ps "^e write 8' (Modulus of safe shearing) we have
for the safe resilience of the shaft
U'=4r^ -. . . . (7)
If the torsional elasticity of an originally unstrained shaft
is to be the means of arresting the motion of a moving
mass whose weight is O, (large compared with the parts
intervening) and velocity =v, we write (§ 133)
g 2'
as tlie condition that the shali shall not be injnrecL
238 mecha:^ics of engineering.
21U. roiar Moment of Inertia. — ^For a shaft of circular
cross section (see § 94) /p=i^7rr*; for a hollow cylinder
/p=i^7r(ri* — r^) ', while for a square shaft If=yih^, h being
"the side of the square ; for a rectangular cross-section
sides h and li, I^=lJbh{lf-\-¥). For a cylinder e=r; if hoi-
low, e=r , the greater radius. For a square, e=i^6y'2.
220. Ifon-Circular Shafts. — If the cross-section is not cir-
cular it becomes warped, in torsion, instead of remaining
plane. Hence the foregoing theory does not strictly ap-
ply. The celebrated investigations of St. Tenant, how-
ever, cover many of these cases. (See § 708 of Thompson
and Tait's Natural Philosophy ; also, Prof. Burr's Elas-
ticity and Strength of the Materials of Engineering). His
results give for a square shaft (instead of the
ab'E.^ Pa of eq. (4) of § 217),
Ql
Pa=OMl^t . . . . (1)
and Pa=Jffi^p^f instead of eq. (3) of § 216, 2?s being the
greatest shearing stress.
The elements under greatest shearing strain are found
at the middles of the sides, instead of at the corners, when
the prism is of square or rectangular cross-section. The
warping of the cross-section in such a case is easily veri '
fied by the student by twisting a bar of india-rubber in
his fingers.
221. Transmission of Power. — Fig. 216. Suppose the cog-
wheel B to cause A, on the
same shaft, to revolve uni-
formly and overcome a resis-
tance Q, the pressure of the
teeth of another cog-wheel,
P 5 being driven by still another
Fig. 216. wheel. The shaft AB is un-
der torsion, the moment of torsion being =Pa= Qh. (Pi
and ^1 the bearing reactions have no moment about the
axis of the shaft). If the shaft makes u revolutions per
unit-time, the work transmitted {transmitted ; not expend^
TOESION. 239
ed in twisting the shaft whose angle of torsion remains
constant, corresponding to Fa) per unit-time, i.e. the Power,
is
X/=P.27ra.u=27ruPa . , , (8)
To reduce L to Horse Power (§ 132), we divide by N,
the number of units of work per unit -time constituting
vOne H. P. in the system of units employed, i.e.,
Horse Power =H. p=?!E!^
JSl
For example JSf =33,000 ft. -lbs. per minute, or =396,000
inch -lbs. per minute ; or = 550 ft. -lbs. per second. Usually
the rate of rotation of a shaft is given in revolutions per
minute.
But eq^. (8) happens to contain Fa the moment of torsion
acting to maintain the constant value of the angle of tor-
sion, and since for safety (see eq. (3) § 216) Fa= S' I.^-^ e,
with -ZJ,= y^TiT^ and e=r for a solid circular shaft, we have
for such a shaft
(Safe),H.P.=?f^ . . . (9)
N
which is the safe H. P., which the given shaft can trans-
mit at the given speed. S' may be made 7,000 lbs. per sq.
inch for wrought iron ; 10,000 for steel, and 5,000 for cast-
iron. If the value of Pa fluctuates periodically, as when
a shaft is driven by a connecting rod and crank, for (H. P.)
we put toX(H. p.), m being the ratio of the maximum to
the mean torsional moment; m= about 172 under ordi-
nary circumstances (Cotterill).
With a hollow cylindrical shaft, of outer radius = rj, and inner = r 2
the r^ of eq. (9) must be replaced by (?*i*— /•2*)-^'"i- If> furthermore, the
thickness of metal is small, we may proceed thus, taking numerical data:
Let the radius to the middle of the thickness be ro = 10 in., the thickness
t=\ in., and the (steel) shaft make m=120 revs./min. ; with *S' = 5000
Ibs./in.^; then the total safe shearing stress in the cross-section is-
12' = 27rroi5;' = 27rlOXiX 5000 = 78,540 lbs., whHe the velocity of the
mid-thickness is v' = 27rroU = 27: 10X2 = 125.6 in./sec. = 10.47 ft. /sec. Hence
the (safe) power that may be transmitted at given speed is L = R'v'
= 78,540X10.47 = 822,100 ft.-lbs. per sec; or, (-^550), =1495 H.P.
240
MECHAITICS OF ElifGINEEEING.
222. Autographic Testing Machine. — Tlie principle of Prof
Thurston's invention bearing this name is shown in Fig
• Fie. 217.
217. The test-piece is of a standard shape and size, its
central cylinder being subjected to torsion. A jaw, carry-
ing a handle (or gear-wheel turned by a worm) and a drum
on which paper is wrapped, takes a firm hold of one end
of the test-piece, whose further end lies in another jaw
rigidly connected with a heavy pendulum carrying a pen-
cil free to move axially. By a continuous slow motion of
the handle the pendulum is gradually deviated more and
more from the vertical, through the intervention of the
test-piece, which is thus subjected to an increasing tor-
sional moment. The axis of the test-piece lies in the axis
of motion. This motion of the pendulum by means of a
proparly curved guide, WH, causes an axial (i.e., parallel
to axis of test-piece) motion of the pencil A, as well as an
angular deviation /9 equal to that of the pendulum, and
this axial distance CF,=^sT, of the peiicil from its initial
position measures the momenr of torsion =i^«=:^(? sin )5..
As the piece twists, the drum and paper move relatively
to the pencil through an angle sUo equal to the angle
of torsion a so far attained. The abscissa so and ordinate
sT oi the curve thus marked on the paper, measure,,
when the paper is unrolled, the values of a and Pa through.
TOESrOK
241
all the stages of the torsion. Fig. 218 shows typical
Fig. 218.
Jurves thus obtained. Many valuable indications are
given by these strain diagrams as to homogeneousness of
composition, ductility, etc., etc. On relaxing the strain
at any stage within the elastic limit, the pencil retraces
its path ; but if beyond that limit, a new path is taken
called an " elasticity-line," in general parallel to the first
part of the line, and showing the amount of angular re-
CO very, BC, and the permanent angular set, OB.
2222i.. Torsion Clinometers. — ^When the test-piece used in the
Thurston testing machine is short, the indicated angles of
torsion below the elastic limit are far in excess of the actual
values, on account of the initial yielding of the wedges in the
jaws. By the use of " torsion clinometers," however (see
frontispiece) the angle of torsion can be measured accurately
within one minute of arc.
223. Examples in Torsion. — The modulus of safe shearing
strengtn. S', as given in § 221, is expressed in pounds per
square inch ; hence these two units should be adopted
throughout in any numerical examples where one of the
above values for S' is used. The, same statement applies
to the modulus of shearing elasticity, E^, in the table of
§ 210.
- Example 1.— Fig. 216. With P = 1 ton, a = 3 ft., I ^
10 ft. , and the radius of the cylindrical shaft r=2.5 inches,
required the max. shearing stress per sq. inch, ps, the
shaft being of wrought iron. From eq. (3) § 216
Pae 2,000x36x2.5 o oomi, • -u
^----T^' V..X(2.5)^ =2,930 lbs. per sq. inch,
which is a safe value for any ferrous raetaL
242
MECHANICS OF EIN GINEEEIXG.
Example 2. — What H. P. is the shaft in Ex. 1 transmit-
ting, if it makes 50 revolutions per minute ? Let u =
number of revolutions per unit of time, and N = the num-
ber of units of work per unit of time constituting one
horse-power. Then H. V.^Pu^na-^N, which for the foot»
pound-minute system of units gives
H. P.=2,000x50x27rx3--33,000=57i4: H. P.
Example 3. — What different radius should be given t(-
the shaft in Ex. 1, if two radii at its extremities, originally
parallel, are to make an angle of 2° when the given moment
of torsion is acting, the strains in the shaft remaining con-
stant. From eq. (4) § 217, and the table 210, with a=i|^c;r=*
0.035 radians (i.e. ;7-measure), and I^=^j^T^^ we have
y^
2,000x36x120
>^7r0.035x 9,000,000
-—=17.45 .-. r=2.04 inches.
(This would bring about a different p,, but still safe.) The
foregoing is an example in stiffness.
Example 4. — A working shaft of steel (solid) is to tran:^-
mit 4,000 H. P. and make 60 rev. per minute, the maximum
twisting moment being 1^ times the average; requireil
its diameter. • c^=14.74 inches. Ans.
Example 5. — In example 1, p = 2,930 lbs. per square
inch ; what tensile stress does this imply on a plane at 45°
with the pair of planes on which Ps acts ? Fig. 219 shows
p,dx
dx^'Ps
'dx^ps
Via. 220.
TOESiOE^. 243
a small cube, of edge =dx, (taken from the outer helix of
Fig. 215,) free and in equilibrium, tbe plane of the paper
being tangent to the cylinder ; while 220 shows the portion
BD 0, also free, with the unknown total tensile stress jorfa;^,^/^
acting on the newly exposed rectangle of area =dxxdx^%
p being the unknown stress per unit of area. From sym-
metry the stress on this diagonal plane has no shearing
component. Putting 2' [components normal to^-D]=0,
we have
pdx^^2=2dx'^p^Gos4:5°=dx^p^^/2.'.p=ps . (1)
That is, a normal tensile stress exists in the diagonal
plane BD of the cubical element equal in intensity to the
shearing stress on one of the faces, i.e., =2,930 lbs. per sq.
in. in this case.
Similarly in the plane AG will be found a compressive
stress of 2,930 lbs. per sq. in. If a plane surface had been
exposed making any other angle than 45° with the face of
the cube in Fig. 219, we should have found shearing and
normal stresses each less than p^ per sq. inch. Hence the
interior dotted cube in 219, if shown " free " is in tension
in one direction, in compression in the other, and with
no shear, these normal stresses having equal intensities.
Since S' is usually less than T' or C, ii Ps is made = S'
the tensile and compressive actions are not injurious. It
follows therefore that when a cylinder is in torsion any
helix at an angle of 45° with the axis is a line of tensile,
or of compressive stress, according as it is a right or left
handed helix, or vice versa.
Example 6. — A solid and a hollow cylindrical shaft, of
equal length, contain the same amount of the same kind
of metal, the solid one fitting the hollow of the other.
Compare their torsional strengths, used separately.
The solid shaft has only ^ the strength of the hollow
one, Ans.
Example 7. — Compare the shafts of Example 6 as to tor-
sional stiffness (i. e. , the angles of torsion due to equal moments) .
The solid shaft is only one-third as stiff as the other ; an equal
moment produces three times the angle. Ans.
244 MECHANICS OF El!fGrNEi:KrN:G.
CHAPTER in.
FliEXURE OF HOMOGENEOUS PRISMS UHDEK
PERPENDICULAK FORCES IN ONE PLANE.
224. Assumptions of tlie Common Theory of Plexure. — Wlien
a prism is bent, under tlie action of external forces per-
pendicular to it and in tlie same plane witli each otlier, it
may be assumed tliat the longitudinal fibres are in tension
on the convex side, in compression on the concave side,
and that the relative stretching or contraction of the ele-
ments is proportional to their distances from a plane in-
termediate between, with the understanding that the flex-
ure is slight and that the elastic limit is not passed in any
element. i
This " common theory " is sufficiently exact for ordinary
engineering purposes if the constants employed are prop-
erly determined by a wide range of experiments, and in-
volves certain assumptions of as simple a nature as possi-
ble, consistently with practical facts. These assumptions
are as follows, (for prisms, and for solids with variable cross
sections, when the cross sections are similarly situated as
regards a central straight axis) and are approximately
borne out by experiment :
(1.) The external or " applied " forces are all perpendicu-
lar to the axis of the piece and lie in one plane, which may
be called the force-plane ; the force-plane contains the
axis of the piece and cuts each cross-section symmetri-
cally ;
(2.) The cross-sections remain plane surfaces during
flexure ;
(3.) There is a surface (or, rather, sheet of elements)
which is parallel to the axis and perpendicular to the
force-plane, and along which the elements of the solid ex-
FLEXURE.
245
perience no tension nor compression in an axial direction,
this being called tlie Neutral Surface;
(4.) The projection of the neutral surface upon the force
plane (or a || plane) being called the Neutral Line or Elastic
Curve, the bending or flexure of the piece is so slight that
an elementary division, ds, of the neutral line may be put
^dx, its projection on a line parallel to the direction of
the axis before flexure ;
(5.) The elements of the body contained between any
two consecutive cross-sections, whose intersections with
the neutral surface are the respective Neutral Axes of the
sections, experience elongations (or contractions, accord-
ing as they are situated on one side or the other of the
neutral surface), in an axial direction, whose amounts are
proportional to their distances from the neutral axis, and
indicate corresponding tensile or compressive stresses ; ,
(6.) E,=E,;
(7.) The dimensions of the cross-section are small com-
pared with the length of the piece ;
(8.) There is no shear perpendicular to the force plane
on internal surfaces perpendicular to that plane.
In the locality where any one of the external forces is
Applied, local stresses are of course induced which demand
separate treatment. These are not considered at present.
225. Illustration. — Consider the case of flexure shown in
Fig. 221. The external forces are three (neglecting the
Fig. 22i.
246
MECHANICS OF EXGIXEBRING.
weight of the beam), viz.: P^, Pg, and P3. P^ and P3 are
loads, P2 the reaction of the support.
The force plane is vertical. N^L is the neutral line or
elastic curve. NA is the neutral axis of the cross-section
at m / this cross-section, originally perpendicular to the
sides of the prism, is during flexure ~| to their tangent
planes drawn at the intersection lines ; in other words, the
side view QNB, of any cross-section is perpendicular to
the neutral line. In considering the whole prism free we
have the system Pj, P2, and P3 in equilibrium, whence
from 2^=0 we have P2=Pi+P3j and from 2" (mom. about
P) =0, P3?3=Pi?i. Hence given Pi we may determine the
other two external forces. A reaction such as Pg is some-
times called a supporting force. The elements above the
neutral surface NiOLS Sive in tension ; those below in com-
pression (in an axial direction).
226. The Elastic Forces. — Conceive the beam in Fig. 221
separated into two parts by any transverse section such
as QA, and the portion NiOJSf, considered as a free body
in Fig. 222. Of this free body the surface QAB is one of
^^dx
T«e. 222.
FLEXURE. 247
i
tlie bounding surfaces, but was originally an internal sur-
face of tlie beam m Fig. 221. Hence in Fig. 222 we must
put in the stresses acting on all the dF^^ or elements of area
of QAB. These stresses represent the actions of the bodj
taken away upon the body which is left, and according to
assumptions (5), (6) and (8) consist of normal stresses (ten-
sion or compression) proportional per unit of area, to th©
distance, z, of the cZi^'s from the neutral axis, and of shear-
ing stresses parallel to the force-plane (which in most
cases will be vertical).
The intensity of this shearing stress on any dF varies
with the position of the dF with respect to the neutral
axis, but the law of its variation will be investigated later
(§§ 253 and 254). These stresses, called the Elastic Forces
of the cross-section exposed, and the external forces Pj and
P2, form a system in equilibrium. We may therefore ap-
ply any of the 3onditions of equilibrium proved in § 38.
227. The Neutral Axis Contains the Centre of Gravity of the
Cross-Section. — Fig. 222. Let e— the distance of the outer-
most elem.ent of the cross-section from the neutral axis, and
the normal stress per unit of area upon it be =p, whether
tension or compression. Then by assumptions (5) and (6),
§ 224, the intensity of nprmal stress on any dF is = -1 p
and the actual
normal stress on any dFis= — pdF , {1}
This equation is true for dF's having negative «'s, i.e.
on the other side of the neutral axis, the negative value
of the force indicating normal stress of the opposite char-
acter ; 'for if the relative elongation (or contraction) of two
axial fibres is the same for equal g's, one above, the other
below, the neutral surface, the stresses producing the
changes in length are also the same, provided ^t=:^^; see§§
184 and 201.
For this free body in equilibrium put 2'X=0 (Xis a
horizontal axis). Put the normal stresses equal to their
X components, the flexure being so slight, and the X com-
248 MECHANICS OF ENGINEEEIKG.
ponent of the shears = for the same reason. This gives
(see eq. (1) )
r± pdF= ; i.e. Z PdFz^ ; or, ^ i^i=0 (2)
Ih which z— distance of the centre of gravity of the cross-
section from the neutral axis, from which, though un-
known in position, the g;'s have been measured (see eq.
(4) § 23).
In eq. (2) neither p-^e nor F can be zero .•. z must = ;
i.e. the neutral axis contains the centre of gravity. Q. E. D.
[If the external forces were not all perpendicular to the
beam this result would not be obtained, necessarily.]
228. The Shear. — The " total shear," or simply the
'* shear," in the cross-section is the sum of th.e vertical
shearing stresses on the respective dF's. Call this sum
J, and we shall have from the free body in Fig. 222, by
putting ^y=0 (F being vertical)
P,—F,—J=0.:J=F,—Pi . . (3)
That is, the shear equals the algebraic sum of the ex-
ternal forces acting on one side (only) of the section con-
sidered. This result implies nothing concerning its mode
of distribution over the section.
229. The Moment. — By the "Moment of Flexure" or
simply the Moment, at any cross- section is meant the sum
of the moments of the elastic forces of the section, taking
ihe neutral axis as an axis of moments. In this summa-
tion the normal stresses appear alone, the shear taking no part,
having no lever arm about the axisiVA. Hence, Fig. 222, the
moment of flexure (or "moment of resistance")
=J(ipdF).=f/dF.^=£^ (*)
This function, CdFz^, of the cross-section or plane figure
FLEXURE. 249
is the quantity called Moment of Inertia of a plane figure,
§ 85. For the free body in Fig. 222, by putting 2'(mom.3
about the neutral axis NA)=0, we have then
^ — PiX^-]rP^X2=Q, or in general^ lL=zM . (5)
e e
in which M signifies the sum of moments,* about the neutral
axis of the section, of all the forces acting on the free body
considered, exclusive of the elastic forces of the exposed
section itself. M is also called the "Bending Moment."
Example. — In Fig. 222 let Pi = 3 and P2 = 4 tons, Xi = l ft. 8 in. and
^2 = 5 in.; the section of the beam being a rectangle, with NA=b = 3 in.
and QB=^h = Q in. Then I about axis NA is, (p. 94), fo/i^n- 12 = 54 in.*;
and e=3 in. Hence the "bending moment," M, =3X20-4X5 = 40
in. -tons. Equating M to the "moment of resistance" [or moment of
the "stress couple" (see § 230)] we obtain, from eq. (5), p = Me-^I =
40 X 3 H- 54 = 2.22 tons/in.^ for the unit normal stress in the outer
fibre at Q, or B. We find also, for the shear at section QB, / = 4— 3 = 1 ton.
230. Strength in Flexure. — Eq. (5) is available for solving
problems involving the Strength of beams and girders, since
it contains p, the greatest normal stress per unit of area to
be found in the section.
In the cases of the present chapter, where all the exter-
nal forces are perpendicular to the prism or beam, and
have therefore no components parallel to the beam, i.e. to
the axis X, it is evident that the normal stresses in any
section, as QB Fig. 222, are equivalent to a couple ; for the
condition I!X=0 falls entirely upon them and cannot be
true unless the resultant of the tensions is equal, parallel,
and opposite to that of the compressions. These two equal
and parallel resultants, not being in the same line, form a
couple (§ 28), which we may call the stress -couple. The
moment of this couple is the " moment of flexure " '~ , and
it is further evident that the remaining forces in Fig. 222,
viz.: the shear J and the external forces Pj and Pg* are
equivalent to a couple of equal and opposite moment to
the one formed by the normal stresses,
* It is evident, therefore, that J!f (ft.-lbs., or in. -lbs.) is numerically equal
to the "moment of flexure," or moment of the " stress couple " ; so that
occasionally it maybe convenient to use "Jf" to denote the value of the
latter momeut also.
250
MECHAKICS OF EXGHirBEIiriirG.
231. Flexural Stiffness. — Tiie neutral line, or elastic curvo^
containing the centres of gravity of all tlie sections, was
originally straight ; its radius of curvature at any point,
as N, Fig. 222, c'uring flexure may be introduced as fol-
lows. QB and U'V are two consecutive cross -sections,
originally parallel, but now inclined so that the intersec-
tion G, found by prolonging them sufficiently, is the centre
of curvature of the ds (put =dx) which separates them, at
JSf, and CG=p= the radius of curvature of the elastic
curve at N. From the similar triangles U' TIG and GNG we
have dk'.dx'.:e;Pf in which dX is the elongation, U' U^ of a
portion, originally =c?cc, of the outer fibre. But the rela-
tive elongation £=-t— of the latter is, by §184, within the
elastic limit, =^.\ -:^ =— and eq. (5) becomeL
E E p ^ ^
EI
=M
(6)
AXIS X
From (6) the radius of curvature can be computed. E~
the value of E^—E^, as ascertained from experiments in
bending.
~ To obtain a differential equation of the elastic curve, (6)
may be transformed thus, Fig. 223. The curve being very
flat, consider two consecutive
(is's with equal dx's ; they may
be put = their c^x's. Produce
the first to intersect the dy of the
second, thus cutting off the d^y^
i'e. the difference between two
^■^Jfy consecutive dy'^. Drawing a per-
pendicular to each ds at its left
extremity, the centre of curva-
ture G is determined by their in-
tersection, and thus the radius
of curvature p. The two shaded
Fig. 223. triangles have their small angles
equal, and d^y is nearly perpen-
dicular to the prolonged ds ;
hence, considering them sim-
ilar, we have
\p,dx:'.dx'.d^y :.-^J^^,
FLEXURE. 251
and hence from eq. (6) we ) , , ^A^V
may write | (approx.) ±EI^=M . (7)
as a differential equation of the elastic curve. From this
the equation of the elastic curve may be found, the de-
flections at different points computedj and an idea thus
formed of the stiffness. All beams in the present chap-
ter being prismatic and Jiomogeneous both jE' and / are the
same (i.e. constant) at all points of the elastic curve^ In
using (7) the axis Xmust be taken parallel to the length
of the beam before flexure, which must be slight ; the
minus sign in (7) provides for the case when d^y-r-dx^ ises"=
sentialiy negative.
232. Resilience of Flexure. — If the external forces are made
to increase gradually from zero up to certain maximum
Yalues,. some of them may do work, by reason of their
points of application moving through certain distances
due to the yielding, or flexure, of the body. If at the be-
ginning and also at the end of this operation the body is
at rest, this work has been expended on the elastic resis-
tance of the body, and an equal amount, called the work
of resilience (or springing-back), will be restored by the
elasticity of the body, if released from the external forces,
provided the elastic limit has not been passed. The energy
thus temporarily stored is of the potential kind; see §§
148, 180, 196 and 218,
232a. Distinction. Between Simple, and Continuous, Beams (or
■** Girders "). — The external forces acting on a beam consist
generally of the loads and the " reactions " of the sup*
ports. If the beam is horizontal and rests on two supports
only, the reactions of those supports are easily found by
elementary statics [§ 36] alone, without calling into ac-
count the theory of flexure, and the beam is said to be a
Simple Beam, or girder ; whereas if it is in contact with
more than two supports, being " continuous,'* therefore,
over some of them, it is a Continuous Girder (§ 271). The
Temainder of this chapter will deal only with simple
252
MECHANICS or E:NGi:tfEEIl]:NG.
ELASTIC CURVES.
233. Case I. Horizontal Prismatic Beam, [Supported at Both
Ends, With a Central Load, Weight of Beam Neglected. — Fig.
224. First considering the whole beam free, we find eack
k-
-Vd-
%
.—x-
-I-
:^
Fig. 324. § 233.
reaction to be =%P. AOB is the neutral line ; required
the equation of the portion OB referred to as an origin,
and to the tangent line through as the axis of X To
do this consider as free the portion mB between any sec-
tion, m on the right of and the near support, in Fig.
225 The forces holding this free body in equilibrium
Fig. S25.
Fis. S26.
nre the one external force ^P, and the elastic forces act-
ing on the exposed surface. The latter consist of J, the
shear, and the tensions and compressions represented in
the figure by their equivalent " stress-couple." Selecting
N, the neutral axis of m, as an axis of moments (that J
may not appear in the moment equation) and putting
2 (mom) =0 we have
P (I
2V"2
— X j
rd^y-i
'y -P (I
dx" dx' 2 \2 /
(1)
Fig. 226 shows the elastic curve OB in its purely geomet-
rical aspect, much exaggerated. For axes and origin as in.
figure d^y-^doc^ is positive.
ELASTIC CURVES. 253
Eq. (1) gives the second a;-deriYative of y equal to a
function of x. Hence tlie first fl?-derivative of y will be
equal to tlie a?-anti-derivative of tliat function, plus a con-
stant, (7. (By anti -derivative is meant tlie converse of de-
rivative, sometimes called integral though not in the sense
of summation). Hence from (1) we have (^/ being a con-
stant factor remaining undisturbed)
M^=~(Lx — -\+G . . (2)*
dx 2 V2 2r
(2)' is an equation between two variables c?2/-i-c?a; and a?, and
holds good for any point between and B; dy-^dx de-
noting the tang, of a, the slope, or angle between the tan-
gent line and X At the slope is zero, and x also zero ;
nencs at (2)' becomes
^7x0=0— 0+C
which enables us to determine the constant C, whose value
must be the same at as for all points of the curve.
Hence C=0 and (2)' becomes
EI
ay _ r ( I xf\ ..^
^~2"i-2'^2j • • ' ^^'
from which the slope, tan. «, (or simply a, ir jt -measure:
since the angle is small) may be found at auy point. Thus
at B we have x=}4l and dy-^dx=ai, and
. _ 1 PI'
••"^""16" M
Again, taking the cp-anti-derivative of both, members of eq,
(2) we have
^i2/=-f-(^-^)+C" . . . (3)'
and since at both x and y are zero, G' is zero. Hence
the equation of the elastic curve OB is
254 MECHANICS OF ENGIXEEEIFG.
^^^=f (^f) • • • <«'
To compute the deflection of from the right line joiii'^
ing A and ^ in Fig. 224, i.e. BK, =c?, we put x^}^lm{^), a
being then =d, and obtain
^^=■^=©•5 • • • <**
Eq. (3) does not admit of negative values for x ; for if
the free body of Fig. 225 extended to the left of 0, the ex-
ternal forces acting would be P, aownward, at ; and y^P,
upward, at B, instead of the latter alone ; thus altering
the form of eq. (1). From symmetry, however, we know
that the curve AO, Fig. 224, is symmetrical with OB about
the vertical through Q.
Numerical Illustration. — Let [the beam shown in Fig. 224, resting
on two unyielding supports at the same level, be of white oak timber
and bear a load of P = 200 lbs. at the middle, its length being Z=12 ft.
and cross-section rectangular with a width (horizontal) of 6 = 2 in. and
height /i = 6 in. The modiilus of elasticity E will be taken as 1,600,000
lbs./ in. ^ Required the radius of curvature, p, or the elastic curve at
a point 4 ft. from the right-hand pier (or left).
From the free body in Fig. 225 we have, using the form El-i-p for
the moment of the stress-couple in the section, and putting i'(moms.)j\r
= 0, with x = 2 ft., £7 -^J0= 100X48, the inch and pound being selected
as units. Now I=bh^-irl2 (p. 94) which = 36 iu.^j whence, solving,
(0 = 1,600, 000 X 36 -H 4800 = 4000 in. The cin-ve is evidently very flat.
The smallest radius of curvature is found at the middle of the beam
and is 2666 in.; at either extremity, A or B, it is infinite, since at each
of these points the moment of the stress-couple is zero.
At the same point (4 ft. from B) the "slope" of the elastic curve,
viz., dy^dx, is found by putting x = 2 ft. = 24 in!, in eq. (2) from which
is derived tan a = dy I dx = Q.Q025, corresponding to an angle of 0° 8' 36".
At the extremity B we find, from ai = PP-7-16£'/, the slope of the tangent
line to be ai = 0.0045; which is the tangent of 0° 15' 29".
The deflection of the middle point is known from eq. (4), viz.,
d = PP^48EI; i.e., d=(200X 144 X 144 X 144) -v- (48X1,600,000X36) =
0.216 in.
It now remains to ascertain if the elastic limit is passed in any fibre
of the beam. If we put the form p/-^e (for moment of stress-couple)
in place of the present left-hand member of eq. (1), and solve for the
unit (normal) stress in outer fibre, we findp=JPe(J Z— 2;)-f-/, which
shows that p is greatest in the outer fibre of the section for which ^l—x
is greatest, within the limits of the half-length; and this occurs at the
middle of the beam, where x = 0. With this substitution we obtain
p(max.) = pm = Pie h- (47) ; or pm = (200 X 12 X 12 X 3) -f- (4 X 36) = 600
lbs./ in. ^, which is well within the elastic limit, for tension or compression
in white oak.
ELASTIC CURVES. 255
233a. Load Suddenly Applied. — Eq. (4) gives the deflection
d corresponding to the force or pressure P applied at the
middle of the beam, and is seen to be proportional to it
If a load G hangs at rest from the middle of the beam,
P=G', but if the load G, being initially placed at rest
.upon the unbent beam, is suddenly released from the ex-
ternal constraint necessary to hold it there, it sinks and
dehects the beam, the pressure P actually felt by the beam
varying with the deflection as the load sinks. What is
the maximum deflection d^ ? and what the pressure P^^
between the load and the beam at the instant of maximum
deflection? In this motion of the body, or "load," it is
acted on by two forces, the constant downward force G (its
weight) and the variable upward force P, whose average vahie
is |Pni ; while its initial and final kinetic energy are each zero.
G does the work Gd^, while the work done upon P is ^Pmdm \
hence, by the theorem of '' Work and Energy " (p. 138), v/e
have
Gd^^hPrJra + 0-Q (5)
That is, Pm = 2(r. Since at this instant the load is sub-
jected to an upward force of 2 (r and to a downward force
of only G (gravity) it immediately begins an upward mo-
tion, reaching the point whence the motion began, and
thus the oscillation continues. We here suppose the elas-
ticity of the beam unimpaired. This is called the " sud-
den " application of a load, and produces, as shown above,
double the pressure on the beam which it does when grad-
ually applied, and a double deflection. The work done
by the beam in raising the weight again is called its re-
silience.
Similarly, if the weight G is allowed to fall on the mid-
dle of the beam from a height Ji, we shall have
Gx(h+d^, or approx., Gh,= ^P^d^i
and hence, since (4) gives d,^ in terms of P^,
e;i=i .^P oreA=2i^^ . (6)
256 MECHANICS OIT EXGINEEEIXG.
This theory suppcs5es the mass of the beam small com-
pared with the falling weight.
234. Case II. Horizontal Prismatic Beam, Supported at Both
End? Bearing a Single Eccentric Load. Weight of Beam Neg-
p p lected. — Fig. 227. The reactions
4 t . of the -points of support, Pn and
O I AXIS X I B "■ . X i ' >^
y'^^^^i^^J?^ Jvm ^i^^^^^^^^fe -^1' ^^^ easily found by consider-
j, — -]Ei^._-_J^^^^^^'^ I ing the whole beam free, and put-
j Ip j ting first 2'(mom.)o=0, whence i'l
^\ '^' 1 =PZh-Zi, and then J(mom.)B=0,
ri«227. whence Po= An— O^^i- i'o and
Pi will now be treated as known quantities.
The elastic curves 0(7 and OP, though having a comm on
tangent line at (and hence the same slope a^, and a com-
mon ordinate at 0, have separate equations and are both
referred to the same origin and axes, as shown in the
figure. The slope at 0, «o> and that at P,«i, are unknown
constants, to be determined in the progress of the work.
Eq[uation of OC. — Considering as free a portion of the
beam extending from P to a section made anywhere on
OC, X and y being the co-ordinates of the neutral axis of
that section, we conceive the elastic forces put in on the
exposed surface, as in the preceding problem, and put
2'(mom. about neutral axis of the section) =0 which gives
(remembering that here d?y-~dx^ is negative.)
Ei^^=p{y-x)—p,{k—x)', , . (1)
(X OC
whencO;. by taking the x anti-derivatives of both members
■ M ^ =P(lx—^)-F,{lx— -^)+ C
ax 2 2
To find 0, write out this equation for the point 0, where
dy-^dx=aQ and a;=0, and we have 0=P/«o> hence the
equation for slope is
FLEXURE ELASTIC CUKVES. 257
EI^=P{lx—^)-P,{l,x-^)+EIa^ .. (2)
Again taking the x anti-derivatives, we have from (2)
Ely =P (^^|1_|^._P,^^|^_^ YEIa,x+{C'=0) (3)
'^at Oboth X and 2/ are —0 .°. C'=0). In equations (1), (2),
and (3) no value of x is to be used <0 or >Z, since for
points in CB different relations apply, thus
Equation of CB. — Fig. 227. Let the free body extend
from ^ to a section made anywhere on (7^.2'(moms.), as
before, =0, gives (see foot-note on p, 322)
^^^=-^^^1-^) . . . (4)
(N.B. In (4), as in (1), Eld^y—dx^ is written equal to a neg-
ative quantity because itself essentially negative ; for the
curve is concave to the axis X in the first quadrant of the
co-ordinate axes.)
From (4) we have in the ordinary way (aj-anti-deriv.)
EI"^ =-Pil,x -J^)+C" . . (5X
ax 2
To determine C", consider that the curves CB and OG
have the same slope (dy-r-dx) at G where x=l; hence put
x~l in the right-hand members of (2) and of (5)' and
equate tha results. This gives C" = %PV-\-EIaQ and .-.
^it-^ + ^I'^PS.^t^ . (5)
A A o
258 MECHANICS OE E2fGlNEEIlIN&.
At C, where J5 = ?, botli curves have the same ordinate;
hence, by putting x — l in the right members of (3) and (6)'
and equating results, we obtain C'"— — }iPl^. .'. (6)' bo
comes
Mly = y2Pl'x+EIa^7>—P^
~2 6"
~6"
(6)
as the Equation of CB, Fig. 227. But ag is still an unknown
constant, to find which write out (6) for the point B where
X = li, and y = 0, whence we obtain
- 1 ^Fl'—3Fl\-{-2P,l,^] . . , (7)
" 6m\
«!= a similar form, putting Pq ^^t P,, and (l^ — I) for I.
235. Maximum Deflection in Case II — Fig. 227. The or-
dinate ?/„ of the lowest point is thus found. Assuming
^> /4 k> it will occur in the curve G. Hence put the
dy-h-dx of that curve, as expressed in equation (2), =0.
Also for O.Q write its value from (7), having put Pi=P?-r-Zij
and we have
whence [a? for max. y] = ^yi(2k—l) ■
Now substitute this value of x in (3), also ao from (7), and
putPi =P?-T-Zi, whence
Max. Deflec.=2/max=^% . -^ ll'—3l%+W,'] ^M^^O-
236. Case III. Horizontal Prismatic Beam Supported at Both
Ends and Bearing a Uniformly Distributed Load along its Whole
Length. — (The weight of the beam itself, if considered,
FLEXUEE. ELASTIC CUltVES.
259
constitutes a load of this nature.) Let 1= the length
of the beam and w= the weight, per unit of length,
of the loading ; then the load coming upon any length x
will be =ivx, and the whole load ^=ui. By hypothesis w
is constant. Fig. 228. From symmetry we know that the
W=«)?
Ulli I 1 1 1 lU
Fig. 228.
reactions at A and B are each =}4iol, that the middle of
the neutral line is its lowest point, and the tangent line at
is horizontal. Conceiving a section made at any point
m of the neutral line at a distance x from 0, consider as
free the portion of beam on the right of m. The forces
holding this portion in equilibrium are yz'^h ^^^ reaction
at B ; the elastic forces of the exposed surface at m, viz.:
the tensions and compressions, forming a couple, and J
the total she?r ; and a portion of the load, iv(^/2l — x). The
sum of the mc ments of these latter forces about the neu-
tral axis of m, is the same as that of their resultant; (i.e.,
their sum, since they are parallel), and this resultant acts in
the middle of the length ^Z — x. Hence the sum of these
moments =w(}4l — x)^[}4l — x). Now putting 2' (mom.
about neutral axis of w)=0 for this free body, we have
BI
dx^
}4wl{}4l—x)—}^w(}41^xy
i.e.,^/g- = >
^t^(l^?2_^)
(1)
260 MECHANICS OF ENGINEEEIN&.
Taking tlie cc-anti-derivative of both sides of (1),
^^Tx =yM}{l^'^—}i^')+{G=0) (2)
as the equation of slope. (The constant is =0 since at
both dy-i-dx and x are =0.) From (2),
my=-^i}il'x'-%x')+[C'=0] . . (3)
which is the equation of the elastic curve ; throughout,
i.e., it admits any value of x from x=-\-y2^ to x= — yil.
This is an equation of the fourth degree, one degree high-
er than those for the Curves of Cases I and II, where
there were no distributed loads. If w were not constant,
but proportional to the ordinates of an inclined right line,
eq. (3) would be of the fifth degree ; if lo were propor-
tional to the vertical ordinates of a parabola with axis
vertical, (3j would be of the sixth degree ; and so on.
By putting x=y^l in (3) we have the deflection of be-
low the horizontal thro ugh A and B, viz.: (with W=^ total
load ^wl)
384 ' m S84: ' FI ' ' ^ ^
237. Case IV. Cantilevers. — A horizontal beam whose only
support consists in one end being built in a wall, as in
Fig. 229(a), or supported as in Fig.
229(&) is sometimes called a canti-
lever. Let the student prove that in
Fig. 229(a) with a single end load P,
the deflection of^ below the tangent
at Ois d=j/Pl^-i-£^I;the same state-
ment applies to Fig. 229(&), but the
tangent at is not horizontal if the
beam was originally so. It can also
be proved that the slope at B, Fig.
229(a) (from the tangent at 0) is
FLEXUEE ELASTIC CURVES. 261
«i=
2^7"
The greatest deflection of the elastic curve from the right
line Joining AB, in Fig. 229(6), is evidently given by the
equation for y max. in § 235, by writing, instead of P of
that equation, the reaction at in Fig. 229(&). This assumes
that the max. deflection occurs between A and 0. If it
occurs between and B put (li—l) for I.
If in Fig. 229(a) the loading is uniformly distributed
along the beam at the rate of w pounds per linear unit,
the student may also prove that the deflection of B below
the tangent at is
238. Case V. Horizontal Prismatic Beam Bearing Equal Ter-
minal Loads and Supported Symmetrically at Two Points.—
Fig. 231. Weight of beam neglected. In the preceding
cases we have made use of the approximate form Eld'^y-r-dx^
in determining the forms of elastic curves. In the present
ris~T%
p\
1-^ '- < 1-
Pig. 231. Fig. )i^Z.
case the elastic curve from to (7 is more directly dealt
with by employing the more exact expression EI-^f> (see
§ 231) for the moment of the stress-couple in any sectioUo
The reactions at and Care each =P, from symmetry.
Considering free a portion of the beam extending from A
to any section m between and C (Fig. 232) we have, by
putting 2 (mom. about neutral axis of m)=0,
P{i+x)- ^~Px==o .-. p^ 4r
262 MECHANICS OF EXGINEBKIJfG.
That is, the radius of curvature is the same at all points
of OG ] in other words 0(7 is the arc of a circle with the
above radius. The upward deflection of F from the right
line joining and G can easily be computed from a knowl-
edge of this fact. This is left to the student as also the
value of the slope of the tangent line at (and G). The
deflection of D from the tangent at G=^l^Pf-^EL as ip
Fig, 229(a),
SAFE LOADS IIS^ FLEXUKE.
239. Maximum Moment. — As we examine the different sec-
tions of a given beam undar a given loading we find differ-
ent values oi p, the normal stress per unit of area in the
outer element, as obtained from eq. (5) § 229, viz.:
^=il/. . . . , (1)
e
in which I is the " Moment of Inertia " (§ 85) of the plane
figure formed by the section, about its neutral axis, e the
distance of the most distant (or outer) fibre from the neu<
tral axis, and ilf the sum of the moments, about this neu-
tral axis, of all the forces acting on the free body of which
the section in question is one end, exclusive of the stresses
on the exposed surface of that section. In other words
Jf is the sum of the moments of the forces which balance
the stresses of the section, these moments being taken
about the neutral axis of the section under examination.
For the prismatic beams of this chapter e and /are the
same at all sections, hence p varies with M and becomes a
maximum when J/ is a maximum. In any given case the
location of the " dangerous section" or section of maximum
M, and the amount of that maximum value may be deter-
mined by inspection and trial, this being the only method
(except by graphics) if the external forces are detached.
FLEXUEE SAFE LOADS. 263
If, however, the loading is continuous according to a de-
finite algebraic law the calculus may often be applied,
taking care to treat separately each portion of the beam
between two consecutive reactions of supports^ or detached
loads.
As a graphical representation of the values of 31 along
the beam in any given case, these values may be conceived
laid off as vertical ordinates (according to some definite
scale, e.g. so many inch-lbs. of moment to the linear inch
of paper) from a horizontal axis just below the beam. If
the upper fibres are in compression in any portion of the
beam, so that that portion is convex downwards, these or-
dinates will be laid off below the axis, and vice versa ; for
it is evident that at a section where ilf=0, p also =0, i.e.,
the character of the normal stress in the outermost fibre
changes (from tension to compression, or vice versa) when
if changes sign. It is also evident from eq. (6) § 231 that
the radius of curvature changes sign, and consequently the
curvature is reversed, when J/ changes sign. These mo-
ment ordinates form a Moment Diagram, and the extremities
a Moment Curve.
The maximum riioment, ilf^, being found, in terms of
the loads and reactions, we must make the p of the " dan-
gerous section," where M= M^^ equal to a safe value R',
and thus may write
^=M^ . . . o (2)
e
Eq. (2) is available for finding any one unknown quanti-
ty, whether it be a load, span, or some one dimension of
the beam, and is concerned only with the Strength, and not
with the stiffness of the beam. If it is satisfied in any
given case, the normal stress on all elements in all sections
is known to be = or ^X 60+1x84+4x120] ^3.3 tons .-. Po=5.5—
3.3=2.2 tons.
The shear anywhere between O and ^is J= Po=2.2 tons.
^ and i^ is e/ =2.2— 1^=1.7
tons.
The shear anywhere between i^ and His J =2.2 — ^ — 1 =
0.7 tons.
The shear anywhere between H and B is J = 2.2 — }4 — 1
—4 =—3.3 tons.
Since the shear changes sign on passing H, .-. the max.
moment is at ^; whence making HO free, we have
M at H=M,,, =2.2 x 120— ^^ x 60—1 x 36 =198 inch -tons.
For safety M,„ must = , in which B'='^ ton per sq.
inch, e = }4^ — }4 of unknown depth of beam, and /, §90, =
I bM, with & = 10 inches
,vi. >^ .|-Xl0.¥^198; or 71^-237.6 .-. h=15A inches.
245. Comparative Strength of Rectangular Beams. — For such
a beam, under a given loading, the equation for safe load-
ing is
^=3i;„ i. e. ye E bh'=M^ .... (1)
«
FLEXUEE. SAFE LOADS. 273
whence the following is evident, (since for the same length,
mode of support, and distribution of load, M^ is propor-
tional to the safe loading.)
For rectangular prismatic beams of the same length,
same material, same mode of support and same arrange-
ment of load :
(1) The safe load is proportional to the width of beams
having the same depth (A).
(2) The safe load is proportional to the square of the
depth of beams having the same width (h).
(3) The safe load is proportional to the depth of beams
having the same volume (i. e. the same hh]
(It is understood that the sides of the section are hori-
zontal and vertical respectively and thai the materia] \^
homogeneous.)
246. Comparative Stiflfness of Rectangular Beams.— Taking tli*.
deflection under the same loading as an inverse me^-sure
of the stiffness, and noting that in §§ 233, 235, and 236,
this deflection is inversely proportional to I—k hh^ =
the " moment of inertia "of the section about its neutral
axis, we may state that :
For rectangular prismatic beams of the same length,
same material, same mode of support, and same loading .•
(1) The stiffness is proportional to the width for beams
of the same depth.
(2) The stiffness is proportional to the cube of the
height for beams of the same width (&).
(3) The stiffness is proportional to the square of the
ciepth for beams of equal volume (hhl),
(4) It the length alone vary, the stiffness is inversely
proportional to the cube of the length.
247. Table of Moments of Inertia. — These are here recapitu-
lated for the simpler cases, and also the values of *?. the
distance of the outermost fibre from the axis.
Since the stiffness varies as /(other things being equal).
274
MECHANICS OF ENGINEERING.
while tlie strength, varies* as I~-e, it is evident that a
square beam has the same stiffness in any position (§89),
while its strength is greatest with one side horizontal, for
then e is smallest, being —^6.
Since for any cross-section 1= j dF z^^ in which «=the
distance of any element, dF, of area from the neutral axis,
a beam is made both stiffer and stronger by throwing
most of its material into two flanges united by a vertical
web, thus forming a so-called " I-beam " of an I shape. But
not without limit, for i;he web must be thick enough to
cause the flanges to act together as a solid of continuous
substance, and, if too high, is liable to buckle sideways^
thus requiring lateral stiffening. These points will be
treated later.
SECTION.
/
e
Rectangle, width = b, depth = h (vertical)
Vm bh^
%h
Bollow Rectaiigle, symmet. about neutral axia. See 1
Fig. 238 (a) f
Vi» [6i h,»-b^ h\^
%h,
•Triangle, width =6, height = h, neutral axis parallel ^_
to base (horizontal). )
Vse M3
%h
Circle of radius r
%^r^
r
Eing of concentric circles. Fig. 238 (b)
}in(r\~r*^)
Ti
Ehombus; Fig. 238 (c) h = diagonal which is vertical.
V4e 5AS
%h
Square with side b vertical.
Via b*
%b
" " " 6 at 45° with horiz.
Vn **
HbVS
248. Moment of Inertia of I-beams, Box-girders, Etc. — In
common with other large companies, the Cambria Steel
* This function, /-r-e, of the plane figure formed by the cross-section
of a beam is evidently of three dimensions of length (cubic inches, for
example), and is tabulated in the handbooks of the steel companies for
different shapes of section; it is called the "section-modulus." See
next page.
FLEXURE. SAFE LOADS.
275
Co. of Johnstown, Pa., manufactures prismatic rolled beams
and other "shapes," of structural steel, which are variously
called I-beams, deck-beams (or " bulb -beams "), rails, angles,
T-bars, channels, Z-bars, etc., according to the form of their
sections. See Fig. 239 for some of these forms. The company
d-
^
T
CHANNEL. DECK-BEAM. RAIL.
Fie. 239.
publishes a pocket-book giving tables of quantities rela-
ting to the strength and stiffness of beams, such as the
safe loads for various spans, moments of inertia of their
sections in various positions, etc., etc„ The moments of
inertia of /-beams and deck -beams are computed accord-
ing to §§ 92 and 93, with the inch as linear unit. The
/-beams range from 4 in. to 24 inches deep, the deck-
beams being about 7 and 8 in. deep. (See foot-note, p. 274.)
For beams of still greater stiffness and strength com-
binations of plates, channels, angles, etc., are riveted to-
gether, forming " built-beams,"' or " plate girders." The
proper design for the riveting of such beams will be ex-
amined later. For the present the parts are assumed to
act together as a continuous mass. For example, Fig. 240
shows a " box -girder," formed of two " channels " and
two plates riveted together. If the axis of symmetry, JV,
h h \ is to be horizontal it becomes the neu-
3 ff"* tral axis. Let (7= the moment of iner-
tia of one channel (as given in the
pocket-book mentioned) about the axis
iV perpendicular to the web of the chan-
nel. Then the total moment of inertia oj
the combination is (nearly)
^
m
^
Fig. 340.
4 =W^'iUd?—ld't'{d^y2if
(1>
276
MECHANICS OF EKGIXEBRING.
In (1), &, t, and d are the distances given in Fig. 240 {d ex-
tends to the middle of plate) while d' and t' are the length
and width of a rivet, the former from head to head
(i.e., d' and t' are the dimensions of a rivet-hole).
For example, a box-girder of structural steel is formed of
two 15 -in. channels (35 lbs, per foot) and two plates 10 in.
wide and 1 in. thick ; the rivet-holes | in, wide and If in.
long. That is, 5-10; i = l; d=8; f = f; and d' = l| in.
Also from the hand-book we find that for the channel in
question C==320 in,-* (i.e., biquad. in.). Hence, eq. (1),
7^=640 + 2X10X1X64-4x1. 1(8-^)^ = 1625 in.4^
In this instance e=8^in, ; and if 15,000 lbs, /in.^ ( = 7,5
tons/in.2) be taken as the value of R' (greatest safe normal
stress in the extreme fibre of any section) as used by the
Cambria Steel Co. for box-girders in buildings, we have
" R'l 15000X1625 . . .,
— = ^^ = 2, 867, 500 inch-lbs.
e 8.5 ' '
That is, the max, safe ^'moment of resistances^ of the box-
girder is M^ = 2,867,00O inch-lbs. = 1433.7 inch-tons; this
quantity having to do with normal stresses in the section. The
greatest " hending-momenf^ due to the amount, and mode, of
loading on the beam, must not exceed this. Proper provision
for the shearing stresses in the section, and in the rivets, wiU
be considered later,)
249. Strength of Cantilevers. — In Fig. 241 with a single
, I ^ w^w? concentrated load P at the
i::!::^:::::::;;^ Q_i_J_J_J_J_Jl projecting extremity, we
r° easily find the moment at
71 to be Jf =Px, and the
/ max. moment to occur at
the section next the wall,
j ^f its value being M^=Pl.
The shear, J", is constant,
Fi«-24i. Fig. 242. ^mj = P at all sections.
The moment and shear diagrams are drawn in accordance
with these results.
FLEXUKE. SAFE LOADS.
277
If the load W= wli^ uniformly distributed on tlie can-
tilever, as in fig. 242, by making nO free we have, putting
-2'(mom. about n) = 0,
pi
^-^=ivx . I .'.M=}iwaP.'. J4=;^w;Z2= ^
Wl.
Hence the moment curve is a parabola, whose vertex is at
0' and axis vertical. Putting I (vert, compons.) = we
obtain J = ivx. Hence the shear diagram is a triangle,
and the max. J = wl = TV.
250. Resume'ofthe Four Simple Cases. — The following table
shows the values of the deflections under an arbitrary
load P, or W, (within elastic limit), and of the safe load ;
Deflection
J Safe load (from ?!I
I = Mm)
Relative strength
j Relative stiffness
1 under same load
j Relative stiffness
I under safe load
J Max. shear = Jm,(.wa.d
t location,
Cantilevers.
With one end
loadP
Fii?. 241
EI
B'l
P, (at wall)
With unif . load
Fig. 242
Beams with two end supports.
Load P in
middle
Fig. 234
EI
B'l
J.
3
W, (at wall)
iS'EI
,B'l
ViP, (at supp).
Unif. loaa
W=wl
Fig. 235
5 y\/fi
384' EI
8
128
6
16
5
W, (at suppi)
also the relative strength, the relative stiffness (under the
same load), and the relative stiffness under the safe load,
for the same beam.
The max. shear will be used to determine the proper
web-thickness for /-beams and " built-girders." The stu-
dent should carefully study the foregoing table, noting
especially the relative strength, stiffness, and stiffness
under safe load, of the same beam.
Thus, a beam with two end supports will bear a double
278 MECHANICS OF ENGINEERING.
load, if uniformly distributed instead of concentrated in
the middle, but will deflect ^;( more ; whereas with a given
load uniformly distributed the deflection would be only
5/^ of that caused by the same load in the middle, provided
<-he elastic limit is not surpassed ii? either case.
261. E', etc. For Various Materials.— The formula& = Jf^,
e
from which in any given case of flexure we can compute
the value of p^, the greatest normal stress in any outer
element, provided all the other quantities are known,
holds good theoretically within the elastic limit only.
Still, some experimenters have used this formula for the
rupture of beams by flexure, calling the value of p^ thus
obtained the Modulus of Rupture, B. R may be found to
differ considerably from both the ^ or C of § 203 with
some materials and forms, being frequently much larger.
This might be expected, since even supposing the relative
extension or compression (i.e., strain) of the fibres to be
proportional to their distances from the neutral axis as
the load increases toward rupture, the corresponding
bLi'esses, not being proportional to these strains beyond the
elastic limit, no lono^'er vary directly as tlie distances from the
neutral axis ; and the neutral axis does not pass through the
centre of gravity of the section, necessarily.
The following table gives average values for R, R', R",
and E for the ordinary materials of construction.'^ E, the
modulus of elasticity for use in the formulsB for deflection,
is given as computed from experiments in flexure, and is
nearly the pame as E^^ and E^.
In any example involving R', e is usually written equal
to the distance of the outer fibre from the neutral axis,
whether that fibre is to be in tension or compression ;
since in most materials not only is the tensile equal to the
compressive stress for a given strain (relative extension
or contraction) but the elastic limit is reached at about
the same strain both in tension and compression.
* Wet, or unseasoued, timber is very cousiderably weal^er than that (such as
ordinary " dry" timber) containing only 12 per cent, of moisture. Large pieces
of timber talie a much longer time to season than small ones. (Johnson.)
FLEXUKE. SAFE LOADS.
279
Table foe Use in Examples in Flexure.
Timber.
Cast Iron.
Wro't Iron,
Structural
Steel.
Max. safe stress in outer fi- )
bre— if'dbs. per sq. inch). )
600
to
1,200
6,000 in tens.
12,000 in comp.
12,000
15,000
Stress in outer fibre at Elas. )
limit = j?''(lbs. per sq. in.) )
17,000*
to
35,000
30,000
and upward.
" Modul. of Rupture " 1
=i?=lbs. per sq. inch. )
4,000
to
10,000
40,000
50,000
60,000
E^Mod. of Elasticity, j
=lbs. per sq. inch. )
1,000,000
to
2,000,000
17,000,000
25,000,000
29,000,000
In the case of cast iron, however, (see § 203) the elastic
limit is reached in tension with a stress =9,000 lbs. per
sq. inch and a relative extension of ^^ of one per cent.,
while in compression the stress must be about double to
reach the elastic limit, the relative change of form (strain)
being also double. Hence with cast iron beams, once
extensively used but now largely replaced by rolled beams
of structural steel, an economy of material was effected
by making the outer fibre on the compressed side twice
as far from the neutral axis as that on the stretched side.
Thus, Fig. 243, cross-sections with unequal flanges were
used, so proportioned that the centre of
gravity was twice as near to the outer
fibre in tension as to that in compression,
i.e., e2=2ej; in other words more material
J is placed in tension than in compression.
The fibre A being in tension (within elas-
tic limit), that at B, since it is twice as far from the neu-
tral axis and on the other side, is contracted twice as much
as A is extended ; i.e., is f.nder a compressive strain
double the tensile strain at A, but in accordance with the
above figures its state of stress is proportionally as much
within the elastic limit as that of A.
* In the tests by U. S. Gov. in 1879 with I-beams, B" ranged from 25,000
to 38,000, and tlie elastic limit was reached with less stress in the large
than in the smaller beams. Also, for the same beam, U" decreased with
larger spans.
1
N
i
1 ^
A
f.
Fig. 243.
280 MECHANICS OP ENGIJfEERING.
The great range of values of R for timber is due not
only to the faet that the various kinds of wood differ
widely in strength, while the behavior of specimens of
any ona kind depends somewhat on age, seasoning, etc.,
but also to the circumstance that the size of the beam un-
der experiment has much to do with the result. The ex-
periments of Prof. Lanza at the Mass. Institute of Tech-
nology in 1881 were made on full size lumber (spruce), of
dimensions such as are usually taken for floor beams in
buildings, and gave much smaller values of R (from 3,200
to 8,700 lbs. per sq. inch) than had previously been ob-
tained. The loading employed was in most cases a con-
centrated load midway between the two supports.
These low values are probably due to the fact that in
large specimens of ordinary lumber the continuity of it&
substance is more or less broken by cracks, knots, etc.,
the higher values of most other experimenters having
been obtained with small, straight-grained, selected pieces,
from one foot to six feet in length. See footnote p. 278*
Yaluable information and tables relating to timber beams
may be found in the hand-book of the Cambria Steel Co.
The value R' = 1^,000 lbs. per sq. inch is employed by the
Cambria Steel Co. in computing the safe loads for their
rolled I-beams of structural steel ; but with the stipulation
that the beams (which are high and of narrow width) must
be secure against yielding sideways. If such is not the
case the ratio of the actual safe load to that computed with
i2' = 16,000 is taken less and less as the span increases.
The lateral security referred to may be furnished by the
brick arch-filling of a fire-proof floor, or by light lateral
bracing with the other beams.
252. Numerical Examples. — Example 1. — A square bar of
wrought iron, 1^ in. in thickness is bent into a circular :|:
arc whose radius is 200 ft,, the plane of bending being par-
allel to the side of the square. Bequired the greatest nor^
mal stress p^ in any outer fibre.
Solution. From §§ 230 and 231 we may write
—t =£— .'. p=eU-T-p, i.e., is constant.
FLEXURE. SAFE LOADS. 281
For the units incli and pound (viz. tliose of the table in §
251) we have e=% in., /> =2,400 in., and ^=25,000,000 lbs.
per sq. inch, ani .•.
i>=i>m=^x 25,000,000-^2,400 =7,812 lbs. per sq. in.^
which is quite safe. At a distance of ^ inch from tne
neutral axis, the normal stress is =\_%-^}i.^Pm — %Pm—
5,208 lbs. per sq. in. (If the force-plane (i.e., plane of
bending) were parallel to the diagonal of the square, e
would =}4x 1.5^^2 inches, giving p^ = \l ,S12x ^/2 ] H^s.
per sq. in.) § 238 shows an instance where a portion, 0C7,
Fig. 231, is bent in a circular arc.
Example 2. — A hollow cylindrical cast-iron pipe of radii
3 y2 and 4 inches* is supported at its ends and loaded in
middle (see Fig. 234). Eequired the safe load, neglecting
the weight oi the pipe. From the table in § 250 we have
for safety
P=4^
le
From § 251 we put i?'= 6,000 lbs. per sq. in.; and from §
247/=^(ri* — rf)\ and with these values, r2 being =4> '"'i —
4l, e=ri=4, 7i=-B- and Z=144 inches (the inch must be the
unit of length since i?' = 6,000 lbs. per sq. inch) we have
7>=4x6,000x;^- ^(256-150)-r-[144x4] .-. P=3,470 lbs.
The weight of the beam itself is (r= Vy, (§ 7), i.e.,
Q^^^r,'^ri)lr= f-(16-12i^)144Xjg=448 lbs.
(Notice that y, here, must be lbs., per cubic inch). This
weight being a uniformly distributed load is equivalent to
half as much, 221 lbs., applied in the middle, as far as the
strength of the beam is concerned (see § 250), .*. P must be
taken =3,249 lbs. when the weight of the beam is consid-
ered.
* And length of 13 feet, should be added.
282 MECHANICS OF ENGINEERING.
Example 3. — A Cambria I-beam, of structural steel, is to
be placed horizontally on two supports at its extremities and
is to be loaded imiformly (Fig. 235), the span being ^ = 20 ft.
5g„_^ Its cross-section. Fig. 244, has a depth
T ^ V^ parallel to the web, of 15 in. In the
lA
W f\
^
Fig. 244.
handbook of the Cambria Steel Co. it
-\ n" is designated as B 53, 15 in. in depth,
and weighing 42 lbs. per foot of length;
its section having a moment of inertia
/i = 442 in.4 about a gravity axis per-
pendicular to the web (for use when the web is vertical; the
strongest position) and 72 = 14.6 in.^ about a gravity axis
parallel to the web (i.e., when the web is placed horizontally).
First, placing the web vertically, we have from § 250,
7->/7
Wi = safe load, distributed, = 8-^. With R' = 16,000,
1 1 = 442, I = 240 inches, and ei --^ 7 J inches, this gives *
TFi = (8 X 16,000 X 442) - (240 X 7.5) - 31,430 lbs.
But this includes the weight of the beam, =20x42 = 840
lbs.; hence a distributed load of 30,600 lbs., or 15.3 tons,
may be placed on the beam (secured against lateral yielding).
The handbook of the Cambria Steel Co. referred to gives 15.7
tons as the safe load.)
With the web placed horizontally, we find as safe load
Tf 2 = 8^-^^ = (8 X 16,000 X 14.6) ^ (^240 X ^) = 2830 lbs. ;
or less than 1/10 of Wi. Hence in this position the beam
could carry safely only 1990 lbs. above its own weight.
Example 4. — Required the deflection at the middle in the
first case of Ex. 3. From § 250 this deflection is
, _A
^'"384
* The handbook of the Cambria Steel Co. also gives in a separate
column the quantity /j^ej, called the "section-modulus," S, (cub. in.
or in.^); so that the formula for the safe load would be TFj = 8JS'*S -e- i,
S having the value 58.9 in.' in the present instance.
Wil^ 5 8R'h
l^
5
R' Z2
Ell "384" lei
'Ell
"48
E ' ei
FLEXUliE. SAFP] LOADS. 283
. , 5 16,000 (240)2
'-'•' ^^^48 • 29,000,000 ' V ^ ^'^^
.-. d,=0. 4:4:1 in.
Example 5. — A rectangular beam of yellow pine, of widtli
6=4 inches, is 20 ft. locg, rests on two end supports, and is
to carry a load of 1,200 lbs. at the middle ; required the
proper depth h. From § 250
le T 12 ' ^h
.: h?=6Pl-^4:B'b. For variety, use the inch and ton. For
this system of units P=0.60 tons, i?'=0.50 tons per sq. in.,
1=24:0 inches and 6= 4 inches.
.-. /i2=(6x0.6x240)4-(4x0.5x4)=108sq. in. .-. ^=10.4 in.
Example 6. — Suppose the depth in Ex. 5 to be deter-
mined by the conditien that the deflection shall be = Ygoo
•f the span or length. "We should then have from § 250
d= k 1=1 ^'
600 48 EI
Using the inch and ton, with ^=1,200,000 lbs. per sq. in.,
which = 600 tons per sq. inch, and /=Yl2^^^ we have
;^3^ 500x0.60x240x240x12 ^^^ _. ^^^^^ ^
48x600x4
As this is > 10.4 the load would be safe, as well.
Example 7. — Required the length of a wro't iron pipe
supported at its extremities, its internal radius being 2}^
in., the external 2.50 in., that the deflection under its oivn
weight may equal Yioo of the length. 579.6 in. Ans.
Example 8. — Fig. 245. The wall is 6 feet high and one
foot thick, of common brick work
(see § 7) and is to be borne by an
7-beam in whose outer fibres no
n
greater normal stress than 8,000
*£ lbs. per sq. inch is allowable. If
Pio. 245. a number of I-beams is available,
284 MECHANICS OF ENGINEERLNG.
ranging in height from 6 in. to 15 in. (by whole inches),
which one shall be chosen in the present instance, if their
cross-sections are Similar Figures, the moment of inertia of
the 15-inch beam being 800 biquad. inches ?
The 12-inch beam. Ans.
SHEARIXa STRESSES IN FLEXURE.
253. Shearing Stresses in Surfaces Parallel to the Neutral
Surface. — If a pile of boards (see Fig. 246) is used to sup-
port a load, the boards being free to slip on each other, it
is noticeable that the ends overlap, although the boards
Fig. 246.
are of equal length (now see Fig. 247) ; i.e., slipping has
occurred along the surfaces of contact, the combina-
tion being no stronger than the same boards side by
side. If, however, they are glued together, piled as in the
former figure, the slipping is prevented and the deflection
is much less under the same load P. That is, the com-
pound beam is both stronger and stiifer than the pile of
loose boards, but the lendency to slip still exists and is
known as the " shearing stress in surfaces parallel to the
neutral surface." Its intensity per unit of area will now
be determined by the usual " free-body " method. In Fig.
248 let AN' be a portion, considered free, on the left of any
N N
FiQ. S48.
SHEAB, US FLEXURE,
285
section N', of a prismatic beam slightly bent under forces
in one plane and perpendicular to the beam. The moment
equation, about the neutral axis at JSf', gives
■^=M' ; whence «'= — =^
e 1
(1)
Similarly, with AN as a free body, NN' being =^dx,
t—=M; whence ^=- .
e I
. (2)
p and p' are the respective normal stresses in the outer
fibre in the transverse sections N and N' respectively.
Now separate the block NN', lying between these two
consecutive sections, as a free body (in Fig. 249). And
^W
%f^^ I
BART OF J
furthermore remove a portion of the top of the latter block,
the portion lying above a plane passed parallel to the neu-
tral surface and at any distance z" from that surface. This
latter free body is shown in Fig. 250, with the system of
forces representing the actions uj)on it of the portions taken
away. The under surface, just laid bare, is a portion of a sur-
face (parallel to the neutral surface) in which the above men-
tioned slipping, or shearing, tendency exists. The lowfer por-
tion (of the block NN') which is now removed exerted this
286 MECHANICS OF ENGINEEEIKG.
rubbing, or sliding, force on the remainder along the under
surface of the latter. Let the unknown intensity of this
shearing force be X(per unit of area) ; then the shearing
force on this under surface is =Xy"dx, (y",= oa in figure,
being the horizontal width of the beam a.t this distance z"
from the neutral axis of N') and takes its place with the
other forces of the system, which are the normal stresses
between , and portions of J and J', the respective
_z=z"
total vertical shears. (The manner of distribution of J
over the vertical section is as yet unknown ; see next arti-
cle.)
Putting 2 (horiz. compons.) = in Fig. 250, we have
P -p'dF— r -pdF—Xy"dx=0
,'.Xy"dx=P'—P fzdF
^ z"
But from eqs. (1) and (2), p'— p = (iHf — Jf)J-=^ dM,
while from § 240 dM = Jdx ;
.■.Xrd.=^-^jlaF.:X=^fliF (3)
z " z
as the required intensity per unit of area of the shearing
force in a surface parallel to the neutral surface and at a
distance z" from it. It is seen to depend on the " shear " J
and the moment of inertia I of the whole vertical section;
upon the horizontal thickness* y'' of the beam at the sur-
face in question ; and upon the integral / zdF,
^«"
which (from § 23) is the product of the area of that part of
the vertical section extending from the surface in question to
the outer fibre, by the distance of the centre of gravity of that
part from the neutral surface.
* Thickness of actual substance.
SHEAR IN FLEXURE.
287
It now follows, from § 209, that the intensity (per unit
area) of the shear on an elementary area of the vertical
cross section of a bent beam, and this intensity we may call
Z, is equal to that X, just found, in the horizontal section
which is at the same distance (z") from the neutral axis.
254. Mode of Distribution of J, the Total Shear, over the Verti-
cal Cross Section. — The intensity of this shear, Z (lbs. per
sq. inch, for instance) has just been proved to be
Z=X=^, CzdF
(4)
ly
To illustrate this, required the
value of Z two inches above the neu-
tral axis, in a cross section close to
the abutment, in Ex. 5, § 252. Fig.
251 shows this section. From it we
have for the shaded portion, lying
above the locality in question, y" =
4 inches, and C ~ ' sdF = (area
^ z"= 2
of shaded portion) X (distance of
its centre of gravity from J^A) = Fia.aoi.
(12.8 sq. in.) x (3.6 in.) = 46.08 cubic inches.
The total shear J = the abutment reaction = 600 lbs.,
while 1= L bM = ^ X 4 X (10.4)^ = 375 biquad. inches.
Boih J Siud I refer to the whole section.
600x46.08 ift.oiK
= 18.42 lbs. per sq. m..
..Z-
375x4
•qui+e insignificant. In the neighborhood of the neutral
axis, where z" = 0, we have y'" = 4 and
r^ zdF= r^ zdF=^ 20.8 X 2.6=54.8,
J z"=0 J
z''=0 J
wh__e J and I of course are the same as before,
for z" =0
Hence
288
MECHANICS OF ENGINEERING.
Z=^o^ 21.62 lbs. per sq. in.
At the outer fibre since f^ zdF^O, %" being = e, ^ is =
for a beam of any shape.
For a solid rectangular section like the
above, Z and 2" bear the same relation to
each other as the co-ordinates of the para-
bola in Fig. 252 (axis horizontal).
Since in equation (4) the horizontal
thickness, y" , from side to side ef the sec-
tion of the locality where Z is desired, ^iq, 25?
occurs in the denominator, and since / %dF
increases as g" grows numerically smaller, the following
may be stated, as to the distribution of J, the shear, in
any vertical section, viz.:
The intensity (lbs. per sq. in.) of the shear is zero at
the outer elements of the section, and for beams of ordi-
nary shapes is greatest where the section crosses the neu-
tral surface. For forms of cross section having thin webs
its value may be so great as to require special investiga-
tion for safe design.
Denoting by Z^ the value of ^at the neutral axis, (which
=Xo in the neutral surface where it crosses the vertica
section in question) and putting the thickness of the sub-
stance of the beam = &o at the neutral axis, we have,
Zq—Xq —
J
Iho
X
area above
neutral axis
(or below)
X
the dist. of its cent,
grav.from that axis
(5)
255. Values of Zo for Special Forms of Cross Section. — From
the last equation it is plain that for a prismatic beam the
value of Zo is proportional to J, the total shear, and hence
to the ordinate of the shear diagram for any particular
case of loading. The utility of such a diagram, as obtain-
SHEAR ll>r FLEXURE.
289
ed in Figs. 234-237 inclusive, is therefore evident, for by-
locating tlie greatest shearing stress in the beam it
enables us to provide proper relations between the load-
ing and the form and material of the beam to secure safety
against rupture by shearing.
The table in § 210 gives safe values which the ^^s-^
maximum Zq in any case should not exceed. It is \
only in the case of beams with thin webs (see Figs.
238 and 240) however, that Zq is likely to need at-
tention.
For a Rectangle we have, Fig. 253, (see eq. 5, §
!
l4
Fig. 253.
254) 6o=&, I=>/xib'h\ and C\dF=%'h'h , yih^yiW
.'.Zn — Xn-
2
~ (total shear) -7- (whole area)
Hence the greatest intensity of shear in the cross-section
is A as great per unit of area as if the total shear were
uniformly distributed over the section.
Fig. 254. Fig. 233. Fig. ^5
For a Solid Circular section Fig. 254
Fig. 257.
Z,= -^f^dF =
IhffJo l^nr^ . 2r
Stt 3 Tir^
[See § 26 Prob. 3].
For a Hollow Circular section (concentric circles) • Fig.
255, we have similarly,
290
Zo=-
MECHATTICS OF BKGIKEERIIJ^G.
J
Xri*-r2*)2(ri-n)
J{r{'—r.i')
3 ;r(ri*— r2*)(ri— rg)
A-pplying this formula to Example 2 § 252, we first have
as the max. shear J„, = ^P =1,735 lbs., this being the abut-
ment reaction, and hence (putting tc = (22 -=- 7))
^0 max. =
_ 4x7xl735[64-42.8]
3x22[256-150](4-3.5)
= 294 lbs. per sq. in.
which cast iron is abundantly able to withstand in shear-
ing.
For a Hollow Rectangular Beam, symmetrical about its
neutral surface, Fig. 256 (box girder)
;7_ 3}i{hJh'-hJi i) _3 J[5,V-5,V ]
' %ihji,'-hjh%h-h,) 2* [6A-^-M/][6;-&,]
The same equation holds good for Fig. 257 (I-beam with
square corners) but then &2 denotes the sum of the widths
of the hollow spaces.
256. Shearing Stress in the Web of an I-Beam. — It is usual to
consider that, with I-beams (and box-
beams) with the web vertical the shear J,
in any vertical section, is borne exclusively
by the web and is uniformly distributed
■ over its section. That this is nearly true
may be proved as follows, the flange area
being comparatively large. Fig. 258.- Let
Fi be the area of one flange, and F^ that of
the half web. Then since
_i =
N
b<
Fig. 258.
/=^ W+2i^, (f )
SHEAE IK PLiEXUEE. 291
(tlie last term approximate, ^ /iq being taken for the radi-
us of gyration of jPi,) while
r zdF=Fi ±.-{-Fq-^, (the first term approx.) we have
J r\dF
Jo _ J%K{^F,+F,) J
if we write (2i^i+i^o) ^ (6i^i+2i^o)=K • ^ut &o^o is the
area of the whole web, .'. the shear per unit area at the
neutral axis is nearly the same as if J were uniformly dis-
tributed over the web. E. g., with jPi = 2 sq. in., and Fq
= 1 sq. in. we obtain Zq = 1.07 (J-r-hoh^.
Similarly, the shearing stress per unit area at n, the
upper edge of the web, is also nearly equal to e7-f- JqAo (see
eq.,4(§254) for then \ T {zdF)'\ ^ F^.y^K nearly,
while / remains as before.
The shear per unit area, then, in an ordinary I-beam 1&
obtained by dividing the total shear J by the area of the
web section.*
Example. — It is required to determine the proper thick-
ness to be given to the web of the 15- inch structural steel
rolled I-beam of Example 3 of p. 282, the height of web
being 13 in., and the maximum safe shearing stress being
taken as 8750 lbs. /in. ^ (as prescribed by the Philadelphia
building laws for mild steel). The web is vertical.
The greatest total shear, J^^ which occurs at either support,
is equal to half the load, i.e., to 15,715 lbs.; and hence,
with 6o denoting the thickness of web, we have
J 15 715
Zomax.=^; i.e., 8750= ,-^—j^ ; .-. 6o = 0.138 in.
* That, is, for the vertical, or horizontal, section of web. The shear
on bome oblique plane may be somewhat larger than this. . See §§ 270a
and 314.
292
MECHANICS OF ENGINEERING.
(Units, inch and pound). The 15-incli I-beam in question of
the Cambria Steel Co.), weighing 42 lbs. to the linear foot,
has a web 0.41 in. thick, which provides a very ample resist-
ance to shearing stress.
In the middle of the span, Zo = 0, since J = 0. '
257. Designing of Riveting for Built Beams. — The latter are
generally of the I-beam and box forms, made by riveting
together a number of continuous shapes, most of the ma-
terial being thrown into the flange members. E. g., in fig.
259, an I-beam* is formed by riveting together, in the
manner shown in the figure, a " vertical stem plate " or
web, four "angle-bars," and two "flange-plates," each of
Fig. 259.
Fig. 260.
these seven pieces being continuous through the whole
length of the beam. Fig 260 shows a box-girder. If the
riveting is well done, the combination forms a single rigid
beam whose safe load for a given span may be found by
foregoing rules ; in computing the moment of inertia, how-
ever, the portion of cross section cut out by the rivet
holes must not be included. (This will be illustrated in
a subsequent paragraph.) The safe load having been com-
puted from a consideration of normal stresses only, and
the web being made thick enough to take up the max.
total shear, J",,, with safety, it still remains to design the
riveting, through whose agency the web and flanges are
caused to act together as a single continuous rigid mass.
It will be on the side of safety to consider that at a given
* Such a built I-beam is usually designated a " plate-girder. '\ See
handbook of the Cambria Steel Co.
SHEAK m FLEXURE. 293
locality in the beam the shear carried by the rivets con-
necting the angles and flanges, per unit of length of beam,
is the same as that carried by those connecting the angles
and the web ("vertical stem -plate"). The amount of this
shear may be computed from the fact that it is equal to
that occurring in the surface (parallel to the neutral sur-
face) in which the web joins the flange, in case the web
and flange were of continuous substance, as in a solid I-
beam. But this shear must be of the same amount per
horizontal unit of length as it is per vertical linear unit in
the web itself, where it joins the flange ; (for from § 254 Z
=X) But the shear in the vertical section of the web,
being uniformly distributed, is the same per vertical linear
unit at the junction with the flange as at any other part
of the web section (§ 256,) and the whole shear on the ver-
tical section of web = J, the " total shear " of that section
of the beam.
Hence we may state the following :
The riveting connecting the angles with the flanges, (or
the web with the angles) in any locality of a built beam,
must safely sustain a shear equal to J on a horizontal length
eqtial to the height of web.
The strength of the riveting may be limited by the re-
sistance of the -rivet to being sheared (and this brings
into account its cross section) or upon the crushing resist-
ance of the side of the rivet hole in the plate (and this in-
volves both the diameter of the rivet and the thickness of
the metal in the web, flange, or angle. In its hand-book, the
Cambria Steel Co. gives tables for the safe strength of rivets,
and compressive resistance of plates ; based on unit shearing
stresses from 6,000 to 10,000 Ibs./sq. in. for shearing stress in
the rivets, and 12,000 to 20,000 Ibs./sq. in. compressive re-
sistance, in the side of the rivet hole, the axial plane section of
the hole being the area of reference.
In fig. 259 the rivets connecting the web with the angles
are in double shear, which should be taken into account in
considering their shearing strength, which is then double ;
those connecting the angles and the flange plates are in
294 MECHANICS OF ENGINEERING.
single shear. In fig. 260 (box-beam) where the beam is
built of two webs, four angles, and two flange plates, all
the rivets are in single shear. If the web plate is very high
compared with its thickness, vertical stiffeners in the form
of "angles " may need to be riveted upon them laterally
[see § 314].
Example. — ^A built I-beam of structural steel (fig. 259)
is to support a uniformly distributed load of 40 tons, its ex-
tremities resting on supports 20 feet apart, and the height
and thickness of web being 20 ins. and J in. respectively.
How shall the rivets, which are | in. in diameter, be spaced
between the web and the angles which are also ^ in. in thick-
ness? Let the unit stresses taken be 7500 for shearing, and
12,500 for side compression (Ibs./in.^). Referring to fig. 235
we find that J = | W = 20 tons at each support and diminishes
regularly to zero at the middle, where no riveting will there-
fore be required. Each rivet, having a sectional area of J7r(|)2
= 0.60 sq. inches, can bear a safe shear of 0.60x7500 = 4500 lbs.
in single shear, and hence of 9000 lbs. in double shear, which is
the present case. But the safe compressive resistance of the
side of the rivet hole in either the web or the angle is only
I in. Xj in. X 12500 = 5470 lbs., and thus determines the spacing
of the rivets as follows :
Near a support the riveting must sustain a shear equal
to 40,000 lbs. on a horizontal length equal to the height of
web, i.e., to 20 ins., and the safe compression for each rivet
is 5470 lbs. Hence 4000 h- 5470, or 7.2, rivets will be needed
for the 20-inch length. In other words, they must be spaced
20-^7.2 = 2.7 in. apart, center to center, near the supports;
5.4 in. apart at ^ the span from a support; none at all in the
middle. By the Cambria handbook, this distance apart
should never be less than 3 diameters of the rivet; and, in
connecting plates in compression, should not exceed 16 times
the thickness of the plate.
As for the rivets connecting the angles and flange plates,
being in two rows and opposite (in" pairs) the safe shear-
FLEXURE, BUILT BEAM.
295
ing resistance of a pair (eacli in single shear) is 9,000 lbs.,
while the safe compressive resistance of the sides of the
two rivet holes in the angle bars (the flange plate being
much thicker) is =10,940 lbs. Hence the former figure
(9,000) divided into 40,000, gives 4.44 as the number of
pairs of rivets for 20 in. of length of the beam; i.e., the
rivets in one row should be 20^-4.44 = 4.5 in. apart, centre
to centre, near a support ; the interval to be increased in
inverse ratio to the distance from the middle of span,
(^bearing in mind the practical limitation just given).
If the load is concentrated in the middle of the span,
instead of uniformly distributed, e/is constant along each
half-span, (see fig. 234) and the rivet spacing must accord-
ingly be made the same at all localities of the beam.
SPECIAL, PROBLEMS IN FLEXURE.
258. Designing Cross Sections of Built Beams. — The last par-
agraph dealt with the riveting of the various plates ; we
now consider the design of the plates themselves. Take
for instance a plate-girder, fig. 261 ; one vertical stem=
Fig. 261.
296 MECHANICS OF ENGINEERING,
plate, four angle bars, (each of sectional area = A, re-
maining after the holes are punched, with a gravity axis
parallel to, and at a distance = a from its base), and two
flange plates of width = h, and thickness = t. Let the
whole depth of girder = 7^, and the diameter of a rivet
hole =f. To safely resist the tensile and compressive
forces induced in this section by M,^ inch-lbs. (itf^ being
the greatest moment in the beam which is prismatic) we
have from § 239,
ifn. = — - (1)
e
E' for mild steel = 15,000 lbs. per sq. inch, e is = ^ ^
while i, the moment of inertia of the compound section,
is obtained as follows, taking into account the fact that
the rivet holes cut out part of the material. In dealing
with the sections of the angles and flanges, we consider
them concentrated at their centres of gravity (an approx-.
imation, of course,) and treat their moments of inertia
about N as single terms in the series fdF z^
(see § 85). The subtractive moments of inertia for the
rivet holes in the web are similarly expressed ; let 6o =
thickness of web.
j It, for web = ^h, (h—2tf—2b,t' [^—t—a'Y
„•, ■} I^ for four angles = 4 A ['l — t — aY
( In for two flanges = 2(6— 2f) t ('^f
the sum of which makes the ^ of the girder. Eq. (1) may
now be written
which is available for computing any one unknown quan-
tity. The quantities concerned in /^ are so numerous and
they are combined in so complex a manner that in any
numerical example it is best to adjust the dimensions of
the section to each other by successive assumptions and
FLEXUEE BUILT BEAM. 297
trials. (The hand-book of the Cambria Steel Co. gives tables
of safe loads of beam box-girders and plate-girders for a large
variety of sizes and distances between supports ; but attention-
is called to the fact that the loads given in the tables are based
on the assumption that the girder is supported laterally, and
that otherwise a proper reduction in the allowable safe load
must be made, as explained elsewhere in the hand-book. The
value of 15,000 Ibs./sq. in. for R' has been used in computing
these tables.)
Example. — (Units, inch and pound) . A plate-girder with
end supports, of span = 20 ft. = 240 inches, is to support
a uniformly distributed load of 45 tons = 90,000 lbs. If f
inch rivets are used, angle bars 3" X 3'' X 2'% vertical
web I" in thickness, and plates 1 inch thick for flanges, how
wide (6 = ?) must these flange-plates be ? taking h = 22
inches = total height of girder.
Solution. — From the table in § 250 we find that the max.
31 for this case is ^ Wl, where W = the total distributed
load (including the weight of the girder) and I = span-
Hence the left hand member of eq. (2) reduces to
Wl h 90000 X 240 X 22
16 • R' - 16 X 15000 "■^^^^•
That is, the total moment of inertia of the section must
be = 1,980 biquad. inches, of which the web and angles
supply a known amount, since &o = >^", t = l''> t'= )'i" ,
a' = 1^", A= 2.0 sq. in., a = 0.9', and h = 22", are
known, while the remainder must be furnished by the
flanges, thus determining their width b, the unknown
quantity.
The elective area. A, of an angle bar is found thus :
The full sectional area for the size given, = 3 X ^ +
2>^ X % = 2.75 sq. inches, from which deducting for two
rivet holes we have
A= 2.75—2 X ^ X >^_ 2.0 sq. in.
The value a = 0.90" is found by cutting out the shape
298 MECHANICS OF ENGINEEKING.
X3
of two angles from sheet iron, tlius : I
and balancing it on a knife edge* (The
gaps left by the rivet holes may be ignored,
without great error, in finding or). Hence, fig!^ a
substituting we have
Ih for web =A. . 1^x20^— 2x>^ . ^ l8}(Y=282.d
In for four angles =4x2x [9.10]2=662.5
In for two flanges=2(6— |)xlx(10>^)2=220.4(&— 1.5)
.-. 1980=282.3+662.5+(Z^1.5)220.4
whence b = 4.6 + 1.5 = 6.1 inches
the required total width of each of the 1 in. flange plates.
This might be increased to 6.5 in. so as to equal the
United width of the two angles and web.
The rivet spacing can now be designed by § 257, and
the assumed thickness of web, )4 in., tested for the max,
total shear by § 256. The latter test results as follows ;
The max. shear J^„ occurs near either support and =
)4 ^=45,000 lbs. .-., calling 6'o the least allowable thickness
of web in order to keep the shearing stress as low as 8, 000
lbs. per sq. inch,
6'o X 20" X 8000 = 45000 .-. 6'o=0.28 in.
showing that- the assumed width of )4 in. is safe.
This girder will need vertical stiffeners near the ends,
as explained subsequently, and is understood to be sup-
ported laterally, f Built beams of double web, or box-
form, (see Fig. 260) do not need this lateral support,
259. Set of Moving Loads. — When a locomotive passes over
a number of parallel prismatic girders, each one of which
experiences certain detached pressures corresponding to
the dijfferent wheels, by selecting any definite position of
the wheels on the span, we may easily compute the reac-
tions of the supports, then form the shear diagram, and
finally as in § 243 obtain the max. moment, Jf^s and the
* The Cambria hand-book gives values of / and a for sections of angle-
bars.
t See § 314.
FliEXUEE. MOVING LOADS.
299
max. shear J^, for this particular position of the wheels.
But the values of Jf^ and J^ for some other position may
be greater than those just found. We therefore inquire
"which will be the greatest moment among the infinite
number of {M^Js, (one for each possible position of the
wheels on the span). It is evident from Fig. 236 from the
nature of the moment diagram, that when the pressures or
loads are detached, the 31^ for any position of the loads,
which of course are in this case at fixed distances apart,
must occur under one of the loads (i.e. under a wheel).
We begin .*. by asking : What is the position of the set of
moving loads when the moment under a given wheel is
greater than will occur under that wheel in any other po-
sition? For example, in Fig. 262, in what position of the
Fig 263.
loads Pi, P2> stc. on the span will the moment ilfa* i-6.,
under Pn, be a maximum as compared with its value under
Pg in any other position on the span. Let P be the resultant
of the loads which are now on the span, its variable distance
from be = cc, and Unfixed distance from Pg = a'', while
a, h, c, etc., are the fixed distances between the loads
(wheels). For any values oi x , as the loading moves
through the range of motion within which no wheel of the
set under consideration goes off the span, and no new
wheel comes on it, we have Pi =--^ P, and the mament
under Pg
=M^=RS-(x-a'y]—P^h-Pih-\-c) '
i.e. M2=j(l^—'^^a'x)—P,b—P,(b-\-c)
(1)
300 MECHANICS OF EXGINEBRING.
In (1) we liave M2 as a function of x, all tlie other quan-
tities in tlie right hand member remaining constant as the
loading moves ; x may vary from x=^a-\-d to
x=l—{c-\-h—a). For a max. M2, we put dMi-h-dx=0, i. e.
m
j{l-2x+a)=0 .'. x{ioT Max M.,)=}4l+}^(^'
(For this, or any other value of x, (FM^-^daf is negative,
hence a maximum is indicated). For a max. M2, then, B
must be as as far {%Oj') on one side of the middle of the
span as P2 is on the other ; i.e., as the loading moves, the
moment under a given wheel becomes a max. when that
wheel and the centre of gravity of all the loads {then on
the span) are equi-distant from the middle of the span.
In this way in any particular case we may find the-
respective max. moments occurring under each of the
wheels during the passage, and the. greatest of these is the
3I„^ to be used in the equation ilt/,„ =^R'I-^e for safe loading:.*
As to the shear J, for a given position of the wheels this
will be the greatest at one or the other support, and
equals the reaction at that support. When the load moves
toward either support the shear at that end of the beam
evidently increases so long as no wheel rolls completely
over and beyond it. To find J" max., then, dealing with
each support in turn, we compute the successive reactions
at the support when the loading is successively so placed
that consecutive Avheels, in turn, are on the point of roll-
ing ofi^ the girder at that end ; the greatest of these is the
max. shear, J^^. As the max. moment is apt to come under
the heaviest load it may not be necessary to deal with
more than one or two wheels in finding M,„.
Example. — Given the following wheel pressures,
^< . . 8' . . >B< . . 5' . . >C< . . 4 . . as I2 or Z^ is the
greater segment.
It is also evident that for a given span and given beam
the safe load P', as computed from eq. (1) above, becomes
very large as its point of application approaches a sup-
port ; this would naturally be expected but not without
limit, as the shear for sections between the load and the
support is equal to the reaction at the near support and
may thus soon reach a limiting value, when the safety of
the web or the spacing of the rivets, if any, is considered.
Secondly, considering the weight of the heam, or any
uniformly distributed loading, weigliing w lbs. per unit of
length of beam, in addition to P, Fig. 264, we have the
reactions '
iJ,=^+|'; and B,=i^+^
Let 1-2 he >?i ; then for a portion Om of length a?<^
moments about m give
FLEXURE. SPECIAL PROBLEMS.
303
^ — BoX + ivx.~ =0
e " 2
Le., on 00, M—B-iX — y^ lox^ . . . , (2)
Evidently for x = (i.e. at 0) M = 0, while for x = Iz (i.e.
at (7) we have, putting w = W -r- I
Mr=B>,l,— y:
wll-
Z 2 ^^ J
(3)
(4)
it remains to be seen whether a value of M may not exist
in some section between- and G, (i.e., for a value of x
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) Zi and < ?i+?2
Px-P {x--\)—^=0 .: M=P\
(1>
(2)
That is, see (1), ilf varies directly with x between and C,
while between G and D it is constant. Hence for safe
loading
i.e., — ^Pl , . (3)
FLEXURE. SPECIAL PROBLEMS.
305
a(
i
11
illlllllk.i
MOMS.
illlllllllllllllllll
1
SHEARS
Tlie construction of the
B moment diagram is evident
^l^ ^1 ^^ p from equations (1) and (2).
\ ! As for J", tlie shear, the
same free bodies give, from
I, (vert. forces)=0.
On OG . J=^P ... (4)
On CD . J=P—P==zerol5)
(4) and (5) might also be ob-
Pig 265 tained from (1) and (2) by-
writing J=d M-T-dx, but the
former method is to be preferred in most cases, since the
latter requires M to be expressed as a function of x while
the former is applicable for examining separate sections
without making use of a variable.
If the beam is an I-beam, the fact that J is zero any-
where on G D would indicate that we may dispense with
a web along G D io unite the two flanges ; but the lower
flange being in compression and forming a " long column "
would tend to buckle out of a straight line if not stayed by
a web connection with the other, or some equivalent brac-
ing.
282. XTniform Load over Part of the Span. Two End Supports.
Fig. 266. Let the load= W, extending from one support
over a portion =c, of the span, (on the left, say,) so that
W= IOC, w being the load per unit of length. Neglect
wei ght o f beam. For a 'free body Dm of any length
X <, B (i.e. < c), 2" moms^=0 gives
pi
wx-
2
-^icc=0 .'.M=
(1)
which holds good for any section on B. As for sections
on B (7 it is more simple to deal with the free body m'G,
of leiigth
' x' < G B from which we have M^R^ x' . (2)
MECHANICS OF ENGINE EEH^TG.
Fig. 2G6.
wMch shows the moment
curve for B G tohe a. straight
line DC, tangent at D to the
parabola 0' D representing
eq. (1.) (If there were a con-
centrated load at B, CD
would meet the tangent at
D at an angle instead of co-
inciding with it ; let the stu-
dent show why, from the
shear diagram).
The shear for any value of
ic on -S is :
On 5
while on B C
. e/= Bo= constant
(3)
The shear diagram is constructed accordingly.
To find the position of the max. ordinate of the para-
bola, (and this from previous statements concerning the
tangent at the point D must occur on B, as will be seen
and will .'. be the M^ for the whole beam) we put e/=--0 in
eq (3) whence
X (for JC)=
i?i_JF[?— |] ^_(?
w
tu
(5)
W
and is less than c, as expected. [The value oi Bi^--j- (l — '^\
—[wc ~-T) (I — 2), (the whole beam free) has been substi-
tuted]. This value of x substituted in eq. (1) gives
is the equation for safe loading.
The max. shear J^ is found at and is
evidently >i?2j at C.
Bx, which is
FLEXUEE. SPECIAL PROBLEMS.
30^
263. XTniforin Load Over Whole length With Two Symmetricj
Supports. Fig. 267. — With the notation expressed in the fig-
ure, the following results may be obtained, after having
divided the length of the beam into three parts for sepa-
rate treatment as necessitated by the external forces, which
are the distributed load W, and
and the two reactions, each =
}^ W. The moment curve is
made up of parts of three dis-
tinct parabolas, each with its
axis vertical. The central par-
abola maj sink below the hori-
zontal axis of reference if the
supports are far enough apart,
in which case (see Fig.) the elas-
tic curve of the beam itself becomes concave upward be-
tween the points E and F of " contrary flexure." At each
of these points the moment must be zero, since the radius
of curvature is co and M = EI ^ p (see § 231) at any sec-
tion ; that is, at these points the moment curve crosses its
horizontal axis.
As to the location and amount of the max. moment M^,
inspecting the diagram we see that it will be either at H,
the middle, or at both of the supports B and C (which from
symmetry have equal moments), i.e., (with I = total length,)
Fig. 267.
w
Mr
.[and.-.— J=
( either ~ \ %li-l,^-] at ^
or
Ell' at ^ and a
2Z
according to which is the greater in any given case ; i.e.
according as I2 is > or < l^ y'g.
The shear close on the left oi B = ivl^, while close to the
right oiBit=}4 W — id^. (It will be noticed that in this
case since the beam overhangs, beyond the support, the
shear near the support is not equal to the reaction there,
as it was in some preceding cases.)
308
MECHANICS OF ENGINEEEHsTG.
Hence (/„=
wli
/2 ^-t^Zi P^^^^^^^g ^^ ^1 <^
264, Hydrostatic Pressure Against a. Vertical Plank. — From
elementary hydrostatics we know that the pressure, per
unit area, of quiescent water against the vertical side of a
tank, varies directly with the depth, x, below the surface,
and equals the weight of a prism of water whose altitude
= X, and whose sectional area is unity. See Fig. 268.
Fig. 26S.
*Tt& plank is of rectangular cross section, its constant
breadth, — b, being r~ to the paper, and receives no sup-
port excepi at its two extremities, and B, being level
with the water surface. The loading,' or pressure, per unit
of length of the beam, is here variable and, by above defini-
nition, is = w= yxb, where ;' = weight of a cubic unit
(i.e. the heaviness, see § 7) of water, and x = Om =■ depth
of any section m below the surface. The hydrostatic pres-
sure on dx = ivdx. These pressures . for equal dx's, vary
as the ordinates of a triangle ORiB.
Consider Onti free. Besides the elastic forces of the ex-
posed section m, the forces acting are the reaction Bq, and
the triangle of pressure OEm. The total of the latter is
W.
0(?
= I wdx = yb I xdx = 'fb-^
(1)
and the sum of the moments of these pressures about m is
equal to that of their resultant ( = their sum, since they
are parallel) about m, and .% ==: jb -^ , ^,
A o
rLEXHRB. SPECIAL PEOBLEMS. 309
[From (1) wh«n x==1,wq have for tlie total water pres-
sure on the beam Wi = jb ^ and since one-third of this
will be borne at we have i?o =^}i T^^^-~\
Now putting i'( moms, about the neutral axis of wi)=0,
for Om free, we have
Box—JK . ^—^=0 .-. 31= /eyb {Vx—:j(?)
O 6
(2)
(which holds good from x = Oto x — I). From I (horiz.
forces) = we have also the shear
J=R,— W^=% yh {P—Sx') .... (3)
as might also have been obtained by differentiating (2),
since J = dM -^ dx. By putting e7 = (§ 240, corollary)
we have for a max. M, x = I -i- V3, which is less than I
and hence is applicable to the problem. Substitute this
in eq. 2, and reduce, and we have
Efl ,, . R'l 1 1
-^=Ji^, i.e. — =g "^^•rbV' . (4)
as the equation for safe loading.
265. Example. — If the thickness of the plank is h, re-
quired 7i = ?, if B' is taken = 1,000 lbs. per sq. in. for
timber (§ 251), and I = 6 feet. For the inch-pound-second
system of units, we must substitute B' = 1,000 ; ? = 72
inches ; y = 0.036 lbs. per cubic inch (heaviness of water
in this system of units); while I =h¥ -4- 12, (§ 247), and e
— }i h. Hence from (4) we have
1000 &A3 0.0366x723 ,„ ..^ . n 07 •
^rs 7T— — n /- 1 ••• ^^=5.16 .'. h = 2.27 m.
It will be noticed that since x for ilfm = I -^ Vs, and not
^ I, ifm does not occur in the section opposite the resul-
tant of the water pressure ; see Fig. 268. The shear curve
is a parabola here ; eq. (3).
310
MECHANICS OF ENGINEERING.
^ ^ ^ ^'^ ^T
£r ^ 1 wTtr-
Fig. 289.
Fig. 269a.
266. Flat Circular Plate, Homogeneous and of Uniform Thick-
ness, Supported all Eound its Edge and Subjected to Uniform Fluid
Pressure of w lbs. per sq. in. A strict treatment of this case
being very complicated, an approximate method, due to Prof.
C. Bach, will be presented."* Fig. 269 shows a top view of the
circular plate, in a horizontal position and covering a circular
I opening, its edge
being supported
C^) continuously on
the edge of the
opening (but not
clamped to it).
Let the radius of
the plate be r and
its thickness h.
Under the plate
is the atmosphere,
while on its upper surface, acting uniformly over the whole of
that surface, is a fluid pressure whose excess over that of the at-
mosphere is w Ibs./sq. in. The particles near the upper surface
are under compressive stress, which is obviously greater near the
center of the circle ; those near the lower surface are in tension.
Let now the half-plate, CODE, (cutting along the diameter
CD) be considered as a "free body" in Fig. 269a; the tensile
and compressive stresses in the section COD being assumed to
form a stress-couple, as in previous case of flexure, the unit-
stress varying as the distance from the middle of the thick-
ness, with the stress in the outermost fiber denoted by p.
Then the moment of this couple will be written pi — e, as
before, where e = ^h and I =2rh^ -i- 12. On the upper surface
of the free body we find a total pressure of | W7rr^ lbs., covering
a semicircular surface ; so that (p. 22) the distance of the re-
sultant from is 4 r -^ Stt. The upward reaction from the
supporting edge is also | wirr'^ lbs., but its resultant acts 2r/7r
in. from (center of gravity of a semicircular "wire," p. 20).
Taking moments, then, about we have
wrrr^ r2 r 4 rl _ prh^
~2~ [V~3^J^ ""3"
* Elasticitaet und Festigkeit, by C. Bach ; Berlin, 1898.
or, tv = -,«
(0)
FLEXUEE. PLATES.
311
Notwithstanding the imperfections of this analysis, the
experimental work of Prof. Bach shows that a modification of
eq. (0), viz. : — -
(1)
5 K" „,
r
may be used with safety for the design of a plate under these
circumstances ; R% a safe unit working stress for the material,
having been substituted for p.
For example, let the plate (e.g., cylinder-head of a loco-
motive) be of mild steel with h = 1 in. and r = 8 in. Putting
R' = 16,000 lbs. /sq. in., we have from eq. (1) a safe w
= 1 (16000) X (1 H- 8)' =. 208 lbs. per sq. in.
[N.B. If the plate is clamped all round the edge, we may
write f instead of the |. (Bach.) ]
266a. Homogeneous Circular Plate of Uniform Thickness, h,
Supported all Round the Edge, with Concentrated Load (P lbs.) in
Center. By the same method as before we may here derive
P = 1 Trh^p ; but from his experiments in this case Prof. Bach
concludes that the formula for safe design should be written
P
lirh'R'.
o
(2)
It is seen from eq. (2) that the value of P is independent of
the radius of the plate; depending only on the material and
the thickness, h.
266b. Homogeneous Elliptical and Rectangular Plates of Con-
stant Thickness, h , Supported all Round the Periphery. According
to Prof. Bach's approximate analysis, as supplemented by his
experimental researches, we may use the following formulae for
Fig. 269&.
safe design of elliptical and rectangular plates, supported (not
clamped) around the whole periphery. See Fig. 2696 for
notation of dimensions ; h being the uniform thickness in each
312 MECHANICS OF ENGINEERING.
case, and a being > 6. R' = max. safe unit stress for the
material.
For the elliptical plate under unifornily distributed pressure
(over whole area) of w lbs. / sq. in., denoting the ratio 6 -r- a by
m, we have
w = ^ {1 +m')J^,.R'; .... (3)
and under a central concentrated load of P lbs.,
3^ 3 + 2m^ + 3w^ , ...
(N.B. If the edge is clamped down all round we may use
values of w and P about 50 per cent, greater than the above.)
With rectangular plates under a uniformly distributed pressure,
denoting the ratio 6 -;- a by m, we have
w = l{l +m')f,.R' ..... (5)
and for a concentrated central load P, with n denoting the ratio
P = i- (1 + n') . h'R' .'.... (6)
on
In the particular case of the square plate, the side of the
square being a, eqs. (5) and (6) reduce to
7 2
(uniform pressure) w = S.6 — .R'; (7)
(central load) P =o h'R' (8)
o
266c. Homogeneous Flat Circular Plate, of TTniforin Thickness, used as Piston
of an Engine. In such a case we have fluid pressures ou both sides of the plate
or disc, neither of which is necessarily one atmosphere ; while at the center
we have acting the concentrated pull or thrust, P lbs., of the piston rod. (Fric-
tional forces around the edge may be disregarded.) If w denote the greatest
difference between the (uniform) fluid pressures (per sq. in.) on the two sides,
we may write (according to Grashof's analysis, as quoted by Unwin), for safe
design in this case : —
^-If.-^' . (9)
(E', h, and r, have the same meaning as before.)
FLEXUEE SPECIAL PROBLEMS. 313
267. ResilienceofBeamWithEndSupports.— Fig. 270. If a
9g mass whose weight is G {G large com-
!^ I pared with that of beam) be allowed to
^^ J __^^'- l_p fall freely through a height = h upon
g J I - 1^ ^j^^ centre of a beam supported at its
a-.y TPm extremities, the pressure P felt by the
Fig. 270. beam increases from zero at the first
instant of contact up to a maximum P^, as already stated
in §233a, in which the equation was derived, d^ being
small compared with h,
The elastic limit is supposed not passed. In order that
the maximum normal stress in any outer fibre shall at most
be=^', a safe value, (see table §251) we must put
=-7^ [according to eq. (2) §241,] i.e. in equation (a)
above, substitute F^= [4 Ii'l]-^Ie, which gives
having put I==FJi? {h being the radius of gyration §85)
and Fl= V the volume of the (prismatic) beam. From
equation (&) we have the energy, Gh, (in ft. -lbs., or inch-
lbs.) of the vertical blow at the middle which the beam of
Pig. 270 will safely bear, and any one unknown quantity
can be computed from it, (but the mass of G shaiili not
be small compared with that of the beam.)
The energy of this safe impact, for two beams of the
same material and similar cross-sections (similarly placed),
is seen to be proportional to fheii volumes; while if further-
more their cross-sections are the same and similarly
placed, the safe Gh is proportional to their lengths. (These
same relations hold good, approximately, beyond the elas'
tic limit.)
It will be noticed that the last statement is just the re-
314
MECHANICS OF ENGINEEEING.
verse of wliat was found in §245 for static loads, (the
pressure at tlae centre of the beam being then equal to
the weight of the safe load) ; for there the longer the beam
(and .°. the span) the less the safe load, in inverse ratio.
As appropriate in this connection, a quotation will be
given from p. 186 of " The Strength ' of Materials an^
Structures," by Sir John Anderson, London, 1884, viz.:
"It appears from the published experiments and state-
ments of the Railway Commissioners, that a beam 12 feet
long will only support )4 of the load that a beam 6 feet
long of the same breadth and depth will support, but that
it will bear double the weight suddenly applied, as in the
case of a weight falling upon it," (from the same height,
should be added) ; " or if the same weights are used, the
longer beam will not break by the weight falling upon it
unless it falls through twice the distance required to frac-
ture the shorter beam."
268. Combined Flexure and Torsion. Crank Shafts. Fig. 271.
Let OiB be the crank, and NOi the portion projecting
beyond the nearest bearing
N. P is the pressure of the
connecting-rod against the
crank-pin at a definite in-
stant, the rotary motion be-
ing uniform. Let a= the
perpendicular dropped from
the axis OOi of the shaft
upon P, and 1= the distance
of P, along the axis Oj from
the cross-section iV^TwiV^' of the
Let NW be a diameter of this
In considering the portion
NOiB free, and thus exposing the circular section iVmZV^,
we may assume that the stresses to be put in on the ele-
ments of this surface -are the tensions (above NN') and
the compressions (below NN') and shears "| to NN', due
to the bending action of P ; and the shearing stress tan=
shaft, close to the bearing,
section, and parallel to a.
FLEXURE. SPECIAL PROBLEMS.
315
gent to tlie circles which have as a common centre, and
pass through the respective dF's or elementary areas,
these latter stresses being due to the twisting action of P.
In the former set of elastic forces let p = the tensile
stress per unit of area in the small parallelopipedical ele-
ment m of the helix which is furthest from NN' (the neu-
tral axis) and /= the m.oment of inertia of the circle about
NN'-, then taking moments about NN' for the free body,
(disregarding the motion) we have as in cases of flexure
(see §239)
pT T.7 . .. . .. Plr
.= PI
; i.e., p-
(«)
[None of the shears has a moment about iVW.] Next
taking moments about OOi, (the flexure elastic forces, both
normal and shearing, having no moments about OOi) we
have as in torsion (§216)
■^^^-i^= Pa ; i.e., p^=
Par
~I7
Q>)
in which p^ is the shearing stress per unit of area, in the
torsional elastic forces, on any outermost dF, as at m ;
and 7p the polar moment of inertia of the circle about its
centre 0.
Next consider free, in Fig. 272, a small parallelopiped
taken from the helix at m (of Fig. 271.) The stresses [see
§209] acting on the four faces p" to the paper in Fig. 272
are there represented, the dimensions (infinitesimal) being
n " to NN, & II to 00,, and d -] io the paper in Fig. 272.
/pnd
pM'
^Pgticl
— "pM
p^na
pnd
- H
-J) M
P,M^,
-p/id
Fig. 272.
qcd-
./""
Fiff. 273.
pnd
Sl(i MECHANICS OF EXGINEEKllJTG.
By altering the ratio of 6 to % we may make the angle 6
what we please. It is now proposed to consider free the
triangular prism, GUT, to find the intensity of normal
stress q, per unit of area, on the diagonal plane GH, (oi
length — c,) which is a bounding face of that triangulai'
prism. See Fig, 273. By writing 2" (compons. in direc-^
tion of normal to GII)=0, we shall have, transposing,
qcd=pnd sin d+pjbd sin d+pjid cos d ; and solving for q
q=jp -- sin d+p, -sin^. To find 6 for q max. we put
tJ, =j9sin2^4-2^sCos2(?, . . (4)
= 0, and obtain: tan[2(^ for q max)]=> — ~ . . • (5)
Call this value of 6, 6'. Since tan 2d' is negative, 2d' lies
either in the second or fourth quadrant, and hence
sin2^^=± , ^" and cos 2^'= rp— 7=^= (6)
[See equations 28 and 29 Trigonometry, O. W. J.] The
FliEXUBE. CEANK SHAFT. 317
apper signs refer to tlie second quadrant, the lower to tlie
fourth. If we now differentiate (4), obtaining
^=2i)cos2^-4p,sin2^ . . . (7)
W8 laote that if the sine and cosine of the [2^'] of the 2nd
quadrant [upper signs in (6)] are substituted in (7) the re-
sult is negative, indicating a maximum ; that is, g is a max-
imum for 6= the d' of eq. (6) when the U2>per signs are taken
(2nd quadrant). To iind q max., then, put 6' for 6 in (3)
substituting from (6) (upper signs). We thus find *
g-max =;^[p+Vy+4^] . . (8)
A similar process, taking components parallel to GH,
Fig. 273, will yield q^ max., i.e., the max. shear per unit of
area, ^hich for a given p and p^ exists on the diagonal
plane GH in any of its possible positions, as d varies.
This max. shearing stress is
g^max =^yp2_|_4^^2 ^ ^ (9j
In the element diametrically opposite to m in Fig. 211, p
is compression instead of tension ; q maximum will also
be compression but is numerically the same as the q max.
of eq. 8.
269. Example.— In Fig. 271 suppose P=2 tons = 4,000
lbs., a=Q in., 1=5 in., and that the shaft is of wrought
iron. Required its radius that the max. tension or com-
pression may not exceed i^' = 12,000 lbs. per sq. in.; nor the
max. shear exceed /S" = 7,000 lbs. per sq. in. That is, we
put 5'=12,000 in eq. (8) and solve for r : also ^,,=7,000 in
(9) and solve for r. The greater value of r should be
taken. From equations (a) and (5) we have (see §§ 219 and
247 for /p and i)
* According to the conceptions of § 405&, safe design would require
that we put the max. '^ strain" in this case equal to a safe value, as
determined by simple tensile or compressive tests. Here the max.
strain (tensile) is £=[|p + |-\/p^ + 4ps^]-^-E'- (Grashof's method.)
318 MECHANICS OF EKGINEEEIITG.
\P= r ^^cl p^= -.
•which in (8) and (9) give
mas. g=>^~ [4?+|/(4^)=+4(2«)"] ^ . . (8^
p
and max. g's=^— 3A/(4!)H4(2a)2 . . • (9«)
With max. g= 12,000, and the values of P, a, and Z, already
given, (units, inch and pound) we have from (8a), r^=2.72
cubic inches .*. r=1.39 inches.
Next, with max. 5's=7,000; P, a, and I as before; from
(9a), r^=2.84 cubic inches .*. r=1.41 inches.
The latter value of r, 1.41 inches, should be adopted. It
is here supposed that the crank-pin is in such a position
(when P= 4,000 lbs., and a=Q in.) that q max. (and q^
max.) are greater than for any other position ; a number
of trials may be necessary to decide this, since P and a are
different with each new position of the connecting rod. If
the shaft and its connections are exposed to shocks, H and
S' should be taken much smaller.
270. Another Example of combined torsion and flexure is
shown in Fig. 274. The
'" ' /^^ "^^ "^i< ^B 'wo^k of the working force
Pi( vertical cog-pressure) is
B expended in overcoming the
resistance (another vertical
cog-pressure) Q^.
^la- 27'4. That is, the rigid body
consisting of the two wheels and shaft is employed to
transmit power, at a uniform angular velocity, and since
it is symmetrical about its axis of rotation the forces act-
ing on it, considered free, form a balanced system. (See
§ 114). Hence given Pi and the various geometrical quan-
TLEXUEE. CEAXK SHAFT. 319
titles «!, 5i, etc., we may obtain Q^, and the reactions Pq and
Pr, in terms of Pj. The greatest moment of flexure in the
shaft will be either FJi, at G; or PJ3, at B. The portion
CD is under torsion, of a moment of torsion =Piai= Qih^.
Hence we proceed as in the example of § 269, simply put-
ting Poll (or Pb4, whichever is the greater) in place of Fl,
and PiCTi in place of Pa. We have here neglected the
weight of the shaft and wheels. If Qi were an upuard ver-
tical force and hence on the same side of the sh:it as Pj,
the reactions Pq and Pg would be less than before, and on©
or both of them might be reversed in directioji.
270a. Web of I-Beam. Maximum Stresses on an Oblique
Plane. — The analysis of pp. 315, 316, etc., also covers the
case of an element of the web of a horizontal I-beam under
stress, when this element is taken near the point of junction
with the flange. Supposing that the thickness of web has
already been designed such that the shearing stress on the
vertical (and therefore also on the horizontal) edges of such
an element is at rate of 8000 lbs. per sq. inch ; and that the
horizontal tension at each end of this element (since it is
not far from the outer fibre of the whole section) is at rate
of 10,000 lbs. per sq. in.; we note that Fig, 272 gives us a.
side view of this element, with p^ = 8000, and p = 14,000,
lbs. per sq. inch. GTis the lower edge of the upper flange,
corresponding (in an end view) to the point n in Fig. 258 on
p. 290. (We here suppose the upper flange to be in tension ;
of course, an illustration taken from the compression side
would do as well.)
Substitution in equations (8) and (9) of p. 317 results in
giving as maximum stresses on internal oblique planes :
q max. = 17,630 lbs. per sq. in. tension;
and g^ max. =10,630 " " " " shearing.
These two values are seen to be considerably in excess of
the respective safe values for shearing and tensile stresses in
the case of structural steel, and the necessity is therefore em-
phasized of adopting values for shearing stress in webs some-
what lower than the 8000 lbs./in.2 used above ; to avoid the
occurrence of excessive stress on internal oblique planes. See
p. 291.
320 MECHANICS OF ENGINEEHIiN^a.
CHAPTER IV.
FLEXURE, CONTINUED.
CONTINUOUS GIRDERS.
271. Definition. — A continuous girder, for present pur«
poses, may be defined to be a loaded straight beam sup-
ported in more than two points, in which case we can no
longer, as heretofore, determine the reactions at the sup-
ports from simple Statics alone, but must have recourse
to the equations of the several elastic curves formed by its
neutral line, which equations involve directly or indirect-
ly the reactions sought ; the latter may then be found as
if they were constants of integration. Practically this
amounts to saying that the reactions depend on the man-
ner in which the beam bends ; whereas in previous cases,
with only two supports, the reactions were independent of
the forms of the elastic curves (the flexure being slight,
however).
As an Illustration, if the straight beam of Fig. 275 is placed
on three supports 0, B, and (7, at the same level, the
reactions of these supports seem at first sight indeterm-
inate ; for on considering the p -i ^
whole beam free, we have three \'^_~^~ 1. '^ j;* $
unknown quantities and only bZT""^ Z^° ^ — -^
two equations, viz : S (vert. fig. 275.
compons.) = and S (moms, about some point) = 0. If
now be gradually lowered, it receives less and less pres-
FLEXUBE. CONTIJJUOUS GIKDEBS. 321
sure, until it finally readies a position where the beam
barely touches it ; and then O's reaction is zero, and B and
C support the beam as if were not there. As to how
low must sink to obtain this position, depends on the
stiffness and load of the beam. Again, if be raised
above the level of B and C it receives greater and greater
pressuTt., until the beam fails to touch one of the other
supports. Still another consideration is that if the beam
were tapering in form, being stiffest at 0, and pointed at
B and (7, the three reactions would be different from their
values for a prismatic beam. It is therefore evident that
for more than two supports the values of the reactions de-
pend on the relative heights of the supports and upon the
form and elasticity of the beam, as well as upon the load.
The circumstance that the beam is made continuous over
the support 0, instead of being cut apart at into two
independent beams, each covering its own span and hav-
ing its own two supports, shows the significance of the
term " continuous girder."
All the cases here considered will be comparatively
simple, from the symmetry of their conditions. The
beams will all be prismatic, and all external forces (i.e.
loads and reactions) perpendicular to the beam and in the
same plane. All supports at the same level,
272. Two Equal Spans; Two Concentrated Loads, One in the Mid-
^e of Each Span. Prismatic Beam. — Fig. 275. Let each half-
Bpan = i^ /i. Neglect the weight of the beam. Required
the reactions of the three supports. Call them P^, Pq and
p
\.. From symmetry P^ = Pc, and the tangent to the
elastic curve at is horizontal ; and since the supports
are on a level the deflection of C (and B) below O's tangent
is zero. The separate elastic curves OD and DC have a
common slope and a common ordinate at D.
For the equation of OD, make a section n anywhere be-
tween and Z>, considering n(7 a free body. Fig. 276 (a)
322
MECHANICS OF ENGINEERIXG.
Y
—X ^>|
(6)
Fig. 276.
•with origin and axes as there indicated. * From H (moms
about neutral axis oi n) = we have (see § 281)
Ei'^^=p{y2i—x)—Pc{i—x)
eA =F{y2ix—%
dx
(1
(2)
The constant = 0, for at both x, and dy -^ dx, = 0.
Taking the x-anti-derivative of (2) we have
^/2/=P(^_^')-Pe[^-|'] . . (3)
Here again the constant is zero since at 0,x and y both =0.
(3) is the equation of OD, and allows no value of cc <0
or>^. It contains the unknown force P^.
For the equation of BC, let the variable section n be made
anywhere between D and C, and we have (Fig. 276 ih\ j x
may now range between J^Z and T)
^^^^— ^^(^-)
^jdy_
dx
Ix-t^+C
(4)
(5)'
To determine C\ put x = }4l both in (5)' and (2), and
equate the results (for the two curves have a common
tangent line at D) whence C" = ^ PV
Elp.^yiP¥—Pjl
0[?\
2~j
(5)
* These are such that XOY is our "first quadrant"; in which, for points
in a part of a curve convex toward the axis of X, d^yldx^ is essentially
positive; and vice versa. It will be seen that both eqs. (1) and (4) are
on this basis. They must be on the same basis; otherwise, later com-
parisons of equations would result in error.
FLBXUEE. CONTINUOUS GIRDEES. 323
Hence Ely ^ % PT?x-PA^^—^'\j^O" . . (6)'
At D tlie curves have the same y, hence put a? = i- in the
right hand member both of (3) and (6)', equating results,
and we derive C"= — ^ Pf
EIy = y,PVa>~P^\^__p^^XPf . . (6)
which is the equation of DC, but contains the unknown
reaction P^. To determine P^ we employ the fact that O's
tangent passes through G, (supports on same level) and
hence when a? = Hn (6), y is known to be zero. Making
these substitutions in (6) we have
Q=y,pf-y,p^f-±pi^ ... P=^P
From symmetry P^ also = —P, while Pq must = ~P,
since P^ + P^ + P<7 = 2 P (whole beam free). [Note. —
If the supports were not on a level, but if, (for instance)
the middle support were a small distance = Ag below
the level line joining the others, we should put x = I and
y = — Iiq in eq. (6), and thus obtain P^ = Pc= -^^ P +
SET—, which depends on the material and form of the
prismatic beam and upon the length of one span, (whereas
with supports all on a level, P^ — P^ = -| P is independent
of the material and form of the beam so long as it is ho-
mogeneous and prismatic.) If Pq, which would then =
?| P — 6 EI {Jiq-^F'), is found to be negative, it shows that
requires a support from above, instead of below, to
cause it to occupy a position 7^o below the other supports,
i.e. the beam must be " latched down " at 0.]
The moment diagram, of this case can now be easily con-
structed ; Fig. 277, For any free body nC, n lying in BG,
we have
324
MECHANICS OF ENGINEEEING.
i.e., varies directly as cc, un-
c til X passes D wlien, for any
point on DO,
I wliicli is =0, (point of in-
flection of, elastic curve)
I for .T=yii? (note that x is
Fig. 277. measured from C in this
figure) and at 0, where x= I, becomes — ^^Pl
•'• K=—lPl; M^^O; 3I^=LPl; andif„=0
Hence, since if max. =^Ply the equation for safe loading
is
B'l 6
-PI
(7)
The shear at (7 and anywhere on CD=~Pf while on DO
it =^^P in the opposite direction
.•.j;„=;ip . " . . . (8)
The moment and shear diagrams are easily constructed,
as shown in Fig. 277, the former being svmmetrical about
a vertical line through 0, the latter about the point 0"
Both are bounded by right lines.
273. Two Equal Spans. IJniformly Distributed Load Over
„, , Whole Length. Prismatic Beam.
^ y\ ^_:^ ^ c —Fig. 278. Supports B, 0,
bQIM I I 1 \ [\\ 111 C, on a level. Total load
V~- — o| 1 ^ -- ^ ^---1 = 2 W= ^icl and may include
I |po j ^'^k^'^ j that of the beam
P.
^ , , "w IS con-
I I li I I 11 stant. Asbefore, from sym-
metry P^=P^, the unknown
i Pc| reactions at the extremi-
VMt. 278. ties.
FLBXUKE. CONTINUOUS GIRDERS.
125
Let On=x ; then with wC free, 2" moms, about n= gives
rdy
IV
EI'^^- lFx-lx'+^]-F,[lx- ^]+[Coiist=0] (2)
[Const. =0 for at both dy-i-dx the slope, and x, are =0]
... EIy= '^[}4Fc^-j4M+Vi2^]-P.[}4lx^-y6^]+{G=0) (3)
[Const. =0 for at both x and y are =0]. Equations (1),
(2), and (3) admit of any value of x from to I, i.e., hold
good for any point of the elastic curve OC, the loading on
■which follows a continuous law (viz. : w= constant). But
when x=l, i.e., at G, y is known to be equal to zero, since
0, B and G are on the axis of X, (tangent at 0). "With
these values of x and y in eq. (3) we have
0= |L . t-y^pj? ... p,=y8wi=yQW
.-. PB=^^and Po=27r— 2Pe=? W
The Moment and Shear Diagrams can now be formed since
;j jovv all tli6 external forces are
Lw^ known. In Fig. 279 meas-
ure X from G. Then in any
section n the moment of the
"stress-couple " is
M^yQWoo-
wxr
. (1)
j which holds good for any
value of x on GO, i.e., from
07=0 up to x=l. By inspec-
PiG S79. tion it is seen that for 07=0,
M=0 ; and also for x=yi, M=0, at the in/lection point' G,
beyond which, toward 0, the upper fibres are in tension
326 MECHANICS or engixeert^tg.
the lower in compression, whereas between C and G the;^
are vice versa. As to the greatest moment to be found on
CG, put dM-i-dx—0 and solve for x. This gives
^ W—wx=0 .'. [X for if max.]=^Z . (2)
i^rhich in eq. (1) gives
Jfu(at JV, seefigure)=+^jr? . . (2)
But this is numerically less than Mo{=—}i Wl) hence the
stress in the outer fibre at being
T/ Wle /Q\
Po=%—j-, ... (3)
the equation for safe loading is
B'l _.,
Wl . . . . (4)
the same as if the beam were cut through at 0, each half,
of length I, retaining the same load as before [see § 242 eq.
(2)]. Hence making the girder continuous over the mid-
dle support does not make it any stronger under a uni-
formly distributed load ; but it does make it considerably
stiffer.
As for the shear, J, we obtain it for any section by tak-
ing the x-derivative of M in eq. (1), or by putting ^(ver-
tical forces) =0 for the free body nG, and thus have for
any section on GO
J=z/qW—wx ... (5)
j/is zero for x='^l (where M reaches its calculus maxi-
mum M^ ; see above) and for x=l it = — Yq fF" which is nu
merically greater than yi W, its value at G. Hence
Jm=y8w . . . ". (6)
FLEXURE. CONTINUOUS GIRDERS. 327
The moment curve is a parabola (a separate one for each
span), the shear curve a straight line, inclined to the hor-
izontal, for each span.
Problem. — How would the reactions in Fig. 278 be
changed if the support were lowered a (small) distance
Iiq below the level of the other two ?
274. Prismatic Beam Fixed Horizontally at Both Ends (at
Same Level). Single Load at Middle. — Fig. 280. [As usual
j /^~x p-| the beam is understood to
1^^ — ■ V^py ■ — — - ' I be homogeneous so that E
E ^P~ I '-' ^ is the same at all sections].
IJ I * j The building in, or fixing,
i lyij ji of the two ends is supposed
objr^-Jt:;;™--— 1^ — --—^^^ — (c* to be of such a nature as to
Yi ' Br — —- — Vi-T--^ cause no 'horizontal con-
FiG. 280. straint ; i.e., the beam does
not act as a cord or chain, in any manner, and hence the
sum of the horizontal components of the stresses in any
section is zero, as in all preceding cases of flexure. In
other words the neutral axis still contains the centre of
gravity of the section and the tensions and compressions
are equivalent to a couple (the stress-couple) whose mo-
ment is the " moment of flexure."
If the beam is conceived cut through close to both wall
faces, and this portion of length=Z, considered free, the
forces holding it in equilibrium consist of the downward
force P (the load) ; two upward shears J^ and J^ (one at
each section) ; and two " stress-couples " one in each sec-
tion, whose moments are 31^ andJ/g. From symmetry we
know that J, — J„ and that M^=M,. From I Y=Q for the
free body just mentioned, (but not shown in the figure),
and from symmetry, we have «/„= % P and J^-= % P ', but
to determine M^ and M„ the form of the elastic curves
B and B G must be taken into account as follows :
Equation of OB, Fig. 280. I [mom. about neutral axis
of any section n on 5] = (for the free body nC which
528
MBCHAXics or EXGi:^rEEEi:srG.
lias a section exposed at each end, n being tlie variable
section) will give
BI^y-= P(y2 l-x) + M,-
-y2P{i~x)
(1)
J^Note. In forming this moment equation, notice that
M^ is the sum of the moments of the tensions and com-
pressions at G about the neutral axis at n, just as much as
about the neutral axis of 0', for those tensions and com-
pressions are equivalent to a couple, and hence the sum of
their moments is the same taken about any axis whatever
"I to the plane of the couple (§32).]
Taking the a:-anti-derivative of each member of (1),
EI^=P(% I x—% a^)-f- if, x—y PQ x—% x")
ax
(2)
(The constant is not expressed, as it is zero). Now from
symmetry we know that the tangent-line to the curve B
s>i B is horizontal, *.e., for x^y^l, dy-^dx^Q, and these
values in eq. (2) give us
0=yi Pf+ y^I^l—f^PV; whence M,=M,=}i PI , (3)
Safe Loading. Fig. 281. Having now all the forces which
act as external forces in straining the beam 00, we are
ready to draw the moment diagram and find M^^. For con-
venience measure x from 0. For the free body nO, we
have [see eq. (3)]
y2Px-M, + P~=0.'.M=}iPl-}4Px ... (4)
p Eq. (4) holds good for any
J section on OB. By put-
f7^ ting x=0 we have M=M^=
y% PI; \&j oEEO'=M, to
scale (so many inch-pounds
moment to the inch of pa-
per). At B, for x^y I,
M^= — y^ PI ; hence lay
offB'I)=ys PI on theop-
FiG. 281. posite side of the axis O'O'
c
FLEXUEE. CONTIGUOUS GIEDERS.
329
from HG', and join DH. DK, symmetrical witli i>^ about
B'D, completes the moment curves, viz.: two riglit lines.
The max. iHf is evidently =yi Fl and the equation of safe
loading
?Li=upi
(5)
Hence the beam is twice as strong as if simply supported
at the ends, under this load ; it may also be proved to be
four times as stiff.
The points of inflection of the elastic curve are in the
iniddles of the half-spans, while the max. shear is
J.n = y2P
(8)
275. Prismatic Beam Fixed Horizontally at Both Ends [at Same
Level]. Uniformly Distributed Load Over the Whole Length.
Pig. 282. As in the preceding problem, we know from
symmetry that e/o=^c=/^ ^=/^ *^^> ^^^ tl^s-^ Mq=M^, and
determine the latter quantities by the equation of the
curve OG, there being but one curve in the present in-
stance, instead of two, as there is no change in the law of
loading between and C. "With nO free, I (mom^)=0
.gives
ax 2 o
X
9
(1)
(2)
^N^wl
&
i
C \n
J
opr
J i I 11 I I H i L
Fig. 282.
MBgHAITlCS OF ENGIXEEKIlira.
The tangent line at being horizontal we have for x=0,
dx
0, .'. C=0. But since the tangent line at C is also hori-
zontal, we may for x=l put dy-^dx=0, and obtain
O^—i^Wl'+lIol+yewV; whence Mo=^Wl
(3)
a.s the moment of the stress-couple close to the wall at
and at 0.
Hence, Fig. 283, the equation of the moment curve (a
single continuous curve in this case) is found by putting
2' (moma)=0 for the free body nO, of length x, thus
obtaining
w
y^=wl
lUi I ] j lull
Fig. 283.
lL+i4Wx-Mo
iva^
=0
I.e.
M=lWl+ ^-}4Wx
, .(4)
an equation of the second degree, indicating a conic. At 0,
M=Mq of course,= 4- ^^/ ati?by putting a; = i^ Z in (4), we
have M^— — }^ Wl, which is less than Jig, although M^ is the
calculus max. (negative) for 31, as may be shown by writ-
iijg the expression for the shear {J=% W — wx) equal to
zero, etc.
FiiEXUKE. coxTrsruous giudees. 331
Hence 31^=^ Wl, and tlie equation for safe loading is
--^Wl (5)
B'l __^
Since (with this form of loading) if the beam were not
built in but simply rested on two end supports, the equa-
tion for safe loading would be \_R'I-^e\ = yi Wl,isee §242),
it is evident that with the present mode of support it is 50
per cent, stronger as compared with the other ; i.e., as re-
gards normal stresses in the outer elements. As regards
shearing stresses in the web if it has one, it is no stronger,
since t/m = j^ JFin both cases.
As to stiffness under the uniform load, the max. deflec-
tion in the present case may be shown to be only i- of that
in the case of the simple end supports. Eiieiit
It is noteworthy that the shear diagram in Fig. 283 is
identical with that for simple end supports §242, under
uniform load ; while the moment diagrams differ as fol-
lows : The parabola KB'A^ Fig. 283, is identical with tha*-
in Fig. 235, but the horizontal axis from which the ordi-
nates of the former are measured, instead of joining the
extremities of the curve, cuts it in such a way as to have
equal areas between it and the curve, on opposite sides
i.e., areas [^C"^'+^i6^'0']=area R'G'B'
In other words, the effect of fixing the ends horizontally
is to shift the moment parabola upward a distance = 3Ic
(to scale), = i Wl, with regard to the axis of reference,
0'^', in Fig. 235.
276. Remarks. — The foregoing very simple cases of con-
tinuous girders illustrate the means employed for deter-
mining the reactions of supports and eventually the max.
moment and the equations for safe loading and for deflec-
tions "When there are more than three supports, with
spans of unequal length, and loading of any description
the analysis leading to the above results is much more
complicated and tedious, but is considerably simplified
332 MECHAXTCS OF ENGINEERING.
and systematized by tlie use of tlie remarkable theorem of
three moments, the discovery of Clapeyron, in 1857. By
this theorem, given the spans, the loading, and the vertical
heights of the supports, we are enabled to write out a rela-
tion between the moments of each three consecutive sup-
ports, and thus obtain a sufficient number of equations to
determine the moments at all the supports [p. 641 Eankine'a
Applied Mechanics.] From these moments the shears
close to each side of each support are found, then the
reactions, and from these and the given loads the moment
at any section can be determined ; and hence finally the
max. moment ilf^,,, and the max. shear J^„.
The treatment of the general case of continuous girdera
hy algebraic methods founded on the properties of familiar geo-
metrical figures, however, is comparatively simple ; and will
be developed and applied in another part of this book. (See
Chap. XII, pp. 485, etc.)
THE DANGEROUS SECTIOIS^ OF ]S^O]?^-PRIS-
MATIC BEAMS.
277. Eemarks. By " dangerous section " is meant that sec-
tion (in a given beam under given loading with given mode
of support) where p, the normal stress in the outer fibre,
at distance e from its neutral axis, is greater than in the
outer fibre of any other section. Hence the elasticity of
the material will be first impaired in the outer fibre of
this section, if the load is gradually increased in amount
(but not altered in distribution).
In all preceding problems, the beam being prismatic, I,
the moment of inertia, and e were the same in all sections,
hence when the equation P—=M [§289] was solved for »,
e
Me ....
giving i>=— . . . . (1)
we found that p was a max., = p^, for that section whose
ili" was a maximum, since p varied as M, or the moment
FLEXUEE NON-PEISMATIC BEAMS.
333
of the stress-couple, as successive sections along the beam
were examined.
But for a non-prismatic beam Zand e change, from sec-
tion to section, as well as 31, and the ordinate of the
moment diagram no longer shows the variation of p, nor
is ^ a max. where ilf is a max. To find the dangerous
section, then, for a non-prismatic beam, we express the ilf,
the I, and the e of any section in terms of x, thus obtain-
ing ^=func. {x), then writing dp-~dx=0, and solving for x.
278. Dangerous Section in a Double Truncated Wedge. Two
End Supports. Single Load in Middle. — The form is shown in
Fig. 284. Neglect weight of beam ; measure x from one sup-
port 0. The
r e a c tion a t
each support
is i^ F. The
width of the
beam == 5 at
all sections, while its height, v, varies, being = h at 0.
To express thee = }4 v, and the /= 1 hv^ (§247) of any
section on 0(7, in terms of x, conceive the sloping faces
of the truncated wedge to be prolonged to their intersec-
tion A, at a known distance = c from the face at 0. We
then have from similar triangles
[Tpl
Fig. 284.
V : X -{- c : : h
3, .: V = ~ (x -\- c)
c
and
e =
h
(x-{-G) and I = K^
-^x-^rcf
For the free body nO, H (moms.^)
Px—tL±
e
[That is, the M = )4 Fx.]
p=SF
and^= 3F ^
ax
hH ' {x+cf '
By putting dp -t- dx =
(1)
(2)
(3)
we find X = + c\ showing a
gives
Fxe
'~w • • •
But from (2), (3) becomes
dp_ o-p & {x-\-cy — 2a;(a;-t-c)
p-
334:
MECHAXICS OF ENGINEERING.
maximum for p (since it will be found to give a negative
result on substitution in d^p 4- dotf).
Hence tlie dangerous section is as far from tlie support
0, as the imaginary edge^ A, of tlie completed wedge, but
of course on the opposite side. This supposes that the
half -span, )4l, is, > ^; if not, the dangerous section will
be at the middle of the beam, as if the beam were
prismatic.
tx -xi, ) the equation for safe { ■Dfh'h'2
Hence, with L ^^.^1^ .^^ ^f^>^^,^^^,\ ^-%Pl (5)
A'' <^ ) at middle) ( ^
while with )tl^e equation for safe j ^,j ^A]^ , , p ,..
1/7 ^ r h loading is : (put x=c ■{ ^-^= V2 Pc (6)
/^^ > ^ ) and_p=i?'in [3]) ( ^
(see §239.)
279. Double Truncated Pyramid and Cone. Fig. 285. For
Fig. 285.
the truncated pyramid both width = u, and height = v^
are variable, and if h and Ji are the dimensions at 0, and
c = QJ[ = distance from to the imaginary vertex A, we
shall have from similar triangles u=~ (a;+c)and v= ~ (x-\-c).
G c
Hence, substituting 6=^^ and 7=1 uv^, in the moment
equation
£^_^=0.weW^=34,.^-|-,. . (7)
. dp ^ op <^ (x+c)^ — 3x [x+cy
' * dx bW" ' {x-\-cf
(8)
FLEXURE NON-PRISMATIC BEAMS. 335
Putting the derivative =0, for a maximum p, we liave x =
-h ^ c, hence the dangerous section is at a distance a? = ^ c
from 0, and the equation for safe loading is
either :?^= 14: PZ if >^ Z is < >^ c . . . , (9)
(in which V and h' are the dimensions at mid-span)
or
MM)lhlf=y^P,,iy^i
6
is > >^ c ... (10)
For the truncated cone (see Fig. 285 also, on right) where
e = the variable radius r, and / = i^ ;r r*, we also have
/=[Const.] .,—^.3 (11)
and hence j9 is a max. for a? = ^ c, and the equation for
safe loading
either 5£i^ = % Fl,iox %l <% c , . , , . (12)
(where r' = radius of mid-span section) ;
^^ ^-R' (l^o)' ^%Fc,ioxy2l> %c (13)
(where r^ = radius of extremity.)
IS^ON-PMSMATIC BEAMS OF "UNIFORM
. STRE]?^GTH."
380. Eemarks. A beam is said to be of " uniform
strength " when its form, its mode of support, and the dis-
tribution of loading, are such that the normal stress ^ has
the same value in all the outer fibres, and thus one ele-
ment of economy is secured, viz. : that all the outer fibres
may be made to do full duty, since under the safe loading,
p will be = to B' in all of them. [Of course, in all cases
of flexure, the elements between the neutral surface and
336 MECHANICS OF EIS^GIJ^JEEEmG.
fclie outer fibres being under tensions and compressions
less than R' per sq. inch, are not doing full duty, as
tegards economy of material, unless perhaps with respect
to shearing stresses.] In Fig. 265, §261, we have already
had an instance of a body of uniform strength in flexure,
viz. : the middle segment, CD, of that figure ; for the
moment is the same for all sections of CD [eq. (2) of that
§], and hence the normal stress p in the outer fibres (the
beam being prismatic in that instance).
In the following problems the weight of the beam itself
is neglected. The general method pursued will be to find
an expression for the outer -fibre-stress p, at a definite sec-
tion of the beam, where the dimensions of the section are
known or assumed, then an expression for p in the varia-
ble section, and equate the two. For clearness the figures
are exaggerated, vertically.
' 281. Parabolic Working Beam. UnsymmetricaL Fig. 286
i.
Pig. 286,
CBO is a working beam or lever, B being the fixed fulcrum
or bearing. The force P^ being given we may compute P^
from the mom. equation Pq^o = PJ^u while the fulcrum
reaction is P^^P^-^-P^^. All the forces are ~\ to the beam.
The beam is to have the same width h at all points, and is
to be rectangular in section.
Ilequir6*d first, the proper height hx, at B, for safety.
From the free body BO, of length = l^, we have I (momss)
= ; i.e.,
-^ rX,oxp^- -— ... (1)
FLEXUEE. IfON-PIlISMATIC BEAMS. 337
Hence, putting jo^ = B', h^ becomes known from (1).
Required, lecondly, tlie relation between the variable
height V (at any section n) and the distance x ot n from 0.
For the free body nO, we have (2 momSu = 0)
3iL=F,x ; or ^" ^^ ^^' =P,x and .-. p, =^l^ (2)
But • for " uniform strength " p^ must = p^ \ hence
equate their values from (1) and (2) and we have
^ = — 1, which may be written {% vj = .>'^p' x (3)
so as to make the relation between the abscissa x and the
ordinate }4 v more marked; it is the equation of a para-
bola, whose vertex is at 0.
The parabolic outline for the portion BC is found simi-
larly. The local stresses at G, B, and must be proper-
ly provided for by evident means. The shear J = Pq, at
0, also requires special attention.
This shape of beam is often adopted in practice for the
working beams of engines, etc.
The parabolic outlines just found may be replaced by
trapezoidal forms, Fig. 287, without using much more ma-
terial, and by making the slop-
ing plane faces tangent to the
parabolic outline at points Tq
and Ti, half-way between and ^^"^^b^^"'^'^ °
B, and C and B, respectively. It fis. 287.
can be proved that they contain minimum volumes, among
all trapezoidal forms capable of circumscribing the given
parabolic bodies. The dangerous sections of these trape-
zoidal bodies are at the tangent points Tq and Ti. This is
as it should be, (see § 278), remembering that the subtan-
gent of a parabola is bisected by the vertex.
338
MECHANICS OF ENGINEEKING.
283. Rectang. Section. Height Constant. Two Supports (at Ex-
tremities). Single Eccentric Load.
— Fig. 289. h and Ji are tlie
dimensions of the section at
B. "With BO free we have
Pah
■Folo=0.\pj,
■ 6Po?o
(1)
Fig. 289.
At any other section on BO, as n, where the width is u,
the variable whose relation to x is required, we have for
wOfree
P^^.F,x;ovPll/^=P,x
QP,,x
Pn =
Equating pu and ^„ we have u :h :: x :Iq „
That is, BO must be wedge-shaped ; edge at 0, vertical.
(2)
(3)
■k- 1 ^k-^wl
Fig. 289 a.
283a. Sections Rectangular and Similar. Otherwise as Before. — Fig. 289a.
The dimensions at B are b and h; at any other section n, on BO, the
height V, and width u, are the variables whose relation to x is desired,
and by hypothesis are connected by the relation u:v::b:h (since the
section at m is a rectangle similar to that at B). By the same method
as before, putting pB = Pn, we obtain lf^^bh'^ = x-i-uv^; in which placing
u^bv^h, we have finally
v^=(h^^lf))x; and similarly, u^={b^-^lQ)x; . . . (4)
i.e., the width u, and height v, of the different sections are each pro-
portional to the cube root of the distance x from the support. (The
same relation would hold for the radii, in case all sections were circular.)
283b. Beam of Uniform Strength under Uniform Load. Two End Supports.
Sections Rectangular with Constant Width. — Fig. 289&. WeigM of beam
neglected. How should the height v vary, (the height and width
at middle being h and b) ? As before, we equate pB and pn ', whence
finally
(ivy = [h^^P](lx-x') (5)
This relation between the half-height ^v (as ordinate) and the abscissa.
X is seen to be the equation to an ellipse with origin at vertex.
FLEXURE OF BEINFOKCED CONCRETE BEAMS. 339
CHAPTER V.
Flexure of Reinforced Concrete Beams.
284. Concrete and "Concrete-Steel" Beams. Concrete is an
artificial stone composed of broken stone or gravel (sometimes
cinders), cement and sand, properly mixed and wet beforehand
and then rammed into moulds or " forms " and left to harden or
" set." This material, after thorough hardening or " setting,"
thouglT fairly strong in resisting compressive stress is compara-
tively weak in tension. When it is used in the form of beams
to bear transverse loads (i. e., under " transverse stress ") the
side of the beam subjected to tensile stress is frequently " re-
inforced " by the imbedding of steel rods on that side. In this
way a composite beam may be formed which is cheaper than a
beam of equal strength composed entirely of concrete or one
composed entirely of steel.
Of course the steel rods are placed in the mixture when wet,
and previous to the ramming and compacting, and their aggre-
gate sectional area may not need to be more than about one per
cent, of that of the concrete.
No reliance being placed on the tensile resistance of the
concrete (on the tension side of the beam) it is extremely
important that there should be a good adhesion, and consequent
resistance to shearing, between the sides of the steel rods and
the adjacent concrete, for without this adhesion the -rods and
the concrete would not act together as a beam of continuous
substance.*
In some specifications, for instance, it is required that the
shearing stress, or tendency to slide, between the steel rods and
the concrete shall not exceed 64 lbs. per sq. in. Sometimes
the steel rods are provided with projecting shoulders, or ridges,
or corrugations, along their sides, to secure greater resistance to
sliding. *
* For an account of tests of this adhesion see Engineering News, Aug. 15,
1907, p. 169, and also p. 120 of the Engineering Record for Aug. 3, 1907.
340
MECHANICS OF ENGINEERBsTG.
Fig. 290 gives a perspective view of a concrete-steel beam
of rectangular section, placed in a horizontal position on two
supports at its extremities, and thus fitted to sustain vertical
loads or weights ; while Fig. 291 shows a concrete-steel beam
flange-
teeb, or stem-
of T-section, in which the flange is intended to resist compres-
sion, while the steel rods in the lower part of the " stem " are
to take care of the tension. These two shapes of beam will be
the only ones to be considered here, in a theoretical treatment.
The ratio of the Modulus of Elasticity of steel (viz. — about
30,000,000 lbs. per sq. in.) to that of concrete (say, from
1,000,000 to 4,000,000 lbs. per sq. in., according to the propor-
tions of ingredients used) is of great importance in the theory,
since in general the stresses induced in two materials for a given
percentage change of length are directly proportional to the
modulus of elasticity (for same sectional area).
Generally the diameter of a steel rod is so small compared
with the full height of the beam that the stress in the rod is
taken as uniform over the whole of its section.
285. Concrete-Steel Beam of Rectangular Section. Flexure
Stresses. — As in the common theory of flexure of homogeneous
beams, it will be assumed that cross-sections plane before
flexure are still plane when the beam is slightly bent, so that
changes of length occurring in the various fibers are propor-
FLEXUEE OF EEINFORCED CONCRETB BEAMS.
541
tional to the distances of those fibers from a certain neutral axis
of the cross-section, and upon the amount of any such change
of length (relative elongation) can be based an expression for
the accompanying stress. Now in the case of concrete it is not
strictly true that stresses are proportional to changes of length
(" strains " or deformations) ; in other words its modulus of
elasticity, E, is not constant for different degrees of shortening
under compressive stress. Nevertheless, since this modulus
does not vary much, within the limits of stress to which the
concrete is subjected in safe design, it' will be considered con-
stant, the resulting equations being sufficiently accurate for
practical purposes.
Let us now take as a " free body " any portion, ON, of the
beam in Fig. 290, extending from the left-hand support to any
section, at any distance x from that support. In the plane
section terminating this body on the right, BNS (see now Fig.
292, in which we have also, at the right-hand, an end-view of
< 6 >
t
1
]
h
\
1
-4--
1
AS-;>:-iy:V;-:;?il!
N. axis
— • — • — •—
End View.. "
the body), we note that the fibers of concrete from Z> down to a
neutral axis iV^are in a state of compression, while below iVthe
steel rod alone is considered as under stress, viz., a total tensile
stress of F'p', where F' is the aggregate sectional area of all
the steel rods, these rods being at a common distance a' above
the lower edge of the section, and p' is the unit (tensile) stress
in the steel rods.
The distance BN, or " ^," of the neutral axis N below the
" outer fiber " i), is to be determined. Let p denote the unit
compressive stress in the fiber at I) (outer fiber) of the concrete ;
then the unit stress in any fiber of the concrete at distance z
from iV will be - j9, lbs. per sq. in., and the total stress on any
342 MECHANICS OF ENGLNEERING.
such fiber is — »c?i^, lbs. (where c?jPis the sectional area of the
e
fiber). All the (horizontal) fibers between the two consecutive
cross sections DS and D' 8' were originally dx inches long, but
now (during stress), we find that the fiber at D has been shortened
an amount d\ and the steel rod "fibers " elongated an amount
d\'^ so that we have the proportion d\ : d\' : :e: a — e;
d\ e , ,».
or, ■-— = (0)
dV a- e ^ ^
For the free body in Fig. 292 we have, for equilibrium, the
sum of horizontal components of forces = (the shear J" has no
horizontal component); that is, remembering that below iVno
tensile forces are considered as acting on the concrete, but
simply the total tensile stress F'p' in the steel rods,
t/n
- pdF - F'p' = 0.
e
But here =^ is a constant ; and for the rectangular cross -section,
dF = b . dz, and
'iI>'^-'T4-'-'T-^y • • • W
But from the definition of modulus of elasticity (F for the
concrete and E^ for the steel), we have (§ 191)
F = p —- (relat. elongation), or F = p -i- [dX/dx) ; and
similarly, F' = p' -^ {dV /dx) ; whence
d\' p'' E ^ ^
But, from eq. {\.)^ p -— p' = IF' -^ he, combim'ng which
with eqs. (0) and (2), and denoting the ratio F' -^ Fhj n,
we mid = — ^ — ■. . (3)
a — e be
The ratio n may have a value from 10 to 25 for " rock-
concrete," and still higher for " cinder-concrete ; " see § 284.
Now solve eq. (3) for the distance e, obtaining
F'nfj2ab ^ A
FLEXURE OF REINFORCED CONCRETE BEAMS.
34S
This locates the neutral axis, iV". [See, later, eq. (29), §
291.]
Returning to the free body OJV in Fig. 292 above, we note
that the resultant compression in the concrete between iV^
and D, viz., ^ p .he, lbs. [see eq. (1)], is equal in value to
the total tension F'p', lbs., in the steel rods at G', and that they
are parallel. Consequently they form a couple (the " stress-
couple " of the section) whose moment is equal to the product
of one of these forces, say
F'p\ by the perpendicular | j^ \
distance = a", between Gr'
and a point Cr (see now Fig.
293) whose distance from the
" outer fiber " i> is one-third
of e. The "arm" of this
couple is a'
a- -■ For
Fig. 293.
equilibrium of the free body ON in Fig. 292 the shear J and
the two forces V (reaction) and P^ (load) must be equivalent
to a couple of opposite and equal moment to that of the stress
couple. Call this moment M [in this case it has a value of
Yx — P^{x — a;J]; it is the " bending moment " of the section
at DS. We may therefore write (see Fig. 293) :
M = F'p' [« — i e] ; and .-. p' =
M
F\a-\e)
(5)
which will give the unit-stress p\ induced in the steel rods at
section BS. It is seen to depend on the position of the neutral
axis N (i.e., upon e); upon the bending moment, M, at that
section; upon the sectional area F' of the steel rods (aggre-
gate); and on the distance, a, at which they are placed from
the compression edge, B, of the beam.
But since the resultant compression, h p -he, is equal to the
resultant tension, .F'p', we may also write
2M
M= ^p .be [« — i e] and .-. p =
he (a — I e)
(6>
which gives the unit-stress (compression) in the outer " fiber "
at i), of the concrete, for this section BS.
1
e
..i
G
■^—
N
N"
V'(p + clp'l
s"
344 MECHANICS OF ENGINEERING.
286. Horizontal Shear in the Foregoing Case (Rectangular Sec-
tion). The shear per sq. in. along the sides of the steel rods,
I ^^ and also along the horizontal
,/' " Neutral Surface,'' NN" (see
Fig. 294), may be obtained as
follows : — Let dx be the length
of a small portion of the beam
,, , , „ (of Fig'. 290) situated between
■^ r^^^^n '- — two vertical sections Do and
D"S". Fig. 294 shows this
^^*^- 294. portion as a " free body." The
forces acting consist of the tension F'p' on the left-hand end of
the steel; the tension on the right-hand end of these rods
[being something greater (say) and expressed by F' (^p' -\- dp'^
in which dp is the difference between the unit-tensions at the
two ends of the steel " re-inforcement "] ; the resultant com-
pression, i he.p, in the concrete on the left; and that,
1 5g . (j? -f dp^, on the right ; and, finally, the two vertical shears,
J'and J". Here p is the unit compressive stress (lbs. per sq. in.)
in outer fiber of concrete at the left-hand extremity of the same,
while p -\- dp expresses the unit compressive stress in the same
outer fiber at the right-hand extremity.
Evidently the difference between the total tensile stresses at
the extremities of the steel rods will give the total horizontal
shearing stress on the sides of those rods and this may be
written pjl^dx (lbs.), where pj = unit shearing stress between
the steel and concrete and l^ == aggregate perimeter of the steel
rods (so that l^dx = total area of the outside surface of rods in
Fig. 294);
hence p/l^dx ^ F' {p' j^- dp') — F'p' .... (7)
But if, for the free body of Fig. 294, we put 2 moms. =
about the point Cr (a distance ^ e from upper fiber)
we find Jdx = [F' (p' + dp') — F'p'']{a -\ e)\ . . (8)
and hence ) , J" /q\
see (7), \ ^' ^ l^ (a -^ e) ^ ^
FLEXURE OP EElNFOliCED CONCRETE BEAMS.
346
p bdx
Also, if we let p^ denote the unit shearing stress (or tendency
to slide) along the horizontal sur-
face NN" or neutral surface, the
total amount is pjbdx (lbs.).
In Fig. 295, which shows as a
free body the portion NN"S"S of
Fig. 294, we see this horizontal
force (of concrete on concrete) act-
ing toward the left. The other ^^°- ^^s.
forces acting on the free body are as shown in Fig. 295 and, by
putting 2 horiz. compons. = 0,
we find
and finally,
see eq. (8),
F' {f + df) — F'p' = pJbdx;
J
Ps
(9a)
(10)
5(a-ie) •
This (unit) shearing stress in the concrete along NN", the
"neutral surface," should nowhere exceed a certain value \Q.g.,
64 lbs. per sq. in.). For horizontal planes above NN" it is
smaller than along NN". Similarly, the unit stress pj should
not exceed a proper limit.
287. Numerical Example of a Concrete-Steel Beam of Bectangular Section.
(See foregoing equations.)
Fig. 296 sliows the section [8 by 11 inches] of the beam. Four round steel
rods are imbedded near the under (tension) side, their centers being 10 in. from
W=600 Ihs.
.- d=0.45
4L
p-^?
Fig. 296.
the top of section (a = 10 in.). This beam is to be placed on two supports at
the same level and 8 feet apart, and is to support a concentrated load P, lbs.,
at the middle of the span as well as its own weight, which is K = 600 lbs.
P is to be determined of such a safe value that the greatest stress in the
steel rods shall not exceed 16,000 lbs. per sq. in. The compressive stress in
concrete is not to exceed 700 lbs. per sq. in., nor the greatest shear either in the
concrete or between the steel and the concrete, 64 lbs. per sq. in.
Each steel rod is continuous throughout the whole span and has a diameter
of 0.45 in., from which we easily compute the aggregate perimeter of the rods
346 MECHANICS OF ENGHSTEERING.
to be 5.65 inches (=io)> ^^^ ^^^ aggregate sectional area to be 0.64 sq. in
(= Fy
The ratio of the modulus of elasticity for steel to that of the concrete will
be taken as 15 to 1 ; i.e., n = 15.
The first step is to locate the neutral axis by finding the value of e from
eq. (4), thus: —
64 15 / / 2 X 10 X 8 , , , \ „ J, . .
^-iOQ-T(V .64X15 +l-l)=3.84m.
Next, if for p' we write 16,000 (using inch and pound) and substitute in
eq. (5), solving for M, we obtain the greatest bending moment to which any
section of the beam should be exposed, so far as the steel is concerned, viz: —
M = p'F'la- I) = 16,000 x 0.64 (10 - 1.28) = j
e\ ..nnn.. n«..in 1 oe^ _ i 89,000
in.-lbs.
i.e., max. moment is to be 89,300 in.-lbs.
For the mode of loading of the present beam the max. moment occurs at
the section at the middle of the span and has a value (with I denoting the span,
PI Wl
or 96 in.) of — + -tt- • We therefore write
PX96 ^ 6_00x9_6 ^ gg 3^^_ ^^^^^ ^ ^ g^^^O lbs.
4 o
To find the accompanying maximum compressive stress in the concrete,
eq. (6) gives (for outer "fiber")
2Jf 2 x 89,300 __ -,
P = T-, ; — -s = 5 — FTTH s-^ = 666 lbs. per sq. m.,
•^ le{a-\e) 8 x 3.84 x 8.72 ±' ^ >
which is within the limit set (700 lbs. per sq. in.).
As for the max. shearing unit stresses Ps and ps, they are greatest where
the vertical shear, J, is a max., which is close to one of the supjjprts. Here we
note that J is equal to J of 3,420 + i of 600 = 2,010 lbs. Hence, from eq. (9),
2,010 2,010
^' - 5.65 X (10 - 1.28) = 5.65 x 8.72 " = ^^"^ ^^'- ^^^ "I- '''■'
while ) 2,010
from (10) i^^ = 8"3r8^ = ^^'^ ^^^- P^^ ^*1- ^"•'
These shearing stresses are seen to be well within the limit set, of 64 lbs. per
sq, in. As to compressive stress, the building laws of most cities put 500 lbs.
per sq. in. as max, safe limit forp, the compressive stress in concrete.
288. Concrete-Steel Beam of T-Form Section. See Fig. 297.
In this form of beam, to secure simplicity in treatment, it will
be considered that the flange {TK') alone is subjected to com-
pressive stress [although strictly a small portion of "stem"
between the flange and the neutral axis of a section is under
that kind of stress] . The part of stem below the neutral axis
(as before) is not considered to offer any tensile resistance, all
FLEXUEE OF HEINFOECED CONCEETE BEAMS.
347
tension being borne by the steel rods or " re-inforcement." Fig.
297 shows a side view and also an end-view of a portion of the
beam in Fig. 291 extending from the left-hand support np to
any section DjS (or up to W in Fig. 291). - As before, sections
plane before flexure are considered to be still plane during
flexure, so that the elongations or shortenings of any horizontal
*' fiber," w^hether steel or concrete, are proportional to the dis-
Fig. 297.
tances from a neutral axis iV, at some distance e from the top
fiber of the flange, where the unit compressive stress has some
value p.
Also, since the U for concrete in compression is to be taken
as constant the stresses in the concrete will also be proportional
to the distances of the " fibers " from H the neutral axis. Let
p'^ denote the unit-stress in the concrete at H, the bottom fiber
of the flange ; then, by proportion, p :p^' : : e: e — d, where d is
the thickness of the flange. Since the compressive stresses in
the concrete between H and D are distributed over a rectangle
their average unit-stress is (p-\-p'')/2, and their resultant,
which acts horizontally through some point Gr, has a value of
bd.(p -\- p")l2\ or, as it may be written (see above for y ),
'{^p {1 e - d) .Id) -^ {1 e\ .
The total tensile stress in the steel rods will be ¥'p', as
before, where ¥' is the aggregate sectional area of the rods and
p' the unit stress in them at section DS. Besides the stresses
just mentioned the other forces acting on the free body in Fig.
297 are all vertical ; viz., the shear J'and the pier reaction and
certain loads between and D ; hence by summing the horizon-
tal components we note that the compressive stress in the con-
crete is equal to the tensile stress F'p' in the steel, so that this
348 MECHANICS OF ENGINEERING.
tensile force F'p' and the resultant compressive stress form a
couple (^^ stress-couple '^ of the section; with an arm = 6r6r',
= a"), and we have
pi2e-d)bd ^ •
2e -^ ^ ^
Consequently the shear and the other vertical forces acting on
the free body form a couple also, and the moment of this couple
(equal to that of the " stress-couple ") will be called M. In the
figure these vertical forces are not shown, but simply an equiva-
lent couple (on the left).
If at this part of the beam a length dx of the steel has
stretched an amount dX' and an equal length, dx, of the outer
fiber at D has shortened an amount dX, we have from eq. (2)
of previous work
dX'-p'"E' ^^^^'
where E' and E are the moduli of elasticity of the steel and
concrete, respectively. But, from (11),
p' (e - ^d)bd' ^ ^
and from similar triangles dX : dX' : : e : (a — e) . . . .(14)
Eqs. (33), (14), and (12), with E' -i- E = n, give
F-n.a+—-
'= bd + En ' ^^^^'
and thus the neutral axis, iV, is located.
It will now be necessary to locate the point of application,,
between E and E, of the resultant compressive stress on EE ;
that is, the point (r in Fig. 298 which gives a side view of
these stresses alone, forming, as they do, a trapezoidal figure
whose center of gravity, U, projected horizontally on EE gives-
the desired point, G: The lower base EC^^ of this trapezoid
FLEXURE OF JREINFOECED CONCRETE BEAMS.
349
represents the unit stress jw"; the upper, DC", represents the
unit stress p. The distance, call it c,
of G- from N, is to be determined.
Let the trapezoid be divided into a
rectangle BD'" C" H and a triangle
D'"G"'Q". The center of gravity of
the latter is at a vertical distance of \ d
from a line WW" drawn horizontally
at distance \ d from D . H" H'" passes
through the center of gravity of the
rectangle. Let us now find the distance GrR" by writing the
moment of the resultant stress about point W equal to the
sum of those of its two parts, or components, represented by
the rectangle and the triangle ; whence we have
Fig. 298.
\{p^p").uy. aw = o + ^P-^P^'^ .'^
(16)
Noting that ji?'' =
d-"
- d
p, we have, solving,
aw=\
6 2e- d
NCr, i.e., c, = e
, and therefore, measuring from N,
d 1
2 "^ 6 ■ 2e - cZ
(17)
Now that both e and c have been determined in any given
case it remains to find expressions for the unit stresses p' and j?
(in steel and fiber I) of concrete}.
Since (r is the point of application of the resultant compres-
sion in flange, the arm of the stress-couple, a" (Fig. 297), is
the distance from G- to Gr' (see Fig. 297); that is,
a" = c -{- [a— e); and hence we may write
Pf(c + a-e)~M;.:f = ^,^/^^_^^ .(18)
Also, by eliminating the ratio d\:d\' from eqs. (12) and
(14) we have, solving for^,
p =
p e
n (a — e)
(19)
350 MECHANICS OF ENGINEERING.
289. Shearing Stresses in T-Form Concrete-Steel Beams. As
regards the unit shearing stress, p'^ induced on the sides of the
steel rods, in this case of the concrete steel beam of T-form
section, an analysis similar to the corresponding one in the case
of the beam of rectangular section leads to the result
J ( where «7is the total vertical shear at ) ,oa\
Iq{c -\- a— e) I the section BS, and l^ the aggregate )
perimeter of the steel rods.
And, similarly, for the unit shearing stress on the horizontal
surface separating the flange from the "web" or "stem" (see
Fig. 297 ) at H, where the width of the web is h", we find for this
unit horizontal shear, p^,
P'= V-(c + a-e) (^^'
290. Deflections of Concrete-Steel Beams. The deflection of a
loaded prismatic concrete-steel beam resting on two supports at
its extremities, may be obtained for the cases dealt with in
§§ 233-236 inclusive, in connection with homogeneous beams ;
provided the product EI occurring in the expressions for these
deflections be replaced by ■ [ a— -^), for concrete-steel beams
of rectangular section; and by E' F' {a — e) (c + a — e), for
those of T-form section.
291. Practical Formulae and Diagrams for use with Concrete-Steel Beams
of Rectangular Section. The equations of the foregoing theory will now be put
into convenient form for practical use in designing these
beams. Let us denote the ratio of p' (stress in steel at
section of max. moment) to p (stress in outer fiber of
concrete) by r ; i.e., r = p'/p ; while n = E'/E, as before.
Also let ?n, = M -^h, denote the max. bending moment per
inch of width (6) of beam ; and let F' (area of steel) -h & be
called/, i.e., steel areajjer mc/i of width {b). In other words,
we have the notation
r =-?:_; n= -=-;™ = ^; and /'= -=-; . . . . (22)
Fig. 299. p E
If we now substitute e = 2 r/, from eq. (1), in eq. (3),
we have 2Tf {r + n)=an (23)
Now e = 2 r/, which from (23) =an-^{r + n) ; Hence eq. (5)
-will give 3?n(r + Ti) = a/'p'(3r + 2n) (24)
REINFORCED CONCRETE BEAMS
DIAGRAM 1
^ rectangular').
20 25 30 40
L/ 1 r^ M 1 ir^ III .!,•
;
- -r ->
1 H -
■i::
:-_!_: '.!.'.
.-H ::f.
i-t.. 60
s =
/«-
»
:;-r-:-?:
- T-- ^^
~ «->«./'„ . „ \ '
- _l- _
:.-^y-.
in which ^c
::z5:::t:
,?-i- -i-
I'.Tl'i
- -\-^
■ J-.
y
y
..2f_
-■^•- 50
—
— 1
/
— 1 — ^.
1 ^1
-
- -L V-
^ y J
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y
/'^ 1
.,2^
y
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\ d ah 1
xc
/
: _Ly
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^5 ^ ^ n Q
'"i""
'/
/
-
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1 I'' '
---1---
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y
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CO
p°^
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in
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y
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m ^'
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y
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1
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T ■ 35
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y
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^
^
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' ^.>'
' 1 ■■ 20
1 1
/
/y
/^
y
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y
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y-
.-^ ;-
4:j s
/
y j
/
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y
y-
1
y
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y
' *;'x.
1 {
/
y/^
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^
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1
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y
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X
y
y
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-y
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y^-1..
..4... ]g
■ r ' ■
/..
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X
y
y
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y
^
y
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- -I- -
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-H -- -1-
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l_ 1
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1 1
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> 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 -t-2^
^
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
N
s
^
\
s 1
\ ^^ i ••
!..i.|..
\
t
\
s
^^'
^4.
r'ii
IS
\
I)
.^
\>
N i S,
: '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
1
^^<,
kH 's
1
1
1^
L_.
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 = n-h[2r{r+n)] (25)
3 J'4-2 n
Asain, let ^— r- be denoted by Q - . . . (26)
° 3(r + n) ^ '
Prom (24), (25) and (26) we may then derive
M=qSp'a^b .... (27); andi m = QSp'a^ ...... (28)
for practical use in design; in connection with Diagrams I and II (opp. pp.
350 and 351) from which we may obtain values of Q and 8, respectively, for
any given values of r and n.
As to unit shearing stresses, which should not exceed 64 lbs. per sq. in.
(say), use is made of eqs. (9) and (10), § 286, in which the maximxim total vertical
shear Jm should be substituted. In computing "e " for use in (9) and (10), it is
simplest to employ the relation [derived from eqs. (1), (23) and (25)]
e = 2 Sra (29)
292. Numerical Examples. Rectangular Section. Let us suppose that in the
cross-section of maximum moment the stresses in both materials are to attain
their greatest safe values; viz., 16,000 lbs. per sq. in. tension for the steel, and
600 for the compressive stress in outer fiber of the (rock) concrete. Also suppose
that ^' = 30,000,000 and £' = 2,000,000 lbs. per sq. in. That is, we have n=15
and r = 16,000 H- 500, = 32 ; and from Diagram I find S = 0.005. In other words
the necessary area of steel section is J of one per cent, of the area of the concrete
(above steel rods). "We also find, from Diagram II, that Q = 0.894,
If now " a " be taken as 10 in., eq. (28) gives : —
m = 0.894 X 0.005 X 16,000 X 100, = 7,152 inch-lbs. bending moment that
could safely be withstood by each inch in the width &; so that if "&" were 8
inches we should have M = mb = 7,152 x 8 = 57,216 inch-lbs., safe bending
moment for the section of the beam.
Again, with r and n still equal to 32 and 15, respectively, and hence with
Q and /S as before, viz., 0.894 and 0.005, if b is assumed as 10 in., and the max.
bending moment to be sustained is If = 80,000 inch-lbs. (so that m = 8,000),
we find from eq. (28)
= \/^
« = ^/ -.894x0^07x16,000 = ^^''^ ^^^^^^'
as necessary value of a ; while the total area of steel section needed is
F', = S . a &= 0.005 X 10. 58 X 10 = 0. 5290 sq. in.
which is seen to be one half of one per cent, of the area [10x10.58] of the
concrete above steel (i.e., S = 0.005, as found from Diagram I originally).
293. Cost of Beams of Rectangular Section. While in a general sense
economy in cost is favored by having the width " &" of the rectangular section
small compared with the height " a," a limit to narrowness of width is set by
the unit shearing stress in the neutral surface which would be found to exceed
a safe limit if the beam were too narrow. The thickness "a'" of concrete
below the steel rods (see Fig. 292) might be made -^5 of " a."
362
MECHANICS OF ENGINEERING.
CHAPTER VI.
Flexure. Columns and Hooks. Oblique Loads.
294. Oblique Prismatic Cantilever. In Fig. 301, at (a),
(on p. 354) we have a prismatic beam built in at K, projecting-
out obliquely, and carrying -a vertical load P at upper end ; the
line of action of P passing through the center of gravity of the
upper base of the prism. In such a case the fibers of the beam
where they cross any transverse plane mg will evidently be
subjected to compressive stress (called a ^'■thrust'''') due to the
component of P parallel to the axis OKoi the prism ; to a shear
J" due to the component of P at right angles to that axis ; and
also to additional stresses, both tensile and compressive, formings
a " stress-couple^^ due to the moment of P (i.e., Pu) about ^, the
center of gravity of the cross-section m'm.
More in detail, consider in Fig. 300 a portion AB of the
prism, being the part lying above a cross-section mm' near the
top, so that the portion gO of
axis is practically perpen-
dicular to the section mm'
which is a plane both before
and after flexure, g being
the center of gravity of the
plane figure formed by the
cross-section.
Let the unit stress on the
end of the extreme fiber at
m be represented by the
length sm and that [also com-
pression (say)] on the other
extreme fiber, at m', by s'm'.
Draw the straight line ss' ;
then by the common theory of flexure the stress on any inter-
mediate fiber, at c, would be the intercept, or ordinate, ac to this
line. Now the unit stress p^, on the fiber g at the center of
gravity of cross-section, being gr, draw through r a line t'rt
Fia. 300.
FLEXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 353
parallel to m'm, and we now have the stress on any fiber as o
divided into two parts be, or p^, the same for all the fibers ; and
ab, different for the different fibers but proportional to the
distance z of the fiber from g. Hence we have :
the unit stress on any fiber c is
P = Pi -\ P2 (ibs. per sq. in.) . . . . (1)
where p^ is st and e the distance of the extreme fiber ??? from g ;
and hence the total stress on fiber c is pdF =p^dF H — Pnli^F, lbs. ;
where dF is the area (sq. in.) of section of fiber, or element of
area of the cross-section, F being the total area of the cross-
section, mm' . Geometrically, we note that while the system of
normal stresses on all the fibers forms a trapezoid, m's'sm in
this side-view, and that they are all compressive, they are
equivalent to a rectangle, m't'tm, of stress of uniform compressive
unit-stress p^ ; and two triangles, one, rst, of compressive stress,
and the other, rs't', of tensile stress.* It will now be shown that
the sum of the moments of the stresses of the rectangle about
center g is zero, and that the two triangles of stress form a couple.
^(moms.) of stresses in triangle = I (p^dF)z = pi dFz
=p^Fz = zero ; since z =zero, the 2's being measured from the
center of gravity, g, of section mm' [§ 23, eq. (4)].
Again, if we sum (algebraically) the stresses of the two
triangles,
we have / - p dF = — I zdF = ^Fz, = zeYo
Jz = -e' e e J e
that is, the resultant of the compressive stresses in rts equals
that of the tensile stresses in rs't' ; hence they form a couple.
If, therefore, we have occasion to sum the moments about
g, of all the stresses acting on the fibers in section wm' we are
to note that this moment-sum involves the stresses of the triangles
alo7ie (that is, of the couple), and is
in.-lbs. ; where I^ is the " moment of inertia " of the cross-section
* These plane figures are the side views of geometric solids.
354
MECHANICS OF ENGINEERING
referred to an axis through g (its center of gravity) and perpen=
dicular to the "force plane " (plane of paper here).
If, again, we sum the components of all the stresses (on plane
mm') parallel to the axis gO we note that this sum is zero for
the couple and also for the shear J and hence reduces simply to
fp^dF =_Pj CdF=p^F, lbs. (the Thrust) . . (3)
(corresponding to the rectangle, fm).
The sum of components perpendicular to axis ^ is of course
simply the shear, J, lbs.
Evidently the unit stress (normal) in fiber at m is expressed
e'
as jp,„ = j9j 4-^2' ^^^^ ^^^^ ^^ *^' '^^^ Pm'=Pi P2' ^^ ^^ ^^^J
case the latter is negative it indicates that the actual stress in
this fiber is tension.
295. Oblique Cantilever. Fig. 301, (a) and (5). At (h) is shown
as a " free body," a portion [of the cantilever at (a)] of any
length X from top. The
forces acting are the vertical
load P at 0, and the stresses
on the ends of the fibers in
the section m'm.; and these
stresses are now indicated
as consisting of a thrust, T,
of uniform intensity p^, the
total thrust being ^9^^', lbs.,
(where Y is the total area of
section) ; of a stress-couple,
' '" C, whose moment is -^ in.-
FiG. 301. e
lbs., in which pj = Pm — P^ ^^^ I is the " moment of inertia "
of the cross-section (about an axis through its center of gravity
g at right angles to the plane (" force plane ") containing Og
and force P ; the same I that has been used in previous cases of
flexure) ; and the total shear, J, lbs., parallel to force plane and
perpendicular to gO. The lever arm of P about g i^ u which
practically = x sin a (unless the beam is considerably bent or
is nearly vertical).
FLEXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 355
For this free bjdy (in order to find p^, p^ and J )
Xxr A • Ti -n f\ Pcosa
Jl = Ogives: /^ cos a— p^i^ =0 ; .•. p^ =
F
X^ X A Vol r> A -^^^
(moms.)^ = . .-^ Fu = 0; .-. p^ = — — •
(4)
. (5)
and
Xi'=o
P sin a — J" = (9 ; .-.J — P sin a, (6)
As X varies, from to Z, we note that p^ and J remain
unchanged but tliat p., increases ivith u ; so that the maximum
value of the unit stress jo,,,, Avhicli = p^ + p,' will be found in
the section at K, where x = I ; and if this stress is not to exceed
a safe value, R', for the material, we put p^i,^^ K) +p^ = R',
(as the equation of safe loading) ;
^^nan ^, .... (7)
or,
P
"cos a
~'f~
Pn
(N. B. For a cross-section of unusual shape the stress
e'
, = p^ P2, at K, might happen to be numerically greater
than Pj^„ and thus govern the design).
296. Experimental Proof of Foreg^oing. A
stick or test piece of straight-grained pine
wood, 12 inches in length and of square
cross-section (one inch square), originally
straight and planed smooth and with bases
perpendicular to ^the length, was placed in
a testing machine ; steel shoes, with (outside)
spherical bearing surfaces, being centered
on the ends. See Fig. 302, where AB is
the stick and S, S\ the two steel shoes. The
stick was gradually compressed between
the two horizontal plates B,B'^ of the machine
and bent progressively in a smooth curve
under increasing force. From the nature of
the "end conditions," as the stick changed
form, the line of action of the two end
pressures P,P, always passed through the
centers of gravity, a and h, of the respective bases.
When the force P had reached the value 4500 lbs. a fine
wrinkle was observed to be forming on the right-hand surface
Fie. 302.
356 MECHANICS or ENGINEERING.
of the stick at the outside fiber m of the middle section gm. The
other tibers of this section were evidently uninjured. At m then,
the unit-stress must have been about 8000 lbs. per sq. in., the
crushing stress (as known from previous experiments with sticks
of similar material and equal section but only three or four
inches long; these were too short to bend, and wrinkles formed
around the whole 'perimeter^ showing incipient crushing in all the
fibers). The distance gc at this time was found to be \ in. ;
i.e., the lever arm, w, of the force P about g, the center of gravity
of the section. In this case, then, it is to be noted that the
value of %i was entirely due to the bending of tit e piece.
Substituting, in eqs. (4) and (5) of § 295, the values w = |- in.,
a = 0, cos a = l, e = e', =i inch, i^=l sq. in.,
hh^ 1x1^ 1
and i; = — , = ^^ = — in/, we find p^ = 4500
lbs. per sq. in. and p^= 3375 lbs. per sq. in.
Hence stress at m, = Pi-\- P21 == 7875 lbs. per sq. in., which is
about 8000, as should be expected. On the fiber at 0, how-
ever, we find a stress of p^ — p., or of only 1125 lbs. per sq. in.
compression.
We find, then, that in the section om, when P reached the
value of 4500 lbs., there was a total-thrust (p,F^ of 4500 lbs.;
a unit-thrust (w^) of 4500 lbs. per sq. in. ; and a stress-couple
pi
having a moment of Pw, = ^— , = 562.5 in .-lbs., (implying a
separate stress oi p^^^'^l^ lbs. per sq. in. in the outer fibers,
to be combined with that due to the thrust). Also that /, the
shear, was zero.
297. Crane-Hooks. First (Imperfect) Theory. Fig. 303 shows
a common crane-hook of iron or steel. Early writers (Brix and
others) treated this problem as follows : —
The load being P, if we make a horizontal section at AB^
about whose gravity axis, gr, P has its greatest moment, and con-
sider the lower portion C as a free body, in Fig. 304), we find,
using the notation and subdivision of stresses already set forth
in § 294 for an oblique prism, that the uniformly distributed
pull (or " negative thrust ") on the fibers is p^F = P, lbs. ;
FI^EXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 357
P
while the moment of the stress-couple is ^- = Pa ft.-lbs.; and
e
that the shear, /, is zero.
Hence on the ex-
treme fiber at B we
have a total unit
tensile stress of
Pae
T'
which for safe de-
sign must not ex-
ceed the safe unit-
stress for the ma-
terial, R' lbs. per sq. in. ; whence we should have
p[l + f] = iS' . .
as the equation of safe loading.*
Example: Safe P = ?, if section AB is a circle of radius
2 in., while a = 4 in. ; the material being mild steel for which
(in view of the imperfection of the theory) a low value, say
6000 lbs. per sq. in., should be taken for R'. With these data
we obtain: —
1 4x2"
Fig. 303.
Fig. 304.
Fig. 305.
(8)
P = 6000
-[
12.56 ^ 50.24
h
25130 lbs.
The simple crane in Fig. 305, being practically an inverted
hook, may be treated in the same manner.
298. Crane-Hooks. Later, More Exact, Theories. The most
exact and refined theory of hooks yet produced is that of
Andrews and Pearson,! but it is very complicated in practical
application and far too elaborate and extended to be given
here.
The next best (and fairly satisfactory) treatment is that of
Winkler and Bach, of which the principal practical features
and results will now be presented.
* See experiments by Prof. Goodman, in Engineering, vol. 72, p. 537. Re-
sults are irregular, due probably to the use of this imperfect theory.
t Drapers' Company Research Memoirs. Technical Series I. London,
1904.
358
MECHANICS OF ENGINEERING.
In AB, Fig. 306, we have again the free body of Fig. 304,
but the vertical stresses acting on the cross-section m'm are
proportional to the ordinates of a curve instead of a straight
line. The imperfection of the early theories lies in the fact
that the sides of a hook are curved, and not straight and par-
allel as in the prismatic body of Fig. 301 ; and the variation of
stress from fiber to fiber on the cross-section must follow a dif-
ferent law, as may thus be illustrated :
As preliminary, the student should note, from the expres-
P EX
sion— = — -of p. 209, that in the case of two fibers under ten-
F I
sion, with the same sectional area F, the unit-stress P -^ F (or
p) is not proportional to the elongation }. of the fiber unless the
two lengths I are equal. In Fig. 306 the center of gravity of
the cross-section is g, and is the center of curvature of the
curved axis gk of this part of the hook (or other curved body).
The two consecutive radial sections m'm and ft are assumed
to remain plane during stress, and hence the changes of length,
due to stress, of the (verti-
cal) fiber lengths between
them are proportional to
the ordinates of a straight
line ; and if these fiber
lengths were equal in
length (as would be the
case for a prismatic beam)
the unit-stresses acting
would also be proportional
to the ordinates of a
straight line (this is the
case in Fig. 301).
But in the present case
these fiber-lengths are un-
equal, so that the unit-
stresses in action are (in
general) proportional to
the ordinates of a curved line. Such a curved line we note in
vCi Fig. 306, the ordinates between which and the horizontal
line hi represent the unit-stresses, p, acting on the upper ends
Fig. 306.
FLEXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 359
of the vertical fibers from m' to m. Tlius, the stress on the
fiber mt is p^ = ei (tension); and that on the other extreme
fiber, (at m') is p^, = hv (compression).
If now we compute the average unit-stress p^ =^ P -i- F and
lay it off, == is, upward from hi, and draw the horizontal 6s, we
thereby re-arrange the stresses into a uniformly distributed pull
(or " negative thrust ") p^F ^hs., represented by the rectangle
hsih, and a stress-couple formed by the ordinates lying between
the curve and the axis hs.
It will be noted in Fig. 306 that there is a fiber at some
point n (on right of g) where the stress is zero ; i.e., the " neu-
tral axis " of the section is at n, ~1 to paper. Also, at some
point n', the actual stress is equal to the average, p^, and an
axis ~] to paper through this point would be the neutral axis if
the forces acting on this free body, other than the fiber stresses,
consisted, not of a single force P, but of a couple, with a mo-
ment = Pa. This axis through n' might be called the neutral
axis for " pure bending ", since then the whole system of fiber
stresses would reduce to a couple and the stresses would be
measured by the ordinates between hs and the curve.
299. Crane-Hooks. Winkler-Bach Theory. Formula for Stress. In Fig.
306, let F be the area of the plane figure formed by the section m'm, dF an
element of this area, and z its distance (reckoned positive toward the right)
from the gravity axis, g^ of the section. The radius of curvature of gk is r,
and a is the lever arm of P, the load, about g. Let gm = e and gm' = e' (dis-
tances of extreme fibers) and let
/Vr r'=+^ I dF
S denote the quantity ( t, / (
an abstract number depending on the area, shape, and position, of the cross-
section m'ni ; and upon the radius of curvature r. Its value may be obtained
by the calculus (or Simpson's Eule) for ordinary cases. For instance, if the
section is a rectangle of width b, and altitude = A, = m'm, we find
-1 ; ■ (1)
'-jh-'^l)-' (^:
From the Winkler-Bach theory it results that the unit-stress on any fiber
between m and m', at a distance z from the gravity axis g (on the right, toward
the center of curvature, 0; if on the left, z is negative) is
a /. Z 1"'
J"
^^F
I
r\ r — z S/.
(3)
lbs. per sq. inch. A positive result from (3) indicates tension; a negative, com-
pressive stress. Of course, for P -=- Pwe might write the symbol p^, or " aver-
age stress." If p were set = zero, a solution of (3) for z would locate the neu-
360 MECHANICS OF ENGINEERING.
tral axis, n, of Fig. 306; while by placing p — Pj = 0, a solution for z would
locate the point n', or neutral axis for "pure bending."
300. Numerical Example. Let the cross-section be a trapezoid, of base
6 = 3 in. at m, and upper base 6' = 1 in. at ?/i', both | to paper ; the altitude
/i, =• m/m, being 4 in. This brings g f in. (= e ) from m and | in. (= e') from
m'. Let N be in the same vertical as and Om = 2 in. Hence r=a = 2 + |.=
y in. The material is mild steel and the load P is 8 tons ; find p^ and -pm' .
From above dimensions we find area ^=8 sq. in,, while from eq. (1),
(using the calculus), S= 0.0974. For p^ w^ put z = + f in. in eq. (3) ; and for
Pm', 3 = - I in. ; obtaining, finally, p,« = 17,120 lbs. per sq. in. (tension) ; and
Pm' = — 7,980 (compression). Evidently the elastic limit is not passed.
Using the imperfect theory of § 297, we should have obtained pm = 12,000
lbs. per sq. in., only ; which is seen to be about 30 per cent, in error, compared
with the above value of 17,120. The reason for taking a low value for the safe
unit-stress, B', in the example of § 297 is now apparent, an additional reason
being the fact that loads are sometimes "suddenly applied " on hooks.
301. By "column" or "long column" is meant a straight
beam, usaally prismatic, which is acted on by two com-
pressive forces, one at. each extremity, and whose length
is so great compared with its diameter that it gives way
(or " fails ") by buckling sideways, i.e. by flexure, instead
of by crushing or splitting like a short block (see § 200).
The pillars or columns used in buildings, the compression
members of bridge-trusses and roofs, the " bents " of a
trestle work, and the piston-rods and connecting-rods of
steam-engines, are the principal practical examples of long
columns. That they should be weaker than short blocks
of the same material and cross-section is quite evident, but
their theoretical treatment is much less satisfactory than
in other cases of flexure, experiment being very largely
relied on not only to determine the physical constants
which theory introduces in the formulae referring to them,
but even to modify the algebraic form of those formulae,
thus rendering them to a certain extent empirical.
302. End Conditions. — The strength of a column is largely
dependent on whether the ends are free to turn, or are
fixed and thus incapable of turning. The former condi-
tion is attained by rounding the ends,' or providing them
with hinges or ball-and-socket-joints ; the latter by facing
off' each end to an accurate plane surface, the bearing on
which it rests being plane also, and incapable of turning.
In the former condition the column is spoken of as having
FLEXURE. LONG COLUMNS.
361
round ends ; * Fig. 311, (a) ; in the latter as having fixed ends,
(ov flat bases ; or square ends), Fig. 311, (&).
Fig. 312.
Sometimes a coliimn is fixed at one end while the othei
end is not only round but incapable of lateral deviation from
the tangent line of the other extremity ; this state of end
conditions is often spoken of as "Pin and Square," Fig.
311, (c).
If the rounding * of the ends is produced by a hinge or
** pin joint," Fig. 312, both pins lying in the same plane
and having immovable bearings at their extremities, the
column is to be considered as round-ended as regards flex-
ure in the plane 1 to the pins, but as square-ended as re-
gards flexure in the plane containing the axes of the pins.
The " moment of inertia " of the section of a column will
be understood to be referred to a gravity axis of the sec-
tion which is "I to the plane of flexure (and this corres-
ponds to the " force-plane " spoken of in previous chap-
ters), or plane of the axis of column when bent.
303. Euler's Formula. — Taking the case of a round-ended
column, Fig. 313 (a), assume the middle of the length as
an origin, with the axis X tangent to the elastic curve at
that point. The flexure being slight, we may use the form
EI (Py-^dx^ for the moment of the stress-couple in any
* With round ends, or pin ends, it should be understood tliiit the force
at each end must be so applied as to act through the centre of gravity of the
base (plane figure) of the prismatic column at that end ; and continue to do
so as the column b^ nds.
562
MECHANICS OF ENGINEERING.
dp
dy
dy
dx-
—'r-
y
J/c
dx-
" /
/
dx
7
f
-a-y.
1
if
4+-
-Or—
1
"1
\x
1
1
1
1
1
J
1
I
Y
Fig. 313.
Fig. 314.
^section w, remembering tliat with this notation the axis X
must be || to the beam, as in the figure (313). Considering
the free body nC, Fig, 313 (h), we note that the shear is
zero, that the uniform thrust =P, and that 2'(moms.n)=0
gi'ves (a being the deflection at 0)
EI
d'y
dx^
--F{ar-y)
Multiplying each side by dy we have
El
dx"
dy (Fy=Pa dy — Fy dy
(1)
(2)
' Since this equation is true for the y, dx, dy, and d^y of any
element of arc of the elastic curve, we may suppose it
written out for each element from where ?/=0, andc''y=0,
up to any element, (where dy=dy and y=y) (see Fig. 314)
and then write the sum of the left hand members equal to
itliat of the right hand members, remembering that, since
dx is assumed constant, l-^dsc^ is a common factor on the
left. In other words, integrate between and any point
of the curve, n. That is.
f[dy]d[dy] =Fa f dy—P T ydy (3)
The product dy d^y has been written {dy)d(dy\ (for d^y m
EI
da?
FLEXURE. LONG COLUMNS. 363
the differential or increment of dy) and is of a form like
xdx, or ydy. Performing the integration we have
EI d_l y^ .... (-1)
dx' 2^2 ■ ^
which is in a form applicable to any point of the curvej
and contains the variables x and y and their increments
dx and dy. In order to separate the variables, solve for dxy
and we have
di
dx=l^-JL==^OTdx=^ I EI, \aJ ...
d(y)
'^ (X \C(/ /
i.e.,a!=±y-p- (vers, sin ^^j , , . (6)
(6) is the equation of the elastic curve DOG^ Fig. 313 (a),
and contains the deflection a. If P and a are both given,
y can be computed for a given cc, and vice versa, and thus
the curve traced out, but we would naturally suppose a to
depend on P, for ineq. (6)whena7=^Z, y should —a. Mak-
ing these substitutions we obtain - ^
'A^= V^ (^^^"- ^^^ "' ^-^^^ ' ^•^- >^^= 7^ I ^^^
Since a has vanished from eq. (7) the value for P ob-
tained from this equation, viz.:
i\=EI ^ .... (8)
is independent of a, and
is ,\ to be regarded as that force (at each end of the round'
ended column in Fig. 313) which will hold the column at
any small deflection at which it may previously have been
set.
364
MECHANICS OF ENGINEERING.
In other words, if the force is less than Pq no flexure at
all will be produced, and hence P,, is sometimes called the
force producing " incipient flexure." [This is roughly ver-
ified by exerting a downward pressure with the hand on
the upper end of the flexible rod (a T-squai e-blade for in-
stance) placed vertically on the floor of a room ; the pres-
sure must reach a definite value before a decided buckling
takes place, and then a very slight increase of pressure oc-
casions a large increase of deflection.]
It is also evident that a force slightly greater than P^
would very largely increase the deflection, thus gaining for
itself so great a lever arm about the middle section as to
cause rupture. For this reason eq. (8) may be looked
upon as giving the Breaking Load of a column with round
ends, and is called Euler^ s fornfiula.
Referring now to Fig. 311, it will be seen that if the three
parts into which the flat-ended column is di-
vided by its two points of inflection A and B
are considered free, individually, in Fig. 315,
the forces acting will be as there shown, viz.:
At the points of inflection there is no stress-
couple, and no shear, but only a thrust, =P,
and hence the portion AB is in the condition
of a round-ended column. Also, the tangents
to the elastic curves at and G being pre-
served vertical by the f rictionless guide-blocks
and guides (which are introduced here simply
as a theoretical method of preventing the ends
from turning, but do not interfere with verti-
cal freedom) OA is in the same state of flex-
ure as half of AB and under the same forces.
Hence the length AB must = one half the
total length I of the flat-ended column. In
other words, the breaking load of a round-
ended column of length =^Z, is the same as
that of a flat-ended column of length —I.
Hence for the I oi eq. (8) write %l and we
have as the breaking load of a column with
flat-ends and of length =1.
}il
f/MWM
Fig. 315.
TLEXURE. LONG COLUMNS. 365
r.^4.m^ .... (9)
Similar reasoning, applied to tlie " pin-and-square "
mode of support (in Fig. 311) where the points of inflec-
tion are at B, approximately y^ I from G, and at the
extremity itself, calls for the substitution of ^ I for I in
eq. (8), and hence the breaking load of a ^'pin-and-square "
column, of length = I, is
P^=l ^/^ . . . (10)
Comparing eqs. (8), (9), and (10), and calling the value of
Pi (flat-ends) unity, we derive the following statement :
The breaking loads of a given column are as the numbers
1
flat-ends
9/16
pin-and-square
y^ j according to the
round-ends \ mode of support.
These ratios are approximately verified in practice.
Euler's Formula [i.e., eq. (8) and those derived from it,
(9) and (10)] when considered as giving the breaking load
is peculiar in this respect, that it contains no reference to
the stress per unit of area necessary to rupture the material
of the column, but merely assumes that the load producing
" incipient flexure ", i.e., which produces any bending at
all, will eventually break the beam because of the greater
and greater lever arm thus gained for itself. In the canti-
lever of Fig. 241 the bending of the beam does not sensibly
affect the lever-arm of the load about the wall-section, but
with a column, the lever- arm of the load about the mid-
section is almost entirely due to the deflection produced.
It is readily seen, from the form of eqs. (8), (9) and (10),
that when I is taken quite small the values obtained for Po, Pi,
and P2 become enormous, and far exceed what would be
found from the formula for crushing load of a short block,
viz., P = FC (see p. 219), with F denoting the area of section
of the prism and C the crushing unit-stress of the material.
The degree of slenderness a column must have to justify the
use of Euler's relations will appear in the next paragraph.
36G
MECHANICS OF ENGINEERING.
304. Euler's Formula Tested by Experiment. — Since the
"moment of inertia," /, (referred to a certain axis) of the cross-
section of the column may be written I = Fk^, where k is the
"radius of gyration " (see p. 91), and F the area of the plane
figure, eq. (8), for ''round 1 Pq tz^E
ends," may be written \ F~~(l^ky ' *
Here Po^F is the average unit-stress (compressive) on the
cross-section and l^k is a ratio measuring the slenderness of
the column. (Of course, when the column actually gives way
by buckhng, the unit-stress on the concave side at the middle
of the length is much greater than the average). In the ex-
periments by Christie, described on p. 112 of the Notes and
Examples, the value of the ratio l-i-k ranges from 20 to 480.
As an example consider a 3"x3"Xi" angle-bar (or "angle")
of wrought iron, with Z = 15 ft., to be
used as a column. Fig. 315a shows
the cross-section of this shape, with di-
mensions. Q is the center of gravity
of this plane figure. Let the force be
applied at each end of the column
according to Christie's mode of "round
ends," i.e., by a ball-bearing device. Fig. 3l5a.
the force always passing through the point C of the section at each
extremity of the column. Since the ends are free to turn in any
plane, the axis
of the column
will deflect in
the plane CN 1
to the axis 2 ... 2
(of the plane
figure) about
which the values
of / and of k are
least. For this
shape, we find
from the hand-
book of the Cam-
bria Steel Co.,
that k about
axis 2 ... 2 is the least radius of gyration and =0.58 in. ; also that
FLEXURE. LONG COLUMNS. 367
the area of the figure is F = 2.75 sq. in. Hence the "slender-
ness-ratio-;' l^k, is 180" -^ 0.58" = 310; and from eq. (11) we
have, with E for wrought iron taken as 25,000,000, lbs./in.2
(p. 279),
I (Po -rF)=Ti^X 25,000,000 -^ (310)2 = 2570 lbs. / in.2 ;
while from the Christie experiments we find (Po^P)=2650
lbs. /in.2 as the average unit-stress at rupture; a fairly close
agreement with the Euler result. The total rupturing load
would then be Po = 2570X2.75 = 7070 lbs., and the safe load,
with the "factor of safety " of 8 recommended in the Christie
report, would be 884 lbs.
In this way it may be ascertained that for values oi l^k
from 200 to 400 for "round ends " and from 300 to 400 for
fixed ends there is an approximate agreement between
Euler's equations and the Christie experiments. But most
of the columns used in engineering practice involve values
oil^k less than 200, so that Euler's formulae are not adapted
to actual columns (though used to some extent in Germany).
A formula of such nature as to be available for all degrees
of slenderness has therefore been established (Rankine's,
see next paragraph), based partly on theory and partly on
experiment, which has obtained a very wide acceptance
among engineers.
In Fig. 3156 is shown a curve, Er, resulting from plotting as abscissa
and ordinate the values of Po-Hi^ and Z-h A:, as related in Euler's formula
(8) for columns with round ends, for "medium" structural steel; with
£' = 30,000,000 lbs./ in- ^ Ej is a similar curve plotted from Euler's formula
(9) for fixed ends for the same material. Each of these Euler curves is
tangent to both axes at infinity. The other curves wUl be referred to later.
305. Rankine's Formula for Columns. — The formula of this
name (some times called Gordon's, iu some of its forms) has
a somewhat more rational basis than Euler's, in that it in-
troduces the maximum normal stress in the outer fibre and
is applicable to a column or block of any length, but stili
contains assumptions not strictly borne out in theory, thus
introducing some co-efficients requiring experimental de-
termination. It may be developed as follows :
Since in the flat-ended column in Fig. 315 the middle
portion AB, between the inflection points A and B, is
acted on at each end by a thrust = P, not accorapanied by
any shear or stress-couple, it will be simpler to treat thai
368
MECHANICS OF ENGINEERING.
p.,
portion alone Fig. 316, (a), since the thrust and stresa-
couple induced in tlie section at
R, the middle of AB, will be equal
to those at the flat ends, and G,
in Fig. 315. Let a denote the de-
flection of R from the straight line
AB. Now consider the portion
AR as a free body in Fig. 316, (b),
putting in the elastic forces of the
section at R, which may be clas-
sified into a uniform thrust =
PiF, and a stress couple of moment Fiq. sie.
294). (The shear is evidently zero, from
= L:_, (see
e
I (hor comps.) = 0). Here p^ denotes the uniform pres-
sure (per unit of area), due to the uniform thrust, and jpg
the pressure or tension (per unit of area), in the elastic
forces constituting the stress-couple, on the outermost
element of area, at a distance e from the gravity axis (~|
to plane of flexure) of the section. F is the total area of
the section. / is the moment of inertia about the said
gravity axis, g
1 (vert, comps.) = gives P == p^F , . (Tj
2' (moms.j,) = gives Pa = ■^— .... (2)
For any section, n, between A and R, we should evidently
have the same^j as at R, but a smaller pi, since Py < Pa
while e, /, and F, do not change, the column being pris-
matic. Hence the max. (pi+JJa) is oil the concave edge at
R and for safety should be no more than G -^ n, where G
is the Modulus of Crushing (§ 201) and w is a " factor of
safety." Solving (1) and (2) for j9i and^gj and putting their
sum = C -V- %, we have
P.Pae G
(3)
We might now solve for P and call it the safe load, biat a§
FLEXURE. LONG COLUMNS. 369
is customary to present the formula in a form for giving
the breaking load, the factor of safety being appHed after-
ward. Hence, we shall make n=l, and solve for P, calling
it then the breaking load. Now the deflection a. is unknown,
but may be expressed approximately, as follows, in terms
of e and l.
If we now consider ARB to be a circular arc, of radius = |0,
we have from geometry (similar triangles) a=(Z-^- 4)^^2/9;
and if we equate the two expressions for the moment of the
EI V2I
stress-couple at R there results — = — (see pp. 249 and 250) .
A combination of these two relations gives ae=(p2^S2E)P,
Now under a safe load the total stress, pi + p2, in the outer
fibre (concave side) at R will have reached a safe value, R',
for the material, and is therefore constant for this material,
and if the rude assumption is made that the portion p2 of
this stress is also constant, it follows that the fraction (p2 h- S2E)
= a constant; which may be denoted by /?, (an abstract number).
Let us also write, for convenience, I = Fk^, (k being the radius
of gyration of the cross-section about a (gravity) axis through
^ 1 to paper). Hence finally, we have, from eq. (3),
Breaking load 1 FC
forflatends J ^^r+^5(m)2 • • • (4)
By the same reasoning as in § 303,' for a round-ended
column we substitute 21 for I; for a column with one end round
and the other '^fiat " or ''fixed " (i.e., for a " pin-and-square "
column), ^l for I; and obtain
Breaking load for a round- 1 FC ^
ended column \^^^TTW(hW' • • • (^)
Breaking load for a ''pin- 1 FC
and-square " column J ^^^l + 1.78/9(Z-^/c)^' ' ' • ^^^
Each of these equations (4), (5), and (6), is known as Ran-
kine's Formula, for the respective end-conditions mentioned.
They find a very extended use among engineers in English-
speaking countries; with some variation, however, in the
370
MECHANICS OF ENGINEERING.
numerical values used for quantities C and /?, which are con-
stants for a given material; and also in the fraction of the
breaking load which should be taken as the safe, or working,
load (the reciprocal of this fraction being called the "factor
of safety,") =n. A set of fair average values for these con-
stants, as recommended by Rankine and others, is here pre-
sented :
Hard
steel.
Medium
Steel.
Soft
Steel.
Wrought
Iron.
Cast
Iron.
Timber,
C (lbs./in.2)
70,000
50,000
45,000
36,000
70,000
7,200
/? (abstract number) ....
1
1
1
1
36,000
1
1
25,000
36,000
36,000
6,400
3,000
The factor of safety, n, usually employed with the fore-
going formulae and constants, is n = 4 for wrought iron and
steel in quiescent structures; and 5 under moving loads, as
in bridges; while n = 10 should be used for timber and 8 for
cast iron.
In Fig. 315?) are two dotted curves, plotted for round ends
(Rr) and fixed ends (Rj) in the case of medium steel; the above
equations (Rankine), with the above values of C and /?, having
been used. The ''slenderness ratio," l^k, i& the abscissa;
and Po^F, or Pi^F, (the average breaking unit-stress), is
the ordinate, of any point. These curves may now be com-
pared with the Eule'r curves, E^. and £/, (in the same figure),
.already mentioned as having been plotted for structural steel
(of modulus of elasticity £? = 30,000,000 lbs./in.2)
306. Examples; under the Rankine Formulae. — Example 1.
Let it be required to compute the breaking load of awrought-
iron solid cylinder, used as a column, of length 1 = 8 ft. and
diameter, =d, =2.4 inches; with round ends, i.e., the pressure
acting at each end at the center of the circular base, the ends
being free to turn in any direction.
The "end conditions " call for the employment of the
''least k," but here k is the same for any gravity axis of the
circular section. That is we have
A;2 = / ^^ = 17,^4 ^ ;,^2_i ^2 _ 1(1, 2)2 _ 0.36 in.2; .-. /(; = 0.6 in.
FLEXURE. LONG COLUMNS. 371
and (Z-^A;) = "slenderness-ratio " = 96-j-0.6 = 160. Hence from
eq. (5)
■KT^C 1
Po = j^pX^Tieop' ^^^^ '^^ 36 000 ^^^ <^ = 36,000 lbs./in.2; Le.,
;r( 1.2)236,000 162,800 ,<, .nn ik
It is seen that, on account of the degree of slenderness of
the column, the breaking load is about one quarter of what
it would be for a short prism of same section.
With a factor of safety of 5 we should take 5 of 42,300,
i.e., 8460 lbs., as safe load.
Example 2. — It is required to compute the diameter, d, .
of a solid cast-iron cylinder, 16 ft. in length, to serve as a
column with fiat ends, whose safe load is to be 6 tons, the
factor of safety being 6. This calls for the use of eq. (4) in
which we put Pi = 6x12,000 = 72,000 lbs., the required break-
ing load ; with C = 70,000 lbs. / in.2 and /3 = 1 -^ 6400. The least
radius of gryation should be used, but in this case the k"^ is
constant for all axes of the section, viz., k'^ ^\7tii^ ^nr^ ^d^ ^IQ.
Hence from eq. (4) we have (for inch and pound)
„ \Kd^C 54,980^2 ^onnmu
^' = l + 7ra-/cl2 =- r^ = 72,000 lbs.
This on reduction leads to the bi-quadratic equation
#-1.309^2 = 120.7;
which being solved for d^ gives d2 = o.645± 11.01. The upper
sign being taken we have, finally, c? = 3.41 in. as the required
diameter.
The "slenderness ratio," therefore, proves to be 192-^0.85
= 225, which though seemingly high is not extreme for a flat-
ended column; corresponding, as it does, to 112 for a round-'^
ended column.
Example 3. — A prism of medium steel, of uniform rec-
tangular section (solid) with dimensions 6 = 3 in. and /i = l in.,
is to be subjected to a thrust (connecting-rod of a steam-
engine). Its ends are provided with pins (see Fig. 312) capable^
of turning in firm bearings, the axis of each pin being T to
372 MECHANICS OF ENGINEERING. '
the "b" dimension of the rectangular section. The length
between axes of pins, is Z = 6 ft. It is required to find the
breaking load by the Rankine formula).
Since the end conditions would be ''round-ends" if the
axis of the column were to bend in a plane T to the axes of
the pins (as in Fig. 312), but ''flat-ends" [Fig. 311(6)] in
case it bent in the plane containing the axes of the pins;
and since the k of the section is different for the two cases,
it will be necessary to make each supposition in turn and
take the smaller of the two results for breaking load (i.e.,
as the one to which the factor of safety should be applied).
For round-ended buckling the value of k^ is I^F =
[hh^^l2]^hh = 0.75 in.^; and, with the values of C and ^
for medium steel, we have from eq. (5),
p 50,000X3.0 150,000 ^,^^^,.
^36,000' 0.75
while for flat-ended buckling, in the other plane, the P
to be used would be A;^ = [6/^3^ 12] -6/i = 0.0833 in.2, and
hence from eq. (4)
p 50,000X3.0 150,000 -, ^.q lu
^^36,000' 0.0833
It is seen that Pi is smaller than Pq, so that with a factor
of safety of 6 we have for the safe, or working, load, I of
54,933, =9,155 lbs.
307. Radii of Gyration. — The following table, taken from
p. 523 of Eankine's Civil Engineering, gives values of ^'^,
the square of the least radius of gyration of the given cross-
eection about a gravity-axis. By giving the least value oi
h^ it is implied that the plane of flexure is not determined
by the end-conditions of the column (i. e., it is implied
that the column has either flat ends or round ends). If
either end (or both) is a pin-Joint the column may need to
be treated as having a flat-end as regards flexure in a plane
containing the axis of the column and the axis of the pin,
if the bearings of the pin are firm ; while as regards flexure
in a plane perpendicular to the pin it is to be considered
round-ended at that extremity.
FLEXURE. LONG COLUMNS.
373
In the case of a " thin cell " the value of h"^ is strictly
true for metal infinitely thin and of uniform thieJcness ; still,
if that thickness does not exceed ^ of the exterior diame-
ter, the form given is sufficiently near for practical pur-
poses; similar statements apply to the branching forms.
f f^mm
h i
h
<—-h >
(a) (5)
wmmm,
Fig. 317.
*— -..
\W
(6) (cj
Fia. 818.
Solid Eectangle.
%— least side.
Thin Squfire Cell.
Side— In.
Thin Kectangular Cell. Yia 317 fc> -" /i^ /i+36
h^=- least side.
Solid Circular Section.
Diameter —d.
Thin Circular Cell.
Exterior diam. = d.
Fig. 317(a). ]z''=l^}i^
Fig. 317(6). A;2 = |/i2
p =
12'/i+6
Fig. 317 (c?). p^i^2
Id
Fig. 317(e). A;2 = Jd2
Angle-Iron of Equal j^ig. 317^^) A;2=^-62
ribs
F:
62^*
A^ngle-Iio-n of unequal , -^-g g^g^^^^ ^= 12(F+3?)
Cross of equal arms. Fig. 318 (6). ^=4^'
I-Beam as a pillar.
Let area of web =5, j.- 313 (c). F= -. . -^-j-^
« « &o^A flanges & ^ ^ 12 A-^-H
=A.
Channel ^ig. 318(^). ^=^^^ [l2T^)+iT^^]
Let area of web =B; of flanges =A (both). ^ extends
from edge of flange to middle of web.
374
MECHANICS OF ENGINEERING.
308. Built Columns. — The "compression members" of
bridge trusses, and columns in steel framework buildings are
generally composed of several pieces of structural steel riveted
together, each column being thus formed of a combination of
plates, channels, angles, Z-bars, etc. In Figs. 319 and 320
PHCENIX COLUMN.
Fig. 319. ^^**-
are shown examples of these compound shapes. The Phoenix
column is seen to consist of four quadrantal segments riveted
together. In Fig. 319 is a combination of two channels and
one plate, these three pieces being continuous along the whole
length of the column. On the side opposite to the plate are
seen lattice bars, arranged in zig-zag, which serve to stiffen
the column on that side. The center of gravity of the cross-
section of this column is nearer to the edge carrying the plate
than to the lattice edge; and if the ends of the column are
provided with pins 1 to the webs of the channels the axis of
each of these pins should be so placed as to contain the center
of gravity of the cross-section of the column at that point.
The handbooks of the various steel companies present
formulae and tables enabling the breaking loads to be found
for their various designs of built columns, and for single I-beams
used as columns. For example, the tables given in the hand-
book of the Cambria Steel Co. for built columns of "medium
steel " are stated to be computed from the following formulae
(which are evidently of the Rankine type).
The breaking load for a column of length I and with cross-
section of area F and least radius of gyration k is (in pounds) :
Square Bearing, Pin and Square Bearing. Pin Bearings.
50,000i^ „ 50,000i^ _ 50,000i^
Pi =
1 +
36,000
W
P2 =
1 +
24,000
ar
Po =
1+
18,000VA:
FLEXURE. LONG COLUMNS. 375
In these formulae I and k should be in the same unit (both
feet, or both inches; since (l^k) is a ratio) and the proper
k to be used for the case of "pin and square bearing " (i.e.,
one end provided with a pin and the other with a square
bearing) should be ascertained as in example 3, p. 371.
To obtain the total safe load for the column: "For quiescent
loads, as in buildings, divide by 4. For moving loads, as in
bridges, divide by 5."
Considerable variety will be found among the formulae
of the Rankine type proposed by different engineers as best
satisfying the results of experiment. For accounts of ex-
periments beyond those already quoted in the author's
"Notes and Examples in Mechanics," the reader is referred
to special works. Kent's Pocket Book for Mechanical
Engineers contains much valuable matter on the subject
of columns. The handbooks of the Carnegie Steel Co., the
Pencoyd Iron Works, and the Phoenix Iron Co., give ex-
tensive data relating to steel columns. Osborne's Tables of
moments of inertia and radii of gyration of compound sec-
tions is a valuable book in this connection.
309. Moment of Inertia of Built ColumiL Example.— It is pro-
posed to form a column by joining two I-beams by lattice-
work, Fig. 321, (a). (While the lattice-work is relied upon
to cause the beams to act together as one piece, it is not
regarded in estimating the area F, or the moment of iner-
tia, of the cross section). It is also required to find the
proper distance apart = x, Fig. 321, at which these beams
must be placed, from centre to centre of webs, that the
liability to flexure shall be equal in all axial planes, i.e.
that the 1 of the compound section shall be the same
about all gravity axes. This condition will be ful-
filled if Iy can be made ~i^* (§89), being the centre
of gravity of the compound section, and X perpendicular
to the parallel webs of the two equal I-beams.
Let F' = the sectional area of one of the I-beams, Fx
Tsee Fig. 321(a) its moment of inertia about its web-axis,
that about an axis ~[ to web. (These quantities can 1)8
* That is, with flat ends or ball ends ; but with pin ends, Fig. 313, if the
pin is II to X. put 4/y = Ixi if II to Y, put 47x = Ir .
376
MECHANICS OF ENGINEERING.
found in tlie hand-book of the iron company, for each size
of rolled beam).
Then the
total 7x = 2rx ; and total I^ = 2ri'v -f W-Yl
(see §88 eq. 4.) If these are to be equal, we write them so
and solve for a?, obtaining
X
V jr,
(1)
310. Numerically; suppose each girder to be a 10}4 inch
light I-beam, 105 lbs. per yard, of the N. J. Steel and Iron
Co., in whose hand-book we find that for this beam I'x =
185.6 biquad. inches, and I'r = 9.43 biquad. inches, while
F' = 10.44 sq. inches. "With these values in eq. (1) we
have
„^J± (185.6-9.43) Vera = 8.21 inchea.
V 1 0.4-4.
n^
V
7^
■^
r'l^
-a;— H
^
iai
^
^
^^
-P-
^11:=.
(6)
Fig. 331.
The square of the radius of gyration will be
F=2Px-^2i^'= 371.2 -^20.88=17.7 sq. in. . (2)
and is the same for any gravity axis (see § 89).
As an additional example, suppose the two I-beams united
by plates instead of lattice. Let the thickness of the plate
■= t, Fig. 321, (&). Neglect the rivet-holes. The distance
a is known from the hand-book. The student may derive
a formula for x, imposing the condition that (total /x)= /y-
FLEXURE. LONG COLUMNS. 377
310a. Design of Columns. — General considerations governing
economy and efficiency in the design of built columns are
that the various pieces, besides being continuous for the whole
length, should be placed as far from the axis of the column
as possible, in order to increase the value of k the (least) radius
of gyration, thus leading to a larger value of the safe load for
a given amount of material, or to a minimum amount of material
for a given required safe load; and that the parts should be well
fastened together by rivets, preventing all relative motion. The
economy secured by placing the material as far from the center
as possible also holds, of course, for single pieces used as columns.
For example, if the safe load of a hollow cylindrical cast-iron
flat-ended column, 20 ft. long, is to be 40 tons, i.e., 80,000 lbs.,
and the thickness of metal is not to be less than \ in., we find,
after a few trials with Rankine's formula eq. (4), p. 369, taking
a factor of safety of 8 (so that the breaking load would be
640,000 lbs.) that an outside diameter of d = 8 in. is the largest
permissible. Thus, taking the least k"^, {^(F^8), from p. 373,
for a thin cylindrical cell, with Z = 240 in., with the sectional
area, F, as the quantity to be solved for, we have
\^'^^^.94QN2 =640,000 lbs.; .-. i^ = 19.43 sq. in.
1 + :
6400 [82-8]
Let ^2 denote the internal diameter of the section; then
j(82 — d22) = 19.43; whence ^2 = 6.26 in.; i.e., the thickness
of metal==^(d — ^2) =0.87 in., or practically | in.
310b. The Merriman-Ritter-Formula for Columns was de-
rived independently by Professors Merriman and Ritter (see
Engineering News, July 19, 1894) and has a mathematical
basis as follows. In Fig. 315& curves have been plotted for
the Euler and Rankine formulae for medium steel, both for
flat and round ends; and it is seen that each of the Rankine
curves is tangent to the horizontal line through V and is roughly
parallel to, and not very distant from, the corresponding
Euler curve on the extreme right. Professor Merriman de-
rives the equation (of the same form as Rankine's) for a curve
which has a horizontal tangent at V, and is exactly tangent
378 MECHANICS OF ENGINEERING.
to the Euler curve at some point on the extreme right (at
infinity, in fact) and thus secures a more rational value for
the constant called ^ in Rankine's formula.
With P' denoting the safe load for the column and C the
safe compressive unit-stress for the material, this
formula may be written . . . P' = — Tvrpp^i • • '^ • (M)
where C" denotes the unit compressive stress at elastic limits
E the modulus of elasticity, F the sectional area, and n an
abstract number whose value (as before, in the Rankine for-
mulae) is 1, 16/9 (or 1.78), and 4, for flat ends, pin-and-square,
and round ends, respectively.
If for Q we write P, the breaking load, and correspondingly
C for C, and plot values oi P-^F and l-^k, the curve would
not differ greatly from the Rankine curve in Fig. 120 for medium
steel; and similarly for wrought iron; but for timber and cast
iron the variation is considerable, and hence Prof. Merriman
does not recommend the use of his formula for the latter two
materials. (Crehore's formula differs from the above only
in replacing C" by C.)
310c. The "Straight-Line Formula."— It will be noticed
that in Fig. 315& the straight line connecting points A and C
(medium steel, round ends) or A' and C (medium steel, flat
ends) would not vary widely from the Rankine curve, so that
on account of its simpKcity, when restricted to proper Hmiting
values of the ratio l-^k,& straight line, or hnear relation, between
the quantity P-^F and ratio l-i-k was proposed by Mr. T. H.
Johnson (see Transac. Am. Soc. C. E., 1886, p. 530) for the
breaking loads of columns of various materials. Among them
are the following :
Wrought iron: Hinged ends, Po = [42,000- 157/'^]li^;
" Flat ends. Pi = [42,000- 128(|-)lP;
Mild steel : Hinged ends, Po = [52,000 - 220(77)]^;
" " Flat ends, Pi = [52,000 -179(^)lp.
FLEXURE. LONG COLUMNS. 379
In these formulae Pq, or Pi, is breaking load in lbs., F=
sectional area (in sq. in.), Z = tlie length, and k is the least
radius of gyration of the cross-section for flat ends (as for
hinged ends, see example 3, § 306) ; I and k in same unit.
310d. The J. B. Johnson Parabolic Formula for Columns.
— If in. Fig. 315a a parabola be plotted with its axis vertical
(and downward) and vertex at the point V of the two Rankine
curves, and also made tangent to the Euler curve for the end
conditions concerned, the points on such a curve for values of
l^k between zero and the point of tangency to the Euler curve
are found to agree fairly well with experiment; and the corre-
sponding formula, or the equation to the curve, is of much
simpler form than that of the Rankine types, being almost
as simple as the straight line formula. Such a formula was
proposed by the late Prof. J. B. Johnson, those for mild steel
and wrought iron being given below (breaking load in lbs.).
Mild steel:
Pin ends, Pq = [42,000 -0.97(^HF; U not >150
Flat ends. Pi = [42,000- 0.62(^ MP; 1^ not >190
Wrought iron:
Pin ends, Po = [ 34,000 -O.erQHp; (^ not >170
Flat ends. Pi = [34,000 -0.43(^ MP; U not >210
The notation is the same as in the preceding article. The
limiting values mentioned for l^k refer to the points of tan-
gency with the Euler curve. In Fig. 3156 the curve FTT^A^
is a parabola fulfilling the above mathematical condition for
medium steel, with flat ends.
311. Solid Wooden Columns and Posts. Formula of U. S.
Dept. of. Agriculture, Division of Forestry. — This formula was
derived by Johnson from the results of experiments sonducted
by the Division of Forestry and appUes to solid wooden columns
provided with '' square ends," the constraint due to which,
however, is not to be considered as fully equivalent to that
of "fixed ends." The breaking load being denoted by Pi,
380 MECHANICS OF ENGINEERING.
the sectional area by F, the ratio of length I to the ^^ least
dimension,'^ d,' oi the cross section, by m (i.e., l^d = m), and
the unit crushing stress for the material by C, the formula is
J700 + 15m)FC
^ 700 + 15m + m2 ^^
The values of C to be used for different kinds of timber
are given as follows :
White oak and Georgia yellow pine 5000 Ibs./in.^
Douglas fir and short-leaf yellow pine 4500 ''
Red pine, spruce, hemlock, cypress, chestnut, CaU-
fornia redwood, and Cahfornia spruce 4000 ' '
White pine and cedar 3500 ' '
The fraction of Pi to be taken as the safe load depends
on the wood and the degree of moisture present, four classes
being designated in this respect; from Class A (18 per cent
of moisture; timber exposed to weather), to Class D (10 per cent;,
timber at all times protected from the weather). For yellow
pine the safe load should be from 0.20Pi for Class A to O.SlPi
for Class D. For all other timbers, from 0.20Pi for- Class A
to 0.25Pi for Class D.
312. Column under Eccentric Loading. — In Fig. 322 let the load P be
applied at i, at a distance or "eccentricity" =c from the center of gravity
1^1 A oi the upper base of the column, the reaction at
j^\ j the. other end (at k) having an equal eccentricity
yj i from B; the ends of the column being free to turn.
^^ / j (In an extreme case Ai and Bk might be brackets
/ [ "^ fastened to the ends of the column.)
/ . I AOB is the elastic curve, or bent condition of the
1^1 \ j^ axis of the column, originally straight. With as
)"r" I I origin, any point n in the elastic curve has a vertical
I I I co-ordinate x and a horizontal co-ordinate y. The
\ I I unknown lateral deflection of the point from AB
\ I T is a. With n cs any point in the elastic curve, and
\i I nAi as free body, we have for the moment of the
■^ Bj I stress couple in section at n £'/[d2?/^dx^] = P(c -I- a—?/);.
!^_c_J which is seen to differ from eq. (1) of p. 362 only in
I having the constant c + a in place of the constant a.
Fig. 322. y^^ ^^j therefore use eq. (6) of p. 363 for the present
case, after replacing a by c + a; and hence, denoting 's/P^EI by h, remem-
bering that vers. sin. = 1 — cos, we may write, as the equation to the elastic
curve, y=(c + a)[l-cos (6x)] (1)
For x = ^l, y should = the deflection o; on substituting which values in.
(1) there results finally
FLEXURE. LONG COLUMNS.
381
o=c sec (-^1—1 . . (2); and c+o=c- sec (-^j. •
Hence the moment of the stress couple at is M^ = P{c+ a) = Pc- sec I -^\
(3)
bl\
and the unit stress in outer fibre on concave side at is
p-
P_ M,e
P Pc secQftZ)
"y^ I ■
(4)
(In this case of eccentric loading, then, the deflection a is not indeter-
minate as was the case in deriving Euler's formula on p. 363. Note that
^bl is an angle in radians.)
Example. — Let the value of P be 10,000 lbs., the length of the colunin
be Z=20 ft. = 240 in., and the cross-section be a square cell [see Fig. 317 (6)]
4 inches being the side of the outer square ; area F = 7 m? and / = 14.58 in.^ Let
the eccentricity be c = 2 in., each force P being applied in the middle of a side
of the 4 in. square. Let £' = 30,000,000 Ibs./in.^; material, medium steel.
"With this position of the force plane, e = 2 in.
Here we have hhl
=*(/^
10,000
) X 240 = 0.5736 radians, corre-
V
30, 000, 000 X 14. 58^
spending to 32° 52', whose sec. = 1.190; and therefore a = 2X (1.190-1)
= 0.380 in., and Mo=10,OOOX2X 1.190 = 23,800 in.-lbs. Finally
■p-.
10,000 23,800X2
■ + -
= 1430 + 3265 = 4695 lbs./ in. ^
7 14.58
With P= 20,000 lbs., we should obtain iW= 0.811 radians (46° 30'),
a=0.906 in., Mo = 57,120 in.-lbs., and p = 2860 + 7835 =10,695 Ibs./in.^
This latter unit stress is seen to be only moderate in value for the mate-
rial, leading to the conclusion that 20,000 lbs. for P is a safe load; but on
account of the possible original lack of straightness in the column, and of
lack of homogeneity, both of which causes might increase a and Mq, it
would be better to limit the load to 15,000 lbs. ; considering, also, the fact
that Rankine's Formula for round ends (with a safety factor of 4) applied
to this column for the case of no eccentricity would give about 22,000 lbs.
as safe load.
Fig. 322a.
318. Beam or Column with Eccentric End Pressures and also under Uniform
Transverse Loading. — For example, in Fig. 322a let AB he the -bent axis
of a beam, or column (originally straight), the longitudinal forces P and
P being applied at an eccentricity c from A and B, while there is at the
same time a vertical loading W, =wl, uniformly distributed along the
382 MECHANICS OF ENGINEERING.
whole length at rate of w lbs. per running inch. The reactions of the two
end supports will therefore be each ^W. The ends of the column are free
to turn. It is required to find the deflection a, the moment Mq of the couple
at middle section 0, and the unit stress p on the concave side at O. Take
the free body iAn, n being any point of the elastic curve AOB, with co-
ordinates X and y referred to the horizontal and vertical axes through
as an origin, as shown. Then the moment of stress couple at n is
EI{d'y^dx') = P{c + a-y) + {i)w{P-4x^) .... (5)
Since (d^y-r-dx^) is a variable, let us denote i—EI-i-P)(d^y-T-dx^) by u,
as an auxiliary variable; and eq. (5) wUl now read
y-u=c+a+[{i)w{P-4:X^)]^P (6)
Differentiating (6) twice, with respect to x, we have
d^y d^u w . d^u P w
= ; that is, — = u-\ — (7)
dx' dx' P' ' dx' EI P ^^
Multiplying (7) by 2du, and denoting P-i-EI by b^ and 2w-^P by h, we
have by integration, {dx'^ is a constant, x being the independent variable),
(dM)^-7-(dx)^= — 6^M^ + /iM+C, where C is a constant of integration; and
hence dx = dM-f- (VC + Zim— 6V), which integrates into
a;=A.sm-M , — +C^ (8)
where C is a constant. Transfo rmation of (8) gives
(Vh' + ACb') sin [b(x-C')]+h = 2b^u (9)
Eliminating u by aid of eqs. (6) and (9) we have
2b'y==\/h' + 4Cb'-sin [b(x-C')] + h + 2b\c+a) + (i)b^h{P-4x^) (10)
from which
2b\dyldx) = bVh^ + 4C6^ • cos [b {x- C')]-b^hx . . . (11)
To determine the three constants C, C", and a, we now make use of the
facts that in (10) when x = 0, y also =0, and for x = ^l, y = a; and that in
(11) for x=0, dy/dx, must =0. The three equations thus obtained, con-
taining constants only, enable us to determine C, C, and a, and insert their
values in (10) ; thus giving us as the equation to the elastic curve AOB,
2/ =
(, h \ri-cos (6a: "1 ,, , ,^„^
^+26-^jL^os(i60-J"*^'^' ^^'^
as also the value of the deflection
a^[c+{h^2b^)lsec{^bl)-l]-{i^)hV .... (13)
To find the moment of stress couple, Mq, at 0, we have now only to sub-
stitute a; = and y = in eq. (5), and for a its value from (13); and thus
obtain
[fc+^Ysec(i60-^j (14)
With F as the sectional area of the cross-section of the (prismatic) column
(or beam, as it might also be called in this connection), and e as the dis-
tance of the outer fibre from the gravity axis of the section, we now have
for p, the stress in outer fibre on concave side at 0,
V-l^^f (15)
M„=P
FLEXURE. LONG COLUMNS.
383
Since 6 and h denote VP-i-EI and 2w-^P, respectively, it is seen
that when w is zero, h is zero and eq. (13) reduces to eq. (2) of the
previous article. Again, if the two forces P are central, i.e., [applied at
A and B, we put c = 0; in which case an approximate result may be
reached by writing for the deflection a the value it would have if the
5 WP
end forces P were not present, i.e., ^^ • -^j, as due to the uniform load
W alone (see p. 260). On this basis the value of M^ is Pa+ (,^)Wl.
(In case the vertical load on the beam or column in Fig. 322a is a
single load Q concentrated in the middle at 0, a treatment similar to
the foregoing may be applied, but is somewhat more complicated. For
details of such a case the reader is referred to Mecanique AppUquee,
by Bresse, Tome I. p. 384.)
314. Buckling of Web-Plates in Built Girders. — In §257 men-
tion was made of the fact that very high web plates in
built beams, such as /beams and box-girders, might need
to be stiffened by riveting " angles " on the sides of the web.
(The girders here spoken of are horizontal ones, such as
might be used for carrying a railroad over a short sj^an of
20 to 50 feet.
An approximate method of determining whether such
stiffening is needed to prevent lateral buckling of the web,
may be based upon Rankine's formula for a long column
and will now be given.
In Fig. 323 we have, free, a portion of a bent I-beam,
between two vertical sections at a distance apart= Ai =
the height of the web. In such a beam under forces L ^o
its axis it has been proved (§256) that we may consider
the web to sustain all the shear, J, at any section, and the
flanges to take all the tension and compression, which
form the " stress -couple" of the section. These couples
and the two shears are shown in Fig. 323, for the two
exposed sections. There is supposed to be no load on this
portion of the beam, hence the shears at the two ends are
Fig. 324.
384 MECHANICS OF ENGINEERING,
equal. Now tlie shear acting between eacli flange and tlie
horizontal edge of the web is equal in intensity per square
inch to that in the vertical edge of the web ; hence if the
web alons, of Fig. 323, is shown as a free body in Fig. 324,,
we must insert two horizontal forces = J, in opposite
directioii^,, on its upper and lower edges. Each of theM
^ J since we have taken a. horizontal length hi = height
of web. In this figure, 324, we notice that the effect oi
the acting forces is to lengthen the diagonal BD ancj
shorten the diagonal AG, both of those diagonals making
an angle of 45° with the horizontal.
Let us now consider this buckling tendency along ^(7,
by treating as free the strip ^(7, of small width = \. This
is shown in Fig. 325. The only forces acting in the direc-
tion of its length AG&ie the components along AG oi the
four forces J' at the extremities. "VVe may therefore treat
the strip as a long column of a length I = hi ^2, of a sec-
tional area F = bb^, (where b is the thickness of the web
plate), with a value of F = Yjg 6^ (see § 309), and with
fixed (or flat) ends. Now the sum of the longitudinal
components of the two J'.'s &t A is. Q = 2 J' ]/?, V2
= J' V2 ; but J' itself =■ rr. b j4 bi \'% since the small
rectangle on which J' acts has an area = b )4 h^ ^2, and
ihe shearing stress on it has an intensity of (J -r- bh{) per
unit of area. Hence the longitudinal force at each end of
this long column iis
'i-r/ w
According to eq. (4) and the table in § 305, the safe load
(factor of safety = 4) for a medium steel column of this form,
with flat ends, would be (pound and inch)
ib6i50,000 _ 12,500&&i
1 1 2/ii2 1 h^ • • (2)
^36,000* V1262 1 + 1,500' 62
If, then, in any particular locality of the girder (of medium
steel) we find that Q is >Pi, i.e.
FLEXUEE. LONG COLUMNS.
385
12,500&
l+rl,.-'^
(pound and inch).
(3)
W =40 TONS
1,500 62
then vertical stiffeners will be required laterally.
When these are required, they are generally placed at inter-
vals equal to hi, (the depth of web), along that part of the
girder where Q is >Pi.
Example Fig. 326.— Will stiffening pieces be required in
a plate 'girder of 20 feet span, bearing a uniform load of
40 tons, and having a web 24 in. deep
and I in. thick?
From § 242 we know that the -|t
greatest shear, J max., is close to
either pier, and hence we investigate
that part of the girder first.
J max. = iTF = 20 tons -40,000 lbs.
.'. (inch and lb.), see (3),
J _ 40,000
hi
-10^ — -^
Trn
Fig. 326.
24
= 1666.6
while, see (3), (inch and pound),
12,500X1
1 + .
242
1270
(4)
(5)
1,500 (1)2
which is less than 1666.66.
Hence stiffening pieces will be needed near the extremities
of the girder. Also, since the shear for this case of loading
diminishes uniformly toward zero at the middle they will
be needed from each end up to a distance of ^ of 10 ft.
from the middle.
^86 MECHANICS OF ENGLNEEBIlfG.
CHAPTER Vn.
UJTEAK ARCHES (OF BLOCKWOKI^.
815. A Blockwork Arch is a structure, spanning an openicg
or gap, depending, for stability, upon the resistance to
compresssion of its blocks, or voussoirs, the material ot
which, such as stone or brick, is not suitable for sustain-
ing a tensile strain. Above the voussoirs is usually
placed a load of some character, (e.q. a roadway,) whose
pressure upon the voussoirs will be considered as vertical,
only. This condition is not fully realized in practice,
unless the load is of cut stone, with vertical and horizontal
joints resting upon voussoirs of corresponding shape (see
Fig. 327), but sufficiently so to warrant
its assumption in theory. Symmetry
of form about a vertical axis will also
be assumed in the following treatment.
316. Linear Arches. — For purposes of
theoretical discussion the voussoirs of
Fig. 327 may be considered to become
Fig. 327. infinitely small and infinite in number,
thus forming a " linear arch," while retaining the same
shapes, their depth "1 to the face being assumed constant
that it may not appear in the formulae. The joints
between them are "1 to the curve of the arch, i.e., adjacent
voussoirs can exert pressure on each other only in the
direction of the tangent-line to that curve.
LINEAR AKCHES.
387
317. Inverted Catenary, or Linear Arch Sustaining its Own
Weight Alone. — Suppose tlie infinitely smalJ voussoirs to
have weight, uniformly distributed along the curve, weigh-
ing q lbs. per running linear unit. The eqiii]ibrium of
such a structure, Fig. 328, is of course unstable but theo-
retically possible. Required the form of the curve when
equilibrium exists. The conditions of equilibrium are,
obviously : 1st. The thrust or mutual pressure T between
any two adjacent voussoirs at any point. A, of the curve
must be tangent to the curve ; and 2ndly, considering a
portion BA as a free body, the resultant of Hq the pres-
FiQ. 328.
Fig. 329.
Fig. 330.
sure at B the crown, and T &i A, must balance R the re-
sultant of the il vertical forces (i.e.,weights of the elementary
voussoirs) acting between B and A.
But the conditions of equilibrium of a flexible, inexten-
sible and uniformly loaded cord or chain are the very
same (weights uniform along the curve) the forces being
reversed in, direction. Fig. 329. Instead of compression
we have tension, while the || vertical forces act toward in-
stead of away from, the axis X. Hence the curve of equi-
librium of Fig. 328 is an inverted catenary (see § 48) whose
equation is
y+c=
e -\- e
. (1)
See Fig. 330. e = 2.71828 the Naperian Base. The "par-
ameter " c may be determined by putting x = a, the half
span, and y= Y, the rise, then solving for c by successive
388
MEOHAJSriCS OF ENGINEBKING.
approximations. The " horizontal thrust" or H^^, is = yc,
while if s = length, of arch OA, along the curve, the thrust
T at any point A is
T=^I Riffs' (2..)
From the foregoing it may be inferred that a series ot vcui»'
soirs of finite dimensions, arranged
so as to contain the catenary curve,
with joints "I to that curve and of
equal weights for equal lengths of
arc will be in equilibrium, and
moreover in stable equilibrium on
account of friction, and the finite
width of the joints ; see Fig. 331.
FIG. 331.
318. Linear Arches under Given Loading. — The linear arches
to be considered further will be treated as without weight
themselves but as bearing vertically pressing loads (each
voussoir its own).
Problem. — Given the form of the linear arch itself, it is
required to find the law of vertical depth of loading under
which the given linear arch will be in equilibrium. Fig.
332, given the curve ABC, i.e., the linear arch itself, re-
quired the form of the curve MON, or upper limit of load-
ing, such that the linear arch ABC shall be in equilibrium
under the loads lying between the two curves. The load-
ing is supposed homogeneous and of constant depth "^ to
paper ; so that the ordinates z between the two curves are
proportional to the load per horizontal linear unit. Assume
a height of load z^ at the crown, at pleasure ; then required
the z of any point m as a function of ^ and the curve
ABC.
LINEAB ARCHES.
389
Practical Solution. — Since a linear arcli under vertical
pressures is nothing more than the inversion of the curve
assumed by a cord loaded in the same way, this problenj
might be solved mechanically by experimenting with a
light cord, Fig. 333, to which are hung other heavy cords,
or bars of uniform weight per unit length, and at equal
horizontal distances apart ivhen in equilibrium,. By varying
the lengths of the bars, and their points of attachment, we
may finally find the curve sought, MON. (See also § 343.)
Analytical Solution. — Consider the structure in Fig. 334
A number of rods of finite length, in the same plane, are in
equilibrium, bearing the weights P, P^ etc., at the con-
FiG. 334.
Tig. 335.
necting joints, each piece exerting a thrust T against the
adjacent joint. The joint A, (the " pin " of the hinge), im-
agined separated from the contiguous rods and hence free,
is held in equilibrium by the vertical force P (a load) and
the two thrusts T and T', making angles = d and d' with
the vertical ; Fig. 335 shows the joint -4 fi'^e. From 2'(hor«^
izontal comps.)=0, we have.
That is, the horizontal component of the thrust in any ro J
is the same for all ; call it H^. ,\
T^
H.
Bin
(1)
390
MECHANICS OF ENGINEEKING.
Now draw a line As *i to T' and write 2* ( compons. I to
As)=0; whence F sin ^'=2^ sin ^, and [see (1)]
. p_ jgp sin /?
sm 6
sm
(2)
Let the rods of Fig. 334 become infinitely small a,nd infi-
nite in number and the load continuous. The length of
each rod becomes =ds an element of the linear arch, fi is
the angle between two consecutive ds's, d is the angle be-
tween the tangent line and the vertical, while P becomes
the load resting on a single dx, or horizontal distance be-
tween the middles of the two cZs's. That is, Fig. 336, if
Y= weight of a cubic unit of the
loading, P-=yzdx. (The lamina of
arch and load considered is unity,
1 to paper, in thickness.) -Ho=a
constant = thrust at crown ;
6=6', and sin /3=ds-^p, (since the'
angle between two consecutive tan-
gents is = that between two con-
secutive radii of curvature). Hence
eq. (2) becomes
Yzdx =
Kds
p BID? 6
but dx—ds sin 6y
Fia. 336.
,\yz-
H.
p siii^/?
(3)
Call the radius of curvature at the crown. /?o» and since
there z=Zq and ^*=90°, (3) gives x^qPq~3^', hence (3) may
be written
sin^ d
(4)
This is the law of vertical depth of loading required. For
a point of the linear arch where the tangent line is verti-
cal, sin 6 =0 and z would == oo ; i.e., the load would be in-
LINEAR ARCHES.
391
finitely high. Hence, in practice, a full semi-circle, for in-
stance, could not be used as a linear arcli.
319. Circular Arc as Linear Arch. — ^As an example of the
preceding problem let us ap-
ply eq. (4) to a circular arc,
Fig. 337, as a linear arcb.
Since for a circle p is con-
stant — r, eq. (4) reduces
to
sin^ 6
(5)
Fig. 337.
Hence tlie deptb of loading
must vary inversely as the cube of tbe sine of the angle d
made by the tangent line (of the linear arch) with the ver-
tical.
To find the depth z by construction.— Having z^ given, C
being the centre of the arch, prolong Ga and make ob =
go ; at 5 draw a 1 to Gb, intersecting the vertical through a
at some point d ; draw the horizontal dc to meet Ga at
some point c. Again, draw ce "| to Gc, meeting ad m e\
then ae= z required ; a being any point of the linear arch.
For, from the similar right triangles involved, we have
z„=ab=ad sin 0=ac sin d. sin ^=ae sin d sin d sin d
ae= — ^— ; i.e., ae=2. Q.E.D.
mn'd [-gee (5.)]
320. Parabola as Linear Arch. — To apply eq. 4 § 318 to a
parabola (axis vertical) as linear arch, we must find values
of p and po the radii of curvature at any point and the
crown respectively. That is, in the general formula.
-M
dy\
dx) _
dx
we must substitute the forms for the first and second dif-
ferential co-efficients, derived from the equation of the
392
MECHANICS OF EI*f G [2f EERIN^G.
Fig. 338.
Fig. 339.
curve (parabola) in Fig. 338, i.e. from x^ =^ 2 py; whence
we obtain
~2.,or cot 0,= — ana-^=—
ax p dor p
Hence ^=i3^°M=^
^ l^P
cosec.
I.e. p
sin^^
. (6)
At tbe vertex d = 90** ,*. />„ = p. Hence by substituting
for p and p^ in eq. (4) of § 318 we obtain
g=s^= constant [Fig. 339) (7)
for a parabolic linear arcli. Therefore tbe depth of homo-
geneous loading must be the same at all points as at the
crown ; i.e., the load is uniformly distributed with respect
to the horizontal. This result might have been antici-
pated from the fact that a cord assumes the parabolic
form when its load (as approximately true for suspension
bridges) is uniformly distributed horizontally. Sae § 46
in Statics and Dynamics.
321. Linear Arch for a Given Tipper Contour of Loading, the
arch itself being the unknown lower contour. Given the
upper curve or limit of load and the depth z^ at crown, re-
quired the form of linear arch which will be in equili-
brium under the homogenous load between itself and that
upper curve. In Fig. 340 let MON be the given upper
contour of load, z^ is given or assumed,s' and z" are the
respective ordinates of the two curves -S^ (7 and MON,
Required the eqation of BAG.
LINEAR ARCHES.
393
Fig. mo.
Fig. 341.
As before, tlie loading is homogenous, so that the
weights of any portions of it are proportional to the
corresponding areas between the curves. (Unity thick-
ness "I to paper.) Now, Fig. 341, regard two consecutive
ds's oi the linear arch as two links or consecutive blocks
bearing at their junction w the load dP =y (^z -\- z"} dx in
which Y denotes the heaviness of weight of a cubic unit of
the loading. If T and T' are the thrusts exerted on these
two blocks by their neighbors (here supposed removed)
we have the three forces dP, T and T', forming a system
in equilibrium. Hence from IX =0,
T cos <^>
as a relation holding good for any point of the linear arch
which is to be in equilibrium under the load included
between itself and the given curve whose ordinates are «",
Fig. 340.
322. Example of Preceding. Tipper Contour a Straight Line.—
Fig. 342. Let the upper contour be a right line and hor-
izontal ; then the a" of eq. 5 becomes zero at all points of
ON. Hence drop the accent of z' in eq. (5) and we have
dot? €^
Multiplying which by dz we obtain
dz dh 1
do(?
zdz (6)
This being true of the z, dz, d?z and dx of each element of
the curve O'B whose equation is desired, conceive it writ-
ten out for each element between 0' and any point m, and
put the sum of the left-hand members of these equations
= to that of the right-hand members, remembering that
a,^ and dx'^ are the same for each element. This gives
dz=dz z=z
d^ I « / ** 2 al2 2j
nJ (te— %/ z=Zo
adz ^ [ zj .... (7.)
d
.'. da; = -T^==««
LINEAR ARCHES.
395
Fig. 342. Pis 343
Integrating (7.) between 0' and any point m
/,
f =«[:iog..(^+,/(-i)-i) . . (8)
i.e., fl?=a log.
D-^]=
or %=
gg r i -E.-1
(8.)
(9.)
This curve is called the transformed catenary since we may
obtain it from a common catenary by altering all the ordi-
nates of the latter in a constant ratio, just as an ellipse
may be obtained from a circle. If in eq. (9) a were = z^
the curve would be a common catenary.
Supposing Sj and the co-ordinates x^ and gj of the point
B (abutment) given, we may compute a from eq. 8 by put-
ting X =Xi and z = g„ and solving for a. Then the crown-
thrust H = ya^ becomes known, and a can be used in eqs.
(8) or (9) to plot points in the curve or linear arch. From
eq. (9) we have
(10)
area
00' mn
Fig. 343.
Call this area, A. As for the thrusts at the different
joints of the linear arch, see Fig. 343, we have crown-
thrust = ZT = ^a' . . . ; • - . • (11)
and at any joint m the thrust
T^VH'+irAf =rV^^^' .... (12}
396 MECHANICS OF ENGINEERING.
323. Remarks. — The foregoing results may be utilized
with arches of finite dimensions by making the arch-ring
contain the imaginary linear arch, and the joints 1 to the
curve of the same. Questions of friction and the resist-
ance of the material of the voussoirs are reserved for a
succeeding chapter, (§ 344) in which will be advanced ^
more practical theory dealing with approximate linear
arches or " equilibrium polygons " as they will then be
called. Still, a study of exact linear arches is valuable on
many accounts. By inverting the linear arches so far pre-
sented we have the forms assumed by flexible and inexten-
sible cords loaded ini the same way-
GliAPHICAIi STATICS, tS9?
CHAPTER VrX
ELEMENTS OE GRAPHICAIi STATICS.
324. Definition. — In many respects graphical processes
titve duvantages over tlie purely analytical, whicli recom-
mend their use in many problems where celerity is desired
without refiiied accuracy. One of these advantages is that
gross errors are more easily detected, and another that
the relations of the forces, distances, etc., are made so
apparent to the eye, in the drawing, that the general effect
of a given change in the data can readily be predicted at
a glance.
Graphical Statics in the system of geometrical construc-
tions by which prt^blems in Statics may be solved by
the use of drafting iixsiruments, forces as well as distances
being represented in amount and direction by lines on the
paper, of proper length and position, according to arbi-
trary scales ; so many fest of distance to the linear inch of
paper, for example, for distances ; and so many pounds or
tons to the linear inch of paper for forces.
Of course results should be interpreted by the same
scale as that used for the data. The parallelogram of
forces is the basis of all constructions for combining and
resolving forces.
325. Force Polygons and Concurrent Forces in a Plsuae. — If a
material point is in equilibrium under three forces Pi P,
P3 (in the same plane of course) Fig. 344, any one of them,
398
MECHANICS OF ENGIXEERING.
as Pi, must be equal and opposite to B the resultant of
the other two (diagonal of their parallelogram). If now
we lay off to some convenient scale a line in Fig. 345 =
Pi and II to Pi in Fig. 344 ; and then from the pointed end
of Pi a line equal and || to Pg and
laid off pointing the same ivay, we
note that the line remaining to
p close the triangle in Fig. 345 must
be = and || to Pg, since that tri-
angle is nothing more than the
left-hand half -parallelogram of
Fig. 345. Fig. 344. Also, in 345, to close
the triangle properly the directions of the arrows must
be continuous Point to Butt, round the periphery. Fig.
345 is called a force polygor ; of three sides only in this
case. By means of it, given any two of the three forces
which hold the point in equilibrium, the third can be
found, being equal and 1| to the side necessary to " close "
the force polygon.
Similarly, if a number of forces in a plane hold a mate-
rial point in equilibrium, Fig. 346, their force polygon.
FiG.344.
Fig. 347, must close, whatever be the order in which its
sides are drawn. For, if we combine Pj and P2 into a re-
sultant Oa, Fig. 346, then this resultant with P3 to form a
resultant Oh, and so on ; we find the resultant of Pi, P2, Ps?
and P4 to be Oc, and if a fifth force is to produce equilib-
rium it must be equal and opposite to Oc, and would close
the polygon OdabcO, in which the sides are equal and par-
GEAPHICAL STATICS.
399
allel respectively to the forces mentioned. To utilize tliis
fact we can dispense witli all parts of tlie parallelograms in
Pig. 346 except tlie sides mentioned, and tlien proceed as
follows in Fig. 347 :
If P5 is the unknown force which is to balance the other
four (i.e, is their anti-resultant), we draw the sides of the
force polygon from A round to B, making each line paral-
lel and equal to the proper force and pointing the same
way ; then the line BA represents the required F^ in
amount and direction, since the arrow BA must follow
the continuity of the others (point to butt).
If the arrow BA were pointed at the extremity B, then
it gives, obviously, the amount and direction of the result^
ant of the four forces Pj . . . P4. The foregoing shows
that if a system of Concurrent Forces in a Plane is in equi-
librium, ii^ force polygon must close.
326. Non-Concurrent Forces in a Plane. — Given a system of
non-concurrent forces m a plane, acting on a rigid body,
required graphic means of finding their resultant and anti-
resultant ; also of expressing conditions of equilibrium.
The resultant must be found in amount and direction ; and
also in position (i.e., its line of action must be determined).
E. g., Fig. 348 shows a curved rigid beam fixed in a vise
at T, and also under the action of forces Pi P2 P3 and P^
{besides the action of the vise); required the resultant of
By the ordinary
parallelogram of
forces we com-
bine Pi and P2 at
a, the intersection
of their lines of
PjQ 34g action, into a re-
sultant Pa, ; then Pa with Pg at b, to form PbJ and finally P,,
with P4 at c to form B^ which is .*. the resultant required,
ie., of Pi . . . . P4 ; and c . , . P is its line of action.
400
MSCHAXICS OF EXGIXEEKIXG.
Fig. 349.
The separate force triangles (half-parallelograms) by
wliich. the successive partial resultants B^^, etc., were found,
are again drawn in Fig. 349. Now since B^ acting in the
line C..F, Fig. 348,
is the resultant of
Pi . . Fi, it is plain
that a force FJ
equal to B,. and act-
ing along c . . i^.but
in the opposite di-
rection, would balance the system Pi . . . P4, (is their anti-
resultant). That is, the forces Pi P2 P3 P4 and BJ would
form a system in equilibrium. The force B^' then, repre-
sents the action of the vise T upon the beam. Hence re-
place the vise by the force B/ acting in the line . . . F . . .c •
to do which requires us to imagine a rigid prolongation of
that end of the beam, to intersect F . . . c. This is shown in
Fig. 350 where the whole beam is free, in equilibrium, under
the forces shown, and in precisely the same state of stress,
part for part, as in Fig. 348. Also, by combining in one
force diagram, in Fig. 351, all the force triangles of Fig. 349
(by making their common sides coincide, and putting B/
instead of B^., and dotting all forces other than those of
Fig. 350), we have a figure to be interpreted in connection
with Fig. 350.
A "poL^^iQH J'
SPACE DIAGRAM
Fig. 350.
FORCE DIAGRAM
Fig. 351.
Here we note, first, that in the figure called a force-dia-
gram, P1P2P3P4 and R/ form a closed polygon and that
Gr^APHlCAli STATICS. 401
their arrows follow a continuous order, point to butt,
around the jperimeter ; which proves that one condition of
equilibrium of a system of non-concurrent forces ir^ a, plane
is that its force polygon must close. Secondly, note that ah
is II to Oa', and be to Oh' ; hence if the force-diagram has
been drawn (including the rays, dotted) in order to deter-
mine the amount and direction of HJ, or any other one force,
we may then find its line of action in the space-diagram, as
follows: (N. B. — By space diagram is meant the figure show-
ing to a true scale the form of the rigid body and the lines
of action of the forces" concerned). Through a, the intersec-
tion of Fi and F-j, draw a line || to Oa' to cut P3 in some point
b ; then through b a line || to Ob' to cut F^ at some point c; cF
drawn || to Oc' is the required line of action of RJ, the anti-
resultant of Pi, F2, P3, and P4.
abc is called an equilibrium polygon; this one having but
two segments, ab and bo (sometimes the lines of action of F^
and RJ may conveniently be considered as segments.) The
segments of the equilibrium polygon are parallel to the respect-
ive rays of the force diagram.
Hence for the equilibrium of a system of no;ti-conciirrent
forces in a plane not only must its force polygon close,
but also the first and last segments of the corre-
sponding equilibrium polygon must coincide with
the resultants of the first two forces, and of the last
two forces, respectively, of the system. E.g., ab coin-
cides with the line of action of the resultant of F^ and F^, I
he with that of F^ and E'c- Evidently the equil. polygon
"will be different with each different order of forces in
the force polygon or different choice of a pole, 0. But if
the order of forces be taken as above, as they occur along
the beam, or structure, and the pole taken at the " butt " of
the first force in the force polygon, there will be only one j
(and this one will be called the special equilibrium polygon
in the chapter on arch-ribs, and the " true linear arch " in
dealing with the stone arch.) After the rays (dotted in
Fig. 351) have been added, by joining the pole to each
402
MEOn Allies OF ENGrKEEEi:srG.
vertex with wliicli it is not already connected, tJbe finai
figure may be called the/brce diagram.
It may sometimes be convenient to give tlie name of
rays to tlie two forces of tlie force polygon which, meet
at the pole, in which case the first and last segments of
the corresponding equil. polygon will coincide with the
lines of action of those forces in the space-diagram (as we
may call the representation of the body or structure on
which the forces act). This " space diagram " shows the
real field of action of the forces, while the force diagram,
which may be placed in any convenient position on the
paper, shows the magnitudes and directions of the forces
acting in the former diagram, its lines being interpreted
on a scale of so many lbs. or tons to the inch of paper ; in
the space-diagram we deal with a scale of so many/ee^ to
the inch of paper.
We have found, then, that if any vertex or corner of the
closed force polygon be taken as a pole, and rays drawn
from it to all the other corners of the polygon, and a cor-
responding equil. polygon drawn in the space diagram., the
first and last segments of the latter polygon must co-incide
with the first and last forces according to the order
adopted (or with the resultants of the first two and last
two, if more convenient to classify them thus). It remains
to utilize this principle.
327. To Find the Resultant of Several Forces in a Plane. — This
might be done as in § 326, but since frei^^uently a given set
of forces are parallel, or nearly so, a special method will
now be given, of great convenience in such cases. Fig. 352.
Let Pi Pg and
Pa be the given
forces whose
resultant is re-
quirsd. Let us
first find their
and -' resultant,
or force which
Fig. 352. Pm. 353. will balance
GEAPHICAL STATICS. 403
them. This anti-resultant may be conceived as decom-
posed into two components P and P' one of which, say P,
is arbitrary in amount and position. Assuming P, then,
at convenience, in the space diagram, it is required to lind
F'. The live forces must form a balanced system ; hence
if beginning at Oi, Fig. 353, we lay off a line O^A = P by
scale, then Al = and || to P,, and so on (point to butt), the
line POi necessary to close the force polygon is = P' re-
quired. Now form the corresponding equil. polygon in
the space diagram in the usual way, viz.: through a the
intersection of P and P^ draw ab || to the ray 0, . . . 1
(Avhich connects the pole Oi with the point of the last force
mentioned). From h, where ab intersects the line of Pg*
draw he, || to the ray O^ . . 2, till it intersects the line of Pg.
A line mc drawn through c and || to the P' of the force
diagram is the line of action of P'.
Now the resultant of P and P' is the anti-resultant of
Pi, P2 and P3; .'. d, the intersection of the lines of P and
P', is a point in the line of action of the anti-resultant re-
quired, while its direction and magnitude are given by the
line BA in the force diagram ; for BA forms a closed poly-
gon both with Pi P2 P3, and with PP'. Hence a line
through (i || to BA, viz., de, is the line of action of the anti-
resultant (and hence of the resultant) of Pj, P2, P3.
Since, in this construction, P is arbitrary, we may first
choose Oi, arbitrarily, in a convenient position, i.e., in such
a position that by inspection the segments of the result-
ing equil. polygon shall give fair intersections and not
pass off the paper. If the given forces are parallel the
device of introducing the oblique P and P' is quite neces-
sary.
328. — The result of this construction may be stated as
follows, (regarding Oa and cm as segments of the equil.
polygon as well as ah and he): If any tivo segments of an
equU. polygon he prolonged, their intersection is a point in
the line of action of the resultant of those forces acting at
404
MECHANICS OF ENGINEERING.
the vertices intervening between the given segments,
the resultant of Pi P2 P3 acts through d.
Here,
329. Vertical Reaction of Piers, etc. — Fig. 354. Given the
vertical forces or loads Pj P2 and P3 acting on a rigid body
(beam, or truss) which is supported by two piers having
smooth horizontal surfaces (so that the reactions must be
vertical), required the reactions Fq and V^ of the piers.
For an instant suppose V^ and V^ known ; they are in
iVn
Fig. 354.
equil. with Pi Pg and P3. The introduction of the equal
and opposite forces P and P' in the same line will not dis-
turb the equilibrium. Taking the seven forces in the
order P Vq Pj Pg P3 V^ and P', a force polygon formed with
them will close (see (h) in Fig. where the forces which
really lie on the same line are slightly separated). With
Oy the butt of P, as a pole, draw the rays of the force dia-
gram OA, OB, etc. The corresponding equil. polygon
begins at a, the intersection of P and V^ in {a) (the space
diagram), and ends at n the intersection of P' and V^.
Join an. Now since P and P' act in the same line, an
must be that line and must be || to P and P' of the force
diagram. Since the amount and direction of P and P' are
arbitrary, the position of the pole is arbitrary, while
Pi, P2, and P3 are the only forces known in advance in the
force diagram.
Hence Vq and V^ may be determined as follows : Lay off
the given loads Pi, P2, etc., in the order of their occur-
rence in the space diagram, to form a " load-line " AD
GRAPHICAL STATICS.
405
(see (h.) Fig. 854) as a beginning for a force-diagram ; take
any convenient pole 0, draw the rays OA, OB, 00 and
OD. Tlien beginning at any convenient point a in the
vertical line containing the unknown Vq, draw ab || to OA,
be II to OB, and so on, until the last segment [dn in this
case) cuts the vertical containing the unknown V„ in some
point n. Join an (this is sometimes called a closing line)
and draw a || to it through 0, in the force-diagram. This
last line will cut ths " load-line " in some point n', and
divide it in two parts n' A and i>w', which are respectively
Vq and Va required.
Corollary. — Evidently, for a given system of loads, in given
vertical lines of action, and for two given piers, or abut-
ments, having smooth horizontal surfaces, the location of the
point n' on the load line is independent of the choice of a
•pole.
Of course, in treating the stresses and deflection of the
ligid body concerned, P and P' are left out of account, as
being imaginary and serving only a temporary purpose.
330. Application of Foregoing Principles to a Roof Truss-
Fig. 355. Wi and W.^ are wind pressures. Pi and P. are
loads, while the lemaining external forces, viz., the re-
406 MECHANICS OF ENGINEERING.
actions, or supporting forces. To, F'„ and H^i niay be fonnd
by preceding §§. (We here suppose that the right abut-
ment furnishes all the horizontal resistance ; none at the
left).
Lay off the forces (known) Wi, W2, Pi, and P2 in the
usual way, to form a portion of the closed force polygon.
To close the polygon it is evident we need only draw a
horizontal through 5 and limit it by a vertical through 1.
This determines H^ but it remains to determine ?^' the
point of division between F^ and V^. Select a convenient
pole Oi, and draw rays from it to 1, 2, etc. Assume a con-
venient point a in the line of V„ in the space diagram, and
through it draw a line || to Oil to meet the line of W^ in
some point b ; then a line || to Oi2 to meet the line of W2
in some point c ; then through c || to OjS to meet the line
of Pi in some point d ; then through d || to Oi4 to meet the
line of P2 in some point e, (e is identical with d, since Pi
and P2 are in the same line) ; then efW to Oi5 to meet Hj^
in some point/; then fg \\ to OS to meet V^ in some
point g.
abcdefg is an equilibrium polygon corresponding to the
pole Oj.
Now join ag, the " closing-line," and draw a || to it
through Oi to determine n', the required point of division
between Vo and V„ on the vertical 1 6. Hence F^ and V^
are now determined as well as H^^.
[The use of the arbitrary pole Oi implies the temporary
employment of a pair of opposite and equal forces in the line
ag, the amount of either being = Oiti'].
Having now all the external forces acting on the truss,
and assuming that it contains no " redundant parts," i.e.,
parts unnecessary for rigidity of the frame-work, we proceed
tc find the pulls and thrusts in the individual pieces, on
the following plan. The truss being pin-connected, no
piece extending beyond a joint, and all loads being con-
sidered to act at joints, the action, pull or thrust, of each
piece on the joint at either extremity will be in the direction
of the piece, i.e., in a knoivn direction, and the pin of each
GKAPHICAL STATICS. 407
joint is in equilibrium under a system of concurrent forces
consisting of the loads (if any) at tlie joint and the pulls
or thrusts exerted upon it by the pieces meeting there..
Hence we may apply the principles of § 325 to each joint
in turn. See Fig. 356. In constructing and interpreting
the various force polygons, Mr. E.. H Bow'g convenient
notation will be used; this is as follows: In the space
diagram a capital letter [ABC, etc.] is placed in each tri-
angular cell of the truss, and also in each angular space in
the outside outline of the truss between the external forces
and the adjacent truss-pieces. In this way we can speak of
the force Wi as the force BC, of W2 as the force C-E, the
stress in the piece a/3 as the force QI), and so on. That
is, the stress in any one piece can be named from the
letters in the spaces bordering its two sides. Corresponding
to these capital letters in the spaces of the space-dia~
gram, small letters will be used at the vertices of the closed
force-polygons (one polygon for each joint) in such a way
that the stress in the piece CD, for example, shall be thQ
forc3 cd of the force polygon belonging to any joint in
which that piece terminates ; the stress in the piece FO
by the force fg, in the proper force polygon, and so on.
In Fig. 356 the whole truss is shown free, in equili-
brium under the external forces. To find the pulls or
thrusts (i.e., tensions or compressions) in the pieces, con-
sider that if all but two of the forces of a closed force
polygon are known in magnitude and direction,, while the
directions, only, of those two are known, the wliole force
polygon may he drawn, thus determining the amounts of
those two forces by the lengths of the corresponding
sides.
We must .'. begin with a joint where no more than two
pieces nieet, as at a ; [call the joints a, /9, y, d, and the cor-
corresponding force polygons a', /9' etc. Fig. 356.] Hence
at a' (anywhere on the paper) make oh \\ and = (by scale)
to the known force AB (i.e., V^) pointing it at the upper end,
and from this end draw he — and || to the known force BG
(i.e., Wj) pointing this at the lower end.
iU8
MECHANICS OF E:i^GINEEBraG.
Fig. 356.
To close the polygon draw througli c a || to the piece
CD, and through a a || to AD ; their intersection deter-
mines d, and the polygon is closed. Since the arrows
must be point to butt round the periphery, the force with
which the piece CD acts on the pin of the joint a is a
force of an amount = cd and in a direction from c toward
d ; hence the piece CD is in compression ; whereas the
action of the piece DA upon the pin at a is from d toward
a (direction of arrow) and hence DA is in tension. Notice
that in constructing the force polygon «' a right-handed
(or clock-wise) rotation has been observed in considering
in turn the spaces ABC and D, round the joint a. A
similar order will be found convenient in each of the other
joints.
Knowing now the stress in the piece GD, (as well as in
DA) all but two of the forces acting on the pin at the joint
/? are known, and accordingly we begin a force polygon, /3',
for that joint by drawing dc,= and || to the dc of polygon
a', hut pointed in the opposite direction, since the action of
OD on the joint /? is equal and opposite to its action on
the joint a (this disregards the weight of the piece).
Through c draw ce = and || to the force CE (i.e., W^ and
GRAPHICAL STATICS. 409
pointing tlie same way ; tlien ef, = and || to tlie load EF
(i.e. Pj) and pointing downward. Througli f draw a || to
tlie piece FG and through d, a || to the piece OB, and the
polygon is closed, thus determining the stresses in the
pieces FG and GT>. Noting the pointing of the arrows,
we readily see that FG is in compression Avhile GD is in
tension.
Next pass to the joint (5, and construct the polygon o' ,
thus determining the stress gli in GB. and that ad in AD ;
this last force ad should check with its equal and oppo-
site ad already determined in polygon a'. Another check
consists in the proper closing of the polygon y', all of
whose sides are now known.
[A compound stress-diagram may be formed by super-
posing the polygons already found in such a way as to
make equal sides co-incide ; but the character of each
stress is not so readily perceived then as when they are
kept separate].
In a similar manner we may find the stresses in any pin-
connected frame-work (in one plane and having no redun-
dant pieces) under given loads, provided all the support-
ing forces or reactions can be found. In the case of a
braced-arch (truss) as
shown in Fig. 357, hinged
to the abutments at both
ends and not free to slide
laterally upon them, the
reactions at and B de-
FiG. 357. pend, in amount and direc-
tion, not only upon the equations of Statics, but on the
form and elasticity of the arch-truss. Such cases will be
treated later under arch -ribs, or curved beams.
332. The Special Equil. Polygon. Its Relation to the Stresses
in the Rigid Body. — Eeproducing Figs. 350 and 351 in Figs.
358 and 359, (where a rigid curved beam is in equilibrium
under the forces P^ Pg, P3, P4 and B'^) we call a . . b . ,
MECHANICS OF Eis^GINEERIKG.
tiie special equil. polygon because it corresponds to a force
diagram in which the same order of forces has been ob-
S3rved as that in which they occur along the beam (from
left to right here). From the relations between the force
SPACE DIAGRAM
Fig. 358.
FORCE DIAGRAM
Pig, 359.
diagram and equil. polygon, this special equil. polygon in
the space diagram has the following properties in connec-
tion with the corresponding rays (dotted lines) in the force
diagram.
The stresses in any cross-section of the portion O'A of
the beam, are due to P^ alone ; those of any cross-section
on AB to Pi and P2, i.e., to their resultant R , whose mag-
nitude is given by the line Oa' in the force diagram, while
its liLe of action is ah the first segment of the equil. poly-
gon. Similarly, the stresses in BC are due to P^, P^ and
P., i.e., to their resultant R^, acting along the segment &c,
its magnitude being =^0h' in the force diagram. E.g., if
the section at m be exposed, considering O'ABm as a free
body, we have (see Fig. 360) the elastic stresses (or inter-
P3
Fig. 360. Fig. 361.
nal forces) at m balancing the exterior or " applied forces "
Pi, Pj and P3. Obviously, then, the stresses at m are just
GEAPHIOAL STATICS. 411
the same as if B^, tlie resultant of Pj, P^ and Pg, acted upon
an imaginary rigid prolongation of tlie beam intersecting
he (see Fig. 361).i?i, might be called the " anti-stress-resuU-
ant" ior the portion PC of the beam. We may .•. state
the following : If a rigid body is in equilibrium under a sys-
tem of Hon-Concurrent Forces m a plane, and the special equi-
librium polygon has been draivn, then each ray of tlie force
diagram is the anti-stress-resultant of that portion of the beam
which corresponds to the segment of the equilibrium polygon
to which the ray is parallel ; and its line of action is the seg-
ment just mentioned.
Evidently if the body is not one rigid piece, but com-
posed of a ring of uncemented blocks (or voussoirs), it may
be considered rigid only so long as no slipping takes place
or disarrangement of the blocks; and this requires that the
" anti-stress-resultant " for a given joint between two
blocks shall not lie outside the bearing surface of the
joint, nor make too small an angle with it, lest tipping or
slipping occur. For an example of this see Fig. 362, show-
ing a line of three blocks in equilibrium under five forces.
The pressure borne at the
s^2 joint MN, is = Pa ^^ the
force -diagram and acts in
the line ab. The con-
struction supposes all
the forces given except
Fig. 362. oue, in amount and posi-
tion, and that this one could easily be found in amount, as
being the side remaining to close the force polygon, while
its position would depend ox^ I;he equil. polygon. But in
practice the t^m forces Pj and B\ are generally unknown,
hence the point 0, or pole of the force diagram, can not
be fixed, nor the special equil. polygon located, until other
considerations, outside of those so far presented, are
brought into play. In the progress of such a problem, as
will be seen, it will be necessary to use arbitrary trial po-
sitions for the pole 0, and corresponding ^rmZ equilibrium
polygons.
412
MECHANICS OF BJUGmBBlWSGc,
CHAPTER IX.
GRAPHICAL. STATICS OF VERTICAL. FORCES,
333. Remarks. — (Witli the excoption of § 378 a) in prob-
lems to be treated subsequently (either the stiff arch-rib,
or the block-work of an arch-ring, of masonry) when the
body is considered free all the forces holding it in equil.
will be vertical (loads, due to gravity) except the reactions
at the two extremities, as in Eig, 363 ; but for convenience
each reaction will be replaced by its horizontal and verti-
cal components (see Fig. 364). The two fi^'s are of course
pqual, since they are the only horizontal forces in the
system. Henceforth, aU equil. polygons under discussion will
he understood to imply this kind of system of forces. Pi, Pz,
r r f
A t\
U
v„
Fig. 363.
Fig. 364a.
Fig. 364.
etc. , will represent the ' ' loads " ; Vq and F„ the vertical
components of the abutment reactions; H the value of
either horizontal component of the same. (We here sup-
pose the pressures To and Tn resolved along the horizontal
and vertical.)
JRAPHICAIi STATICS.
413
334. Concrete Conception of an Equilibrium Polygon. — Any
equilibrium polygon has this property, due to its mode
of construction, viz.: If the ab and be of Fig. 358 were im-
ponderable straight rods, jointed at b without frictioiij they
would be in equilibrium under the system of forces there
given. (See Fig. 364a). The rod ah suffers a compression
equal to the H^ of the force diagram, Fig. 359, and be a
-^compression = B^^. In some cases these rods might be in
tension, and would then form a set of links playing the
part of a suspension-bridge cable. (See § 44).
335. Example of EcLuilibrium Polygon Drawn to Vertical Loads
— Fig. 365. [The structure bearing the given loads is not
shown, but simply the imaginary rods, or segments of an
equilibrium polygon, which would support the given loads
in equilibrium if the abutment points A and B, to which
the terminal rods are hinged, were firm. In the present
case this equilibrium is unstable since the rods form a
standing structure ; but if they were hanging, the equilibri-
um would be stable. Still, in the present case, a very light
bracing, or a little friction at all joints would make the
equilibrium stable.
2 FT. TO aNOH
Fig. S65.
Given three loads Pi, F2, and P3, and two " abutment
verticals " A' and B', in which we desire the equil. poly-
gon to terminate, lay off as a "load-line," to scale, Pj, P2,
and P« end to end in their order. Then selecting any pole.
414 MBCHAXICS OF EjS^GI:N^EEIII]^G.
0, draw the rays 01, 02, etc., of a force diagram (tlie F's
and P's, though really on the same vertical, are separated
slightly for distinctness ; also the H's, which both pass
through and divide the load-line into V^ and F^). We
determine a corresponding equilibrium polygon by draw-
ing through A (any point in A') a line || to . . 1, to inter-
sect Pi in some point b ; through 6 a || to . . 2, and so ou>
until B'' the other abutment-vertical is struck in some
point B. AB is the " abutment-line " or " closing -line."
By choosing another point for 0, another equilibrium
polygon would result. As to which of the infinite
number (which could thus be drawn, for the given loads
and the A' and B' verticals) is the special equilibrium poly'
gon for the arch-rib or stone-arch, or other structure, on
which the loads rest, is to be considered hereafter. In
any of the above equilibrium polygons the imaginary
series of jointed rods would be in equilibrium.
336. Useful Property of an Equilibrium Polygon for Vertical
Loads. — (Particular case of § 328). See Fig. 366. In any
equil. polygon, supporting vertical loads, consider as free
any number of consecutive segments, or rods, with the
loads at their joints, e. g., the 5th and 6th and portions of
C/r^.^ the 4th and 7th which, we sup-
/g i ,> ^ ^6"--. ^ -^^ pose cut and the compressive
— ~S<^^ forces in them put in, T^ and
^^ Tj, in order to consider 4 5 6 7
"^^ ^ as a free body. For equil,,
~^~[-^\ according to Statics, the lines
' ' 'Pe "\ of action of Ti and Ty (the com-
i^iG. 366. pression in those rods) must in-
tersect in a point, C, in the line of action of the resultant
of Fi, P5, and Pq ; i.e., of the loads occurring at the inter-
vening vertices. That is, the point C must lie in the ver-
tical containing the centre of gravity of those loads. Since
the position of this vertical must be independent of the
particular equilibrium polygon used, any other (dotted
lines in Fig. 366) for the same loads will give the same re-
GKAPHICAL STATICS. 415
suits. Hence tlie vertical CD, containing the centre of
gravity of any number of consecutive loads, is easily found
by drawing tlie equilibrium polygon corresponding to
any convenient force diagram having the proper load-line.
This principle can be advantageously applied to finding
a gravity -line of any plane figure, by dividing the latter
into parallel strips, whose areas may be treated as loads
applied in their respective centres of gravity. If the strips
are quite numerous, the centre of gravity of each may be
considered to be at the centre of the line joining the mid-
dles of the two long sides, while their areas may be taken
as proportional to the lengths of the lines drawn through
these centres of gravity parallel to the long sides and lim-
ited by the end-curves of the strips. Hence the " load-
line " of the force diagram may consist of these lines, or
of their halves, or quarters, etc., if more convenient (§ 376).
USEFUL, RELATIONS BETWEEN FORCE DIA-
GRAMS AND EQUILIBRIUM POLYGONS,,
(for vertical loads,)
237. Il6sum6 of Construction.— Fig. 367. Given the loads
Pi, etc., 'their verticals, and the two abutment verticals ^4'
and B', in which the abutments are to lie ; we lay off a
load-line 1 ... 4, take any convenient pole, 0, for a force-
diagram and complete the latter. For a corresponding
equilibrium polygon, assume any point A in the vertical"
A', for an abutment, and draw the successive segments
Al, 2, etc., respectively parallel to the inclined lines of the
force diagram (rays), thus determiDiDg finally the abut-
ment B, in B', v/hich {B) will not in general lie in the hor-
izontal through A.
Now join AB, calling AB the abutment-line, and draw a
parallel to it through 0, thus fixing the point n' on the
416
MECHANICS or ENGINEERIKG.
Pi
P. t
l'
Fig.
load-line. This point %' , as above determined, is indepen'
dent of the location of the pole, 0, (proved in § 329) and
divides the load-line into two portions ( V'o = 1 . . . n\ and
V'n = n' .. .4:) which are the vertical pressures which two
supports in the verticals A' and B' would sustain if the
given loads rested on a horizontal rigid bar, as in Fig. 368.
See § 329. Hence to find the point n' we may use any
convenient pole 0.
[N. B.— The forces V^ and V^ of Fig. 367 are not identi-
cal with F'o and V'„, but may be obtained by dropping a
"I from to the load-line, thus dividing the load-line
into two portions which are V^ (upper portion) and F^.
However, if A and B be connected by a tie-rod, in Fig.
367, the abutments in that figure will bear vertical press-
ures only and they will be the same as in Fig, 368, while
the tension in the tie -rod will be = On'.^
338. Theorem. — The vertical dimensions of any two equili-
brium polygons, drawn to the same loads, load-verticals, and
abutment -verticals, are inversely proportional to their H^s {or
"pole distances "). We here regard an equil. polygon and
its abutment-line as a closed figure. Thus, in Fig. 369,
we have two force-diagrams (with a common load-line, for
convenience) and their corresponding equil. polygons, for
the same loads and verticals. From § 337 we know that
On' is II to AB and OqW' is || to A^B^. Let CD be any ver-
tical cutting the first segments of the two equil. polygons.
GRAPHICAL STATICS.
417
SB|
Denote the intercepts thus determined by z' and %\, respect-
gC r ively. From the
parallelisms just
mentioned, and
others more famil~
,/ iar, we have the
triangle \n' sim*
ilar to the triangle
Az' (shaded), and
the triangle O^n'
similar to the tri-
angle Ajz,^,. Hence
c
1
P.
u-^
/
hi
1 .
p^ ^.
A.
^y
1
—- 1-..
-^-N^
Fig. 369.
the proportions between ( \n'
bases and altitudes (
H h
and
.*. z' : z\ : : H^ '■ H. The same kind of proof may easily
be applied to the vertical intercepts in any other segments,
e. g., 2" and z'\. Q. E. D.
339. Corollaries to the foregoing. It is evident that :
(1.) If the pole of the force-diagram be moved along a
vertical line, the equilibrium polygon changing its form
in a corresponding manner, the vertical dimensions of the
equilibrium polygon remain unchanged ; and
(2.) If the pole move along a straight line which con-
tains the point n\ the direction of the abutment-line
remains constantly parallel to the former line, while the
vertical dimensions of the equilibrium polygon change in
inverse proportion to the pole distance, or H, of the force-
diagram, [^is the "1 distance of the pole from the load-
line, and is called the pole-distance].
§ 340. Linear Arch as Equilibrium Polygon. — (See § 316.)
If the given loads are infinitely small with infinitely small
horizontal spaces between them, any equilibrijim polygon
becomes a linear arch. Graphically we can not deal with
these infinitely small loads and spaces, but from § 336 it
is evident that if we replace them, in successive groups.
418
MECHAXICS OF ENGINEERING.
Fig. 370.
bj finite forces, eacli of wliicli = tlie STim of those com^.
I I I I I I I I P°''^^ """^ ^^""^P ^""^ ^'
.M i V ..M M. .^^ M I ,M I + /, applied tlirougli the cen-
tre of gravity of that
group, we can draw an
equilibrium polygon
whose segments will be
tangent to the curve of
the corresponding linear
arch, and indicate its posi-
tion with siifficient exactness for practical purposes. (See
Fig. 370), The successive points of tangency A, m, n, etc..
lie vertically under the points of division between the
groups. This relation forms the basis of the graphical
treatment of voussoir, or blockwork, arches.
341. To Peas an Equilibrium Polygon Through. Three Arbitrary
Points. — (In the present case the forces are vertical. For
a construction dealing with any plane system of forces see
construction in § 378a.) Given a system of loads, it is re-
^ quired to draw
r /^l ^^ equilibrium
polygon for
t h e m through
-anythree points,
two of which
may be consid-
ered as abut-
ments, outside of the load-verticals, the third point being
between the verticals of the first two. See Fig. 371. The
loads Pi, etc., are given, with their verticals, while A, p,
and B are the three points. Lay oft the load-line, and
with any convenient pole, Oj, construct a force-diagram,
then a corresponding preliminary equilibrium polygon
beginning at A. Its right abutment P,, in the vertical
through B, is thus found. Oj n' can now be drawn || to AB^,
to determine n\ Draw n'O \\ to BA. The pole of the
required equilibrium polygon must lie on n'O (§ 337}
Fig. 371.
GEAPHICAL STATICS.
Draw a vertical throiigli jp. The E. of tlie required equili-
brium polygon must satisfy the proportion H : H^ : : rs i
pm. (See § 338). Hence construct or compute H from
the proportion and draw a vertical at distance H from
the load-line (on the left of the load-line here) ; its inter-
section with n' gives the desired pole, for which a
force diagram may now be drawn. The corresponding
equilibrium polygon beginning at the first point A will
also pass through p and B ; it is not drawn in the figure.
342. Symmetrical Case of the Foregoing Problem.— If two
points A and B are on a level, the third, p, on the middle
vertical between them ; and the loads (an even number)
symmetrically disposed both in position and magnitvde, about
p, we may proceed more simply, as follows : (Fig. 372).
From symmetry n'
must occur in the mid-
dle of the load-line, of
which we need lay off
only the upper half.
Take a convenient pole
■piG. 372. Oi, in the horizontal
through n', and draw a half force diagram and a corres-
ponding half equilibrium polygon (both dotted). The up-
per segment he of the latter must be horizontal and being
prolonged, cuts the prolongation of the first segment in a
point d, which determines the vertical CD containing the
centre of gravity of the loads occurring over the half -span
on the left. (See § 336). In the required equilibrium poly-
gon the segment containing the point p must be horizon-
tal, and its intersection with the first segment must lie in
CD. Hence determine this intersection, C, by drawing the
vertical CD and a horizontal through p ; then join CA,
which is the ^rst segment of the required equil. polygon.
A parallel to GA through 1 is the Jirst ray of the corres-
ponding force diagram, and determines the pole on the
horizontal through n'. Completing the force diagram foi
420
MECHAXIOS OF El^J^GINEBEIN^G.
Fig. 373.
this pole (half of it only here), the required equil. poly-
gon is easily finished afterwards.
343. To Find a System of Loads Under Which a Given Equi-
librium Polygon Would be in Equilibrium, — Fig. 373. Let AB
he the given equilibrium polygon. Through any point
as a pole draw a parallel to each
segment of the equilibrium polygon.
Any vertical, as V, cutting these
lines will have, intercepted upon it,
a load-line 1, 2, 3, whose parts 1 . . 2,
2 . . 3, etc., are proportional to the
successive loads which, placed on
ih@ corresponding joints of the equilibrium polygon would
be supported by it in equilibrium (unstable).
One load may be assumed and the others constructed. A
hanging, as well as a standing, equilibrium polygon may be
dealt with in
hke manner,
but will be
in stable
equilibrium.
The problem
in § 44 may
be solved in
this way, the various steps and final re-
sults being as follows (Fig. 50 is here re-
peated) : —
Let weight Gi be given, =66 lbs., and
the positions of the cord segments be as in
Fig. 50. We first lay of! (see Fig. 373a)
vertically, a6 = 66 lbs., by some convenient
scale, and prolong this vertical fine indefi- d
nitely downward. aO is then drawn parallel to
and bO parallel to 1 ... 2. Their intersection determines a
pole, 0, through which Oc and Od, parallel respectively to
2 ... 3 and 3 . . . n, are drawn, to intersect ad in c and d.
We also draw the horizontal On, in Fig. 373a. By scaling,
we now find the results: — G2 = bc = 42 lbs.; G3 = cd = 50 lbs.;
H = 58.5 lbs., ( = Ho and'^„ of Fig. 50); while 70 = ^^=100
lbs. and y„ = 58 lbs.
Fig. 373a.
.1 of Fig. 50,
AECHES OF MASOifBT.
421
CHAPTER X.
RIGHT ARCHES OF MASONRY.
Note. — The treatment given in this chapter is by many engineers
considered sufficiently exact for ordinary masonry arches, the mOie
refined methods of the "elastic theory" being reserved for arches of
fairly continuous material, such as those of metal and of concrete (re-
inforced or otherwise); and is accordingly retained in this revised
edition.
844. — In an ordinary "right" storce-arcli (i.e., one in
which the faces are "[ to the axis of the cylindrical soffit,
or under surface), the successive blocks forming the arch-
ring are called voussoirs, the joints between them being
planes which, prolonged, meet generally in one or more
horizontal lines; e.g., those of a three-centred arch in three
II horizontal lines ; those of a circular arch in one, the axis
of the cylinder, etc. Elliptic arches are sometimes used. The
inner concave surface is called the soffit, to which the radiat-
ing joints between the voussoirs are made perpendicular.
The curved line in which the soffit, is intersected by a plane
Fig. 374.
H to the axis of the arch is the Intrados. The curve in the
same plane as the intrados, and bounding the outer ex-
tremities of the joints between the voussoirs, is called the
Extrados.
Fig. 374 gives other terms in use in connection with, a
422
MECHAKICS OF ElSGIXEEIxiKG,
stone arch, and explains those already given.
" springing-line."
AB is the
345o Mortar and Friction. — As 'common mortar hardens
very slowly, no reliance should be placed on its tenacity
as an element of stability in arches of any considerable
size ; though hydraulic mortar and thin joints of ordinary
mortar can sometimes be depended on. Friction, however,
between the surfaces of contiguous voussoirs, plays an
essential part in the stability of an arch, and will there-
fore be considered.
The stability of voussoir-arches must .•. be made to
depend on the resistance of the voussoirs to compresssion
and to sliding upon each other ; as also of the blocks
composing the piers, the foundations of the latter being
firm.
346. Point of Application of the Eesultant Pressure between
two consecutive voussoirs ; (or pier blocks). Applying
Navier's principle (as in flexure of beams) that the press-
ure per unit area on a joint varies uniformly from the
extremity under greatest comj)ression to the point of least
compression (or of no compression); and remembering
that negative pressures (i.e., tension) can not exist, as they
might in a curved beam, we may represent the pressure
per unit area at successive points of a joint (from the intra-
dos toward the extrados, or vice versa) by the ordinates of
a straight line, forming the surface of a trapezoid or tri-
angle, in which figure the foot of the ordinate of the cen-
tre of gravity is the point of application of the resultant
pressure. Thus, where the least compression is supposed
Fig. 575.
Fig. 376.
Fig. 377.
Fig. 378,
MASONEY ARCHES.
423
to occur at the intrados A, Fig. 375, tlie pressures vary as
tlie ordinates of a trapezoid, increasing to a maximum value
at B, in the extrados. In Fig. 376, where the pressure is zero
at B, and varies as the ordinates of a triangle, the result-
ant pressure acts through a point one-third the joint-
length from A. Similarly in Fig. 377, it acts one-third
the joint-length from B. Hence, when the pressure is not
zero at either edge the resultant pressure acts within the
middle third of the joint. Whereas, if the resultant press-
ure falls loitliout the middle third, it shows that a portion
-4m of the joint, see Fig. 378, receives no pressure, i.e., the
joint tends to open along Am.
Therefore that no joint tend to open, the resultant press-
ure must fall within the middle third.
It must be understood that the joint surfaces here dealt
with are rectangles, seen edgewise in the figures.
347. Friction. — By experiment it has been found the
angle of friction (see § 156) for two contiguous voussoirs
of stone or brick is about 30° ; i.e., the coefficient of fric-
tion is / = tan. 30°. Hence if the direction of the press-
ure exerted upon a voussoir by its neighbor makes an
angle a less than 30° with the normal to the joint surface,
there is no danger of rupture of the arch by the sliding
of one on the other. (See Fig. 379).
348. Resistance to Crushing. — When the resultant pressure
falls at its extreme allowable limit, viz. : the edge of the
middle third, the pressure per
unit of area at n, Fig. 380, iy
double the mean pressure per
unit of area. Hence, in de-
signing an arch of masonry,
we must be assured that at
every joint (taking 10 as a
factor of safety)
( Double the mean press- | ^^^^ ^^ j^^^ ^^^^ y g
I ure per unit oi area \ '
Fig. 379.
Fig. 380.
424 MECHA]srics of engineekikg.
C being tlie ultimate resistance to crushing, of tlie material
emj)loyed (§ 201) (Modulus of Crushing).
Since a lamina one foot thick will always be considered
in what follows, careful attention must be paid to the units
employed in applying the above tests.
Example. — If a joint is 3 ft. by 1 foot, and the resultant
pressure is 22.5 tons the mean pressure per sq. foot is
p=22.5-^3=7.5 tons per sq. foot
.'. its double=15 tons per sq. foot=208.3 lbs. sq. inch,
which is much less than '/lo of C for most building stones ;
see § 203, and below.
At joints where the resultant pressure falls at the middle,
the max. pressure per square inch would be equal to the
mean pressure per square inch ; but for safety it is best to
assume that, at times, (from moving loads, or vibrations)
it may move to the edge of the middle third, causing the
max. pressure to be double the mean (per square inch).
Gem Gillmore's experiments in 1876 gave the following
results, among many others :
NAME OF BUILDING STONE. C IN LBS. PER SQ. INCH.
Berea sand-stone, 2-inch cube, - - - - 8955
4 " " - - - - 11720
Limestone, Sebastopol, 2-inch cube {chalk)^ - - 1075
Limestone from Caen, France, - - . . 3650
Limestone from Kingston, ]^. Y., - - . - 13900
Marble, Vermont, 2-inch cube, - - 8000 to 13000
Granite, New Hampshire, 2-inch cube, 15700 to 24000
349. The Three Conditions of Safe E(iiiilibriiim for an arch of
uncemented voussoirs.
Recapitulating the results of the foregoing paragraphs,
we may state, as follows, the three conditions which must
be satisfied at every joint of arch -ring and pier, for each
of any possible combination of loads upon the structure :
(1). The resultant pressure must pass within the middle-
third,
(2). The resultant pressure must not make an angle >
30° with the normal to the joint.
(3). The m'^.an pressure per unit of area on the surface
AKCH OF MASOK^RT.
425
of tlie joint must not exceed Ygo of the Modulus of crush-
ing of the material.
350. The True Linear-Arch, or Special Equilibrium Polygon;
and the resultant pressure at any joint. Let the weight
of each voussoir and its load be represented by a vertical
force passing through the centre of gravity of the two, as
in Fig. 381o Taking any
two points A and JB, A
being in the first joint and
B in the last ; also a third
point, p, in the crown
joint (supposing such to
be there, although gener-
ally a key-stone occupies |
the crown), through these fig. ssi.
three points can be drawn [§ 341] an equilibrium polygon
for the loads given ; suppose this equil. polygon nowhere
passes outside of the arch-ring (the arch-ring is the por-
tion between the intrados, mn, and tha (dotted) extrados
m'n') intersecting the joints at h, c, etc. Evidently if such
be the case, and small metal rods (not round) were insert-
ed at A, h, c, etc., so as to separate the arch-stones slight-
ly, the arch would stand, though in unstable equilibrium,
the piers being firm ; and by a different choice of A, p, and
B, it might be possible to draw other equilibrium poly-
gons with segments cutting the joints within the arch-
ring, and if the metal rods were shifted to these new inter-
sections the arch would again stand (in unstable equilib-
rium).
In other words, if an arch stands, it may be possible to
draw a great number of linear arches within the limits of
the arch-ring, since three points determine an equilibrium
polygon (or linear arch) for given loads. The question
arises then : luMch linear arch is the locus of the actual re-
sultant pressures at the successive joints ?
[Considering the arch-ring as an elastic curved beam
inserted in firm piers (i.e., the blocks at the springing-line
426 MEOHAKIOS OF ENGIKEEEING.
are incapable of turning) and Jbaving secured a close fit at
all joints before the centering is lowered, the most satisfac-
tory answer to this question is given in Prof. Greene's
" Arches," p. 131 ; viz., to consider the arch-ring as an
arch rib of fixed ends and no hinges ; see § 380 of next
chapter;, but the lengthy computations there employed
(and the method demands a simple algebraic curve for the
arch) may be most advantageously replaced by Prof.
Eddy's graphic method (" New Constructions in Graphical
Statics," published in Van Nostrand's Magazine for 1877)„
which applies to arch curves of any form.
This method will be given in a subsequent chapter, on
Arch Eibs, or Curved Beams ; but for arches of masonry a
much simpler procedure is sufficiently exact for practical
purposes and will now be presented].
If two elastic blocks
of an arch-ring touch at
one edge. Fig. 382, their
adjacent sides making a
small angle with each
•"iG- ^82. Fig. 383. other, and are then grad-
ually pressed more and more forcibly together at the edge
m, as the arch-ring settles, the centering being gradually
lowered, the surface of contact becomes larger and larger,
from the compression which ensues (see Pig. 383); while
the resultant pressure between the blocks, first applied at
the extreme edge m, has now probably advanced nearer the
middle of the joint in the mutual adjustment of the arch-
stones. With this in view we may reasonably deduce the
following theory of the location of the true linear areh
(sometimes called the " line of pressures " and " curve of
pressure ") in an arch under given loading and with ^rm
piers. (Whether the piers are really unyielding, under the
oblique thrusts at the springing-line, is a matter for sub-
sequent investigation.
351. Location of the True Linear Arch. — Granted that the
voussoirs have been closely fitted to each other over the
ARCH OF MASOXKT. 427
■centering (sheets of lead are sometimes used in tlie joints
to make a better distribution of pressure); and tliat the
piers are firm ; and that the arch can stand at all without
the centering ; then we assume that in the mutual accom-
modation between the voussoirs, as the centering is low-
ered, the resultant of the pressures distributed over any
joint, if at first near the extreme edge of the joint, advances
nearer to the middle as the arch settles to its final posi-
tion of equilibrium under its load ; and hence the follow-
ing
352. Practical Conclusions.
I. If for a given arch and loading, with firm piers, an
•equilibrium polygon can be drawn (by proper selection of
the points A, p, and B, Fig. 381) entirely within the mid^
die third of the arch ring, not only will the arch stand, but
the resultant pressure at every joint will be within the
middle third (Condition 1, § 349) ; and among all possible
equilibrium polygons which can be drawn within the mid-
dle third, that is the " true " one which most nearly coin-
•cides with the middle line of the arch-ring.
II. If (with firm piers, as before) no equilibrium poly-
rgon can be drawn Avithin the middle third, and only one
within the arch-ring at all, the arch may stand, but chip-
ping and spawling are likely to occur at the edges of the
joints. The design should .*. be altered.
III. If no equilibrium polygon can be drawn within
the arch-ring, the design of either the arch or the loading
.must be changed ; since, although the arch may standi
from the resistance of the spandrel walls, such a stability
must be looked upon as precarious and not countenanced
in any large important structure. (Very frequently, in
small arches of brick and stone, as they occur in buildings,
the cement is so tenacious that the whole structure is vir-
tually a single continuous mass).
When the " true " linear arch has once been determined,
the amount of the resultant pressure on any joint is given
by the length of the proper ray in the force diagram.
428
MECHANICS OF EXGIXEEHING.
ARRANGEMENT OF DATA FOR GRAPHIC
TREATMENT.
353. Cli-aracter of Load. — In most large stone arch bridges
the load (permanent load) does not consist exclusively of
masonry up to tlie road-way but partially of earth filling
above the masonry, except at the faces of the arch where
the spandrel walls serve as retaining walls to hold the
earth. (Fig. 384). If the intrados is a half circle or half-
Fig. 385.
Fig. 384.
ellipse, a compactly -built masonry backing is carried up
beyond the springing-line to AB about 60° to 45° from the
crown. Fig. 385 ; so that the portion of arch ring below
AB may be considered as part of the abutment, and thus
AB is the virtual springing-line, for graphic treatment.
Sometimes, to save filling, small arches are built over
the haunches of the main arch, with earth placed over
them, as shown in Fig. 386. In any of the preceding cases
Fig. 387.
it is customary to consider that, on account of the bond-
ing of the stones in the arch shell, the loading at a given
distance from the crown is uniformly distributed over the
width of the roadway.
AECHES OF MASONET. 429
354, Reduced Load-Contour. — In the graphical discussioa
of a proposed arch we consider a lamina one foot thick,
this lamina being vertical and "| to the axis of the arch ;
i.e., the lamina is || to the spandrel walls. For graphical
treatment, equal areas of the elevation (see Fig, 387) of
this lamina must represent equal weights. Taking the
material of the arch-ring as a standard, we must find for
each point "p of the extrados an imaginary height z of the
arch-ring material, which would give the same pressure
(per running horizontal foot) at that point as that due to
the actual load above that point. A number of such or-
dinates, each measured vertically upward from the extra-
dos determine points in the "Reduced Load-Contour," i.e.,
the imaginary line, AM, the area between which and the
extrados of the arch-ring represents a homogeneous load
of the same density as the arch -ring, and equivalent to the
actual load (above extrados), vertical hy vertical.
355. Example of Reduced Load-Contour. — Fig, 388. Given
an arch-ring of granite (heaviness = 170 lbs. per cubic
foot) with a dead load of rubble (heav. = 140) and earth
(heav. = 100), distributed as in figure. At the point p, of
the extrados, the depth 5 feet of rubble is equivalent to a
depth of [^^ x5]=4.1 ft. of granite, while the 6 feet of earth
is equivalent to [l°?x6]=3.5 feet of granite. Hence the
Reduced Load-Contour has an ordinate, above p, of 7.6 feet.
That is, for each of several points of the arch -ring extrados-
reduce the rubble ordinate in the ratio of 170 : 140, and
the earth ordinate in the ratio 170 : 100 and add the re-
sults, setting off the sum vertically from the points in the
■extrados*. In this way Fig. 389 is obtained and the area
*TUs is most conveniently done by graphics, thus : On a right-line set off 17 equal.
parts (of any convenient magnitude.) Call this distance OA. Through t> draw another
right line at any convenient angle (30° to 60°) with OA, and on it from O
set off OB equal to 14 (for the ruhble ; or 10 for the earth) of the eame egaal
parts. Join AB. From O toward A set off* all the rubble ordiaates to be reduced^
(each being set off from 0} and through the other extremity of each draw a Bne par-
allel to AB. The reduced ordinates will be the respective lengths, from O, along OB,
to the intersections of these parallels ynth OB.
* Witli the dividers.
430
MBCHAISICS OF ENGINEERING.
:/EART.HV; Av,*//%V;*i*»'v5i;'i!lV;?V/;i;;*'uVf/-^;;-';^^
there given is to be treated as representing liomogeneous
granite one foot thick. This, of course, now includes the
arch-ring also. AB is the " reduced load- contour."
356. Live Loads. — In discussing a railroad arch bridge
the " live load " (a train of locomotives, e.g., to take an ex-
treme case) can not be disregarded, and for each of its po-
sitions we have a separate Reduced Load-Contour.
Example. — Suppose the arch of Fig. 388 to be 12 feet
wide (not including spandrel walls) and that a train of lo-
comotives weighing 3,000 lbs. per running foot of the track
covers one half of the span. Uniformly * distributed later-
ally over the width, 12 ft., this rate of loading is equiva-
lent to a masonry load of one foot high and a heaviness of
250 lbs. per cubic ft., i.e., is equivalent to a height of 1.4
ft. of granite masonry [since ^[|- X 1.0 — 1.4] over the half
span considered. Hence from Fig. 390 we obtain Fig. 391
in an obvious manner. Fig. 391 is now ready for graphic
treatment.
Fig. 390. Fig. 391.
357. Piers and Abutments. — In a series of equal arches
the pier between two consecutive arches bears simply the
weight of the two adjacent semi-arches, plus the load im-
* If the earth-filling is sLallcw, the Icminse directly under the track prob*
aibly receive a greater pressure than the others.
AKCHES OF MASONRY.
431
mediately above the pier, and .-. does not need to be as
large as the abutment of the first and last arches, since
these latter must be prepared to resist the oblique thrusts
of their arches without help from the thrust of another on
the other side.
In a very long series of arches it is sometimes customary
to make a few of the intermediate piers large enough to
act as abutments. These are called " abutment piers," and
in case one arch should fall, no others would be lost except
those occurring between the same two abutment piers as
the first. See Fig. 392. A is an abutment-pier.
Fig. 39;^.
GRAPHICAL. TREATMENT OF ARCH.
358. — Having found the " reduced load-contour," as in
preceding paragraphs, for a given arch and load, we are
ready to proceed with the graphic treatment, i.e., the first
given, or assumed, form and thickness of arch-ring is to be
investigated with regard to stability. It may be necessary
to treat, separately, a lamina under the spandrel wall, and
one under the interior loading. The constructions are
equally well adapted to arches of all shapes, to Gothic as
well as circular and elliptical.
359.— Case I. Symmetrical Arch and Symmetrical Loading.—
(The " steady " (permanent) or " dead " load on an arch is
usually symmetrical). Fig. 393. From symmetry we need
Fig.
Fig. 394.
Fis. 395.
i33 MECHAJTICS OF ENGINEERmG.
deal witli only one half (say the left) of tlie arch and load.
Divide this semi-arch and load into six or ten divisions
by vertical lines ; these divisions are considered as trape-
zoids and should have the same horizontal width = 6 (a
convenient whole number of feet) except the last one, LKN,
next the abutment, and this is a pentagon of a different
v\ridth hy, (the remnant of the horizontal distance LC). The
weight of masonry in each division is equal to (the area
of division) X (unity thickness of lamina) x (weight of a cu-
bic unit of arch-ring). For example for a division having
an area of 20 sq. feet, and composed of masonry weighing
160 lbs. per cubic foot, we have 20x1x160=3,200 lbs.,
applied through the centre of gravity of the division.
The area of a trapezoid. Fig. 394, is^&(7ii+7i2), audits cen-
tre of gravity may be found. Fig. 395, by the construction
of Prob. 6, in § 26 ; or by § 27a. The weight of the pen-
tagon LN, Fig, 393, and its line of application (through
centre of gravity) may be found by combining results for
the two trapezoids into which it is divided by a vertical
through K. See § 21.
Since the weights of the respective trapezoids {excepts
ing LN) are proportional to their middle vertical in-
tercepts [such as ^(^1+7^2) Fig- 394] these intercepts (trans-
ferred with the dividers) may be used directly to form the
load-line, Fig. 396, or proportional parts of them if more
convenient. The force scale, which this implies, is easily
computed,, and a proper length calculated to represent the
weight of the odd division LN ; i.e., 1 ... 2 on the load-
line.
Now consider A, the middle point of the abutment joint.
Fig. 396, as the starting point of an equilibrium polygon
(or abutment of a linear arch) for a given loading, and re-
quire that this equilibrium polygon shall pass through j>,
the middle of the crown joint, and through the middle of
the abutment joint on the right (not shown in figure).
Proceed as in § 342, thus determining the polygon Ap
for the half-arch. Draw joints in the arch-ring through
those points where the extrados is intersected by the ver-
ABCHES OF MASONRY.
433
Jig. 396. Fig. 397.
Heal separating tlie divisions (not the gravity verticals),
Tlie points in which these joints are cut by the segments
of the equilibrium polygon, Fig. 397, are (very nearly, if
th« joint is not more than 60° from jp, the crown) the points
of application in these joints, respectively, of the resultant
pressures on them, (if this is the " true linear arch " for
this arch and load) while the amount and direction of each
such pressure is given by the proper ray in the force -dia-
gram.
If at any joint so drawn the linear arch (or equilibrium
polygon) passes outside the middle third of the arch-ring,
the point A, or p, (or both) should be judiciously moved
(within the middle third) to find if possible a linear arch
which keeps within limits at all joints. If this is found
impossible, the thickness of the arch -ring may be increased
at the abutment (giving a smaller increase toward the
crown) and the desired result obtained ; or a change in the
distribution or amount of the loading, if allowable, may
gain this object. If but one linear arch can be drawn
within the middle third, it may be considered the " true "
one ; if several, the one most nearly co-inciding with the
middles of the joints (see §§ 351 and 352) is so considered.
360.— Case II Unsymmetrical Loading on a Symmetrical Arch,'
(e.g., arch with live load covering one half -span as in Figs.
390 and 391). Here we must evidently use a full force
diagram, and the full elevation of the arch -ring and load*
434
MBCHAXICS OF EXGINEEELNG.
See Fig. 398. Select three points A, p, and B, as follows,
to determine a trial equilihriu'm ])6lygon : '
Select A at the Joicer limit of the middle third of tLa
Fig. 398.
abutment-joint at the end of the span -vihich is the more
heavily -loaded ; in the other abutment-joint take B at tht
upper limit of the middle third ; and take p in the middle
of the crown-joint. Then by § 341 draw an equilibrium
polygon (i.e., a linear arch) through these three points for
the given set of loads, and if it does not remain within the
middle third, try other positions for A, p, and B, within
the middle third. As to the " true linear arch " alterations
of the design, etc., the same remarks apply as already
given in Case I. Very frequently it is not necessary to
draw more than one linear arch, for a given loading, for
even if one could be drawn nearer the middle of the arch-
ring than the first, that fact is almost always apparent on
mere inspection, and the one already drawn (if within
middle third) will furnish values sufiiciently accurate for
the pressures on the respective joints, and their direction
angles.
360a. — The design for the arch-ring and loading is not
to be considered satisfactory until it is ascertained that for
the dead load and any possible combination of live-load
'(in addition) the pressure at any joint is
ARCHES OF MASONRY. , 435
1.) Witliin the middle third of that joint ;
{2.) At an angle of < 30° with the normal to joint-
SYirface.
(3.) Of a mean pressure per square inch not > thanVa)
of the ultimate crushing resistance. (See § 348.)
§ 361. Abutments. — The abutment should be compactly
and solidly built, and is then treated as a single rigid mass.
The pressure of the lowest voussoir upon it (considering
a lamina one foot thick) is given by the proper ray of the
force diagram (0 .. 1, e. g., in Fig. 396) in amount and direc-
tion. The stability of the abutment will depend on the
amount and direction of the resultant obtained by com-
bining that pressure P^ with the weight G of the abutment
and its load, see Fig. 399. Assume a probable width RS
for the abutment and compute the weight G
of the corresponding abutment OBRS and
MNBO, and find the centre of gravity of the
whole mass G. Apply G in the vertical
through C, and combine it with P„ at their in-
tersection D. The resultant P should not cut
the base R8m. a point beyond the middle third
p/^ / " (or, if this rule gives too massive a pier, take
I / / ° such a width that the pressure per square
I// inch at 8 shall not exceed a safe value as
^ Fis. 399. computed from § 362.) After one or two
trials a satisfactory width can be obtained.
We should also be assured that the angle PD G is less
than 30°. The horizontal joints above RS should also be
tested as if each were, in turn, the lowest base, and if
nscessary may be inclined (like mn) to prevent slipping.
On no joint should the maximum pressure per square inch
be > than y^o the crushing strength of the cement. Abut-
ments of firm natural rock are of course to be preferred
where they can be had. If water penetrates under an
abutment its buoyant effort lessens the weight of the lat-
ter to a considerable extent.
436
MECHANICS OF ENGINEEKIifG.
362. Maximum Pressure Per Unit of Area When the Resultant
Pressure Falls at Any Given Distance from the Middle ; according
to Navier's theory of the distribution of the pressure ; see
§ 346. Case I. Let the resultant pressure P, Fig. 400, (a).
Fig. 400.
Fig. 401.
fall within the middle third, a distance = wc? ( < ^ d)
from the middle of joint [d = depth of Joint.) Then we
have the following relations :
p (the mean press, per.- sq. in.,),^,,, (max. press, persq. in.),
and p^ (least press, per sq. in.) are proportional to the lines
h (mid. width), a (max. base), and c (min. base) respectively,
of a trapezoid. Fig. 400, (&), through whose centre of gravity
P acts. But (§ 26) •
nd=---. i.e., n = y^ — = — or a=h (6w+l)
6 a-\-c n .
''• Pm—JP (6w+l). Hence the following table:
If 7id= j4> d Ya d
press. Pn,= 2 y^
Vs
then the max.
times the mean pressure.
Case II. Let P fall outside the mid. third, a distance =
"nd {^ )4 d) from the middle of joint. Here, since the
joint is not considered capable of withstanding tension,
we have a triangle, instead of a trapezoid. Fig. 401. First
compute the mean press, per sq. in.
V -
P (lbs.)
(1— 2w) 18 d inches
foot thick).
or from this table : (lamina ona
AKOHES OF MASOlfEY.
437
For nd =
^d
■hd
■hd
T\d
^d
^d
P =
1 P
10* d
1 P
8 * d
1 P
6 * d'
1 P
4 <^
1 P
2 6^
infinity.
{d in inches and P in lbs.; with arch lamina 1 ft. in thickness.)
Then the maximum pressure (at A, Fig. 401) />,„, = 2p,
becomes known, in lbs. per sq. in.
362a. Arch-ring under Uon- vertical Forces. — An example of
this occurs when a vertical arch-ring is to support the pressure
of a liquid on its extrados. Since water-pressures are always
at right angles to the surface pressed on, these pressures on the
extradosal surface of the arch-ring form a system of non-paral~
lei forces which are normal to the curve of the extrados at;
their respective points of application and lie in parallel
vertical planes, parallel to the faces of the lamina. "We here
assume that the extradosal surface is a cylinder (in the most
general sense) whose rectilinear elements are 1 to the faces of
the lamina. If, then, we divide the length of the extrados,
from crown to each abutment, into from six to ten parts, the
respective pressures on the corresponding surfaces are obtained
by multiplying the area of each by the depth of its centre of
gravity from the upper free surface of the liquid, and this
product by the weight of a unit of volume of the liquid ; and
each such pressure may be considered as acting through the
centre of the area. Finally, if we find the resultant of each
of these pressures and the weight of the corresponding portion
of the arch-ring, these resultants form a series of non-vertical
forces in a plane, for which an equilibrium polygon can then
be passed through three assumed points by § 378a, these three
points being taken in the crown-joint and the two abutment-
joints. As to the " true linear arch" see § 359.
As an extreme theoretic limit it is worth noting that if the
extrados and intrados of the arch-ring are concentric circles ; if
the weights of the voussoirs are neglected ; and if the rise of
tb« arch is very small compared with the depth of the crown
^/elow the water surface, then the circularGentre-line of the
wrch-ring is the " true linear archP
4:38 MECHAIiflCS OF ENGLNiiJEJiLNG.
CHAPTER XI.
ARCH-RIBS.
Note. — The methods used in this chapter for the treatment of the
"elastic arch" are practically the equivalent of those based on the theory
of "Least Work."
364. Definitions and Assumptions. — An arcli-rib (or elastic-
arch, as distinguished from a block-work arcb) is a rigid
curved beam, either solid, or built up of pieces like a
truss (and then called a braced arch) the stresses in which,
under a given loading and with prescribed mode of sup-
port it is here proposed to determine. The rib is sup-
posed symmetrical about a vertical plane containing its
axis or middle line, and the Moment of Inertia of any cross
section is understood to be referred to a gravity axis of
the section, which (the axis) is perpendicular to the said
vertical plane. It is assumed that in its strained condi-
tion under a load, the shape of the rib differs so little
from its form when unstrained that the change in the ab-
scissa or ordinate of any point in the rib axis (a curve)
may be neglected when added (algebraically) to the co-
ordinate itself ; also that the dimensions of a cross-section
are small compared with the radius of curvature at any
part of the curved axis, and with the span.
365. Mode of Support. — Either extremity of the rib may be
hinged to its pier (which gives freedom to the end-tangent-
line to turn in the vertical plane of the rib when a load is
applied); or may 'hef,xed, i.e., so built-in, or bolted rigid-
ly to the pier, that the en^-tangent-line is incapable of
changing its direction when a load is applied. A hinge
may be inserted anywhere along the rib, and of course
ARCH BIBS.
439
destroys the rigidity, or resistance to bending at that
point. (A hinge having its pin horizontal "1 to the axis of
the rib is meant). Evidently no more than three such
hinges could be introduced along an arch- rib between two
piers ; unless it is to be a hanging structure, acting as a
suspension-cable.
366. Arch Rib as a Free Body. — In considering the whole
rib free it is convenient, for graphical treatment, that no
section be conceived made at its extremities, if fixed ; hence
in dealing with that mode of support the end of the rib
will be considered as having a rigid prolongation reach-
ing to a point vertically above or below the pier junction,
an unknown distance from it, and there acted on by a force
of such unknown amount and direction as to preserve the
actual 'extremity of the rib and its tangent line in the same
position and direction as they really are. As an illustra-
tion of this Fig. 402
shows free an arch rib.
ONB, with its extremi-
ties and BJixed in the
piers, with no hinges, q
and bearing two
loads P. and P^. The
other . :ces of the sys-
tem holding it in equi-
librium are che horizontal and vertical components, of the
pier reactions {H, V, H,„ and V^), and in this case of fixed
ends each .of these two reactions is a single force not in-
tersecting the end of the rib, but cutting the vertical
through the end in some point F (on the left ; and in G on
the right) at some vertical distance c, (or d), from the end.
Hence the utility of these imaginary prolongations OQF,
and BRG, the pier being supposed removed. Compare
Figs. 348 and 350.
The imaginary points, or hinges, F and G, will be called
ctbutments being such for the special equilibrium polygon
Fig. 402.
440 MBCHAlSriCS OF ENGINEEKING.
(dotted line), while and B are the real ends of the curved
beam, or rib.
In this system of forces there are five unknowns, viz.: V,
V,„ H = H^, and the distances c and d. Their determina-
tion by analysis, even if the rib is a circular arc, is ex-,
tremely intricate and tedious ; but by graphical statics
(Prof, Eddy's method ; see § 350 for reference), it is com-
paratively simple and direct aiid applies to any shape of
rib, and is sufficiently accurate for practical purposes.
This method consists of constructions leading to the loca-
tion of the " special equilibrium polygon " and its force
diagram. In case the rib is hinged to the piers, the re-
actions of the latter act through these hinges, Fig. 403,
i e., the abutments of the special
equilibrium polygon coincide with
the ends of the rib and B, and for
a given rib and load the unknown
quantities are only three V, F'n, and
H; (strictly there are four ; but IX "^^
= gives H^ = H). The solution fig. 403.
by analytics is possible only for ribs of simple algebraic
curves and is long and cumbrous ; 'whereas Prol Eddy's
graphic method is comparatively brief and simple and ia
applicable to any shape of rib whatever.
367. Utility of the Special Equilibriun Polygon and its force
diagram. The use of locating these will now be illustrated
[See § 832]. As proved in §§ 332 and 334 the compres-
sion in each " rod " or segment of the '* special equilibrium
polygon" is the anti-stress resultant of the cross sections in
the corresponding portion of the beam, rib, or other struc-
ture, the value of this compression (in lbs. or tons) being
measured by the length of the parallel ray in the force
diagram. Suppose that in some way (to be explained sub-
sequently) the special equilibrium polygon and its force
diagram have been drawn for the arch -rib in Fig. 404 hav-
ing fixed ends, and B, and no hinges ; required the elastic
stresses in any cross-section of the rib as at m. Let the
ARCH RIBS.
Ml
FiG. 404.
of the force-diagram on the right be 200 lbs. to the
inch, say, and that of the space-diagram (on the left) 30 ft.
to the inch.
The cross section m lies in a portion TK, of the rib, cor-
responding to the rod or segment he of the equilibrium
polygon; hence its anti-stress-resultant is a force R2 acting
in the line 6c, and of an amount given in the force-diagram.
Now i?2 is the resultant of V, H, and Pj, which with the
elastic forces at m form a system in equilibrium, shown in
?ig. 405 ; the portion FO Tm being considered free. Hence
Pig. 405. Fio. 406.
taking the tangent line and the normal at m as axes we
should have I (tang, comps.) = ; -T (norm, comps.) = j
and 2* (moms, about gravity axis of the section at w) = Oj
and could thus find the unknowns pi, "p^, and J", which ap-
pear iu the expressions 'p^F the thrust, ^ the moment* of
442 MECHANICS OF ENGINEERING.
the stress-couple, and J the shear. These elastic stresses
are classified as in § 295, which see. p^ and jpa are ^hs. per
square inch, J is lbs., e is the distance from the horizontal
gravity axis of the section to the outermost element of
area, (where the compression or tension is p^ lbs. per sq.
in., as due to the stress-couple alone) while I is the " mo-
ment of inertia " of the section about that gravity axis.
[See §§ 247 and 295 ; also § 85]. Graphics, however, gives
us a m.ore direct method, as follows : Since i?2> i^ the line
he, is the equivalent of V, H, and Pj, the stresses at m will
be just the same as if ^2 acted directly upon a lateral pro-
longation of the rib at T (to intersect ScFig. 405) as shown
in Fig, 406, this prolongation Tb taking the place of TOF
in Fig. 405. The force diagram is also reproduced here.
Let a denote the length of the "] from m's gravity axis
upon he, and 2 the vertical intercept between m and Jc.
For this imaginary free body, we have,
from I (tang. compons.)=0, i?2 cos a=^piF
and from 2' (norm. compons.)=0,i?2 sin «=«/
while from J' (moms, about) ) ^ ^ rj P2I
,-, ., ^- 4; \ A }-we have ixott =^ -^ •
the gravity axis 01 to)=0, j ^ e
But from the two similar triangles (shaded ; one of them
is in force diagram) a :z :; Zf:i?2 .•. R^a^ Hz, wIh&tigq we
may rewrite these relations as follows (with a general state-
ment), viz.:
If the Special Equilibrium Polygon and Its Force Diagram Have
iBeen Drawn for a given arch-rib, of given mode of support,
p.nd under a given loading, then in any cross-section of the
J ib, we have {F = area of section):
The projection of the proper
i \ \ rri.^T>.,m.f i^ ^ rr- J ^«2/ (of tlie force diagram) up-
(L) The Thrust. i.e.,i>,i^-^ ^^^^^ ^^^^^^^ ^.^^ ^^ ^-^^ ^i^,
drawn at the given section.
ARCH KIBS. 443
(2.) The Shear, i.e., J", = C
/ 1-11 1 J.1 The proieetion of me proper
(upon which dependstne , » ,i « -,. ^
^, . , . ,1 ray (oi the lorce diaarram) up-
shearmg stress m the-^ "^ .; ? , i^i -i
web). (See §§ 253 and
256).
on the normal to the rib curve
at the given section.
(3.) The Moment of the
stress couple, i.e.,-^ , = "
6
The product {Hz) of the fl
(or pole-distance) of the force-
diagram by the vertical dis-
tance of the gravity axis of the
section from the spec, equilib-
rium polygon.
By the ** proper ray " is meant that ray which is parallel
to the segment (of the equil. polygon) immediately under
or above which the given section is situated. Thus in
Fig. 404, the proper ray for any section on TK is B2 ; on
KB, i?3 ; on TO, Bi. The projection of a ray upon any
given tangent or normal, is easily found by drawing through
each end of the ray a line ^ to the tangent (or normal) ;
the length between these "I's on the tangent (or normal) is
the force required (by the scale of the force diagram). We
may thus construct a shear diagram, and a thrust diagram
for a given case, while the successive vertical intercepts
between the rib and special equilibrium polygon form a
moment diagram. For example if the s of a point m is ^
inch in a space diagram drawn to a scale of 20 feet to the
inch, while Zf measures 2.1 inches in a force diagram con-
structed on a scale of ten tons to the inch, we have, for the
moment of the stress-couple at m, J!f=^s= [2.1x10] tons
X [ ^ X 20] ft. =210 ft. tons.
368. — It is thus seen how a location of the special equili
Ibrium polygon, and the lines of the corresponding iovoi
diagram, lead directly to a knowledge of the stresses in al
the cross-sections of the curved beam under consideration,
bearing a given load; or, vice versa, leads to a stateme?^^
of conditions to be satisfied by the dimensions of the rir
for proper security.
It is here supposed that the rib has sufficient latei
444
MECHANICS OF ENGlNEElil^rG,
bracing (witli others wliicli lie parallel witli it) to prevent
buckling sideways in any part like a long column. Before
proceeding to the complete graphical analysis of the differ-
ent cases of arch-ribs, it will be necessary to devote the
next few paragraphs to developing a few analytical rela-
tions in the theory of flexure of a curved beam, and to
giving some processes in " graphical arithmetic."
369. Change in the Angle Between Two Consecutive Rib Tan-
gents when the rib is loaded, as compared with its value
before loading. Consider any small portion (of an arch
rib) included between two consecutive cross- sections ; Fig.
407. KHG W is its unstrained form. Let EA, = ds, be
the original length of this portion of the rib axis. The
length of all the fibres (ii to rib-axis) was originally =ds
(nearly) and the two consecutive tangent-lines, at E and Ay
made an angle = dO originally, with each other. While
under strain, however, all the fibres are shortened equally
an amount dX^, by the uniformly distributed tangential
thrust, but are unequally shortened (or lengthened, accord-
ing as they are on one side or the other of the gravity axis
E, or A, of the section) by the system of forces making
what we call the " stress couple," among v/hich the stress
at the distance e from the gravity axis A of the section is
called p-i per square inch ; so that the tangent line at A'
now takes the direction A'D ~j to H'A'G' instead of A'G
(we suppose the section at E to remain fixed, for coUTezii-
^6! =^
t/"/^'^' cIp.^'
.-r-
"'v-^.^f**
^pd?
AECH lilBS. 445
ence, since tlie change of angle between tlie two tangents
depends on the stresses acting, and not on tlie new posi-
tion in space, of this part of the rib), and hence the angle
between the tangent-lines at E and A (originally = dd) is
now increased by an amount GA'D = d(p (or O'A'R = dip);
G'H' is the new position of GH. We obtain the value of
d(p as follows : That part {dk^ of the shortening of the
fibre at Q, at distance e from A due to the force p.^dFy is
§ 201 eq. (1), dX.2 = ^ft' But, geometrically, J^ also ~edf,
Eedcp-^pzds ■ (1.)
But, letting ilf denote the moment of the stress-couple
at section A (ilf depends on the loading, mode of support,
etc., in any particular case) we know from § 295 eq. (6) that
Jf=-^j and hence by substitution in (1) we. have
•, Mds r^x
'^^^I . • . . (2)
[If the arch-rib in question has less than three hinges,
the equal shortening of the due to the thrust (of
the block in last figure) p^F, will have an indirect effect on
the angle d(p. This will be considered later.]
370. Total Change i.e. CcU in the Angle Between the End
Tangents of a Rib, before and after loading. Take the ex-
ample in Fig. 408 of a rib fixed at one end and hinged at
Fig. 408.
446
MECHAT^ICS OF ENGINEEEIJIG.
the other. When the rib is unstrained (as it is supposed
to be, on the left, its own weight being neglected ; it is not
supposed sprung into place, but is entirely without strain)
then the angle between the end-tangents has some value
6' = j dd— the sum of the successive small angles dd for
do
each element ds of the rib curve (or axis). After loading.,
[on the right, Fig. 408], this angle has increased having
now a value
d'-\- r d(p, i.e., a value = d'+ C -—jr
(I.)
Fig. 409.
There must oe no hinge between
and B.
§ 371. Example of Eq[iiation (I.) in Anal"
ysis. — ^A straight, homogeneous, pris-
matic beam, Fig. 409, its own weight
neglected, is fixed obliquely in a wall.
After placing a load P on the free end,
required the angle between the end-
tangents. This was zero before load-
ing .'. its value after loading is
=o+f'=o+ 4r r'-^^^^
UIJo
By considering free a portion between and any da of the
beam, we find that M=Fx=mom.. of the stress couple.
The flexure is so slight that the angle between any ds and
its dx is still practically =a (§ 364), and .*. ds=dx sec a.
Hence, by substitution in eq. (I.) we have
^'=A rms= l^rxdx=
^ EI Jo EI Jo
P sec ar'*'^*
L2 '
EI
... ^'=?^^^l [Compare with § 237].
ARCH EIBS.
447
It is now apparent that if hoth ends of an arcli rib are
fixed, wlien unstrained, and the rib be then loaded (within
elastic limit, and deformation slight) we must have
r {Mds-^Eiy
zero, since (p'=0.
372. Projections of the Displacement of any Point of a Loaded
Uib Relatively to Another Point and the Tangent Line at the Lat-
ter. — (There must be no hinge between and B). Let
be the point whose displacement is considered and B the
other point. Fig. 410. If ^'s tangent-line is fixed while
the extremity is not supported in any way (Fig. 410)
then a load P put on, is displaced to a new position 0^,
Fig. 410.
Fig. 411,
Fig; 413.
With as an origin and OB as the axis of X, the projec-
tion of the displacement OOj, upon X, will be called Ja?,
that upon Y, Ay.
In the case in Fig.. 410, O's displacement with respect to
B and its tangent-line BT, is also its absolute displacement
in space, since neither B nor BT has moved as the rib
changes form under the load. In Fig. 411, however, the
extremities and B are both hinged to piers, or supports,
the dotted line showing its form when deformed under a
load. The hinges are supposed immovable, the rib being
free to turn about them without friction. The dotted line
is the changed form under a load, and the absolute dis-
placement of is zero ; but not so its displacement rela-
tively to B and j5's tangent BT, for BT has moved to a
new position BT'. To find this relative displacement con-
ceive the new curve of the rib superposed on the old in
a way that B and BT m&j coincide with their original po-
448
MECHANICS OF ENGIIsTEEEING.
sitions. Fig. 412. It is now seen that O's displacement
relatively to B and BT is not zero but =00„, and lias a
small Jx but a comparatively large zly. In fact for this
case of hinged ends, piers immovable, rib continuous be-
tween 'them, and deformation slight, we shall write Jx =
zero as compared with Jy, the axis X passing through OB).
373. Values of the X and Y Projections of O's Displacement Rela-
tively to Band B's Tangent; the origin being taken at 0.
Fig. 413. Let the co-
ordinates of the dif-
ferent points jE, I), G,
etc., of the rib, re-
ferred to and an
arbitrary X axis, be
X and y, their radial
distances from be-
. ing u (i.e., u for G, u'
for D, etc.; in gener-
al, ^0- OEDG is the J
unstrained form of the
rib, (e.g., the form it
would assume if it lay flat on its side on a level platform,
under no straining forces), while 0,,E"B'GB is its form
under some loading, i.e., under strain. (The superposi-
tion above mentioned (§ 372) is supposed already made if
necessary, so that BT i^ tangent at B to both forms).
Now conceive the rib OB to pass into its strained condi-
tion by the successive bending of each ds in turn. The
straining or bending of the first ds, BC, through the small
angle d(p (dependent on the moment of the stress couple
at G in the strained condition) causes the whole finite piece
OG io turn about (7 as a centre through the same small
angle d(p ; hence the point describes a small linear arc
00'=ov, whose radius = u the hypothenuse of the x and
y of G, and whose value .*. is dv=ud(p.
Next let the section B, now at D\ turn through its
proper angle d(p' (dependent on its stress-couple) carrying
Fig. 413.
AECH KIBS. ' 449
with it the portion D'O', into the position D'O", making
0' describe a linear arc O'O" = {dvy =u'd(p', in which u' =
the hypothenuse on the x' and y' (of D), (the deformation
is so slight that the co-ordinates of the different points
referred to and X are not appreciably affected). Thus,
each section having been allowed to turn through the an-
gle proper to it, finally reaches its position, 0„, of dis-
placement. Each successive dv, or linear arc described by
0, has a shorter radius. Let dx, {dx)', etc., represent the
projections of the successive (^v)'s upon the axis X; and
similarly dy, (dy)' etc., upon the axis Y. Then the total X
projection of the curved line . . . . 0^ will be
Jx= / (5j? and similarly J?/ = / dy , , , (1)
But d V = u d (f, and from similar -right-triangles,
3 x: dv : : y : u and dy : 8v :: x : u .'. 8x = yd
"m.as .j ^
tangent-lines and ^ [ ~ Jo ^ ' o • • W
The X-Projection of O's Displacement "]
Relatively to B and B's tangent- I A^Myds /tt ^
line ; {the origin being at 0) I — / — ^fjr- • • • (JJ-J
and the axes X and F 1 to [ ^o i^i
each other) •
450
MECHAKICS OF EKGINEEKING.
The Y-Prqjection of O's Displacement, | _
etc., as above.
. (m.)
Hviie X anv.. y are the co-ordinates of points in the rib-
curve, ds an element of that curve, M the moment of the
stress-couple in the corresponding section as induced by
the loading, or constraint, of the rib.
(The results already derived for deflections, slopes, eto„,
for straight beams, could also be obtained from these
formulae, I., 11. and III. In these formulae also it must
be remembered that no account has been taken of the
shortening of the rib-axis by the thrust, nor of the effect
of a change of temperature.)
374a. R^sumfe of the Properties of Ec^mlibiiiim Polygons and
their Force Diagrams, for Systems of Vertical Loads. — See §§ 335
to 343. Given a system of loads or vertical forces, P^, P2,
1 etc., Eig. 414, and
two abutment verti-
cals, F' and G' ; if
"we lay off, vertically,
to form a " load-
line," 1 .. 2 - P^, 2. . .
8=P2> etc., select any
Pole, Oi, and join 0^
... 1, Oi . . . 2, etc. ;
also, beginning at
any point F^ in the
vertical P', if we draw i^i . . . a I| to Oj . . 1 to intersect the
line of Pi ; then ah \\ to Oi . . 2, and so on until finally a
point G]y in G', is determined; then the figure Pj ,ahc G^iis
an equilibrium polygon for the given loads and load verti-
cals, and Oi . . . 1234 is its ". force diagram." The former
is so called because the short segments PjCt oh, etc., if
considered to be rigid and imponderable rods, in a vertical
plane, hinged to each other and the terminal ones to abut-
ments Pi and G^, would be in equilibrium under the given
loads hung at the joints. An infinite number of equilil>
Fig. 414.
AECH-HIBS. 451
rium polygons may be drawn for tlie given loads and
abntment-verticals, by choosing different poles in the force
diagram. [One other is shown in the fignie ; O2 is its
pole. {Fi Gi and F2 U-^ are abutment lines.)] For all of
these the following statements are true :
(1.) A line through the pole, i| to the abutiifint line cutii
the load-line in the same point n', whichever equilibrium
polygon be used ( /. any one will serve to determine n'
(2.) If a vertical GI) he drawn, giving an intercept z' in
each of the equilibrium polygons, the product Hz' is the
same for all the equilibrium polygons. That is, (see Fig.,
414) for any two of the polygons we have
H,:H,:: z/ : z,' ; or H,z,' = H, z,'.
(3.) The compression in each rod is given by thai
" ray " (in the force diagram) to which it is parallel.
(4.) The " pole distance " H, or ~| let fall from the pole
upon the load-line, divides it into two parts which are the
vertical components oi the compressions in the abutment-
rods respectively ( the other component being horizontal) ;
H is the horizontal component of each (and, in fact, of
each of the compressions in all the other rods). The
compressions in the extreme rods may also be called the
abutment reactions (oblique) and are given byti^e extreme
rays.
(5.) Three Points [not all in the same segment (or rod)]
determine an equilibrium polygon for given loads. Hav-
ing given, then, three points, we may draw the eaailibrium
polygon by §341.
«
375. Summation of Prcducts. Before proceedini^ to treat
graphically any case of arch-ribs, a few processes in
graphical arithmetic, as it may be called, must be pre-
sented, and thus established for future use.
To make a summation of products of two factors in each
by means of an equilibrium polygon.
452
MECHANICS OF ENGINEERING.
Construction, Suppose it required to make the summa-
tion I {x z) {. e., to sum the series
Xi %+ X2 Z2 + x^z^ 4.
bj graphics.
Having first arranged the terms in the order of magnf-
tude of the ic's, we proceed as follows : Supposing, for
illustration, that two of the s's (% and z^ are negative
(clotted in figure) see Fig. 415. These quantities x and z
may be of any nature whatever, anything capable of being
represented by a length, laid off to scale.
First, in Fig
416, lay off the
s's in their
order, end to
end, on a ver-
tical load-line
taking care to
"^ lay off % and
.. % upuard in
^« their turn.
Take any con-
FiG. 416. venient j-ola
; draw the rays ... 1, ... 2, etc.; then, having pre-
viously drawn vertical lines whose horizontal distances
from an extreme left-hand vertical F' are made = x^, x-,
Xs, etc., respectively, we begin at any point F, in the verti-
cal F', and draw a line 11 to ... 1 to intersect the Xi ver-
tical in some point ; then 1' 2' II to . . . % and so on, fol-
lowing carefully the proper order. Produce the last seg-
ment (6' ... (x in this case) to intersect the vertical F' in
some point K. Let KF =k (measured on the same scale
as the i»'s), then the summation required is
J/ {xz) = m.
H is measured on the scale of the 2's, which need, not be
the same as that of the aj's ; in fact the 2's may not be the
same kind of quantity as the a;'s.
[Peoof. — From similar triangles H: z^v.x^^: h^, .'. x^z^ — IHc^ \
and " " " H\Zo :: x^ : ^2> •*• x.^Zi=Hki .
Fig. 415.
AECH-KIBS.
453
and so on. But H {h,+h+eiG.)^HxFK=Hh'].
376. Gravity Vertical. — From the same construction in
Fig. 415 we can determine tlie line of action (or gravity
vertical) of tlie resultant of the parallel vertical forces 2i,
Z2, etc. (or loads); by prolonging the first and last segments
to their intersection at
0. The resultant of the
system of forces or loads
acts through C and is
vertical in this case ; its
value being = ^ (2),
that is, it = the length
1 ... 7 in the force dia-
gram, interpreted by the
proper scale. It is now
supposed that the 2's
represent forces, the x'b,
being their respective
lever arms about F. If
the ?'s represent the
areas of small finite por-
tions of a large plane
figure, we may find a
gravity -line (through C)
of that figure by the
above construction; each
z being-applied through
the centre of gravity of
its own portion.
Calling the distance
X between the verticals
through C and F, we
have also x . I [z) =
I (xz) because I (z) is
the resultant of the || z's.
^' This is also evident from
the proportion (similar
triangles)
H : (1. .7)::x:Jc.
454 MECHANICS OF ENGINEERING.
376a. Moment of Inertia (of Plane Figure) by Graphics.* —
rig. 416a. /n= ? First, for the portion on right. Divide OR
into equal parts each = ^x. Let «i, Z2, etc., be the middle
ordinates of the strips thus obtained, and x^, etc. their
abscissas (of middle points).
Then we have approximately
/n for 0R=Ax.ZyX^-\-Ax.Z2xi-{-
=Ax[{z^Xi)x^+{z2X^X2-\- ...]..(!)
But by §375 we may construct the products ZyXi,Z2X2, etc.,
taking a convenient H\ (see Fig. 416, (&)), and obtain \, ki,
etc., such that z^x^ = H'ki, z^x^ = H'k2i etc. Hence eq. (1)
becomes :
li,ioT OR a.-p-prox.=II'^x[kiXi-\-k2X2-{- ...]... (2)
By a second use of § 375 (see Fig. 416 c) we construct Z,
such that kiX^ + kzXz +....= £["l \^H" taken at con-
venience]. .'. from eq. (2) we have finally, (approx.),
In for OR=H'H"lAx (3)
For example if OR — 4 in., with four strips. Ax would =
1 in.; and if ^' = 2 in., H" = 2 in., and I = 5.2 in., then
Jn for OR = 2x2x5.2x1.0=20.8 biquad. inches.
The 7x for OL, on the left of N, is found in a similar
manner and added to 7^ for OR to obtain the total I^. The
position of a gravity axis is easily found by cutting the
shape out of sheet metal and balancing on a knife edge ; or
may be obtained graphically by § 336 ; or 376.
377. Construction for locating a line vw (Fig. 417) at (a), in
the polygon FG in such a position as to satisfy the two
following conditions with reference to the vertical inter-
cepts at 1, 2, 3, 4, and 5, between it and the given points
1„ 2, 3, etc., of the perimeter of the polygon.
* Another graphic method for this purpose will be found in § 76 (p. 80),
of the author's Notes and Examples in Mechanics.
ARCH-EIBS.
455
Condition I. — (Calling these intercepts u^, u^, etc., and tlieir
horizontal distances from a given vertical F, x^, x.^, etc.)
2" (u) is to = ; i.e., the sum of the positive u's must be
numerically — - that of the negative (which here are at 1
and 5). An infinite number of positions of vm will satisfy
condition I.
Condition II. — 2* (ux) is to = ; i.e., the sum of the
ij 1 1 . r-— ;^?n moments of the positive u'^
• — ■' ■ — "^^^^ ' about F must = that of the
'' negative -m's. i.e., the moment
of the resultant of the posi-
tive w's must = that of the
resultant of the negative ;
and .*. (Condit. I being
already satisfied) these two
resultants must be directly
opposed and equal. But the
ordinates u in (a) are indi-
vidually equal to the difiFer-
ence of the full and dotted
ordinates in (&) with the
same cc's .'. the conditions
may be rewritten :
I. 2 (full ords. in (6))=
2" (dotted ords. in (&))
II. 2 [each full ord. in (h)
X its £c] = - [each dotted
ord. in (b) x its x] i.e., the
Fig. 4ir. Centres of gravity of the full
and of the dotted in (6) must lie in the same vertical
Again, by joining ^(x, we may divide the dotted ordi-
nates of (b) into two sets which are dotted, and broken, re-
spectively, in (c) Then, finally, drawing in (d),
B, the resultant of full ords. of (c)
T, " " " broken " " "
T', " " " dotted " " "
456 MECHANICS OF ENGINEERING.
we are prepared to state in still another and final form tlie
conditions wliicli vm must fulfil, viz. :
(I.) T+T must = i?; and (II.) The resultant of T
and T' must act in the same vertical as R.
In short, the quantities T, T', and R must form a bal-
anced system, considered as forces. All of which amounts
practically to this : that if the verticals in which T and T'
act are known and R be conceived as a load supported by
a horizontal beam (see foot of Fig. 417, last figure) resting
on piers in those verticals, then T and T' are the respec-
tive reac^'^ons o/" ^Aose jjiers. It will now be shown that the
verticals of T and T' are easily found, being independent of
the position of vm; and that both the vertical and the mag-
nitude of R, being likewise independent of vm, are deter-
mined with facility in advance. For, if v be shifted up
or down, all the broken ordinates in (c) or {d) will change
in the same proportion (viz. as vF changes), while the
dotted ordinates, though shifted along their verticals, do
not change in value ; hence the shifting of v affects neither
the vertical nor the value of T', nor the vertical of T.
The value of T, however, is proportional to vF. Similar-
ly, if m be shifted, up or down, T' will vary proportionally
to mG, but its vertical, or line of action, remains the same.
T is unaffected in any way by the shifting of m. R, de-
pending for its value and position on the full ordinates of
(c) Fig. 417, is independent of the location of vm. We
may .*. proceed as follows :
1st. Determine R graphically, in amount and position,
by means of § 376.
2ndly. Determine the verticals of T and T' by any trial
position of vm (call it v-im.,), and the corresponding trial
values of T and T' (call them T, and T',).
3rdly. By the fiction of the horizontal beam, construct
(§ 329) or compute the true values of T and T', and then
determine the true distances vF and w6^ by the propor-
tions
vF : v.F :: T : T. and mG : m,G : : T' i T^.
AECH-EIBS.
457
Example of this. Fig. 418. (See Fig. 417 for s and t.)
From A tovi^ard B in (e) Fig. 418, lay off the lengths (or
lines proportional
to them) of the full
ordinates 1, 2, etc.,
of (/). Take any
pole Oi, and draw the
equilibrium poly-
gon {/y and pro-
long its extreme seg-
ments to find C and
thus determine ^'s
vertical. JR is repre-
sented by AB. In
(g) [same as (/) but
shifted to avoid
complexity of lines]
draw a trial VoWi and
join V2 G2. Deter-
mine the sum T2 of
the broken ordi- TFig. 418.
nates (between V2G2 ana ^^2^2) and its vertical line of ap-
plication, precisely as in dealing with B ; also T'2 that of
the dotted ordinates (five) and its vertical. Now the true
T=Btj-{s+t) and the true T'=Bs^(s+t). Hen ce com-
pute vF={T^T2) ^2 and ?^^=(T'-^^^) m^G^., and by
laying them off vertically upward from F and G respec-
tively we determine v and m, i.e., the line vm to fulfil the
conditions imposed at the beginning of this article, rela-
ting to the vertical ordinates intercepted between vm and
given points on the perimeter of a polygon or curve.
Note (a\ If the verticals in which the intercepts lie are
equidistant and quite numerous, then the lines of action
of T2 and T'2 will divide the horizontal distance between
F and G into three equal parts. This will be exactly true
in the application of this construction to § 390.
Note (b). Also, if the verticals are symmetrically placed
about a vertical line, (as will usually be the case) VjWg is
458
MECHANICS OF ENGINEERING.
best drawn parallel to FG, for then T^ and T'^ will be
equal and equi-distant from said vertical line.
378, Classification of Arch-Eibs, or Elastic Arches, according
to continuity and modes of support. In the accompany^
ing figures Htxefull curves show the unstrained form of the
rib (before any load, even its own weight, is permitted to
come upon it) ;the dotted curve shows its shape (much ex-
aggerated) when bearing a load. For a given loading
Three Conditions must be given to determine the special
equilibrium polygon (§§ 366 and 367).
Class A. — Continuous rib, free to slip laterally on the
piers, which have smooth horizontal surfaces. Fig. 420.
This is chiefly of theoretic interest, its consideration
being therefore omitted. The pier reactions are neces-
sarily vertical, just as if it were a straight horizontal
beam.
Class B. Rib of Three Hinges, two at the piers and one
intermediate (usually at the crown) Fig. 421. Fig. 36 also
is an example of this. That is, the rib is discontinuous
and of two segments. Since at each hinge the moment of
tlie stress couple must be zero, the special equilibrium
polygon must pass through the hinges. Hence as three
points fully determine an equilibrium polygon for given
load, the special equilibrium is drawn by § 341.
Fig. 420.
Fig. 421.
[§ 378a will contain a construction for arch-ribs of three
hinges, when the forces are not all vertical.]
Class C. Rib of Two Hinges, these being at the piers, the
rib continuous between. The piers are considered im-
movable, i.e., the span cannot change as a consequence of
loading. It is also considered that the rib is fitted to its
AKCH RIBS.
459
hinges at a definite temperature, and is tlien under no con-
straint from the piers (as if it lay flat on the ground), not
even its own weight being permitted to act when it is fi-
nally put into position. When the " false works "
or temporary supports are removed, stresses are in-
duced in the rib both by its loading, including its
own weight, and by a change of temperature. Stresses
due to temperature may be ascertained separately and
then combined with those duo to the loading. [Classes
A and B are not subject to temperature stresses.] Fig.
422 shows a rib of two hinges,
at ends. Conceive the dotted
curve (form and position un-
der strain) to be superposed
on the continuous curve
(form before strain) in such
a way that B and its tangent
line (which has been dis-
placed from its original position) may occupy their pre-
vious position. This gives us the broken curve O^B. 00,^
is .*. O's displacement relatively to B and -S's tangent,
Now the piers being immovable OqB (right line) =05 ; i.e.,
the X projection (or Jx) of OOn upon OB (taken as an axis
of X) is zero compared with its Jy. Hence as one condi-
tion to fix the special equilibrium polygon for a given load-
ing we have (from § 373)
Fig. 422.
r^[Myds-^EI^=0
(1)
The other two are that the [ must pass through . (2)
special equilibrium polygon ) " " " B . (3)
Class D. Bill with Fixed Ends and no hinges, i.e., continu-
ous. Piers immovable. The ends may be ^xed by being
inserted, or built, in the masonry, or by being fastened to
large plates which are bolted to the piers. [The St. Louis
Bridge and that at Coblenz over the Rhine are of this
class.] Fig. 423. In this class there being no hinges we
460
MECHANICS OF ENGINEERING.
Fig. 423.
have no point given in advance througli whicli the special
equilibrium polygon must pass. However, since O's dis-
placement relatively (and absolutely) to B and ^'s tangent
is zero, both z/a:; and z/^[see § 373] = zero, AIsq the tan-
gent-lines both at and B being
fixed in direction, the angle be-
tween them is the same under
loading, or change of temperature,
as when the rib was first placed
in position under no strain and at
a definite temperature.
Hence the conditions for locating the special equilibrium
polygon are
p^ Mds _ Q . p Myds ^ ^ . n^ Mxds _ q
Jo ^jT ' Jo '~m~ ' Jo EI
In the figure the imaginary rigid prolongations at the
ends are shown [see § 366].
Other designs than those mentioned are practicable
(such as : one end fixed, the other hinged ; both ends fixed
and one hinge between, etc.), but are of unusual occur-
rence.
378a. Eib of Three Hinges, Forces not all Vertical,* If the
given rib of three hinges upholds a roof, the wind-press-
ure on which is to be considered as well as the weights of
the materials composing the roof-covering, the forces will
not all be vertical. To draw the special equil, polygon in
such a case the following
construction holds : Re-
quired to draw an equilib-
rium polygon, for any
plane system of forces,
through three arbitrary
xs^ points. A, p and B ; Fig,
B 423a. Find the line of
action of B^, the resultant
of all the forces occurring
between A and p; also,
Fig. 423a.
* See p. 117 of the author's "Notes and Examples io Mechanics" for >
detailed example of the following construction.
ARCH-EIBS. 4C1
that of R,, tlie resultant of all forces between ^p and B ;
also the line of action of B, tlie resultant of B^ and B.2, [see
§ 328.] Join any point iH^ in ^ witli A and also witli B,
and join the intersections iVand 0. Then A iV will be the
direction of the first segment, B that of the last, and
NO itself is the segment corresponding to p (in the de-
sired polygon) of an equilibrium polygon for the given
forces. See § 328. If A N' p 0' B are the corresponding
segments (as yet unknown) of the desired equil. polygon,
we note that the two triangles MNO and M'N' O, having
their vertices on three lines which meet in a point [i.e., B
meets Bi and B^ in C], are homological [see Prop. YII. of
Introduc. to Modern Geometry, in Chauvenet's Geometry,]
and that . • . the three intersections of their corresponding
sides must lie on the same straight line. Of those inter-r
sections we already have A and B, while the third must be
at G, found at the intersection of AB and NO. Hence by
connecting C and p, we determine N and 0'. Joining
N'A aiid O'B, the first ray of the required force diagram will
be II to NA, while the last ray will be || to O'B, and thus
the pole of that diagram can easily be found and the cor-
responding equilibrium polygon, beginning at A, will pass
through p and B.
(This general case includes those of §§ 341 and 342.)
379. Arch-Rib of two Hinges; by Prof. Eddy's Method.*
[It is understood that the hinges are at the ends.] Re-
quired the location of the special equilibrium polygon. "VVe
here suppose the rib homogeneous (i.e., the modulus of
Elasticity E is the same throughout), that it is a " curved
prism " (i.e., that the moment of inertia / of the cross-
section is constant), that the piers are on a level, and that
the rib-curve is symmetrical about a vertical line. Fig.
424. For each point m of the rib
curve we have an x and y (both
known, being the co-ordinates of
the point), and also a z (intercept
between rib and special equilib-
FiG. 424. ^ rium polygon) and a z' (intercept
*P. S5 of Prof. Eddy's book ; see reference in preface of this work.
462
MECHANICS OF ENGINEEEING.
between tlie spec. eq. pol. and the axis X (whicli is OB).
The first condition given in § 378 for Class C may be
transformed as follows, remembering [§ 367 eq. (3)] that
M = Hz at any point m of the rib (and that EI is con-
stant).
1^
EI
H
r Myds = 0, i.e., — C zyds = . • . f zyds
do El c/o t/o
^^y _^, \-''J^(y~ ^')yds=0', i.e., J^ yyds =J^ yz'ds . (1)
In practical graphics we can not deal with infinitesimals ;
hence we must substitute As a small finite portion of the
rib-curve for ds', eq. (1) now reads I^ yy As = 2'^ yz' As.
But if we take all the As's equal, As is a common factor
and cancels out, leaving as a final form for eq. (1)
I^\yy) = I^^{yz') . . . (1/
The other two conditions are that the special equilibrium
polygon begins at and ends at B. (The subdivision of
the rib-curve into an even number of equal As's will be ob-
served in all problems henceforth.)
379a. Detail of the Construction. Given the arch-rib B,
Fig. 425, with specified loading. Divide the curve into
Fig. 425.
ARCH RIBS.
4G;
eight equal ^s's and draw a vertical through the middle
of each. Let the loads borne by the respective ^s's be
Pi, P2, etc., and with them form a vertical load-line A C to
some convenient scale. With any convenient pole 0"
draw a trial force diagram 0" AC, and a corresponding
trial equilibrium polygon F G, beginning at any point in
the vertical F. Its ordinates %", 22", etc., are propor-
tional to those of the special equil. pol. sought (whose
abutment line is OB) [§ 374a (2)]. We next use it to de-
termine n^ [see § 374a]. We know that OB is the " abut-
ment-line " of the required special polygon, and that . ' .
its pole must lie on a horizontal through n'. It remains
to determine its H, or pole distance, by equation (1)' just
given, viz. : IJ^ yy = Sfyz'. First by § 375 find the value
of the summation Ii{yy), which, from symmetry, we may
write = 2i'/(2/2/) =2 [2/12/1+2/22/2+2/32/3+2/42/4]
Hence, Fig. 426, we obtain
11 {yy)=2 [HM
Next, also by § 375, see Fig.
427, using the same pole dis-
tance Ho as in Fig. 426, we
find
I\{yz")=HA"; i.e.,
+2/22:2'' + 2/3%" +2/4^'.=
Again, since II {yz") = ysz/'
+ 2/7^7" + 2/6^6" + 2/5^5" which
from symmetry (of rib)
=2/i%"+2/2^7"+2/326"+2/'<',
we obtain, Fig. 428,
l"! (2/O = ^oV', (same /^,);
and .-.
Jf {yz")=ff, {\"+hJ'). If now we find that /fc/'+/b/'=2yfc.
464 MECHANICS or engineeking.
the condition 2^1 (yy) = II {y^'') is satisfied, and the pole
distance of our trial polygon in Fig. 425, is also that of
the special polygon sought; i.e., the z" 's.are identical in
value with the s"s of Fig. 424. In general, of course, we
do not find that ky'~{-k/' = 2A;. Hence the z" 's must all
be increased in the ratio 2k: {lc^"-\~k/') to become equal to
the g"s. That is, the pole distance H of the spec, equil*
polygon must be
7j-_ ki'-\-'k/' jT,, (in which W = the pole distance of the
2^c trial polygon) since from §339 the ordi-
nates of two equilibrium polygons (for the same loads)
are inversely as their pole distances. Having thus found
the if of the special polygon, knowing that the pole must
lie on the horizontal through n', Fig. 425, it is easily
drawn, beginning at 0. As a check, it should pass through
B.
For its utility see § 367, but it is to be remembered that
the stresses as thus found in the different parts of the
rib under a given loading, must afterwards be combined
with those resulting from change of temperature and the
shortening of the rib axis due to the tangential thrusts,
before the actual stress can be declared in any part.
Note. — Variable Moment of Inertia. If the / of the rib section is dif-
ferent at different sections we may proceed as follows: For eq. (1); we
PB ds i'B ds
now write I yy y = I yz'—-. Taking the I of the crown section (say)
Jo i Jo I
as a standard of reference, denoting it by /', we may write for any other
section I = nl', where n is a variable ratio, or abstract number; whence
eq. (1) becomes, after putting Js for ds, y / 2/J/— ="77 / V^—-
If now the length of each successive Js, from the crown down, be made
directly prop>ortional to the number n at that part of the rib, the quantity
^ s^n will have the same value in all the terms of each summation and
may be factored out ; and we then have a relation identical in form with
eq. (1)', but with the understanding that the j/'s and 2"s concerned are
those in the successive verticals drawn through the mid-points of the
unequal -s's, or subdivisions along the rib, obtained by following the
above plan that each As is proportional to the value of the moment of
inertia at that part of the rib. For instance, if the / of a section near
the hinge 0, or B, is three times that (/') at the (3rown, then the length
of the Js at the former point must be made three tim,es the length of
the As first assumed at the crown when the subdivision is begun. By
a little preliminary investigation, a proper value for this crown , s may
be decided upon such that the total .number of As's shall be sufficient
for accuracy (sixteen or twenty in all)
ABCn-KIBS.
465
f-$BO. Arch Rib of Fixed Ends and no Hinges,— Example of
Class D. Prof. Eddy's Method.* As before, E and / are
constant along tlie rib Piers immovable. Rib curve
symmetrical about a vertical line. Fig. 429 shows such a
rib under any loading. Its span is OB, wliicli is taken as
an axis X. The co-ordinate of any point m' of the rib
curve are x and y, and z is the vertical intercept between
w' and the special equilibrium polygon (as yet unknown,
but to be constructed). Prof. Eddy's method will now be
■given for finding tha spe-
cial equil. polygon. The
three conditions it
must satisfy (see § 378,
Class D, remembering
that E and /are constant
and that M — ITz from
§ 367) are
H-« ^^
Fig. 429.
/ zds=^ ; / xzds— ; and / yzds =0
e/o e/o e/o
(1)
Now suppose the auxiliary reference line (straight) vm
to have been drawn satisfying the requirements, with
respect to the rib curve that
/ z'ds—0 ; and / xz'ds=Q
e/o c/o
(2)
in which z' is the vertical distance of any point m' from
vm and x the abscissa of mf from 0.
From Fig. 429, letting z" denote the vertical intercept
(corresponding to any m') between the spec, polygon and
the auxiliary line vm, we have z=z'—z", hence the three
conditions in (1.) become
r{z'-z")ds=0; i.e., see eqs. (2) C^. z"ds=0 , .. (3)
* p. 14 of Prof. Eddy's book ', see reference in preface of this work.
466 mecha:nics of engineeeing.
B B
fx {z'—^')ds=0 ; i.e., see eqs. (2) f xz"ds=o (4/
^nifh'-^)ds=;0,^7^^Zl-f}.'ds=fkds . (5)
provided vm has been located ^s prescribed.
For graphical purposes, having subdivided the rib curve
into an even number of small equal J.s's, and drawn a verti-
cal through the middle of each, we first, by § 377, locate
vm to satisfy the conditions
ll{z')=0 and l^,{xz')=0 . . (6)
(see ec[. (2) ; the di cancels out) ; and then locate the
special equilibrium polygon, with vm as a reference-line,
by making it satisfy the conditions.
:EI{z'')=0 . (7); Il{xz")=Q . (8); I^Xyz")^l^Xyz') . (9)
(obtained from eqs. (3), (4), (5) by putting ds = ^s, and can-
celling).
Conditions (7) and (8) may be satisfied by an infinite
number of polygons drawn to the given loading. Any one
of these being drawn, as a trial polygon, we determine for it
the value of the sum l'f^(yz") by § 375, and compare it with
the value of the sum l'^,{yz') which is independent of ihe
special polygon and is obtained by § 375. [N.B. Itmist
be understood that the quantities (lengths) x, y, z, z\ and z" ,
kere dealt with are thosa pertaining to the verticals drawn
through the middles of the respective ^s's, which must be
sufficiently numerous to obtain a close result, and not to
the verticals ia which the loads act, necessarily, since these
latter may be few or many according to circumstances, see
Fig. 429]. If these sums are not equal, the pole distance
of the trial equil. polygon must be altered in the proper
ratio (and thus change the 2;'"s in the inverse ratio) neces-
sary to make these sums equal and thus satisfy conditicn
(9). The alteration of the 2'"s, all in the same ratio, will
AECH-EIBS.
4G7
aot interfere with conditions (7) and (8) whicli are alreadj^
satisfied.
381. Detail of Construction of Last Problem. Symmetrical Arch-
Rib of Fixed Ends. — As an example take a span of the St.
Louis Bridge (assuming /constant) "with. " live load'' cov-
sring the half span on the left. Fig. 430, where the verticaJ
feME:^^d==Ui
Fig. 430.
scale is much exaggerated for the sake of distinctness*.
Divide into eight equal Js's. (In an actual example sixteen
or twenty should be taken.) Draw a vertical through the
* Each arch -rib of the St. Louis bridge is a built up or trussed, rib of steel about 53i
ft. span and 52 ft. rise, ia the form of a segment of a circle . Its moment of inertia,
however, is not strictly constant, the portions near each pier, of a length equal to one
twelfth of the span, having a value of / one-half greater than that of the remainder at
the arc.
468 MECHANICS OF ENGINEERING.
middle of each ^s. P^ , etc., are the loads coming upon
the respective Js's.
First, to locate vm, by eq. (6) ; from symmetry it must
be horizontal. Draw a trial vm (not phown in the figure),
and if the (-{- 8')'^ exceed the (— 2')'s by an amount z^, the
true vm will lie a height —z' above the trial vm (or below,
if vice versa) ; n = the number of z/s's.
Now lay off the load-line on the right (to scale),
take any convenient trial pole 0'^' and draw a correspond-
ing trial equil. polygon F'"G"\ In r"G"', by §377,
locate a straight line v"'m!" so as to make 2^(2'") = and
^l(xz!") = (see Note (&) of § 377).
[We might now redraw F'" G'" in such a way as to bring
v"'m!" into a horizontal position, thus : first determine a
point n'" on the load-line by drawing 0"'n"' \ to v"'m"' ,
take a new pole on a horizontal through n'" , with the same
II'" , and draw a corresponding equil. polygon ; in the lat-
ter v"'m"' would be horizontal. We might also shift this
new trial polygon upward so as to make v"'m!" and vm.
coincide. It would satisfy conditions (7) and (8), having
the same %'"'''& as the first trial polygon ; but to satisfy con-
dition (9) it must have its 2""s altered in a certain ratio,
which we must now find. But we can deal with the individ-
ual 2""s just as well in their present positions in Fig. 430.]
The points ^and L in vm, vertically over E'" and L'" in
v"'m'", are now fixed ; they are the intersections of the special
polygon 7'equired, ivith vm.
The ordinates between v"'m"' and the trial equilibrium
polygon have been called z'" instead of z" ; they are pro-
portional to the respective g"'s of the required special
polygon.
The next step is to find in what ratio the (s'")'s need to
be altered (or H'" altered in inverse ratio) in order to be-
come the {z"^^ ; i.e., in order to fulfil condition (9), viz. :
AUCH-EIBS.
469
^.{yz")=I\{yz') . (9)
This may be done pre-
cisely as for tlie rib with
two hinges, but the nega-
tive (s'")'s must be prop-
erly considered (§ 375)
See Fig. 431 for the de-
tail. Negative 2;"s or g""s
point upward.
From Fig. 431a
[j ' .*. from symmetry
I\{yz')=2H^h
From Fig. 4315 we have
riyz"')=HX
Pig. 431.
and from Fig. 431c
Il{yn=Ho^
[The same pole distance H^ is taken in all these construc-
tions] .♦. I\yz")=H,{k,-\-\).
If, then, Ho {\-\-k,) = 2HJc condition (9) is satisfied by the
z""s. If not, the true pole dista-nce for the special equil.
polygon of Fig. 430 will be
2k '
With this pole distance and a pole in the horizontal through
n'" (Fig. 430) the force diagram may be completed for the
required special polygon ; and this latter may be con-
structed as follows : Beginning at the point E, in vm,
through it draw a segment || to the proper ray of the force
diagram. In our present figure (430) this " proper ray "
would be the ray joining the pole with the point of meet-
ing of P2 and Pi on the load-line. Having this one seg-
470 MECHANICS OF EXGIXEEEING.
ment of the special polygon the others are added in an
obvious manner, and thus the whole polygon completed.
It should pass through L, but not and B.
For another loading a different special equil. polygon
would result, and in each case we may obtain the tkrusty
shear, and moment of stress couple for any cross-section of
the rib, by § 367. To the stresses computed from these,
should be added (algebraically) those occasioned by change
of temperature and by shortening of the rib as occasioned
by the thrusts along the rib. These " temperature
stresses," and stresses due to rib-shortening, will be con-
sidered in a subsequent paragraph. They have no exist-
ence for an arch-rib of three hinges.
Note. — If the moment of inertia of the rib section is
variable, instead of dividing the rib axis into equal Js's,
we should make them unequal, following the plan indicated
in the note on p. 464, the As being made proportional to
the values of the moment of inertia along the rib. After such
subdivision is made, and a vertical drawn through the mid-
point of each Js, the various ^'s, z^'s, etc., in these verticals
are dealt with in the same manner as just shown for the
case of constant moment of inertia.
381a. Exaggeration of Vertical Dimensions of Both Space and
Force Diagrams. — In case, as often happens, the axis of the
given rib is quite a flat curve, it is more accurate (for find-
ing M) to proceed as follows :
After drawing the curve in its true proportions and pass-
ng a vertical through the middle of each of the equal
z/s's, compute the ordinate (y) of each of these middle points
from the equation of the curve, and multiply each y by
four (say). These quadruple ordinates are then laid off
from the span upward, each in its proper vertical. Also
multiply each load, of the given loading, by four, and then
with these quadruple loads and quadruple ordinates, and
the upper extremities of the latter as points in an exagge-
rated rib-curve, proceed to construct a special equilibrium
polygon, and the corresponding force diagram by the
proper method ( for Class B, C, or D, as the case may be)
for this exaggerated rib -curve.
The moment, Hz, thus found for any section of the ex-
AECn-KIBS. 471
aggerated rib-curve, is to be divided by four to obtain the
moment in tlie real rib, in tlie same vertical line. To find
the thrust and shear, however, for sections of the real rib,
besides employing tangents and normals of the real rib W9
must draw, and use, another force diagram, obtained from
the one already drawn (for the exaggerated rib) by re-
ducing its vertical dimensions (only), in the ratio of four
to one. [Of course, any other convenient number besides
four, may be adopted throughout.]
382. Stress Diagrams. — Take an arch -rib of Class D, § 378,.
i.e., of fixed ends, and suppose that for a given loading (in-
cluding its own weight) the special , ^^^^ ^thrust
equil. polygon and its force diagram
have been drawn [§ 381]. It is re- "^^^^ " —coopte-
quired to indicate graphically the
variation of the three stress-elements
for any section of the rib, viz., the
thrust, shear, and mom. of stress-
couple. / is constant. If at any
point TO of the rib a section is made, then the stresses in
that section are classified into three sets (Fig. 432). (See
§§ 295 and 367) and from § 367 eq. (3) we see that the ver-
tical intercepts between the rib and the special equil.
polygon being proportional to the products Hz or
moments of the stress-couples in the corresponding sec-
tions form a moment diagram, on inspection of which we
can trace the change in this moment, Hz = ^ , and
e
hence the variation of the stress per square inch, jJjj (as.
due to stress couple alone) in the outermost fibre of any
section (tension or compression) at distance e from the
gravity axis of the section), from section to section along
the rib.
By drawing through lines On' and OV parallel re-
spectively to the tangent and normal at any point m of the
rib axis [see Fig. 433] and projecting upon them, in turn,
the proper ray (B^ in Fig. 433) (see eqs. 1 and 2 of § 367)
tJ
Fig. 432.
472
MECHANICS OF ENGINEElimG.
we obtain the values of the thrust and shear for the sec-
tion at m. When found in this way for a number of points
along the rib their values may be laid off as vertical lines
from a horizontal axis, in the verticals containing the re-
spective points, and thus a thrust diagram and a shear dia-
gram may be formed, as constructed in Fig. 433. Notice
that where the moment is a maximum or minimum the
shear changes sign (compare § 240), either gradually or
Fig. 433.
suddenly, according as the max. or min. occurs between
two loads or in passing a load ; see m', e. g.'
Also it is evident, from the geometrical relations involv-
ed, that at those points of the rib where the tangent-line
is parallel to the " proper ray " of the force diagram, the
thrust is a maximum (a local maximum) the moment (of
ARCH RIBS 47S
stress couple) is either a maximum or a minimum and the
shear is zero.
From the moment, Hz = ^, p2 — — -
e 1
may be computed. From the thrust = Fp^^, pi=- , (F
= area of cross-section) may be computed. Hence the
greatest compression per sq. inch (Pi+p^) may be found in
each section. A separate stress-diagram might be con-
structed for this quantity (pi+p^)- Its max. value (after
adding the stress due to change of temperature, or to rib-
shortening, for ribs of less than three hinges), wherever it
occurs in the rib, must be made safe by proper designing
of the rib. The maximum shear J,,^ can be used as in §256
to determine thickness of web, if the section i?^ I-shaped,
or box-shaped. See § 295.
383. Temperature Stresses.— In an ordinary bridge truss
and straight horizontal girders, free to expand or contract
longitudinally, and in Classes A and B of § 378 of arch-
ribs, there are no stresses induced by change of tempera-
ture ; for the form of the beam or truss is under no
constraint from the manner of support ; but with the arch-
rib of two hinges (hinged ends, Class C) and of fixed ends
(Class D) having immovable piers which constrain the dis-
tance between the two ends to remain the same at all tem-
peratures, stresses called " temperature stresses '* are in-
duced in the rib whenever the temperature, t, is not the
same as that, t^, when the rib was put in place. These
may be determined, as follows, as if they were the only
ones, and then combined, algebraically, with those due to
the loading.
384. Temperature Stresses in the Arch-Rib of Hinged Ends,—
(Class C, § 378.) Fig. 434. Let E and /be constant, with
i74
MECHANICS OF ENGINEBErNG.
Fig. 434.
oilier postulates as in § 379.
Let t^, = temperature of
erection, and i — any other
temperature ; also let I =
length of span = OB (in-
variable) and 7^ "CO -efficient
of linear expansion of the
material of the curved beam or rib (see § 199), At tempeia-
fcure t there must be a horizontal reaction H at each hinge
to prevent expansion into the form O'B (dotted cuive),
which is the form natural to the rib for temperature t and
without constraint. We may /. consider the actual form
OB as having resulted from the unstrained form O'B by
displacing 0' to 0, i.e., producing a horizontal displace-
ment O'O =1 {t-Q-/j.
But O'O = Jx (see §§ 373 and 374) ; (KB. B'% tangent
has moved, but this does not affect Jx, if the axis X is
horizontal, as here, coinciding with the span ;) and the
ordinate y of any point m of the rib is identical Avith its
z or intercept between it and the spec, equil. polygon,
which here consists of one segment only, viz. : OB, Its
force diagram consists of a single ray Oi n' • see Fig. 434.
Now (§ 373)
.J B
Ja? = -A j3Iyds ; and M=Hz = in this case, Hy
H
.:l{t-Qrj=—Jfds;
hence for graphics, and
equal Js's, we have
Ell{t-t,)y^=HJs I^y' . . . . (1)
From eq. (1) we determine H, having divided the rib-curve
into from twelve to twenty equal parts each called Js .
For instance, for wrought iron, t and t^,, being expressed
In Fahrenheit degrees, -/j = 0.0000066. If E is expressed
in lbs. per square inch, all linear quantities should be la
inches and H will be obtained in pounds.
2'o?/^ may be obtained by § 375, or may be computed. B
being known, we find the moment of stress-couple = Hy,
AKCH-KIBS.
475
at any section, while the thrust and shear at that section
are the projections of //, i.e., of O^n' upon the tangent and
normal. The stresses due to these may then be determined
in any section, as already so frequently explained, and
then combined with those due to loading.
385. Temperature Stresses in the Arch-Ribs with Fixed Ends,—
See Fig, 435. (Same postulates as to symmetry, E and J
constant, etc., as in § 380.) t and t^ have the same meaning
as in § 384.
Here, as before, we
consider the rib to
have reached its ac-
tual form under tem-
perature t by having
had its span forcibly
shortened from the
length natural to
temp, t, viz. : O'B',
to the actual length OB, which the immovable piers compel
it to assume. But here, since the tangents at and B are
to he the same in direction under constraint as before, the two
forces H, representing the action of the piers on the rib,
must be considered as acting on imaginary rigid prolonga-
tions at an unknown distance d above the span. To find
H and d we need two equations.
From § 373 we have, since M=Hz=H {y—d),
Ax, i.e., WO-VBW, i.e., \t-t:)r^,=-^J{y-^yds . (2)
or, graphically, with equal As's
Fig. 435.
EIl{t-
-Qr-
-HAs
I-f-dSly
(3)
Also, since there has been no change in the angle betweeij
end-tangents, we must have, from § 374,
^_rMds=0; i.e., — / 2c?s=0;i.e., ny-d)ds=0
476 MECHANICS OF ENGINEEEING.
or for graphics, witli equal jU 's, I'^y = nd . , • (^\
in wMcli n denotes tli6 number of J.s's. From (4) wq
determine d, and tlien from (3) can compute U. Drawing
the horizontal F G, it is the special equilibrium polygon
(of but one segment) and the moment of the stress-couple
at any section = Hz, while the thrust and shea\' are the
projections of H=^0{ii' on the tangent and normal respect-
ively of any point m of rib.
For example, in one span, of 550 feet, of the St. Louis
Bridge, having a rise of 55 feet and fixed at the ends, the
force H of Fig. 435 is = 108 tons, when the temperature is
80° Fahr. higher than the temp, of erection, and the en-
forced span is 3^ inches shorter than the span natural to
iliat higher temperature. Evidently, ;f the actual temp-
■erature I is lower than that ^„, of erection, ^must act in a
direction opposite to that of Figs. 435 and 434, and th&
"'thrust " in any section will be negative, i.e., a pull.
386. Stresses Due to Rib-Shortening — In § 369, Fig. 407, the
shortening of the element AE to a length A'E, due to the
uniformly distributed thrust, PiF, was neglected as pro-
ducing indirectly a change of curvature and form in the
rib axis ; but such will be the case if the rib has less than
three hinges. This change in the length of the different,
portions of the rib curve, may be treated as if it were due
to a change of temperature. For example, from § 199 we-
see that a thrust of 50 tons coming upon a sectional area.
of i^ = 10 sq. inches in an iron rib, whose material has a
modulus of elasticity — E = 30,000,000 lbs. per sq. inch,
and a coefficient of expansion yj = .0000066 per degree
Fahrenheit, produces a shortening equal to that due to a
fall of temperature {to—t) derived as follows: (See § 199)
(units, inch and pound)
^° ^ FEri 10 X 30,000,000 X. 0000066"
Fahrenheit.
Practically, then, since most metal arch bridges of
©lasses G and D are rather flat in curvature, and the thrusts.
AECH-RIBS. 477
due to ordinary modes of loading do not vary more than 20
or 30 per cent, from each other along the rib, an imagin-
ary fall of temperature corresponding to an average thrust
in any case of loading may be made the basis of a con-
struction similar to that in § 384 or § 385 (according as the
ends are hinged, ov fixed) from which new thrusts, shears,
and stress-couple moments, may be derived to be combin-
ed with those previously obtained for loading and for
change of temperature.
387. Resume — It is now seen how the stresses per square
inch, both shearing and compression (or tension) may be
obtained in all parts of any section of a solid arch-rib or
curved beam of the kinds described, by combining the re-
sults due to the three separate causes, viz.: the load,
change of temperature, and rib-shortening caused by the
thrusts due to the load (the latter agencies, however, com-
ing into consideration only in classes G and D, see § 378).
That is, in any cross-section, the stress in the outer fibre
is, [letting J',/, T-^", T^"', denote the thrusts due to the
ihree causes, respectively, above mentioned ; {H&)', {Hz)"y
{Hz)'", the moments]
^T}}.^I}l'^I^±tUHz)'±{Hzy'±{Hzy"'\ . . . (1)
i.e., lbs. per sq. inch compression (if those units are used).
The double signs provide for the cases
where the stresses in the outer fibre, due
to a single agency, may be tensile. Fig.
436 shows the meaning of e (the same
used heretofore) /is the moment of in-
ertia of the section about the gravity
axis (horizontal) (7. i^ = area of cross-
section. [Ci = e ; cross section symmet-
rical about (7]. For a given loading we
may find the maximum stress in a given rib, or design the
rib so that this maximum stress shall be safe for the ma-
terial employed. Similarly, the resultant shear (total, not
478
MECHANICS OF ENGINEERrNG.
per sq. inch) = «/' ± J" ± 3'" is obtained for any section
to compute a proper thickness of web, spacing of rivets,
etc.
388 The Arch-Truss, or braced arch. An open-work
truss, if of homogeneous design from end to end, may be
treated as a beam of constant section and constant moment
of inertia, and if curved, like the St. Loi*is Bridge and the
Coblenz Bridge (see § 378, Class D), may be treated as an
arch-rib.* The moment of inertia may be taken as
r=2i^,
A
(I)
where F^ is the sectional area of one of the pieces II to the
curved axis midway between them. Fig. 437, and h = dis«
fcance between them.
Fig. 438.
Fig. 437.
Treating this curved axis as an arch-rib, in the usual
way (see preceding articles), we obtain the spec, equil. pol.
and its force diagram for given loading. Any plane ~| to
the rib -axis, where it crosses the middle m of a " web-
member," cuts three pieces, A^ B and 6', the total com-
*The St Louis Bridge 13 not strictly of constant moment of inertia, being somewha*
strengthened near eaoli pier,
ARCH-EIBS. 479
pressionB (or tensions) in which are thus found : For the
point m, of rib-axis, there is a certain moment = Hz, a
thrust = Th, and a shear = J, obtained as previously ex-
plained. We may then write Psin/9 = J . . . • (1)
and thus determine whether P is a tension or compres-
sion ; then putting P'+P" ± P cos /? = T,, 2
(in which P is taken with a plus sign if a compression, and
mlQus if tension), and
(P'-P")^=Rz ...... (3)
we compute P* and P", whi(5h are assumed to be both com-'
pressions here. /9 is the angle between the web member
and the tangent to rib-axis at m, the middle of the piece.
See Fig. 406, as an explanation of the method just
adopted.
Circular Ribs akd Hoops.
389. Deflections and Changes of Slope of Curved Beams. Analyt-
ical Method. For finding these quantities we may use eqs. (I.),
(II.) and (III.) of § 374. For example, we have in Fig. 439,
a curved beam of the form of the ox k —
quadrant of a circle, fixed vertically ^
at lower extremity p, and carrying
a single concentrated load, P, at
the free end 0. [Its own weight
neglected.]
As a consequence of the load- i /^^^' \q \
ing, the extremity is displaced to kfl__ 1__^J^
some position, ^\, but the bending M^^^
is slight. Required, the projections ^^^' ^^^'
Ax and Ay of this displacement and also the angle OKOn or <^,
which the tangent-line at 0„ makes with its former fhorizon-
tal) position OX. The beam is homogeneous and of constant
cross-section ; i.e., E and / are constants.
To use the equations for Ax and Ay we must take as an
origin (since is the point whose displacement is under con-
480 MECHANICS OF ENGINEERING.
sideration). Hence the co-ordinates x and y of any point, m, of
the axis of the beam are as shown in the figure. Taking now
polar co-ordinates, as shown, we note tliat x = r cos_fi-|
y = r ( 1 — sin ^) ; and ds = rdO. We must also put down
the following integral forms for reference ; viz. : —
fsin ^ . d^ = — cos ^ ; f^ . cos 6* . di9 = e . sin ^ + cos ^ ;
fcos e.de = + sin d ; fsin^ 6 .dd = ^ 6 ~ \s,m2 d ;
fsin e.cos e .de = \ sin^ e ; C c,o%^ 6 ,de = ^ 6 + \sva.20.
Taking the portion 0„m {m being any point on curved axis
of beam) as a free body, we have, for the moment of the stress
couple a.t m, M — Px, = Pr cos 6, and hence derive, for the
angle ^,
Also 5
and z z
1 r^ Pr^ r /'^ z*^ 1 Pr^
^^'=^X^-^''^' = ^[Jo «<^«^-^^-X -^^•--^•^^] = 2^z- (3>-
It must be understood that the elastic limit is not passed in
any fiber and that the bending is very slight. A simple curved
crane and a ship's davit are instances of this problem, provided
the cross-section has the same moment of inertia, /, about a
gravity axis perpendicular to the plane of the paper in Fig.
439, at all parts of the beam.
390. Semi-Circular Arch-Rib. Hinged at the Two Piers or Sup-
ports, and Continuous Between. Fig. 440. The supports are at the
same level. The arch-rib, or curved beam, is homogeneous
and has a constant I at all sections. (It is a " curved prism".)
It is stipulated that no constraint is necessary in fitting the rib
upon the hinges at the piers before any load is placed on the
rib ; that is, that the distance apart of the piers (which are
unyielding') is just equal to the distance between the ends of
the rib when entirely free from strain. In other words, after
the rib is in position it is under no stress until a load is put
upon it. Its own weight is neglected and the load is a concen-
trated one of 2 P lbs. placed at the " crown ", B. As a conse-
CIRCULAR RIBS AND HOOPS.
481
quence of the gradual placing of the load the crown B settles
slightly, but on account of symmetry the tangent-line to the
curved axis at B remains horizontal. Also the extremities
and A tend to spread further apart, but this is prevented by
the fact that the piers are immovable (or we may express it
"the span is invariable"). Hence the reaction at each hinge
support will have a horizontal component ^as well as a ver-
tical component, V, lbs. Fig. 2 shows the axis of the rib.
Taking the whole rib as a
free body we easily find (by
putting ^ vert, comps. = zero,
and from symmetry) that
each y=P; the whole load
being called 2 P ; but for de-
termining the value of H
(same at each hinge ; from
2 (horiz. comps.) =zero) we
must have recourse to the
theory of elasticity ; i.e., must depend on the following fact,
viz. : — that in the gradual settling of point B under the load,
B remains in the same vertical, and the tang, line at B remains
horizontal, and hence (since moves neither horizontally nor
vertically in actual space) the horizontal projection of O's dis-
placement relatively to B and ^B's tangent is zero (or Ax^O),
while the vertical projection of O's displacement relatively to
B and 5's tangent (A?/) equals the distance B has settled in
actual space. Here we must take as origin for x and y (as
in figure) for any point m between and B\ and note that
the X = r {1 — cos 6), and y = r mi 6; while ds = r. d6.
With Om as a free body (m being any point between and
5) we have for the moment of the stress couple at m,
M.^Vx- Hy, = Px- Hy.
1 rB r^
Ax, = — j Myds, =0 ; .'. I [P(X- rco^ 6) - Hr sm e'jr^ sm Odd =^0',
Fig. 440.
.-. P C sin0.de - P C smdcosddd-H C sin'6'.d^ = 0:
Jo Jo Jo
and hence, (see integrals in § 389),
482
MECHANICS or ENGINEERING.
(-[-
COS 6
Slll^
H
■0 _ sin 2 6h\"^
2"" 4
= 0.
Inserting the limits, we have
pr_0 + l-i + 0J-//|^|-0-0-(-0)"
= 0;
.:H =
2 P load
Also we may obtain, for the settlement of the crown, at B
1 C^ Pr'^ r3 TT 1 "I
Aw of relatively to 5, = =7 / Mxds = -77^ -. — 2
^ "^ EI Jo EI I 4: TT j
while the tangent-line at 0, originally vertical, now makes with
the vertical (on the outside)
'eT
This is a '■'statically indeterminate structure " ; that is, one in
which a solution is impossible by ordinary statics but must
depend on the theory of the elastic change of form of the beam
or body in question.
If the load were not placed at the crown, or highest point,
we should be obliged to put
an angle
EI Jo
Mds =
- +1-
jj£Myds^O
EI
for the Ax of relatively to A (instead of to jB).
391. Cylindrical Pipe Loaded on Side. A cylindrical pipe of homogeneous
material and small uniform thickness of pipe-wall, i, and length I, (so that
the moment of inertia of the cross-section of wall is for present purposes
I = It^ -i- 12) rests in a horizontal position on a firm horizontal floor and bears
a concentrated load of 2P at the highest point, or crown, jB. See Fig. 441.
It is to be considered as a continuous curved beam or " hoop ", without hinges.
We neglect the weight of the pipe itself. The dotted circle shows the original
unstrained form of the pipe-wall, or hoop, while the full line is its (slightly
deformed) shape when it bears the load. The elastic limit is of course not to
be passed. The upward force 2P at iV"is the reaction of the floor. Required,
the maximum moment of stress-couple ; and also the increase in the length of
the horizontal diameter, and the decrease in that of the vertical diameter.
Consider as a free body the upper left-hand quadrant of the hoop, viz.,
OB, in Fig. 442, cutting just on the loft of the load at L', a horizontal section
being made at 0. At each end of this body we must indicate a stress-couple,
a shear, and a thrust. But at it is evident, after a little consideration, that
the shear (which would be horizontal) must be zero ; there being at 0, .•. only
CIECULAE EIBS AND HOOPS.
483
a thrust Tg and a stress-couple of unknown moment M^. At the other section
the shear must he equal to one half of the load 2P (from considerations of
symmetry) i.e., J" at B = P ; while the thrust at B is soon shown to be zero
(since S (horiz. compons.) must = zero, and this thrust if it existed would b»
Fig. 441.
Fig 442.
Fig. 443.
the only horiz. force besides those formhig the stress-couple at B). At B,
therefore, we find only a stress-couple, of an unknown moment Ms, and a
shear Jb of direction shown in Fig. 442. By writing S (vert, compons.)
= zero for this free body we find that the thrust, Tg, at 0, must have a value P.
To determine M^ we make use of the fact (evident from Fig. 441) that in
the deformed condition of the ' ' hoop ' ' the tangent-lines at points and B are
still vertical and horizontal, respectively ; in other words that the angle be-
tween them has not changed, i.e., is still' 90°. Hence the value of 0, or change
of angle between tangents at and B is zero. Apply this fact to Fig. 442. Take
as origin for the x and y of any point m on OB (using 6 later). From a
consideration of the free body Om shown in Fig. 443 we have for the stress-
couple-moment M at any section m the value M = Px — M^. We have also
a; = r (1 — cos d) ; y = r sin 6 ; and ds = rdd.
Since
1 r^ r^
0, = ^ j Jfds, = 0, .-. j \_Pr^ - Pr^ cos 6 - Mf\ dd = ;
i.e., {Pr^ [0 - sin e\ - M^rd) 2 = ; or, Pr^ f^ ~ ^ 1 ~ -^^^l ^ ^ ''
whence, finally, we have Jlfg = Pr 1 I
(1)
Now that Jlfg is known, we may find Mb by taking moments about the
lower section, 0, in Fig. 442, with OB as fj-ee body whence Mb = (2 -=- tt) Pr,
which is greater than M^. Hence the equation for safe loading is {R'l -=- e)
= 2Pr -^ TT, where R' is the maximum safe unit-stress for the material, and e
the distance of the extreme fiber from the gravity axis of a section. (If, how-
ever, the radius, r, of the cylinder is not large compared with the radial thick-
ness of the section, see §§ 298 and 299.)
Evidently the horizontal diameter has been lengthened by an amount 2 Ax,
if Ax denote the horiz. proj. of O's displacement relatively to B and 5's tan-
gent ; and similarly, the shortening of the vertical diameter is 2 Ay, if Ay denote
the vert. proj. of O's displacement with regard to B and JS's tangent-line.
484 MECHANICS OF ENGINEERING.
Hence ^
ix =. =y j Myds = J- j [Pr^ sin 6 .dO - Pr^ cos 6 sin edd — M^r'' sin edd];
from which we have, with M^ = Pr [1 — {2 -t- tt)],
TT
^y = ^C -^a^s, = -^ r ^ [Pr3 (1 - COS ey de - M,r^ {dd - COS edd)-],
Pr^ ^2 _ 8
It will be noted that the results obtained in this problem apply also to the
case where the hoop is a circular link of a chain under a tension 2 P, except
that the moments will be of opposite character and shears and thrusts of oppo-
site direction. Also, the change of length 2 Ax of the horiz. diameter will be a
shortening, that of the vertical diameter, a lengthening. (See Prof. Filkins'
article on p. 99 of Vol. IV of the Transac. of Assoc. C. E. of Cornell Univ.
and Engineering News, Dec. 1904, p. 547.)
BTumerical Example. Fig. 441. The length of a cast iron pipe is 10 ft., the
thickness of wall ^ inch, and the radius of the pipe (measured* to the middle
of the thickness) is 6 inches. Kequired, the value of the safe load at crown, 2 P
when the pipe is supported horizontally on a firm smooth bed or floor; the
max. safe unit-stress being taken at the low figure 2000 lbs. per sq. inch.
Solution. "We have only to substitute these values in M^ = R' I -i- e and
*9000 V 120 V C^y
Obtain (since I =1. P ^ 12), frrr_iL^Lil^== (0.6366 P X 6) ; hence safeload
(^) X (2) X 1^
= 2 P, = 5236. lbs. ; that is, 43.63 lbs. per running inch of pipe length.
If now the thickness be doubled, i.e., t = 1", with other data unchanged,
we find the safe load to be four times as great, i.e., 2 P = 20,944 lbs. ; or 174.5
lbs. per running inch of pipe-length.
Although the load is called "concentrated" as regards the end-view of the
pipe, it must be understood to be uniformly distributed along the length.
FLEXURE OF BEAMS: GEOMETEICAL TREATMENT. 485
CHAPTER XII.
Flexure of Beams ; both Simple and Continuous.
Geometrical Treatment.
392. By Geometrical Treatment is meant making use of the
properties of geometrical figures to deduce algebraic relations.
This does not necessitate the use of drafting instruments ; but the
graphic ideas involved greatly simplify the algebraic detail of
finding deflections, angles, moments, shears, etc., in the case of
horizontal beams originally straight and slightly bent under
vertical loads and reactions. In the case of " continuous
beams", or "girders", (p. 320), this mode of treatment leads to
conceptions and methods which are remarkably clear and simple.
393. Angle Between End-Tangents of a Portion of a Bent Beam.
If the cantilever beam of Fig. 443a (slender and originally
o
|C B
.ax I
norvuilto CD '_ ^^
— T" ^. ,,
\d(p ^■'' »
' -* 1
^1
Fig. 443a. Fig. 4436.
straight) be loaded as shown, and the beam thus slightly bent,
the two cross-sections, AH and CD, at the two ends of any dx
of the axis of beam, are no longer parallel but become in-
clined at a small angle d<^ which is also the angle between the
normals to these sections, in their new (relative) position (see
now Fig. 4436). AZT now occupies the position A'W (relatively
to (7Z)). The outer fiber AC (originally of length = dx) is
longer by some amount dX ; and evidently the value of angle
d^ may be written = dX -j- g. But, by the definition of the
modulus of elasticity of the material, J5', we have also
E = p H- — , (p. 209) ; whence d^ = '^-^
dx hiQ
(1)
486 MECHANICS OF ENGINEERING. '
Now if M denote the moment of the stress-couple to which
the tensions and compressions on the ends of the fibers in the
section A^ H' are equivalent (M would equal Px in this simple
case) we may combine the relation, (§ 229), M = pi -r- e with
eq. (1) and thus derive, as
a fundamental relation : , . . d4> = — — (2)
EI
for the angle between two tangents to the elastic curve,
one at each end of the elementary length, dx, of the curve ;
since the two normals to the sections A'H' and C D in Fig.
4436 are tangents to the ends of the short length dx of the
elastic curve. (This value of the angle d^ is in 7r-measure ;
i.e., radians.)
It follows, therefore, that when the cantilever of Fig. 443a
is gradually bent from its original condition (in which the
tangent lines at the two extremities and B Avere coincident,
i.e., made with each other an angle of
zero) into its final form, by the gradual
application of the load P at 0, the
angle between the tangent to elastic
curve at 0^ (the final position of 0)
and that at B (which tangent, in this
,,. case, has not moved) will have a value
Fig. 444. \ _ -^
obtained by summing up all the small
values of d^, one for each of the dx'^ between and B (these
dx'^ making up the length of curve between those points).
Or, in general, if 0„ and B are any two points of an elastic
curve (of axis of bent beam, originally straight and now only
slightly bent, x beiag measured along the beam) we have for the
angle between the ^ _ ., _ T^ Mdx ,on
tangents at 0„ and B\^ Jo EI
(See Fig. 444 for case of cantilever.) This may. be called
the angle hstween end-tangents of any portion of such elastic
curve. The beam must be continuous between these two
j)oints and only slightly bent. Usually the beam in question
is homogeneous and then E may be taken outside of the integral
sign. Also, if the beam be prismatic in form (i.e., sides par-
allel to a central axis, originally straight) the moment of inertia,
FLEXURE OF BEAMS ; GEOMETEICAL TREATMENT,
487
/, of the cross-section is the same for each dx, and may be
placed outside of the / sign.
394. (Relative) Displacement of any Point, 0, of Elastic Curve
of a Bent Beam. In the case of the simple cantilever of
Fig. 443a let us consider that the axis of the beam, originally
straight and in position OB, passes gradually into its final form
or elastic curve 0„ A'" A'' . . . B hj the successive change of
form of each small block, or elementary length dx ; beginning
Successive bending of each
dx of Cantilever.
Fig. 444a.
at the end B. When the section at A^ turns through its angle
d(f)^, as due to the lengthening and shortening of the fibers
forming the block (i.e., to the stress-couple in section A', of
moment M') it carries with it all the portion OA' (still straight)
into position O^A' so that the extremity describes a small
distance (practically vertical) 00 ^ = OA' . d(f>^. Similarly
when, next in order, the section at A^' turns through its small
angle d(f>^, the left-hand end of the beam describes a further
small distance Ofi^ ^ ^^" • ^^2 ' ^^^ ^° ^^^ ' "i^^til finally the
extremity has arrived at its final position 0„, having executed
a total (vertical) displacement OOn-, which will be called Ay.
If, now, any one of the elementary vertical displacements
(like OjOj? as typical) be called 8y, we note that Ay is the sum
of all these small 8i/'s, each of which is practically a small cir-
cular arc described with a radius x swinging through a small
angle d(fi, (the successive x's being successively smaller for the
Sy's lower in the series), so that By = xd(}>; hence
Ay, =- fSy, = Cxd(j>. But, from eq. (2), d(f> = Mdx -^ AT;
( Displacement of point 0) _ _ T ^ Mxdx
I relatively to 5's tangent ) Jo -^-^
(4)
488 MECHANICS OF ENGINEEEING.
(N.B. In the use of this relation the x of each dx must he
measured from the point whose displacement is desired.)
Although the special case of the cantilever has been in mind
in the figure used in this connection, this result in eq, (4) may-
be generalized by stating that it gives the displacement A?/ of
any point from the tangent-line drawn at any other point, B,
of the elastic curve formed by the axis of a beam originally
straight and slightly bent under the action of vertical forces
and reactions. In order to use it, the value of the moment M
of the stress-couple in each successive dx must be expressed as
a function of x. If, in addition, the beam has a constant moment
of inertia, /, of the cross-sections, the " I " may be taken outside
of the sign of integration. An integration is then generally
possible. (For example, in the above cantilever, for M we
should write Px.)
395. Deflections and Slopes of Straight Homogeneous Prismatic
Beams Slightly Bent under Vertical Loads and Eeactions. (Beam
Horizontal.)
If the beam is a prism and homogeneous, both U and I are
constant along its length and may be taken outside of the in-
D^ tegral sign in eqs. (3)
B,.-*':^^'' and (4), and these two
J \ '^T^v----^c'""°''^ equations may now be
I [dxi
applied to a portion of a
beam situated between
any two points and B
Fig. 4446. of the elastic curve as-
sumed by the (originally straight) axis of the beam (Fig. 4446)
under some load. The tangent-lines at 0„ and B were origi-
nally coincident, and hence the angle between these tangents
when the beam is bent is the total change in angle between the
1 r^
tangents, and consequently may be written (^ = __ / Mdx and
is 0„ 00 in the figure. Again, if a vertical be drawn through
the point 0„ to B's tangent-line OB, the length 00,^ is evi-
dently O's displacement relatively to 5's tangent-line, since
originally the point 0„ was situated in 5's tangent itself.
1 r^
That is, 00„, or A?/, = -— I Mxdx,\n which M is the mo-
hil Jo
PLEXUBE OF BEAMS : GEOMETRICAL TREATMENT.
489
ment of the stress-couple in the cross-section at any distance, x,
from 0. Note that in general Mis a variable; also that the x
must be measured from the point whose displacement is
under consideration.
Example I. Simple cantilever (Fig. 444,), huilt Inliorizontally at B and bear-
ing a concentrated load = P lbs. at the free extremity. Both E and / are con-
sfcant (homogeneous prism). Pind the deflection 00« and the slope (p.
Solution. From the free body OhVI (m being any point between and B)
we have M = Px as mom. of stress-couple at ?n.
PP
2ET
1 /.5 p p!
the "slope" at On- For Jy
we have
Pl^
Fig. 444i. Fig. 4442-
Example II. Prismatic beam on two end-supports. Concentrated load P,
V)S., in middle, Fig. 4442- The two supports being at same level we note that
from symmetry the tangent at the middle point B of the elastic curve is hori-
zontal. Hence the displacement OOn of the extremity from this tangent is
«qual to the deflection of B itself below the horizontal line 0„G. To find OOn
or Jy,
Pl^
40/
Example III. Prismatic beam on two end-supports at same level, the load
being uniformly distributed over the whole span, I. Fig. 4443. That is, W = wl,
1 r-B 1 /-S pP -| P /•■!■= 2
Jy = -=-:- I Mx dc = -r— 1 TT X \ xdx = -— , ^ I x^dx ^
^ ElJo EI Jo L2 J 2ElJx=^
W = ivl
Fig. 4443. Fig. 4444.
w being the load per running inch. As before, the tangent-line at middle
point B of the elastic curve must be horizontal, so that the displacement of
extremity On from this tangent will also give the deflection of B from the
horizontal OnC. Measuring x from (as must always be done in these cases)
we note that M at any point m
W
wx-
I pBr-WX'^ WX3 -| 1 r WX^ WXn 5
°^« = MJo I ^^"- -2-^^J= ^[-6-- Xjo =
5
584
EI
490 MECHANICS OF ENGINEEEIKG.
Example IV. Prismatic beam on end-supports, hearing two equal loads, each
= P, symmetrically placed on the span. Fig. 444^. Required, tlie deflection
of the middle point, B, of the elastic curve, below the horizontal OnC.
Length = 4a.
Solution. In previous problems of this article the expression for M, the
mom. of stress-couple for any point m betvreen the points O and B, has been a
single function of x, applying to all such points m. But in the present problem,
having found the reaction at O to be = P lbs., we note by considering a free
body Onin (where m is any point between On and B) that the value of 31 is
M= Pz ; whereas if the free body extends into the portion DB the expression
for M (the free body being now Onm') is M=Px —P{x — a) which reduces to
M=Pa, a, different function of x; (in fact a constant). Therefore, in making
the summation 00« = (l-=-^J) ( Mx dx for all the dx^s between and B, this
summation must be divided into two parts, viz. : one from to D, involving for
X the limits x=0 and x=a; and the other from D to B, for which the limits for
X are x=a and x=2a. Hence
(The student should verify all details of this operation, noting that each sum-
mation or integi'al contains the proper value of M, as a function of x, for the
proper portion of the elastic curve. As before, it should be said that on ac-
count of symmetry the tangent-line at the middle point B is horizontal, and
parallel to OnC. Otherwise OnO would not he equal to the deflection of B.)
396. Non-prismatic Beam. VariableMoment of Inertia, I. If the J is vari-
able, (e.g., if the beam tapers) it must be retained on the right of the integral
sign in the expressions for cj) and Jy and then expressed as a function of x be-
fore the integration can be proceeded with. In some cases I may be constant
within the limits of definite portions of the beam and then the procedure is
simple. For instance, if the beam in Fig. 444^ has a constant value, = I^, for
3
the portions OB and FC, and a larger (but constant) value, of J,) = h ^i»
for all the sections from B to F, the following takes the place of eq. (1) above :
. P fa Pa /'2a 4 Pa^
397. Properties of Moment Diagrams (Moment-Areas and Cen-
ters of Gravity). Prismatic Beams in Horizontal Position. Vertical
Loads and Reactions. In Fig. 445 let AI) be the bent condition
(i.e., elastic curve) of the axis of a straight prismatic homo-
geneous beam supported on supports at, or nearly at, the
same level (so that all tangent lines to the elastic curve deviate
but slightly from the horizontal. That is, the bending is slight).
Also, let A"D"B"' 0'" be the corresponding moment diagram (as
defined and illustrated on pp. 265 to 309). For instance, for
any point m of the elastic curve the moment of stress-couple (or
"bending moment ") in that section of the bent beam is repre-
FLEXURE OF BEAMS; GEOMETRICAL TREATMENT, 491
sented (to scale) by the ordinate m"m"' or M, in the same
vertical as m.
If now a small horiz. distance dx^ or m'V, be laid off
from m" and a vertical r . .8
be drawn through r, the pro-
duct M . dx would be proj)or-
tional to, and may be repre-
sented by, the area of the
vertical strip m"rsm"'. Now
dx being inches (say) and M
being inch-lbs., this product
might be called so many "sq.
inch-lbs." of moment-area (as
it will be called). But the
angle <^ between the tangent-
i^.
,^ ^ Mom. Diagraml
Fig. 445.
lines drawn at an}- two points On and B of the elastic curve is
1 r^
equal to — - / Mdx ; and hence we may write
[total " moment-area
between and B
]'
EI
(la]
or, for brevity, ((> ={A^^ ) ^ EI . . . . . . . . . . (1)
This " moment-area," then, between and B is the pro-
duct of the base O'^B" (inches) by the average moment be-
tween and B regarded as the average altitude of the figure,
0"B"B"V", this altitude being inch-lbs.
Again, if the elementary " moment-area " Mdx be multiplied
by x, its horizontal distance from 0" (i.e. from and 0„), and
these products summed up for all the dx'^ between and 5,
there results the expression I (M , dx) . x which may be
written {A^^.x, where x denotes the horiz. distance of the
center of gravity of the moment-area O^'B'" from 0''0"' (since,
from the theory of the center of gravity, the sum of the pro-
ducts of each ^tri]} of an area by its x co-ordinate is equal to
the product of the whole area by the distance of its center of
gravity from the same axis). In the figure the center of grav-
ity of the moment-area 0"B"' is shown at C" •, and the cor-
responding X is marked.
492
MECHANICS OF ENGINEERING.
But we have
^y, or OOn, -
\J Mxdx 1 -5- E/ =r r Mdx ,x'\^ EI\
Sv — -
and hence we may write
00,,, or AV, = [(A^) ,i] -^ E7 . . . . (2)
which furnishes us with a simple means of determining the dis-
placement of any point 0^ in the elastic curve of the bent beam
from the tangent-line at any other point B in that elastic curve.
Evidently, from equations (1) and (2) we have A?/, = 00^^
= j>x ; and can therefore state that the intersection of the two
tangent-lines, one drawn at 0, the other at B, lies in the same
vertical as the center of gravity of the intervening moment-area-
(N.B. Instead of the product {Ao)'X, we may, of course,
use the algebraic sum of similar products for any component
parts into which it may be convenient to subdivide the total
moment-area.)
398. Examples of TJse of Eqs. (1) and (2) of Preceding Paragraph.
Example I. Simple Cantilever. Concentrated load at free end. Fig. 444i.
Constant E and I. {Prism.) Here the moment-diagram for whole length is a
triangle (§ 249) whose base is I inches and whose
altitude is PI inch-lbs.
• Hence, with and B taken as in Fig.
445,, we note that
AB-.
'?=('•?).
...,= [.?].
2
and that x = ^ Z.
o
E.I.==
at On', while OOn =
T^ 1
Fig. 445i.
I.e., OOn^^
EI
['•"]
PP ^
2 EI'
1_
EI
pn 2
3
for the slope
i^l)' ^>
Z =
PP
3. EI
Example II. Prismatic Beam on Two End-Supports, Load Uniformly dis-
tributed over the whole span or length, I', W = wl. From p. 268 we know that
the moment-diagram (Fig. 4462) is a symmetrical segment of a parabola with
axis vertical, and that the moment at the middle section is Wl -^ 8. Also,
from p. 12 of Notes, etc., in Mechanics, we find the x of the left-hand half of this
moment-figure, measured from the left-hand extremity On, is ^ Z — | of 1 1; i.e.,
i = f of 1 1.
The area of this semi-pardbolic-segment is two-thirds that of the circum-
scribing rectangle. From symmetry, the tangent-line drawn at B, the middle
point of the elastic curve, is parallel to On D, so that the displacement OnO of
from that tangent is equal to the deflection of B from OnD. Hence
-2 IWl 5 I-
3 •2* 8
OnO,
(Ao).x
EI '
a^
384 EI
ELBXUEE OF BEAMS: GEOMETRICAL TREATMENT. 493
Example III. Prismatic Beam. Ends Supported. Two Concentrated Loads
Equidistant from Supports. Fig. 4453.
Here, as before, from symmetry the tangent at B is horizontal, parallel to
V yvYvVVvVYYVVYVVVYVV
On C • D.
-^
j^-
FiG. 4452.
Fig. 4453.
OnD; so that On equals the deflection of B from C (its position before load-
ing of beam). Each load P is in middle of a half -span. Required OnO ;, i.e.,
CB=?
In this case the moment-diagram is easily shown to consist of a triangle at
each end with a central rectangle of altitude = Pa (inch-lbs.). To find OnO
we need the product (A^).x. But this A^ consists of the triangle 0"A"N
with its center of gravity distant f of a from and of the rectangle A"B"KN
whose center of gravity is at a distance of | of a from 0. Utilizing, therefore
the principle stated in the N.B. of § 397, we write
Q-Q _ (^)O-
1 r Pa
2a -r. 3 a
--l-a.Pa-^
n Pa^
~(S-Er'
Example IV. Prismatic Beam on End Supports. Single Eccentric Load, P.
Fig. 445^. Here a tangent drawn to the elastic curve at the load-point B, not
being horizontal, is not par-
allel to OnCn, and hence OnO
does not = the deflection, 5,
otB.
However, the displace-
ments ( = (^1 andd.,), of Ofrom
5's tangent and of C from
B's tangent, are easily found,
the moment-diagram 0"NC"
having been drawn, in which
'B"N = {Pa,a, -^ I) inch-lbs.
(§260). Call" the "moment- Fig. 445,.
area" of triangle 0"B"N, A' ■ and that on right of load, viz. of C"B"N, A".
Then, from eq. (2) of § 397, we may write
Eld^ = A'z,; and Eld., = A"x.,.
If now we draw a horizontal line, HI)., through the point B of the elastic
curve, we note, from the similar triangles thus formed, the proportion
-J — -r = — . From these three equations d^ and d., may be eliminated and d
494
MECHANICS OF ENGlISEEillNG.
obtained; (since A
and Xj = I a, ■)
Pa{a^ ~ I, and A" = ^ Pa^^ -i- I; while Xj = ja^,
We thus obtain 5 = (J Pa^^a.^-) -^ {EI, I)
(3)
This is for the load-point. For the maximwn deflection see next example.
Example V. Maximum Deflection of Prismatic Beam. End Supports.
Single Eccentric Load. Fig. 4i5g. To locate the lowest point D of elastic
curve and determine its deflection,
d, below the horizontal 0„B.
Draw a tangent at D, also at
B whose distance n from D is, as
yet, unknown. Note that the tan-
gent at D is horizontal.
The moment-diagram is a tri-
angle of altitude ik' ; (M' = Pah^l);
denote the moment at B by m'.
We have m' =.(n -h 6) .M'. Now
the angle (p =d' -h I, and
d' = (AI) .x-^EI =
Fig. 4455. ^ M'l (a+| [J (a + &) - a])--E'I.
. •. 6 EI4> = M' (2 a -j- 6). But0, = (^^) -- ^I, = n . ?n' ^ 2EI, and ^ = 0^ ;
.-., finally, we have n = \^\b{2a+ b), which locates the point R.
Now note that the intersection C lies in the vertical through the center of
gravity of the shaded triangle (§ 397). Hence CB = In and therefore from
similar triangles B8 = \ns. But RB, =d, = BS- BS, and BS = .\n. Hence d = |0n and finally by substitution, and with M' placed =
Pah -H I, we have (with h> a) .
Pah
^-\ EI
[2a + 6] Vifi (2a + h)
same as
on page 258
399. The "Normal Moment Diagram.'' If a portion, OB, of a
horizontal beam carrying loads, be conceived separated from
the remainder of beam and placed on two supports at its
extremities and B, while carrying the loads [say P^ and PJ
originally lying between and B, the corresponding moment-
diagram, 0'"TB"' of Fig. 446 may be called the "normal-moment
diagram" for portion* OB (of original beam) and its load. If
Vq is the pier reaction at left, we have for any section t (say
between P^ and P^) x ft. from 0, the moment of stress-couple
[call it Mn or " normal moment "]
M„
Vf^x — P,(x — a)
(1)
Now consider OB in its original condition (see lower part
of Fig. 446) when forming part of a much longer beam sup-
FLEXURE OF BEAMS: GEOMETKICAL TREATMENT. 495
ported in any manner. If we consider OB, now, as a " free
body," M^e must put in, besides the loads P^ and P^, a shear Jq
and a stress-couple of moment Mq in section at 0, and Jq and
couple of moment Mb at B. The moment in any section t of
OB is now M = Mo + /o^ — Pi (x — a). Let 7 = difference
between J^ and Vq,
I.e.,
then
M = Mo + 7x + [7o:c-P,(a;-a)] . (2)
i.e. [see (1)],
M = M^ + Vx + M„
(3)
N, ~T^° :-
Hence the moment M {=kwin. Fig. 446) of any section of OB
is made up of a constant
part Mw a part proportional
to X, and a third part equal
to the " normal moment " of
th&,t section. Therefore, if,
in the moment-diagram
0'B'B"wO" for OB we join
0" and B" by a straight
line, and also draw a hori-
zontal through 0'^ the ver-
tical intercepts [such as
uvo] between the line 0"B"
[or "chord"] and the broken
line 0"wB" are the normal
moments for OB and its
load, and the area (mom.-
area) of the figure formed by
these intercepts is equal to
that of the normal moment
diagram.
It is also evident that the center of gravity of the figure
0"B"w lies in the same vertical as that of the normal moment
diagram.
(In the next paragraph the trapezoid 0'B'B"0" will be
divided into two triangles, instead of into a triangle and a
rectangle. )
Fig. 446.
496
MECHANICS OF ENGINEERING.
400. The Theorem of Three Moments. Let 0, B, and C,
Fig. 446a, be any three points in the elastic curve of a
homogeneous, continuousy
and prismatic beam, origi-
nally straight and hori-
zontal but now slightly
bent under vertical forces
(some of which are reac-
tions of supports; no loads
or forces are shown in the
figure).
Let Mq, Mj, and M^ be
the moments of the couples
in sections 0, B, and C.
The moment-diagram for
portion OBC is 0'C'C"T^B"TP". Join 0"B' and C"B' ; also
0"B" and C"B" . At the point B of elastic curve draw a tan-
gent mjn^ and join OC. Then Omo, or d^, is the displacement
of point from 5's tangent, and d^ = Cm^, is the displacement
of Cfrom the same tangent; while S is the deflection of Bfrcm
the straight line joining and C. ■
Now the vertical displacement d^ = [mom .-area O'B'T^ X
distance of its cent. grav. from 00"'\ -^ El. But the moment-
figure O'B'T^, under OB, is composed of the two triangles shown
and the '■'■normal moment-diagram ''for OB, viz. : 0"B"T^, whose
mom.-area may be called A^ and whose center of gravity is x^ ft.
from 00", while the corresponding distances for the triangles
are i a and | a .
Hence, from eq. (2), § 397, we have:
(1)
and similarly, with corresponding notation, for the right-hand
portion, or- segment, BC-, of OBC (denoting the " normal mom.-
area " C"B"T^ by A^ and reckoning x^, etc., from CC"),
Eld. = i M^a^
¥ ^2 +
i M,a, . 7 a, + A.x^.
(2)
If now a straight line be conceived to be drawn 'through B
parallel to OC, we have, from the similar triangles so formed.
FLEXURE OF BEAMS ; GEOMETRICAL TREATMENT. 497
(as ill Fig. 4454), {d, -h) ^ a, = {h- d,) -^ a,
this with eqs. (1) and (2) we have finally
Combining
Mpg, M^(a, + aS) M^a^ A^ A^x
+
+
+
+
EI8
(4)
6 ' 3 '6
1
which is the "Theorem of Three Moments."
E is the modalus of elasticity of material of beam, I the
" moment of inertia " of its cross-section ; M^, M^, M^, the
moments of stress-couples (" bending-moments ") at 0, B, and
C respectively. Distances a^ and a^ are shown in Fig. 446a,
while A^, A^, x^, and x^ are as above ; 8 being the deflection of
point B from the straight line joining and C.
U.B. It should be carefully noted that eq. (4) does not
apply unless the part of beam from to C is continuous and pris-
matic ; also that in its derivation, the elastic curve is considered
concave upward throughout ; hence if a negative number is ob-
tained for Mq, M,, or M^, in any example by the use of eq.
(4), it implies that at that section the beam is convex upward,
instead of concave ; in other words that the upper fibers are in
tension and the lower in compression (instead of the 'reverse,
as in Fig. 446a).
401. Values of ^,1-, and A2X2 iii Special Cases. The Theorem of Three
Moineuts involves the use of the (imaginary) normal mom.-area of each of the
two portions (left and right "spans", or "panels"), OB and BC ; i.e. of the
products ^jXi and A^x.^, where Xj is measured from the left end of the left panel,
and x, from the right end of the right panel. "We are now to determine values
of ^jX, and A^x., for several ordinary cases of loading.
I. Single Cen-
tral Concentrated Load,
P, Fig. 446j. Here, for
a left-hand panel,
PI I I
4 2 2'
A^x,=
— ; and for a right-
hand panel,
AoX., =
PP
16"'
1^ X
Fig. 446i.
Case II. Single Non-central Concentrated Load, P.
case as a left-hand panel,
Pbc (I + c)
A,x^= [-
Fig. 4462.
Fig. 4463. For this
while, as a right-hand
panel, -42X2=
498
MECHANICS OF ENGIKEEKING.
Case III. Two (or more) Concentrated Loads. Fig. 4463.
A^, = I [P'b'C (I + b') + P"b"c" {I + &")] ; and for each load more than
two add a proper term in the bracket.
For A.^2 interchange b' and c', b" and c", etc.
' B o
-hi
^%-
Fig 446,.
Fig. 4464.
Case IV. Any Continuous Load over a Part or the Whole of the Span; of w
ibs. per linear foot, w being variable or constant. Fig. 446^. The load on a
length dx (of loaded part) is wdx lbs. ; comparing which with the P of Case II,
(or one of the P's of Case III), we note that x corresponds to b, and l—x to c ;
— 1 /^X=Ci 1 /»Cl
hence AjX^ = ^ t wdx {I — x)x{l + x) =g- t wx {P — x') dx.
If w is variable it must first be expressed in terms of x. (For A^^ we
measure x from the right-hand end, B.)
Case V. Uniformly Distributed Load over Whole Span ; (i. e. , w is constant).
Let W, = wl, = whole load, lbs. Fig. 4465.
o liilUIIIUiUlB oliiiUUU
A^x, = A.x^
-2 I Wl I
^WP
~ 24*
Case VI. Uniformly
Distributed Load Ad-
joining one End of
Span; (left end for
example). Fig. 446g.
Total load =W= wb. Applying method of Case IV, withCj = &, and b^ =0,
we have A^^ = J^ Wbd^ — \ b^). Also from Case IV, now measuring x from
B,A.j^, = ^\W{l'-c^) (l + c).
Parabola
Fig. 446s.
FLEXUJRE OF BEAMS ; GEOMETRICAL TREATMENT. 499
Case VII. Uniformly Distributed Load Not Adjoining either End of the
Span. Fig^. 446. Whole load = W= w (e- 6). By Case IV, we tind
A,x, —
W{e + b)
-[''-
e2 + 62 ■
■A^2 ~~
12 , L ' 2 J '
12
■D'-^l
It is now seen how A^x^^ and ^2*2 ™^y
be obtained for any loading.
402. Continuous Girders Treated
by the Theorem of Three Moments.
This theorem is of special advan-
tage in solving continuous beams
(p. 271) ; and examples will now
be given.
Example I. Fig. 447^. A straight, homogeneous, prismatic
beam or girder, 35 feet long, is placed upon three supports at
the same level, forming two spans of 15' and 20'; two concen-
trated loads in the left span, a uniformly distributed load on part
of right span. Required the maximum moment, and maximum
shear. (Neglect weight of beam.)
Take 0, B, and C, as the three sections where the three
moments Mo, M^, and M^ are situated [respectively] used in the
theorem of § 400. But both Mo and M^ are zero in this case,
and 8 (deflection of point B from line joining and C) is also
zero (since the supports are on the same level). Hence M^ (i.e.,
at B) is the only unknown quantity in applying the theorem of
§ 400 (eq. (4) ) to this problem.
Taking the A^x^ from Case III, and A^x^ from Case VII (with
/= 0), of § 398, we obtain (using the foot and ion as units),
Mr^^ + ^O) _^ ^ _^_ _1 1-6x4x11x19+8x10x5x251
3 6 X 15 •- -"
+
16x16
12x20
202-
16^
1=0; and .'. M, = - 39.2 ft.-tons.
The negative sign shows that at section B the elastic curve
is convex on its upper side (see N. B. in § 400). To follow up
the solution from this point, let us draw the actual moment-
diagram somewhat differently from that in Fig. 446a, (which
see), where the actual moment for any section is measured from
500
MECHANICS OF ENGINEERING.
a continuous horizontal line, O'C, as an axis. Let the
« chords " 0"B" and B"C" of the two normal moment fig-
AVo
>'E
-f.^>-^6'-
>'F
Vb
10 urns
Hy i y V I i l i i I i i I i
AVo
k
-^—i5'-j- — M-t--
—16'— >-l:
Fig. 447i.
ures, 0"B"T^
and C"B"T^, be
made a contin-
uous horizontal
line by an up-
ward shifting
{each in its own
vertical) of the
intercepts of
those figures.
The intercepts
0"0', B"B',
and C"C', and
the four tri-
angles involved
with them, now
extend upward
from that hori-
zontal line. But in our present problem both M^^ and M^ are
zero; hence the two upper triangles disappear and the two inner
triangles project above the horizontal line, with M^ as a com-
mon base. The actual moments of the points of the elastic
curve are now measured (in general) by the vertical intercepts
between the lower boundary of the normal moment figures and
the upper edges of the two triangles ; but since in the present
case ilf J is negative, M^ must be laid off below the (new) hori-
zontal line so that the lines 0"B" and C"B" will cross the
lower boundaries of the normal figures ; the actual moments being
now measured by the vertical intercepts between these two oblique
lines and the curved (or broken^ boundaries of the normal figures.
This re-arrangement has been observed in the moment-diagram
0"B" G" of Fig. 447j, where line-shaded areas correspond to the
parts of the elastic curve which are concave upward; and the dot-
shaded areas, to parts convex upward (or upper fibers in tension).
Before determining the shears, J, along the beam, we must
first determine the reactions at the support, viz. Vq, Fgj and V^
Consider the portion OB as a " free body", cutting just on left
FLEXURE OF BEAMS: GEOMETRICAL TREATMENT. 501
of support B, and put ^ (moments) = about the neutral
axis of section at B -, deriving
6x11 + 8x5- 39.2 _ 7^ x 15 = 0; and .-. Vo= 4.5 tons.
Similarly, with EC as free body, taking moments about B,
16 x 12 - 39.2 - y^, X 20 = 0; and .-. Vo = 7.6 tons; and
hence, since V o + Vb + Vc = ^^ tons, Vg = l'^-^ tons. The
shear-diagram is now easily formed (see Fig. 417^) ; the Jiiax-
imum shear being evidently 9.55 tons, occurring in the section
just on the left of support B.
We note that the shear changes sign three times, corre-
sponding to the three (local) maximum moments (at E, B, and
■K). To locate, and determine, M^, note that the change of sign
of the shear at I) is gradual and that hence the shear is exactly
zero at K; which requires that "the load between K and the
support be equal to the reaction at C, (from the free body
concerned). Since w along HC is one ton per foot, the dis-
tance ^Cmust be 7.64 tons -v- 1.00, = 7.6 ft. From this free
body, KC, we now find, by moments, that ilf^ = 29.2 ft.-tons.
As to the other maximum moment, i.e., at E, we note that
the moment at E in the normal moment^figure would be E^'' E"
= 28.2; from which by deducting t*^ of Mb (i.e., of 39.2) we
obtain M^, = 17.7 ft.-tons. Hence, the greatest moment to be
found in the beam is that at B, viz. 39.2 ft.-tons, and upon
this depends the choice of a safe and economical beam. .
Example II. Fig. 4472, Con-
tinuous prismatic beam OG.
Three supports at same level.
Find maximum moment, etc.,
under given loading, the 12 tons
being uniformly distributed over
whole of right-hand span. Neg-
lect weight of beam.
i/ max. =ilfi,= 16.6 ft.-tons. Ans.
rj ions
-^5 ■
"■B-
us 20-
-40-
''10 if 10 tons
Fig. 4473.
■^—16'— 4^
> ' 16 tons
Fig. 4472.
Example III. Continuous
prismatic beam on three supports
O, B, C, at same level. Three
concentrated loads. Neglect
weight of beam. Find Mb and
maximum moment, etc.
Mb= - 92.6 ft.-tons. Max.
if = 116 ft.-tons, at D. Ans.
502
MECHANICS OF ENGINEERING.
Example IV. Continuous prismatic beam, 40 ft. long and extending ovei
four supports at the same level. The loading is symmetrical, as shown (Fig.
lUlUiUiUl
UllilUllUio
4474). Here we note that from symmetry Jf^ must = Mc; also Mo and Mj)
each = 0. Applying the Three-Moment Theorem to O, B, and C, (with 5=0)
we find
+ 1 Jf5X26+lilfBXl2+ l.?|xl7+ J-^.i X 12^ =0;
and . •. Ms = — 23.7 ft. -tons, ( = Mc, also).
Completing the mom. -diagram we find that Mb(= Mc), or 23.7, is greater
than any other moment along the beam; .-. M max. =23.7 ft.-tons. The
reactions of the supports are found to be : Fq (and Vd) =8.3 tons ; and Vb
25 tons (and Vc) = 16.6 tons. Evi-
illliJ,!! dently the elastic curve is con-
vex up, over both B and C.
' 10 tons
-\
^-6':
\<—7'-
S! ^
Fig. 4476-
The maximum shear is 11.8
tons, close on the left of sup-
port B, (or close on right of C) .
Example V. Continuous
prismatic beam, 38 ft. long, on
three supports at the same level.
Fig. 447^. Uniformly distributed loads over portions of the length. Find the
maximum moment and maximum shear, (J).
Max. if = - 39.6 ft.-tons (at B) ;
Max. J = 10.2 tons (close at right of B)
Example VI. Fig. 447„. Continuous prismatic beam, 50 ft. long, on /our
supports at same level ; but the arrangement of loading and span-lengths is non-
symmetrical. Fig. 4476. Find the max. M and max. J. In this case Mo and
Md are each = 0, but Mb is not = Mc- We are therefore compelled to apply
the Theorem of Three Moments twice, viz. : first to the three points 0, B, and
C; and then to the three points B, C, and B ; whence we have
- Mb (14 + 20) McX 20 16 X 6 X 8 (14 + 6) 20 x 2D»
Ans.
6 X 14
24 X20
Mb X 20 Mc (20 + 16)
+ +
20 X 203 16 X 7 X 9 qe + 7) _
24 X 20
6X16
FLEXURE OF BEAMS; GEOMETRICAL TREATMENT. 503
(1)
(2)
or, 68Jlf5 + 20 if c+ 1097.1 + 2000 = 0; ....
and 20 Mb + 72 if c + 2000 + 1358.3 = ; ....
two simultaneous equations, foi- determining ifs and Mc-
Solving, we find if 5 = - 34.6, and Mo = - 37.0, ft.-tons.
The three "normal mom. -diagrams " having been drawn to the common
base 0"B"C"D" in the figure, we lay off B"B' downward from B" and
aiiiuiiiiuiiiic
■''-\-s'-
20'.
9^
—7^
Iq" Ic" Id"
1^— -:J4
16'-\-
Fig. 4476.
= 34.6 ft.-tons; and C"C', also downward, = 37.0; and draw the straight
lines 0"B\ B'C, and C'B" ; thus completing the mom. -diagram, in which,
as before, the differently shaded portions show whether the elastic curve is con-
cave up (line-shading), or convex up (dot-shading), in the corresponding part of
beam.
The four reactions of supports are then found, viz. : To, F^, Fc, and Vd,
= 6.7, 19.2, 19.0, and 6.1 tons, respectively. Shears are now easily found and
are shown in the shear diagram, the max. J being 9.8 tons, occurring just on
the right of support B. The max. if is found to occur at E, and to have a value
of 42.7 ft.-tons.
402a. Continuous Beam with One Span Unloaded. In Fig. 448
we have a continuous prismatic beam supported at three points
0, S, and C at the same level ; but the support at is above
the beam, instead of below; since that end tends to rise, there
being no load in the left-hand span. (Case of a drawbridge with
the left-end " latched down "). Neglect weight of beam. By
the Theorem of Three Moments applied to 0, B, and C, with
Mq and Mc = 0, and Mb unknown, we find Mb = — 4.9 ft.-tons.
In forming the moment-diagram here, we note that since there is
no load from to B the lower edge of the "normal mom.-diagram"
604
MECHANICS OF ENGINEERING.
J), tons
■ Fig. 448.
for OB coincides with its upper edge, i.e., with the axis itself,
viz. 0"B". The "normal mom.-diagram " for BG is, a. triangle,
with. B"Q" as base. Laying off B"B' = 4.9 downward from
B', and joining 0" B' and
B'C'^-, we complete the
actual (shaded) mom.-
diagram, as shown.
Max. moment is found
under the load and = +
9.55 ft. -tons.
To find the reaction
at 0, take OB as " free
body", cutting close on
left of ^. (See (i^) in
the figure. Note the
position of stress-couple
at right-hand end of this
body.) By moments about B we have V a X 10 — 4.9 = 0,
whence V q = (say) 0.5 tons. The other reactions and the shear
diagram are now easily d'etermined.
403. Supports out of Level. In the foregoing examples the
quantity S has been zero in each instance of the application of
the Theorem of Three Moments ; but when such is not the
case the quantities E and I are brought into play. In this con-
nection it must be remembered that any unequal settling of the
supports (originally at same level) after the beam has been put
in place, may cause considerable changes in the values of the
various moments and shears, and consequent stresses in the
material. (See lower half of p. 323.)
404. Continuous Beam with " Built-in" Ends. Fig. 449. As
a case for illustration take the prismatic beam in Fig. 449,
" huilt-in " or clamped, horizontally, at B and at C ; at the
same level. A load P is placed as
shown. On account of the mode of
support the tangents to the elastic
curve at B and will be horizontal
and are coincident ; so that portions
of the curve near the ends are convex
up
a->
Fig. 449.
Now conceive the beam to be sustained at i? by a simple
FLEXURE OF BEAMS: GEOMETRICAL TREATMENT. 505
support underneath and to extend toward the left a length Qq
at the end of which a support, 0, is placed above the beam, and
at same level as B and C (allowing for thickness of beam).
This makes an additional span (with M q = zero) ; and the tan-
gent to elastic curve at B will no longer be horizontal. But it
may he made as nearly horizontal as we please, by taking ttg small
•enough (supposing no limit to strength of beam). When a^ =
zero the tangent at B will be in its actual position (horizontal).
We may therefore apply the Theorem of Three Moments (§400)
to 0, B, and C, [noting that there is no load on 5], if we
write both Mq, and a^, = 0, whence (see also Case II of § 401),
a(Sa + 2a)
^^ Ms (0 + 3a) ^ Mc^ ^ Q ^ P^a
0.
3 ' 6 ' ' 6 x,3a
Similarly, by conceiving the beam extended to tlie right, a dis-
tance a' to a point D, for another support, etc., we may apply
the theorem to the three points, B, C, and D in like fashion,
with Mo and a' = 0, obtaining
Ms^a , M^(3a + 0)
6
3
+ +
P2a. a (3a + a)
6 X 3a
+ = 0.
Elimination gives Mb = — t: Pa-, and Mc =
V
-Pa, ft. -tons.
y
405. Deflections found by the Theorem of Three Moments for
Prismatic Beams. Since this theorem contains h (see Fig. 446a)
the deflection of the point B
of the elastic curve of a con-
tinuous prismatic beam from
the line joining the two
others, and C, we may
use the theorem in many
cases for determining deflec-
tions when the three mcments are known.
Example I. Fig. 450. Case of two end-supports and a
single non-central load, P ; (with weight of beam neglected).
Taking 0, B, C, as the three points (i.e., OB is the left-hand,
and BO the right-hand, span: icith no lead on either span) we
have, with Mq and Mc = 0, and Mg = Pa^a, -i- I,
Fig. 450.
+ ~ i^ - + -h
I 3
= ElS
a J
d =
Pa,%^
SEI '
506 MECHANICS OF ENGINEERING.
Example n. If n is any point between and 5, at x ft.
from 0, and 0, n, and C are taken as the three points for the
theorem, we may find 8„, the deflection of n below OC^ That
is, with Mo and Mc = and Mn = P{a^ ^ I) x,
+^^L£-ii + 4- + -P^2(c^t -^)i{l -^) +^ 2]
3Z ' Q{1 — x)
= EI8„
X I — xj
or, after reduction,
^"^eif^ [z^-<-a:^> (4)
Now if ttj > a^, we may find the distance x\ from 0, of the
point of maximum deflection, by putting " ■ = : whence is
dx
obtained x' = V ^ {l^ - a,') C^)
and this substituted in (4) gives
Pa
)EI
(Compare with pp. 258 and 494.)
max. deflection, = -^^ (f - a}) V\ {I' - a/) ... (6)
(The following example is the one referred to at the loot of page 514.)
Example. — A hollow sphere of mild steel, of thickness 2 in. and internal
radius of ?'o = 4 in., contains fluid at a pressure of 2 tons/in.^ Find max.
stress and max. strain; with £'=15,000 tons/in.^ and A; = 0.30. Here
»i = 1.5; and by substitution in eq. (30) we obtain max. hoop stress to be
§■0= — 2.26 tons/in.^ (tension), while from eqs. (22) and (23) the tangential,
or hoop strain, at inner surface is found to be £3=" 0-000145 (elongation),
and the radial strain to be £1= +0.000224 (shortening).
The latter strain, £■^, is seen to be the greater and the ideal ^'equivalent
simple stress" (see § 4056) is ££1=4-3.35 tons/in.^, compression, i.e., much
larger than the actual max. stress (2.26) in this case. On the "elongation
theory" (see § 4056), the 3.36, and not the 2.26, tons/in.^, is the figure that,
for safety, should not pass a prescribed unit stress ais Liferred from com-
pressive tests with an ordinary testing machine.
THICK HOLLOW CYLINDERS AND SPHERES.
507
A' d\
CHAPTER XIII.
THICK HOLLOW CYLINDERS AND SPHERES.
405a. General Relations between Stress and Strain. — (Elas-
tic limit not passed.) If a small cube of homogeneous and
isotropic material, dx inches long on each edge, is subjected
to a compressive stress of pi Ibs./in.^ on two opposite faces,
not only is its length in direction of the stress diminished,
and by an amount dX, but its lateral dimensions are increased
by an amount d/' which is a certain fraction (from 0.20 to
0.35 for metals) of dL This ratio, or fraction, is called Poisson's
Ratio, and will be denoted by k. Thus, in Fig. 450a we have
such a cube, AD being its unstrained
form. Axes 1 and 2 are in the plane of
the paper while axis 3 is 1 to the paper.
On the left and right faces is shown acting
the compressive unit-stress pi Ibs./in.^,
A'D being the form of the cube under
this stress. If now E represent the
modulus of elasticity (Young's) of the
material, we have (see p. 203) denoting
the ratio dX-^dx, or relative decrease in length, by si, ei = px -r-E;
so that if dX" is the increase in length of the vertical edges we
havedX" -i-dx (call this ratio £2=^—kpi^E; while the relative
increase of length in the horizontal edges 1 to paper will
be an equal amount, viz., e^^—kpi-^E. These ratios £i, £2,
and £3 are called the strains along the three axes 1, 2, and
3, respectively, and are abstract numbers. Hence the three
strains produced by the stress pi acting alone are
P\ kpi kpx .
£i = ^; ^2=-~-^; and £3=--^. . . (1)
Now if while pi is still in action a compressive stress of
P2 lbs./in.2 acts on the two horizontal faces, and also a com-
pressive stress of ps on the two vertical faces which are parallel
to the paper, the total strain in the direction of axis 1 (that is,
the relative shortening of the cube in that direction) will be,
by superposition, £r
1- dx
Fig. 450a.
Pi k{p2 + Pz) J • M 1 • +U
-^ ^ — ; and similarly, m the
directions of the other two axes, we have
£0 — ^ —
P2 kipi + ps)
E
E
and £3 = Ts —
Ps k(pi + p2)
E E
(2)
508 MECHANICS OF ENGINEERING.
(This form of stress-strain relation is due to Grashof.)
Note that if either p^, p2, or ps is a tensile stress, a negative
number must be substituted for it; and that if a negative num-
ber is obtained for si, £2, or £3, in any problem, it indicates a
lengthening instead of a shortening. Similarly, if the con-
dition is imposed that the strain £2 (say) shall be a relative
elongation of 0.00020,-0.00020 must be substituted for it in
above relation.
405b. "Elongation Theory" of Safety. — In all preceding
chapters the criterion of safety has been that the unit-stress
in the element of the elastic body where the stress is highest,
regardless of stress on the side faces, should not exceed a cer-
tain value, or working stress, =R' Ibs./in.^, as determined
upon by a consideration of the stress at "elastic limit; " this
"elastic limit " being itself determined by the ordinary ex-
periments on "simple " tension or compression of rods of the
material in question, there being no stress on the sides of the
rod. In such experiments, however, an element with four
faces parallel to the axis is subjected to stress, say p, on two
(end) faces only; and the question naturally arises whether
the elastic limit would be reached for the same value of p as
before, in case there were also present tensile or compressive
stress acting on the side faces of the element. Experiments
which would throw much light on this point are unfortunately
wanting, and some authorities,, notably on the continent of
Europe, contend that the extreme limit of safety, as regards
state of stress in isotropic materials, is when the greatest rela-
tive strain (elongation or shortening), say £1, is as great as
would be produced at the elastic limit in an experiment involving
only "simple " tension, or compression (as above described),
in an ordinary testing machine. This view would make the
greatest "strain," or deformation (change of form), the cri-
terion of safety instead of the greatest stress. Now if a stress
of "simple " tension, =p', (no side stresses) acts on an element,
the highest strain produced is in thQ direction of this stress
and has a value e' = p' ^E, since Young's modulus, E, is deter-
mined by experiments of this very nature; that is, p' = Ee'.
Hence if the greatest strain in an element in some compound
state of stress, as in § 405a, is £1, and it is desired to place it
equal to | (say) of the £1 in simple tension at elastic limit,
we may write £i = f £' = |(p'-^£'); or E£i = lp'. If now we
THICK HOLLOW CYLINDEES AND SPHERES.
509
denote |p' by p" we may write Eei==p" and describe j/' , or Eei,
as the ideal tensile stress which would produce a strain, or relative
elongation, equal to £i in case there were no side stresses; Cotterill
calls this ideal stress (Esi) the ^'equivalent simple stress."
For instance, if on an element of the shell of a cylindrical
steam-boiler of soft steel the "hoop stress " (p. 537) is pi on
two end faces and the stress on two of the other faces is p2,
= -|pi, (the stress on the remaining two opposite faces being
ps = practically zero in this connection) we have for the strain
in direction 1, by eq. (2), Eei = pi-k{hpi + 0). . . . (2')
Let p", =—15,000 lbs./in.2, tension, be the safe working
stress for the metal in simple tension; with £' = 30,000,000
lbs./in.2, and Poisson's ratio = A; = 0.30. Then according to
the view of preceding chapters the greatest safe value for the
stress pi would be —15,000 Ibs./in.^ But according to the
new view now being illustrated the safe value of pi must be
determined by limiting the strain si to a value which would
be produced by 15,000 Ibs./in.^ in simple tension, i.e., —0.00050,
(which =^p" -^E); which amounts to the same thing as re-
quiring that the ''equivalent simple stress" shall =15,000
lbs./in.2 Hence, substituting in eq. (2'), we have
- 15,000 = pi(l- J. (0.30)); i.e., pi = -17,600 lbs./ in.2
tension; which is considerably greater than the 15,000 allowed
by the older theory. The relation thus brought out in this
case that the tenacity of a material is increased by the presence
of lateral tension "can hardly be considered as intrinsically
probable, and such direct experimental evidence as exists is
against the supposition " (Cotterill).
But in many cases the results of this "elongation theory "
are more probable than those based on the older theory ; hence
the former is much favored by continental writers.
405c. Thick Hollow Cylinder. Stresses and Strains. — Fig.
4506 shows a longitudinal section of a thick hollow cyUn-
der of homoge-
neous and isotropic
material (say steel
or iron) provided
with end stoppers
(^no iriciion nor vj/,,,,,,,^,,,,,,. „y ,,^yy„,,^//,/,,///////,///
leakage); and Fig. Fig. 4506.
450c a transverse section, giving dimensions.
Fig. 450c.
ro is the inner
510 MECHANICS OF ENGINEERING.
radius and nro the outer radius {n is a ratio). Fig. 450c also
shows (dotted Knes) an elementary hoop, or shell, of inner
radius r and outer radius r + dr. The interior of the cylinder
is filled with fluid under a high pressure, po Ibs./in.^; and it
is required to determine the stresses and strains in a cubic
element in any elementary hoop or sign such as ABC, Fig. 450c.
Let the half hoop ABC of Fig. 450c be considered as a "free
body " in Fig. 450d, showing also at 3 a small cubic element,
as mentioned above. The compressive stress
P+dpk \ A-j-,'^ on the inner surf ace of the hoop isp (radial),
\l jK^^'^^jT exerted on it by the adjacent inner hoop;
while on the outer surface of the elementary
\ I hoop, and exerted on it by the adjacent
r___i-Ji— a'- outer hoop, is the compressive stress p+dp.
The thickness of the hoop is dr. The stress
on the edges, A and C, of this free body
(half hoop), will be taken as compressive
/\ ^h-y^l" at first, of intensity q Ibs./in.^ Let the
21 / / f s ^ hoop or thin shell have a length =Z, 1 to
Fig. 450d. ^^^^^ -^ ^ ^ longitudinally (see Fig. 4506).
Now for this free body put IX = and we have (see pp. 525
and 526) {p + dp)i2r + 2dr)l-2ql . dr~p{2rl)=0 . .(3)
i.e., pr + r . dp + p . dr + dp . dr — q . dr—pr = . . (4)
and hence, omitting the term dp . dr of the second order,
r . dp + p . dr^q . dr (5)
which is a differential equation of stress. Next consider the
relations of stress and strain to be found in the small cube at
3, Fig. 450d. It is subjected to a compressive stress p along
the radial axis 1 ; to a compressive stress q a long the tangential
axis 2; while on the front and back faces the stress is p3 = zero,
parallel to axis 3 (t to paper). Let now oi denote the radial
strain, £2 the tangential, and £3 the axial strain, this latter
being parallel to the axis of the cylinder. All of these strains
are supposed to be shortenings for the present; and from the
circumstances of the case the third strain £3 (axial) is con-
sidered constant (i.e., the same for all values of the variable
r), since the cylinder is under no constraint as to longitudinal
change of form.
We may therefore write the relations (see § 405a)
Ee, = p-k.q, (6); Ee2 = q-k.p, (7); Ee3 = 0-k(p + q), (g)
THICK HOLLOW CYLINDERS AND SPHERES. 511
From (8) we have q=—'p—{Eez-^k), which in (5) gives
r .dp + 2pr .dr=-{Ee3^k)dr .... (9)
Now multiply by r (integrating factor) and denote Eos-^k
by A, an unknown constant, (unknown since it contains the
strain £3) and we have
r2. dp + 2pr . dr=—Ar . dr; . . . (10)
that is to say, d[r^p]= —Ar . dr; . .' . (11)
which may be integrated; giving, r^p=—^Ar^-\-C; . . (12)
where C is a constant of integration. The two constants A
and C may now be determined by substituting in (11) the
values To and po which the two variables r and p assume at
the inside surface of the cylinder. Similarly, at the outside
surface r and p have the values nro and (atmospheric pressure
relatively small and hence neglected); which being placed in
(11) give rise to a second equation, which like the first contains
constants only. From these two equations we easily find
. 2po , „ n^ro^po
A = -^ — r; and C = ^ — r-;
n^ — r n^ — 1
and hence finally, from equations (12) and (8),
P=^[^-'l (13);and,= -J^j[5>Vl]. (14)
From (13) and (14) we may find the stresses p and q for
any value of the variable distance, r, from the axis. The
negative sign for q shows that it is in reality a tensile stress,
the reverse of the character assigned to it at first; i.e., it is
a "negative compressive " stress for this case of fluid pressure
acting inside the cylinder. Both p and q have maximum values
at the inner surface and diminish toward the outside.
Example. — With inner radius rQ = 4 in.; and outer, =5 in. (hence w = 5/4
or 1.25); and j9o = 800 lbs. /in. ^; we find q max. (or q^), at inner surface,
to be 3644 Ibs./in.^; while at outer surface g= — 2844 lbs. /in. ^, tension.
If the metal is cast iron we may put ^ = 0.23 (see p. 230) and £' = 15,000,000
lbs. /in. ^, and thus obtain for the radial strain at the inner surface, e,=
+ 0.000109, indicating a shortening; and for the "hoop" strain (or tan-
gential strain) at the same place, £2= ~ 0-000255, i.e., a relative elongation
of about 2^ parts in 10,000. (The student should verify all the details of
this example, carefully noting the proper signs to be used).
405(1. Thick Hollow Cylinder under External Fluid Pressure. — If the cylinder is
siirrounded by fluid under high pressure, pn lbs. /in. ^, the internal pressure
Po being practically zero (atmosphere) in comparison, we must determine
the constants A and C of eq. (12) on the basis that p = for r = ro and
that p = pn for r = nr^ ; whence, finally, we obtain
^=5^('-'^)^ ■ ■ ■ ('« "-^^ii^^')) a«
for the stresses at any distance r from the axis. Here both p and q are com-
512 MECHANICS OF ENGINEERING.
pressive stresses; the latter increasing, and the former diminishing, toward
the center. Evidently if the cylinder were not hollow, but solid, r^ would = 0,
and n = cc ; and both p and q would be constant, =pn, at all points.
405e. Approximate equalization of the tensile hoop stress in
successive rings of a thick hollow cylinder under internal
fluid pressure may be brought about by using two or more
separate, cylinders of which fhe outer ones are successively
"shrunk on" before the fluid is introduced. For instance,
with only two cylinders, the outer one is first heated to such
an extent that it barely fits over the inner one, the latter being
cold. When the compound cylinder has cooled the outer one
has shrunk and is in a state of hoop tension, while the inner
one is in a state of hoop compression. The radial pressure
between them, at their siirface of contact, is an internal or
bursting pressure for the outside cylinder, and an external
or collapsing pressure for the inside cylinder. The original
and final temperatures being known, we are able to make
use of the foregoing equations [(6) to (16)] to compute all
stresses and strains thus induced before the fluid is intro-
duced into the interior of the smaller cylinder. When the
internal pressure is eventually produced, the hoop stresses
in the smaller cylinder, initially compressive, will be con-
verted into moderate tensions and the tensile stresses in the
external cylinder will be increased; but the maximum ten-
sion is not so great as if the complete cyhnder had been origi-
nally continuous.. (See Prof. Ewing's Strength of Materials).
In the case of thick hollow cylinders subjected to the severe
internal pressures needed in the manufacture of lead pipe, to
produce ''flow" of the metal, it is well known (Cotterill)
that a permanent increase in the internal diameter takes place,
showing that in the inner layers of the cylinder the elastic
limit has been passed. In this way an approach is made to
equalization in the hoop stresses of all the layers and the above
formuliE no. longer hold; but the cylinder as a whole is not
injured, having simply adapted itself better to its function.
Cast-iron hydraulic press cylinders are sometimes subjected
to the high internal pressure of 3 tons/in.^ If the cylinder
is short, its resistance is doubtless much increased by connec-
tion with the end plates, or ''domes."
405f. Equation of Continuity for Thick Hollow Cylinder under Stress. — (Cotterill.)
In Fig. 450e we have mABCD the form and position of a " cubical" element (of
an elementary hoop) between the two radial planes ABO and Z)CO, in its con-
THICK HOLLOW CYLINDERS AND SPHERES.
513
dition of no stress. After it is subjected to stress it will still be iound*between
the same radial planes and will occupy some position A'B'C'D'. The radial'
thickness AB = dr, or t, f i
will have changed to t', / /L ? '
BC has changed to B'C,
etc. With axes 1 and 2
as shown, 1 being radial,
and 2 tangential (or cir- I"
cumferential), we note
that the whole circumfer-
ence of which BC is a part
has shortened from a
length 2Tzr to 27rs, so that 27rs=2ffr(l— Sj), £2 being the value of the tan-
gential strain, or "hoop" strain, at distance r from center 0; and similarly
2ns' = 27zr'{l— £'2), where £3' is the hoop strain at distance r' (i.e., r+dr)
from 0. But £2 varies with r, so £2'= ^2 + t ■ -j^- Hence, subtracting.
FiG. 450e.
t' = t-t£,-r't
s=(r'-
ds2
dr
■r){l~e2)-
dr'
-r't-r.
dr
or.
whence
£1 — e. = r
d£.
-r't'^-
dr '
dr
(17)
(18>
.(19)
which is the "equation of continuity of substance," of the cylinder.
Since in the foregoing cases of thick hollow cylinder, under bursting or'
collapsing fluid pressure, both s^ and $2 may be expressed as functions of^
r, it is a simple matter for the student to show that (19) is verified in those
cases, and that hence the solutions given are adequate.
Evidently eq. (19) also holds in the case of the thick hollow sphere, where
there is a hoop strain £3, q to paper in Fig. 450, equal to that, £2? along axis 2.
lOcg. Thick Hollow Sphere under Internal Fluid Pressure. — (For thin-
walled spheres see p. 536). As thick hollow spheres are sometimes used'
to hold fluids under high pressure, and the halves of such spheres fre-
quently form the ends of thick hollow cylinders, the following treatment
will be of practical value. In Fig. 450/ we have a cross-section of the sphere
through the center. The inner radius is Tq; the
outer, nro (where n is a ratio) ; while the (variable)
r is the inner radius of
any thin spherical shell,
Fig. 450/. Fig. 450?.
of thickness dr, an infinite number of which make up the complete sphere.
Though each of these shells is under a "hoop tension," when the internal
fluid pressure is in action, we shall at first deal with this stress as if
compressive, for uniformity in applying the principles of § 450a.
514 MECHANICS OF ENGINEERING.
Consider as a free body any hemispherical shell as shown in Fig. 450g.
The pressure (radial) on the inside, from the adjacent shell, is p lbs. /in.*;
and that from the adjacent shell on the outside is p + dp, or p'. The radius
of the outside will be called r', { = r + dr). The "hoop compression" on the
thin edge of the shell is q Ibs./in.^ These three quantities, p, r, and q,
are variables, i.e., refer to any infinitesimal shell of the sphere. The strains
affecting any small "cubical" block in any shell are p, radial; q, tangential;
and 53, =q, along a tangent T to the first mentioned.
Taking an axis X through the center and T to sectional plane AB, and
putting 2 (X-components) = 0, for equilibrium, we have
7ir'^p' — 7rr^p—27:r.dr.q = 0; i.e., r'^p' — r^p = 2qr . dr.
But the difference, r'^p' — r^p, is nothing more than the increment accruing
to the product r^p when r takes the increment dr, and is therefore the
differential of the quantity or product r^p ; hence we may write
d{r^p)=^2qr . dr; or, r^ . dp+2pr . dr = 2qr . dr . . . (20)
We next consider the relations of stress and strain in the small cubical
element shown in Fig. 450/i, having four radial and two tangential faces.
The unit stresses on the faces are as there shown; p on the two tan-
gential, and q on the four radial, faces. Radial strain = e^ and hoop strain
(tangential) = £2 = same for any tangent. It has already been proved in
§405/ that e,-e, = r.^; (21)
and we also have, from § 405a, Eei = p—2kq; (22)
and Ee2 = q-k{p + q) (23)
From the four equations, (20), (21), (22) and (23), containing the five
variables p, q, £1, £2) and r, we may by elimination and integration finally
determine p and q, each as a function of r; as follows: (C, C", Cj, C^, etc.,
will denote constants of integration, or, constant products involving E and k.
From (22) and (23) we obtain a value for £1— £n which in (21) gives rise
to an expression for p—q. Another expression for p—q is obtained from
(20) . Equating these two expressions we derive —dp = Ci . dsi', that is,
by integration, — p = Ci£2 + C2 (24)
By elimination of £2 between (24) and (23) we obtain q in terms of p which
substituted in (20) produces r^ . dp + 2>pr . dr = C^r . dr .... (25)
Eq. (25) is made integrable by multiplying by r (integrating factor);
whence r^ . d'p + 3pr^ . dr = C^r'^ . dr
The left-hand member is evidently d[r^p\ Therefore (^[r^^?] = C^r^ . dr ;
C"
or, integrating, Hp = ^Cor^ + C^ ; that is, p =C'-\ — 3-, . . (26)
dp 3C" C"
whence, also, -f^= ^; which in (20) gives rise to q = C' — -^. . (27)
We may now determine the two constants C and C" substituting in eq. (26),
first p = P(, and r = rg ; and then p = with r = nr^. The values of C" and C" so
obtained are placed in (26) and (27), resulting finally in the relations
P=„-^5?^-l)^ • (28) and ,_--.?i-.("^+l). . (29)
The negative sign of q shows that it is a tensile stress, or "hoop tension."
It is evidently a maximum for r=rf), this maximum value being
Pi
^(^+0 <^«
(For a numerical example see foot of p. 506).
go, =gmax.,= . ^^
PART IV.
HYDRAULICS.
CHAPTEE I.
DEFINITIONS— FLUID PRESSURE-HYDROSTATICS BEGUN.
406. A Perfect Fluid is a substance the particles of which
are capable of moving upon each other with the greatest free
dom, absolutely without friction, and are destitute of mutual
attraction. In other words, the stress between any two con-
tiguous portions of a perfect fluid is always one of comjpression
and normal to the dividing surface at every point ; i.e., no
shear or tangential action can exist on any imaginary cutting
plane.
Hence if a perfect fluid is contained in a vessel of rigid ma-
terial the pressure experienced by the walls of the vessel is
normal to the surface of contact at all points.
For the practical purposes of Engineering, water, alcohol,
mercury, air, steam, and all gases may be treated as perfect
fluids within certain limits of temperature.
407. Liq[uids and Gases. — A fluid a definite mass of which
occupies a definite volume at a given temperature, and is in-
capable both of expanding into a larger volume and of being
compressed into a smaller volume at that temperature, is called
a Liquid, of which water, mercury, etc., are common examples;
whereas a Gas is a fluid a mass of which is capable of almost
indefinite expansion or compression, according as the space
within the confining vessel is made larger or smaller, and al-
ways tends to fill the vessel, which must therefore be closed in.
every direction to prevent its escape.
515
616 MECHANICS OF ENGINEERING.
Liquids are sometimes called inelastic fluids, and gases
elastic fluids.
408. Eemarks. — Though practically we may treat all liquids
as incompressible, experiment shows them to be compressible
to a slight extent. Thus, a cubic inch of water under a pres-
sure of 15 lbs. on each of its six faces loses only fifty millionths
(0.000050) of its original volume, while remaining at the same
temperature ; if the temperature be sufficiently raised, how-
ever, its bulk will remain unchanged (provided the initial tem-
perature is over 40° Fahr.). Conversely, by heating a liquid in
a rigid vessel completely filled by it, a great bursting pressure
may be produced. The slight cohesion existing between the
particles of most liquids is too insignificant to be considered in
the present connection.*
The property of indefinite expansion, on the part of gases,
by which a confined mass of gas can continue to fill a confined
space which is progressively enlarging, and exert pressure
against its walls, is satisfactorily explained by the " Kinetic
Theory of Gases," according to which the gaseous particles are
perfectly elastic and in continual motion, impinging against
each other and the confining walls. Nevertheless, for prac-
tical purposes, we may consider a gas as a continuous sub-
stance.
Although by the abstraction of heat, or the application of
great pressure, or both, all known gases may be reduced to
liquids (some being even solidified); and although by con-
verse processes (imparting heat and diminishing the pressure)
liquids may be transformed into gases, the range of tempera-
ture and pressure in all problems to be considered iu this work
is supposed kept within such limits that no extreme changes of
state, of this character, take place. A gas approaching the
point of liquefaction is called a Vapor.
Between the solid and the liquid state we find all grades of
intermediate conditions of matter. For example, some sub-
stances are described as soft and plastic solids, as soft putty,
moist earth, pitch, frosh mortar, etc.; and others as viscous and
sluggish liquids, as molasses and glycerine. In sufficient bulk,
* Water has recently been subjected to a pressure of 65,000 Ibs./in.^;
resulting in a reduction of 10 per cent in the volume. See Engineering
News, Oct. 1900, p. 236.
DEFINITIOlSrS — FLUID PRESSURE — HYDROSTATICS. 517
however, the latter may still be considered as perfect fluids.
Even water is slightly viscous.
409. Heaviness of Fluids. — The weight of a cubic unit of a
homogeneous fluid will be called its heaviness,* or rate of
weight (see § 7), and is a measure of its density. Denoting it
hy y, and the volume of a definite portion of the fluid by Vy
Ive have, for the weight of that portion,
6^= Fr.
a)
This, like the great majority of equations used or derived in
this work, is of homogeneous form (§ 6), i.e., admits of any sys-
tem of units. E.g., in the metre-kilogram-second system, if y
is given in kilos, per cubic metre, Y must be expressed in
cubic metres, and G will be obtained in kilos.; and similarly
in any other system. The quality of 7, = 6^ -f- "F, is evidently
one dimension of force divided by three dimensions of length.
In the following table, in the case of gases, the temperature
and pressure are mentioned at which they have the given
heaviness, since under other conditions the heaviness would be
different ; in the case of liquids, however, for ordinary pur-
poses the effect of a change of temperature may be neglected
(within certain limits).
HEAVINESS OF VARIOUS FLUIDS.*
[In ft. lb. sec. system; y = weight in lbs. of a cubic foot.]
Liquids.
Freshwater, y =^ 63.5
Sea water 64.0
Mercury 848.7
Alcohol 49.3
Crude Petroleum, about 55.0
(N.B. — A cubic inch of water
weighs 0.0361 lbs.; and a cubic
foot 1000 av. oz.)
p(„„„„ J At temp, of melting ice; and 14.7
^^tdbet, j jjjg pgj, gq jj^_ tension.
Atmospheric Air 0807C
Oxygen 0.0893
Nitrogen ...0.0786
Hydrogen 0.0056
Illuminating ) from 0. 0300
Gas, )to 0.040(^
Natural Gas, about 0.0500
* Sometimes called "specific weight;" while its reciprocal, or \^x
may be styled the "specific volume" of the substance, i.e., the volume
of a unit of weight.
518
MECHANICS OF ENGINEEEING.
For use in problems where needed, values for the heaviness
of pure fresh water are given in the following table (from
Rossetti) for temperatures ranging from freezing to boiling ;
as also the relative density, that at the temperature of maxi-
mum density, 39°. 3 Fahr. being taken as unity. The temper-
atures are Fahr., and y is in lbs. per cubic foot.
Temp.
Rel.
Dens.
7-
Temp.
Rel.
Dens.
V-
Temp.
Rel.
Dens.
V-
32°
.99987
62.416
60°
.99907
62.366
140°
.98338
61.886
35°
.99996
62.421
70°
.99802
62.300
150°
.98043
61.203
39°. 3
1.00000
62.424
80°
.99669
62.217
160°
.97729
61.006
40°
.99999
62.423
90°
.99510
62.118
170°
.97397
60.799
43°
.99997
62.422
100°
.99318
61.998
180°
.97056
60.586
45°
.99992
62.419
110°
.99105
61.865
190°
.96701
60.365
50°
.99975
62.408
120°
.98870
61.719
200°
.96333
60.135
55°
.99946
62.390
130°
.98608
61.555
212°
.95865
59.843
From D. K. Clark's] for temp. =
" Manual." \ y =^
230°
59 4
250° 270°
58.7 58.2
290°
57.6
298°
57.3
338°
56.1
366°
55.3
390°
54.5
Example 1. What is the heaviness of a gas, 432 cub. in. of
which weigh 0.368 ounces? Use ft.-lb.-sec. system.
432 cub. in. = \ cub. ft. and 0.368 oz, = 0.023 lbs.
.*. y =. -z=: -'— — = 0.092 lbs. per cub. foot.
y i
Example 2. Required the weight of a right prism of mer-
cury of 1 sq. inch section and 30 inches altitude.
30
F=30 X 1 = 30 cub. in
table, y for mercury = 848.Y lbs. per cub. ft.
30
^ ^^ - cub. feet : while from the
1Y28. '
its weight = ^ = Yy
17!
X 848 r ■= 14.73 lbs.
410. Definitions. — By Hydraulics we understand the me-
<'iianics of fluids as utilized in Engineering. It may be divided
i7Jto
Hydrostatics^ treating of fluids at rest ; and
Hydrokinetics^ which deals with fluids in motion. (The
name Pneumatics is sometimes used to cover both the statics,
and kinetics of gaseous fluids.)
DEFINITIONS — FLUID PRESSURE — HYDROSTATICS. 519
411. Pressure per Unit Area, or Intensity of Pressure. — As in
§ 180 in dealing with solids, so here with fluids we indicate the
pressure per unit area between two contiguons portions of
fluid, or between a fluid and the wall of the containing vessel,
by J?, so that if dP is the total pressure on a small area dF^
we have
dP ...
^^dF (^)
as the pressure per unit area, or intensity of pressure (often,
though ambiguously, called the tension in speaking of a gas)
on the small surface dF. If pressure of the same intensity
exists over a finite plane surface of area = F, the total pres-
sure on that surface is
P = fpdF=pfdF= Fp, 1
P ^ .... (2)
or i? = p. J
(N.B. — For brevity the single word " pressure" will some-
times be used, instead of intensity of pressure, where no am-
biguity can arise.) Thus, it is found that, under ordinary con-
ditions at the sea level, the atmosphere exerts a normal pressure
(normal, because fluid pressure) on all surfaces, of an intensity
of about p = 14:7 lbs. per sq. inch (= 2116. lbs. per sq. ft.).
This intensity of pressure is called pne atmosjphere. For ex-
ample, the total atmospheric pressure on a surface of 100 sq.
in. is [inch, lb., sec]
P=i^p=. 100X14.7 = 1470 lbs. ( = 0.735 tons.)
The quahty of p is evidently one dimension of force
divided by two dimensions of length.
By one ^' atmosphere,^' then (or "standard atmosphere;"
an arbitrary unit), is to be understood a unit-pressure of
14.70 Ibs./sq. in., or 2116.8 Ibs./sq. ft. This would be the
weight of a prismatic column of w^ater one sq. in. in section
and 33.9 ft. high (commonly considered 34 ft. for ordinary
computations); or of a prismatic column of mercury 30 in.
high and one sq. in. section. These numbers, 14.70, 34,
520
MECHANICS OF ENGINEERING.
and 30, with their meanings as above, should be memorized
by the student; as they are to be of very frequent service
in this study.
At high altitudes the actual pressure of the air is much
smaller than the conventional "atmosphere." E.g. (see
p. 621) at 7000 ft. above the sea the height of a mercury
column measuring the air pressure- is only about 24 in.,
instead of the 30 in. above cited; varying somewhat, of
course, with the weather.
412. Hydrostatic Pressure; per Unit Area, in the Interior of a
Fluid at Rest. — In a body of fluid of uniforin heaviness, at
rest, it is required to find the mutual pressure per unit area be-
tween the portions of fluid on opposite sides of any imaginary
cutting ])]ane. As customary, we shall consider portions of
the fluid as free bodies, by supplying the forces exerted on
them by all contiguous portions (of fluid or vessel wall), also
those of the earth (their weights), and then apply the condi-
tions of equilibrium.
First, cutting plane horizontal. — Fig. 451 shows a body of
homogeneous fluid confined in a rigid
vessel closed at the top with a small air-
tight but frictionless piston (a horizontal
disk) of weight = G and exposed to at-
mospheric pressure {=Pa per ^ii^it area)
on its upper face. Let the area of piston-
face be = ^. Then for the equilibrium
of the piston the total pressure between
its under surface and the fluid at must
be
WSM
Fm. 451.
F=G + Fpa,
and hence the intensity of this pressure is
' G
i^o
F
-\-Pa'
(1)
It is now required to find the intensity, p, of fluid pressure
between the portions of fluid contiguous to the horizontal cut-
ting plane BCa.t a vertical distance = A vertically below the pis-
ton 0. In Fig. 452 we have as a free body the right parallel©-
FLUID PEESSUEE.
521
piped OBC oi Fig. 451 with vertical sides (two || to paper and
four ~j to it). The pressures acting on its six faces are normal
to them respectively, and the weight of the prism is = vol.
X;k = FhVi supposing y to have the same value at all parts of
the column (which is practically true for any height of liquid
and for a small height of gas). Since the
prism \2 in equilibrium under the forces
shown in the figure, and would still be so
were it to become rigid, we may put (§ 36) >i
2 (vert, compons.) = and hence obtain *" j
Fp-F:p,- Fhy = 0. . . (2) r ! ^^^
>JT-i-T— rx"
(In the figure the pressures on the ver- ^
tical faces i| to paper have no vertical com- ^P
ponents, and hence are not drawn.) From fig. 452,
(2) we have
JP =P. + hy.
(3)
{hy, being the weight of a column of homogeneous fluid of unity
cross-section and height A, would be the total pressure on the
base of such a column, if at rest and with no pressure on the
upper base,, and hence might be called intensity due to weigJd.)
Secondly, cutting plane oblique. — Fig. 453. Consider free
an infinitely small right triangular prism bed, whose bases are
li to the paper, while the three side
faces (rectangles), having areas = dF,
dF^ , and dF^ , are respectively hori-
zontal, vertical, and oblique ; let angle
cbd = a. The surface he is a portion
^_ V^h of the plane BC oi Fig. 452. Given
H — V j? (= intensity of pressure on dF) and
" ) Of, required ^2? the intensity of pressure
on the oblique face hd, of area dF^.
SJS,. B. — The prism is taken very small
in order that the intensity of pressure may be considered con-
stant over any one face ; and also that the weight of the prism
may be neglected, since it involves the volume (three dimen-
Fm. 453.
522 MECHANICS OF ENGINEEEIFG.
sions) of the prism, while the total face pressures involve onlj
two, and is hence a differential of a higher order.]
From ^ (vert, compons.) = we shall have
p^dF^ cos a —j>dF= ; but dF-^ dF^ = cos or ;
I>.=P, (4)
which is independent of the angle a.
Hence, the mtensity of fluid 'pressure at a given point is
the same on all imaginary cutting planes containing the
point. This is the most important property of a fluid, and is
true whether the liquid is at rest or has any kind of motion ;
for, in case of rectilinear accelerated motion, e.g., although the
sum of the force-components in the direction of the accelera-
tion does not in general = 0, but = mass X ace, still, the
mass of the bodj in question is = weight -i- g, and therefore
the term mass X ace. is a dijfferential of a higher order than
the other terms of the equation, and hence the same result
follows as when there is no motion (or uniform rectilinear
motion).
413. The Intensity of Pressure is Equal at all Points of any
Horizontal Plane in a body of homogeneous fluid at rest. If
we consider a right prism of the fluid in Fig. 451, of small
vertical thickness, its axis lying in any horizontal plane £0^
its bases will be vertical and of equal area dF. The pressures
on its sides, being normal to them, and hence to the axis, have
no components |1 to the axis. The weight of the prism also
has no horizontal component. Hence from 2 (hor. comps.
II to axis) = 0, we have,^i smdp^ being the pressure-intensi-
ties at the two bases,
p,dF-p,dF=0; .:p=p,, .... (1)
which proves the statement at the head of this article.
It is now plain, from this and the preceding article, that
the pressure-intensity p at any point in a homogeneous fluid
at rest is eqiial to that at any higher point, plus the weight
FLUID PRESSURE.
523
{hy) of a column of the jiuid of section unity and of altitude
\fi) = vertical distance between the joints.
^.(^.,
p =p. + hy,
(2)
whether they are in the same vertical or not^ and whatever he
the shajpe of the containing ^
vessel {or pipes), provided the
fluid is continuous hetween
the two points I for, Fig, 454,
bj considering a series of
small prisms, alternately ver-
tical and horizontal, ohcde, we
know that
Pd=Pc — Ky ; and Pc—Pd'i
hence, finally, by addition we have
(in which A = Aj — h^.
If, therefore, upon a small piston at 100 feet (for, gas being compressible, the lower
i^^trata are generally more dense than the upper), but in (3) the
pistons must be fixed, and P^ and P„ refer solely to the in-
terior pressures.
524 MECHANICS OF EISTGIlSrEERIlS'G.
Again, if A is small or jp^ very great, the term hy may be
omitted altogether in eqs. (2) and (3) (especially with gases,
since for them y (heaviness) is usually small), and we then
have, from (2),
i^=i?o; (4)
being the algebraic form of the statement: A l)ody of Jkiid
at 7'est transmits pressure with equal intensity in every direc-
tion and to all of its parts. [Principle of "Equal Transmis.
sion of Pressure."]
414. Moving Pistons. — If the fluid in Fig. 454 is inelastic
and the vessel walls rigid, the motion of one piston (c) through
a distance s^ causes the other to move through a distance s^ de-
termined by the relation F^s^ = F^s^, (since the volumes de-
scribed by them must be equal, as liquids are incompressible) ;
but on account of the inertia of the liquid, and friction on the
vessel walls, equations (2) and (3) no longer hold exactly, still
are approximately true if the motion is very slow and the
vessel short, as with the cylinder of a water-pressure engine.
But if the fluid is compressible and elastic (gases and vapors ;
steam, or air) and hence of small density, the effect of inertia
and friction is not appreciable in short wide vessels like the
cylinders of steam- and air-engines, and those of air-compres-
sors ; and eqs. (2) and (3) still hold, practically, even with high
piston-speeds. For example, in the space ABy
Fig. 455, between the piston and cylinder-head
of a steam-engine (piston moving toward the
right) the intensity of pressure, ^, of the
steam against the moving piston B is prac-
FiG. 455. tically equal to that against the cylinder-head
A at the same instant.
415. An Important Distinction between gases and liquids
(i.e., between elastic and inelastic fluids) consists in this:
A liquid can exert pressure against the walls of the contain-
ing vessel only by its weight, or (when confined on all sides)
by transmitted pressure coming from without (due to piston
pressure, atmospheric pressure, etc.); whereas —
FLUID PRESSUEE. 625
A gas, confined, as it must be, on all sides to prevent dif-
fusion, exerts pressure on the vessel not onlj by its v^eiglit,
but by its elasticity or tendency to expand. If pressure from
without is also applied, the gas is compressed and exerts a still
greater pressure on the vessel walls.
416. Component, of Pressure, in a Given Direction. — Let
A BCD, whose area = c.Zi^ be a small element of a surface,
plane or curved, and^ the intensity of A
fluid pressure upon this element, then ^ap\ /i\
the total pressure upon it is pdJF, and is \/^ I \q
of course normal to it. Let A'B' CD be / ---'''^^ >
the projection of the element dJc upon cc X \|b'^,--'
a plane CDM making an :?.ngle a with y" ^X,-^-'"''
the element, and let it be required to j '
find the value of the component oi jpdF ^^^' '*°^"
in a direction normal to this last plane (the other component
being understood to be 1| to the same plane). We shall have
Compon. ofpdF ~\ to CDM = pdF cos a =j>{dF. Goa a). (1)
But dF . cos a = area A'B' CD ^ the projection of „ 4- A;j/ = 25236 + 3 X 62.5 = 25423 lbs. per sq. ft.
= 175.6 lbs. per sq. inch or 11.9 atmospheres, and the total
upward pressure at e on base of plunger is
P = FePe =7t'±-p, = i 7r{iy X 25423 = 81194 lbs.,
or almost 16 tons (of 2000 lbs. each). The compressive force
upon the block or bale, C, = P less the weight of the plunger
and total atmos. pressure on a circle of 15 in. diameter.
528
MEGHAlSriCS OF ENGINEERING.
419, The Dividing Surface of Two Fluids (which do not mix) in
Contact, and at Sest, is a Horizontal Plane, — For, Fig. 459, sup-
posing any two points e and O of this, sur-
face to be at different levels (the pressure
at being ^o, that at ejCg, and the teavi-
nesses of the two fluids 7, and y^ respec-
tively), we would have, from a considera-
tion of the two elementary prisms eb an
to (vertical and horizontal;, the relation
Fig. 459.
while from the prisms eo and gO^ the relation
These equations are conflicting, hence the aoove supposition
is absurd. Therefore the proposition is true.
For stable equilibrium, evidently, the heavier fluid must oc-
cupy the lowest position in the vessel, and if there are several
fluids (which do not mix), they will arrange
themselves vertically, in the order of their den-
sities, the heaviest at the bottom, Fig. 460. On
account of the property called diffusion the par-
ticles of two gases placed in contact soon inter-
mingle and form a uniform mixture. This fact
gives strong support to the " Kinetic Theory of
Gases" (§ 408).
Fig. 460,
420. Free Surface of a Liquid at Rest. — The surface (of a
liquid) not in contact with the walls of the containing vessel
,...,._.,...,,.., is called a free surface^ and is necessarily
^fi^'^^^^^^^^m^:^'^ T' horizontal (from § 419) when the liquid is at
:..;,(.:,.- w-i;: .,•' yq%\.. Fig. 461. (A gas, from its tendency
to indefinite expansion, is incapable of hav-
ing a free surface.) This is true even if the
space above the liquid is vacuous, for if the
surface were inclined or curved, points in the
body of the liquid and in the same horizon-
tal plane would have different heights (or " heads") of liquid
Fig. 461.
TWO LIQUIDS IIS'^ BEISTT TUBE.
529
between ttein and the surface, producing different intensities
of pressure in the plane, which is contrary to § 413.
When large bodies of liquid like the ocean are considered,
grayitj can no longer be regarded- as acting in parallel lines ;
consequently the free surface of the liquid is curved, being ~\
to the direction of (apparent) gravity at all points. For ordi-
nary engineering purposes (except in Geodesy) the free surface
of water at rest is a horizontal plane.
421. Two Liquids (whicli do not mix) at Rest in a Bent Tube
open at Both Ends to the Air, Fig. 460 ; water and mercury, for
instance. Let their heavinesses be y^
and y^ respectively. The pressure at e
may be written (§ 413) either
or
according as we refer it to the water
column or the mercury column and
their respective free surfaces where the
pressure j?Oj =i?Og = Pa = atmos. press.
€ is the surface of contact of the two liquids.
_.-Xv_>«
Hence we have
l>a + Kr,=Pa + Kn; i.e., ^, : K-'-n- r^- • (3)
le., the heights of the free surfaces of the two liquids above the
surface of contact are inversely proportional to their respec-
tive heamnesses.
ExamJ'le. — If the pressure at ^ = 2 atmospheres (§ 896) we
shall have from (2) (inch-lb.-sec. system of units)
KYx = JPe — /?a = 2 X 14.7 — 14.7 = 14.7 lbs. per sq. inch.
.-. \, must = 14.7 -j- [848.7 -H 1728] = 30 inches
(since, foi mercury, y^ = 848.7 lbs. per cub. ft.). Hence,
from (3), .
, h,y, 30 X [848.7 -- 1728] , „g . , „ . ._^
^' = 7r" 6275-^1728 = ^^^ ^^'^"' = ^^ ^-
530
MECHANICS OF ENGINEEKING.
i.e., for equilibrium, and that j?e may = 2 atmospheres, k^ aud
Aj (of mercarj and water) must be 30 in. and 34 feet respec-
tively.
422. City Water-pipes. — If h = vertical distance of a point
^ of a water-pipe below the free surface of reservoir, and tlie
water be at rest, the pressure on the inner surface of the pipe
at B (per unit of area) is
p =p^-\- hy ; and here j!?o =^„ = atmos. press.
Example. — If h = 200 ft. (using the inch, lb., and second)
P = 14.7 + [200 X 12][62.5 -=- 1T28] = 101.5 lbs. per sq. in.
The term hy, alone, = 86.8 lbs. per sq. inch, is spoken of as the
'hydrostattG press-ure due to 200 feet height, or "Head," of
water. (See Trautwine's Pocket Book for a table of hydro-
static pressures for various depths.)
If, however, the water {^flowing through the pipe, the pres-
sure against the interior wall becomes less (a problem of Hy-
drokinetics to be treated subsequently), while if that motion
is suddenly checked, the pressure becomes momentarily much
greater than the hydrostatic. This shock is called '■ water-
ram" and " water-hammer," and may be as great as 200 to 300
lbs. per sq. inch.*
423. Barometers and Manometers for Fluid Pressure. — If a
tube, closed at one end, is filled with water, and ihe other ex-
tremity is temporarily stopped and afterwards
opened under water, the closed end being then
a (vertical) height = h above the surface of
the water, it is required to find the intensity,
jp^ , of fluid pressure at the top of the tube, sup-
posing it to remain filled with water. Fig.
463. At E inside . the tube the pressure is
14.7 lbs. per sq. inch, the same as tljat outside
at the same level (§ 413) ; hence, from Pe=^ PE-hy.
w
* See pp. 203-214 of the author's "Hydraulic Motors."
BAROMETERS. 531
Example. — Let A = 10 feet (with inch-lb.-sec. system) ; then
f^ = 14.Y - 120 X [62.5 -^ 1T28] = 10.4 lbs. per sq. inch,
or about f of an atmosphere. If now we inquire the value
of A to make 'p^ = zero, we put j?^ — hy = and obtain h =z
408 inches, = 34 ft., which is called the height of the water-
haroineter. Hence, Fig. 463^^, ordinary atmospheric pressure
will not sustain a column of water higher than 34 feet. If
mercury is used instead of water the height supported by one
atmosphere is
I = 14.Y -^ [.848.7 -=- 1723] = 30 inches,
Fig. 463a.
= 76 cefatims. (about), and the tube is of more manageable
proportions than with water, aside from the ad-
vantage that no vapor of mercury forms above
the liquid at ordinary temperatures. [In fact, the
water-barometer height 5 = 34 feet has only a
theoretical existence since at ordinary tempera-
tures (40° to 80° Fahr.) vapor of water would
form above the column and depress it by from
0.30 to 1.09 ft.] Such an apparatus is called a
Barometer^ and is used not only for measuring
the varying tension of the atmosphere (from 14.5
to 15 lbs. per sq. inch, according to the weather and height
above sea-level), but also that of any body of gas. Thus, Fig.
464, the gas in D is put in communication with
the space above the mercury in the cistern at
(7; and we have jp = hy^ where y = heav. of
mercury, and p is the pressure on the liquid in
the cistern. For delicate measurements an at-
tached thermometer is also used, as the heavi-
ness y varies slightly with the temperature.
If the vertical distance CD is small, the ten-
sion in Cis considered the same as in D.
For gas-tensions greater than one atmosphere,
the tube may be left open at the top, forming an open ma-
n.
Fig. 464.
532
MECHANICS OF ENGINEERITiG.
nometevy Fig. 4C5. In this case, tlie tension of the gas above
the mercury in lh% cistern is
c
-.^
AjR-:-':,-'
h
m
■;;;;
—
■ ■■'■ Y ■
i
D
1
'■'■'■'■-CS
Fig. 465.
J? = (A -f h)r
(1)
in which 5 is tlie height of mercury (abf>ut 30
in,) to which the tension of the atmosphere above
the mercury column is equivalent.
Example. — If A = 51 inches, Fig. 465, we
have (ft., lb., sec.)
p = [4.25 ft. + 2.5 ft.] 848.Y = 5728 lbs. per sq. foot
= 39.7 lbs. per sq. incli = 2.7 atmospheres.
Another form of the open manometer consists of a U tube,
Fig. 464, the atmosphere having access to one branch, the gas
to be examined, to the other, while the
mercury lies in the curve. As before, we
•^ ' AIR
nave -^ ^ -"
jp = qi-^l)y = hy -\-2>^
(2)
r^:^
where j^ei = atmos. tension, and h as above.
The tension of a gas is sometimes spoken , ^j
of as measured by so many inches of 7ner- •"- ^^^^
G^iry. For example, a tension of 22.05 fig. 466.
lbs. per sq. inch {1^ atmos.) is measured by 45 inches of mer-
cury in a vacuum manometer (i.e., a common barometer),
Fig. 464. With the open manometer this tension (1-|- atmos.)
would be indicated by 15 incbes of actual mercury, Figs. 465
and 466. An ordinary steam-gauge indicates the eaaoess of
tension over one atmosphere ; thus " 40 lbs. of steam" implies
a tension of 40 + 14:.7 = 54.7 lbs. per sq. in.
The Bourdon steam-gauge in common use consists of a
curved elastic metal tube of flattened or elliptical section
^^with the long axis ~] to the plane of the tube), and has one
end fixed. The movement of the other end, which is free and
DIFFERENTIAL MANOMETER.
533
closed, bj proper mechanical connection gives motion to the
pointer of a dial. This movement is caused by any change of
tension in the steam or gas admitted, through the fixed end, to
the interior of the tube. As the tension increases the ellip-
tical section becomes less flat, i.e., more nearly circular, caus-
ing the tv70 ends of the tube to separate more widely, i.e., the
free end moves away from the fixed end ; and vice versa.
Such gauges, however, are not always reliable. They are
graduated by comparison with mercury manometers ; and
should be tested from time to time in the same way.*
424. The Differential Manometer.— In Fig. 467 OO'NK is a
portion of a pipe with the upper wall 00' horizontal. In
this pipe water is flowing from left to right in so called
"steady flow; " that is, there is no change, as time goes on,
in the velocity or internal pressure of the water at a given
section, as at or 0'. At 0' the velocity is greater than at
0, since the sectional area is smaller and the pressure po is
smaller than that, po, at 0; (as explained later) (p. 654).
The U-tube dmm contains mercury weighing 7-^ lbs. /eft.
in its lower part and is connected by the tubes aO and hO',
as shown, with holes in the upper wall of the pipe at and
0'. Air previously contained in these
tubes has been expelled through the
cocks a and 6, which are now closed.
The water columns Oac and dhO' and
the mercury in cm have adjusted them-
selves to a state of rest and are therefore
in a hydrostatic condition. The water
in the pipe exerts an upward pressure, yo,
as it flows by, against the base of the
stationary Uquid in tube Oa; and at 0' a
smaller upward pressure, p^ , against
the base 0' of the stationary water
column O'h. If the height h (between summits of the
mercury columns) be read from a scale, we are enabled to
compute the value of the difference, po — p^ , of the pressures
at and 0'; as may thus be shown (with p^ and p^ denoting
the pressures at c and d, respectively) : —
Since between and c we find stationary and continuous
water (heaviness = 7-) , we have po = Pc+^r- • • • (1)
* Of late years gauges have come into use constrncted of boxes with cor-
rugated sides of thin metal like the aneroid barometer. Motion of the sides,
under varying internal fluid pressure, causes movement of a pointer on a dial.
534
MECHANICS OF ENGINEERING.
Similarly, between 0' and d, po' -=pd+(hi-x)y;
while between d and c we have, for the mercury,
By elimination there easily results
(2)
(3)
po-
^-^ = /if^-l
^0-*
•■>f«.-v.-?-V-V..:V.>
■Po'=h{rm-r); or ^-^^=h[^j-ij. . (4)
Evidently, if in place of the mercury we use a Hquid only
slightly heavier than water and that
does not mix with the latter, h would
be quite large for a small value of
po — po'', i-G-, the instrument would
be more sensitive. If kerosene oil,
which is a little lighter than water,
were used instead of mercury, an
arrangement of tubes like that in
Fig. 467a would be necessary, and
similar analysis, (if yk denote the
heaviness of kerosene) gives rise to
the formula.
Po
-po'^Hr-n); or 2»-^'=a(i-Q),
(5)
■c— ^
:is
■E 4~yi
^■•■- -r I-'- ■
Fig. 468.
425. Safety-valves. — Fig. 468. Eequired the proper weight
G to be huDg at the extremity of the horizontal lever AB,
with fulcrum at B, that the flat
disk- valve ^sliall not be forced
upward by the steam pressure,^',
until tlie latter reaches a given
value =p. Let tlie weight of
the arm be G^ , its centre of grav-
ity being at 0, a distance = o
from JB ; the other horizontal distances are marked in the
figure.
' Suppose the valve on the point of rising; then the forces
acting on the lever are the fnlcrum-reaction at B, the weights
G and G^ , and the two fluid-pressures on the disk, viz. : JPp^
(atmospheric) downward, and I^p (steam) upward. Hence,
from ^(moms. ^) =0,
Gb + G,G + F;p^a - Fpa = 0. ... (1)
BUKSTING OF PIPES.
635
Solving, we have
(^ = IF{P-I>a)-G,t.
(2)
Example. — "With a = 2 inches, 5 = 2 feet, c — 1 foot
G^=: 4c Ibs.,^ = 6 atmos., and diam. of disk = 1 inch; with
the foot and pound,
G = I. . ^'fAVce X 14.7 X IM - 1 X 14.7 X 144] - 4 X^.
24*4 \12
.-. G = 2.81 lbs.
[Kotice the cancelling of the 144; for J^{p —j)a) h pounds,
being one dimension of force, if the pound is selected as the
unit of force, whether the inch or foot is used in both fac-
tors.] Hence when the steam pressure has risen to 6 atmos.
(= 88.2 lbs. per square inch) (corresponding to 73.5 lbs. persq.
in. by steam gauge) the valve will open if 6^^ = 2.81 lbs., or be
on the point of opening.
426. Proper Thickness of Thin Hollow Cylinders (i.e., Pipes
and Tubes) to Resist Bursting by Fluid Pressure.
Case I. Stresses in the G?vss-section due to End Pressure j
Fig. 469. — Let AB be the circular cap clos-
ing the end of a cylindrical tube containing
fluid at a tension =:^. Let i>^ = internal
radius of the tube or pipe. Then considering
the cap free, neglecting its weight, we
have three sets of || forces in equilibrium
in the figure, viz. : the internal fluid pres-
sure =:: nr'^jp ; the external fluid pressure
= nr^'Pa ; while the total stress (tensile) on
the small ring, whose area now exposed is
^Tcrt (nearly), is = '^■rcrtj^^ , where t is the thickness of the pipe,
aTid^?j the tensile stress per unit area induced by the end-pres-
sures (fluid).
Fig. 469.
536
MECHANICS OF ENGINEERING.
For equilibrium, therefore, we may put ^(hor. comps.) = ;
1.6.,
KP -Pa)
.i>i =
^t
(1)
(Strictly, the two circular areas sustaining the fluid pressures
are different in area, but to consider them equal occasions but
a small error.)
Eq. (1) also gives the tension in the central section of a thin
hollow sphere, under bursting pressure.
Case II. Stresses in the longitudinal section of pipe, due to
radial 'fluid pressures.^ — Consider free the half (semi-circular)
of any length I of the pipe, be-
tween two cross-sections. Take an
axis X (as in Fig. 470) ~\ to the
longitudinal section which has been
made. Let p^ denote the tensile
stress (per unit area) produced in
the narrow rectangles exposed at A
and B (those in the half-ring edges,
having no X components, are not
drawn in the figure). On the in-
ternal curved surface the fluid pres-
sure is considered of equal intensity
=.p at all points (practically true even with liquids, if 2r io
small compared with the head of water producing p). The
fluid pressure on any dF or elementary area of the internal
curved surface is =. pdF. Its X component (see § 416) is
obtained by multiplying j? by the projection of dF on the ver-
tical plane ABO, and since p is the same for all the dF^% of
the curved surface, the sum of all the JT components of the in-
ternal fluid pressures must = 2^ multiplied by the area of rect-
angle ABCD, = 'iirlp I and similarly the X components of the
Fig. 470.
* Analytically this problem is identical with that of the smooth cord on
a smooth cylinder, § 169, and is seen to give the same result.
BUESTING OF PIPES. 637
external atraos. pressures = '^rljp^ (nearly). The tensile stresses
( II to X) are equal to 'ilt])^ ; hence for equilibrium, '2X =
gives
mp^ - 2rZ^ + 2r^j9a = ;
"Pi — ^ \^)
This tensile stress, called hoop tension, p^, opposing rupture by
longitudinal tearing, is seen to be double the tensile stress ^:>i
induced, under the same circumstances, on the annular cross'
section in Case I. Hence eq. (2), and not eq. (1), should be
used to determine a safe value for the thickness of metal, t, or
any other one unknown quantity involved in the equation.
For safety against rwpture, we must put p^ = T', a safe
tensile stress per unit area for the material of the pipe or tube
(see §§ 195 and 203) ;
,,t = TiPp^ (8)
(For a thin hollow sphere, t may be computed from eq. (1) ;
that is, need be only half as great as with the cylinder, other
things being equal.)
Example. — A pipe of twenty inches internal diameter is to
contain water at rest under a head of 340 feet ; required the
proper thickness, if of cast-iron.
340 feet of water measures 10 atmospheres, so that the in
ternal fluid pressure is 11 atmospheres ; but the external pres
sure Pa being one atmos., we must write (inch, lb., sec.)
{p —pa) = 10 X 14. Y = 147.0 lbs. per sq. in., and r = 10 in.,
while (§ 203) we may put T' =^oi 9000 = 4500 lbs. per sq.
in. ; whence
. 10 X 147 ^ Qo« • 1
^ = — Tir^ — == 0.326 mches.
4500
638 MECHANICS OF ENGINEERING.
But to insure safety in handling pipes and iraperviousnesB to
the water, a somewhat greater thickness is adopted in practice
than given by the above theory.*
Thus, Weisbach recommends (as proved experimentally also)
for
Pipes of sheet iron, t = [0.00172 rA + 0.12] inches;
" " cast " t = [0.00476 rA + 0.34] "
" " copper t = [0.00296 rA 4- 0.16] "
" " lead t = [0.01014 rA + 0.21] "
" " zinc t = [0.00484 rA + 0.16] "
in which t =■ thickness in inches, r = radius in inches, and A
= excess of internal over external fluid pressure (i.e., p — Pa)
expressed in atrnmjpheres.
For instance, for the example just given, we should have
(cast-iron), ^-.00476x10X10 + 0.34 = 0.816 in.
With riveted steel pipes, if the longitudinal seams are pro-
vided with two rows of rivets, a value of 10,000 Ibs./ini^
may be used for the T' of eq. (3). This makes a fair allow-
ance for the weakening of the steel plates by the rivet holes.
The East Jersey Water Co. uses such pipes 2r — 4 ft. in
diameter, with a thickness of i = f in., under a head of 340 ft.
At the Mannesmann Works in Hungary, special steel tubes
4 in. in diameter and | in. thick have been made, safely
withstanding an internal fluid pressure of 2000 Ibs./in.^
Water Ram. — When water flowing in pipes is subject to sudden arrest
of motion, a high bursting pressure, called '"water ram,", or "water
hammer," may be produced. See pp.- 204-211 of the writer's Hydraulic
Motors.
In thick hollow cylinders, on account of the thickness of the walls,
the stress in the nietal is not uniformly distributed. See pp. 507, etc.,
of this book.
427. Collapsing of Tubes under Fluid Pressure. (Cylindrical
boiler-flues, for example.) — If the external exceeds the internal
fluid pressure, and the thickness of metal is small compared
with the diameter, the slightest deformation of the tube . or
pipe gives the external pressure greater capability to produce
a further change of form, and hence possibly a final collapse;
just as with long columns (§ 303) a slight bending gives great
advantage to the terminal forces. Hence the theory of § 426
COLLAPSE OF TUBES. 539
is inapplicable. According to Sir Wm. Fairbairu's experi-
ments (1858) a thin wrouglit-iron cylindrical (circular) tube
will not collapse until the excess of external over internal
pressure is
^(in lbs. per sq. in.) = 9672000^. . . (1) . . (not homog.)
(f, I, and d must all be expressed in the same linear unit.)
Here t = thickness of the wall of the tube, d its diameter, and
I its length ; the ends being understood to be so supported aa
to preclude a local collapse.
Example. — ^With 1 = 10 ft. = 120 inches, d = 4: in., and t =
^ inch, we have
p = 9672000
J
-^ H- (120 X 4) = 201.5 lbs. per sq. inch.
For safety, ^ of this, viz. 40 lbs. per sq. inch, should not be
exceeded ; e.g., with 14.7 lbs. internal and 54.7 lbs. external.
[Note. — For simplicity the power of the thickness used in eq. (1) above
has been given as 2.00. In the original formula it is 2.19, and then all
dimensions must be expressed in incJies. A discussion of the experiments
of Mr. Fairbairnwill be found in a paper read by Prof. Unwin before the
Institute of Civ. Engineers (Proceedings, vol xlvi.). See also Prof. Unwin's
" Machine Design," p. 66. It is contended by some that in the actual con-
ditions of service, boiler-flues are subjected to such serious straining
actions due 1o unequal expansion of the connecting parts as to render the
above formula quite unreliable, thus requiring a large allowance in ita
application.]
437a. Collapsing Pressure of Steel Tubes. — Recent experiments by Prof.
R. T. Stewart (see Engineering News of May 10, 1906, p. 528) on Bes-
semer steel lap-welded tubes of 8§- in. in diameter and all commercial
thicknesses of wall and in lengths of 2]-, 5, 10, 15, and 20 ft.; and also on
single lengths of 20 ft. (between end connections) in seven sizes from 3 to
10 inches outside diameter and in all commercial thicknesses obtainable;
have shown that length has practically no influence on the strength, if the
length is greater than six times the diameter. From these experiments
Prof. Stewart has deduced the following formulae in which p is the col-
lapsing unit-pressure in lbs. per sq. inch, d the outside diameter of the
tube in inches, and t the thickness of wall of tube, also in inches : —
p = 1000(l-/j/l-1600|^), (4)
p = 86670 J— 1386.0 (5)
Eq. (4) is for use where f-j-d is less than 0.028, and eq. (5) for larger
T^alues of that ratio.
540 MECHANICS OF ENGINEEKIKCh.
CHAPTEE II.
HYDROSTATICS (Cow^mwecZ)— PRESSURE OF LIQUIDS IN TANKS.
AND RESERVOIRS.
428. Body of Liquid in Motion, but in Relative Equilibrium.—
By relative equilibrium it is meant that the particles are not
changing their relative positions, i.e., are not moving among
each other. On account of this relative equilibrium the fol-
lowing problems are placed in the present chapter, instead of
under the head of Hydrodynamics, where they strictly belong.
As relative equilihriwm is an essential property of rigid bodies,
we may apply the equations of motion of rigid bodies to bodies
of liquid in relative equilibrium.
Case L All the particles moving in parallel right lines
with equal velocities ^ at any given instant (i.e., a motion of
translation.) — If the common velocity is constant we have a
uniform translation, and all the forces acting on any one par-
ticle are balanced, as if it were not moving at all (according to
iNewton's Laws, § 54); hence the relations of internal pressure^
free surface, etc., are the same as if the liquid were at rest.
Thus, Fig. 471, if the liquid in the moving tank is at rest rel-
^ atively to the tank at a given instant, with
" ?____ _n its free surface horizontal, and the motion
^^^^^^^ of the tank be one of translation with a uni-
y^J \ y^ ' '" form velocity, the liquid will remain in this
mdmm^^ condition of relative rest, as the motion
Fig. 471. ,
proceeds.
But if the velocity of the tank is accelerated with a consta/nt
acceleration ^=p (this symbol must not be confused with p
for pressure), the free surface will begin to oscillate, and finally
come to relative equilibrium at some angle oc with the horizon-
tal, which 18 thus found, when the motion is horizontal. See
Fig. 472, in which the position and value of a are the same,
whether the motion is uniformly accelerated from left to rigbt
EELATIVE EQUILIBRIUM OF LIQUIDS.
541
Let be the lowest
-->p
Fig. 473.
or uniformly retarded from right to left,
point of the free surface, and Oh a
small prism of the lig^uid with its
axis horizontal, and of length = x ;
rib is a vertical prism of length =
0, and extending from the extremity
of Oh to the free surface. The
pressure at both and n \% jp^i=.
atmos. pres. Let the area of cross-
section of both prisms be = dF.
l!^ow since Oh is being accelerated in direction ^(horizont.),
the difference between the forces on its two ends, i.e., its ^Xy
must = its mass X accel. (§ 109).
.'.jp^dF-jpadF^ixdF.y — g']]). . . . (1)
{y = heaviness of liquid ; pi, = press, at h) ; and since the ver-
tical prism nh has no vertical acceleration, the -2(vert. com-
pons.) for it must = 0.
.\pSF-padF-zdF.y=^, ..... (2)
From (1) and (2),
xv —
-^.p = zy\
X g°
(3)
... (4)
Hence On is a right line, and therefore
z V
tan a, or — , =-^. . .
X g
[Another, and perhaps more direct, method of deriving this
result is to consider free a small particle of the liquid lying in
the surface. The forces acting on this particle are two : the
first its weight = dG ^ and the second the resultant action of
its immediate neighbor-particles. ]^ow this latter force (point-
ing obliquely upward) must be normal to the free surface of
the liquid, and therefore must make the unknown angle a with
the vertical. Since the particle has at this instant a rectilinear
accelerated motion in a horizontal direction, the resultant of the
two forces mentioned must be horizontal and have a value =
mass X acceleration. That is, the diagonal formed on the two
542
MECHANICS OF ENGINEERING,
forces must be horizontal and have the value mentioned, =
{dG -V- g)jp ; while from the nature of the figure (let the stu-
dent make the diagram for himself) it must also = dG tan a.
dG
P
Q. E. D.
.*. dG tan a = — .p ; or, tan a =
9 9
If the translation were vertical, and the acceleration xijpward
[i.e., if the vessel had a uniformly accelerated upward motion
or a uniformly retarded downward motion], the free surface
would be horizontal, but the pressure at a depth = h below the
surface instead oi jp =^Pa + ^7 would be obtained as follows:
Considering free a small vertical prism of height = Ji with
upper base in the free surface, and putting 2(vert. compons.)
= mass X acceleration, we have
hdF. y
dF . p — dF . jpa — hdF . y
9
p
'P=I>a-[-hy
' 9+P
L g ..
(5)
If the acceleration is downward (not the velocity necessarily)
we make J? negative in (5). If the vessel falls freely, j} =— g
and .'.p =pai in all parts of the liquid.
Query : Suppose jp downward and > g.
Case II. Uniform Rotation about a YerticalAxis. — If the
narrow vessel in Fig. 473, open at top and containing a liquid,
^c be kept rotating at a uniform angu-
lar velocity cl> (see § 110) about a
vertical axis Z, the liquid after some
oscillations will be brought (by fric-
tion) to relative equilibrium (rotat-
ing about Z, as if rigid). Required
the foi-m of the free surface (evi-
dently a surface of revolution) at
each point of which we know
JP=Pa-
Let be the intersection of the
axis Z with the surface, and n any point in the surface ; J being
V
1
71
/
1
/
\ 1 /
-'k. n /
?\:~~-^^f
3S —
/
Fig. 473.
TJlSriFOEM EOTATION- OF LIQUID IIST VESSEL. ^43
a point vertically under n and in same horizontal plane as 0.
Every point of the small right prism nh (of altitude = s and
sectional area 6^^) is describing a horizontal circle ahont Z. nnd
has therefore no vertical acceleration. Hence for this prism,
free, we have 2Z = 0; i.e.,
dF.j?t, - dF.pa - zdF. y = (1)
!Now the horizontal right prism Oh (call the direction ...h^
^) is rotating uniformly about a vertical axis through one ex-
tremity, as if it were a rigid body. Hence the forces acting
on it must be equivalent to a single horizontal force, — ca/Mp.
(§122«,) coinciding in direction with JT. IM= mass of prism
= its weight -^ g, and p = distance of its centre of gravity
from ; here p = ^x= ^ length of prism]. Hence the 2X
xdF
of the forces acting on the prism Oh must = — ta^ y^x.
But the forces acting on the two ends of this prism are their
own ^components, while the lateral pressures and the weights
of its particles have no Xcompons. ;
JTT JTT —Gifx^dF.y ,^.
,'.dF.jc>a — dF.pt — o ~' • • (2)
From (1) and (2) we have
^--27 -2^' ...... (3)
where v = axe = linear velocity of the point n in its circular
path.
[As in Case T, we may obtain the same result by considering
a single surface particle free, and would derive for the resultant
force acting upon it the value dG tan n' in a horizontal direc-
tion and intersecting the axis of rotation. But here a is dif-
ferent for particles at different distances from the axis, tan a
being the -j- of the curve On. As the particle is moving uni-
formly in a circle the resultant force must point toward the
544 MECHANICS OF ENGIKEERINa.
eectre of the circle, i.e., horizontally, and have a value — . — ,
g 10
where x is the radius of the circle [§ 74, eq. (5)] ;
jj^ . dG (goxY ^ dz Go^x
.*. aijr tan a = -^ — -^ ; or tan «,=—-,= — ;
g X ax g
.*. / dz — -~ I xdxi or, z = — . -. . Q. E. D.
Hence any vertical section of the free surface through the
axics of rotation Z is a, parabola, with its axis vertical and vertex
at 0; i.e., the free surface is 2i paraboloid of revolution, with
Z as its axis. Since cox is the linear velocity v of the point
h in its circular path, ^ = " height due to velocity" v [§ 52],
Example. — If the vessel in Fig, 4Y3 makes 100 revol. per
minute, required the ordinate s at a horizontal distance of a? =
4 inches from the axis (ft.-lb.-sec. system). The angular veloc-
ity OS =. [^Tt 100 -V- 60] radians per sec. [K. B. — A radian =
the angular space of which 3.1415926 . . . make a half-re vol.,
or angle of 180°]. With a? = i f t. and g — 32.2,
and the pressure ht b (Fig. 473) is (now use inch, lb., sec.)
62 5
P& —Pa + ^r— 14:.7 +^X j;^ = 14.781 lbs. per sq. in.
Picf. Mendelejeff of Russia has recently utilized the fact an-
nounced as the result of this problem, for forming perfectly
true paraboloidal surfaces of plaster of Paris, to receive by
galvanic process a deposit of metal, and thus produce specula
of exact figure for reflecting telescopes. The vessel contain-
ing the liquid plaster is kept rotating about a vertical axis
at the proper uniform speed, and the plaster assumes the de-
sired shape before solidifying. A fusible alloy, melted, may
also be placed in the vessel, instead of liquid plaster.
EELATIYE EQUILIBKIUM.
545
top, ex-
Remaek. — If the vessel is quite full and closed on
except at 0' where it communicates
bj a stationary pipe with a reser-
voir, Fig. 474, the free surface
cannot be formed, but the pres-
sure at any point in the water is
just the same during uniform rota-
tion, as if a free surface were formed
with vertex at ;
i.e., p-„=p^ + (A„ 4- z)y. . (4)
See figure for h^ and z. (In subse-
quent paragraphs of this chapter
the liquid will be at rest.)
Fig. 474.
428a. Pressure on the Bottom of a Vessel containing Liquid at
Rest. — If the bottom of the vessel is plane and horizontal, the
intensity of pressure upon it is the same at all points, being
^ISi^
Fig. 475.
p—ip^J^hy (Figs. 475 and 476), and the pressures on the ele-
ments of the surface form a set of parallel (vertical) forceSo
This is true even if the side of the vessel overhangs, Fig. 476,
the resultant fluid pressure on the bottom in both cases being
P = Fp-F2?a = Fhy.
(1)
(Atmospheric pressure is supposed to act under the bottom.)
It is further evident that if the bottom is a rigid homogeneous
plate and has no support at its edges, it may be supported at a
-U^='-
^6 MECHANICS OF ENGINEEEllSTG.
single point (Fig. 477), which in this case (horizontal plate)
is its centre of gravity. This point is calle*]
the Centre of Pressure, or the point of apjili-
cation of the resuliant of all tlie fluid pressures
acting on the plate. The present case is such
that tliese pressures reduce to a single result-
ant, but this is not always practicable.
Example. — In Fig. 476 (cylindrical vessel
containing water), given A = 20 ft., h^ =
15 ft., r^ = 2 ft., )\ = 4 ft., required the pressure on the bot-
tom, the vertical tension in the cylindrical wall CA^ and the
hoop tension (^§ 426) at G. (Ft., lb., sec.) Press, on bottom =
Fhy= Ttr^hy = vtlQ X 20 X 62.5 = 62857 lbs.; while the
upward pull on CA =
{m;' - 7rr;)h,y = ;r(16 - 4)15 X 62.5 = 35357 lbs.
If the vertical wall is t = -^-q inch thick at C'this tension will
be borne by a ring-shaped cross section of area = 27T7\t (nearly)
= 27r48 X tV = 30.17 sq. inches, giving (35357 -^ 30.17) =
about 1200 lbs. per sq. inch tensile stress (vertical).
The hoop tension at C is horizontal and is
p" = r^P —Pa) -^ i (see § 426), where ^ =p^ -f h,y ;
„ 48 X 15 X 12 X (62.5 ^ 1728) __„ ,,
.\j}" z= ^ i = 3125 lbs. per sq. in.,
(using the inch and pound).
429. Centre of Pressure. — In subsequent work in this chapter,
since the atmosphere has access both to the free surface of
liquid and to the outside of the vessel walls, and p^ would can^
eel out in finding the resultant fluid pressure on any elemen,
tary area d.F of those walls, w^e shall write :
' The res^dtant fluid pressure on any dF' of the vessel wall is
normal to its surface and is dP z= pdF "= zydF. in which s
is the vertical distance of the element below the free surface
of the liquid (i.e., s = the '■'head of water"). If the surface
pressed on is plane, these elementary pressures form a system
of parallel forces, and may be replaced by a single resultant
CENTRE OF PEESSURE.
547
(if the plate is rigid) which will equal their sum, and whose
point of application, called the Centre of Pressure, may be
located by the equations of § 22, put into calculus form.
If the surface is curved the elementary pressures form a sys
tem of forces in space, and hence (§ 38) cannot in general he
reduced to a single resultant, but to two, the point of applica-
tion of one of which is arbitrary (viz., the arbitrary origin,
§38).
Of course, the object of replacing a set of fluid pressures by
a single resultant is for convenience in examining the equi-
librium, or stability, of a rigid body the forces acting on which
include these fluid pressures. As to their effect in distorting
the rigid body, the fluid pressures must be considered in their
true positions (see example in § 264), and cannot be replaced
by a resultant.
430. Resultant Liquid Pressure on a Plane Surface forming
Part of a Vessel "Wall. Co-ordinates of the Centre of Pressure. —
Fig. 478. Let JL5 be a portion (of any shape) of a plane
surface at any angle with the
horizontal, sustaining liquid
pressure. Prolong the plane
of AB till it intersects the free
surface of the liquid. Take
this intersection as an axis Y,
being any point on Y. The
axis X, ~1 to Y, lies in the
given plane. Let a = angle
between the plane and the free
surface. Then x and y are the
co-ordinates of any elementary
area dF oi the surface, referred to Xand Y. z = the "head
of water," below the free surface, of any dF. The pressures
are parallel.
The norinal pressure on any dF =■ zydF", hence the swn of
these, = their resultant.
Fig. 478.
=.P, = yfsdF=Fzy',
(1)
648 MECHANICS OF ENGINEEEIJSTG,
in which z = the " mean s," i.e., the b of the centre of gravity
G of the plane figure AB, and ^= total area of AB \_F z —
fzdF^ from eq. (4), § 23]. y = heaviness of liquid (see § 409).
That is, the total liquid py^essure on a plane figure is equal
to the weight of an imaginary jprism of the liquid hamng a.
'base = area of the given figure and an altitude = vertical
depth of the centre of gravity of the figure helow the surface
of the liquid. For example, if the figure is a rectangle with
one base (length = V) in the surface, and lying in a vertical
plane,
P = lh. \hy = ^hhy.
Evidently, if the altitude be increased, P varies as its square.
From (1) it is evident that the total pressure does not de-
fend on the horizontal extent of the water in the reservoir.
Now let a?c and y^ denote the co-ordinates, in plane YOX^
of the centre of pressure., G, oy point of application of the re-
sultant pressure P, and apply the principle that the sum of
the moments of each of several parallel forces, about an axis "1
to them, is equal to the moment of their resultant about the
same axis [§ 22]. First taking OY 2iS> an axis of moments,
and then OX, we have
Px^ = J* {zydF)x, and Py^ = f {zydF)y. . (2)
But P = Fzy = Fx{sm a)y, and the z of any dF= X sin a.
Hence eqs. (2) become (after cancelling the constant, y sin a)
F X Fx Fx
in which Iy = the " tnojn. of inertia''^ of the plane figure re-
ferred to Y (see § 85). [_'N. B. — The centre of pressure as
thus found is identical with the centre of oscillation (§ IIY)
and the centre of percussion [§ 113] of a thin homogeneous
plate, referred to axes X and Y, Y being the axis of suspen-
sion.]
Evidently, if the plane figure is vertical a = 90°, a? = ^ for
CENTRE OF PRESSUEE.
549
all 6?^'s, and a? = s. It is also noteworthy that the position
of the centre of pressure is independent of a.
JSToTE. — Since the pressures on the equal cZ^'s lying in any
horizontal strip of the plane figure form a set of equal parallel
forces equally spaced along the strip, and are therefore equiva-
lent to their sum applied in the middle of the strip, it follows
that for rectangles and triangles with horizontal bases, the
centre of pressure must lie on the straight line on which the
middles of all horizontal strips are situated.
431. Centre of Pressure of Rectangles and Triangles with Bases
Horizontal. — Since all the dF''% of one horizontal strip have
the same a?, we may take the area of the strip ^
for B
Fia. 479.
while (see note, § 430) y^ = i^-
Whe7i the tipjyer hase lies in the surface, h^ = 0, and x^ =
1^2 =: ^ of the altitude.
For a triangle loith its hase horizontal and vertex tip. Fig.
480, the length %i of a horizontal strip is variable and dF=-
udx. From similar triano;les it = -{x — h^ \ therefore
\ — A,
K
I x^{x — h}jdx
lh{K
K)Uh-^Wh-K)]
x'ix — h^dx =
U,V4
'2 /a?' 7 X
-6--B >
Fig. 480.
J, 3A/ + 2AA + A;
(2)
650 MECHANICS OF ENGINEEEING.
Also, since the centre of pressure must lie on the line AB join-
ing the vertex to the middle of base (see note, § 430), we easilji
determine its position.
Evidently for A, = 0, i.e., when the vertex is in the surface,
o
^ jY Xc = f Aj. Similarly, /br a triangle with
] y*i hase horizontal and vertex doion^ Fig. 481,
mV 7~i /D we find that
\l....±..^ If tli6 base is in the surface, A^ = and
Fig. 481. (3) rcduccs to Xc = ^h^.
It is to be noticed that in the case of the triangle the value
of Xc is the same whatever be its shape, so long as h^ and k^
remain unchanged and the base is horizontal. If the base is
not horizontal, we may easily, by one horizontal line, divide
the triangle into two triangles whose bases are liorizontal and
whose combined areas make up the area of the first. The re-
sultant pressure on each of the component triangles is easily
found by ths foregoing principles, as also its point of applica-
tion. The resultant of the two parallel forces so determined
will act at some point on the line joining the centres of pres-
sures of the component triangles, this point being easily found
by the method of moments, while the amount of this final re-
sultant pressure is the sum of its two components, since the
latter are parallel. An instance of this procedure will be
given in Example 3 of § 483. Similarly, the rectangle of Fig.
479 may be distorted into an oblique parallelogram with hori-
zontal bases without affecting the value of x^ , nor the amount
of resultant pressure, so long as h^ and h^ remain unchanged.
432. Centre of Pressure of Circle. — Fig. 482. It will lie on
the vertical diameter. Let r = radius. From eq. (3), § 430,
° 1 i ] ^ /^ L^Fla" iTtr' -^ Ttr'a"
■ ^4-1 — ^ -T a? jT a? Tvrx
M 1^l\ C^®^ ®
A
1
--4. - /)■
— I — i —
1
^^\=\x,=
i'
-1 1
9*
1
=i-^p-
B
E
\Ei
1
1
1
— 1 —
1
<-
^
v//////y////y////w///
Mf;
•— ^
Fig. 486.
435. Stability of a Vertical Rectangular Wall against Water
Pressure on One Side. — Fig. 486. All dimensions are shown in
the figure, except /, which is the length
of wall 1 to paper. Supposing the wall
to be a single rigid block, its weight G'
= Vh'ly' iy' being its heaviness (§ 7),
and I its length). Given the water
depth = A, required the proper width
h' for stability. For proper security :
First, the resultant of G' and the
water-pressure P must fall within the
base BD (or, which amounts to the same thing), the moment
of G' about D, the outer toe of the wall, must be numerically
greater than that of P ; and
Secondly, P must be less than the sliding friction /'6r' (see
§ 156) on the base BD.
Thirdly, the maximum pressure per unit of area on the
base must not exceed a safe value (compare § 348).
ISTowP = Fzy ^ hi ~ y ^ ^^^^^Vfiv = heaviness of water) ;
and Xg = \h.
Hence for stability against tipping about Z>,
P\h must le < G'W ; i.e., ^Uly < ^V'h'ly' ; . (1)
while, as to sliding on the base,
P must le m-> af P to have a definite safe value (for the pressure j^m at
D^ or quite near D^ will evidently be greater than elsewliere
on BD ; i.e., it is the maximum pressure to be found on BD).
This may be done by the principles of §§ 346 and 362.
First, assume that H cuts BD otdside of the middle third',
i.e., that
YE\ = nb\>\V{pxn>\y,
'where n denotes the ratio of the distance of E' from the mid-
dle of the base to the whole width, Z*', of base. Then the pres-
sure (per unit area) on- small equal elements of the base BD
(see § 346) may be considered to vary as the ord mates of a
triangle MND (the vertex M. being within the distance BD),
9^^WD^\\\ = \WD\ i.e.,
* The student should note that Fig. 487 only shows the lower jmH
of the wall of Fig. 486 ; and that the resultant R, now applied at E' , is
there decomposed into its original components P and G'.
556 MECHANICS OF ENGINEEEING.
The mean pressure per unit area, on MD^
and hence the maximum pressure (viz., at D), being double
the mean, is
J?^3.26^'-[3J7(i-7^)]; . . . . (0)
and if j[)„jis to equal C'{^ee §§ 201 and 203), a safe value ifor the
crushing resistance, per unit area, of the material, y^e shall
have
57(i -n)C'= W = Wh'l/,
•° 2 3"^' ^^^
To find h', knowing n, we put the JS'(moms.) of the G' and P"
at ^, about ^', = zero (for the only other forces acting on
the wall are the pressures of the foundation against it, along
JUD ; and since the resultant of these latter passes through E\
the sum of their moments about £'' is already zero) ; i.e.,
G'nh' - P\h = ; or, nh"h'ly'= ^h . ^hHy ;
Having obtained h\ we must also ascertain if P is 0, the angle of friction, with the ver-
tical.
If ?i, computed from (1), should prove to be < ^, our first
assumption is wrong, and we therefore assume n < -|-, and pro-
ceed thus :
Secondly, n being < | (see §§ 346 and 362), we have a
STABILITY OF KESERVOIR WALLS. 567
trapezoid of pressures, instead of a triangle, on BD. Let the
pressure per unit area at D be j?^ (the maximum on base).
The whole base now receives pressure, the mean pressure (per
unit area) being = 6r' -f- [57] ; and therefore, from § 362,
Case I, we have
p^ = [6^ + l]^; ..... (Oa)
and since, here, G' = Vh'ly', we may write
Pm = {Qn + l)h'y'.
For safety as to crushing resistance we put
{Qn + l)h'y' = C'\ whence n = ^ f^ ~ ^1 ' ' ^^
Having found n from eq. (la), we determine the proper
width of base V from eq. (2), in case the assumption n. < -^ is
verified.
Example.— In Fig. 486, let h! — 12 ft., A = 10 ft., while
the masonry weighs {j' =) 150 lbs. per cub. ft. Supposing
it desirable to bring no greater compressive stress than 100 lbs.
per sq. inch (=::? 14400 lbs. per sq. ft.) on the cement of the
joints, we put C = 14400, using the ft.-lb.-sec. system of units.
Assuming n > \^ we use eq. (1), and obtain
_ 1 _ 2 12 X 15 _ 5_
**~2 3* 14400 ~~ 12'
which is >-|-; hence the assumption is confirmed, also the
propriety of using eq. (1) rather than (la).
Passing to eq. (2), we have
7, H. / 62.5 X 10 ' o/TiT .
6' = 10 X A / 1 -.S^ -.^A = 3-^ feet.
^ Y I X 12 X 150
But, as regards frictional stability, we find that, with/* = 0.30,
a low value, and V = 3.7 ft. (ft., lb., sec),
558 MECHANICS OF EISTGINEEEING.
P _ ^liHy _ 100 X 6 2.5 -15.
JG' "fhWiy ~ 2 X 0.3 X 3.7 X"T2 X 150 ~ ' '
which is greater than unity, showing the friction to be insuf-
ticient to prevent sliding (with y=0.30); a greater width
must therefore be chosen, for frictional stability.
If we make n = 1-, i.e., make li cut the base at the outer
edge of middle third (§ 362), we have, from eq. (2),
z/ -.A / 62.5 X 10 ^ Koaf ,
and the pressure at D is now of course well within the safe
limit ; while as regards friction we find
i> H-/(?' = 0.92, < unity,
and therefore the ^vall is safe in this respect also.
With a width of base = 3.7 feet lirst obtained, the portion
3ID, Fig. 487, of the base which receives pressure [according
to Navier's theory (§ 346)] would be only 0.92 feet in length,
or about one fourth of the base, the portion .i> JIT tending to
open, and perhaps actually suffering tension, if capable (i.e., if
cemented to a rock foundation), in which case these tensions
should properly be taken into account, as with beams (§ 295),
thus modifying the results.
It has been considered safe by some designers of high
masonry dams, to neglect these possible tensile resistances, as
has just been done in deriving h' = 3.7 feet ; but others, in
view of the more or less uncertain and speculative character of
Javier's theory, when applied to the very wide bases of such
structures, prefer, in using the theory (as the best available),
to keep the resultant pressure within the middle third at the
base (and also at all horizontal beds above the base), and thus
avoid the chances of tensile stresses.
This latter plan was favored by Messrs. Church and Fteley,
as engineers of the proposed Quaker Bi^idge Dam in connec-
tion with the IsTew Croton Aqueduct of JSTew York City, in
their report of 1887. See § 439.
RESEKVOIR WALLS.
559
437. Wall of Trapezoidal Profile. Water-face Vertical. —
Economy of material is favored by using a trapezoidal profile,
A^— -6-'— >
Fig. 488. With this form the
stability may be investigated in
a corresponding manner. The
portion of wall above each
horizontal bed should be ex-
amined similarly. The weight
G' acts through the centre of
gravity of the whole mass.
Detail. — Let Fig. 488 show
the vertical cross-section of a
trapezoidal wall, with notation
for dimensions as indicated ; the
portion considered having a length == ^, "] to the paper. Let
Y = heaviness of water, y' that of the masonry (assumed homo-
geneous), with ?^ as in § 436.
For a triangle of ^ressure^ MD, on the base, i.e., with ^ > -g-,
or resultant falling outside the middle third (neglecting pos-
sibility of tensile stresses on left of M\ if the intensity of
pressure j?m at 2> is to = ^' (§ 201), we put, as in § 436,
ni^ -n-\C' = %G\ i.e., = ¥h' . W + V')y',
whence
1 \h'y' J'_|_
n = — — — -
2 3 (7' h'
(1)'
For r. trapezoid of pressure, i.e. with n <^,oy the resultant
of P and G' falling within the middle third, we have, as be-
fore (§ 362, Case I),
G'
whence
n
irc'h'i r\ . ir ^c'h' ,n ., y
660 MECHANICS OF ENGINEERIISTG.
From the geometry of the figure, having joined the middles
of the two bases, we have
(§ 26,Prob. 6), and, by similar triangles, OV : KV '.'.gO'.h',
whence
^^=#it^'-^''^'
The lines of action of G' and P meet at E, and their result-
ant cuts the base in some point E'. The sum of their moments
about E' should be zero, i.e., P . ^h=^ G' . 0E'\ that is, (see
eq. {a) above, and eq. (1), § 430,)
^my = ih'y'^q)' + h")\^-. y^y , + ^^ J ; (^)
i.e., cancelling,
hy =^h'y'[{V + %"){h' - h') -t- Qnl)'{h' + h")\ (2)'
Hence we have two equations for finding two unknowns
viz.: (l)'and (2)' when n y \\ and (la)' and (2)' when n i, . . . . (1)''
and
i?r. = lAV[6^ + l]for^• ..7? =z ■ (ly
and neglecting tension ) * '^"^ S.C'J) .l(^— n)
For £:' inside middle third I ..p^ = (^^_j~ ^)^ • naY^'
' CD. I
where I = length of wall 1 to paper, usually taken = one foot,
or one inch, according to the unit of length adopted; ior n,
see § 436.
]^or, when the reservoir is emjpty and the water pressure
lacking, must the weight G resting on each bed, as (7j9, cut
the bed in a point E" so near the edge (7 as to produce exces-
sive pressure there (computed as above). The figure shows
the general form of profile resulting from these conditions.
The masonry should be of such a character, by irregular bond-
ing in every direction, as to make the wall a monolith.
440. The New Croton Dam* (completed in Feb. 1906).—
Attempts, by strict analysis, to determine the equation of the
curve BN, AM. being assumed straight, so as to bring the
point E' at the outer edge of the middle third of its joint, or
to make the pressure at D constant below a definite joint, liave
failed, up to the present time ; but approximate and tentative
methods are in use which serve all practical purposes. As an
illustration the method set forth in the report on the Quaker
Bridge Dam will be briefly outlined ; this method confines E'
to the middle third.
The width AB = h" is taken = 22' for a roadway, and h" =
T ft. The profile is made a vertical rectangle from A down
to a depth of 33 ft. below the water surface {o^eservoir full).
Combining the weight of this rectangle of masonry with the
corresponding water pressure (for a length of wall = one foot),
we find the resultant pressure comes a little within the outer
edge of the middle third of the base of the rectangle, while
p^ is of course small.
* This dam was built across the Croton River (see foot of p. 558) about
a mile above the site chosen (in 1884) for the proposed "Quaker Bridge"
Dam (which was not built). The profile and plan adopted were based
entirely upon the plans prepared for the Quaker Bridge Dam. See
Mr. Wegmann's paper on the New Croton Dam in vol. Ivii (June, 1907)
of Transac. Am. Soc. Civil Engineers; also his. book on Construction of
Masonry Dams, which gives an analytical treatment to replace the
tentative graphic method originally used, and described in the report
on the Quaker Bridge Dam.
HIGH MASONRY DAMS. 565
The rectangular form of profile might be continued below
this horizontal joint, as far as complying with the middle
third requirement, and the limitation of pressure-intensity, is
concerned ; but, not to make the widening of the joints too
abrupt in a lower position where it would be absolutely re-
quired, a beginning is made at the joint just mentioned by
forming a trapezoid between it and a joint 11 ft. farther down,
making the lower base of the latter of some trial width, which
can be altered when the results to which it gives rise become
evident. Having computed the weight of this trapezoid and
constructed its line of action through the centre of gravity of
the trapezoid, the value of the resultant G of this weight and
that of the rectangle is found (by principle of moments or by
an equilibrium polygon) in amount and position, and combined
with the water pressure of the corresponding 44 ft. of water to
form the force R^ whose point of intersection with the new
joint or bed (lower base of trapezoid) is noted and the value of
jp^ computed. These should both be somewhat nearer their
limits than in the preceding joint. If not, a different width
should be chosen, and changed again, if necessary, until satis-
factory. Similarly, another layer, 11 ft. in height and of
trapezoidal form, is added below and treated in the same way ;
and so on until in the joint at a depth of 66 ft. from the
water surface a width is found where the point E' is very
close upon its limiting position, while j&^ is quite a little under
the limit set for the upper joints of the dam, 8 tons per square
foot. For the next three 11 ft. trapezoidal layers the chief
governing element is the middle-third requirement, E' being
kept quite close to the limit, while the increase of jp^ to 7.95
tons per sq. ft. is unobjectionable ; also, we begin to move
the left-hand edge to the left of the vertical, so that when the
reservoir is empty the point E" shall not be too near the up-
stream edge C.
Down to a depth of about 200 ft. the value of j9^ is allowed
to increase to 10.48 tons per sq. ft., while the position of E'
gradually retreats from the edge of its limit. Beyond 200 ft.
depth, to prevent a rapid increase of width and consequent
extreme flattening of the down-stream curve, p^ is allowed
to mount rapidly to 16.63 tons per sq. ft. (=231 lbs. per
sq. in.), which value it reaches at the point N oi the base of
566 MECHANICS OF ENGINEERING.'
the dam, which has a width = 216 ft., and is 258 feet below
the water surface when the reservoir is full.
The heaviness of the raasonrj is taken as y' = 156.25 lbs.
per cubic foot, just f of ;/ = 62.5 lbs. per cub. foot, the heavi-
ness taken for water.
"When the reservoir is empty, we have the weight G of the
superincumbent mass resting on anj bed CD, and applied
through the point ^J" ; the pressure per unit area at C can
then be computed by eq. (la)"', § 439, n being the quotient of
{^CD — GE"^-^ GT) for this purpose. In the present case
we find E" to be within middle third at all joints, and the
pressures at G to be under the limit.
For further details the reader is referred to the report itself
(reprinted in Engineering News, January, 1888, p. 20). The
graphic results were checked by computation, Wegmann's
method, applied to eacli trapezoid in turn.*
441. Earthwork Dam, of Trapezoidal Section, f — Fig.4:91. Itis
stability. With the dimensions of
the figure, y and y' being the heavi-
FiG. 491. nesses of the water and earth respec-
tively (see § Y), we have
Weight of dam = 6^, = vol. X r' = ^^i[^ + i(<^: + ^W - (1)
Kesultant water press. = P =F zy = OA X I X ^hy, , (2)
Horiz. comp. oi P = H = P Bm a
= [OA sin a-]ihly = ^hHy. . . . (3)
From (3) it is evident that the horizontal component of P is
just the same, viz., = hi . ^hy, as the water pressure would be
* See the Engineering Record of Dec. 30, 1899, for a description of the
Assuan Dam across the river Nile; also the issue of April 22, 1905, for
mention of Mr. Atcherley's paper on masonry dams, etc.
f Mr. Bassell's "Earth Dams" is a recent publication; New York, 1904.
EAKTHWORK DAM. 567
on a vertical rectangle equal to the vertical projection of OA
and with its centre of gravity at the same depth (^h). Com-
pare §416. Also,
Vert. comp. oi I* = V= P cos a
= [OA cos a]^hly = ^ahly, ... (4)
and is the same as the w^ater pressure on the horizontal projec-
tion of OA if placed at a depth = O'G = ^h.
For stability against sliding, the horizontal component of P
must be less than the friction due to the total vertical pressure
on the plane A£^, viz., G^-\- V; hence if /"is the coefficieEt of
friction onA£J, we must have ^)- GcosdS -<^)"l4-esiii(^-(|))cos(j8 4-6-(^)
d4> J
-dG
Sm i H -(- O — CD 1 1
dP
di> ~ sin^ [|8 + 5 - (/)]
For P to be a maximum we must put
numerator of above = {a)
dG
To find a geometrical equivalent of -y— - , denote A C by L,
and draw ^^, making an angle = ^0 with AC. Now the
area AOI =: AI X kUE={L + dL)\Ld(l> = iZ'd(p . . .
(neglecting infinitesimal of 2d order). Now
dG
dG = y X area ACT X unity ; .-. -^- = ^yP'; .*. (a) becomes
sin (/? + d— ^)lyP sin (y^— 0)- sin (/? + (^ — 0)6^ cos(/S-0)
+ 6^ sin (yS - 0) cos (/? + (^ - 0) = ;
i.e., (9= =
l-^/Z" sin (/? — 0) sin (yg -[- (^ - 0)
sin (/? + (^ — 0) cos (y5 — 0) — cos (yS + (^ — 0) sin (/? — 0)
576 MECHANICS OF ENGINEERING.
when 7^ is a maximum ; and hence, calling O' and 4*' and I)
the values of O^ 0, and Z, for max. P^ we have
ff/ _ ^ x^n sin (^ -00 sin (^ + ^ - 00 ^ ^ /gx
sin (^ J • • \ /
and therefore from (1) P max. itself is
P' = ^yL'\^y^Slz:^ (3)
sin d
447. Geometric Interpretation and Construction. — If in Fig.
496 we draw GF^ making angle S with AD^ C being anj
point on the ground surface BP, we have
CF=L ^^" (^ ~ ^\
sin d
Drop a perpendicular FH from Fio AC, and we shall have
FS=^ (7F. sin (/? + <^ - 0), = Z . «in(^-0)sin(/?+cy-0).
sm d
From this it follows that the weight of prism of base^CZ'
and unit height
= ^yL.FE=^yL\'2^^1±l'^^±±:zA. (4)
sm (^
When ^ 6^ (as varies) assumes the position and value AC\
bounding the prism of maximum thrust, Fig. 49Y. Z becomes
= Z'. and 0=0'; and eq. (4) gives the weight of the prism
AC'F'. This weight is seen to be equal to that of the prism
(or wedge) of maximum thru&t ABC ^ by comparing eq. (4)
with eq. (2); that is, AC hisects the area ABCF', and
hence may he determined hy fixing such a point C\ on the
upper profile BD^ as to make the triangular area AC F'
equal to the sectional area of the wedge BCA\ CF' being
drawn at an angle = S with AP.
This holds for any form of ground surface ZZ>, or any
EETAINING WALLS.
677
C is best found graphic-.
values of the constants y5, a^ or d.
ally by trial, in dealing with
an irregular profile BD.
Having found AC\ =
L', P' can be found from
(3), or graphically as fol-
lows : (Fig. 49Y) With F'
as a centre and radius =
C'F\ describe an arc cut-
ting AD in J\ and join
C'J'. The weight of prism
with base C'J'F' and unit height will = P' . For that prism
has a weight
Fig. 497.
but
and
= iy.F'J' .C'H'',
yrrp ^ -pTQ? ^ L' sin (/? - (}>')
sin d '
wei
C'H' = L' sin (/? - 00 ;
ight of prism C'J'F = iyZ' ' !^^^ ;
= P'
[See eq. (3).]
448. Point of Application of the Resultant Earth Thrust. —
This thrust (called P' throughout this chapter except in the
present jparagra])h)\% now known in magnitude and direction,
but not in position ; i.e., we must still determine its line of
action, as follows :
Divide AB into a number of equal parts, «5, 5c, cd^ etc.*
see Fig. 498. Treat ah as a small retaining wall, and find the
magnitude P' of the thrust against it by § MY ; treat ac simi-
larly, thus finding the thrust, P", against it ; then a/7, ae, etc.,
the thrusts against them being found to be P'" ^ P^^, etc. ; and
so on. ISTow the pressure
P' on db is applied nearly at middle of «J,
P" - P' •■' " " " Ic,
pin _p„ « « « u ^^^
f)78
MECHANICS OF ENGINEEEING.
and so on. Erect perpendiculars at the middle points of ab,
he, ed, etc., equal respectively to I*',
P" - P\ P" - P", etc., and join the
ends of the perpendiculars. The per-
pendicular through the centre of gravity
of the area so formed (Fig. 498) will
give, on" AB, the required point of ap-
plication of the thrust or earth pressure
on AB. and this, with the direction and
magnitude already found in § 447, will
completely determine the thrust against the wall AB.
449. Special Law of Loading. — If the material to be retained
consists of loose stone, masses of masonry, buildings, or even
moving loads, as in the case of a wharf or roadway, each can
be replaced by the same weight of earth or other material
which will render the bank homogeneous, situated on the same
verticals, and the profile thus reduced can be treated by §§447
and 448.
Should the solid mass extend below the plane of rupture,
AC , and the plane of natural slope, it will become a retaining
wall for the material beyond, if strong enough to act as such
(limiting the profile ABGD of Fig. 496 to the front of the
mass, or to the front and line of rupture for maximum thrust
above it, if it does not reach the surface); if not strong enough,
or if it does not reach below the plane of natural slope, its
presence is better ignored, probably, except that the increased
weight must be considered.
The spandrel wall of an arch may present two of these
special cases; i.e., the profile may be enlarged to include a
moving load, while it may be limited at the back by the other
spandrel.
If the earth profile starts at the front edge of the top of
wall, instead of from the back as at B, Fig. 496, eq. (3) would
only apply to the portion behind AB prolonged, leaving the
part on the wall (top) to be treated as a part of the wall to aid
in resistmg the thrust.
If the wall is stepped in from the footings, or foundation
EETAINING WALLS. 579
courses, probably the weak section will be just above them ; if
stepped at intervals up the back of the wall, the surface of separa-
tion between the wall and filling, if it is plane, will probably
pass through the first step and incline forward as much as pos-
sible without cutting the wall.
450. Straight Earth-profile. — The general case can be simpli=
fied as follows (the earth-profile BD being straight, at angle
= C with vertical, = DET) : Since the triangles ABC and
Fig. 499.
C'AF' are equal, from § 447, and A C is common, therefore
BS=F'H (both being drawn "1 to AC). Draw AE and
B3£ II to'F'C (i.e., at angle S with AB), cutting BB, pro-
longed, in E. We have
BE EA , CE EA
and
CE EA - CF'' BE EA- BM*
But CF' = BM (since BS = M'F') ;
BE CE
therefore Z^ = ^=^: i.e., BE . BE = CE\
CE BE
which justifies the following construction for locating the de-
sired point C on 5Z>, and thus finding AC ^= L' and the
angle 0': Describe a circle on EB as a diameter, and draw
680
MECHANICS OF ENGINEEEING.
BX "1 to BD^ thus fixing X jn the curve. With centre E
describe a circular arc through X, cutting^/) in C\ required
Having A') cos a ' sin (C — \
where a and h are the respective lever-arms of the two forces
about the front edge of the middle third. {AB is the back of
the wall.) In other words, their resultant will pass through
this point.
The following table is computed on the basis just mentioned,
viz., that the resultant of P' and G shall j>ass through the
front edge of the midde third.
The symbols of eq. (5) and the table are all shown in Fig.
499, except y^ B, and d. y ■= weight of a cubic foot of earth,
here assumed = f that of masonry (e.g., if earth weighs
100 lbs., masonry is assumed to weigh 150 lbs. per cubic foot) ;
6 = angle which the thrust P' makes with the back of the
wall ; and d = a -\- 0, — 6 in this case as the wall is vertical,
or « = 0. <^ is the proper safe thickness to be given to the
wall, of rectangular section, to prevent overturning, as stated
above ; h is the altitude, and A is the fraction shown in eq. (5).
Whether the wall is safe against sliding on its base, and
whether a safe compression per unit area is exceeded on the
front edge of the base, are matters for separate consideration.
The latter will seldom govern with ordinary retaining walls.
a = 0; i e., wall is vertical; also density of wall
= 1 that of the earth.
I.
II.
III.
i = 90°
^ = 90'
^=/3
e = 90°
0=/S
= /3
tan/S
/s
^'
A
d
>'
A
d
4>'
A
d
1.0
45°
22^°
.17
Mfi
26°
.18
.22h
45°
.71
.SSh
1.5
56i°
28°
.39
.Uh
33°
.36
.307^
56°
.83
ASh
2.0
63+°
31f°
.38
.5\h
38°
.33
.367i
63°
.89
.517i
4.0
76°
38°
.61
M7i
45°
.54
.50h
76°
.97
.65h
Infinity
90°
45°
1.00
.82h
90°
1.00
.82h
90°
1.00
.82h
In Case I of table, since « = 0, ^ = 90° and C = 90° ;
.-. d = 90°, and hence O'F' of Fig. 499 is "1 to AD, so that
582
MECHANICS OF ENGINEEEING.
(GiDce the area of /\ABC' = l\AC'F') cp' must = |/?.
These values, in (5), give
P' = ^yh"" tan' i/? ; i.e., A = tan'' i/3. .
(6)
A
In Case II, since C = 90°, ^ = and 6 = j3, .'. d = /3;
and (5) reduces to
^ -^^^ siny5cos>" '•^•'^~^/^cos>'- • ^^^
In Case III, C = /? and ^i> will be || to AD, B being at
^^ infinity. See Fig. 501. Through
' c..^4^# Bdv'dwBB-] to AB, arid BF"
//--^l'//^\\ making angle d with JLD. C is
////"^vsi "^>, now to be located on BB, so as
|V^^^f^'^'^"''"to n^ake (area of) aABC =
|>\ /irVFl\\\-# (area of) aAG'F' (according
^/> ^|^'^//\\\^'8/W to § 447), the angle (7'i^'^ being
' " ' '^' )v^\\ = (J = a: + ^ ; = ^, in this case,
and hence also = y5. Conceive
Fig. 501, B and F' to be joined.
Kow aAG'F' = aABF" + aBF'F".
But A ABC = aBF'F" (equal bases and altitudes).
Hence A ABC cannot = aACF' unless C is moved out
to infinity / and then 0' becomes = /?, and eq. (5) reduces to
P'= ^yhj" sin fS ; i.e., A = sin /?.
(8)
[Increasing a from zero will decrease the thickness d ; i.e.,
inclining the wall inwards will decrease the required thickness,
but diminish the frictional stability at the base, unless the lat-
ter be '1 to AB. The back of the wall is frequently inclined
outwards, making the section a trapezoid, to increase the fric-
tional stability at the base when necessary, as with timber
walls supporting water.]
RETAINING WALLS. 583
452. Practical Considerations. — An examination of the
values of A and d in the table of § 451 will show that in sup-
porting quicksand and many kinds of clay which are almost
fluid under the influence of water, it is important to know
what kind of drainage can be secured, for on that will depend
the thickness of the wall. With well compacted material free
from water-bearing strata, an assumed natural slope of If to 1
(i.e., 1^ hor. to 1 vert.) will be safe ; the actual pressure below
the effect of frost and surface water will be that due to a much
steeper slope on account of cohesion (neglected in this theory).
The thrust from freshly placed material can be reduced by
depositing it in layers sloping back from the wall. If it is not
so placed, however, the natural slope will seldom be flatter
than If to 1 unless reduced by water. In supporting material
which contains water-bearing strata sloping toward the wall
and overlain by strata which are liable to become semi-fluid
and slippery, the thrust may exceed that due to semi-fluid ma-
terial on account of the surcharge. If these strata are under
the wall and cannot be reached by the foundation, or if resist-
ance to sliding cannot be obtained from the material in front
by sheet-piling, no amount of masonry can give security.
Water at the back of the wall will, by freezing, cause the
material to exert an indeflnitely great pressure, besides disinte-
grating the wall itself. If there is danger of its accumulation,
drainage should be provided by a layer of loose stone at the
back leading to "weep-holes" through the wall.
A friction-angle at the back of the wall equal to that of the
filling should always be realized by making the back rough by
steps, or projecting stones or bricks. Its effect on the required
thickness is too great to be economically ignored.
The resistance to slipping at the base can be increased, when
necessary, by inclining the foundation inwards; by stepping
or sloping the back of the wall so as to add to its effective
weight or incline the thrust more nearly to the vertical ; by
sheet-piling in front of the foundation, thus gaining the resist-
ance offered by the piles to lateral motion ; by deeper founda-
tions, gaining the resistance of the earth in front of the wall.
584 MECHANICS OF ENGINEERING.
The coefficient of friction on the base ranges, according to
Trautwine, from 0.20 to 0.30 on wet clay ;
" .50 to .66 " dry earth ;
" .66 to .75 " sand or gravel ;
" .60 on a drj^ wooden platform ; to .75 on a
wet one.
If the wall is partially submerged, the buoyant effort should
be subtracted from G^ , the weight of wall.
453. Results of Experience. — (Trautwine.) In railroad prac-
tice, a vertical wall of rectangular section, sustaining sand,
gravel, or earth, level with the top [p. 682 of Civ. Eng. Pocket
Book] f^nd loosely deposited, as when dumped from carts, cars,
etc., should have a thickness d, as follows :
If of Cut stone, or of first-class large ranged rubble, in mortar. . . . d= .35A
" good common scabbled mortar-rubble, or brick d — AOh
" well scabbled dry rubble .d= .50h
Where h includes the total height, or about 3 ft. of foundations.
{a) For the best masonry of its class h may be taken from
the top of the foundation in front.
{h) A mixture of sand or earth, with a large proportion of
round boulders or cobbles, will weigh more than the backing
assumed above ; requiring d to be increased from one eighth to
one sixth part.
(c) The wall will be stronger by inclining the back inwards,
especially if of dry masonry, or if the backing is put in place
before the mortar has set.
{d) The back of the wall should be left rough to increase
friction.
(e) Where deep freezing occurs, the back should slope out-
ward for 3 or 4 feet below the top and be left smooth.
if) When a wall is too thin, it will generallj' fail by bulging
outward at about one third the height. The failure is usually
gradual and may take years.
(g) Counterforts, or buttresses at the back of the wall, usually
of rectangular section, may be regarded as a waste of ma-
sonry, although considerably used in Europe; the bond will
EETAINING WALLS. ^ 585
seldom hold tliem to the wall. Buttresses in front add to the
strength, but are uot common, on account of expense.
ill) Land-ties of iron or wood, tying the wall to anchors im-
bedded below the Hue of natural slope, are sometimes used to
increase stability.
(/) Walls with curved cross-sections are not recommended.
454. . Conclusions of Mr. B. Baker. — (" Actual Lateral Pressure
of Earthwork.") Experience has shown that d = 0.25A, with
batter of 1 to 2 inches per foot on face, is sufficient when
backing and foundation are both favorable; also that under no
ordinary conditions of surcharge or heavy backing, with solid
foundation, is it necessary for d to be greater than 0.50A.
Mr. Baker's own rule is to make d = 0.33A at the top of
the footings, with a face batter of \\ inches per foot, in ground
of average character; and, if any material is taken out to form
a face-panel, three fourths of it is put back in the form of a
pilaster. The object of the batter, and of the panel if used, is
to distribute the pressure better on the foundation. All the
walls of the "District Kailway" (London) were designed on
this basis, and there has not been a single instance of settle-
ment, of overturning, or of sliding forward.
455. Experiments with Models.— Accounts of experiments
with apparatus on a small scale, with sand, etc., may be found
in vol. Lxxi of Proceedings of Institution of Civil Engineers,
London, England (p. 350) ; also in vol. n of the " Annales des
Fonts et Ohaussees" for 1885 (p. Y88).
The results of those experiments, and the results of experi-
ence given in §§ 453 and 454, when compared with the table
of p. 581, indicate a fairly close agreement between practice
and theory. This agreement is believed to be close enoui^h
so that the general method of §§ 447 and 451, wnth the table
of p. 5S1, can be relied upon in practice. The greatest value of
this method will, of course, be for cases of exceptional loading.
inclined w^alls, etc., where the results of experience do not
furnish so valuable a guide.
Note. — A recent and valuable book in this connection is Tlie Design
of Walls, Bins, and Grain Elevators, by Milo S. Ketchum. Published
by John Wiley & Sons, New York, 1907.
CHAPTEE ly.
HYDROSTATICS (Co7i G', and
Fig. 503.;
588 MECHANICS OF ENGINEERING.
to preserve equilibrium the body is attached by a cord to the
bottom of the vessel. The tension in this cord is
s,= vy-G\ . ■ (1)
At (c) V'y is < G\ and the cord must be attached to a
support above, and its tension is
S,= G'- V'y (2)
If in eq. (2) [(c) in figure] we call 8c the apparent weigJit of
the immersed body, and measure it by a spring- or beam-bal-
ance, v^e may say that
The apparent weight of a solid totally immersed in a liquid
equals its real weight diminished hy that of the amount of
liquid displaced; in other words, the loss of weight = the
weight of displaced liquid.
Example 1. — How great a mass (not hollow) of cast-iron can
be supported in water by a wrought-iron cylinder weighing
140 lbs., if <"he latter contains a vacuous space and displaces
3 cub. feet of water, both bodies being completely immersed ?
[Ft., lb., sec]
The buoyant effort on the cylinder is
FV = 3 X 62.5 = 18Y.5 lbs.,
leaving a residue of 47.5 lbs. upward force to buoy the cast-
iron, whose volume V" is unknown, while its heaviness (§ 7)
is ]/' = 450 lbs. per cub. foot. The direct buoyant effort of
the water on the cast-iron is V"y = [F''X62.5] lbs.,
and the problem requires that this force -|- 4Y.5 lbs. shall
= V^'y" = the weight G'' of the cast-iron ;
.-. y X 62.5 + 4T.5 = V X 450 ;
.-. F" = 0.12 cub. ft., while 0.12 X 450= 54 lbs. of cast-iron.
Ans.
Example 2.^ilequired the volume F', and heaviness y\
of a homogeneous solid which weighs 6 lbs. out of water and
4 lbs. when immersed {apparent weight) (ft., lb., sec ).
" IMMERSION. 589
From eq. (2), 4 = 6 - F' X 62.5 ; .\ V = 0.032 cub. feet ;
... y' ^Q' ^ Y> = Q^ 0.032 = 187.5 lbs. per cub. ft,
and the ratio of y' to y is 187.5 : 62.5 = 3.0 (abstract num-
ber) ; i.e., the substance of this solid is three times as dense,
or three times as heavy, as water. [The buoyant effort of the
air has been neglected in giving the true weight as 6 lbs.]
458. Specific Gravity. — By specific gravity is meant the ratio
of the heaviness of a given homogeneous substance to that of
a standard homogeneous substance; in other words, the ratio
of the weight of a certain volume of the substance to the
weight .of an equal voluvae oi the standard substance. Dis-
tilled water at the temperature of maximum density (4° Centi-
grade) under a pressure of 14.7 lbs. per sq. inch is sometimes
taken as the standaud substance, more frequently, however, at
62° Fahrenheit (16°.6 Centigrade). Water, then, being the
standard substance, the numerical example last given illustrates
a common method of determining experimentally the specific
gravity of a homogeneous solid substance, the value there ob-
tained being 3. The symbol (t will be used to denote specific
gravity, which is evidently an abstract number. The standard
substance should always be mentioned, and its heaviness ;k;
then the heaviness of a substance whose specific gravity is cr is
y'^^cry^ (1)
and the weight G' of any volume V of the substance may be
written
G' = V'y' = Very (2)
Evidently a knowledge of the value of y' dispenses with the
use of cr, though when the latter can be introduced into prob-
lems involving the buoyant effort of a liquid the criterion as
to whether a homogeneous solid will sink or rise, when im-
mersed in the stewf^arc? liquid, is more easily applied, thus :
Being immersed, the volume V of the body = that, V, of
displaced liquid. Hence,
690 MECHANICS OF ENGINEERING.
if G' is > FV, i.e., if V'y' is > Y'y^ -or cr > 1, it sinks ;
while if G' is < Y'y^ . . . . . . or cr < 1, it rises;
i.e., according as the weight G' is > or < than the buoyant
effort.
Other methods of determining the specific gravity of solids,
liquids, and gases are given in works on Physics.
459. Equilibrium of Flotation. — In case the weight G' of an
immersed solid is less than the buoyant effort Y'y (where V is
the volume of displacement, and /the heaviness of liquid) the
body rises to the surface, and after a series of oscillations comes
to rest in such a position. Fig. 503, that its centre of gravity C
and the centre of buoyancy B (the new B^ belonging to the
new volume of displacement, which is limited above by the
horizontal plane of the free surface of the liquid) are in the
same vertical (called the axis of flotation, or line of support),
and that the volume of displacement has diminished to such a
new value F, that
Yy=G' (1)
In the figure, F = vol. AWD, below the horizontal plane
AN, and the slightest motion of the body will change the form
of this 'volume, in general (whereas with
complete immersion the volume of dis-
placement remains constant). For stable
equilibrium it is not essential in every
case that C (centre of gravity of body)
should be below B (the centre of buoy-
ancy) as with complete immersion, since if
Fig. 503. the solid is turned, B may change its posi-
tion in the body, as the form of the Tolume AND changes.
There is now no definite relation between the volume of
displacement Fand that of the body, F', unless the latter is
homogeneous^ and then for G' we may write Y'y\ i.e.
Y'y' = Yy (for a homogeneous solid) ; . . (2)
or, the volumes are inversely proj>ortional to the heavinesses.
FLOTATION".
591
The buoyant effort of the air on the portion ANE may be
neglected in most practical cases, as being insignificant.
If the solid is hollow^ the position of its centre of gravity C
may be easily varied (by shifting ballast, e.g.) within certain
limits, but that of the centre of buoyancy B depends only on
the geometrical form of the volume of displacement AND^
below the horizontal plane AN.
Example. — (Ft., lb., sec.) Will a solid weighing G' — 400
lbs., and having a volume T^' =: 8 cub. feet, without hollows
or recesses, float in water? To obtain a buoyant effort of
400 lbs., we need a volume of displacement, see eq. (1), of
Y=
Q'
400
62^
only 6.4 cub. ft.
Hence the solid will float with 8 — 6.4, or 1. 6, cub. ft. pro-
jecting above the water level.
Query : A vessel contains water, reaching to its brim, and
algo a piece of ice which floats without touching the vessel.
When the ice melts will the water overflow ?
460. The Hydrometer is a floating instrument for determin-
ing the relative heavinesses of liquids. Fig. 504 shows a sim-
ple form, consisting of a bulb and a cylin-
drical stem of glass, so designed and
weighted as to float upright in all liquids
whose heavinesses it is to compare. Let F
denote the uniform sectional area of the
stem (a circle), and suppose that when float-
ing in water (whose heaviness = y) the
water surface marks a point A on the stem ;
and that when floating in another liquid,
say petroleum, whose heaviness, = ;^^, we
wish to determine, it floats at a greater
depth, the liquid surface now marking A'
on the stem, a height = x above A. G' is
the same in both experiments ; but while the volume of dis-
placement in water is V, in petroleum it is F4- Fx. There-
fore from eq. (1), § 459,
Fig. 504.
592 MECHANICS OF ENGINEERING.
in the water G' = Vy, ^(1)
and in the petroleum G' = {V-}- Fx)yp ; . . (2)
from which, knowing G\ F, x, and y, we find Fand ^p, i.e.,
V= — and yp— r,, ^ TP • • • • (3)
y ^ G' -\-Fxy ^ ^
[N.B. — ^is best determined bj noting the additional dis-
tance, = I, through which the instrument sinks in water under
an additional load P, not immersed / for then
G'^P^{Y-\-Fl)y, or F=^?^
Example. — [Using the inch, ounce^ and second, in which
system y = 1000 ^ 1728 = 0.5Y8 (§ 409).] With G' = Z
ounces, and F^= 0.10 sq. inch, x being observed, on the
graduated stem, to be 5 inches, we have for the petroleum
3 X 0.578 „ .„. X- ' X.
y^ = --— — — — — — = 0.525 oz. per cubic inch
^^ 3 + 0.10 X 5 X 0.578 ^
= 56.7 lbs. per cub. foot.
Temperature influences the heaviness of most liquids to
some extent.
In another kind of instrument a scale-pan is fixed to the top
of the stem, and the specific gravity computed from the weight
necessary to be placed on this pan to cause the hydrometer to
sink to the same point in all liquids for which it is used.
461. Depth of Flotation. — If the weight and external shape
of the floating body are known, and the centre of gravity so
situated that the position of flotation is known, the depth of
the lowest point 'below the surface may he determined.
FLOTATION.
593
Case I. Right i^visTn or cylinder with its axis vertical. —
Fig. 505. (For stability in this position,
see § 464a.) Let G' = weight of cylin-
der, i^'^the area of its cross-section (full -^^S
circle), h' its altitude, and h the un- ~^.
known depth of flotation (or di^aught) ; -^^
then from eq. (1), § 426,
G' = Fhr]
h ^' •
(1)
Fig. 505.
in which y = heaviness of the liquid.
If the prism (or cylinder) is homo-
geneous (and then C, at the middle of h', is higher than £)
and y' its heaviness, we then have
ly y
. . (2)
in which cr = specific gravity of solid referred to the liquid as
standard. (See § 458.)
Case II. pyramid or cone with axis vertical and vertex
down. — Fig. 506. Let Y' = volume of
whole pyramid (or cone), and V = vol-
ume of displacement. From similar
pyramids,
V
V
But G' =Vy', or, V=
G
whence
1
A'
s
1
1 A
^' — 1--
~^/n
.zzt
--^
-l_
r:=r ttS.
-_ ■ _
^£4^
Fig
.506.
h=h^^/'.
G'
V'y
(3)
594
MECHANICS OF ENGINEEEING.
Case III. Ditto, hut vertex ujp, — Fig. 507. Let the nota-
tion be as before, for Y and Y' . The
part out of water is a pyramid of volume
= Y" =: Y' — Y, and is similar to the
whole pyramid ;
... Y'- Y: Y' v.h'" : h"
Also,
Y =
Fig. 507.
Y
....■=.:, ^:^^,;^rjL=f.,
,'., finally, h = A' 1 _ ^1 _ [Q' -- F>]
(4)
Case I Y. Sphere. — Fig. 508. The volume immersed is
Y = fl^x')d3 = TtfX^rz - z^)dz = Tth' ^ ^^
Z=zO
r —
and hence, since Yy = 6^' ~ weight
of sphere,
s r'
(5)
^^ From which cubic equation h may be
^- ^E^^S^^JZ ^: ^-^H ^=^ obtained by successive trials and ap-
■^^^^^- =-i^S~' — = : :=^ proximations.*
^i»- 508. [An exact solution of (5) for the
unknown h is impossible, as it falls under the irreducible case
of Cardan's Rule.]
Case Y. Right cylinder with axis horizontal. — Fig. 509.
lers. Jy] = ^^''^ ^^ '^^- ^^^^ X ^
= {^r^a — ir' sin 2«')? ;
G'
Fig. 509.
hence, since Y ■■
y
G'
lr'\a — h sin 2a:l = — . «
e e
(6j
* See p. 224 of the Engineering Record for Feb. 22, 1908, for a diagram
to be used in solving this equation.
EXAMPLES OF FLOTATION". 595
From this transcendental equation we can obtain or, by trial,
in radians (see example in § 428), and finally A, since
h = r{l — cos a) (T)
Example 1. — A sphere of 40 inches diameter is observed to
have a depth of flotation A = 9 in. in water. Required its
weight G\ From eq. (5) (inch, lb., sec.) we have
G' = [62.5 ^ 1728];r9^[20 - ^ x 9] = 156.5 lbs.
The sphere may be hollow, e.g., of sheet metal loaded with
shot; constructed in any way, so long as G' and the volume
T^of displacement remain unchanged. But if* the sphere is
homogeneous, its heaviness (§ Y) y' must be
= O' -^Y'=^G' ^ ^7tr' = (156.5) ^ |;r20'
= .00466 lbs. per cubic inch,
and hence, referred to water, its specific gravity is cr = about
0.13.
Example 2. — The right cylinder in Fig. 509 is homogeneous
and 10 inches in diameter, and has a specific gravity (referred
to water) of cr = 0.30. Required the depth of flotation h.
Its heaviness must be y' = cry ; hence its weight
G' = Very = nr^lcry ;
hoflce, from eq, (6),
r'lla — f sin 2ar] = nrHo; .'. a — ^ sin 2ar = ttct
(lEvolving abstract numbers only). Trying a = 60° ( — |;r in
radians), we have
^Tt —i sin 120° = 0.614 ; whereas Ttcr = .9424
For ct = 70°, 1.2217 — i sin 140° = 0.9003 ;
For a = 71°, 1.2391 - i sin 142° = 0.9313 ;
T^or a = 71° 22', 1.2455 — ^ sin 142° 44' = 0.9428, which ma^
hp considered sufficiently close. ]^ow from eq. (7),
A = (6 in.) (1 - cos 71° 22') = 3.40 in.— J.n«,
MECHAIN'ICS OF ENGINEERING.
462. Draught of Ships. — In designing a ship, espeeiali}^ it oi
a new naode], the position of the centre of gravity is found by
eq. (3) of § 23 (with weights instead of volumes) ; i.e., the sum
of the products obtained by multiplying the weight of each
portion of the hull and cargo by the distance of its centre of
gravity from a convenient reference-plane (e.g., the horizontal
plane of the keel bottom) is divided by the sum of the weights,
and the quotient is the distance of the centre of gravity of the
whole from the reference-plane.
Similarly, the distance from another reference-plane is de-
termined. These two co-ordinates and the fact that the centre
of gravity lies in the median vertical plane of symmetry of the
ship (assuming a symmetrical arrangement of the framework
.^nd cargo) fix its location. The total weight, 6r', equals, of
*^ourse, the sum of the individual weights just mentioned. The
'Oentre of huoyancy, for any assumed draught and correspond-
ing position of ship, is found by the same method ; but more
simply, since it is the centre of gravity of the imaginary homo-
geneous volume between the water-line plane and the wetted
surface of the hull. This volume (of "displacement") is
divided into an even number (say 4 to 8) of horizontal laminae
of equal thickness, and Simpson's Rule applied to find the vol-
ume (i.e., the Y oi preceding formulae), and also (eq. 3, § 23)
the height of its centre of gravity above the keel. Similarly,
by division into (from 8 to 20) vertical slices, 1 to keel (an
even number and of equal thickness), we find the distance of
the centre of gravity from the bow. Thus the centre of buoy-
ancy is fixed, and the corresponding buoyant effort Yy (tech-
nically called the displacemeni and usuahy expressed in tons')
computed, for any assumed draught of ship (upright). That
position in which the " displacement" = (7' = weight of ship
is the position of equilibrium of the ship when floating up-
right in still water, and the corresponding draught is noted.
As to whether this equilibrium is stable or unstable, the fol-
lowing will show.
In most ships the centre of gravity G is several feet above
the centre of buoyancy, JB, and a foot or more below the water
line.
DRAUGHT OF SHIPS.
597
After a ship is afloat and its draught actually noted its total
A^eight G\ = Vy^ can be computed, the values of T^for dif-
ferent draughts having been calculated in advance. In this
way the weights of different cargoes can also be measured.
Example. — A ship having a displacement of 5000 tons id
itself 5000 tons in weight, and displaces a volume of salt water
Y= G'-^ y = 10,000,000 lbs. -f- 64 lbs. per cub. tt. = 156250
cub. ft.
463. Angular Stability of Ships. — If a vessel floating upright
were of the peculiar form and position of
Fig. 510 (the water-line section having an
area = zero) its tendency to regain that
position, or depart from it, when sliglitly
inclined an angle from the vertical is due
to the action of the couple now formed by
the equal and parallel forces Vy and G',
which are no longer directly opposed. This
couple is called a righting coujple if it acts
to restore the first position (as in Fig. 511,
where G is lower than B)^ and an
upsetting couple if the reverse, G
above B. In either case the mo-
ment of the couple is
= Yy . BG sin = Yye sin 0, ^= — ^
Fig. 510.
- ^ — ;.^ .^-j.g-=^
Fig. 511.
and the centre qfhuoyancy B does not
change its position in the vessel, since
the water-displacing shape remains
the same; i.e., no new portions of
the vessel are either immersed or
raised out of the water.
But in a vessel of ordinary form, when turned an angle from
the vertical, Fig. 512 (in which ED is a line which is vertical
when the ship is upright), there is a new centre of buoyancy,
B^ , corresponding to the new shape A^NJ) of the displacement-
volume, and the couple to right the vessel (or the reverse)
598
MECHANICS OF ENGIK^E^^EING.
consists of the two forces G' at C' and Yy at B^ , and has a
moment (which we may call J/j or
''E tnoment of stability) of a value
(§28)
'-^ M =^ Yy . m sin cp.
(1)
!Now conceive put in at B (centre
of buoyancy of the upright posi-
tion) two vertical and opposite
forces, each = Yy = G\ calling
them P and P, (see § 20), Fig. 512.
We can now regard the couple \0-\ Vy'\ as replaced by the
two couples [6r', P] and [Pj, Fk] ; J^or evidently
Fig. 512.
Yy .mCm\ = Yy . BO sin (p -\- Yy . mB sin ;
(§§33 and 34;)
.-. M= Yy BO sin OK, M is nega-
tive, indicating an upsetting cotijple. '^^^- ^^^•
That is, for = the equilibrium is stable, but for = OK,
unstable ; and Jlf = in both positions. From eq. (4) we see
why, ii Cis> above B, instability does not necessarily follow.
464a. Metacentre ef a Ship. — Kef erring again to Fig. 512,
we note that the entire couple \_G', Vy] will be a righting
couple, or an upsetting couple, according as the point m (^the
600 MECH'ANICS OF ENGINEERING.
intersection of the vertical through B^ , the new cantre of
buoyancy, with BG prolonged) is above or below the centre
of gravity C of the ship. The location of this point ra changes
with ^ = yz4>dF, and its moment about OL is (pyz^dF.
Hence the total moment, = Qa, or V^ya, of Fig. 513,
= X /oi
THE METACENTRE. 601
of water-line section, in which /qj, denotes the " moment of
inertia'-' (§ 85) of the plane figure OjlLNO about the axis OL.
Hence from (5), putting = sin 0(true when cf) = Oj, we have
mJJ = Iql -J- V^\ and therefore the distance 7/^,6', of the meta-
centre m above (7, the centre of gravity of the ship, Fig. 512, is
^,=h^,=^ 7o.,^water-line sec.) ^ ^^ ^ ^ ^ ^^^
in which e = BC =^ distance from the centre of gravity to the
centre of buoyancy, the negative sign being used when C is
above B ; while V=- whci^j volume of water displaced by the
ship.
We may also write, ir:^m. eqs. (6) and (1), for small values
Mom. of righting couph \= M= Vy sin -~ ±.eL . {1}
or
M=^ V sin cplIoL ± Ve]. . . . , (7)'
cf—
•B-
i
Eqs. (Y) and (7)' will give close approximations for < 10° or
15° with ships of ordinary forms.
Example 1. — A homogeneous right paralleiopiped. of
heaviness y', floats upright as in
Fig. 516. Find the distance ~
mC = hjn for its metacentre in this
position, and whether the equilibrium ^^
is stable. Here the centre of gravity, :£\e
C, being the centre of figure, is of '£:)j^
course above B, the centre of buoy- 3-1
ancy ; hence e is negative. B is the ^^j
centre of gravity of the displacement, ^^^ -" -
and is therefore a distance ^h below
the water-line. We here assume that I is greater than ¥.
From eq. (2), § 461,
--h^,=
h =
hy\
"605 MECHANICS OF ENGINEERING.
and since CD = \Ti\ and BD — |A, .-. e ~ ^Ji' — h)\
i.e., e = ^h'
L r J
while (§ 90) /oi, of the water-line section AM, = ^'^^
Also,
r
and hence, from eq. (6), we have
Hence if 5'' is > 6^'" ^ fl - ^ V the position in Fig. 516 is
one of stable equilibrium, and vice versa. E.g., if y' = ^yy
y = 12 inches and A' = 6 inches, we have (inch, pound, sec.)
h^ = ^^= J^ [144 - 6 X ^(1 - i)j = 2.5 in.
The equilibrium will be unstable if, with y' = ^y, h' is made
less than 1.225 A'; for, putting m(7 = 0, we obtain h' =.
1.225 A'.
Example 2. — (Ft., lb., sec.) Let Fig. 517 represent the half
water-line section of a loaded ship of G' = Yy =■ 1010 tons
Fig. 51?.
displacement ; required the height of the metacentre above the
centre of buojancj, i.e., 7i%B = ? (See equation just before eq.
(6).) Now the quantity /ox, , of the water-line section, may,
from symmetry, (see § 93,) be written
JoL = ^f\fdx, (1)
METAC'ENTEK. 603
in which y = the ordinate ~| to the axis OL at any point ; and
this, agaiu, bj Simpson's Kule for approximate integration,
OL being divided into an even nnmber, n. of equal parts, and
ordinates erected (see ligure), may be written
2 OL-0
^^^-3- 3n
+ 2(y/ + 2//+.:.+2/l_2) + 2/l
From which, by numerical substitution (see figure for dimen-
sions ; n = 8),
/oi. = I . gi^g [(0.5' + 4(5^ + 12^ + 13' + T)
+ 2(9= + 14:' + ir) + 0.5'
or.
/^^ = ^[0.125+4 X 4393 + 2 X 4804+ 0.125]
loL 120801
125
1728
729
2197
2744
343
1331
= 120801 biquad. ft. ; .-. m^ = ±^ =.- ^^^^^ _ ^^^
^ ' V [2020000^64]
= 3.8 feet.
That is, the metacentre is 3.8 feet above the centre of buoyancy,
and hence, if £C=2 feet, is 1.90 ft. above the centre of
gravity. [See Johnson's Cyclopgedia, article Naval Architec-
465. Metacentre for Longitudinal Stability. — If we consider
the stability of a vessel with respect to pitching, in a manner
similar to that just pursued for rolling, we derive the position
of the metacentre for pitching or for longitudinal stability —
and this of course occupies a much higher position than that
for rolling^ involving as it does the moment of inertia of the
water-line section about a horizontal gravity axis ~\ to the keel.
"With this one change, eq, (6) holds for this case also. In
large ships the height of this metacentre above the centre of
gravity of tlie ship may be as great as 90 feet.
|i" m' f I ill" Till iirr
Fig. 518.
CHAPTER Y.
HYDROSTATICS {Continued)— QAS'EOVQ FLUIDS.
466,, Thermometers. — The temperature, or " hotness" of
■liquids has, within certain limits, but little influence on tlsair
statical behavior, but with gases must always be taken into
account, since the three quantities, tension, temperature, and
volume, of a given mass of gas are connected by a nearly in-
variable law, as will be seen.
An air-thermometer, Fig. 518, consists of a large glass bulb
filled with air, from which projects a line straight tube of
^„_-,,^^ , even bore (so that equal lengths
A'''.:\.'.\\ j j -V: represent equal volumes). A
small drop of liquid, A, sepa-
rates the internal from the ex-
ternal air, both of which are
at a tension of (say) one at-
mosphere (14.7 lbs. per sq. inch). When the bulb is placed
in melting ice (freezing-point) the drop stands at some point F
in the tube; when in boiling water (boiling under a pressure
of one atmosphere), the drop is found at B, on account of the
expansion of the internal air under the influence of the heat
imparted to it. (The glass also expands, but only about y^-^
as much ; this will be neglected.) The distance FB along the
tube may now be divided into a convenient number of equal
parts called degrees. If into one hundred degrees, it is found
that each degree represents a volume equal to the ytu'^w^
(.00367) part of the total volume occupied by the air at freez-
ing-point ; i.e., the increase of volume from the temperature of
freezing-point to that of the boiling-point of water = 0.36Y of the
volume at freezing, the pressure being the same, and even having
any value whatever (as well as one atmosphere), within ordi-
nary limits, so long as it is the same both at freezing and boil-
604
THERMOMETERS, 605
ing. It must be understood, however, that by teinjjerature of
toiling is always meant that of water boiling under one at-
mosphere pressure. Another way of stating the above, if one
hundred degrees are used between freezing and boiling, is as
follows : That for each degree increase of temperature the in-
crease of volume is -g^-g of the total volume at freezing ; 2Y3
being the reciprocal of .0036Y.
As it is not always practicable to preserve the pressure con-
stant under all circumstances with an air- thermometer, we use
the common mercurial thermometer for most practical pur-
poses. In this, the tube is sealed at the outer extremity, with
a vacuum above the column of mercury, and its indications
agree very closely with those of the air-thermometer. That
equal absolute increments of volume should imply equal incre-
ments of heat imparted to these thermometric iiuids (under
constant pressure) could not reasonably be asserted without
satisfactory experimental evidence. This, however, is not al-
together wanting, so that we are enabled to say that within a
moderate range of temperature equal increments of heat pro-
duce equal increments of volume in a given mass not only of
atmospheric air, but of the so-called " perfect" or "permanent"
gases, oxygen, nitrogen, hydrogen, etc. (so named before it was
found that they .could be liquefied). This is nearly true for
mercury also, and for alcohol, hut not for water. Alcohol
freezes at — 200° Fahr., and hence is used instead of mercury
as a thermometric substance to measure temperatures below
the freezing-point of the latter.
The scale of a mercurial thermometer is fixed ; but with an
air-thermometer we should have to use a new scale, and in a
new position on the tube, for each value of the pressure.
467. Thermometric Scales.— In the Fahrenheit scale the tube
between freezing and boiling is marked o£E into 180 equal
parts, and the zero placed at 32 of these parts below the freez-
ing point, which is hence -f- 32°, and the boiling-point +212°.
The Centigrade, or Celsius, scale, which is the one chiefiy
used in scientific practice, places its zero at freezing, and 100"
at boiling-point. Hence to reduce
606 MECHANICS OF EJN^GINEEEIlSrG.
Fahr. readings to Centigrade, subtract 32° and multiply by -| ;
Cent. " " Fahrenheit, multiply by f and add 32°.
468. Absolute Temperature. — Experiment also shows that if
a mass of air or other perfect gas is confined in a vessel whose
volume is but slightly affected by changes of temperatui'e,
equal increments of temperature (and therefore equal incre-
ments of heat imparted to the gas, according to the preceding
paragraph) produce equal increments of tension (i.e., pressure
per unit area) ; or, as to the amount of the increase, that wlien
the temperature is raised by an amount 1° Centigrade, the ten-
sion is increased ^rs ^^ ^^^ value at freezing-point. Hence,
theoretically, an ideal barometer (containing a liquid unaffected
by changes of temperature) communicating with the confined
gas (whose volume practically remains constant) would by
its indications serve as a thermometer,
Fiff. 519, and the attached scale could be
graduated accordingly. Thus, if the col-
imn stood at A when the temperature
was freezing, A would be marked 0° on
^^ the Centigrade system, and the degree
^ spaces above and below A would each
^^°- si^- = sfs of the height AB, and therefore
the point B (cistern level) to which 'the column would sink if
the gas-tension were zero would be marked — 273° Centi-
grade.
But a zero-pressure, in the Kinetic Theory of Gases (§ 408),
signifies that the gaseous molecules, no longer impinging
against the vessel walls (so that the press. = 0), have become
motionless; and this, in the Mechanical Theory of Heat, or
Tiierinodynarrdcs^ implies that the gas is totally destitute of heat.
Hence this ideal temperature of — 273° Centigrade, or — 460°
Fahrenheit, is called the Ahsolute Zero of Temperature^ and by
reckoning temperatures from it as a starting-point, our formulae
will be rendered much more simple and compact. Tempera-
ture so reckoned is called ahsolute temperature, and will be
denoted by the letter T. Hence the following rules for re-
duction :
GASES AND VAPORS. 607
Absol. temp. T in Cent, degrees = Ordinary Cent, -j- 273° ;
Absol. temp. T in Fahr. degrees = Ordinary Fahr. \ 460°.
For example, for 20° Cent., T = 293° Abs. Cent.
469. Distinction Between Gases and Vapors. — All known
gases can be converted into liquids by a sufficient reduction oi
temperature or increase of pressure, or both; some, however,
with great difficulty, such as atmospheric air, oxygen, hydro-
gen, nitrogen, etc., these having been but recently (187S) re-
duced to the liquid form. A vapor is a gas near the point of
liquefaction, and does not show that regularity of behavior
under changes of temperature and pressure characteristic of a
gas when at a temperature much above the point of liquefac-
tion. All gases treated in this chapter (except steam) are sup-
posed in a condition far removed from this stage. The fol-
lowing will illustrate the properties of vapors. See Fig. 520.
Let a quantity of liquid, say water, be intro- THfRM.
duced into a closed space, previously vacuous, J:?==^^F=^*/^'
of considerably larger volume than the water, /^"''.■■'•■' '• \^^^^
continues to do so until its pi-essure attains a ^ i/>m
definite value dependent on the temperature, ^^^' ^~^'
and not on the ratio of the volume of the vessel and the origi-
nal volume of water ; e.g., if the temperature is 70° Fahren-
heit, the vapor ceases to form when the tension reaches a value
of O.-SG lbs. per sq. inch. If heat be gradually applied to raise
the temperature, more vapor will form (with ebullition ; i.e ,
from the body of the liquid, unless the heat is applied very
slowly), but the tension will not 7'ise .above a fixed value for
each temperature (independent of size of vessel) 80 long as
there is any liquid left. Some of these corresponding values,
for water, are as follows : For a
Fahr. temp. = 70° 100° 150° 212° 220° 287° 300°
Tension (lbs ) ^ q gg q 93 3^9 -^^^^ -^^ ^ 55 q g/^2
persq. in.) \
= one atm.
At any such stage the vapor is said to be saturated.
608 MECHANICS OF ENGINEERING.
Finally, at some temperature, dependent on the ratio of the
original volume of water to that of the vessel, all of the water
wall have been converted into vapor (i.e., steam); and if the
temperature be still further increased, the tension also increases
and no longer depends on the temperature alone, hut also on
the heaviness of the vapor when the loater disappeared. The
vapor is now said to be superheated, and conforms more in its
properties to perfect gases.
470. Critical Temperature. — From certain experiments there
seems to be reason to believe that at a certain temperature,
called the critical temperature, different for different liquids,
all of the liquid in the vessel (if anv remains, and supposing
the vessel strong enough to resist the pressure) is converted
into vapor, whatever be the size of the vessel. That is, above
the critical temperature the substance is necessarily gaseous,
in the most exclusive sense, incapable of liquefaction by pres-
sure alone ; while below this temperature it is a vapor, and lique-
faction will- begin if, by compression in a cylinder and conse-
quent increase of pressure, the tension can be raised to a value
corresponding, for a state of saturation, to the temperature
(in such a table as that just given for water). For example, if
vapor of water at 220° Fahrenheit and tension of 10 lbs. per
sq. inch (this is superheated steam, since 220° is higher than
the temperature which for saturation corresponds to ^ = 10
lbs. per sq. inch) is compressed slowly (slowly, to avoid change
-of temperature) till the tension rises to 17.2 lbs. per sq. in.,
which (see above table) is the pressure of saturation for a tem-
perature of 220° Fahrenheit for water-vapor, the vapor is satu-
rated, i.e., liquefaction is ready to begin, and during any fur-
ther slow reduction of volume the pressure remains constant
and some of the vapor is liquefied.
By " perfect gases," or gases proper, we may understand,
therefore, those which cannot be liquefied by pressure unac-
companied by great reduction of temperature; i.e., whose
" critical temperatures" are very low. The critical temperature
of ]S^2^, or nitrous oxide gas, is between — 11° and -\- 8° Cen-
tigrade, while that of oxygen is said to be at — 118° Centi-
LAW OF CHARLES. 609
grade. [See p. 471, vol. 122 of the Journal of the Iranklin
Institute. For an account of the liquefaction of oxygen, etc.,
see the same periodical, January to June, 18T8.]
471. Law of Charles (and of Gay Lussac). — The mode of gradu-
ation of the air-thermometer may be expressed in the follow-
ing formula, which holds good (for practical purposes) within
the ordinary limits of experiment for a given mass of any
perfect gas, the tension remaining constant :
Y=Y,-\- 0.00367 Y,t = F„(l + .003670 ;••(!)
in which F^ denotes the volume o«cupied by the given mass
at freezing-point under the given pressure, Y its volume at
any other temperature t Centigrade under the same pressure,
Now, 273 being the reciprocal of .00367, we may write
Y- Yi^:il±J). ie Z_ Z i press. ) . .g)
^~ ^^ 273 ' '•^•' Y,~ T, ' 'I const, p ^""^
(see § 468 ;) in which T^ = the absolute temperature of freezing-
point, = 273° absolute Centigrade, and T the absolute tem-
perature corresponding to t Centigrade. Eq. (2) is also true
when T and T^, are both expressed in Fahrenheit degrees (from
absolute zero, of course). Accordingly, we may say that, the
pressure remaining the same, the volume of a given mass of
gas varies directly as the absolute temperature.
Since the weight of the given mass of gas is invariable at a
given place on the earth's surface, we may
always use the equation Yy = Y^y^ , (3)
pressure constant or not, and hence (2) may be rewritten
V T
— = -=-. . . (press, const.) ; . (4)
Y T^
i.e., if the pressure is constant, the heaviness (and therefore
the specific gravity^ varies inversely as the absolute temperO'
ture.
610
MECHANICS OF EISTGHSTEERUSTG.
Experiment also shows (§468) that if the vohime [and there-^
fore the heaviness, eq. (3)] remains constant, while the tem-
perature varies, tlie tension p will change according to the
following relation, in which ^„ = the tension when the tem=
perature is freezing :
i>==^o+¥^i?o^
273 -\-t
273
(5)
t denoting the Centigrade temperature
as before, we have
p _ T
Hence transforming.
vol., and .
heav., const. ^ '
(6)
or, the volume and heaviness remaAning constant^ the tension
of a given mass of gas vai'ies directly as the ahsolute tempera-
ture. Tliis is called the Law of Charles (or of Oay Lussac).
472. General Formulae for any Change of State of a Perfect Gas,
— If any two of the three quantities, viz., volume (or heavi-
ness), tension^ and temperature^ are changed, the new value of
the third is determinate trom those of the other two, according
to a relation proved as follows (remember-
ing that henceforth the absolute temperature
only will be used, T, § 468) : Fig. 521.
At A a certain mass of gas at a tension of
j?o, one atmosphere, and absolute tempera-
ture T^ (freezing), occupies a volume V^ .
Let it now be heated to an absolute temp.
= T', without change of tension (expanding
behind a piston, for instance). Its volume will increase to a
value V which from (2) of § 471 will satisfy the relation
A
B
V,
C
V,
Fig. 531.
Y_
t:
CO
(See B io figure.)
Let it now be heated without change of volume to an abso-
lute temperature T {^C in figure). Its volume is still Y, but
LAWS OF PEEFECT GASESo . 611
the tension has risen to a value jp^ such that, on comparing B
and G bj eq. (6), we have
jrT' (^)
Combining (Y) and (8), we obtain for any state in which the
tension is_^, volume Y, and absohite temperature T, in
7} I' 7? V 7? r^
[General) . . . -=-^ =.-Li^ j or ^--=- = a constant ; . (9)
or
{General). , , . V^^^Pj^^ ...... (10)
which, since
{General) . , Vy = V,y, = V„,r^ = VnVn , = . . (H)
is true for any change of state, we may also write
. . > c . (12)
[General) . . .
F _ J^o
' yT - y.r:
or
X>m _ JPn
y-m-'-m, yn-'-n
(13)
These equations (9) to (13), inclusive, hold good for any state
of a mass of any perfect gas (most accurately for air). The
subscript refers to the state of one-atmosphere tension and
freezing-point temperature, m and oi to any two states what-
ever (within practical limits) ; y is the heaviness, §§ Y and 409,
and T the absolute temperature, § 468.
If^, y, and T oi equation (9) be treated as variables, and
laid off to scale as co-ordinates parallel to three axes in space,
respectively, the surface so formed of which (9) is the equation
is a hyperbolic paraboloid.
473. Examples. — Example 1. — What cubic space will be
occupied by 2 lbs. of hydrogen gas at a tension of two atmos-
pheres and a temperature of 27° Centigrade?
612 MECHANICS OF EJSTGINEEEING.
With the inch-lh.-seo. system we have p^ = 14.7 lbs. per sq.
inch, Y, = [.0056 -^ 1728] lbs.* per cubic inch, and T, = 273°
absolute Centigrade, when the gas is at f reezing-poinf at one
atmosphere (i.e., in state suh-zero). In the state mentioned ^V:
the problem, we have jt? = 2 X 14.7 lbs. per sq. in., -
;7^ - 273 + 27 = 300° absolute Centigrade,
while y is required. Hence, from eq. (12),
2 X 14.7 14/r .
y 300 ~ (.0056 -^ 1728)273 '
0102
.-. y = '——-- lbs. per cub. in. = .0102 lbs. per cub. foot ; and if
^ 1728 ^ ^ '
the total weight, = 6^, = Yy, is to be 2 lbs., we have (ft., lb.,
sec.) F = 2 -^ 0102 = 196 cubic iQet.—Ans.
Example 2. — A mass of air originally at 24° Centigrade
and a tension indicated by a barometric column' of 40 inches
of mercury has been simultaneously reduced to half its
former volume and heated to 100" Centigrade ; required its
tension in this new state, which we call the state n, m being the
original state. Use the inch, lb., sec. We have given, there-
fore, p^ = f f X 14.7 lbs. per sq. inch, T^ = 273 + 24 = 297°
absolute Centigrade, the ratio
Y^\ Vn = 2:l, and T^ = 273 + 100 =373° Abs. Cent.;
while pn is the unknown quantity. From eq. (10), hence,
^, = -^.^-.^^=2xW.f^Xl4.7 = 49.221bs.per3q,m,,
which an ordinary steam-gauge would indicate as
(49.22 — 14.7) = 34.52 lbs. per sq. inch.
(That is, if the weather barometer indicated exactly 14.7 lbs.
per sq. inch.)
* See table on p. 517.
EXAMPLES. PERFECT GASES.
613
Example 3.— A mass of air, Fig. 522, occupies a rigid closed
vessel at a teuiperature of 15° Centigrade (equal to that of sur-
^
•■v«-
Vr
Fig. 522.
rounding objects) and a tension
of four atmospheres [state m].
By opening a stop-cock a few
seconds, thus allowing a portion
of the gas to escape quickly, and
then shutting it, the remainder
of the air [now in state ti] is found to have a tension of only
2.5 atmospheres (measured immediately) ; its temperature can-
not be measured immediately (so much time being necessary
to affect a thermometer), and is less than before. To compute
this temperature, T^, we allow the air now in the vessel to
come again to the same temperature as surrounding objects
(15° Centigrade) ; find then the tension to be 2.92 atmospheres.
Call the last state, state ?' (inch, lb., sec). The problem then
stands thus :
2)m = 4 X 14.7
Pn = 2.5 X 14.7
rm = ?
rn = -i
r™ = 388° Abs. Cent.
m _ j principal
" ~~ I unknown
Pr = 3.92 X 14.7
rr = rn (since Vr= Vn)
Tr = Tm = 388° Abs. Cent.
In states n and r the heaviness is the same ; hence an equa-
tion like (6) of § 4Y1 is applicable, whence
orZ.
2.5 X 14.7
2.92 X 14.7
X 288 = 246° Abs. Cent.
or — 27° Centigrade ; considerably helow freezing, as a result of
allowing the sudden escape of a portion of the air, and the con-
sequent sudden expansion, and reduction of tension, of the re-
mainder. In this sudden passage from state tn to state n^ the
remainder altered its heaviness (and its volume in inverse ratio)
in the ratio (see eqs. (11) and (10) of § 472)
Yn _
v^
_p^ T^ 2.5X14.7 288
0.73.
Vm
Yn
i>^' r, 4X14.7 '246
Kow the heaviness in state m (see eq. (12), § 472) was
614
MECHANICS OF ENGINES KING.
^ ^^ ro^o ^ ^ X i4.r .0807
^'^ r^ " ^„ 288 ■ 1728
lbs. per cub. in. = .306 lbs. per cub. ft.
273
14.7
.306
1728
.-.. y^ = 0.73 X Vm = 0.223 lbs. per cub. ft.,
and also, since Vm = 0.73 Y^ , about -^-q of the original quan-
tity of air in vessel Las escaped.
[II^OTE. — Bj numerous experiments like this, the law of
cooling, when a mass of gas is allowed to expand suddenly (as,
e.g., behind a piston, doing work) has been determined ; and
vice versa, the law of heating under sudden compression' ; see
§ 487.]
474. The Closed Air-manometer. — If a manometer be formed
of a straight tube of glass, of uniform cylindrical bore, which
is partially filled with mercury and then inverted in a cistern
of mercury, a quantity of air having heen left hetween the
mercury and the upper end of the.
txibe, which is closed, the tension of
this confined air (to be computed
from its observed volume and tem-
perature) must be added to that due
to the mercury column, in order to
obtain the tension j?' to be measured.
See Fig. 523. The advantage of this
kind of instrument is, that to meas-
ure great tensions the tube need not
be very long. Let the temperature
Tj of whole instrument, and the tension p^ of the air or gas
in the cistern, be known when the mercury in the tube stands
at the same level as that in the cistern. The tension of the
air in the tube must now be^^ also, its temperature T^ , ?.nd its
volume is V^ = 2^h^ , ^Z^being the sectional area of the bore of
die tube ; see on left of figure. When the instrument is used,
gas of unknown tension j9' is admitted to the cistern, the tem-
perature of the whole instrument being noted (= T), and the
heights A and h" are observed {h-\-h" cannot be put,= A,
CLOSED AIR-MANOMETEE. • 615
anless the cistern is very large), p' is then computed as fol-
lows (eq. (2), § 413) :
P'=^h'ym+p; (1)
in which p = the tension of the air in the tube, and y^n ^^6
heaviness of mercury. But from eq. (10), § 472, putting
F, = i^%, and V=Fh,
^. T K T ,^.
^^^^T'Tr'h'T.^ ^^^
Hence finally, from (1) and (2),
h T
p'^h'y^ + f.YPt (3)
Since T^^p^^ and A, are fixed constants for each instrument,,
we may, from (3), compute j?' for any observed values of h and
T (N.B. T and T^ are absolute temperatures)^ and construct
a series of tables each of which shall give values of p)' for a
range of values of A, and one special value of T,
Example. — Supposing the fixed constants of a closed air
manometer to be (in inch-lb.-sec. system) p^ = 14.Y (or one
atmosphere), T^ = 285° Abs. Cent, (i.e., 12° Centigrade), and
A, = 3' 4" = 40 inches ; required the tension in the cistern
indicated by k'^ = 25 inches and h = 15 inches, when the
temperature is — 3° Centigrade, or T = 270° Abs. Cent.
For mercury, y^ = [848.7 -^ 1728] (§ 409) (though strictly
it should be specially computed for the temperature, since it
varies about .00002 of itself for each Centigrade degree).
Hence, eq. (3),
lbs. per sq. inch, or nearly 3^ atmospheres [steam-gauge would
read 34.7 lbs. per sq. in.].
475. Ms.riotte's law, (or Boyle's,) Temperature Constant ; i.e..
Isothermal Change. — If a mass of gas be compressed, or al-
616 • MECHANICS OF ENGIJS^EEEINe.
lowed to expand, isothermally, i.e., without change of fcera.
peratare (pricticallj this cannot be done unless the walls ^l the
vessel are conductors of heat, and then the motion must be
slow), eq. (10) of § 472 now becomes (since T^ = T^)
Mariott^s Law^ \ y r) ~ V m
Temjp. constant \ ' ^'^^'^- *^^^«
jPn, i' TO
1.605 the tem/perature remaining unchanged^ the tensions are
inversely ^proportional to the volumes^ of a given mass of a
perfect gas ^ or, the product of volume hy tension is a constant
quantity. Again, since YmYm = ^nVn for any change of
state,
j Mariott^s Law, \ Pm ^^Vm qj, Pb = ^' Cg\
I Temp, constant ) ' ' Pn Vn' Ym Vn'
!.e;, the pressures {or tensions are directly proportional to tJie
{^first power of the) heamnesses, if the temperature is the same.
This law, which is very closely followed by all the perfect
gases, was discovered by Boyle in England and Mariotte in
France more than two hundred years ago, but of course is only
a particular case of the general formula, for any change of
state, in § 4:72. It may be verified experimen-
tally in several ways. E.g., in Fig. 524, the
tube OM being closed at the top, while PiV is
open, let mercury be poured in at P until it
reaches the level A'B'. The air in OA' is now
at a tension of one atmosphere. Let more mer-
ciiry be slowly poured in at P, until the aii
confined in has been compressed to a volume
,-]a"' OA" = i of 0A\ and the height B"E' they
measured ; it will be found to be 30 inches ; i.e„,
" ^ ^ the tension of the air in is now two atmoS'
pheres (corresponding to 60 inches of mercury)
Fig. 524. Again, compress the air in to ^ its original
volume (when at one atmosphere), i.e., to volume OA'" =^
^0A% and the mercury height B"'E"' will be 60 inches, show-
ing a tension of three atmospheres in the confined air at (90
II
B
b'
-
/
e'"
E^~
B II
A
^
—
^^—:=^
N
-=^- r=z-
maeiotte's law.
617
inches of mercury in a .barometer). It is understood that the
temperature is the same, i.e., that time is given the compressed
air to acquire the temperature of surrounding objects after
being heated by the compression, if sudden.
[Note, — The law of decrease of steam-pressure in a steam=
engine cylinder, after the piston has passed the point of " cut
off " and the confined steam is expanding, does not materially
differ from Mariotte's law, which is often applied to tli© case
of expanding steam ; see § 479.]
While Mariotte's law may be considered exact for practical
purposes, it is only approximately true, the amount of the
deviations being different at different temperatures. Thus,
for decreasing temperatures the product Vp of volume by
tension becomes smaller, with most gases.
Example 1. — If a mass of compressed air expands in a
cylinder behind a piston, having a tension of 60 lbs. per sq.
inch (45.3 by steam-gauge) at the beginning of the expansion,
which is supposed slow (that the temperature may not fall) ;
then when it has doubled in volume its tension will be only
30 lbs. per sq. inch ; when it has tripled in volume its tension
will be only 20 Ibs^ per sq. inch, and so on.
Example 2. Diving-iell. — Fig. 525. If the cylindrical
diving-bell AB is 10 ft. in height, in what
depth, A = ?, of salt water, can it be let down
to the bottom, without allowing the water to
rise in the bell more than a distance 6« = 4 ft. ?
Call the horizontal sectional area, i^. The
mass of air in the bell is constant, at a constant
temperature. Firsts algebraically j at the
surface this mass of air occupied a volume
F^ = Fh" at a tension p^ = 14.T X 144 lbs.
per sq. ft., while at the depth mentioned it is
compressed to a volume Vn = F{h" — a), and
is at a tension p^ —p^ + (A — cC)y^ , in which /^j^^/^/j////^
?/,p = lieavineiBS of salt water. Hence, from st&sss.
-A-=-?
■^l°lq^
'f
^-T
■ • v,,.--:
n,
-t—
^! -^
—1-
— ^ —
— f-
'V'
^^ -B
^
V^ Fh" = [i>«» +(A - a)y:\F{h" - «);
(^
618 MEOHAlSllCS OF ENGIJN'EERING.
h =■ a
1 1 P-n
{h" — a)y^_
hence, numerically, (ft., lb., sec.,)
14.7 X 144
(4)
A = 4x
(10-4)X 64_
= 26.05 feet.
476. Mixture of Gases. — It is sometimes stated that if a vessel
is occupied by a mixture of gases (between which there is no
chemical action), the tension of the mixture is equal to the sum
of the pressures of each of the component gases present; or,,
more definitely, is equal to the sum of the pressures which the
separate masses of gas would exert on the vessel if each in turn
occupied it alone at the same temperature.
This is a direct consequence of Mariotte's law, and may be
demonstrated as follows :
Let the actual tension be j?, and the capacity of the vessel V.
Also let y^, V^^ etc., be the volumes actually occupied by the
separate masses of gas, so that
F, + F,+ ...= F; ..... (1)
and ^j5^2? 6tc., the pressures they would individually exert
when occupying the volume V alone at the same tempera-
ture. Then, by Mariotte's law,
^, = y,P ; -Vp. = y,P ; etc. ; ... (2)
whence, by addition, we have
F(i7,+^,+ ...) = (F,+ r,+ ...)^;
i.e., i^=p,+i?,+ (S)
Of course, the same statement applies to any number of
separate parts into which we may imagine a mass of homo-
geneous gas to be divided.
For numerical examples and practical questions in the solu-
tion of which this principle is useful, see p. 239, etc., Ean-
kine's Steam-engine. (Rankine uses 0.365, where 0.367 has
been used here.)
BAEOMETEIC LEVELLIJSTG.
619
477. Barometric Levelling. — By measuring with a barometer
tlie tension of the atmosphere at two different levels, simul-
taneously, and on a still day, the two localities not being widely
separated horizontally, we may compute their vertical distance
apart if the temperature of the stratum of air between them
is known, being the same, or nearly so, at both ^^
stations. Since the heaviness of the air is
different in different layers of the vertical
column between the two elevations iVand M^
Fig. .526, we cannot immediately regard the
whole of such a column as a free body (as was
done with a liquid, § 412), but must consider
a horizontal thin lamina, Z, of thickness
= dz and at a distance = z (variable) below
J!/, the level of the upper station, N being
the lower level at a distance, A, from M.
The tension, ^, must increase from M
downwards, since the lower laminae have to support a greater
weight than the upper ; and the heaviness y must also increase,
proportionally toj?, since we assume that all parts of the col-
umn are at the same temperature, thus being able to apply
Mariotte's law. Let the tension and heaviness of the air at
the upper base of the elementary lamina, Z, be jp and y re-
spectively. At the lower base, a distance dz below the upper,
the tension is ji? -f- djp. Let the area of the base of lamina be
F\ then the vertical forces acting on the lamina are Fp^ down-
ward ; its weight yFdz downward ; and F{]) -\- dp) upward.
For its equilibrium .5'(vert. compons.) must = ;
Ml
'."•;• r-r.v
:-~
■ ■' z
--T--I— i?+c
Ip
;• }.\ .:• Jr.
Fig. 526.
/. F{p -\-dp)- Fp- Fydz = 0;
i.e., dp = ydz, " . (1)
which contains three variables. But from Mariotte's law,
§ 4T5, eq. (2), if p^ and y^, refer to the air at iV, we niay
substitute y = —p and obtain, after dividing by ^, to separate
Pn
the variables^ and s.
620
MECHANICS OF ENGINEEEING.
Vn
(2)
Summing equations like (2), one for each lamina between
J^ (where J? ■=-jPm and s = 0) and ir(where^ =^^and^ = A),
we have
.e.,A = ^ log.,
Vn
'Jp.
.Pm_
. (3)
which gives A, the difference of level, or altitude, between M
and iV, in terms of the observed tensions p^ and p^^, and of y^ y
the heaviness of the air at H, which may be computed from
eq. (12), § 472, substituting from which we have finally
h=i^.i^. log.e r^'
. . (4)
in which the subscript refers to freezing-point and one at-
mosphere tension ; T^, and T^ are absolute temperatures. For
the ratio j?^ : j?^ we may put the equal ratio h^ '. h^ of the
actual barometric heights which measure the tensions. The
lag. e(or I^aperian, or natural, or hyperbolic, log.) = (common
log. to base 10) X 2.30258. From § 409, ;/„ of air = 0.080T6
lbs. per cub. ft., and j?^ = 14.701 lbs. per sq. inch ; T„ = 273°
Abs. Cent.
If the temperatures of the two stations (both in the shade)
arj not equal, a mean temp. = ^{T^-\- T^) may be used for
Tn in eq. (4), for approximate results. Eq. (4) may then be
written -
A (in feet) = 26213^. log.
T
-p.
_Pm_
(5)
i^o _
The quantity ^ = 26213 ft., just substituted, is called the
height of the homogeneotis atm,osp)here^ i.e., the ideal height
which the atmosphere would have, if incompressible and non-
BAROMETRIC LEVELLING — ADIABATIC LAW. 621
expansive like a liquid, in order to exert a pressure of 14.701
lbs. per sq. inch upon its base, being th'*^"gll':*"+ "f a constant
heaviness = .08076 lbs. per cub. foot.
By inversion of eq. (4) we may also write
p.^e^^ ^- =^„, ..... (6)
where e = 2.71828 = the Naperian Base, which is to be raised
to the power whose index is the abstract number — . ^ . A,
and the result multiplied by^^ to obtain ^^.
Example. — Having observed as follows (simultaneously) :
At lower station iV, h^ = 30.05 in, mercury ; temp. = 77.6° F. ;
"upper " J[/,A„, = 23.66 « • " « =70.4°F.;
required the altitude h. From these figures we have a mean
absolute temperature of 460° + K'^'^'-^ + 'i'0.4) = 534° Abs.
Fahr, ; hence, from (5),
h = 26213 X m X 2.30258 X log. ,, \ ^
= 6787.9 ft.
(Mt. Guanaxuato, in Mexico, by Baron von Humboldt.)
Strictly, we should take into account the latitude of the place,
since y^ varies with g (see § 76), and also the decrease in the
intensity of gravitation as we proceed farther from the earth's
centre, for the mercury in the barometer weighs less per cubic
inch at the upper station than at the lower.
Tables for use in barometric levelling can be found in Traut-
wine's Pocket-book, and in Searles's Field-book for Railroad
Engineers, as also tables of boiling-points of water under dif-
ferent atmospheric pressures, forming the basis of another
method of determining heights.
478. Adiabatic Change — Poisson's Law. — By an adiahatic
change of state, on the part of a gas, is meant a compression
or expansion in which work is done wpon the gas (in compress-
622 MECHANICS OF ENGIN BERING.
ing it) or hy the gas (in expanding against a resistance) when
there is oio trmismission of Jieat between the gas and enclosino
vessel, or surrounding objects, bj conduction or radiation.
This occurs when the volume changes in a vessel of non-con-
ducting material, or when the compression or expansion takes
place so quickly that there is no time for transmission of heat
to or from' the gas.
The experimental facts are, that if a mass of gas in a cylinder
be suddenly compressed to a smaller volume its temperature is
raised, and its tension increased more than the change of vol-
ume would call for by Mariotte's law ; and vice.versd, if a gas
at high tension is allowed to expand in a cylinder and drive a
piston against a resistance, its temperature falls, and its tension
diminishes more rapidly than by Mariotte's law.
Again (see Example 3, § 473), if y^g of the gas in a rigid
vessel, originally at 4 atmos, tension and temperature of
15° Cent., is allowed to escape suddenly through a stop-cock
into the outer air, the remainder, while increasing its volume
in the ratio 100 : 73, is found to have cooled to — 27° Cent.,
and its tension to have fallen to 2.5 atmospheres; whereas, by
Mariotte's law, if the temperature had been kept at 288° Abs.
Cent., the tension would have been lowered to -^^^ of 4, i.e.,
to 2.92 atmospheres only.
The reason for this cooling during sudden expansion is, ac-
cording to the Kinetic Theory of Gases, that since the " sensi-
ble heat" (i.e., that perceived by the thermometer), or ''^ hot-
fiess'^ of a gas depends on the velocity of its incessantly moving
molecules, and that each molecule after impact with a receding'
piston has a less velocity than before, the temperature neces-
sarily falls; and vice versa, w\ie,u an advancing piston com-
presses the gas into a smaller volume.
If, however, a mass of gas expands without doing worh, as
when, in a vessel of two chambers, one a vacuum, the other
full of gas, communication is opened between them, and the
gas allowed to fill both chambers, no cooling is noted in the
mass as a whole (though parts may have been cooled tem-
porarily).
By experiments similar to that in Example 3, § 473, it has
ADIABATIC CHANGE — EXAMPLES. 623
been found that for air and the ' ' perfect gases, " in an
adiabatic change of vohnne [and therefore of heaviness], the
tension varies inversely with the 1.411h power of the volume.
This is called Poisso'n/s Law. That is to say.
Change] ' ' p^ Vn/ Pn \Vj'
and combining this relation with the general eqs. (10) and
(13), § 4:Y2, we have also
Change} ' ' Pn xTn' T^ \pnl'
i.e., the tension varies directly as the 3.41:th po"wer of the
absolute temperature ; also,
Adiahat \ (IAJTjlT^ or InJlA'''^ f^)
Change] ' ' \vJ-\tJ rm~\Tj ' ^^ ^
i.e., the volume is inversely, and the heaviness directly, as the
2.44:th power of the absolute temperature.
Here m and n refer to any two adiabatically related states.
Tis the absolute temperature.
Example 1. — Air in a cylinder at 20° Cent, is suddenly
compressed to \ its original volume (and therefore is six times
as dense, i.e., has six times the heaviness, as before). To what
temperature is it heated % Let m be the initial state, and n the
final. From eq. (3) we have
= (y); .-. r„=611°Abs. Cent.,
293
or nearly double the absolute temperature of boiling water.
Example 2. — After the air in Example 1 has been given
time to cool again to 20° Cent, (temperature of surrounding
obiects) it is allowed to resume, suddenly, its first volume, i.e.,
624 MECHANICS OF ENGINEERING.
to increase its volume sixfold by expanding beliind a piston.
To what temperature has it cooled? Here r^ = 293° Abs.
Cent., the ratio Vm'-yn=h ^^^ ^« is required. Hence,
from (3),
^=Q; .-. T„=293X0.4Y96 = 140.5° Abs. Cent.,
or=— 132,5° Cent., indicating extreme cold.
From these two examples the principle of one kind of ioe-
making apparatus is very evident. As to the work necessary
to compress the air in Example 1, see § 483. It is also evident
why motors using compressed air expansively have to encoun-
ter the difficulty of frozen watery vapor (present in the air to
some extent).
Example 3. — "What is the tension of the air in Example 1
(suddenly compressed to ^ its original volume) immediately
after the compression, if the original tension was one atmos-
phere ? That is, with F„ : F^ = 1 : 6, and prn — 14.7 lbs. per
sq. inch, /»« = 2 From eq. (1), (in., lb., sec.,)
p„=14.7x(6)i-*i = 14.rxl2.52 = 184
lbs. per sq. inch ; whereas, if, after compression and without
change of volume, it cools again to 20° Cent., the tension is
only 14.T X 6 = 88.2 lbs. per sq. inch (now using Mariotte's
law).
479. "Work of Expanding Steam following Mariotte's Law.—*
Although gases do not in general follow Mariotte's law in es*
panding behind a piston (without special provision for sup-
plying heat), it is found that the tension of saturated steam
(i.e., saturated at the beginning of the expansion) in a steaui
engine cyhnder, when left to expand after the piston has
passed the point of ■" cut-off^^^ diminishes very nearly in
accordance with Mariotte's law, which may therefore be ap
phed in this case to find the work done per stroke, and thence
tile power. In Fig. 627 a horizontal steam-cylinder k
, EXPANDING STEAM„
pikown in which the piston is making its left-to-rigl
The " back- pressure" is con-
stant and = I^q, F being the
area of the piston and ^ the
intensity (i.e., per unit area)
of tiie back or exhaust pres-
sure tin the right side of the
pston 5 while the forward
pressure on the left face of the
pigton = Fjp, in which j? is the
steam-pressure per unit area,
and is different at different
points of the stroke. While the
piston is passing from 0" to
D'\]) is constant, being — ^e, = the boiler-pressure, since the
iteam-port is still open. Between D" and C" ^ however, the
steam being cut off (i.e., the steam-port is closed) at D'\ a dis-
tance a from (?'', ^ decreases with Mariotte's law (nearly), and
its value is {Fa -r- Fx)pi, at any point on C" D'\ x being the
distance of the point from 0".
Above the cylinder, conceive to be drawn a diagram in
which an axis OX\q \\ to the cylinder-axis, OY an axis 1 to
the same, while is vertically above the left-hand end of the
cylinder. As the piston moves, let the value of ^ correspond-
ing to each of its positions be laid off, to scale, in the vertical
immediately above the piston, as an ordinate from the axis X.
Make OD' = g- by the same scale, and draw the horizontal
D'C Then the effective work done on the pigton-rod while
it moves through any small distance dx is
dW = force X distance = F\j^ — q)dx.
and is proportional to the area of the strip RS, whose width is
*!« and length =^ — 2; so that the effective work of one
strode is
G
{p~-q)dx,
f\
(1)
626 MECHANICS OF ENGINEEEING.
and is represented graphically by the area A'ARBQ'D'A\
)^Yom. 0" to 3" 'p is constant and =_^6 (while q is constant at
all points), and x varies from to a ;
.-. \w=F{p,-q)Xdx=F^,-q)a, . . (9)
which may ue called the work of entrance, and is represented
by the area of the rectangle A' ADD',
(1
From D" to C'p is variable and, by Mariotte's lav7, = — ^j, ;
[w=J^[ap,log.J-]-q{l-a)] ... (3)
L.D" I— ^O' j J
= the worlc of expansion^ adding which to that of entrance,
we ha^e for the total effective worlc of one stroTce
r=i5>,a[l + log.. (i)]-i^^?. ... (4)
By effective work we mean that done upon the piston-rod ,
and thus transmitted to outside machinery. Suppose the
engine to be " double-acting" ; then at the end of the stroke a
communication is made, by motion of the proper valves, be-
tween the space on the left of the piston and the condenser of
the engine ; and also between the right of the piston and the
boiler (that to»the condenser now being closed). On the return
stroke, therefore, the conditions are the same as in the forward
stroke, except that the two sides of the piston have changed
places as regards the pressures acting on them, and thus the
same amount of effective work is done as before.
If n revolutions of the fly-wheel are made per unit of time
(two strokes to each revolution), the effective work done per
imit of time, i.e., ihiQ power of the engine, is
= 2w Tr= 2nFTapSl + log., (^)l - ^H.
(5)
WORK OF STEAM-ENGINES. 627
For simplicity the above theory has omitted the considera-
tion of " dearmicej^ that is, the fact that at the point of " cut-
ofE " the mass of steam which is to expand occupies not only
the cylindrical volume Fa, but also the " clearance" or small
space in the steam-passages between the valve and the entrance
of the cylinder, the space between piston and valve which is
never encroached upon by the piston. "Wire-drawing" has
also been disregarded, i.e., the fact that during communication
with the boiler the steam-pressure on the piston is a little less
than boiler-pressure. For these the student should consult
special works, and also for the consideration of water mixed
with the steam, etc. Again, a strict analysis should take into
account the difference in the areas which receive fluid-pressure
on tlie two sides of the piston.
Example 1. — A reciprocating steam-engine makes 120 revo-
lutions per minute, the boiler-pressure U 40 lbs. by the gauge
(i.e.,j?j, = 40-1- 14.7= 54.7 lbs. per sq. inch), the piston area
is i^=: 120 sq. in., the length of stroke 1 = 1Q in., the steam
being "cut off" at J stroke (.*. « =: 4 in., and I '. a = 4.00),
and the exhaust pressure corresponds to a " vacuum of 25
inches" (by which is meant that the pressure of the exhaust
steam will balance 5 inches of mercury), whence q = ^ oi
14.7 = 2.45 lbs. per sq. inch. Required the work per stroke,
W, and the corresponding power L.
Since ? : « = 4, we have log., 4 = 2.302 X .60206 = 1.386,
and from eq. (4), {foot, lb., sec ,)
W= m (54.7 X 144) . 1 . [2.386] - i|| (2.45 x 144) . |
= 5165.86 - 392.0 = 4773.868 ft. lbs. of work per ^tiok.
and therefore the power at 2 rev. per sec. (eq. 5) is
Z = 2 X 2 X 4773.87 = 19095.5 ft. lbs. per second.
Hence in horse-powers, which, in ft,,-lb.-8ee. system, =iZT-S50
Power = 19095.5 -^ 550 = 34.7 H. P.
Example 2. — Required the weight of steam consumed per
S28 MECHANICS OF ENGIFEERI]S"G.
second bv the above engine with given data ; assuming with
Weisbach that the heaviness of saturated steam at a definite
pressure (and a corresponding tenvpe7'ature^ §^69) is about.f of
that of air at the same pressure and temperature.
The heaviness of air at 54.7 lbs. per sq. in. tension and
temperature 287° Fahr. (see table, § 469) would be, from sq.,
(12) of § 472 (see also § 409),
£ ^ .0807 X 492 ^ 54^ ^ ^ .^^g
'^ T p, 460 + 287 14.7
lbs. per cub. foot, f of which is 0.1237 lbs. per cub. ft. Now
the volume* of steam, of this heaviness, admitted from the
boiler at each stroke is V = Fa = iff . -i = 0.2777 cub. ft.^
and therefore the weight of steam used per second is
4 X .2777 X 0.1287 = 0.1374 lbs.
Hence, per hour, 0.1374 X 3600 = 494.6 lbs. of feed-watei
tire needed for the boiler.
If, with this same engine,. the steam is used at full boiler
pressure throughout the whole stroke, the power will be
greater, viz. = '^,nFl{pi, ~ q) ~ 33440 ft. lbs. per sec, but
the consumption""" of steam will be four times as great; and
hence in economy of operation it will be only 0.44 as efficient
(nearly).
480. Graphic Hepresentation of any Change of State of a Con«
fined Mass of Gas. — The curve of expansion AB m Fig. 527 is
an equilateral hyperbola, the axes ^and 1^ being its asymp-
totes. If compressed air were used instead of steam its ex-
pansion curve would also be an equilateral hyperbola if its
temperature could be kept from failing during the expansion
(by injecting hot-water spray, e.g.), and then, following
Mariotte's law, we would have, as for steam, (§ 475,) j*? I^= con-
stant, i.Q.^pFx = constant, and therefore poe = constant, M^hich
is the equation of a hyperbola, p being the ordinate and x the
abscissa. This curve (dealing with a perfect gas) is also called
an isothermal, the x and y co-ordinates of its points being pro-
* We here neglect the practical fact that a portion of the fresh steam
entering the cylinder is condensed prematurely, so thai the actual con-
eumption is somewhat greater than as here aerived.
GRAPHICS OF CHANGE OF STATE OF GAS.
629
\ \ \
1
\
V
\
\ \a ^8^,
\
-.593°
283°
i '~-— -[Sk
273°
"
IP ,E ^^
-
X
Fig 528.
portional to the volume and tension, respectively, of a mass of
air (or perfect gas) whose temperature is maintained constant.
Hence, in general, if a mass, of gas be confined in a rigid
cylinder of cross-sec-
tion J^' (area), provided
with an air-tight pis-
ton, Fig. 528, its vol-
ume, 2^x, is propor-
tional to the distance
OD = X (of the piston
from the closed end of
the cylinder) taken as
an abscissa, while its
tension jp at the same
instant may be laid off
as an ordinate from D.
Thus a point A is fixed. Describe an equilateral hyperbola
through A, asymptotic to X and Y^ and mark it with the ob-
served temperature (absolute) of the air at this instant. In a
similar way the diagram can be filled up with a great number
of equilateral hyperbolas, or isothermal curves, each for its
own temperature. Any point whatever (i.e., above the critical
temperature) in the plane angular space YOX will indicate by
its co-ordinates a volume and a tension, while the correspond
ing absolute temperature T will be shown by the hyperbols
passing through the point, since these three variables always
satisfy the relation (§ 472)
^=coBst.;i.e.,5^=i^. . . . (ly
Any change of state of the gas in the cylinder may now be
represented by a line in the diagram connecting the two pointa
corresponding to its initial and final states. Thus, a point
moving along the line AB, a portion of the isothermal marked
293° Abs. Cent, represents a motion of the piston from D to
^, and a consequent increase of volume, accompanied by just
sufficient absorption of heat by the gas (from other bodies) to
maintain its temperature at that figure (viz., its temperature at
630 MECHANICS OF ENGINEEEING.
A). If the piston move from D to £', without transmission
of heat, i.e., adiabatically (§ 478), tlie tension falls more
rapidly, and a point moving along the line AB' represents the
corresponding continuous change of state. AB' is a portion
of an adiahatic curve, whose equation, from § 478, is
Vk
1.41
1.41.
Fxk
-^J, or p:ri-4i ^p^xxi-*i = const. ; . (1)
in which p^ and a?^ refer to the point K where this particular
adiabatic curve cuts the isothermal of freezing-point. Evi-
dently an adiabatic may be passed through any point of the
diagram. The mass of gas in the cylinder may change its
state from A to B' by an infinite number of routes, or lines on
the diagram, the adiabatic route, however, being that most likely
to occur for a rapid motion of the piston. For example, we
may cool it without allowing the piston to move (and hence
without altering its volume nor the abscissa x) until the pres-
sure falls to a value J95' = DL = EB\ and this change is rep-
resented by the vertical path from ^ to Z ; and then allow it
to expand, and push the piston from D to E (i.e., do external
work), during which expansion heat is to be supplied at just
such a rate as to keep the tension constant, =Pb' ^ Pli this
latter change corresponding to the horizontal path LB' from
Z to B'.
It is further noticeable that the worh done by the expanding
gas upon the near face of the piston (or done upon the gas when
compressed) when the space dx is described by the piston, is
= Fpdx, and therefore is proportional to the area pdx of the
small vertical strip lying between the axis X and the line or
route showing the change of state ; whence the total work done
on the near piston-face, being = Ffpdx, is represented by the
area fpdx of the plane figure between the initial and final
ordinates, the axis X and the particular roitte followed be-
tween the initial and final states (N.B. We take no account
here of the pressure on the other side of the piston, the latter
depending on the style of engine). For example, the work
done on the near face of the piston during adiabatic expansion
WOEK OF ADIABATIC EXPANSION".
631
from D \o E is represented by the plane figure AB'EDA^
and is measured by its area.
The mathematical relations between the quantities of heat
imparted or rejected by conduction and radiation, and trans-
formed into work, in the various changes of which the con=
fined gas is capable, belong to the subject of Thermodynamics,^
which cannot be entered upon here.
It is now evident how the cycle of changes which a definite
mass of air or gas experiences when used in a hot-air engine,
compressed-air engine, or air-compressor, is rendered more in-
telligible by the aid of such a diagram as Fig. 528 ; but it
must be remembered that during the entrance into, or exit
from, the cylinder, of the mass of gas used in one stroke, the
distance x does not represent its volume, and hence the locus
of the points in the diagram determined by the co-ordinates^
and X during entrance and exit does not indicate changes of
state in the way just explained for the mass when confined in
the cylinder. However, the work done by or upon the gas
during entrance and exit will still be represented by the plane
figure included by that locus (usually a straight horizontal
line, pressure constant) and the axis of X and the terminal
ordinates.
481. Adiabatic Expansion in an Engine using Compressed Air.
— Fig. 529. Let the compressed air at a tension jp^ acd a?^
absolute temperature T^ be supplied
from a reservoir (in which, the loss is
continually made good by an air-com-
pressor). Neglecting the resistance of
the porty its tension and temperature
when behind the piston are still i?^ and
T^ . Let Xn = length of stroke, and o[
let the cut-off (or closing of communi-
jation with the reservoir) be made at
fiome point D where a? = a?^ , the posi-
tion of ^ being so chosen (i.e., the ratio
^m '■ Xn so computed) that after adia-
batic expansion from D to E the pres-
^are shall have fallen iromj>^ at M {state in) to a value p^ =Pa
632 MECHANICS OF ENGINEERING.
= one atmosphere at N{state ti), at the end of stroke ; so that
when the piston returns the air will be expelled (" exhausted")
at a tension equal to that of the external atmosphere (though
at a low temperature). Hence the back-pressure at all points
either way will be = j)n per unit area of piston, and hence the
total back-pressure = i^/>^, i^ being the piston area.
From to D the forward pressure is constant and =Fj^rn.->
and the effective work, therefore, or work on piston-rod from
Oto D, is
Work of entrance = W = F\_prn —PnJ^mf • • (1)
Lo
represented by the rectangle MMLN' . The cut-off being
made at Z>, the volume of gas now in the cylinder, viz.,
Y^ = Fx^^ is left to expand. Assuming no device adopted
(such as injecting hot-water spray) for preventing the cooling
and rapid decrease of tension during expansion, the latter is
odiabatic, and hence the tension at any point P between M
and H will be
p=Pm[—J''^' . [see§478; V=Fx']; . . (a)
,\ Work of expansion
^ rv= rxp -Pn)dx = Fr^dx - f^jx^ ~ x^\ (2)
and is represented by the area MPNL.
i.e., Ffydx = 2A4:Fp^xJl-(^Y''l. . . (3)
Now substitute (3) in (2) and then add (2) to (1), noting that
OOMPEESSED-AIE ENGINE. 633
F{Pm — I>v)^ra " FfJ^X^ — i»^) = Fp^
1 -
'^mPm,
which furthermore, since n and m are adiabatically related
[see (a)], can be reduced to
and we have finally :
'::r;:tr.r"i='^=3.44.„4i-(i=)°"]. . ,.,
But FXjn = V^, and the adiabatic relation holds good,
therefore we may also write
F = 3.447^p^[l-(^)''']; .... (5)
in which F"^ = the volume which the mass of air used per
stroke occupies in the state m, i.e., in the reservoir, where the
tension is^^ and the absolute temperature = T^.
To find the wor/c done per poicnd of air used (or other unit
of weight), we must divide W by the weight G = V^nYm of
the air used per stroke, remembering (eq. (13), § 4^2) that
ymYm = [ ymPmYoT,'] ^ {T^P,).
Work per unit of weight of \^ ^^^ po [ i ___ (pA^'^^I ^q.
air used in adiahatic working I ' ""ro^oL \Pmf J
The back-pressure j)„ —Pa = one atmosphere.
In (6), Yo = 0.0807 lbs. per cub. it, p, = (14.7 X 144) lbs.
per sq. ft., and T, = 273° Abs. Cent., or 492° Abs. Fahr.
634 MECHANICS OF ENGINEERING.
It is noticeable in (6) that for given tensions jt?^ and p^, the
work per unit of weight of air used \?> proportional to the ab-
solute temperature T^ of the reservoir. The temperature 7^
to which the air has cooled at the end of the stroke is obtained
as in Example 2, §4Y8, and may be far below freezing-point
unless T^ is very high or the ratio of expansion, x^ : x^ , large.
Example. — Let the cylinder of a compressed-air engine have
a section of i^= 108 sq. in. and a stroke a?^ = 15 inches. The
compressed air entering the cylinder is at a tension of 2 atmos.
(i.e., p^ = 29.4 lbs. per sq. in., and p^ -^ Pm. = i)-, and at a
temperature of 27° Cent, (i.e., T^ = 300° Abs. Cent.). Ee-
quired the proper point of cut-off, or cp^ = ? , in order that the
tension may fall to one atmosphere at the end of the stroke ;
also the work per stroke, and the work per pound of air. Use
the foot, pound, and second.
From eq. (a), above, we have
^«» = ^n(— )''-=l-25(i)"-^^ = 1.25X.6112 = 0.764ft.,
and hence the volume of air in state m, used per stroke [eq.
(5)] is
7^ = i?'x^=lgx0.764 = 0.573 cubic feet;
while the work per stroke is
Tf = 3.44X0.573 X 29.4X144 X[l-(i) 29] = 1S19 ft. lbs.,
and the work obtained from each pound of air, eq. (6),
14 7 X 144
= 3.44x300X QQ3Qy^^y3 X[l-(i)°-29] = 18040
ft. lbs. per pound of air used.
The temperature to which the air has cooled at the end of
stroke [eq. (2), § 478] is
COMPRESSED-AIR ENGINE. 635
Tr,= Tj—) ' =300x(i)-29 = 300X.818 = 245°Abs.C.;
i.e., —28° Centigrade.
482. Remarks on the Preceding. — This low temperature ia
objectionable, causing, as it does, the formation and gradual
accumulation of snow, from the watery vapor usually found
in small quantities in the air, and the ultimate blocking of the
ports. By giving a high value to T^^ , however, i.e., by heat-
ing the reservoir, T^ will be correspondingly higher, and also
the worJcpe?' pound of air^ eq. (6). If the cylinder be encased
in a " jacket" of hot water, or if spray of hot water be injected
behind the piston during expansion, the temperatui^e may be
maintained nearly constant, in which event Mariotte's law will
hold for the expansion, and more work will be obtained per
pound of air ; but the point of cut-off must be differently
placed. Thus if, in eq. (4), § 479, we make the back-pressure,
q, equal to the value {Fa — Fl);pb, to which the air
pressure has fallen at the end of the stroke by Mariotte's law,
we have
Worh per stroJce with \ -n, ■^ (l\ ^ , f i\ f-,\
isotherm. exj?ans. \ = ^""^^ ^^S- l^j= ^^^^ ^^g- WJ' ^^^
and hence
Work per unit of weight of air, ^ ~ T ^^ 1 {^\ f9\
with isothermal expansion \~ "^ y T ^^' \^/' " ^^^
Applying these equations to the data of the example, we
obtain
Work per unit of weight of air with iso- \ _ r. n^ jt Pa
thermal expansion ) ~~ ' "y jr '
whereas, with adidbatio expansion, work ) ^ q g^ yi Po
per unit of weight of air is only ) " ^ToTq
636
MECHANICS OF ENGINEERIITG.
■ - ■ 1
■.:••£
1
„
FiQ. 530.
483. Double-acting Air-compressor, with Adiabatic Compres-
sion. — This is the converse of § 481. In Fig. 530 we have the
piston moving from right to left, compressing a mass of air
which at the beginning of the stroke fills the cylinder.- This is
brought about by means of an external
motor (steam-engine or turbine, e.g.)
which exerts a thrust or pull along the
piston-rod, enabling it with the help
of the atmospheric pressure of the
fresh supply of air flowing in behind
it, to first comp7'ess a cylinder-fnll of
air to the tension of the compressed
air in the reservoir, and then, the
port or valve opening at this stage,
to force or deliver it into the reservoir.
Let the temperature and tension of the
cylinder-full of fresh air be Tn^ and
Pny^ , and the tension in the reservoir be Pm . Suppose the
compression adiabatic. As the piston passes from E toward
the left, the air on the left has no escape and is compressed, its
tension and temperature increasing adiabatically until it reaches
a value pm^ = that in reservoir, at which instant, the piston
being at some point D, a valve opens and the further progress
of the piston simply transfers the compressed air into the re-
servoir without further increasing its tension. Throughout
the whole stroke the piston-rod has the help of one atmosphere
pressure on the right face, since a new supply of air is entering
on the right to be compressed in its turn on the return stroke.
The work done from ^to D may be called the work of com^
pression y that from D to 0, the worlc of delivery.
[Sinc'e, here, dx and ^Tr(or increment of work) have cori"
trary signs, we introduce the negative sign as shown.^
TJie work of compression =— J F{p — 2)n^)dx. . . . (Ic)
The worl' of delivery = —J^F{pmi —jpni)dx.
AIR-COMPRESSOE. 637
In these equations only^ and a? are variables. In the sum-
mation indicated in (Ic) jp changes adiabaticallj ; in (l^j? i«
constant = J?ni, as now written.
In the adiabatic compression the air passes from the state w,
to the state m^ (see N^ and M^ in figure).
The summations in these equations being of the same form
as those in equations (1) and (2) of § 481, but with limits in-
verted, we may write immediately,
Workyer8troke=W=ZA^Y^^^A\-{^\ 1, (2)
and
Vor^Vernnitofw ^^j), 2/3 is not only less than y^ , but the dif-
ference is greater than before. We have therefore found
experimentally that, in a general way, when water is flowing
in a pipe it presses less against the side of the pipe than it did
before the flow was permitted, or (what amounts to the same
thing) the pressure between the transverse laminae is less than
the hydrostatic pressure would be.
In the portion HN^O of the pipe we find the pressure less
than one atmosphere, and consequently a manometer register-
ing pressures from zero upward (and not simply the excess
over one atmosphere, like the Bourdon steam-gauge and the
open piezometer just mentioned) must be employed. At N^y
e.g., we find the pressure
= \ atmos., i.e., — - = IT ft.
y
Even below the level BC, by making the sections quite nar-
row (and consequently the velocities great) the pressure may be
made less than one atmosphere. At the surface BC \\xq pres-
sure is of course just one atmosphere, while that in the jet at
N^^ entering the right-hand tank under water, is necessarily
^^ = 1 atmos. + press, due to col. Ti' of water practically at rest;
* N ^ being stopped and L open.
STEADY FLOW OF LIQUIDS. 651
I.e., -^ = pressure-head at iV« = 5 -f" ^'\
(whereas if N^ were stopped by a diaphragm, the pressn re-
head just on the right of the diaphragm would be 5 -j- A', and
that on the left h -\- h^ .)
Similarly, when a jet enters the atmosphere in parallel fila-
ments its particles are under a pressure of one atmosphere, i.e.,
their pressure-head = 5 = 34 ft. (for water) ; for the air im-
mediately around the jet may be considered as a pipe between
which and the water is exerted a pressure of one atmosphere.
491. Recapitulation and Examples. — We have found experi-
mentally, then, that in a steady flow of liquid through a rigid
pipe there is at each section of the pipe a definite velocity and
pressure which all the liquid particles assume on reaching that
section ; in other words, at each section of the pipe the liquid
velocity and pressure remain constant with progress of time.
Example 1. — If in Fig. 532, the flow having become steady,
the volume of water flowing in 3 nainutes is found on meas-
urement to be 134 cub. feet, the volume per second is, from
eq. (1), § 490,
Q = i|4 =: 0.744 cub. ft. per second.
Example 2. — If the flow in 2 miu. 20 sec. is 386.4 lbs., the
volume of flow per second is [ft., lb., sec. ; eqs. (1) and (2)]
Q = — = '- t = ^^ ' . -— - = 0.0441 cub. ft. per sec.
^ t r 62.5 140 ^
Example 3. — In Fig. 532 the height of the open piezometer
at iV^i is 1/^ = 9 feet; what is the internal fluid pressure?
[Use the inch, lb., and sec] The internal pressure is
j}^ =p^-\-y^y = 14.Y + 108 X ^^ = 18.6 lbs. per sq. inch.
The pressure on the outside of the pipe is, of course, one at-
mosphere, so that the resultant bursting pressure at that point
(iV,) is 3.9 lbs. per sq. in.
Example 4. — The volume of flow per second being .0441
662 MECHAisrics of engii^eering.
cub. ft. per sec, as in Example 1, required the velocity at a
section of the (circular) pipe where the diameter is 2 inches.
[Use ft., lb., and sec]
while at another section of the pipe where the diameter is four
inches (double the former) and the sectional area, F^ is there-
fore four times as great, the velocity is ^ of 2.02 = 0.505 ft.
per sec
492. Bernoulli's Theorem for Steady Flow ; without Friction. —
If the pipe is comparatively short, without sudden bends,
elbows, or abrupt changes of cross-section, the effect of friction
of the liquid particles against the sides of the pipe and against
each other (as when eddies are produced, disturbing the paral-
lelism of flow) is small, and will be neglected in the present
analysis, whose chief object is to establish a formula for steady
flow through a short pipe and through orifices.
An assumption, now to be made, (A flow in 'plane layers^ or
laminated flow, i.e., flow in laminae "1 to the .axis of the pipe
at every point, may be thus stated : (see Fig. 533, which shows
a steady flow proceeding, through a
pipe CD of indefinite extent.) All the
liquid particles which at any instant
form a small lamina, or sheet, as AJB^
"I to axis of pipe, Iceep company as a
lamina throughout the whole flow.
Fig. 533. The thickness, 6^6'', of this lamina re-
mains constant so long as the pipe is of constant cross-section,
but shortens up (as at C) on passing through a larger section,
and lengthens out (as at D) in a part of the pipe where the
section is smaller (i.e., the sectional area, F, is smaller). The
mass of such a lamina .is Fds'y -^ g [§ 55], its velocity at any
section will be called v (pertaining to that point of the pipe's
axis), the pressure of the lamina just behind it is Fp, upon the
rear face, while the resistance (at the same instant) offered by
its neighbor just ahead is F{p -f- dp) on the front face ; also
BERTTOTJLLl'S THEOHEM.
653
its weight is tbe vertical force Fds'y. Fig. 534 shows, as a
free body, the lamina which at
any in'stant is passing a point
A of the pipe's axis, where the
velocity is v and pressure ^.
Note well the forces acting ;
the pressures of the pipe wall
on the edges of the lamina have
no components in the direction
of V, for the wall is considered
smooth, i.e., those pressures are
"1 to wall ; in other words, no
friction is considered. To this free body apply eq. (7) of § 74,
for any instant of any curvilinear motion of a material point
/9
''dG = Fdsy
Fis. 534.
vdv = {tang, acceleration) X ds,
(1)
in which ds ^ a, small portion of the path, and is described in
the time dt. 'Now the tang, accel. = .^(tang. com pons, of the
acting forces) -^ mass of lamina, i.e..
tang. ace. = ^:iZ(£±M±Z>:^^«^.
(2)
l^ow. Fig. 535, at a definite instant of time^ conceive the
volume of water in the pipe to be subdivided into a great
number of laminae of equal fnass (which implies equal volumes
Fig. 535.
in the case of a liquid, but not with gaseous fluids), and let the
ds just mentioned for any one lamina be the distance from its
centre to that of the one next ahead ; this mode of subdivision
654 MECHANICS OF ENGINEEKING.
makes the ds of any one lamina identical in value with its
thickness ds', i.e.,
.ds = ds' • (3)
"We have also
ds cos (J)=. — dz, or ds' cos (p =^ — dz\ . . (4)
B being the height of the centre of a lamina above any con-
venient horizontal datum plane. Substituting from (2), (3),
and (4) in (1), we derive finally
1 1
— vdv -\ — dp-\-d3 = (5)
The flow being steady, and the subdivision into laminge
being of the nature just stated, each lamina in some small time
dt moves into the position which at the beginning of dt was
filled by the lamina next ahead, and acquires the same velocity,
the same pressures on its faces, and the sam,e value of z, that
the front lamina had at the beginning of dt.
Hence, considering the simultaneous advance made by all
the laminae in this same dt, we may write out an equation like
(5) for each of the laminae between any two cross-sections n and
m of the pipe, thus obtaining an infinite number of equations,
from which by adding corresponding terms, i.e., hy integra-
tion, we obtain
whence, performing the integrations and transposing,
Vm^ i Pms^ _<' _l_Pn_x_^ j 'Ber7WulWs \ .n^
-2^- + y + ^--2^ + y + ^- • 'I Theorem f" "W
Denoting by Potential Head the vertical height of any section
of the pipe above a convenient datum level, we may state
Bernoulli's Theorem as follows :
In steady flow without friction, the sum of the velocity-
head, j)ressure-he ad, and potential head at any section of the
pipe is a constant quantity, heing equal to the sum of the cor-
responding heads at any other section.
APPLICATIONS OF BERNOULLI'S THEOEEM.
655
It is noticeable that in eq. (T) eacli of the terms is a liueai
quantity, viz., a height, or head, either actual, such as ^„ and
s^ , or ideal (all the others), and does not bring into account the
absolute size of the pipe, nor even its relative dimensions (v^
and Vn, however, are connected bj the equation of continuity
I^mVrn = -F'n'^n\ ^^^ contains no reference to the volume oi
water flowing per unit of time [Q'] or the shape of the pipe's
axis, "When the pipe is of considerable length compared with
its diameter the friction of the water on the sides of the pipe
cannot be neglected (§ 512).
It must be remembered that Bernoulli's Theorem does not
hold unless the flow is steady, i.e., unless each lamina, in com-
ing into the position just vacated by the one next ahead (of
equal mass), comes also into the exact conditions of velocity
and pressure in which the other was when in that position.
[N.B. This theorem can also be proved by applying to all
the water particles between n and m, as a collection of small
rigid bodies (water being incompressible) the theorem of Work
and Energy for a collection of Rigid Bodies in § 142, eq. (xvi),
taking the respective paths which they describe simultaneously
in a single dt.~\
493. First Application of Bernoulli's Theorem without Friction.
— Fig. 536 shows a large tank from which a vertical pipe of
uniform section leads to another tank and dips below the sur-
face of the water in the latter. Both surfaces are open to the
air. The vessels and pipe being filled with
water, and the lower end m of the pipe un-
stopped, a steady flow is established almost
immediately, the surface BO being very
large compared with F, the area of the {uni-
form) section of the pipe.
Given F, and the heights A„ and A, re-
quired the velocity 'y,„ of the jet at m and
also the pressure, p^i ^^^ ^ (i^^ pips i^ear en-
trance of same), m is in the jet, just clear
of the pipe, and practically in the water-
level, AD. The velocity v^ is unknown, fig. ms.
but the pressure p^, is practically =^a = one atmosphere, since
'.•.
•aih:
■B ••..•..■...::••• cr
* —
=-^^^;:l.-_= 1
= E^x;f--^
1
m
Zlg +
ii'i
n
1 ii
ii
••air".
1
1 A ;
m
Dr
;fi^;
®J
656
MECHANICS OF EIS^GINEEKING.
the pressure on the sides of the jet is necessarily the hydro^
static pressure due to a slight depth below the surface AD.
.'. Press.-head at m is^ = ^=^h = ^4: feet. . . (§ 423)
r y
1^0 w apply Bernoulli's Theorem to sections m and ?r, taking
a horizontal plane through m as a datum plane for potential
heads, so that z^ = h and z^n, = 0, and we have
2^
1 4- 5 _l_
V.
Pn
4-^ + A.
(1)
But, assuming that the section of the jpijpe is filled at every
point, we must have
6'*M. — Vm
for, in the eq^uation of continuity
F V — F V
if we put F^ = Fn , the pipe being of uniform section, we ob-
tain Vot = ^w • Hence eq. (1) reduces to
Pn_
y
^ = 5 - A = 34 ft. - A.
. (2)
en
^^
nA
Hence the pressure at n. is less than one atmosphere, and if a
small tube communicating with an air-tight receiver full of air
were screwed into a small hole at n, the air in
the receiver would gradually be drawn off until
its tension had fallen to a value jt?,^. [This is the
principle of SjprengeVs air-jpump, mercury, how-
ever, being used instead of water, as for this
heavy liquid h = only 30 inches.]
If h is made > h for water, i.e. > 34 feet (or
> 30 inches for mercury), j)^ would be negative
from eq. (2), which is impossible, showing that
the assumption of full pipe-sections is not borne
out. In this case, h> h, only a portion, mn\
(in length somewhat less than h,) of the tube will be kept full
Fig. 537.
APPLICATIONS OF BEElSTOtTLLl'S THEOEEM. 667
during the flow (Fig. 537) ; while in the part Kn' vapor of
water, of low tension corresponding to the temperature
(§ 469), will surround an internal jet which does not fill the
pipe. As for the value of ^'„J, Bernoulli's Theorem, applied
to BG and m, in Fig. 536, gives finally -?;„,, = ^'^gh^ .
ExAJviPLE. — If h = 20 feet, Fig. 536, and the liquid is water,
the pressure-head at n is (ft,, lb., sec.)
l^ = b-h = W -^0'= 14 ft.,
r
and therefore
p^=14:X 62.5 = 875 lbs. per sq. ft. = 6.07 lbs. per sq. in.
494. Second Application of Bernoulli's Theorem without Fric-
tion. — Knowing by actual measurement the open piezometer
height y^ at the section ^^ in .
Fig. 538 (so that the pressure- ^ ^^— - — c/"~\]
head, ^ = h -{-y^', at^ ^ is ^^ r^/^^^^^:~ '";'.";''
known) ; knowing also the f*iyjy__-___Trr?^^<-.
vertical distance h^ from n ■:■■.'
to m, and the respective • fig. 538.
cross-sections i^„ and i^ (^ being the sectional area of the
jet, flowing into the air, so that ^ = h), required the volume
of flow per sec; i.e., required Q, which
= F^v^ = F^v^. (1)
The pipe is short, with smooth curves, if any, and friction
will therefore be neglected. From Bernoulli's Theorem [eq.
(7), § 492J, taking m as a datum plane for potential heads, we
have
^' + 5 + = ^-l-(2/« + 5) + A^. ... (2)
But from (1) we have
658 MECHANICS OF ENGnSTEEEHSTG.
^n — L-^ m "^ -*- n\ ^m )
substituting which in (2) we obtain, solving for %
^m — / - 500000 Kvi
and hence the volume per unit of time becomes known, viz.,
Q = F^v^. ....... (4)
KoTE. — If the cross-section Fn^ of the nozzle, or jet, is > ^ ,
v^ becomes imagmarj (unless y^ is negative (i.e.,_^„ < one at-
mos.), and numerically > h„) ; in other words, the assigned
cross-sections are not filled hy the flow.
Example. — If y^ = lY ft. (thus showing the internal fluid
pressure at n to be j?,, = y{yn -\-h) = 1^ atmos.), k^ = 10 ft.,
and the (round) pipe is 4 inches in diameter at n and 3 inches
at the nozzle m, we have from (3) (using ft.-lb.-sec. system of
units in which g = 32.2)
^ 4/2 X32.2(17 + 1 0) ^ ^^_^ ^^_ ^_
[N.B. Since ^ -f- ^ is a ratio and therefore an abstract
number, the use of the inch in the ratio will give the same
result as that of the foot.]
Hence, from (4),
Q = F^D^ =: i^T^)' X 50.4 = 2.474 cub. ft. per sec.
495. Orifictss m Thin Plate. — Fig. 539. When efflux takes
place through an orifice in a thin plate, i.e., a sharjp-edged
orifice in the plane wall of a tank, a contracted vein (or " vena
ORIFICE IN THIN PLATE.
669
/SO'
^'m • ••
Fig. 539.
contracta") is formed, the filaments of water not becoming
parallel until reaching a plane, m, ..••.•..-....
parallel to the plane of vessel wall,
which for circular orifices is at a dis-
tance from the interior plane of vessel
wall equal to the radius of the circular
aperture ; and not until reaching this
plane does the internal fluid pres-
sure become equal to that of the sur-
rounding medium (atmosphere, here),
i.e., surrounding the jet. We assume
orifice is small compared with h
horizontal.
The area of the cross-section of the jet at m, called the con-
tracted section^ is found on measurement to be from .60 to .64
of the area of the aperture with most orifices of ordinary
shapes, even with widely different values of the area of aper-
ture and of the height, or head. A, producing the flow. Call-
ing this abstract number [.60 to .64] the Coefficient of Con-
traction, and denoting it by C, we may write
that the width of the
unless the vessel wall is
Frr.= CF,
in which F = area of the oriflce, and F^ = that of the con-
tracted section. C ranges from .60 to .64 with circular orifices,
but may have lower values with some rectangular forms. (See
table in § 503.)
A lamina of particles of water is under atmospheric
pressure at n (the free surface of the water in tank or reser-
voir), while its velocity at 7i is practically zero, i.e. v^—'O
(the surface at B being very large compared with the area of
orifice). It experiences increasing pressure as it slowly de-
scends until in the immediate neighborhood of the orifice,
when its velocity is rapidly accelerated and pressure decreased,
in accordance with Bernoulli's Theorem, and its shape length-
ened out, until finally at m it forms a portion of a filament of
a jet, its pressure is one atmosphere, and its velocity, = v„j, ,
we wish to determine. The course of this lamina we call a
660
MECHANICS or ETSTGHSTEEKING.
" stream-line,''^ and Bernoulli's Theorem is applicable to it,
just as it it were enclosed in a frictionless pipe of the same
form. Taking then a datum plane through the centre of m,
we have
2^m
y
= h, Sr>i = 0, and v^ = ?;
while
^ also = b, Sn = h, and v^ =
r
Hence Bernoulli's Theorem gives
^9
+ 5 + = + 5 + A;
^9
= A, . .
o o o o o
and
v^= i/2^A.
(1)
That is, the velocity of the jet at m is theoretically the same as
that acquired hy a hody falling freely in vacuo through a
height = A = the " head of water." We should therefore ex-
pect that if the jetweredirectedver-
tically upward, as at m, Fig. 540,
a height ~-
would be actually attained. [See
§§ 52 and 53.] Experiment shows
that the height of the jet (at m)
does not materially differ from h if
h is not > 6 or 8 feet. For A > 8 ft., however, the actual height
reached is < A, the difference being not only absolutely but
relatively greater as A is taken greater, since the resistance of
the air is then more and more effective in depressing and
breaking up the stream. (See § 5Y8.)
At m', Fig. 540, we have a jet, under a head = A', directed
Fm. 540.
ORIFICE IlSr THI]Sr PLATE. 661
at an angle «'„ with the horizontal. Its form is a parabola
(§ 81), and the theoretical height reached is K" = h' sin* a,
(§ 80).
The jet from an orifice in thin plate is very limpid and clear.
From eq. (1), we have theoretically
v^— V'2gh
(an equation we shall always use for efflux into the air through
orifices and short pipes in the plane wall of a large tank whose
water-surface is very large compared with the orifice, and is
open to the air), but experiment shows that for an " oi'ifice in
thin plate'^ this value is reduced about 3^ by frictiou at the
edges, so that for ordinary practical purposes we may write
-y^ = V^gh = 0.9Y V^gh, .... (2)
in which is called the coefficient of velocity.^
Hence the volume of flow, Q, per time-unit will be
Q = F^'^m = CFcf) V^, on the average = O^F V^gh. (3)
It is to be understood that the flow is steady, and that the
reservoir surface (very large) and the jet are hoth under at-
Tnospheric pressure. 067 is called the coefficient of effiux.
Example 1. — Fig. 539. Required the velocity of efllux,
v^ , at m, and the volume of the flow per second, Q, into the
air, if A = 21 ft. 6 inches, the circular orifice being 2 in. in
diam. ; take C = 0.64. [Ft., lb., and sec]
From eq. (2),
v^ = 0.9Y V2 X 32.3 X 21.5 = 36.1 ft. per sec. ;
hence the discharge is
,Q = F^v^ = 0.64 X t(:j|-)'x 36.1 = 0.504 cnb. ft. per second.
Example 2. — [Weisbach.] Under a head of 3.396 metres
the velocity v^ in the contracted section is found by measure-
* See Engineering News for Sept. 27, 1906, p. 326, for an account of
extensive experiments on flow through orifices. Values of (f> as high as
0.99 were obtained.
662 MECHANICS OF ENGINEEKING.
merits of the jet-curve to be 7.98 metres per sec, and the dis-
charge proves to be 0.01^25 cub. metres per sec. Required
the coefficient of velocity (0) and that of contraction {C), if
the area of the orifice is 36.3 sq. centimetres.
Use the 7/ietre-kilogra7n-second system of units, in which
g = 9.81 met. per sq. second.
From eq. (2),
= -4^ = ^-^^ = 0.978;
V 2g/i V2 X 9.81 X 3.396
while from (3) we have
F(p V 2gh Fv^ Tww X 7.98 " ' •
and (7, being abstract numbers, are independent of the sys-
tem of concrete units adopted.
ISToTE. — To fir.d the velocity v^ of the jet at the orifice by
measurements of the jet-curve, as mentioned in Example 2,
we may proceed as follows : Since we cannot very readily as-
sure ourselves that the direction of the jet at the orifice is
horizontal, we consider the angle a^ of the parabola (see Fig.
93 and § 80) as unknown, and therefore have two unknowns
to deal with, and obtain the necessary two equations by meas-
uring- the X and y (see page 84) of two points of the jet, re-
membering that if we use the equation (3) of page 84 in its
present form points of the jet below the orifice will have nega-
tive 2/'s. The substitution of these values a?^ , a?^ , y^ , and y^
in equation (3) furnishes two equations between i^onstants, in
which only o'j, and h are unknown. To eliminate «'„, for
1
we write 1 + tan'^ a. « and taking x^ = 2a?, for coUp
cos a'„
venience, we finally obtain
8 y,-2y. ' "V 4(2/. -22/.) '
in which y^ and y^ are the vertical distances of the two points
BOUNDED OKIFICE. 668
cLosen helow the orifice ; that is, we have already made them
negative in eq. (3) of page 84. The h of the preceding equa-
tion simply denotes v„^ -^ 2^, and must not be confused with
that of the last two figures. For accuracy the second point
should be as far from the orifice along the jet as possible.
496. Orifice with Rounded Approach.* — Fig. 541 shows the
general form and proportions of an orifice or mouth-piece in
the use of which contraction does not
take place beyond the edges, the inner ^ . \
surface being one "of revolution," and \\
so shaped that the liquid filaments are :^:-"
parallel on passing the outer edge m\ ~l~-
hence the pressure-head at m is = 5 -".'^
(= 34 ft. for water and 30 inches for -' /'
mercury) in Bernoulli's Theorem, if i ,
efflux takes place into the air. We fig. 541.
have also the sectional area ^^ = J^= that of final edge of
orifice, i.e., the coefficient of contraction, or O, = unity = 1.00,
so that the discharge per time-unit has a volume
Q = F^ii^ = Iv„,.
The tank being large, as in Fig. 540, Bernoulli's Theorem
applied to m and n will give, as before,
Vm= V2gh
as a theoretical result, while practically we write
v^=(pi/^, (1)
and Q = F(pV~2gh (2)
As an average is found to differ little from 0.97 with this
orifice, the same value as for an orifice in thin plate (§ 495).
497. Problems in Efflux Solved by Applying Bernoulli's
Theorem. — In the two preceding paragraphs the pressure-
heads at sections m and n were each = p^-^ y = height of
* Smooth conical nozzles for fire-streams give (p — .97 with h = press.-
head -f- veloc.-head at base of play-pipe ; see p. 833.
664
MECHANICS OF ENGII^EERIN'&.
the liquid barometer = J; but in tlie following problems this
will not be the case necessarilj. However, efflux is to take
place through a simple orifice in the side of a large reservoir,
whose upper surface (n) is very lafge, so that v^ may be put
5= zero.
Problem I. — Fig. 542. What is the veic»city of efflux, -y^, at
the orifice m (i.e., at the contracted sec-
tion, if it is an orifice in thin plate)
of a jet of water from a steam-boiler, if
the free surface at 7^ is at a height = h
above m, and the pressure of the steam
over the water is j?„ , the discharge tak-
ing place into the air?
Applying Bernoulli's Theorem to sec-
tion m at the orifice [where the pres-
sure-head is b and velocity-head vj -^ 2^ (unknown)] and to
section n at water-surface (where velocity-head = and pres-
sure-head = j)^^ y), we have, takiag m as a datimi iox poten-
tial heads so that ^^ = and 2^ = A,
Fig. 542.
5^4-5 +
2^
O + ^ + A-
y
'Om=\7^9
Pn
S + A
■]■
. . (1)
Example. — Let the steam-gauge read 40 lbs. (and hence
Pn = S4.7 lbs. per sq. inch) and A = 2 ft. 4 in. ; required v^.
Also \i F=^ ^ sq. in., in " thin plate," required the rate of
discharge (volume). The temperature of saturated steam of
the given tension must be 286° Fahr. [see foot of page 607].
The water is practically at the same temperature and hence
of a heaviness, y, of 5Y.7 lbs. per cubic ft. (p. 518).
From eq. (1) above, then, with ft. lb. and sec, noting that
for this case h = [(14.7 X 144) -^ 57.7] feet,
V,n = \/2XS2.2
"54.7 X 144
14.7 X 144 28
+ 12
■]
= 81.1 ft. per sec.
57.7 57.7
, theoretically ; but * practically
• Another practical matter in this case is that some of the hot water will
'flash" into steam on relief from the hig-her pressure.
PKOBLEMS OF EFFLUX — ORIFICES.
665
v^ = 0.97 X 81.1 == 78.6 ft. per sec. ;
so that the discharge begins at the rate of
Q = 0.64 i^y^ = 0.64 x i • tt¥ X 78.6 = 0.174 cub. ft. p. sec.
Problem II. — Fig. 543. "With what velocity, v^ , will wafces
flow into the condenser O of a steam-engine where the tension
of the vapor is p^, < one atmosphere, if
h = the head of water, and the flow takes
place through an orifice in thin plate?
Taking position m in the contracted section
where the filaments are parallel, and the
pressure therefore equal to that of the sur- ^
rounding vapor, viz.,j?^, and. position n in
the (wide) free surface of the water in the
tank, where (at surface) the pressure is one fig. 643.
atmosphere [and /. ^ = 5 = 34 ft.] and velocity practically
zero; we have, applying Bernoulli's Theorem to n and m^ tak-
ing m as a datum level for potential heads (so that s„ — A and
£m = 0),
^9 r
«'«.=y^2^[A + ^-|^], , , . . (1)
and Q = J^m'i>mt . . . * c , (2^
as theoretical results. But practically we must write
^■^Hc^^v'^n
— p. ^>- —
iiil
— ^
^0.97^2g[h-^h-^^,. .
(3)
and
Q = FmV.^ — CFv^ 5 (4)
in which ^= area of orifice in thin plate, and ۥ= coefiicient
of contraction = about 0.62 approximately [see § 495],
666
MECHANICS OF ENGIJSTEEEING.
Example. — If in the condenser there is a " vacuum" of 2Y^
inches (meaning that the tension of the vapor would support
2 J inches of mercury, in a barometer), so that
Vm = [w X 14.7] lbs. per sq. inch, and A = 13 feet,
while the orifice is |- inch in diameter ; we have, using the ft.,
ib., and sec,
= 0.97a /:
2 X 32.2
;i2 + 34-AX^;^]
= 51.1 ft. per sec.
(We might also have written, for brevity,
^ = [2i : 30] X 34 = 2.833,
since the pressure-head for one atmos. = 34 feet^ for water.
Hence, for a circular orifice in thin plate, we have the volume
discharged per unit of time,
Q = CFv = 0.62 X f (^)'x 51.1 = 0.0431 cub. ft. per see.
497a. Efflux through an Orifice in Terms of the Internal and
External Pressures. — Fig. 544. Let efflux take place through
a small orifice from the plane side of a large tank, in which at
the level of the orifice the hydrostatic pressure was =^p' be-
fore the opening of the orifice, that of the medium surround-
ing the jet being =:p". When a steady flow
is established, after opening the orifice, the
pressure in the water on a level with the ori-
fice will not be materially changed, except in .
(o) ^T.^-£^ l:;f the immediate rieighhorhood of the orifice [see
]L-| :zp^_r.zii^:^ § 495] ; hence, applying Bernoulli's Theorem
to m in the jet, where the filaments are parallel,
and a point n^ in the body of the liquid and
at the same level as ?/?, and where the particles
Fig. 544.
are practically at rest [i.e., -y^ = 0] (hence not too near the
FOECE-PUMP.
667
orifice), we shall have, cancelling out the potential heads which
are equal,
^g'^ y ^^ y'
^^ = 0.9Ya/%
V- y A
. . (1)
(In Fig. 544 p would be equal to y^ + hy) Eq. (1) is con-
veniently applied to the jet produced by
a force-pu7np, supposing, for simplicity,
the orifice to be ^Vi the head of the pump-
cylinder, as shown in Fig. 545. Let the
thrust (force) exerted along the piston-
rod be = P, and the area of the piston
he = F'. Then the intensity of internal
pressure produced in the chamber AB
(when the piston moves uniformly) is
y = ^ + ^>.
Fig. 545.
F'
while the external pressure in the air around the jet is simply
Pa (oiie atmos.).
.%^^=0.97^/2<7.^ (1)'
(!N^.B. Of course, at points near the orifice the internal
pressure is < p'\ read ^ 495.)
Example. —'Let the force, or thrust, P, [due i\j steam-pres-
sure on a piston not shown in figure,] be 2000 lbs., and the
diameter of pump-cylinder be c? = 9 inches, the liquid being
salt water (so that ;j/ = 64 lbs. per cubic foot).
Then
F' = \Tt{-^y = 0.442 sq. ft.
and [ft., lb., sec]
6b8 MECHAT^ICS OF EISTGINEEHTTSTG-.
^. = 0.97^^2 X 32.2 X ^j^-q~^ = 65.4 ft. per se«.
If the orifice is well rounded, with a diameter of one inch,
the volume discharged per second is
Q = F^v^, = Fn^ ■= l^-iy X 65.4 = 0.353 cub. ft. per sec.
To maintain steadily this rate of discharge, the piston must
move at the rate [veloc. = v'^ of
v' ^Q-^F' = .353 -^
|Q']= 0.800 ft. per sec,
and the force P must exert 2i power (§ 130) of
L=iPv' = 2000 X 0.800 = 1600 ft. lbs. per sec.
= about 3 horse-power (or 3 H. P.).
If the water must be forced from the cylinder through a
})ipe or hose before passing out of a nozzle into the air, the
velocity of efflux will be smaller, on account of ''^ fluid frio-
tioii^^ in the hose, for the same P ; such a problem will be
treated later [§ 513]. Of course, in a pum ping-engine, by the
use of several pump-cylinders, and of air-chambers, a practically
steady flow is kept up, notwithstanding the fact that the mo-
tion of each piston is not uniform, and must be reversed at the
end of each stroke.
498. Influence of Density on the Velocity of Efflux in the Last
Problem. — From the equation
Vm = \/^9 —
■P
y
of the preceding paragraph, where j?'' is the external pressure
around the jet, and ji?' the internal pressure at the same level
as the orifice but well back of it, where the liquid is sensibly
EELATION OF DEI-fSITY TO VELOCITY OF EFf LUX. 669
at rest, we notice that for the same difference of pressure
Vjp'~ JP"^ 25Ae velocity of efflux is inversely proportiocial to thi
square root of the heaviness of the liqxiid. Hence, for the
same {p' — p"\ mercury would flow out of the orifice with a
velocity only 0.272 of that of water ; for
1000
Again, assuming that the equation holds good for the flow or
gases (as it does approximately when j?' does not greatly exceed
»"; e.g., by 6 or 8 per cent), the velocity of efliux of atmospheric
air, when at a heaviness of 0.807 lbs. per cub. foot, would be
/
62.5
0807
1/775.3 = 27.8
times as great as for water, with the same p' — p'\ (See
§ 548, etc.)
499. Efflux under Water. Simple Orifice.— Fig. 546. Let \
and Aj be the depths of the (small) ori-
fice below the levels of the " head " and
" tail " waters respectively. Then, using
the formula of § 497a, we liave for the
pressure at n (at same level as m, the
J6t)
p' = {K + h)y,
-.'.aIr.- .
_ J
'■.h-:':-"-'--'.'-
_7i,
-^
L^ -f-
-4 -_'^
— n K
Z m >
TT .
and for the external pressure, around
the jet at m,
y = (/*.+%;
whence, theoretically,
Fig. 546.
^^m^W^p' —
-P'
V^g{K-h:)=^V^gh, . (11
where h = difference of level between the surfaces of the two
bodies of water.
670
MECHANICS OF ENGITTEERIKG,
Practically,
= cf) \/%jh;
(2)
but the value of for efflux under water is somewhat uncer-
tain ; as also that of C, the coefficient of contraction. Weis-
bach sajs that //, = 0(7, is -^-^ part less than for efflux into the
air ; others, that there is no ditference (Trautwine). See also
p. 389 of vol. 6, Jour, of Engin. Associations, where it is
stated that with a circular mouth-piece of 0.37 in. diam., and
of '' Dearly the form of the vena contracta,^^ jx was found to be
.952 for discharge into the air, and .945 for submerged dis-
charge.
500. Efflux from a Small Orifice in a Vessel in Motion.
Case I. WJien the motion is a vertical translation and uni-
formly accelerated. — Fig. 547. Suppose the vessel to move up-
ward with a constant acceleration p.
(See § 49a.) Taking m and n as in the
two preceding paragraphs, we know that
p^ ^p" =1 external pressure = one at-
mos. =Pa (and.*. — = h). As to the
internal pressure at n (same level as tn,
but well back of oriiice), p^ j this is not
equal to {h -{- h)y, because of the acceler-
ated motion, but we may determine it by considering free the
vertical column or prism On of liquid, of cross-section = "•:.=.;,• '
(= vj)dm), in which Vy^ = the =^^Ef^Z^^i
velocity of any filament, as m, = — =j^^^'iE:£^-A
in the jet, and bdx = cross-sec- ^^t^^r-^^'^
tion of the small prism which -j^f^ ^^^ C^jT "~1S^}^\ j^i i \\\u}p
passes through any horizontal — i ]~~ Ja-7Jl r^^^^Z~^^^%
strip of the area of orifice, in a "^j^^^.;^;;^^^' ^ ^Sr^^^ \
unit of time, its altitude being ^^r^W^^^^^A \ '"^^ ^
v^ . For each strip there is a fig 549.
different x or " head of water," and hence a different velocity.
Now the theoretical discharge (volume) per unit of time is
v^dF\
i.e., Q = 'bX^^dx (ly
But from Bernoulli's Theorem, if h =. & -^ 'ig =^ the velocity-
head at n, the surface of the channel of approach nC.h being
the pressure-head of n, and x its potential head referred to m as
datum (IT.B. This 5 = 34 ft. for water, and must not be con-
fused with the width h of orifice), we have [see § 492, eq. (7)]
.-. Vrn = V2g Vx-\-k; (2)'
and since dx = d{x + ^), ^ being a constant, we have, from (l)'^
and (2)',
Theoret. Q = l V~2g fix + k)^d{x -f ^),
'^hi + k
674 MECiiAisrics of eitgineering.
or
TJieoret. Q = %h \/^g [{K + ^^ - {K + ^)^]. • (1)
(5 now denotes the width of orifice.) If c is small, the chan-
nel of approach being large, we have
Theoret. Q = %-b V^ {hi -h^) .... (2)
(c being =i Q -^ area of section of nC).
If Aj = 0, i.e., if the orifice becomes a notch in the side, or
an overfall [see Fig 550, which shows the contraction which
actually occurs in all these cases], we have for an overfall *
Theoret.Q = %'bV^g\_{\-\-h)i-'ki].. ... (3)
Note. — Both in (1) and (2) \ and h^ are the vertical depths
.... . • . . -.. •.-, • . . of the respective sills of the orifice
from the surface of the water
three or four feet hack of the plane
of the orifice, where the surface is
comparatively level. This must
be specially attended to in deriv-
Fi»- 550. ing the actual discharge from the
theoretical (see § 503).
If Q were the unknown quantity in eqs. (1) and (3) it would
be necessary to proceed by successive assumptions and ap-
proximations, since Q is really involved in Ic ; for
h = ^ and F,c=Q
(where F^ is the sectional area of the channel of approach nC).
With ^ = (or c very small, i.e., F^ very large), eq. (S) re-
duces (for an overfall) to
Theoret.Q = llhJ^K/ .... (3^)
or f as much as if all parts of the orifice had the same head of
water = h^ (as for instance if the orifice were in the horizontal
bottom of a tank in which the water was h^ deep, the orifice
having a width = 5 and length = h^.
* The most satisfactory mathematical treatment of the flow over an overfall
weir is that of Elamant (see p. 96 of his Hydraulique, Taris, 1900, 2d edition).
Its resulting formula is in remarkable accord with experiment, but is not con-
venient for practical use.
TEIANGULAR ORIFICE.
676
502. Theoretical Efflux tlirough a Triangular Orifice in a Thin
Vertical Plate or Wall. Base Horizontal. — Fig. 551. Let the
channel of approach be so large that the velocity of approach
may be neglected, h^ and J\ = depths of sill and vertex,
which is downward. The analysis differs from that of the
preceding article only in having h-= and the length u, of a
horizontal strip of the orifice, variable ; h being the length of
the base of the triangle. From similar triangles we have
u
h„ — X
I.e., u =
c% Theoret. Q = fv,ndF ^= fi}y,;m^ = 0.608. E'ow n = ii hencec,
frona eq. (7)',
M = 0.608 [1 + 0.155 X i]= 0.6551 ;
hence, eq. (7),
Q = 0.656 X2X-^V2X 32.2(8+^.-^)
= 10.23 cub. ft. per sec.
Case III. Imperfect Contraction.— li there is a submerged
channel of approach, symmetrically
placed as regards the orifice, and of
area (cross-section), = G,
"1 G
C^WlA— i^ much larger than that, = i^, of the
-_^r=- ^ ^i^^^-^^^^ orifice (see Fig. 555), the contraction
Z£E^£:/->_^//.' [ •^•' j is less than in Case I, and is called
imperfect contraction. Upon his
experiments with Poncelet's orifices.
FiO' S55. with imperfect contraction, Weisbach
bases the f oliowiDg formula for the discharge (volume) per
unit of time, viz.,
© = /^«5^/2^(^.+ |)
RECTANGULAR ORIFICES.
681
(see Pig. 553 fci notation), with, the understanding that the co-
efficient
A« = A«.(1 + /S), (sy
where yw^ '^ ^b© coefficient obtained from the tables of Case I
(as if the contraction were perfect and complete), and ^ an ab
stract number depending on the ratio F i G = mfas follows:
^ = 0.0760 [S'^ - l.(
(§r
To shorten computation Weisbach gives the following tabk
for^:
Ex^MPLB.~Iet h, = 4' 9|'' (= 1.46
met.), the dimensions of the orifice
Table A.
width =ib = 8 in. (= OM"^);
height = flj = 5 in. (= 0.126"^) ;
while the ebannel of approach {CJDf
Fig. 555) is one foot square. From
Case I, we have, for the given ^-^
mensions and head,
A«„ = 0.610;
m.
^.
m.
.05
.009
.55
.178 1
.10
.019
.60
.208
.15
.030
.65
.241
.20
.043
.70
.378 '
.85
.056
.75
.319
.80
.071
.80
.365
.35
.088
.85
.416
.40
.107
.90
.473
.45
.138
.95
.537
.go
.153
1.00
.608
G 144 sq. m.
We find [Table A]
/?= 0.062;
and hence /^ = /i„ (1.062), from eq. (8)'. Therefoi'e, ivom 6^
(8), with ft., lb., and sec,
^ = 0.610 X 1.062 X3^.^V2 X 32.2 X 5
= 3.2.2 cub. ft. per sec.
Case IY. Head measured in Moving Water. — See Pigc
556. If the head A, , of the upper sill, cannot be measured to
the level of stiK water, but must be taken to the surface of "x
rfVcvrjol of approach, where the velocity of approach is quite
682
MECHANICS OF ENGINEERING.
appreciable, not only is the contraction imperfect, but
strictly we should use eq. (1) of § 501, in
which the velocity of approach is considered.
Let i^ = area of orifice, and G that of the
cross-section of the channel of approach;
then the velocity of approach is c^Q^G^
and k (of above eq.)=c^^2g = Q^-i-2gG^;
but Q itself being unknown, a substitutionof
k in terms of Q in eq. (1), § 501, leads to an
equation of high degree with respect to Q.
Fis. 556.
Q^nah
yl'^gihi
a
+ 2
(9)
Practically, therefore, it
is better to write
and determine /z by experiment for different values of the
ratio F^G. Accordingly, Weisbach found, for Poncelet's
orifices, that if /.(q is the coefficient for complete and perfect
contraction from Case I, we have
/i = /io(H-/?0, where ^' = ^Ml{F ^Gf . . (90
hi was measured to the surface one metre back of the plane
of the orifice, and F:G did not exceed 0.50.
504. Actual Discharge of Sharp-edged Overfalls (Overfall'
Weirs; or Rectangular Notches in a Thin Vertical Plate).
Case I. Comjplete and Perfect Contraction {the normal
case), Fig. 557 ; i.e., no edge is flush
with the side or bottom of the
reservoir, whose sectional area is
very large compared with that, hh^.
of the notch. By deptn, h^ , of the
notch, we are to understand the
depth of the sill helow the surface
a few feet hacJc of the notch where
it is level. In the plane of the
notch the vertical thickness of the stream is only from f to -^
of Aj . Putting, therefore, the velocity of approach = zero,
and hence ^ = 0, in eq. (3) of § 501, we have for the
Fig. 557.
Actual Q = >Wol^^a ■^2^^ii) •
(10)
DISCHARGE OF OVERFALL-WEIRS.
683
(h = width of notch.) wliere j^^ is a coefficient of efflux to be
cletermined bj experiment.
Experiments with overfalls do not agree as well as might be
desired. Those of Poncelet and Lesbros gave the results in
Table C.
Example 1. — With
A, = 1 ft. 4 in. (= .405%
5 = 2 ft. (= O.eO""-),
we have, from Table C, jWo = -^SB,
and (ft., lb., sec.)
Table C.
For 6 = 0.20^
Forfc = 0.60™. !
metres.
metres.
■
h^
fio
7l2
^•0 1
.01
(5B6
.06
618 !
03
620
08
613 1
03
618
10
609 i
04
610
12
605
06
601
15
600
08
595
30
593
10
592
30
.586
15
589
40
586
20
585
50
586
32
577
60
.585
For appros. results Mn =
BO 1
/. ^=.586X1 X2xf 1/2x32.2x1
= 9.54 cub. ft. per sec.
Example 2. — What width, J,
must be given to a rectangular notch, for which h^ = 10 in.
issL 0.25'"-), that the discharge maj be ^ = 6 cub. feet per sec?
Since h is unknown, we cannot use the table immediately,
but take pi^ = .600 for a first approximation ; whence, eq, (10),-
(ft., lb., sec.,)
I =
6
0.6 X I X It V2 X 32.2 X |i
= 2.46 ft.
Then, since this width does not much exceed 0.60 metre,
we may take, in Table C, for A^ = 0.25 met., /fj, = .589 ;
.589 X I X if V2 X 32.2 X U
= 2.50 ft
Case II. Incomplete Contraction; i.e., hoth ends are flush
with the sides of the tanh, these heing 1 to the plane of the
notch. According to Weisbach, we may write
Q = ^mM, V^gK
(11)
in which /^ = 1.041yWo , /n^ being obtained from Table C for the
normal case, i.e., Case I. The section of channel of approach
is large compared with that of the notch; if not, see Case IV.
684
MECHANICS OF ENGINEERING.
Case m. Imperfect Contraction; i.e., the velocity of wp.
jproacli is appreciable ; the sectional area G
of the channel of approach not being much
larger than that, F, = lli^ = area of notch.
Fig. 558. h = width, and A^ = depth of
notch (see Case I). Here, instead of using
a formula involving
Tc = d'~<^g = iQ~6J^2g
Fig. 558.
(see eq. (3), § 501), it is more convenient to pufc
as before, with
Q=lf^hh,i/^gh,,
/* = /' = '>-'''' + m) • . • (18)
Eq. (17) is homogeneous, i.e., admits of any system of units.
* A valuable resume of " Weir Experiments, Coefficients, and Forraulas,"
by Robert E. Horton, appeared as " Water-supply and Irrigation Paper,"
No. 150, issued by the IF. S. Geological Survey in 1906.
OYERFALLS.
689
Provision was made in these experiments for tlie free en-
trance of air under the sheet (a point of great importance),
wliile the walls of the channel of approach were continued
down-stream, beyond the plane of the weir, to prevent any
lateral expansion of the sheet. The value of jp ranged from
0.20 to 2.00 metres.*
507. Efflux through Short Cylindrical Tubes. — When efflux
takes place through a short cylindrical tube, or " short pipe,'^
at least 2|- times as long as wide,
inserted at right angles in the
plane side of a large reservoir,
the inner corners not rounded
(see Fig. 560), the jet issues
from the tube in parallel fila-
ments and with a sectional area^
F^^ equal to that, F^ of interior
of tube.
To attain this result, however,
the tube must be full of water before the outer end is un-
stopped, and must not be oily ; nor must the head, A, be
greater than about 40 ft. for efflux into the air. Since at m
the filaments are parallel and the pressure-head therefore equal
to h (=: 34 ft. of water, nearly), = that of surrounding medium,
= head due to one atmosphere in this instance ; an application
of Bernoulli's Theorem [eq. (Y), § 492] to positions m and n
would give (precisely as in §§ 454 and 455)
. ...■;air;: •• .-;.
^^^^— ~ —
1
1
r
--N^.V
— . ^>^^^=^^
-=---3^-^^'
:=?r=£3:i£^^
^fsstl
' ^ 1 . . •
.•'■'•■'V'^
Fig. 560.
v^ = veloc. at m = ^'^gh
as a theoretical result; but experiment shows that the actual
value of -y^ in this case is
li^ = 0„ i/2gh =0.815 V2gh,
(1)
0.815 being an average value for 0„ , the coefficient oj^ velocity, for
ordinary purposes. It increases slightly as the head decreases,
* Mr. Eafter's paper, in Vol. 44 (p. 230) of the Trans, Am. Soc. C. E., gives an
account of Bazin's experiments witti weii'S of irregular forms ; as also of similar
experiments made at the Hydraulic Laboratory of the College of Civil Engineer-
ing at Cornell University.
690 MECHANICS OF ENGHSTEEEING.
and is evidently much less than the yalue 0.97 for an orifice in
a thin plate, § 495, or for a rounded mouth-piece as in § 496.
But as the sectional area of the stream where the filaments
are parallel, at ;?^, where v^ = 0.815 V'2iyh, is also equal to that,
Ff of the tube, the coefficient of efflux, /<„ , in the formula
is equal to 0o ? i-s., there is no contraction, or the coefficien't;
of contraction, C , in this case = 1.00.
Hence, for the volume of discharge per unit of time, we
hstNQ practically
^ = 0oi^l/%A = 0.815 i^|/2p: ... (2)
The discharge is therefore about \ greater than through an
orifice of the same diameter in a thin plate under the same
head [compare eq. (3), § 495] ; for although at m the velocity
is less in the present case, the sectional area of the stream is
greater, there being no contraction.
This difference in velocity is due principally to the fact that
the entrance of the tube has square edges, so that the stream
contracts (at m', Fig. 561) to a
section smaller than that of the
tube, and then re-expands to the
full section, F, of tube. The
eddying and accompanying in-
ternal friction caused by this re-
expansion (or "sudden enlarge-
ment" of the stream) is the prin-
cipal resistance which diminishes
the velocity. It is noticeable, also, in ^
this case that the jet is not limpid and
clear, as from thin plate, but troubled
and only translucent (like ground-
glass). The internal pressure in the
stream at mf is found to be less than
one atmosphere, i.e. less than that at m,
as shown experimentally by the suck- fig. 563.
ing in of air wheil a small aperture is made in the tube op
Fig. 561.
INCLINED SHORT PIPES.
691
posite 7n'. If the tube itself were so formed internally as to
lit this contracted vein, as in Fig. 562, the eddying would be
diminished and the velocity at m increased, and hence the
volume Q of efflux increased in the same proportion. (See
§ 509a.)
If the tube is less than 2|- times as long as wide, or if the
interior is not wet hy the water {2& when greasy), or if the head
is over 40 or 50 ft. (about), the efflux takes
place as if the tube were not there, Fig. 563,
and we have
v^ = 0.97 V2gh, as in § 495.
Example. — The discharge through a short
pipe 3 inches in diameter, like that in Fig. 560,
is 30 cub. ft. per minute, under a head of
2' 6", reservoir large. Required the coefficient of efflux
;/^ , = 00 , in this case. For variety use the inch-pound-min-
ute system of units, in which g = 32.2 X 12 X 3600 (see Note,
§ 51). >Mo, being an abstract number, will be the same numer-
ically in any system of units.
From eq. (2),
Fig. 563.
Q
00 = >^« = -ZT~^= =
30 X 1Y28
FV2gh J ^ g, ^2 ^ g^^ x 12 X 60^ X 30
= 0.803.
508. Inclined Short Tubes (Cylindrical).
short tube is inclined at some angle
or < 90° to tlie interior plane of the
reservoir wall, the efflux iis smaller than
when the angle is 90°, as in § 507.
We still use the form of equation
Q = iiF ^^ = 4>F V^\ . (3)
but from Weisbach's experiments /i
should be taken from the following table :
If the
Fie.664.
MECHANICS OF ENGINEERING.
TABLE F, COEFFICIENT OF EFFLUX (Inclined Tube).
Fora-= 90°
take jii^ (f>=^ .815
80°
70°
.783
60°
.764
50°
.747
40°
.731
30°
.719
Example. — With A = 12 ft., d = diam. of tube = 4 ins.
and a = 46°, we have for the volume discharged per sec. (ft,
lb., and sec.)
Tt (1
Q = [0.731 + ^ (.016)] - y 4/64.4X 12 =1.79 cub.ft.per sec.
The tube must be at least 3 times as long as wide, to be
filled.
509. Conical Diverging, and Converging, Short Tuhes. — "With
conical convergent tubes, as at ^, Fig. 565, with inner edges
not rounded, D'Aubuisson and Castel found by experiment
values of the coefficient of velocity, 0, and of that of efflux, /i,
[from which the coefficient of contraction, C = >m -^ 0, may be
Fig. 565.
computed,] for tubes 1.55 centimeters wide at the narrow end,
and 4.0 centimeters long, under a head of A = 3 metres, and
with different angles of convergence. By angle of converg-
ence is meant the angle between the sides CE and DB, Fig.
565. In the following table will be found some values of ^
and founded on these experiments, for use in the formulae
v^ = '^'^gh and Q = jjiFV^igh^
in which ..^denotes the area of the outlet orifice EB,
CONICAL SHOUT PIPES. 693
Table G (Conical Converging Tubes).
Angle of > 30 iQ,
convergence
IX = .895
(p= .894
8°
10° 20'
13° 30'
19° 30'
30°
49°
.930
.932
.938
.951
.946
.963
.924
.970
.895
.975
.847
.984
Evidently yu is a maximum for 12>^°.
With a conically divergent tube as at J/IZT, having the in-
ternal diameter MO =^ .025 metre, the internal diam. NP
= .032 metre, and the angle between JO^and PO = 4° 50',
"Weisbach found that in the equation Q = piFV'igh (where
F =^ area of outlet section NP) P- should be = 0.553; the
great loss of velocity as compared with S^'-lgh being due to the
eddying in the re-expansion from the contracted section at M
(corners not roimded)^ as occurs also in Fig. 549. The jet
was much troubled and pulsated violently.
When the angle of divergence is too great, or the head h
too large, or if the tube is not wet by the water, efflux with
the tube filled cannot be maintained, the flow then taking
place as in Fig. 563.
Yenturi and Eytelwein experimented with a conically di-
vergent tube (called now " Yen-
turVs tube "), with rounded en-
trance to conform to the shape
of the contracted vein, as in
Fig. 566, having a diameter of
one inch at 7nf (narrowest part),
where the sectional area = F'
= 0.7854 sq. in., and of 1.80
inches at 7n (outlet), where area = F; the length being 8 ins.,
and the angle of convergence 5° 9'.
With Q = /^FV2gh they found u = 0.483.
Hence 2|- times as much water was discharged as would have
flowed out under the same head through an orifice in thin
plate with area = F = the smallest section of the divergent
tube, and 1.9 times as much as through a short pipe of sec-
tion = F'. A similar calculation shows that the velocity at
m' must have been v^' =1.55 \^2gh, and hence that the pres-
sure at m' was much less than one atmosphere.
Fia. 566.
694
MECHANICS OF ElSTGIlSrEEKIlSrG.
Mr. J. B. Trancis also experimented with Yenturi's tube
(see " Lowell Hydraulic Experiments''). See also p. 389 of
voL 6 of the Journal of Engineering Societies, for experi-
ments with diverging short tubes discharging under water.
The highest coefficient (/i) obtained by Mr. Francis was 0.782.
509a. New Forms of the Venturi Tube. — The statement made
in § 607, in connection with Fig. 562, was based on purely
theoretic grounds, but has recently (Dec. 1888) been com-
pletely verified by experiments* conducted in the hydraulic
laboratory of the College of Civil Engineering at Cornell
University. Three short tubes of circular section, each 3 in.
in length and 1 in. in internal diameter at both ends, were ex-
perimented with, under heads of 2 ft. and4ft.f Call them A,
B, and C. A was an ordinary straight tube as in Fig. 561;
the longitudinal section of B was like that in Fig. 562, the
narrowest diameter being 0.80 in. [see § 495; (0.8)'^ = 0.64-];
while C was somewhat like that in Fig. 566, being formed
like B up to the narrowest part (diameter 0.80 in.), and then
made conically divergent to the discharging end. The results
of the experiments are given in the following table :
Name of
Tube.
Head.
Number
of Experi-
ments.
Range of Values of m.
Average
Values of ju,.
A
A
B
B
C
C
A = 3 ft.
A = 4 f t.
A = 3 f t.
/t = 4 ft.
/^ == 3 f t.
A = 4 ft.
4
3
5
4
5
4
From 0.804 to 0.833
" 0.819 to 0.833
'* 0.875 to 0.886
" 0.881 to 0.903
" 0.890 to 0.919
" 0.903 to 0.933
0.814
0.831
0.883
0.893
0.901
0.914
The fact that B discharges more than A is very noticeable,
while the superiority of C to B, though evident, is not nearly
so great as that of B to A, showing tiiat in order to increase
the discharge of an (originally) straight tube (by encroaching
on the passage-way) it is of more importance to fill up with
solid substance the space around the contracted vein than to
make the transition from the narrow section to the discharg-
ing end very gradual.
* See Journal of the Franklin Inst., for April, 1889.
t Practically the same co-efficients were obtained later, by Mr. E. M.
Holbrook, with higher heads; up to 18 ft.
*' FLUID FRICTIOlSr." 695
510. " Fluid Friction." — By experimenting witli the flow of
"Water in glass pipes inserted in the side of a tank, Prof. Rey-
nolds of England has found that the flow goes on in parallel
filaments for only a few feet from the entrance of the tube,
and that then the liquid particles begin to intermingle and
cross each other's paths in the most intricate manner. To
render this phenomenon visible, he injected a fine stream of
colored liquid at the inlet of the pipe and observed its further
motion, and found that the greater the velocity the nearer to
the inlet was the point where the breaking up of the parallel-
ism of flow began. The hypothesis of laminated flow is,
nevertheless, the simplest theoretical basis for establishing
practical formulae, and the resistance offered by pipes to the
flow of liquids in them will therefore be attributed to the fric-
tion of the edges of the laminae against the inner surface of
the pipe.*
The amount of this resistance (often called sJcin-fi'iction)
for a given extent of rubbing surface is by experiment found —
1. To be independent of the pressure between the liquid and
the solid ;
2. To vary nearly with the square of the relative velocity ;
3. To vary directly with the amount of rubbing surface^
4. To vary directly with the heaviness [y, § 409] of the
liquid.
Hence for a given velocity v, a given rubbing surface of
area = S, and a liquid of heaviness y, we may write
v"
Amount of friction (force) = fSy ^r— , (1)
^^
m which / is an abstract number called the coefficient of fluid
friction, to be determined by experiment. For a given liquid,
given character (roughness) of surface, and small range of
velocities it is approximately constant. The object of intro-
ducing the '2g is not only because — — is a familiar and useful
function of -y, but that v^ -^ 2^ is a height^ or distance, and there-
fore the product of S (an area) by v"" -i- 2^ is a volume, and this
volume multiplied by y gives the weight of an ideal prism of
* The resistance is really due both to the friction of the water on the sides of
the pipe and to the friction of the water particles on each other. The assump-
tion that it is due to the former action alone simply affects the mathematical
form of our expressions, without invalidating their accuracy, since the value of
/ is in any case dependent on experiment. See Engineei-ing Npws, July-Dec.
1901, pp. 332 and 476.
696
MECHANICS OF ENGINEERING.
1? .
the liquid; lience 8-^ y is a. force and y must be cm abstract
nurriber and therefore the same in all systems of units, in any
given case or experiment.*
In his experiments at Torquay, England, the late Mr. Fronde
found the following values for/*, the liquid being salt water,
while the rigid surfaces were the two sides of a thin straight:
wooden board ^^ of an inch thick and 19 inches high, coated
or prepared in various ways, and drawn edgewise through the
water at a constant velocity, the total resistance being measured
by a dynamometer.
511. Mr. Froude's Results. — (Condensed.) [The velocity
was the same := 10 ft. per sec. in each of the following cases.
For other velocities the resistance was found to vary nearly as
the square of the velocity, the index of the power varying
from 1.8 to 2.16.]
Table H.
Character of Surface.
Value of/ [from eq. (1), § 510].
8 ft. long.
When the
8 ft. long.
board was
20 ft. long.
SO ft. long.
Varnish ./=
Parafflne. "
0.0041
.0038
.0030
.0087
.0081
.0090
.0110
0.0032
.0031
.0028
.0063
.0058
.0062
.0071
0.0028
.0027
.0026
.0053
.0048
.0953
.0059
0.0025
Tinfoil
0025
Calico. . .
.0047
Fine Sand
.0040
Medium Sand
0049
Coarse Sand
N.T5. These numbers multiplied by 100 also give the mean frictional resistance in
lbs. per sq. foot or area of surface in each case {v = 30' per sec), considering the
heaviness of sea water, 64 lbs. per cubic foot, to cancel the 'i,g — 64.4 ft. per sq. sec. of
eq. (1) of the preceding paragraph.
For use in formulae bearing on flow in pipes, /" is best deter-
mined directly by experiments of that very nature, the results
of which will be given as soon as the proper formulae have been
established.
512. Bernoulli's Theorem for Steady Flow, with Friction. — [The
student will now re-read the first part of § 492, as far as eq.
(1).] Consideringfreeanylaminaof fluid. Fig. 567, (according
to the subdivision of the stream agreed upon in § 492 referred
* For very low velocities, below the range of ordinary engineering
practice, the friction varies more nearly as the prst 'power of the velocity,
instead of the square. See p. 42, etc., of Blaine's Hydraulic Machinery.
BEENOTJLLl'S THEOREM WITH FRICTION.
697
to,) the frictions on the edges are the only additional forces as
compared with the system in Fig.
534. Let w denote the length
of the wetted perimeter of the
base of this lamina (in case of a
pipe running full, as we here
postulate, the wetted perimeter
is of course the whole jperimeter^
but in the case of an open chan'
nel or canal, w is only a portion
of the whole perimeter of the
cross-section). Then, since the
area of rubbing surface at the edge is /S'= wds\ the total fric-
tion for the lamina is [by eq. (1), § 510] =fwy (w* — 2ff)ds',
Hence from vdv = (tan. accel.) X ds, and from (tan. accel.) =
[2(tang. compons. of acting forces)^ -r- (mass of lamina), we
have
J^ — I^p + dp) -\- Fyds' cos —fwy
'odv =
?l .ds...{a)
Fyds' -V- g
As in § 492, so here, considering the simultaneous advance of
all the laminae lying between any two sections m and n during
the small time dt, putting ds' = ds, and ds' cos (j)=^ — dz (see
Kg. 568), we have, for any one lamina,
--vdv-\--dj>-\-dz=^f^
• ds.
(1)
W V
g ■ Y ' ' - "^'^
I^ow conceive an infinite number of equations to be formed
like eq. (1), one for each la- ^
mina between n and m, for the
sanne dt, viz., a dt of such ~^:i^^^.
length that each lamina at the
©nd of dt will occupy the
same position, and acquire the
same values of -y, s, and p,
that the lamina next in front
had at the beginning of the
dt (this is the characteristic of a steady fiovj). Adding up
Fig. 568.
MECHANICS OF ENGHSTEERHSTG.
the corresponding terms of all these equations, we have (te^
membering that for a liquid y is the same in all laminaB)j
ie.j after transposition and writing H for J^-^- w, for brevity,
This is Bernoulli's Theorem for steady flow of a liquid in
a pipe of sligMly varying sectional area F, and internal perim-
eter w, taking into account no resistances or friction, except
the " skin-friction," or " fluid-friction," of the liquid and sides
of the pipe.
Resistances due to the internal friction of eddying occasioned
by sudden enlargements of the cross-section of the pipe, by
elbows, sharp curves, valve-gates, etc., will be mentioned later.
The negative term on the right in (3) is of course a height or
head (one dimension of length), as all the other terms are such,
and since it is the amount by which the sum of the three heads
(viz.. velocity-head, pressure-head^ and potential head) at w,
the down-stream position, lachs of being equal to the sura of
the corresponding heads at oi, the up-stream position or section,
we may call it the "Loss of Head" due to skin-friction between
n and 7n\ also called friction-head^ or resistance-head, or
height of resistance.
The quantity R =z F -— w = sectional-area -r- wetted-pe
rimeter, is an imaginary distance or length called the Hydravr
tic Mean Radius, or Hydraulic Mean Depth, or simply
hydraulic radius of the section. For a circular pipe of diam-
eter = d.
R = ^7r<7* -^ nd = ^\
while for a pipe of rectangular section,
a and h are lengths of sides of rectangle.
FRICTION U^ PIPES.
699
I
Vv
■V^D'^-Vr,
Fig. 569.
§13. Problems involving Friction-lieads ; and Examples oS
Bernoulli's Theorem witli Friction.
Problem I. — Let the portion of pipe between n and to \sk
level, and of uniform cir-
cular section and diameter
= d. The jet at m dis-
charges into the air, and
has the same sectional area,
F=z \7td?^2i'& the pipe; then
the pressure-head at m is
^ = 5 = 34 feet (for
Y
water), and the velocitj-
head at m is = that at ti^ since -y^ = v^ . The height of the
water column in the open piezometer at n is noted, and = y^
(so that the pressure-head at n is— = y„-f"^)j while the
length of pipe from n%Q7rh\^=.l.
Knowing Z, • • W
i.e., the open piezometer-height at?i is equal to the loss of head
(all of which is friction-head here) sustained between n and the
mouth of the pipe. (Pipe horizontal.)
Example. — Required tbe value of /", knowing that ^400l<^^ "-"^^
Problem II. HydramliG Accumulator. — Fig. 570. jLet ^Si»
area F^ of the piston on the left be quite large compared with
^*S:A!fiai''«^'
Fie. 570.
that of the pipes and nozzle. The cylinder contains a friction-
FEICTION-HEAD IN PIPES. 701
weighted piston, producing (so long as its downward slow
motion is uniform) a fluid pressure on its lower face of an
intensity ^„= [.0-{-FnPa\ -^ Fn per ^iiit area (j?a = one
atmos.).
Hence the pressure-head at ;» is
Pn_ G^
4-J, (6)
y Kr
where Q = load on piston.
The jet has a section at m = j^ = that of the small straight
nozzle (no contraction). The junctions of the pipes with each
other, and with the cylinder and nozzle, are all smoothly
rounded ; hence the only losses of head in steady flow between
n and m are the friction-heads in the two long pipes, neglect,
ing that in the short nozzle. These friction-heads will be of
the form in eq. (4), and will involve the velocities v^ and -y,
respectively in these pipes {supposed running full), v, and ?;,
may be unknown at the outset, as here.
Knowing G and all dimensions and heights, we are required
to find the velocity v^ of the jet, flowing into the air, and the
volume of flow, Q, per unit of time, assuming / to be known
and to be the same in both pipes (not strictly true).
Let the lengths and diameters be denoted as in Fig. 570,
their sectional areas F^ and F^ , the unknown velocities in them
v^ and v^ .
From the equation of continuity [eq. (3), § 490], we have
v^ = ?jp!L and «,= ^^. . , (T)
To find v^, we apply Bernoulli's Theorem (with friction),
aq. (3), § 512, taking the down-stream position m in the jet
close to the nozzle, and the up-stream position oi just under the
piston in the cylinder where the velocity v^ is practically noth-
ing. Then with m as datum plane we have
^* + J.fO = 04-^4-A-4/A.^_4/A.5^. (8)
2<7 Y '^ d, ^q -^ d^ %g ^^
702 MECHANICS OF ENGIJSTEEEINa.
Apparently (8) contains three unknown quantities, Vm-, Vi,
and V, ; but from eqs. (7) v^ and % can be expressed in terms
of v^ , whence [see also eq. (6)]
%
;i+^4(t)"+'4&)>''"^-^'*^ • ^'^
G
V
;. (10)
i + ^/i(tJ-^^/i(t)'
and hence we ha^e also
Q^Fjo^, (11)
Example. — If we replace the force G of this problem by
the thrust P exerted along the pump-pist n of a steam fire*
engine, we may treat the foregoing as a close approximation
to the practical problem of such an apparatus, the pipes being
consecutive straight lengths of hose, in which (for the probable
character of the internal surface) we shall take /=.0075 (see
Mr. Freeman's experiments on p. 832). (Strictly, / varies
somewhat with the velocity; see § 517.) Let P = 12000 lbs.,
and the piston-area at n = -F„ = 72 sq. in. = | sq. ffc. Also, let
/i = 20 ft., and the dimensions of the hose be as follows:
and similarly, for other values of a (taking from the table,
§ 508), we compute the following values of Q^ (corners not
rounded) for use in the expression for " loss of head," C^ — :
For a = 90°
80°
.565
70°
.635
60°
,713
50°
.794
40°
.870
30°
.987
From eq. (4) we see that the loss of head at the entrance of
the pipe, corners not rounded, with a = 90", is about one half
(.505) of the height due to the velocity v in that part of the
pipe {v being the same all along the pipe if cylindrical).* The
value of. V itself, Fig. 5Y1, depends on all the features of the
design from reservoir to nozzle. See § 518.
If the corners at .fi'are properly rounded, the entrance loss of
head may practically be done away with; still, if v is quite
small (as it may frequently be, from large losses of head
farther down -stream), the saving thus secured, while helping
to increase v slightly (and thus the saving itself), is insignifi
cant.
516. General Form of Bernoulli's Theorem, considering aE
Losses of Head,
In view of preceding explanations and assumptions, we may
write in a general and final form Bernoulli's Theorem for a
steady flow from an up-stream position n to a down-stream
position m, as follows :
f all losses of head )
2g
-\- — -\-'Z^ = -^^-{-—-\-Zn~ \ OGCurringijeUoeen V . {B^
r
%
r
n and m
* If the entrance of the pipe has well-rounded corners (see Fig. 541 on p. 663),
the value of ^e is very small; viz., about 0.05.
LOSSES OF HEAD IIST GKISTERAL. 707
Each loss of head (or height of resistance) will be of the form
C ^— (except skin-friction head in long pipes, viz., 4/* -r-^^jy
the V in each case being the velocity, known or unknown, in
that part of the pipe where the resistance occurs (and hence,
IS not necessarily equal to v^ or -y^).
517. The Co-efficient, f, for Friction of Water in Pipes.—
(See eq. (1), § 510). — Experiments have been made by Weis-
bach, Bossut, Prony, Darcy, Lampe, Stearns, Hamilton Smith,,
Fanning, Herschel, WilHams (with Hubbell and Fenkell),,
Marx (with Hoskins and A^^ng), Saph, Schoder, and otherg,
to deternjine / in cyhndrical pipes of various materials
(wood, tin, glass, zinc, lead, brass, cast and wrought iron),
of diameters from | inch to 72 in. In general, the following
conclusions have been reached:
1st. f decreases when the velocity increases ; e.g., in one
case with the
same pipe/" was = .0070 for •« = 2' per sec,
while/" was = .0056 for v = 20' per see.
2dly. f decreases slightly as the diameter increases (other
things being equal);
e.g., in one experimenty was = .0069 in a 3-in. pipe,
while for the same velocity/' was = .0064 in a 6-in. pipe.
3dly. The condition of the interior surface of the pipe
affects the value oif, which is larger with increased roughness
of pipe.
Thus, Darcy found, with 2^ foul iron pipe with 6?=r 10 in,
and veloe. = 3.67 ft, per sec. the value .0113 for/; whereas
Fanning (see p. 238 of his '• Water-supply Engineering"), with
a cement-lined pipe and velocity of 3.74 ft. per sec. and d =
20 inches, obtained/ = .0052.
The Hazen-Williams formula for new cast iron pipes,
presented by its authors as embodying fairly well the results
of experiment, implies the relation
/= (.00590) ^(d-i66r 15), (1)
where d and v denote diameter and mean velocity, respect-
ively, and the foot and second are used as units ; while for
all fairly smooth pipes, including small brass pipes, Drs. Saph
and Schoder have derived
708
MECHANICS OF ENGINEERING.
/= (.00606) ^(d'2Vi4).
.... (2)
For use with new cast iron pipes Prof. Unwin recommends
the formula /"= (.00538) ^ (d-i^vos) ; (3)-
and values computed from this will be found to differ but
little from those given herewith on p. 709, which (by per-
mission) is condensed from a more extensive table in Mr.
Fanning's Water Supply Engineering.
Let a general expression for / now be written of the form
/=/o-^((i-7;-) (4)
where the factor /o and the exponents m and n are constant
for a given character of pipe surface. According to Prof.
Unwin's conclusions, from a study of the available experi-
ments, the following average values of these quantities may
be used :
Kind of Pipe.
Tin plate
Wrought iron
Asphalted iron
Riveted wrought iron
New cast iron
Cleaned cast iron
Incrusted cast iron . .
/n
.00662
.00565
.00635
.00650
.00538
.00608
.01100
.10
.21
.127
.390
.168
.168
.16
.18
.25
.15
.13
.05
.00
.00
Note. — For riveted steel pipe of 3 ft. diameter and over, f is probably much
larger than as indicated for riveted wrought-iron pipe above. See
Engineering News, Dec. 1895, p. 415; also in Jan., etc., 1896, pp. 59, 74,
193, and 393. In some experiments by Mr. Herschel on new riveted steel
pipe, from 36 to 48 in. diameter, values of / were found about double
those of above table for riveted wrought iron. After four years' use some
of these pipes were found to be discharging considerably less than when
new, probably on account of incrustation and deposits.
Example. — Fig. 572. In the steady pumping of crude
petroleum weighing ^ = 55 lbs. per cubic foot, through a six-
inch pipe 30 miles long,
to a station 700 ft. higher
than the pump, it is
found that the pressure
in the pump cylinder at
n, necessary to keep up
a velocity of 4.4 ft. pp"*
sec. in the pipe, is 1000
Required the coefBcienty in the pipe. A.s
Fm. 572,
lbs. per
all los
the latr.
lie friction-liead in the pipe are insignificant,
'i be considered. The velocity-bead at n may
TABLE OF VALUES OF /.
709
I
Pi
§
I— I
I
P3
o
o
o
I— I
P5
PR
P3
O
I— I
o
Pm
O
O
W
Pm
O
3
bo
a
Ph
B
o
biD
<1
.S :o
i>
eo
OS
CO
«
OS
eo
^
eo
•
CO
CO
s
s
s
CO
CO
eo
eo
03
eo
^
1-1
CO
^ II II
"^
s.si
S!
00
1—1
CO
t-
,^_,
-*
c-
CO
g
C3
1-1
1-1
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00
t-
c^
goM
'^
-*
^
^
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^
CO
eo
CO
CO
^Hl
a.ss'
SI
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d
1—1
C3
03
CD
p
s
00
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CD
la
^
eo
03
1-1
II II
»o
^
^
^
^
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^
■^
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a .si
90
l>
la
o»
OS
00
OS
00
03
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CO
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MH
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10
10
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10
iO
^
iO
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^
^
. II II
a'.s's
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^
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10
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to
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= 00'°.
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d
d
d
tH
1-1
; in parallel filaments, into
the air, and therefore
^■•-•^ has same section as pi pe ;
Fig. 573. — , , £ ^x. . 1
hence, also, ■y^ of the jet
= 'y in the pipe (which is assumed to be running full), and ift
OOEFFICIElSfT OF LIQUID FEICTION. 711
therefore the velocity to be used in the loss of head C^ — at
2^
the entrance ^(§ 515),
Taking m and n as in figure and applying Bernoulli's
Theorem (§ 4Y4), with m as datum level for the potential heads
3^ and Zn , we have
|: + J + = + S + A-C.|-4/J-J. . (1)
Three different problems may now be solved:
First, required the head h to keep up a flow of given volume
= Q per unit of time in a pipe of given length I and diameter
= d.
From the equation of continuity we have
Q = FmVrn = iTtd'Vrr, ',
4:0
.'. veloG. of jet, which = veloc. in pipe, = ij^ = — ^-. . . (2)
Having found v,^ = v, from (2), we obtain from (1) the re-
quired h, thus :
h=''
^g
'l + C^ + 4/|
(3)
Now Cs=z 0.505 if or = 90° (see § 515), while/ may be
taken from the table, § 517, for the given diameter and com-
puted velocity \y^ = v, found in (2)], if the pipe is clean ; if
not clean, see end of § 517, for slightly tuberculated and for
foul pipes.*
Secondly. Given the head h, and the length I and diameter
d of pipe, required the velocity in the pipe, viz., v, = v^, that
of jet; also the volume delivered per unit of time, Q. Solv^
ing eq. (1) for -y^, we have
/"
v„= / ^V2gh', ... (4)
v/i + c.+4/^
* "Hydraulic Tables," for friction-head of water in pipes, by Prof. G.
8. "Williams and Mr. Allen Hazen (New York, 1905, .John' Wiley & Sons),
cover the cases of pipes in various states of tuberculation, etc.
712 MECHANICS OF ENGINEERING.
whence Q becomes known, since
Q = i7rd'v^, .,...,, (5)
[JSToTE. — The first radical in (4) might for brevity be called
a coefficient of velocity^ 0, for this case. Since the jet has the
same diameter as the pipe, this radical may also be called a
coefficient of effiux.']
Since in (4) /" depends on the unknown v as well as on the
known d, we must first put/" = .006 for a first approximation
for v^ ; then take a corresponding value for f and substitute
again ; and so on.
Thirdly, knowing the length of pipe and the head A, we
wish to find the proper diameter d for the pipe to deliver a
given volume Q of water per unit of time. JSTow
"'^''-'^J^' •••••• (6)
which substituted in (1) gives
that iS;
As the radical contains d, we first assume a value for d,
withy=: .006, and substitute in (7). With the approximate
iralue of d thus obtained, we substitute again with a new value
for /based on an approximate v from eq. (6) (with <^ = its
first approximation), and thus a still closer value for d is de-
rived ; and so on. (Trautwine's Pocket-book contains a table
of fifth roots and powers.) If I is quite large, we may put
(i = for a first approximation. In connection with these
examples, see last figure.*
* In Chap. VIII (p. 188) of the Author's " Hydraulic Motors " will be
found additional matter on flow in pipes. In that work are also given fric-
tion-head diagraftis, the use of which saves much computation in solving
problems like those of pp. 713, 714, 731-734 of the present work.
LOJS^G PIPES. 713
Example 1. — What head h is necessary to deliver 120 cub.
ft. of water per minute through a clean straight iron pipe 140
ft. long and 6 in. in diameter'^
FrouQ eq. (2), with ft., lb., and sec, we have
V = i;^ = [4 X WJ ^ 7r(i)^= 10.18 ft. per sec.
Now for -y = 10 ft. per sec. and (^ = -^ ft., we find (in table,
§ 51Y) f^-- .00549 ; and hence, from eq. (3),
(lo.isy
2 X 32.2 L
"l _i_ 505 + ^ X -00549 X 140 '
12.23 ft,
of which total head, as we may call it, 1.60 ft. is used in pro-
ducing the velocity 10.18 ft. per sec. (i.e., i)^ -^ 2^ = l.GO ft.),
while 0.808 ft. ( = C^^J is lost at the entrance E (with a —
90°), and 9.82 ft. (friction-head) is lost in skin-friction.
Example 2. — [Data from Weisbach.] Required the de-
livery, Q^ through a straight clean iron pipe 48 ft. long and
2 in. in diameter, with 5 ft. head (= A). ^;, = -y^, being un-
known, we first take/" = .006 and obtain [eq. (4)]
r
^-= /, , ,n, , 4X.006X 48^2X32.2X5
V 1+-^^^+ — -^ —
= 6.18 ft. per sec.
From the table, § 51Y, for -y = 6.2 ft. per sec. and 6? = 2 in.
/= .00638, whence
/ 1 -
^'"~ ' , ,,. , 4 X .00638 X 48 ^2 >< ^^'2 >^ ^
1 -|- .50o -\
V
=; 6.04 ft. per sec,
which is sufficiently close. Then, for the volume per second,
^ =r - d'v^ = l7t{iy6M = 0.1307 cub. ft. per sec
714 MECHANICS OF ENGINEERING.
[Weisbacli's results in this example are
■y^ = 6.52 ft. per sec.
and Q = 0.1420 cub. ft. per sec,
but his values for/" are slightly different.]
Example 3. — [Data from Weisbach.] What must be the
diameter of a straight clean iron pipe 100 ft. in length, which
is to deliver ^ = ^ of a cubic foot of water per second under
5 ft. head (= A)?
With/= .006 (approximately), we have from eq. (Y), put-
ting ^ = under the radical for a first trial (ft., lb., sec),
4,0
whence v = -^ == 7 ft. per sec.
For d = 3.6 in. and v = 7 it per sec, we find/= .00601 ;
whence, again,
_ e / 1.505 X .30 + 4 X .00601 X 100 /4 X |y_^ go^ft .
Y 2X32.2X 5 • V TT / * '*
and the corresponding v = 6.06 ft.
For this d and -y we find/" = .00609, whence, finally,
d={ /i:S0^X.3O + 4x.0O6O9xlOO/2_V „ 3^3 ^^
Y 2 X 32.2 X 5 \7r/
[Weisbach's result is <^ = .318 ft.]
519. Ch6zy's Formula. — If, in the problem of the preceding
paragraph, the pipe is so long, and therefore I : d so great^
that 4:fl -^ d m eq. (3) is very large compared with 1 -}- Cj^j
we may neglect the latter term without appreciable error;
whence eq. (3) reduces to
A = 4/--.^ . • {pwe very long ; ^ig. hl^), . . (8)
CH:fezy'S FOKMULA.
715
■which is known as Chezy's Formula. For example, if I = 1000
ft. and c? = 2 in. = ^ ft., and/a,pprox. = .006, we have 4/— =
144, while 1 -\- C^ for square corners = 1.505 only.
If in (8) we substitute
(8) reduces to
= §r=Q-^lrtd\
. {very long pijpi).
. . (9)
80 that for a very long pipe, considering / as approximately-
constant, we may say that to deliver a volume = Q per unit
of time through a pipe of given length = I, the necessary head,
A, is inversely jprojportional to the fifth power of the diameter.
And again, solving (9) for Q, we find that the volume con-
veyed per unit of time is directly proportional to the fifth power
of the square root of the diameter ; directly proportional to
the square root of the head ; and inversely proportional to the
square root of the length. (Not true for short pipe ; see above
example.)
520. The Hydraulic Qrade=Iine. — The pipe of Fig. 573 is repeated
in Fig. 573a. It has no nozzle and hence the velocity vm of the jet at
m is equal to that, v, in the pipe. If we conceive of the insertion
of a great number of open piezometers, such as PS, along this pipe,
Fig. 573a. Fig. 5736.
the summits of the respective stationary water columns maintained
in them will lie in the straight line Bm. For a very long pipe the point
B is practically in the plane of reservoir surface _and vertically over
the entrance E. This line Bm is called the " Hydraulic Qrade=line "
or "Hydraulic Gradient." From eq. (3), p. 711, we note that the
total head h is made up of three parts, viz.: vm^-^2g, or velocity head
in jet; l^^(y''-i-2g), or loss of head at entrance; and the friction-head.
hp, in pipe, =(4/Z-T-d)(i)^-f-2gf), which for a very long pipe is practically
equal to h itself (in tliis case of no nozzLeJ. Evidently the vertical drop
716
MECHANICS OF ENGINEERING.
AB={1 + CE)(v'^-^2g). For instance, in example 1 of p. 713, we note
that the vertical distance from A to S is 1.41 ft., while /if = 9.82 ft.
Let us now consider that a conicat nozzle is attached to the end of a
long horizontal pipe whose length is 1200 ft. and diameter 6 in., with
a head of 78 ft., the diameter of the tip of the nozzle (so formed as not
to produce contraction) being 2 in. See Fig. 573&. After a steady
flow has set in, the velocity of the jet is Vm ft. /sec. and that of the water
in the pipe is v, only 1/9 of Vm,; since Q = t(«") '"'^'^'li'o) '"■ "^^^
loss of head in the nozzle may be written = 1/20 of Vni^^^g- Application
of Bernoulli's Theorem between points n and m gives rise to
b + h+0 = b+'"J^+ ^
i-+^4- 27+^4-
(10)
2g ' 20
With / written = .006 at first, and corrected later, we finally derive
v = 6 ft. /sec. and i;m=54 ft. /sec. Hence Q=1.17 cub. ft. /sec. Fig,
573& shows the hydraulic grade-line, BD, for the 6-in. pipe in this prob-
lem. The vertical drop AB = 1.505iv^^2g) = 0.85 ft., while ;i,i;'=30 ft.
The height of the piezometer column at E', at the base of the conical
nozzle, is 47.1 ft. The velocity of the jet being 54 ft. /sec. with Q = 1.17
cub. ft. /sec, the "kinetic power" of the jet (i.e., kinetic energy of the
mass of water passing per second) is i{Qr^g)vm?, =3308 ft.-lbs. per
second, =6.01 H. P.; so that a "jet motor," or impulse wheel, utilizing
80 per cent of this power would develop 4.8 H.P. (See Hydraulic
Motors by the present writer.) The slope of the hydraulic grade-line
in Fig. 5736 is hp-^l, i.e., 30-^1200 or 1/40; that is, it drops one foot
in each 40 ft. of length of pipe.
If a steady flow is proceeding in a pipe of uniform section it may easily
be shown, by Bernoulli's Theorem, that the vertical distance between
the summits of the open piezometers inserted at any two points is equal
to the loss of head occurring between those two points. Even if the
pipe is not of uniform section between the two points the foregoing is
still true if the sectional areas at the two points Ihemselves are equal.
If any part of the pipe, flowing full, projects above the hydraulic grade-
line, the internal pressure in that part is less than atmospheric, and air
previously dissolved in the water may collect after a time, and air may
also enter through imperfect joints, thus causing the pipe to be only
partly full at such points and seriously altering the conditions of flow.
For example, the pipe shown in Fig. 573c discharges into the air at m
and the por-
tion acb rises
above the hy-
draulic grade-
line Babm.
Air may col-
lect in the
summit c to
such a degree
that finally
only the part Eac of the pipe flows full, while in the portion cbm water flows
through only the lower part of each cross-section, with air above. In
such a case the hydra-ulic grade-line for Ec would rise to position Be.
PEESSURE-ENKRGY. 717
621. Bernoulli's Theorem as an Expression of the Conservation
of Energy for the Liq^uid Particles, — In any kind of flow with-
out friction^ steady or not, in rigid imtnovable vessels, the
aggregate potential and kinetic energy of the whole mass of
liquid concerned is necessarily a constant quantity (see §§ 148
and 149), but individual particles (as the particles in the sink-
ing free surface of water in a vessel which is rapidly being
emptied) may be continually losing potential energy, i.e.,
reaching lower and lower levels, without any compensating in-
crease of kinetic energy or of any other kind ; but in a steady
fiow without friction in rigid motionless vessels, we may state
that the stock of energy of a given particle, or small collection
of particles, is constant during the flow, provided we recognize
a third kind of energy which may be called Pressure-energy,
or capacity for doing work by virtue of internal fluid pressure ;
as may be thus explained :
In Fig. 574 let water, with a very slow motion and under a
pressure p (due to the reservoir-head -f- atmosphere-head be-
hind it), be admitted behind a pis-
ton the space beyond which is
vacuous. Let s = length of
stroke, and i^= the area of pis-
^ 7 ton. At the end of the stroke,
^^■^ ^£1^^ I ^ ^;^:^ ■ by motion of proper valves, com-
iZ^lJ T^vAc. munication with the reservoir is
^^«-5^^- cut off on the left of the piston
and opened on the right, while the water in the cylinder now on
the left of the piston is put in communication with the vacu-
ous exhaust-chamber. As a consequence the internal pressure
of this water falls to zero (height of cylinder small), and on
the return stroke is simply conveyed out of the cylinder,
neither helping nor hindering the motion. That is, in doing
the work of one stroke, viz.,
W = force X distance =i Fp X s = Fps,
a volume of water F= Fs, weighing Fsy (lbs. or other unit),
has been used, and, in passing through the motor, has experi-
enced no appreciable change in velocity (motion slow), and
718 MECHANICS OF ENGINEERING.
therefore no change in kinetic energy, nor any change of level,
and hence no change in potential energy, hut it has given up
all its pressure. (See § 409 for y)
!N^ow TF, the work obtained by the consumption of a weight
= ^ = Yy of water, may be written
W=Fps = Fsp=Vp = Yy^=a^. . . (1)
Hence a weight of water = G is capable of doing the worh
G X — = (r X head due to pressure p, i.e., = G X pressure-
head, in giving up all its pressure p • or otherwise, while still
having a pressure j!?, a weight G of water possesses an amou7it
of energy, which we may call pressure-energy, of an amount
= ^• — , where y = the heaviness (§ 7) of water, and-^ = a
y \o y y
height, or head, measuring the pressure p ; i.e., it equals the
pressure-head.
We may now state Bernoulli's Theorem without friction in
a new form, as follows : Multiplying each term of eq. (7),
§ 451, by Qy, the weight of water flowing per second (or other
time-unit) in the steady flow, we have
Qr^+Qy^ + Qr^m=Qr'^-\-Qy^f+QY^n. (2)
V ^ 1 Qy
But Qy -^ = —■^^^'^m = 3 X mass flowing per time-unit X
^ ^ g
square of the velocity = the Mnetic energy inherent in the
volume Q of water on passing the section m, due to the veloc-
ity at m. Also, Qy — = the pressure-energy * of the volume
Q at m, due to the pressure at m ; while QyZm = the potential
energy of the volume ^ at m due to its height z^ above the
arbitrary datum plane. Corresponding statements may be
made for the terms on the right-hand side of (2) referring to
the other section, n, of the pipe. Hence (2) may be thus read :
The aggregate amount of energy (of the three kinds mentioned)
resident in the particles of liquid when passing section m is
* This idea of " pressure energy" in connection with water is artificial,
but is of great convenience in dealing with questions of water power ; it is
of use only when the flow is steady. See p. 8 of the Author's ' ' Hydraulic
Motors," and p. 66 of BUdne's "Hydraulic Machinery."
LOSS OF EINTERGY.
719
equal to that when passing any other section^ as n / in steady
flow without friction in rigid motionless vessels ^ that is, the
store of energy is coustant.
522. Bernoulli's Theorem with Friction, from the Standpoint of
Energy. — Multiply each term in the equation of § 516 by Qy,
as before, and denote a loss of head or height of resistance due
to any cause by h^ , and we have
zg y
Each term QyK{G.g., Qy ^f
I
due to skin-friction in a
long pipe, and Qy Ci
2^
d ^g
due to loss of head at the reservoir
entrance of a pipe) represents a loss of energy, occurring between
any position n and any other position m down-stream from 7?,
but is really still in existence in the form of heat generated by
the friction of the liquid particles against each other or the
sides of the pipes.
As illustrative of several points in this connection, consider
two short lengths of pipe in
Fig. 575, A and B, one offering
a gradual, the other a sudden,
enlargement of section, but
otherwise identical in dimen-
sions. We suppose them to
occupy places in separate lines
of pipe in each of which a
steady flow with full cross-sec-
tions is proceeding, and so reg-
ulated that the velocity and in-
ternal pressure at n, in A, are
equal respectively to those at n F»tt. 575.
in £. Hence, if vacuum piezometers be inserted at n, the
720 MECHANICS OF EISTGINEEEING.
smaller section, the water columns maintained in them bj the
internal pressure will be of the same height, — , for botli A
and B. Since at 7?2, the larger section, the sectional area is the
same for both A and B, and since ^^ in ^ = F^ in _5, so that
Q^ = Q^ , hence v„^ in vl = v^i in B and is less than v^^ .
Now in ^ a loss of head occurs (and hence a loss of energy)
between n and m, bat none in A (except slight friction-head);
hence in A we should find as much energy present at m as at
71, only differently distributed among the three kinds, while at
m in B the aggregate energy is less than that at ti in B.
As regards kinetic energy, there has been a loss between n
and ??^ in both A and B (and equal losses), for -y^ is less than
Vn . As to potential energy, there is no change between 71 and
m either in A or B, since n and m are on a fevel. Hence if
the loss of kinetic energy in B is not compensated for by an
equal gain of pressure-energy (as it is in A), the pressure-head
(£3 \ at m in B should be less than that f^] at m in J.. Ex-
\r h \y U
periment shows this to be true, the loss of head being due to
the internal friction in the eddy occasioned by the sudden en-
largement ; the water column at m in B is found to be of a
less height than that at m in A, whereas at 7i they are equal.
(See p. 467 of article " Hydromechanics" in the Ency. Bri-
tannica for Mr. Fronde's experiments.)
In brief, in A the loss of kinetic energy has been made up
in pressure -energy, with no change of potential energy, but in
B there is an actual absolute loss of energy = Qy/i^ , or
= Qyti -^, suffered by the weight Qy of liquid. The value
2^
of C in this case and others will be considered in subsequent
paragraphs.
Similarly, losses of head, and therefore losses of energy,
occur at elbows, sharp bends, and obstructions, causing eddies
and internal friction, the amount of each loss for a given
weight, G, of water bfeing = Gh^ = GC -- ; hr = C^ being
the loss of head occasioned by the obstruction (p. 704). It is
BUDDElSr ENLAEGEMENTgl US' PIPES. 7iil
therefore very important in transmitting water through pipes
for purposes of power to use all possible means of preventing
disturbance and eddying among the liquid particles. E.g.,
sharp corners, turns, elbows, abrupt changes of section, should
be avoided in tlie design of the supply-pipe. The amount of
the losses of head, or heights of resistance, due to these various
causes will now be considered (except skin-friction, already
treated). Each such loss of head will be wiTtten in the form
C— , and we are principally concerned with the value of the
abstract number C, oi* coefficient of resistance^ in each case.
The velocity v is the velocity, known or unknown, where the
resistance occurs ', or if the section of pipe changes at this
place, then v = velocity on the down-stream section. The late
Professor Weisbach, of the mining-school of Freiberg, Saxony,
was one of the most noted experimenters in this respect, and
will be frequently quoted.
523. Loss of Head Due to Sudden (i.e., Square-edged) Enlarge-
ment. Borda's Formula. — Fig, 576. An eddy is formed in the
1 angle with consequent loss of energy. Since
l^^^^r each particle of water of weight = G^ , arriving
tY%^s^;^ with the velocity -y^ in the small pipe, may be
I u)\><:^^- ^ considered to have an impact against the base
Fig. 576. of the intinitely great and more slowly moving
column of water in the large pipe, and, after- the impact,
moves on with the same velocity, v^, as that column, just as
occurs in inelastic direct central impact (§ 60), we may find
the energy lost by this particle on account of the impact by
eq. (1) of § 138, in which, putting M^= G^-r- g, and M^ =
G^-^ g =: mass of infinitely great body of water in the large
pipe, so that M^ = oo , we have
Energy lost by particle = G^ ^^-i-- — ^, . , 0^
and the corresponding
Zoss of head — -^-^- — —.
•^ 2g '
723
MECHANICS OF ENGINEEEING.
■which, since F^'g^ = F^v^ , may be written
Loss of head in sudden enlargement
jr -IS
±j._i
^g
That is, the coefficient C for a sudden enlargement is
=(S-
(2)
(3)
Eq.
F^ and F^ are the respective sectional areas of the pipes.
(2) is Borda^s Formula.
KoTE. — Practically, the flow cannot always be maintained
with fnll sections. In any case, if we assume the pipes to be
running full (once started so), and on that assumption compute
the internal pressure at F^ , and find it to be zero or negative,
the assumption is incorrect. That is, unless there is some
pressure at F^ the water will not swell out laterally to fill the
large pipe.
Example. — Fig. 57T. In the short tube AB containing a
sudden enlargement, we have given F^ = F^ = 6 sq. inches,
.-.,,,-■-• J^, = 4 sq. inches, and A = 9 feet. Re-
quired the velocity of the jet at 7n (in
the air, so that j!?^^ -i- y =zh = S4: ft.), if
the only loss of head considered is that
due to the sudden enlargement (skin-
friction neglected, as the tube is short ;
the reservoir entrance has rounded cor-
ners^. Applying Bernoulli's Theorem
to mj as down-stream section, and n in reservoir surface as up-
stream position (datum level at m)^ we have
Fm. 577.
2^ 2^
Bn 1;, here, ■y, = -w^ ;
From eq. (3) we have
•'• ^^ + ^^1^ = '^ ° ^^>
C = (!-!)' = 0.25,
SUDDEN EISTLARGEMEISTT IN PIPE. 723
and finally (ft., lb., sec.)
1~
^«. = W ^-^ 1^2 X 32.2 X 9 = 0.895 V2 X 32.2 X 9
= 21.55 ft. per sec.
(The factor 0.895 might be called a coefficient of velocity for
this case.) Hence the volume of flow per second is
' Q = Fr^Vra = tIt X 21.55 = 0.898 cub. ft. per sec.
We have so far assumed that the water fills both parts of the
tube, i.e., that the pressure^!, at F^ , is greater than zero (see
foregoing note). To verify this assumption, we compute p^
by applying Bernoulli's Theorem to the centre of F^ as down-
stream position and datum plane, and n as up-stream position,
with no loss of head between, and obtain
^4_£i4-o = o + & + A-o. . • . . m
"ig y
But since F{0^ = F,v^ , we have
v: = {i)%' = (iYvj,
and hence the pressure-head at F^ (substituting from equations
above) is
and .*. i?i = f|- of 14.7 = 11.6 lbs. per sq. inch, which is
greater than zero ; hence efflux with the tube full in both parts
can be maintained under 9 ft. head.
If, with Fj^ and F^ as before (and .*. C)? we put p^ = 0, and
solve for A, we obtain h = 42.5 ft. as the maximum head
under which efflux with the large portion full can be secured.
524. Short Pipe, Square-edged Internally. — This case, already
734
MECHANICS OF ENGINEERING.
treated in §§ 507 and 515 (see Fig. 578 ; a repetition of 560),
presents a loss of head due to the sudden enlargement from
the contracted section at in' (whose sec-
tional area may be put = CF, C being
an unknown coefficient, or ratio, of
contraction) to the full section F of
the pipe. From § 515 we know that
the loss of head due to the short pipe
is K = C^^ (for a = 90°), in which
2^
^^ = 0,505 ; while from Borda's For-
F
Fig. 5?8.
mula, § 523, we have also C^ = -^^ — 1 . Equating these,
we find the coefficient of internal contraction at m' to be
C =
1 + ^C^ 1 + ^.505
= 0.584,
or abont 0.60 (compare with (7= .64 for thin-plate contrac-
tion, § 495). It is probably somewhat larger than this (.584),
since a small part of the loss of head, A^, is dne to friction at
the corners and against the sides of the pipe.
By a method similar to that pnrsued in the example of
§ 523, we may show that unless h is less than 40 feet, about,
the tube cannot be kept full, the discharge being as in Fig.
551. If the efflux takes place into a "partial vacuum," this
limiting value of h is still smaller. Weisbach's experiments
confirm these statements (but those in the 0. U. Hyd. LaK
seem to indicate that the limiting value for h in the first case
is about 50 ft.).
525. Diaphragm in a Cylindrical Pipe. — Fig. 579. The dia-
phragm, being of "thin plate,"
let
the circular opening in it /^\ z:=£::5.^'lJ>' yjr =^.r— '
(concentric with the pipe) have 1 ^^ ] ^=^"^^y^^^ir^z'- ^^ '
an area = F^ while the sectional '^'^'' ~^ -^8 1 -^^ ^^=— ^^=^ -~ '^ -'
\w-^
area of pipe = F^ . Beyond 7^, the fig. 579.
stream contracts to a section of area = OF = F^ , in enlarging
123456789 10 II 12 123456789 10 l( 12
Fig. 580c. — Chart, showing rates of flow during twenty-four hours.
The ordinates are gallons per hour.
a
.^
^
Fig. 5806. — Chart Recorder; an
attachment to Register. It re-
cords the rate of flow upon
a sheet of paper. (See above,
in Fig. 580c.)
Fig. 580a.- — ^The complete meter; consisting of a
Venturi Tube and a Register. They are con-
nected by two pressure-pipes, and the Register is
driven by weights. The narrowest part of Tube is
the "Throat."
THE VENTURI METER.
[To face page 724.
Fig. 580(?. — Mano-
meter; which may
replace the Re-
gister.
o ^
a
O
P^'
in which
l^ews for November 1887.)
Valve-gate.
Throttle-valve.
-
s
d
Z
a
C
5°
.24
1.0
.00
10°
.52
1
.07
15°
20°
.90
1.54
f
.26
25°
2.51
30°
3.91
i
.81
35°
6.22
40°
10.8
1
2.06
45°
18.7
50°
32.6
#
5.52
55°
58.8
f
17.00
60°
118.0
65°
256.0
i
97.8
70°
751.
-— ^.r ^ «>
Fig. 589.
Example 1. — Fig. 589. What head, = h, will be required
)» ((deliver ^ U. S. gallon (i.e. 231 cubic inches) per second
732 MECHANICS OF ENGHSTEERHSTG.
through the continuous line of pipe in the figure, containing tw^v
sizes of cylindrical pipe {d^ = 3 in., and d^ = 1 in,), and two
90° elbows in the larger. The flow is into the air at m, the
Jet being 1 in. in diameter, like the pipe. At £1, a = 90", and
the corners are not rounded ; at li^ also, corners not rounded.
Use the ft.-lb.-sec. system of units in which g — 32,2.
Since Q = i gal. = ^ • Yij-^-g = .0668 cub. ft. per sec, and
therefore the velocity of the jet
Vm = %= Q-^ 4>^(iV)' = 12.25 ft. per sec;
hence the velocity in the large pipe is to be v^ = {^Yv^ = 1.36
ft. per sec. From Bernoulli's Theorem, we have, with m as
datam plane,
involving six separate losses of head, for each of which there
is no difficulty in finding the proper C or/, since the velocities
and dimensions are all known, by consulting preceding para-
graphs. (Clean iron pipe.)
From § 515, table, for a = 90° we have . . . Ce = 0.505
" § 517, for 6^„= 3 in., and V, =1.36 ft. per sec,/, = .00725
« « « ^^ = 1 in., and V, ^12.25 " " /= .00613
« § 528 (elbows), for «r = 90° .... Q. =0.984
" § 527, for sudden diminution at K we have
[since ^,-T-i'; = r ^3^ = 0.111, .-. (7 = 0.625] .
= f_J_ - iV = 0.360.
\.625 J
Solving the above equation for h, then, and substituting
above numerical values (in ft.-lb.-scc.-system), we have (noting
that Vjn = -y,, and v^ = ^v^)
sK
= m?l [l + (iV (.505 + ^ X .00725 X 50 _^ ^ ^ ^^
64.4 L ^ T ^
+ .360+i><^i^^l!
IT -J
examples: with losses of head.
733
C.e.,
h, = ^^^- [l +(.00623 +.07160 + .0243)+(.360+ 5.8848~];
.-. A = 2.323 X T.3469 = 17.09 H.—Ans.
It is here noticeable how small are the losses of head in the
large pipe, the principal reason of this being that the velocity
in it is so small (v^ = only 1.36 ft. per sec), and that in gen-
eral losses of head depend on the square of the velocity
(nearly).
In other words, the large pipe approximates to being a reser-
voir in itself.
With no resistances a head A = vj' -f- 2^ = 2.32 ft. would be
sufficient.
Example 2. — Fig. 590. With the valve-gate F"half raised
(i.e., s = ^d in Fig. 587), required the volume delivered per
second through the clean pipe here shown. The jet issues
---^80
Fm. 590.
from a short straight pipe, or nozzle (of diameter d, = 1^ in.)
inserted in the end of the larger pipe, with the inner corners
not rounded. Dimensions as in figure. Eadius of each bend
= r = 2 in. The velocity v^ of the jet in the air = velocity
11, in the small pipe ; hence the loss of head at ^is
C^i f ^m
2(7 2^
Kow v^ is unknown, as yet ; but v^ , the velocity in the large
pipe, is ='y^pp ; i.e., «;« = ^^J« . From Bernoulli's The-
734 MECHANICS OF ENGINEEEING.
©rem (m as datum level) we obtain, after transposition,
^g 2g 2g ^g -" d,2g 2g
Of the coefficients concerned, y^ alone depends on the un-
known velocitj v^. For the present [first approximation],
put //= .006
From § 515, with a = 00°, C.b = .505
From § 517, valve-gate with s = ^d, ..... Cf = 2.06
From § 529, with a : r = 0.5, Cb = 0.294
While at IC, from § 527, having
(i^,:i^„) = (F: 2^ = ^^ = 0.562;
we find from table, , .' . C = 0.700
and .-. C^=(^- lj'= (0.428)'. . . . i.e., C^ = 0.183
Substituting in eq. (1) above, with v^^ = {r^Yv^^, we have
/ 1
in which the first radical, an abstract number, might be called
a coefficient of velocity, 0, for the whole delivery pipe ; and
), since in this case Q, = F„^v^ = F^v^ , may be
Q = jxF^ V2gA, it may be named a coefficient of effiux, jx.
Hence
V-'
.505 + 2.06 ^ 8 X .294 +
4 X .006x80"
+ .183
VS>«32.S>c^;
.% «j^= 0.421 V^h = 0.421 4/2 X 32.2 X 25 = 16.89 ft. per set.
(The .421 might be called a coefficient of velocity.) The
volume delivered per second is
Q = \nd*v^ = \nii-J 16.89 = .207 cub. ft. per sec.
(As the section of the jet F^ = F^ , that of the short pipe or
nozzle, we might also say that .421 = ju.=: coefficient of effiiix,
for we may write Q = fxF^ V2gh, whence jj. = .421.)
FLOW THEOITGH SIPH01«^S,
W
532. Siphons.— In Fig. 590a the part HN2O of the pipe is
above the level, BC^ of the surface of the water in the head
reservoir BL, and jet under proper conditions a steady flow
can be maintained with all parts of the pipe full of water, in
eluding ^iVg (7. if the atmosphere exerted no pressure, this
would be impossible; but its average tension of 14.7 lbs. pei
sq. inch is equivalent to an additional depth of nearly 34 feet
of water placed upon BC. With no flow, or a very small
velocity, the pipe may be kept full if iV^ is not more than
33 or 34 feefc above BG\ but the greater v^, the velocity of
flow at iV^2, and the greater and more numerous the losses
of head between L and i\^j, the less must be the height of JSf^
above BC for a steady flow.
The analytical criterion as to whether a flow can be main-
tained or not, supposing the pipe completely fllled at the out-
set, is that the internal pressure must be > at all parts of
the pipe. If on the supposition of a flow through a pipe of
given design the pressure ji? is found < 0, i.e. negative, at any
point [_W^ being the important section for test] the supposition
is inadmissible, and the design must be altered.
For example, Fig. 590a, suppose LN^N^ to be a long pipe of
uniform section (diameter = d^ and length = Z), and that under
the assumption ol filled,
sections we have com-
puted V4,, the velocity of
the jet at A^4; i.e., 2^4 =
^
l\^2gh.
1 + Cl+4/^
To test the snpposition, apply Bernoulli's Theorem to the
snrface jSCand the point iT^ where the pressure isj?, , velocity
vl^^^v^^ since we have supposed a uniform section for whole
pipe), and height above BC=h^. Also, let length of pipe
LNMN^ =■ L . Whence we have
^' + A, = + %=5)-fO-C.i^-4/A!^
2^
d 2^
(2)
[BC being datum plane.l
736
MECHAIS^ICS OF ENGUSTEERIN©,
ing for ^ , we have
y
We note, then, that tor jp^ to be > 0,
. . , (^
h^musthe <
Z4:feet
^9
'^9
i^)
In the practical working of a siphon it is found that atmos-
pheric air, previously dissolved in the water, gradually collects
at iVjjthe highest point, during the flow and idnallj, if not re-
moved, causes the latter to cease. See reference below.
One device for removing the air consists in first allowing it
to collect in a chamber in communication with the pipe be-
neath. This communication is closed by a stop-cock after the
water in it has been completely displaced by air. Another
stop-cock, above, being now opened, water is poored in to re-
place the air, which now escapes. Then the upper stop=cock is
shut and the lower one opened. The same operation is again
necessary, after some hours.
In Engineering News of Nov. 12, 1887, p. 346, is an account of a
siphon used in connection with the water-works at Kansas City. It
is 1350 ft. long and transmits
water from the river to the
artificial "well" from which
the pumping engines draw
their supply. At the highest
point, which is 16 ft. above
low-water level of the river,
is placed a "vacuum cham-
ber " in which the air collects
under a low tension corre-
sponding to the height, and a
pump is kept constantly at
work to remove the air and
The diameter of the
dips 5 ft. below the
-"^ ^2'
Fig. 590 b.
prevent the "breaking" of the (partial) vacuum
pipe is 24 in., and the extremity in the "well
level of low water. See p. 63 of vol. lix (Dec, 1907) of the Transac.
Am. Soc. Civil Engineers, for Mr. Anthony's paper on air in siphons.
532a. Branching Pipes.* — If the flow of water in a pipe is
caused to divide and pass into two others having a common
* Problems of this kind are best solved by tables or diagrams. See
pp. 197, etc., of the writer's Hydraulic Motors.
BEANCHING PIPES. 73'7
junction with the first, or vice versa ; or if lateral pipes lead
out of a main pipe, the problem presented may be very com-
plicated. As a comparatively simple instance, let us suppose
that a pipe of diameter d and length 7 leads out of a reservoir,
and at its extremity is joined to two others of diameters d^ and
d^ and lengthsZj and l^ respectively, and that the further extrem-
ities of the latter discharge into the air without nozzles under
aeads h^ and A, below the reservoir surface. Call these two
pipes ITos. 1 and 2. See Fig. 5906.
Assuming that all entrances and junctions are smoothly
rounded, so that all loss of head is due to skin-friction, it is re-
quired to find the three velocities of fiow, -y, v^^ and v^^ in the
respective pipes. First applying Bernoulli's Theorem to a
stream-line from the reservoir surface through the main pipe
to the jet at the discharging end of pipe No. 1, we have
2^-^^ ^-^ d'^ -^^'2^' • • • ^^>
and similarly, dealing with a stream-line through the main
pipe and ifo. 2,
while the equation of continuity for this case is
ikd'v^iTtd.^ + iTtd.X (3)
From these three equations, assuming/' the same in all pipes
as a first approximation, we can find the three velocities (best
by numerical trial, perhaps) ; and then the volume of discharge
of the system per unit of time
Q = i7td'v (4)
533. Time of Emptying Vertical Prismatic Vessels (or Inclined
Prisms if Bottom is Horizontal) under Variable Head.
Case I. Through an orifice or short pipe in the hottom and
opening into the air. — Fig. 591. As the upper free surface,
738 jmechanics of ei^ghneering.
of area = F\ sinks, F' remains constant. Let z = head ol
water at any stage of the emptying ; it = s^ at the outset, and
= when the vessel is empty. At any
instant, Q, the rate of discharge (= vol-
ume per time-unit) depends on z and is
/
/
A —
,'' 1
^^
j ■^
'-''
— _
^^ 1 z
—
^^
— >-- 1-
— -
'^ ... i
O F =
?
Q = },FV^yz, . . . (1)
)! where jx = coefficient of effiux r= (pC =■
Fig. 591. Coefficient of velocity X coefficient of con-
traction [see § 496, eqo (3)]. We here suppose F' so large
compared with F., the area of the orifice, that the free surface
of the water in the vessel does not acquire any notable velocity
at any stage, and that hence the rate of efflux is the same at
any instant, as for "a steady flow with head = z and a zero
velocity in the free surface. }x is considered constant. From
(1) we have
dY= (vol. discharged in time dt) = Qdt — fxF V^gz dt. . (2
But this is also equal to the volume of the horizontal laraina,
F'dzy through which the free surface has sunk in the same
time dt.
__ —F'
... —F'dz=^iAF^'lgzUt\ .% dt = ——=z-Hz. .(3)
"We have written minus F'dz because, dt being an increment,
dz is a decrement. To reduce the depth from z^ (at the start,
time = ^ = zero) to s„, demands a time
Lo ixF^/^gJz, }aI< V^g
whence, by putting z^ = 0, we have the time necessary to
empty the whole prism
_ 2F'zi _ ^F'z, _ 2 X volume of vessel ^ ,g.
~ fxFV2g ~ jxFV^o ~ ^°^^^^^ ^^*® ^^ discharge '
TIME OF EMPTYING VESSELS.
739
that is, to empty the vessel requires douMe the time of dis-
charging the same amount of water if the vessel had been kept
full (at constant head = 2^ = altitude of prism).
To Jill the same vessel through an orifice in the bottom, the
flow through which is supplied from a
body of water of infinite extent hori-
zontally, as with the (single) canal lock
of Fig. 592, will obviously require the
same time as given in eq. (5) above,
since the efiective head z diminishes
from s„ to 0, according to the same law.
Example. — What time will be needed
to empty a parallelopipedical tank (Fig. 591) 4 ft. by 5 ft. in
horizontal section and 6 ft. deep, through a stop-cock in the
bottom whose coeflicient of efflux when fully open is known
to be /f = 0.640, and whose section of discharge is a circle of
diameter =: -^ in. ? From given dimensions i^^' = 4 X 5 = 20
sq. ft., while So = ^ f^- Sence from eq. (5) (ft.-lb.-sec.)
Fig. 593.
time of ] 2 X 20 X 1/6
emptying] 0.64 X ^^tiiif V2 X 32.2 '
13980 seconds
__ Qhours K Qmin. n sec.
Case II. Two communicathig pynsmatic 've,<:iels. Required
the time for the water to come to a common level ON, Fig.
^ 593, efflux taking place through a small
^ '' ° orifice, of area = F, u Jder water. At
any instant the rate of discharge is
r a; —
i o
— -r^ Cz.
Q = }xF'
as before, z = difference of level. ]S^ow
if F' andi^'' are the horizontal sectional
areas of the two prismatic vessels (axes
vertical) we have F'x = F'^y, and hence z, which = £0-\-y,
= x-^{F'~F")x',
Fig. 593.
X =
1 + ^77
and dx =
F' '
1 + 1^7
MECHANICS OF ENGINEERIISTG.
As before, we have
r- F'F" z'^dz
— F'dx = }xF V2g zHt, or dt— — - , , „,. —^7^ '
Hence, integrating, the time for the difference of level to
change from z^ to z^
2F'F" zh - zj
F' -^F"' }jiFV¥g'
and by making s„ = in (6), we have the
time of coming to a common ievei =■ -nn ^jpn ' ~^
(6)
tt^aZ-S:- i^)
2^
F'
1— -rfl
— -P
O .
---- ^
Algebraic Example. — In the double lock in Fig. 594, let
Z' be full, while in F' the water stands at a level MJV the
same as that of the tail-
water. F' and F" are the
horizontal sectional areas of
the prismatic locks. Let
the orifice, 0, between
them, be at a depth = /\
below the initial level FF
of F, and a height = h^
above that, J/i\^,"of F'.
The orifice at 0, area = F, being opened, efflux from F be-
gins into the air, and the level of F' is gradually raised from
MN to OD, while that of F sinks from KF to AB a distance
= a, computed from the relation vol. F'a = vol. F"h^^ and
the time occupied is [eq. (4)]
Fig. 594.
t.=
2F'
fjiFV^g
= iVh,- Vh, — »]•
(8)
As soon as is submerged, efflux takes place under water, and
we have an instance of Case 11. Hence the time of reaching
a common level (after submersion of 0) (see eq. Y) is
1 =
%F'F'
A.
}xF{F'+F")'y 2g
and the total time i^ = t^-\-t^, with a = F^'h^ -r- F'.
(9)
TIME OF EMPTYING VESSELS.
741
Case III. Emptying a vertical prismatic
rectangular '•''notch'''' in the side, or over-
fall. — Fig. 595. As before, let even the
initial area (= s/) of the notch be small
compared with the horizontal area F' of
tank. Let z = depth of lower sill of notch
below level of tank surface at any instant,
and b = width of notch. Then, at any in-
stant (see eq. 10, § 504),
through a
Fie. 595.
Rate ofdisch. (vol.) = Q — ^fxjbz V^z = ^fjb V2gsK
,\ vol. of disch. in dt = |yu5 V2g B^dt,
and putting this = — F^dz = vol. of water lost by the tank
in time dt, we have
^^ 3 F' .,
whence
2>u5 V2g
3 F'
2>u5 V2gJzo
'Udz = -l ^'
^ -"in'
2yM&|/2^U — t
i.a.
n=J^r^__Ll,.. . . (10)
Lo uhV^qLVzr. Vz„J
as the time in which the tank surface sinks from a height z^
above sill to a height z^ above sill. If we inquire the time t'
for the water to sink to the level of the sill of the notch we
put Zn = zero, whence t' = infinity. As explanatory of this
result, note that as z diminishes not only does the velocity of
flow diminish, but the available area of efflux (= zl)) also grows
less, whereas in Cases I and II the orifice of efflux remained
of constant area = F.
Eq. (10) is applicable to the waste- weir of a large reservoir
or pond.
534, Time of Emptying Vessels of Variable Horizontal Sec-
tions. — Considering regular geometrical forms first, let us take
742
MECHANICS OF EISTGIWEEIJING.
Fig. 59e.
Case I, Wedge-shaped vessel, edge horizontal and under-
neath, orifice F in the edge, so that
^5 the variable head, is always the
altitude of a triangle similar to the
section ^^C'of the body of water
when efflux begins. At any instant
during the efflux the area, 8^ of
the free surface, variable here,
takes the place of F' in eq. (3) ol
§ 633, whence we have,
Ss-hdz
for any case of 'oariable free surface^ dt = z=^. . (11)
In the present case S = ul, and from similar triangles
u : s :: h : B^l
whence
and
^ = his -r- 3„
dt-
U
— tlz^dz ,
Lo
z'^dz =
\9_^^fJ, .(12)
liFz^ V^gJ^^ }JiFz^ V2g
and hence the time of emptying the whole wedge^ putting
s„ = 0, is
__ 4 i-^^So _ 4 '^ol. of wedge
° ^ uF V'iqz 3 initial rate of discharge
(13)
i.e., |- as long as to discharge the same volume of water under
a constant head = s„ . This is equally true if the ends of the
wedge are oblique, so long as they are parallel.
Case II. Right segment of jjaraholoid of
revolution. — Fig. 597. Axis vertical. Ori-
fice at vertex. Here the variable free surface
has at any instant an area, -=8,-= nu^, u be-
ing the radius of the circle and variable.
From a property of the parabola
i*' : 5' :: s : s„ ; /. 8= nVz — s^^ Fra.sgr.
TIME OF EMPTYING VESSELS. 743
saaA hence, from eq. (11),
dt =
IxFz, \f¥g '
3 jxFz, V2g
whence, putting s^ = 0, we have the time of emptying the
whole vessel
_ 4 Tth^-k^^ _ 4: total vol. ,^..
° 3jjiFV2gSg 3 initial rate of disch/ '
same result as for the wedge, in Case I ; in fact, it applies to
any vessel in which the at'eas of horizontal sections vary
directly with their heights ahove the orifice.
Case III. Any jpyrainid or cone ', vertex down j small ori'
fice in vertex. — Fier. 598. Let area of the ^ ^
base = /iS'o , at upper edge of vessel. At f^^CJ ^^^^^^^
any stage of the flow S = area of base of / T^^T^y i
pyramid of water. From similar pyra- s^^^^^^ ^»
mids ^^^ t I
and [eq. (11)]
R 1
So fxFV^g
whence (s^ = 0) the time of emptying the whole vessel is
__ 6 Total volume .. .
' " 6 initial i^ate of disch.
Case IY. Sphere. — Similarly, we may show that to empty
744
MECHANICS OF El^GINEERHSTG.
a sphere, of radius = r, througli a small orifice, of area = F^
in lowest part, the necessary time is
1 = 1"
7cr
8
Vol.
15 uFVgr ^ init. rate of disoh.
635. Time of Emptying an Obelisk-shaped Vessel. — (An obe-
lisk may be defined as a solid of six plane faces, two of which
are rectangles in parallel planes and with sides respectively
parallel, the others trapezoids; a frustum of a pyramid is a
particular case.)
A volume of this shape is of common occurrence; see Fig.
599. Let the altitude = A, the two rectangular faces being
horizontal, with dimensions as in figure. By drawing through
F, G, and H right lines par-
^ ~~~~~ allel to EC, to cut the upper
base, we form a rectangle
KLMG equal to the lower
base. Produce ML to P and
KL to N, and join PG and
NG. We have now sub-
divided the solid into a paral-
lelopiped KLMG- EHGF,
a pyramid PBNL - G, and
two wedges, viz. APLK-HG and LNDM-FG, with
their edges horizontal ; and may obtain the time necessary to
empty the whole obelisk-volume by adding the times which
would be necessary to empty the individual component vol-
umes, separately, through the same orifice or pipe in the bot-
tom plane EG. These have been already determined in the
preceding paragraphs. The dimensions of each component
volume may be expressed in terms of those of the obelisk, and
all have a common altitude = h.
Assuming the orifice to be in the bottom, or else that the
discharging end of the pipe, if such is used, is in the plane of
the bottom EG, we have as follows, F being the area of dis-
charge :
Fig. 599.
TIME OF EMPTYIISTG RESEEVOIES. 745
Time to empty the pai'allelopijyed] ^ __?^A_ rr
separately would he {Case /, § 533) ) ' ' ' ~ uF V^q
Time to empty the two ) o7,/7 i \ \ i ri. i \
wedges separately W, = - . ^A^ ~ ^') + ^A^ ~ ^) ^^/r- ,„-,
For the pyramid} ^ _2 (^ — ^Q (^ — ^i) , ,^
{Case III, § 534) f " ' * '^ " 5 ' ^F V^ Vh. . . {^)
Hence to emptj the whole reservoir we have
i.e.,
t = [SU + 83,^, + 2U, + 2hJ] - ^ ^^ . . . (4)
Example. — Let a reservoir of above form, and with h = 50 ft.,
1=60 ft., h, = 10 ft., I, = 20 ft., and depth of water h = 16
ft., be emptied through a straight iron pipe, horizontal, and
leaving the side of the reservoir dose to the bottom, at an angle
a = 36° with the inner plane of side. The pipe is 80 ft. long
and 4 inches in internal diameter; and of clean surface. The
jet issues directly from this pipe into the air, and hence
F= l7t{^y sq. feet. To find ju, the " coefficient of efflux"
{= 0, the coefiacient of velocity in this case, since there is no
contraction at discharge orifice), we use eq. (4) (the first radical)
of § 518, withy approx. = .006, and obtain
(ISr.B. Since the velocity in the pipe diminishes from a
value
V = .361 V2g X 16 = 11.6 ft. per sec.
at the beginning of the flow to v = zero at the close, /"= .006
is a reasonably . approximate average with which to compute
the average above ; see § 517.
746
MECHANICS OF ENGIIfEEEING.
Hence from eq. (4) of this paragraph (ft.-lb.-sec. system)
_ [3 X 50 X 60 4- 8 X 10 X 20+2(50 X 20+10 X 60)]2 VT&
15x0.361X^(i)V2x32.2
8 hrs. 5 min. | ^'<^^^^^J the ^rath. ""' ^^ ""^
t =
= 29110 sec. =
536. Time of Emptying Reservoirs of Irregular Shape. Simp-
son's Rule. — From eq, (11), § 534, we have, for the time in
which the free surface of water in a vessel of any shape what-
ever sinks through a vertical distance =6?s,
at = , whence
time
j^FV^g
— / Sz-Hz, . . (1)
where S is the variable area of the free surface at any in-
stant, and z the head of water at the same instant, efflux
proceeding through a small orifice (or extremity of pipe) of
area ^ F. li 8 can be expressed in terms of s, we can in-
tegrate eq. (1) (i.e., provided that Sz-^ has a known anti-
derivative) ; but if not, the vessel or reservoir being irregular
in form, as in Fig. 600 (which shows a pond whose bottom
has been accurately surveyed, so that we know the value of S
for any stage of the emptying), we can still get an approximate
solution by using Simpson's
Rule for approximate inte-
gration. Accordingly, if we
inquire the time in which
the surface will sink from
to the entrance Fot the pipe
in Fig. 600 (any point oi ; at
E, or short of that), we
divide the vertical distance
from to n (4 in this figure) into an even number of equal
parts, and from the known form of the pond compute the area
aS' corresponding to each point of division, calling them S^, S^,
etc. Then th§ required time is approximately
Fig. 600.
l-o
TIME OF EMPTYIISTG POND.
^n — 1
747
n. — 1
*4 71—2 "^Jl .J
Example. — Fig. 600. Suppose we have a pipe ^vi of the
same design as in the example of § 535, and an initial head of
^0 = 16 ft., so that the same value of pt, = .361, may be used.
Let Zn — s„ = 8 feet, and divide this interval (of 8 ft.) into
four equal vertical spaces of 2 ft. each. If at the respective
points of division we find from a previous survey that S\, =
400000 sq. ft., S, = 320000 sq. ft., S, = 270000 sq. ft., S, =
210000 sq. ft., and S, = 180000 sq. ft. ; while n = 4, // ^ .361,
and the area ^ = ■kT^HT = .0873 sq. ft., we obtain (ft., lb., sec.)
16-
Lo 0.361 X .0873 V2 X 32.2 X 3 X 4:
"400000 , 4 X 320000
L 1/16
4/14
2 X 270000 , 4 X 210000 , 180000n
4/12
4/10
Vs _
= 2444000 sec.
= 28<^- 6^- 53™- 20^
The volume discharged, V, may also be found by Simpson's
Kule, thus : Since each infinitely small horizontal lamina has
a volume
dV =
or, approximately.
Sds,
n PO
^0
v=
'S'n ^n
S,Jr^S,-{-'2S, + 4.S,-{-...+Sn
_o Sn
Hence with n = 4 we have (ft., lb., sec.)
16 — 8r-^^^nA I A \ 320000, , o vy o^onAA
400000 + 4 j I 210000 f + ^ >< 270000
M) 3X4
+
+ 180000
= 2,160,000 cub. ft.
748
MECHANICS OF ENGINEERING.
537. Volume of Irregular Reservoir Determined by Observing-
Progress of Emptying. — Transforming eq. (11), § 534, we have
8dz = — piFV^jzUt.
But Sdz is the infinitely small volume dV oi water lost bj
t!]^ reservoir in the time dt^ so that the volume of the reser-
voir between the initial and final (0 and n) positions of the
horizontal free surface (at beginning and end of the time t^
may be written
L-O
Y^fxFV^g
z^dt.
(1)
This can be integrated approximately by Simpson's Eule, if
the whole thne of emptying, = z;^, be divided into an even
.:-:..-.-:.. number of equal
parts, and the values
^„ , Sj , 02 , etc., of the
head of water noted
at these equal inter-
vols of time (not of
vertical height). The
corresponding sur-
face planes will not
Whence for the particular case
FiQ. 601.
be equidistant, in general,
when 71 = 4 (see Fig. 601)
y = ^^^y{i^:~^^ [^0^ + 4: (3.4 + ^^) + 2a,* + ^n- • • (2)
-0 O X 4:
537a. Time of Change of Surface-level of Reservoir when In
fiow Exists as well as Outflow. — Solutions of problems of this
nature involve somewhat more extended mathematics than
the foregoing. Eesults andformulse applicable to a number
of such cases Avill be found on pp. 147-1 55 of Mr. Horton's
monograph mentioned in the foot-note of p. 688. Cases of
both constant and variable inflow are treated in that paper.
Sfte also Engineering News of Nov. 14, 1901, pjD, 362, 363 ;
and Dec. 5, 1901, p. 431.
CHAPTEK yil.
HYDROKINETICS (CowfoVi^ec?)— STEADY FLOW OF WATER IN
OPEN CHANNELS.
538. Nomenclature. — Fig. 602. When water flows in an
open channel, as in rivers, canals, mill-races, water-courses,
ditches, etc., the bed
and banks being rigid,
the upper surface is
free to conform in
shape to the dynamic
conditions of each case,
which therefore regu-
late to that extent the
shape of the cross-sec-
tion- ^
In the vertical trans- ^'®- ^^■
verse section AC m figure, the line AC is "called the air-profile
(usually to be considered horizontal and straight), while the
line ABC, or profile of the bed and banks, is called the wetted
perimeter. It is evident that the ratio of the wetted perimeter
to the whole perimeter, though never < -J, varies with the
form of the transverse section.
In a longitudinal section of the stream, JEFGH, the angle
made by a surface filament ^T^ with the horizontal is called
the slope, and is measured by the ratio s = h : I, where I is the
length of a portion of the filament and h = thefiall, or vertical
distance between the two ends of that length. The angle be-
tween the horizontal and the line HG along the bottom is not
necessarily equal to that of the surface, unless the portion of
the stream forms a prism ; i.e., the slope of the bed does not
necessarily = 5 = that of surface.
Examples. — The old Croton Aqueduct has a slope of 1.10
ft. per mile; i.e., s = .000208. The new aqueduct (for !N^ew
749
760
MECHAlSriCS OF EISTGINEEEING.
York) has a slope s = .000132, with a larger transverse section,
Eor large sluggish rivers s is much smaller,
539. Velocity Measurements. — Yarions instruments and
methods may be employed for this object, some of which are
the following :
Surface-floats are small balls, or pieces of wood, etc., so
colored and weighted as to be readily seen, and still but little
affected by the wind. These are allowed to float with the cur-
rent in different parts of the width of the stream, and the sur-
face velocity c in each experiment computed from c ^^l -^ t,
where I is the distance described between parallel transverse
alignments (or actual ropes where possible), whose distance
apart is measured on the bank, iand t = the time occupied.
Douhle-floats. Two balls (or small kegs) of same bulk and
condition of surface, one lighter, the other heavier than water,
,/-:; /^ . a^'G united by a slender chain, their
weights being so adjusted that the
light ball, without projecting notably
above the surface, buoys the other
ball at any assigned depth. Fig. 603.
It is assumed that the combination
moves with a velocity c', equal to the
arithmetic mean of the surface veloc-
ity c„ of the stream and that, c, of the water filaments at the
depth of the lower ball, which latter, c, is generally less than
Cj . That is, we have
Fig. 603.
c' =z ^{c^ -j- G) and ,-. c = 2c' — c.
(1)
Hence, g„ having been previously obtained, eq. (1) gives the
velocity g at any depth of the lower ball, c' being observed.
T\iQ floating stafl' is a hollow cylindrical rod, of adjustable
length, weighted to float upright with the top just visible. Its
observed velocity is assumed to be an average of the velocities
of all the filaments lying between the ends of the rod.
Woltinann^s Mill / or Tachometer / or Ourrent-7neter, Fig.
604, consists of a small wheel with inclined floats (or of a small
FxG. 604a.— The Buff and Buff Current Meter. Held at the end
of a pole. 5-inch wheel.
[To face page 750.
go;
CURRENT-METERS, ETC.
751
Fig. 604.
"screw-propeller" wheel S) held with its plane 1 to the car
rent, which causes it to re-
volve at a speed nearly pro-
portional to the velocity, g,
of the water passing it.
By a screw-gearing W on
the shaft, connection is
made with a counting ap-
paratus to record the num-
ber of revolutions. Some-
times a vane i> is attached,
to compel the wheel to face
the current. It is either
held at the extremity of a pole; or suspended by a weighted
cable, for work in deep water. In some types the counting
device is actuated by electric connection with the revolving
wheel, and hence may be placed in a boat or on shore. With
the screw-gearing, a cord and spring are used for throwing
in and out of gear.
On the opposite sheet are shown four varieties of current
meter: the Buff and Buff meter, for attachment to a pole;
the Haskell meter; the Ritchie-Haskell direction-current
meter (see below) ; and the Price meter. This last resembles
the anemometer of p. 824.
A special form of this instrument has been recently in-
vented, called the Ritchie-Haskell Direction-current Meter,
possessing the following special features: "This meter registers
electrically on dials in boat ........
from which used, the dii'eGtion
and velocity^ simultaneously,
of any current. Can be used
in river, harbor, or ocean cur-
rents."
Bitot'' s Tube * consists in
principle of a vertical tube
open above, while its lower
end, also open, is bent hori-
zontally up-stream; see A in
figure.
-•^r ■''•■:--''
=
s
;'•
/')■'-
■ z-^-iri
=
^^=
=
—
^^^S
--^-^13
i
3
m^=
Fig. 605.
After the oscillations have ceased, the water in the
* A description of Mr. W. M. White's extensive experiments with this
instrument will be found in the Journal of the Assoc, of Engineering Societies
for Aug. 1901, p. 35 (Vol. xxvii. No. 3).
752 MECHAITICS OF en^gineerhstg.
tube remains stationary with its free surface a Leiglit, li\
above that of the stream, on account of the continuous im-
pact oj£ the current against the lower end of the column.
By the addition of another vertical tube (see B in figure)
with the face of its lower (open) end parallel to the current
(so that the water-level in it is the same as that of the cur-
rent), both tubes being provided with stop- cocks, we may,
after closing the stop-cocks, lift the apparatus into a boat
and read off the height Ji' at leisure. We may also cause
both columns of water to mount, through flexible tubes,
into convenient tubes in the boat by putting the upper ends
of both tubes in communication with a receiver of rarefied
air, and thus watch the oscillations and obtain a more
accurate value of h'. As to a theoretical relation between
the velocity c, of the current and the height h' , we have,
[from eq. (7a), p. 804], writing k' for I-hVT, {k' to be found
by experiment)
c = k'\/2^' (1)
The form and position of the "static opening," m, may
be such as to cause (by ''suction") the water level in the
corresponding tube to stand lower than that of the outside
surface; thus affecting the observed h'. This opening is
frequently made in the side- of the tube, or of a thin casing.
Hence each instrument requires a separate rating, for k'.
Mr. W. M. White's experiments show, in case the " impact
tube" is used alone (see A, Fig. 605) that so long as it is a
solid of revolution with its axis exactly on a line of the cur-
rent, the form of the "impact opening," or point, is imma-
terial, and that the value of k' is practically 1.00, (with c
ranging from 3 to 15 ft. per sec); but that when the "static
tube" is used {B, Fig. 605) ¥ may be considerably less
than 1.00. (See also the experiments of Profs. Boyd and
Judd, Engineering News, Mar., 1904, p. 318.)
The Bitot tube is also used for measuring the velocity
of water in free jets, and flowing under pressure in pipes
(see p. 833). In closed pipes conditions are more complicated
than in an open channel, and the instrument itself forms
more of an obstruction.
CURRENT-METERS.
753
Rating of current meters. — The relation between the velocity c in ft. per
second, of the current, and the revolutions per second, n, made by the
wheel, is usually taken as c = c^ + b .n, where Cq is the value of c below
which the wheel (through friction) does not turn at all, and b is another
constant. The determination of these two constants by experiments
is called the "rating" of the meter and is usually accomplished by
moving it uniformly through still water; the instrument being sup-
ported from the bow of a boat which is towed over the course, or from
the overhanging arm of a truck moved along the edge of a canal. The
total time t, total distance s, and total number of revolutions N, made
by the wheel, are recorded. Then c=^s^t and n = N-^t.
If a number of such experiments or "runs" have been made, cover-
ing a fair range of velocities, the most probable values of the two con-
stants, Cq and b, may be found (according to the theory of "least
squares"), by the following formulae, where m denotes the number of
"runs:"
_ mI(nc)-I(n)I{c) , I (c) I (n') - 1 (n) I (nc)
ml{n')-[l{n)f ' ^ » m2 (n') ~[I (n)f '
Example. — In the table below are placed the data of a set of eight
test runs for a current meter, over a course of 200 ft. length; and also
the other quantities needed for computing b and Cq from the above formulae.
No.
N
Total
Revs.
« = Total
time (sec.)
c
ft. /sec.
n
revs. /sec.
nc
nJ
1
118
396.25
0.505
0.298
0.150
0.088
2
158
214.75
0.931
0.736
0.685
0.542
3
170
143.75
1.391
1.183
1.645
1.390
4
177
79.00
2.532
2.241
5.672
5.022
5
180
58.00
3.448
3.104
10.702
9.635
6
184
41.50
4.819
4.434
21.367
19.660
7
187
33.00
6.061
5.667
34.343
32.115
8
187
27.50
7.273
6.800
49.455
46 . 240
Sums .
26.960
24.463
124.019
114.701
Also
[2'(n)P = 598.41
8X124.019-24.463X26.960 332.35
and
8X114.701-598.41 319.21
26.960X114.701 - 24.463X124.019 58.40
1.042,
= 0.183
" 8X114.701-598.41 319.2
For this meter, therefore, we have c = 0.183 + 1.042tc for the inter-
pretation of its indications when in actual service; and it is seen that
the wheel makes about one revolution for each foot of current move-
ment.
540. Velocities in Different Parts of a Transverse Section. —
The results of velocity-measurements made by many experi-
menters do not agree in supporting any very definite relation
between the greatest surface velocity (\ ~= ing relation, y^' being a number dependent
■^1' ~ i ' 2J -=^ \ ' Sr-^ on the force of the wind (from for no
— 1-~ I »j' =^ wind to 10 for a hurricane) :
Z^i^i^/'^S 6^, = [0.31Y±0.06/'']^; . . (3)
~ I — — */__~ where d is the total depth, and the double
^^^^m^ sign is to be taken + for an up stream, -
Fig. 607. for a dowu-strcam, wind. Tlie following
relations were also based on the results of the survey :
(putting, for brevity, B = 1.69 -=- |/c?+ 1.5,) . (^)
VELOCITIES l]Sr OPHN CHANNEL.
765
C = Ca, — VBv
'z — d^
IT
(5)
^m = fCd. + K + ^ (i^o - |Cd), . .
and
Cid = Cm-\--h ^''^•
. . (6)
. . (7)
{These equations are not of homogeneous form^ but call for
the/bo^ and second as units.)
In (5), (6), and (7),
c = velocity at any depth s below the surface ;
c^ = mean velocity in the vertical curve ;
Ca, = max. " " " "
C|(j = " at mid-depth ;
c^ = velocity at bottom ;
V = mean velocity of the whole transverse section.
Flow under Ice. — In current observations made for the TJ. S. Govern-
ment in 1897 by Assistant Engineer E. E. Haskell, C.E., (now Director
of the College of Civil Engineering at Cornell University) on a section
of the St. Mary's River (near Sault Ste. Marie, Michigan) when frozen
over, it was found that in the mean vertical curve of the whole section
(mean of 22 curves; involving 220 separate velocity-observations) the
maximum velocity was 1.250 (occurring at 0.4 the depth from surface)
and the mean, v, 1.087, ft. /sec. The velocity at mid-depth, Cq.s, in
this mean curve was 1.232 ft. /sec, and the ratio v^c^.^ was 0.882.
The friction due to the ice was found to be very nearly 31 per cent of
that due to the bottom. (See report of the Chief of Engineers of U. S.
Army for 1897, Part 6, p. 4100.)
541. Gauging a Stream or River.* — ^Where the relation (eq. (2),
§ 540) V = .83 (
756
MECHANICS OF ENGINEEEING.
With a small stream or ditch, however, we may erect a ver-
tical boarding, and allow
the water to flow through a
rectangular notch or over-
fall, Fig. 609, and after the
head surface has become
permanent, measure h^
(depth of sill below the
level surface somewhat
back of boards), and h
(width) and use the formu-
lae of § 504; see examples
Fig.
in that article.
543. Uniform Motion in an Open Channel. — We shall now
consider a straight stream of indefinite length in which the
flow is steady^ i.e., a state of perTnanency exists, as distin-
guished from a freshet or a wave. That is, the flow is steady
when the water assumes fixed values of mean velocity v. and
sectional area F, on passing a given point of the bed or bank ;
and the
Eq. of continuity . . Q = Fv = F^v^ = F^i\ = constant . . (1)
holds good whether those sections are equal or not.
By uniform motion is meant that (the section of the bed
and banks being of constant size and shape) the slope of the
bed, the quantity of water (volume = Q) flowing per time-
unit, and the extent of the wetted perimeter, are so adjusted
to each other that the mean velocity of flow is the same in all
transverse sections, and consequently the area and shape of the
transverse section is the same at all points ; and the slope of
the surface = that of the bed. We may therefore consider,
for simplicity, that we have to deal with a prism of water of
indefinite length sliding down an inclined rough bed of con-
stant slope and moving with uniform velocity (viz., the mean
velocity v common to all the sections) ; that is, there is no ac-
celeration. Let Fig. 610 show, free, a portion of this prism,
of length = I, and having its bases 1 to the bed and surface.
UNIFOEM MOTION. OPEN CHANNEL. 757"
*S^ hydrostatic pressures at the two ends balance each other
from the identity of conditions. The only other forces having
LL-i-jr-,
FiQ. 610.
components parallel to the bed and surface are the weight
G = Fly of the prism (where y = heaviness of water) making
An angle = 5 (= slope) with a normal to the surface, and the
friction between the water and the bed which is parallel to the
■surface. The amount of this friction for the prism in question
may be expressed as in § 510, viz.:
P=fric.=f8y^=fwly^, ... (2)
in which 8 = wl = rubbing surface (area) = wetted perimeter,
Wj X length (see § 538), and /"an abstract number. Since the
mass of water in Fig. ,610 is supposed to be in relative equili-
brium, we may apply to it the laws of motion of a rigid body,
and since the motion is a uniform translation (§ 109) the com-
j^onents, parallel to the surface, of all the forces must balance.
.*. G sin s must = P =friG. ; .•. Fly y =.fwly -— ;
Jphence
or
^=/^-f ' (^)
in which F-^w \b called B, the hydraulic mean dejpth, or
hydraulic radius. (3) is sometimes expressed by saying that
758
MECHANICS OF EI^GI]S^EEEI]SI G.
the whole fall, or head, A, is (in uniform motion) absorbed in
friction-head. Also, since the slope s = h-rrljwe have
V =
v = A VEs,
(4)
which is of the same form as Chezy's formula in § 519 for a
very long straight pipe (the slope s of the actual surface in this
case corresponding to the slope along piezometer-summits in
that of a closed pipe). In (4) the coefficient A = V2g -^f is
not, like/", an abstract number, but its numerical value depends
on the system of units employed.*
542a. Experiments on the Flow of Water in Open Channels. —
Those of Darcy and Bazin, begun in 1855 and published in
1865 (" Recherches Hydrauliques"), were very carefully con-
ducted with open conduits of a variety of shapes, sizes, slopes,
and character of surface. In most of these a uniform flow was
secured before the taking of measurements. The velocities
ranged between from about 0.5 to 8 or 10 ft. per second, the
hydraulic radii from 0.03 to 3.0 ft., with deliveries as high as
182 cub. ft. per second. For example, the following results
were obtained in the canals of Marseilles and Oraponne, the
quantity A being for the foot and second. The sections were
nearly all rectangular. See eq. (4) above.
No.
(cub. ft.
R.
s.
abs.
V.
(ft. per
A.
(foot and
Character of the masonry
per sec.)
(ft.)
numb.
sec)
sec.)
1
183.73
1.504
.0037
10.26
137.1
Tery smooth.
2
143.74
1.774
.00084
5.55
125.
Quite "
3
43.93
.708
.029
11.23
78.4
4
43.93
.615
.060
13.93
73.5
Hammered stone.
5
43.93
.881
.0121
7.58
73.5
Rather rough.
6
43.93
.835
.014
8.36
77.3
7
167.68
2.871
. 00043
3.54
73.3
Mud and vegetation.
[In Experiment !N"o. 1 the flow had not fully reached a state
of permanency.]
Fteley and Stearns's experiments on the Sudbury conduit at
Boston, Mass. {Trans. A. 8. O. E., '83), from 18Y8 to 1880, are
also valuable. This open channel was of brick masonry with
* Values of this coeflBcient A, for the English foot and second as units,
may be obtained from diagrams in the Appendix of the Author's "Hydraulic
Motors." They are based on Kutter's Formula (see next paragraph).
uisriFOEM MOTION", kuttek's foemula. 759
good mortar joints, and about 9 ft. wide; the depths of water
ranging from 1.5 to 4.5 ft. With plaster of pure cement on
the bed in one of the experiments the high value of ^ = 153.6
was reached (foot and second), with v = 2.805 ft. per second,
R = 2.111 ft., .9 = .0001580, and Q = 87.17 cu. ft. per second.
Captain Cunningham, in his experiments on the Ganges
Canal at Roorkee, India, in 1881, found A to range from 48
to 130 (foot and second).
Humphreys and Abbot's experiments on the Mississippi
River and branches (see § 540), with values of i? = from 2 or
3 ft. to 72 ft., furnish values oi A =■ from 53 to 167 (foot and
second).
542b. I^utter's Formula. — The experiments upon which
Weisbach based his deductions for/", the coefficient of fluid
friction, were scanty and on too small a scale to warrant gener-
al conclusions. That author considered that / depended only
on the velocity, disregarding altogether the degree of rough-
ness of the bed, and gave a table of values in accordance with
that view, these values ranging from .0075 for 15 ft. per sec.
to .0109 for 0.4 ft. per sec; but in 1869 Messrs. Kutter and
Ganguillet, having a much wider range of experimental data
at command, including those of Darcy and Bazin, and those
obtained on the Mississippi River, evolved a formula, known
as JTuttef'^s Formula^ for the uniform motion of water in open
channels, which is claimed to harmonize in a fairly satisfactory
manner the chief results of the best experiments in that direc-
tion. They make the coefficient A in eq. (4) (or rather the
factor contained in A^ a function of R. s, and also n an
abstract number, or coefficient of roughness, depending on the
nature of the surface of the bed and banks ; viz.,
V in
ft.
per
sec.
4L6 + l:^ + :5^
i+(«.6 + £2pi)
VR[mii.)Xs,..{b)
s I 4/^(infeej)_
wnich is Kutter^ s Formula.* The bracket is the A of (4).
* A book of "Diagrams of Mean Velocity based on Eutier's Formula," by
tlie present writer (New York, J. Wiley & Sons, 1902), obviates the necessity
of numerical substitution in Kutter's formula for all v'r!<^tical purposes
760 MECiiAisrics of enghsteeeing.
That is, comparing (5) with (4), we have /*a function of n,
H, and s, as follows :
f=
1 +
r^^ _. 00281 n
S
n
VJi in ft.
5 181 1 '^^^^ 1
.00035
n
s
• (6)
From (6) it appears that f decreases with an increasing H,
as has been also noted in the case of closed pipes (§ 51Y) ; that
it increases with increasing roughness of surface ; and that it
is somewhat dependent on the slope. Values of n may be
taken as follows: •
.009 for well planed timber evenly laid;
.010; plaster in pure cement; glazed surfaces in good order;
.011; plaster in cement with one-third sand; iron and cement pipes in
good order and well laid ;
.012; unplaned timber, evenly laid and continuous.
.013; ashlar masonry and well laid brick work;* also the above cate-
gories when not in good condition nor well laid;
.015; "canvas lining on frames"; brick-work of rough surface; foul
iron pipes; badly jointed cement pipes;
.017; rubble in plaster or cement in good order; inferior brick- work;
tuberculated iron pipes; very fine and rammed gravel;
.020; canals in very firm gravel; rubble in inferior condition; earth of
even surface;
.025; canals and rivers in perfect order and regimen and perfectly free
from stones and weeds ; f
.030; canals and rivers in earth in moderately good order and regimen,
having stones and weeds occasionally;
.035; canals and rivers in bad order and regimen, overgrown with vege-
tation, and strewn with stones and detritus.
Kutter's formula finds wide acceptance among engineers.
To save computation, values of A (for the English foot and
second) may be taken from the following table. The slopes
.010, .001, .0004, .0002, and .00005 are indicated as 10,
1, .4, .2, and .05, ( = 1000s); and may be read "fall of 10 ft.
per thousand ft," etc. Ordinary rules of interpolation
apply ; but it must he remembered that for any slope greater
than .010 the value of A is practically the same as for that
slope (s = .010; or 10 ft. per thousand ft.)
* For ordinary brick sewers Mr. R. F. Hartford claims that n=.014
gives good results. See Jour. Eng. Societies for '84-'85, p. 220.
t See Engineering News for Feb. 1, 1908, p. 122, for experiments with an
open channel; giving n = 0.025.
UNIFOEM MOTION. OPEN CHANNELS.
761
TABLE* OF KUTTER'S COEFFICIENT A.
[See eqs. (4) and (5).]
1000 S
Values of n
R
in ft.
.009
.010
.011
.012
.013
.015
.017
.020
.025 .030
10
129
114
100
90
81
67
57
46
33 27
1
128
113
99
. 88
80
66
56
45
33 27
0.2 ■
.4
125
110
97
87
78
64
54
43
32 26
.2
120
105
93
83
74
61
52
42
30 25
^ .05
100
87
78
68
62
52
43
35
26 22
^10
143
126
111
99
90
76
64
52
39 32
1
142
124
110
98
89
75
63
51
38 31
0.3
.4
138
121
107
96
87
73
62
50
37 30
.2
133
116
103
92
83
69
59
48
36 28
, .05
113
98
88
78
71
58
51
41
31 25
'10
163
144
129
116
105
89
77
63
49 39
1
162
142
128
115
104
88
76
62
48 38
0.6
.4
158
140
125
113
102
87
75
61
47 37
.2
154
137
123
110
100
84
73
59
45 36
.05
139
122
110
98
89
76
65
53
41 33
rio
176
156
141
128
117
99
86
72
56 45
1
175
155
140
127
116
99
86
71
56 45
1.0
.4
173
154
138
125
115
98
84
70
55 44
.2
170
151
136
123
112
96
83
68
53 43
.05
157
140
126
113
104
89
77
63
48 39
rio
192
172
155
142
130
113
99
83
66 54
1
190
172
155
142
130
112
98
83
66 54
2.0 ^
.4
189
170
154
141
129
111
97
83
65 53
.2
188
169
153
139
128
110
96
82
64 53
.05
183
164
148
135
124
107
93
79
62 51
rio
204
184
167
153
142
123
109
93
75 63
1
204
184
168
153
142
123
109
93
75 63
4.0 ■
.4
204
184
168
154
143
124
110
94
76 63
.2
205
185
169
154
143
124
110
94
76 64
.05
208
187
171
157
145
126
112
95
77 64
rio
210
190
173
159
148
129
115
98
81 67
1
211
190
174
160
148
129
115
98
82 67
6.0 i
.4
212
192
175
161
149
130
116
99
82 68
.2
213
193
176
162
151
132
117
100
83 69
.05
220
200
183
168
156
138
122
105
86 72
rio
215
194
178
163
151
133
118
102
83 71
1
216
195
178
164
152
1.33
119
103
84 72
8.0 <
.4
217
196
179
165
153
134
120
104
85 73
;2
218
198
181
167
155
136
122
105
86 74
.05
228
206
190
175
164
145
129
112
92 78
rio
220
198
183
168
156
137
123
107
88 76
1
220
199
183
169
157
1.38
124
108
89 77
12.
.4
222
201
185
171
158
139
126
109
90 78
.2
225
204
188
173
161
143
128
111
92 80
.05
240
217
200
185
174
154
139
121
100 86
* See also foot-notes, pp. 758 and 759.
762 MECHANICS OF ENGINEERING.
Example 1. — A canal 1000 ft. long of the trapezoidal sec-
tion in Fig. 611 is required to deliver 300 cubic ft. of water
, per second with the water 8 ft. deep at all
^A^^==l = = iy " sections (i.e., with uniform motion), the
A— __?.=r A slope of the bank beinaj such that for a depth
s\v\\\\\\\\\^ ^^ ^ ^^' ^^^ width of the water surface (or
Fig. 611. length of air-profile) will be 20 ft.; and the
coefficient for roughness being n = .020. What is the neces-
sary slope to be given to the bed (slope of bed = that of sur-
face, here) (ft., lb., sec.) ?
The maan velocity
V ~ Q -^ F = SOO -^ ^ (20 -\- 8)8 = 2.67 ft. per sec.
[So ^hat the surface velocity of mid-channel in any section
would probcbly be (
and again, by putting v^= Q-r- F^, v,= Q-irF^, we may
write
rl 11 fl{w, + w;) (l 1 \1Q\
IF,' F,''^2' F,-\-F, \F;'^F,yj2g''^ '
whence
Q= , ^^ (6)
/I 1,1 /^K + ^0 r 1 , 11
y F," Fy2' F,-\-F, 'iFy F,'\
From eq. (4), having given the desired shapes, areas, etc., of
the end-sections and the volume of water, Q, to be carried per
nnit of time, we may compute the necessary fall, A, of the eur-
icacC; in length = Z; while from eq. (5), having observed in an
actual water-course the values of the sectional areas F^ and ^,5
the wetted perimeters w^ and ««, , the length, = 4 of the pofc-
770
MECHATflCS OF E:\0INEEEING.
tion considered, we may calculate Q and thus gauge the stream
approximately, without making any velocity measurements.
As to the value of/", we compute it from eq. (6), § 54:2b,
using for R a mean between the values of the hydraulic radii
of the end-sections.
546. Bends in an Open Channel. — According to Humphreys
Abbot's researches on the Mississippi River the loss of
due to a bend may be put
(1)
in which v must be mft. jper sec, and 6, the angle ABC, Fig.
617, must be in ^-measure, i.e. in radians.
The section F must be greater than 100
sq. ft., and the slope s less than .0008. v
is the mean velocity of the water. Hence
if a bend occurred in a portion of a
stream of length I, eq. (3) of § 542 be-
comes
Pig. 617.
while eq. (2) of § 545 for variable motion would then become
2^ 2^^
(t) and d as above.) (For " radian" see p. 544.)
547. Equations for Variable Motion, introducing the DepthSo
—Fig. 618. The slope of the bed being sin a (or simply or,
9r meas.), while that of the surface is
different, viz.,
. sin /? = 5 = A -T- ^,
we may write
Pig. 6ia
VAKIABLE MOTION. OPEN CHANNEL. 771
in which d^ and d^ are the depths at the end-sections of the
portion considered (steady flow with variable motion). With
these substitutions in eq. (4), § 545, we have, solving for I,
d-d -(^-l-\^
fw^ (1 ^\Q' . ' ' ' ' \ )
From which, knowing the slope of the bed and the shape
and size of the end-sections, also the discharge Q, we may
compute the length or distance, I, between two sections whose
depths differ by an assigned amount {d„ — d^). But we can-
not compute the change of deptn for an assigned length I from
(6). However, if the width h of the stream is constant^ and
the same at all depths ; i.e., if all sections are rectangles hav-
ing a common width ; eq. (6) may be much simplified by intro-
ducing some approximations, as follows : We may put
1 \\Q' _ F; - F,' Q' _ (F. - FXK + F,-^ <
■f: F:J2g f:f: 2^
d: 2g '
und, similarly,
F: '2g
approx.
_2{d,~d:) v:
d. '2g'
w^
(1 iw ^ w^{f:+f,\ v:
+-^J^ =
F,+F\F: ' F,^)2g {F, + F,)F; 2g
which approx. = ^-A\
Hence by substitution in eq. (6) we have
{d,-d,)[l-^-.^/-l
y_ L d, 2gJ
'■^-TT^~ Sin or
d,b 2g
fO
547a. Backwater.* — Let us suppose that a s*^eady flow has
been proceeding with uniform motion (i.e., the surface parallel
* The subject of backwater is more fully treated in the author's
"Hydraulic Motors," pp. 226-239.
772
MECHANICS OF EISTGINEEEING.
to the bed) in an open channel of indefinite extent, and that a
vertical wall is now set up across the stream. The water rises
and flows over the edge of the wall, or weir, and after a time
a steady flow is again established. The depth, y^, of the water
close to the weir on the up-stream side is greater than d^ , the
original depth. We now have "■ variable motion " above the
weir, and at any distance x up-stream from the weir the new
depth 2/ is greater than d^. Tliis increase of depth is called
backwater, and, though decreasing up-stream, may be percep-
tible several miles above the weir. Let s be the slope of the.
original uniform motion (and also of present bed), and -y the
velocity of the original uniform motion, and let A;=--.
Then, if the section of the stream is a shallow rectangle of
constant width, we have the following relation (Kankine) :
1
a? = —
s
where is a function of —-, as per following table :
2/o-y + «-2^•)(0-0„)
(1)
For ^ = 1.0
4> = ca
1.10
.680
1.20
.480
1.30
.376
1.40
.304
1.50
.255
1.60
.218
1.70
.189
Fori- =1.80
da
0= .166
1.90
.147
2.00
.132
2.20
.107
2.40
.089
2.60
.076
2.80
.065
3.0
.056
0„ is found from ^-,' precisely as from ^, by use of the table.
With this table and eq. (1), therefore, we can find a?, the dis-
tance ("amplitude of backwater") from the weir of the point
where any assigned depth y (or " height of backwater," y — d^
will be found.
For example. Prof. Bowser cites the case from D'Aubuis-
soTi's Hydraulics of the river Weser in Germany, where the
erection of a weir increased the depth at the weir from 2.5 ft.
to 10 ft., the fiow having been originally "uniform" for 10
miles. Three miles above the dam the increase {y — d^ of
depth was 1.25 ft., and even at four miles it was 75 ft.
CHAPTEE yilL
KINETICS OF GASEOUS FLUIDS.
548. Steady Flow of a Gas. — [N.B. The student should now
review § 492 up to eq. (5).] The differential equation from
which Bernoulli's Theorem was derived for any liquid, with-
out friction, was [eq. (5), § 492]
- 1 1
• - vdv -\-dz-\ — dp=zO^ (A)
and is equally applicable to the steady flow of a gaseous fluid,
but with this difference in subsequent work, that the heaviness,
y (§ ^)-) of the gas passing different sections of the pipe or
stream-line is, or iriay be, different (though always the same at
a given point or section, since the flow is steady). For the
present we neglect friction and consider the flow from a large
receiver, where the great body of the gas is practically at rest,
through an orifice in a thin plate, or a short nozzle with a
rounded entrance.
In the steady flow of a gas, since y is different at different
points, the equation of continuity takes the form
Flow of lueight per time-rmit = F^v^y^ — F^v^y^ = etc. ; . {a)
i.e., the weight of gas passing any section, of area F, per unit
of time, is the same as for any other section, or Fvy = con-
stant, y being the heaviness at the section, and v the velocity.
549. Flow through an Orifice — Remarks. — In Fig. 619 we
have a large rigid receiver containing gas at some tension, j9„
higher than that, p^, of the (still) outside air (or gas), and at
some absolute temperature Tn, and of some heaviness y,^; that
is, in a state n. The small orifice of area F being opened, the
gas begins to escape, and if the receiver is very large, or if the
supply is continually kept up (by a blowing-engine, e.g.), after
773
774
MECHANICS OF EINTGINEERHSTG.
^
'TC-N
•••|V//'
■saw .--':>.
■■■i::'-:i;v.--^-f 1
■' \i^ •■ -i- ■ ■■ - |B ^
Fig. 619.
a very short time the flow becomes steady. Let nm represent
any stream-line (§ 495) of the flow. According to the ideal
subdivision of this stream-line into
laminae of equal mass or weight (not
equal volume, necessarily) in estab-
lishing eq. (A) for any one lamina,
each lamina in the lapse of time dt
moves into the position just vacated
by the lamina next in front, and
assumes precisely the same velocity^
'pressure^ and volume {and there-
fore heaviness) as that front one had at the beginning of the
dt. In its progress toward the orifice it expands in volume,
its tension diminishes, while its velocity, insensible at n^ is
gradually accelerated on account of the pressure from behind
always being greater than that in front, until at m, in the
" throaV of the jet, the velocity has become -y^, the pressure
(i.e., tension) has fallen to a value jprn,-, and the heaviness has
changed to y^.^. The temperature T^ (absolute) is less than
T^, since the expansion has been rapid, and does not depend
on the temperature of the outside air or gas into which efflux
takes place, though, of course, after the effluent gas is once
free from the orifice it may change its temperature in time.
We assume the pressure j!?^ (in throat of jet) to be equal to
that of the outside medium (as was done with flow of water),
so long as that outside tension is greater than .52Tj?„; but if it
is less than .527 p>n aud is even zero (a vacuum), experiment
seems to show that j?„i remains equal to 0.527 of the interior
tension p^: probably on account of the expansion of the
effluent gas beyond the throat, Fig. 620, so
that although the tension in the outer edge,
at «, of the jet is equal to that of the outside
medium, the tension at m is greater because
of the centripetal and centrifugal forces devel-
oped in the curved filaments between a and
(See § 553.)
550. Flow through an Orifice; Heaviness assumed Constant
during Flow. The Water Formula. — If the inner tension j5„ ex-
FiG. 620.
m
STEADY FLOW OF GASES, 775
ceeds cne outer, ^^, but slightly, we may assume that, like
water, the gas remains of the same heaviness during flow.
Then, for the simultaneous advance made by all the laminae or
a stream-line, Fig. 619, in the time dt^ we may conceive an
equation like eq. (-4) written out for each lamina between n
and Wj and corresponding terms added ; i.e.,
{For orifices) .... -^X'^dv -\- Jjz ^ J^ y^^' ' ^^^
In ^;'ei]eral, y is different in the different lamina, but in the
present case it is assumed to be the same in all; hence, with
m as datura level and li =:; vertical distance from n to m, we
have, ham. eq. {B\
'^— ^ + 0^A4-^ — ^ = 0. . . . (1)
2g ^g y y
But we may put v„ = ; while A, even if several feet, is
small compared with-^— — . E.g., with t?,,^ = 15 lbs. per
y y
sq. in. and p^, = 16 lbs. per sq. in., we have for atmospheric
air at freezing temperature, with ;- = (16 -^ 14,7 ) X .0807 = .0880^
lbs. /cub. ft.,
&_P==l^><111.1636 ft.
J y .0880 .08«0
Hence, putting v^ = and A == in eq. (1), we have
'o-J JPn — Pm i Water formula ', for small \ /on
2r; ~ y^ ' ' ' \ difference of pressures^ only. \ ' ' ' ^ '
The interior absolute temperature T^ being known, the y^
(interior heaviness) may be obtained from yn =Pny o^o ~^ ^npo
(§ 472), and the volume of flow per unit of time then obtained
(fir.?t solving (2) for v^^) is
Vni "~ -^mVm j \y)
where Fm is the sectional area of the jet at m. If the mouth-
piece or orifice has well-rounded interior edges, as in Fig. 641,
776 MECHANICS OF EISTGINEERIISTG.
its pectional area ^may be taken as the area F^. But if it is
an orifice in "thin plate," putting the coefficient of contraction
^-. O^ 0.60, we have
F^=GF= 0.60 F', and Q„, = 0.60 Fv^,, . . (4)
This volume, ^„j, is that occupied by the flow per time-unit
when in state m, and we have assumed that x^ = /n j hence
th.i weight of flow per time-unit is
G^ = QmYm — F^'o^y^ = F^v,,,y^. . . . • (5)
Example. — In the testing of a blowing-engine it is found
capable of maintaining a pressure of 16 lbs. per sq. inch in a
large receiver, from whose side a blast is steadily escaping
through a "tbin plate" orifice (circular) having an area ^=4
sq. inches. The interior temperature is 30° Cent, and the out-
side tension 15 lbs. per sq. in.
Required the discharge of air per second, both volume and
weight: The data are : j?^ = 18 lbs. per sq. in., T„ = 303°
Abs. Cent., J^= 4 sq. inches, and j9^ = 15 lbs. per sq. in. Use
ft.-lb.-sec. system.
First, the heaviness in the receiver is
r.-^- Sra=^- fix. 0807 = .079 lbs. per cub. ft.
Then, from eq. (2),
L Vn-Vr. | 2X32.2[144X16-144X151
^-=\2y--— =^( 0:079 =
342.6
feet
J^ ^ U.Wit; I per sec.
(97 per cent of this would be more correct on account of
friction.)
.-. Q^=i^mV^ = .6i^^^ = i|-T!iX342.6 = 5.71 cub. ft. per sec.
at a tension of 15 lbs. per sq. in., and of heaviness (by
hypothesis) = .079 lbs. per cub. ft. Hence weight
=(? = 5.71 X. 079 = 0.461 lbs. per sec.
FLOW OF GASES BY MAKIOTTE'S LAW. 77?
The theoretical power of the air-compressor or blowing-en-
gine to maintain this steady flow can be computed as in Exam-
ple 3, § 483.
651. Flow through an Orifice on the Basis of Mariotte's Lawj
or Isothermal Efiliix.^ — Since in .reality the gas expands during
flow through an orifice, and hence changes its heaviness (Fig.
619), we approximate more nearly to the truth in assuming
this change of density to follow Mariotte's law, i.Co, that the
heaviness varies directly as the pressure, and thus imply that
the temperature remains unchanged during the flow. We
again integrate the terms of eq. (j5), but take care to note that,
now, y is variable (i.e., different in different laminae at the
same instant), and hence express it in terms of the variable p
(from eq. (2), § 475), thus :
r = (rnH-i?„)i?.
_ Therefore the term / -^ of eq. {B) becomes
and, integrating all the terms of eq. {B), neglecting h, and caU-
ing Vn zero, we have
'^ _. Pn I Pn j eflux hy Mariotte's \ ^(^\
2g ~ Yn Pm ' ' ' \ -^^^ through orifice f '
As before, ^n = ^f • ^^ n > and the flow of volume per time-
-^ n Pa
unit at m is
wliile if the orifice is in thin plate, F^ may be put = .60 i^,
and the
weight of the flow per time-unit = G — F^v^y^. . (4)
If the mouth-piece is rounded, F^ = F= area of exit orifice
of mouth-piece.
778 MECHANICS OF ENGINEERING.
Example. — Applying eq. (2) to the data of the example in
§ 550, where yn was found to be .079 lbs. per cub. ft., we
have [ft., lbs., sec]
In ym
= ^2X32.2x^^^^X2.3025xlogio[^] .
= 348.7
ft. p. sec.
• •• Q^ = i^^Vrn = 0.60 X^IjX 348.7 = 5.81 cub. ft. per sec.
Since the heaviness at m is, from Mariotte's law,
V 15
rm =— rn = T^ of .079, i.e., rm = .0741 lbs. per cub. ft.,
hence the weight of the discharge is
G = QmYm = 5.81 X .0741 = 0.430 lbs. per sec,
or about 4^ per cent less than that given by the "water for-
mula." If the difference between the inner and outer ten-
sions had been less, the discrepancy between the results of
the two methods would have been smaller.
552. Adiabatic Efflux from an Orifice. — It is most logical to
assume that the expansion of the gas approaching the orifice,
being rapid, is adiabatic (§ 478). Hence especially when the
difference btween the inner and outer tensions is considerable)
it is more accurate to assume y to vary according to Poisson's
Law, eq. (1), p. 623; i.e., (p-pj = (r-^rJ^'^^; in inte-
grating eq. (B) of p. 775. Then the term
I — Wlll=-^ p -'!Hp= [p^-29-p„-29J
t/n / In Jn In
ADIABATIC FLOW OF GASES THROUGH ORIFICES. 779
ana eq. {B), neglecting h as before, and with v^ = 0, becomes
(See Fig. 619)
'w- = — — ~ 1 1 "~ ( ) • iAdiabatic flow ; orifice?) . (1)
Having observed jp^ and T^ in the reservoir, we compute
y^ = ^^° ° (from § 472). The gas at m,, just leaving the
oritice, having expanded adiabatically from the state n to the
state m, has cooled to a temperature T„i (absolute) found thus
(§ 478),
T^ = TJ^ , (2)
and is of a heaviness
r.-rn(^)'7 (3)
and the flow per second occupies a volume (immediately on
exit)
and weighs
^ = F^'^mYm ' ...... (5)
Example 1. — Let the interior conditions in the large reser-
voir of Fig. 619 be as follows {state n) : p^ = 22^ lbs. per sq.
in., and T^ = 294° Abs. Cent, (i.e., 21° Cent); while ex-
ternally the tension is 15 lbs. per sq. inch, which may be taken
as being ^ p^ = tension at m, the throat of jet. The opening
is a circular orifice in " thin plate" and of one inch diameter.
Required the weight of the discharge per second [ft., lb., sec;
^^ = 32.2].
22 5X144 273
First, Tn = -^2^xl4:4: ' 294 ^•^^^^^^•-'--'^^ ^^^' P^^ ^^^' ^*-
Then, from (1),
4
2X32.2X3.44X22.5X144 ^^
= \ qJY5 [!-(§) -29] = 833 ft. per sec.
780 MECHANICS OF ENGINEERING.
Now F = i7r(l)2 _ ,00546 sq. ft.
.-. Q^ = Ci^v^ = .60Fv^ = 0.60 X. 00546X833 = 2.73
cub. ft. per sec, at a temperature of [eq. (2)]
Trn = 294 (f ) -29 = 261° Abs. Cent. = - 12° Cent.*
and of a heaviness [eq. (3)]
r,^ = 0.115(f)-7i = 0.0862 lbs. per cub. ft.,
so that the weight of flow per sec.
= G = Q^rm = 2.73 X. 0862 = .235 lbs. per sec.
Example 2. — Let us treat the example already solved by the
two preceding approximate methods (§§ 550 and 551) by the
present more accurate equation of adiabatic flow, eq. (1).
The data were (Fig. 619) :
'Pn =16 lbs. per sq. in.; 7'„ = 303° Abs. Cent.;
p^ = 15 '' " " ; and F = 4 sq. inches
[F being the area of orifice], yn was found = .079 lbs. per
cub. ft. in § 550; hence, from eq. (1),
,/2x 3 2.2X3.44X18X144 ,^ is
Vm=\ 079 [l-(l6)] = 348.5 ft. per sec.
From (4),
Qm-F^v„, = .6Fvm=-.QX^XM8.5 = 5.Sl cub. ft. per sec;
and since at m it is of a heaviness
rm = -079 (fj--! = .0755 lbs. per cub. ft.,
we have weight of flow per sec.
= (? = Qmrm = 5.81 X. 0755 = 0.439 lbs. per sec
* By the impact of the effluent air on the outside air, with extinction of
velocity, the temperature rises again.
THEORETICAL MAXIMUM FLOW OF WEIGHT OF GAS. 781
Comparing the three methods for this problem, we see that
By the" ivater formula," . . . (r = 0.451 lbs. per sec.
" isothermal formula, . . G^ = 0.430 " ''
" adiabatic formula, . . G = 0.439 " "
553. Practical Notes. Theoretical Maximum Flow of
Weight. — If in the equations of § 552 we write for brevity
Pm-^Pn = ^ we derive, by substitution from (1) and (3)
in (5),
Weight of flow
perunitof time j -e-<3™r»=J»V6:8%S^l-r^»]ix-. (1)
This function of x is of such a form as to be a maximum for
:C=(p^-Pn) = (.830)3-44 = .527; ... (2)
i.e., theoretically, if the state n inside the reservoir remains
the same, while the outside tension (considered = p^ of jet,
Fig. 619) is made to assume lower and lower values (so that
^'i—Vm^Vni diminishes in the same ratio), the maximum flow
of weight per unit of time will occur when p^ = .527p^, a
little more than half the inside tension.
Prof. Cotterill says (p. 544 of his " Applied Mechanics") :
" The diminution of the theoretical discharge on diminution
of the external pressure below the limit just now given is an
anomaly which had always been considered as requiring ex-
planation, and M. St. Tenant had already suggested that it
could not actually occur. In 1866 Mr. R. D. Napier showed
by experiment that the weight of steam of given pressure dis-
charged from an orifice really is independent of the pressure
of the medium into which efflux takes place ^ ; and in 1872
Mr. Wilson confirmed this result by experiments on the reac-
tion of steam issuing from an orifice."
" The explanation lies in the fact that the pressure in the
* When the difference between internal and external pressures is great,-'
should be added.
78a MECHANICS OF EJSTGINEERIIS^G.
centre of the contracted jet is not the same as that of the sur-
rounding medium. The jet after passing the contracted sec-
tion suddenly expands, and the change of direction of the fluid
particles gives rise to centrifugal forces" which cause the pres-
sures to be greater in the centre of the contracted section than
at the circumference ; see Fig. 620.
Prof. Cotterill then advises the assumption that j?^=.527^„
(for air and perfect gases) as the mean tension in the jet at m
(Fig. 619), whenever the outside medium is at a tension less
than .h'^ilj)^. He also says, "Contraction and friction must
be allowed for by the use of a coefficient of discharge the
value of which, however, is more variable than that of the
corresponding coefficient for an incompressible fluid. Little is
certainly known on this point." See §§ 549 and 554.
For air the velocity of this maximum flow of weight is
Vel. of max. G = 997 a /^1 ft. per sec, . (3)
where T^ = abs. temp, in reservoir, and T^ = that of freezing
point. Rankine's Applied Mechanics ( p. 584) mentions ex-
periments of Drs. Joule and Thomson, in which the circular
orifices were in a thin plate of copper and of diameters 0.029
in., 0.053 in., and 0.084 in., while the outside tension was
about one half of that inside. The results were 84 per cent
of those demanded by theory, a discrepancy due mainly, as
Rankine says, to the fact that the actual area of the orifice was
used in computation instead of the contracted section; i.e., con-
traction was neglected.
554. Coefficients of Efflux by Experiment. For Orifices and
Short Pipes. Small Difference of Tensions. — Since the discharge
through an orifice or short pipe from a reservoir is affected
not only by contraction, but by slight friction at the edges,
even with a rounded entrance, the theoretical results for the
volume and weight of flow per unit of time in preceding para-
graphs should be multiplied both by a coefficient of velocity
and one for contraction C, as in the case of water ; i.e., by a
coefficient of ejflux ju, = cf)C. (Of course, when there is no
COEFFICIENTS OF EFFLUX. GAS. 783
contraction, G =. 1.00, and then /« = as with a well-rounded
mouth-piece, for instance, Fig. 541, and with short pipes.)
Hence for practical results, with orifices and short pipes, we
should write for the weight of flow per unit of tivrifo
Vn
(1)
(from the equations of § 552 for adiabatic flow, as most acca-
rate; j?^ -^Jpn ™a.y range from \ to l^OO). F ^=^ area of orifice,
or of discharging end of mouth-piece or short pipe. ;/„ =
heaviness of air in reservoir and =:■■ T^p^y^ ~- T^p^, eq. (13) of
§ 437 ; and fx = the experimental coeflScient of efflux.
From his own experiments* and those of Koch, D'Aubuis-
8on, and others, Weisbach recommends the following mean
values of /< for various mouthpieces, when p^ is not more than
\ larger than jv^ (i.e., about 17 fo larger), for use in eq. (1):
1. For an orifice in a thin plate, ....... /^=0.56
2. For a short cylindrical pipe (innercorners not rounded),yM=0.75
3. For a well-rounded mouth-piece (like that in Fig. 541), yu=0.98
4:, For a short conical convergent pipe (angle about 6°), /i=0.92
Example,— (Data from Weisbacli's Mechanics.) "If the
Bum of the areas of two conical tuyeres of a blovring-machine
J8 i''^ 3 sq. inches, the temperature in the reservoir 15° Cent.,
the height of the attached (open) mercury manometer (see
Fig. 464) 8 inches, and the height of the barometer In the eX'
temal air 29 inches," we have (ft., lb., sec.)
^=r:-JL_=??; r, = 288° Aba. Cent. ;
p^ 29 + 3 32 ** '
Pn ^ (ft) 1*-^ X 144 lbs. per sq. ft. ;
y^ = |||.|f X 0.0807 = 0.0816 lbs. per cnb. ft,
while F = ^ sq. ft. and (see above) ^ = .92 ; hence G =
•^2(i|^)(g)- V64.4 X 3.44(|) X 14.7 X 144 X .0816[1 - (g)'^^] ;
* See also tlie Engineerii g Record of Oct. 1901, p. 409, where an account
is given of experiments on tlie flow of air through orifices made at the
Mass. Inst, of Technology. EHegner's researches are referred to by Ruehl-
mann in his nydromecJianik . p. 675.
784
MECHANICS OF ENGINEERING.
i.e., (t = .606 lbs. per second; wliicli will occupy a volume
Vo=-G^ro=G-^. 0807 = 7.51 cub. ft.
at one atmosphere tension and freezing-point temperature,
while at a temperature of ^^ = 288° Abs. Cent, and tension of
Pot = 3u of o^^® atmosphere" (i.e., in the state in which it was
on entering the blowing-engine) it occupied a volume
7 = ||-gx7.51==8.20 cub. ft.
(This last is Weisbach's result, very nearly, obtained by an
approximate formula. )
>r Orifices and Short Pipes for a
targe Difierenee of Tension.— -For values > ^ and < 2, of the
ratio Pji : p^, of internal to external tension, "Weisbaeh's ex-
periments with circular orifices in thin plate, of diameters (= d)
from OA inches to 0.8 inches, gave the following results :
Pn'.pm =
ir ^"^ dccreascs slightly for increasing size of oriiice.
With short cylindrical pijpes, internal edges not rounded^
and three times as long as wide, Weisbach obtained /* as
follows :
Pn-Pm
__
1.05
1.10
1.30
1.40
1.70
1.74
diam. = .4''^-
/« =
.73
.77
.83
" ~ .6™-
IJ- =
81
.83
M =
.83
"When the inner edges of the 0.4 in, pipe were slightly
founded, }x was found = 0.93 ; while a well-rounded mouth-
piece of the form shown in Fig. 641 gave a value fx — from
.965 to .968, for p^ : p^ ranging from 1.25 to 2.00. These
values of fi are for use in eq. (1), above.
556. To find the Discharge when the Intemal Pressure la
measured in a Small Reservoir or Pipe, not ninch larger than th«
VELOCITY OF APPEOAOH. GASES. 785
Orifice.-- -Fig. 621. If the internal pressure p^, and tempera,
ture Tnt must be measured in i ^ li
small reservoir or pipe, n^ whose Af fU It
sectional area t\ is not verj large — — ^^^r^~^^-l^ -^■-
corapared with that of the oritice, -^^^^^^^Z^![n^^-^^^n
F, (or of the jet. F^ ,) the velocity ~' ' — \?7^^sss=g^
Vn, at n (velocity of approach) can- fig. esi.
not be put = zero. Hence, in applying eq. {B\ § 550, to the
successive laminse between n and m^ and integrating, we shaiJ
have, ior odiahatiG steady flow.
vj v^ 3.4423,
'^hm • • • • ^^>
25^ 2sr r
instead of eq. (1) of § 552. But from the equation of continuity
for steady flow of gases [eq. (a) of § 548], F,{Vr,rn = F v^Tm)
hence v^ = -^^ ""^ '^n?, while for an adiabatic change from n
" n In
to 711, — ={i-^j ; whence by substitution in (1), solving for
Vm, we have
4
. 3.44j9,
^9 -^ —
v^=^-
Vn
^^_(FA'(vA'-''
^^ n' \Pn
As before, from §§ 472 and 478,
(2)
_'Pn^Q fO\
Tn — ^ m ' To \^)
VO -L n
and rm=(— ^) Tn (4)
Having, observed pn, Pm, and Tn, then, and knowing the
area F of the orifice, we may compute ;-„, 7-^, and i;^, and
finally the
Weight of flow per time-unit = G = nFvjnTm, • • • (5)
t86 MECHANICS OF ENGINEERING.
taking fA from § 554 or 555, In eq. (2) it must be remembered
that for an orifice in " thin plate," F^, is the sectional area of
the contracted vein, and = CF', where C may be put = •— .
Example. — If the diameter of AB, Fig. 621, is 3|- inches.
and that of the orifice, well rounded, = 2 in. ; if p^ = 1-j^ at-
mospheres = if X 14.7 X 144 lbs. per sq. ft., while Pm = -^oi
an atmos., so i;hat ^^ = i^, and T^ = 283° Abs. Cent., — lO'
,Pn
quired the discharge per second, using the ft., lb., and sec
From eq. (3),
;/^ = i|.||| X 0.080T « .08433 lbs. per cub. ft,";
whence (eq. (4))
^^ = (li)-7i^^ = .0749 lbs. per cub. ft."
Then, from eq. (2),v =
■^l
64.4X3.44X15.925X144/^
Mm
= 547.3 ft. per sec;
I-©-") -[/i-(S)^©^-"]
.-. G = 0.98|(^T547.3X. 0749 = .876 lbs. per sec.
567. Transmission of Compressed Air; through very Long
Level Pipes. Steady Flow.
Case I. Whe?i the difference between the tensions in the
reservoirs at the ends of the pipe is small. — Fig. 622. Uodef
\9^«
\,
* * •*«■■• *••■'• .*
:^7^:-.
■ "J i.'' •
.-/...•.•-.-
• '.v.-,-.
U
1-
Fig.
682.
>\
these circumstances it is simpler to employ the form of fornmla
that would be obtained for a liquid by applying Bernouiii*8
Theorem, taking into account the " loss of head " occasioned
TRANSMISSION OF OOMPBESSED AlB. 787
by the friction on the sides of the pipe. Since the pipe is
very long, and the change of pressure small, the mean velocity
in the pipe, -y', assumed to be nearly the same at all points
along the pipe, will not be large ; hence the difference be-
tween the velocity-heads at n and m will be neglected ; a cer-
tain mean heaviness y' will be assigned to all the gas in the
pipe, as if a liquid.
Applying Bernoulli's Theorem, with friction, § 516, to the
ends of the pipe, n and m, we have (as for a liquid)
< + A, + o = | + ^» + 0-4/|-g.. . (1)
Putting (as above mentioned) 'oj — • v^ = 0, we have, more
simply, ,
/ " '^ d'2g ^'
The value of f as coefficient of friction for air in long
'pipes is found to be somewhat smaller than for water ; see next
paragraph.
558. Transmission of Compressed Air.* Experiments in the St.
Gothard Tunnel, 1878.— [See p. 96 of Yol. 24 (Feb. '81), Yan
^Nostrand's Engineering Magazine.] In these experiments,
the temperature and pressure of the flowing gas (air) were ob-
served at each end of a long portion of the pipe whieli delivered
the compressed air to the boring-machines three miles distant
from the tunnel's mouth. The portion considered was selected
at a distance from the entrance of the tunnel, to eliminate the
fluctuating influence of the weather on the temperature of the
flowing air. A steady flow being secured by proper regulation
of the compressors and distributing tubes, observations were
made of the internal pressure {p\ internal temperature {T), as
well as the external, at each end of the portion of pipe con-
sidered, and also at intermediate points ; also of the weight
of flow per second G ^= QoYoi measured at the compressors
under standard conditions (0° Cent, and one atmos. tension).
Then knowing the p and T at any section of the pipe, the
* For experiments at Paris in 1891 see Proc. Inst. Civ. Engineers, vol,
105, p. 180; and Engineering News of Dec. 1889, p. 556. See also foot-note
on next page.
788
MECHANICS OF ENGI]SrEERIIf&.
heaviness y of the air passing that section can be computed
and the velocity v= G -ir Fy, F being
rfrom^- =
L n i?o TA
the sectional area at that point. Hence the nteoM velocity v\^
and the mean heaviness y\ can be computed for this portion
of the pipe whose diameter = d and length = I. In the ex-
periments cited it was found that at points not too near the
tunnel-mouth the temperature inside the pipe was always
about 3° Cent, lower than that of the tunnel. The values of
yin the different experiments were then computed from eq.
(2) of the last paragraph ; i.e.,
Y
(2)
all the other quantities having been either directly observed,
or computed from observed quantities.
THE ST. GOTHARD EXPERIMENTS.
[Concrete quantities reduced to English uniU^
No.
I
(feet.)
d
(ft.)
V
(lbs. cub.
ft.)
Atmospheres.
Pn-Pm
lbs. sq. in.
V
ft. per sec.
mean
temp.
Cent.
*
Pn
Pm
1
2
3
4
5
6
15093
15093
7 5093
1713
1713
1713
f
1
0.4058
0.3209
.2803
.3765
.3009
.3641
5.60
4.35
3.84
5.24
4.13
3.65
5.34
4.13
3.65
5.00
4.06
3.54
5.39
3.23
3.79
3.53
1.03
1.54
19.33
16.30
15.55
37.13
30.83
39.34
21°
31°
31°
26.5
36.5
36.5
.0035
.0038
.0041
.0045
.0034(?)
.0045
In the article referred to ( Yan l^ostrand's Mag.) y is not
computed. The writer contents himself with showing that
Weisbach's values (based on experiments with small pipes and
high velocities) are much too great for the pipes in use in the
tunnel.*
"With small tubes an inch or less in diameter Weisbach
found, for a velocity of about 80 ft. per second,/ =.0060;
for still higher velocities/ was smaller, approximately, in ao*
eordance with the relation
f- .0542
'^v' (in ft. per sec.)
* See -also experiments described in Engineering News, Nov. 3, 1904, p.
387. In that article the quantity called / is the/ of this chapter divided
by 64.4.
TRANSMISSION OF COMPRESSED AIR,
789
On p, 3T0, vol. XXIV, Yan ISTostrand's Mag., Prof. Robinson
of Ohio mentions other experiments with large long pipes.
From the St. Gothard experiments a value oif= .004 may
be inferred for approximate results with pipes from 3 to 8 in.
m diameter.
Example. — It is required to transmit, in steady flow, a supply
of {r = 6.456 lbs. of atmospheric air per second through a pipe
30000 ft. in length (nearly six miles) from a reservoir where
the tension is 60 atmos. to another where it is 5.8 atmos., the
mean temperature in the pipe being 80° Fahr., = 24° Cent,
(i.e. = 29 Y° Abs. Cent.). Required the proper diameter of
pipe ; 6? = ? The value /" = .00425 will be used, and the ft.-
Ib.-sec. system of units. The mean volume passing per second
in the pipe is
Q'=(?-^r' (3)
The mean velocity may thus be written : -y' = -^
/_^'_ Q'
iTtd'
(^)
The mean heaviness of the flowing air, computed for a mean
tension of 5.9 atmospheres, is, by § 472,
^ - llTi^ • Wi ^ -^^^^ = ^-^^^ ^'''' P"" ^"^- ^*- '
and hence, see eq. (3),
at tension of 5.9 atmos., and temperature 29Y° Abs. Cent.
Now, from eq. (2),
JPn— Vm _ 4/ l_
y' 2g' d'
whence
d"
. (5)
790 MECHANICS OF ENGINEEEING.
and hence, numerically,
6 / 4 X .00425 X 0.^31 X 30000 X (14.74)' _ ., qq.
V (•'^854y[14.7 X 144(6.00 - 5.80)]2 X 32.2 ~ •^•^^^e®^-
559. (Case II of § 557) Long Pipe, with Considerable Differ-
ence of Pressure at Extremities of the Pipe. Flow Steady. — Fig.
623. If the difference between the end-tensions is compara-
tively great, we can no longer deal with the whole of the air
r A:
Fig. 623.
in the pipe at once, as regards ascribing to it a mean velocity
and mean tension, but tnust consider the separate lamincBy
such as AB{2k short length of the air-stream) to which we may
apply eq. (2) of § 55 \ A and B corresponding to the n and
m of Fig. 622. Since the^„— ^^, Z, y', and v' of § 557
correspond to the —djp^ ds, y, and v of the present case (short
section or lamina), we may write
(1)
But if ^ = weight of flow per unit of time, we have at any
section, Fvy = G (equation of continuity) ; i.e., v = G -^ Fy^
whence by substitution in eq. (1) we have
dp _ ^ G'ds
I.e.
ydp
^gF'd
ds.
(2)
Eq. (2) contains three variables, y^ p, and s (= distance of
lamina from n'^. As to the dependence of the heaviness y on
the tension j!? in different laminae, experiment shows that in most
cases a uniform temperature is found to exist all along the
pipe, if properly buried, or shaded from the sun ; the loss of
heat by adiabatic expansion being in great part made up by
the heat generated by the friction against the walls of the
GAS IN LONG PIPES. LARGE FALL OF TENSION. 791
pipe. This is due to the small loss of tension per unit of
length of pipe as compared with that occurring in a short dis-
charge pipe or nozzle. Hence we may treat the flow as iso-
thermaL and write j? -=- y =y^^/-^- y^^, (§ 4T5, Mariotte's Law).
Hence ;k = ^-^^, which substituted in eq. (2) enables us to
■■■ - f"p^^=\M^3\P'- ■ • • • (^)
Performing the integration, noting that at n' p=^j[>n>, s = 0,
and at m' p ^ JPm> ^"^^ s ^ I, we have
ir-T) ' — r> >n — ^'^^' — ^ -^ isothermal flow ) ,..
-^l_Pn' I>m \ ^gd' p-"' y^,' ' \ in loug pipes ) • • W
It is here assumed that the tension at the entrance of the pipe
is practically equal to that in the head reservoir, and that at
the end {vi') to that of the receiving reservoir; which is not
strictly true, especially when the corners are not rounded. It
will be remembered also that in establishing eq. (2) of § 557
(the basis of the present paragraph), the "inertia" of the gas
was neglected; i.e., the change of velocity in passing along
the pipe. Hence eq. (4) should not be applied to cases where
the pipe is so short, or the diilerence of end-tensions so great,
as to create a considerable difference between the velocities at
the two ends of the pipe. (8ee Addendum on p. 797.)
Example. — A well or reservoir supplies natural gas at a ten-
sion of pn' = 30 lbs. per sq. inch. Its heaviness ^t 0° Cent,
and one atmosphere tension is .0484 lbs. per cub. foot. In
piping this gas along a level to a town two miles distant, a
single four-inch pipe is to be employed, and the tension in the
receiving reservoir (by proper regulation of the gas distributed
from it) is to be kept equal to 16 lbs. per sq. in. (which would
sustain a column of water about 2 ft. in height in an open
water manometer, Fig. 4 65).
792 MECHANICS OF EFGINEEEING.
The mean temperature in the pipe being 1Y° Cent., required
the amount (weight) of gas delivered per second, supposing
leakage to be prevented (formerly a difficult matter in practice).
Solve (4) for G, and we have
First, from § 472, with T^, = T^, = 290° Abs. Cent., we
compute
&'=£.. ^' = 114x144 _ 290 ^
r„, r. T. .0484 273
Hence with/" = .005,
_ , ,, y / 32.2 X A[(30 X 1 44 )--(16 X 144-)^
T Vi^; Y 4X.005 X 10560 X 46454
= 0.337 lbs. per sec.
(For compressed atmospheric air, under like conditions, we
would have G = 0.430 lbs. per second.)
Of course the proper choice of the coefficient,/ has an im-
portant influence on the result.
From the above result {G = 0.337 lbs. per second) we can
■compute the volume occupied by this quantity of gas in the
receiving reservoir, using the relation ^^/ = — .
/to'
The heaviness y^^' of the gas in the receiving reservoir is
most easily found from the relation -^^ = —', which holds
Ym' Yn'
good since the flow is isothermal. I.e., ^^-^ = 46454 ft.;
/to'
whence y^i = 0.049 lbs. per cubic foot, p^> being 16 X 144
lbs. per sq. ft.
Hence
Q^r = = -^ = 6.794 cub. ft. per sec.
Ym' U.U4ry
FLOW OF GAS IN PIPES. 793
It should be said that the pressure at the up-stream end of
the pipe depends upon the rate of flow allowed to take place.
With no flow permitted, the pressure in the tube of a gas-
well has in some ("ases reached the high figure of 500 or 600
lbs. per sq. in,
560. Rate of Decrease of Pressure along a Long Pipe. — Con-
sidering further the case of the last paragraph, that of a
straight, long, level pipe of uniform diameter, delivering gas
from a storage reservoir into a receiving reservoir, we note
that if in eq. (4) we retain jy,„' to indicate the tension in the
receiving reservoir, but let jp^i denote in turn the tension at
points iti the pipe successively further and further (a distance
xc) from the receiving reservoir mf ^ we may write x for I and
obtain the equation (between two variables, jp^' and x)
Vn^ —Pm^ = Const. Xos (6)
This can be used to bring out an interesting relation men-
tioned by a writer in the Engineering News of July ]887
(p. 71), viz., the fact that in the parts of the pipe more distant
from the receiving end, in\ tlie distance along the pipe in
which a given loss of pressure occurs is much greater than
near the receiving end.
To make a numerical illustration, let us suppose tlat the
pipe is of such size, in connection with other circumstances,
that the tension j:>^/ at A. a distance x == six miles from in\ is
two atmospheres, the tension in the receiving reservoir being
one atmosphere ; that is, that the loss of tension between A
and ml is one atmosphere. If we express tensions in atmos-
pheres and distances in miles, we have for the value of the
constant in eq. (6), for this case.
Const. = (4 — 1) -^ 6 = I ; {for assumed units.) . . (7)
Islow let Pn' = the tension at B^ a point 18 miles from m',
and we have, from eqs. (6) and (7), the tension at ^ = 3.16
atmospheres. Proceeding in this manner, the following set of
values is obtained :
794
MECHANICS OF ENGINEEEING.
Point.
F
E
B
B
A
m'
Total distance L Distance be-
from m" itween consecu-
fromm. tive points.
Tension at
point.
Loss of ten-
sion in eacli in-
terval.
126 mi]
90 '
60 •
86 '
18 '
6 '
'
es.
86 miles.
60 "
34 "
18 "
12 "
6 "
8.00 aim.
6.78 "
5.56 "
4.35 "
3.16 "
2.00 "
1.00 "
1.22 atm.
1.22 "
1.21 "
1.19 "
1.16 "
1.00 "
If the distances and tensions in the second and fourth
columns be plotted as abscissae and ordinates of a curve, the
latter is a parabola with its axis following the axis of the pipe ;
its vertex is not at mf, however.
561. Long Pipe of Variable Diameter. — Another way of stat-
ing the fact mentioned in the last paragraph is as follows : At
the up-stream end of the pipe of uniform diameter the gas is
of much greater density than at the other extremity (the
heaviness is directly as the tension, the temperature being as-
sumed the same throughout the pipe), and the velocity of its
motion is smaller than at the discharging end (in the same
ratio). It is true that the frictional resistance per unit of
iengtli of pipe varies directly as the heaviness [eq. (1), § 510],
but also true that it varies as the sqiiare of the velocity ; so
that, for instance, if the pressure at a point A is double that
at B in the pipe of constant diameter, it implies that the
heaviness and velocity at A are double and half, respectively,
those at B, and thus the gas at A is subjected to only half the
frictional resisting force per foot of length as compared with
that at B. Hence the relatively small diminution, per unit of
length, in the tension at the up-stream end in the example of
the last paragraph.
In the pipe of uniform diameter, as we have seen, the greater
part of the length is subjected to a comparatively high ten-
sion, and is thus under a greater liability to loss by leakage
than if the decrease of tension were more uniform. The
total ^^hoojp-tension^'' (§ 426) in a unit length of pipe, also, is
proportional to the gas tension,* and thinner walls might be
employed for the down-stream portions of the pipe if the gas
* Or, rather, to its excess over that of external air.
FLOW OP GAS IJSr PIPES.
796
tension in those portions could be made smaller than as shown
in the preceding example.
To secure a more rapid fall of pressure at the up-stream end
of the pipe, and at the same time provide for the same delivery
of gas as with a pipe of uniform diameter throughout, a pipe
of variable diameter may be employed, that diameter being
considerably snialler at the inlet than that of the uniform pipe
but progressively enlarging down-stream. This will require
the diameters of portions near the discharging end to be larger
than in the unifoi-m pipe, and if the same thickness of metal
were necessary throughout, there would be no saving of metal,
but rather the reverse, as will be seen ; but the diminished
thickness made practicable in those parts from a less total hoop
tension than in the corresponding parts of the uniform pipe
more than compensates for the extra metal due to increased
circumference, aside from the diminished liability to leakage,
which is of equal importance.
A simple numerical example will illustrate the foregoing.
The pipe being circular, we may replace i^by ^nd' in equation
(4), and finally derive, G being given,
d = Const. X
I
.JP^
' Jftn' —
'=a
I
—_Pn' J/m' _
. . (8)
Let A be the head reservoir, and m' the receiving reservoir,
and B a point half-way between. At A the tension is 10 at-
mospheres ; at m', 2 atmospheres. For transmitting a given
weight of gas per unit-time, through a pipe of constant diam-
ter throughout, that diameter must be (tensions in atmospheres ;
2/^„ being the length), by eq. (8),
CZ„i(.0208)i = 0.46 Cl,^. . (8),
d = Cli
2
.100 - 4
If we substitute for the pipe mentioned, another having a con-
stant diameter d^ from A to B, where we wish the tension to
be 5 atmospheres, and a different constant diameter d, from B
to m', we derive similarly
d. = CLi
100 - 25
= 0.42 CU
796 MECHAlflCS OF ENGINEERING.
and
d^ = CVi
1
.25 -4
0.54 Cl^
It is now to be noted that the sum of d^ and d^ is slightly
greater than the double oi d \ so that if the same thickness of
metal were used in both designs the compound pipe would
require a little more material than the uniform pipe; but,
from the reasoning given at the beginning of this paragraph,
that thickness may be made considerably less in the down-
stream part of the compound pipe, and thus economy secured.
[In case of a cessation of the flow, the gas tension in the
whole pipe might rise to an equality with that of the head-
reservoir were it not for the insertion, at intervals, of auto-
matic regulators, each of which prevents the increase of ten-
sion on its down-stream side above a fixed value. To provide
for changes of length duo to i'lse and tall of temperature, the
pipe is laid with slight undulations.]
It is a noteworthy t eoretical deduction that a given pipe of
variable diameter connecting two reservoirs of gas at specified
pressures will deliver the same weight of gas as before, if
turned end for end. This follows from equation (3)', § 559.
With d variable, (3)^ becomes (with F =^ ^rrd")
l{-I>dp)=G^G"l |; i.e., (?= = i^^J-.. . (9)
tJn' fy
{C" is a constant.)
But /, —is evidently the same in value if the pipe be
turned end for end. In commenting on this circnmstance, we
should remember (see § 559 ) that the loss of pressure along the
pipe is ascribed entirely to friGtional resistance, and in no de-
gree to changes of velocity (inertia).
On p. 73 of the Engineering Nexos of July 1887 are given
the following dimensions of a compound pipe in actual use,
and delivering natural gas. The pressure in the head-reservoir
is 319 lbs. per sq. in.; that in the receiving reservoir, 65. For
2.84 miles from the head-reservoir the diameter of the pipe is
NATURAL GAS. COEFFICIENT OF FLUID FRICTION. 797
8 in.; throughout the next 2.75 miles, 10 in.; while in the
remaining 3.84- miles the diameter is 12 in. At the two
points of junction the pressures are stated to be 185 and 132
lbs. per sq. in., respectively, during the flow of gas under the
conditions mentioned.
561a. Values of the Coefficient of Fluid Friction for Natural
Gas. — In the Ohio Report on Economic Geology for 1888 may
be found an article by Prof. S. W. Robinson of the University
of that State describing a series of interesting experiments
made by him on the flow of natural gas from orifices and
through pipes.* By the insertion of Pitot tubes approximate
measurements were made of the velocity of the stream of gas
in a pipe. The following are some of the results of these ex-
periments, j9j — ^a representing the loss of pressure (in lbs. per
sq. inch) per mile of pipe-length, and /"the coefficient of fluid
friction, in experiments with a six-inch pipe :
Pi -Pi
1.00
1.50
3.25
2.50
5.75
6.25
f
0.0025
0.0037
0.0052
0.0059
0.0070
0.0060
In the flow under observation Prof. Robinson concluded
that / could be taken as approximately proportional to the
fourth root of the cube of the velocity of flow ; though calling
attention to the fact that very reliable results could hardly be
expected under the circumstances.
561b. [Addendum to § 559.] Isothermal Flow in a level pipe,
with consideration of Inertia.— In eq. (1) of p. 697 neglect dz,,
^\xt w -^ F = 4: -^ d, and divide through by v\ In the second
term put G' ^ Fy' for v' and then p {y„, -^Pm) ^or y. We
now find the variables separated, and on integration for steady
flow obtain (after putting v^-^v^= Fy,^ -^ Fy^ = Pm -^Pn)>
1 fPn\ g{ Pn—Pm) ^" Vm __ "¥}_
M^J- 2 •^•^-~ d'
f Notation as in § 557 with Q = FvyA,
* More recent experiments with the Pitot Tube, in measuring ths
velocity of gases in pipes, have been made in Chicago by Messrs. Dreffein
and McBurney. See Engineering News of Dec. 21, 1905, p. 660.
CHAPTEK IX.
IMPULSE AND RESISTANCE OF FLUIDS.
bt)2. The so-called " Eeaction" of a Jet of "Water flowing from
a Vesse.", — In Fig. 624, if a frictionless but water-tight ping B
be inserted in an orifice in the
vertical side of a vessel mounted
on wheels, the resultant action of
the water on the rigid vessel (as a
whole) consists of its weight G,
and a force P' = FJiy (in which
\~^' F= the area of orifice) which is
the excess of the horizontal hydro-
static pressures on the vessel wall
toward the right ( i| to paper) over
those toward the left, since the
pressure P, = Fhy, exerted on the plug is felt by the post C,
and not by the vessel. Hence the post D receives a pressure
Fig. 624.
P' = Fhy.
(1)
Let the plug B be removed. A steady flow is then set up
through the orifice, and now the pressure against the post P is
%FJiy (as will be proved in the next paragraph) ; for not only
is the pressure Fhy lacking on the left, because of the orifice,
but the sum of all the horizontal components ( || to paper) of
the pressures of the liquid filaments against the vessel wall
around the orifice is less than its value before the flow began,
by an amount Fhy. A resistance R = '^Fhy being provided,
nnd the post removed, a slow uniform motion may be main-
iftined toward the right, the working force being ^Fhy = P"
"reactiois^" of a jet.
799
Fig. 625.
(see Fig. 625 ; B is not shown). If an insufficient resistance
be furnished before removing the post D,
the vessel will begin to move toward the
right with an acceleration, which will
disturb the surface of the water and
change the value of the horizontal force.
This force
P" =: 2i#^ ... (2)
is called the " reaction'^ of the water-jet ;
y is the heaviness of the liquid (§ 7).
Of course, as the flow goes on, the
water level sinks and the '• reaction" diminishes accordingly^
Looked upon as a motor, the vessel may be considered to be a
piston-less and valve-less water-pressure engine, carrying its
own reservoir with it.
In Case II of § 500 we have already had a treatment of the
" Reaction-wheel " or " Barker's mill," w^hich is a practical
machine operating on this principle, and will be again con-
sidered in " Hydraulic Motors." *
563. " Eeaction" of a Liquid Jet on the Vessel from which it
Issues. — Instead of slrowing that the pressures on the vessel
close to the orifice are less than they were when there was no
flow by an amount Fhy (a rather lengthy demonstration),
another method will be given, of greater simplicity but some-
what fanciful.
If a man standing on the rear platform of a car is to take np
in succession, from a basket on the car, a number of balls of
equal mass = M, and project each one in turn horizontally
backward with an acceleration =^, he can accomplish this
only by exerting against each ball a pressure = Mjy, and in the
opposite direction against the car an equal pressure = MjJ. If
this action is kept up continuously the car is subjected to a
constant and continuous forward force of P" = Mp.
Similarly, the backward projection of the jet of water in the
case of the vessel at rest must occasion a forward force against
the vessel of a value dependent on the fact that in each small
interval of time M a small mass A3f oi liquid has its velocity
changed from zero to a backward velocity of v = V'igh ; that
* Hydraulic Motors ;
Wiley & Sons.
with related subjects. New York, 1905, John
800 MKCHAlSrrCS of EXGIlSrEETlIjlSrG.
is, has been projected with a mean acceleration of j? =
so that the forward force against the vessel is
F" = mass X ace. = ^-^ (3)
If ^ = the volume of water discharged per unit time, then
^M = -^ Jt, and since also Q = Fv = FV2gh, eq. (3) be-
comes ''Feaction"ofJet = F" = 2Fhy. ... (4)
(A similar proof, resulting in the same value for F", is
easily made if the vessel has a uniform motion with water sur-
face horizontal.)
If the orifice is in " thin plate," we understand by F the
area of the contractedj section. Practically, we have z)=0 '^'igh
(§ 495), and hence (4) reduces to
F" = ^(t>"Fhy (5)
Weisbach mentions the experiments of Mr. Peter Ewart of
Manchester, England, as giving the result F" = 1.7SFhy
with a well-rounded orifice as in Fig. 625. He also found
=: .94 for the same orifice, so that by eq. (4) we should have
F'' =■- %2^yFhy = \niFhy.
"With an orifice in thin plate Mr. Ewart found F" —
1.14:Fhy. As for a result from eq. (4), we must put, for F^
the area of the contracted section .QiF (§ 495), which, with
= .96, gives
F" = 2{My.64.Fhy = 1.18Fhy. . . . (6)
Evidently both results agree well with experiment.
Experiments made by Prof. J. B. Webb at the Stevens
Institute (see Journal of the Franklin Inst., Jan. '88, p. 35)
also confirm the foregoing results. In these experiments the
vessel was suspended on springs and the jet directly down-
ward, so that the "reaction" consisted of a diminution of the
tension of the springs during the flow.
564. Impulse of a Jet of Water on a Fixed Curved Vane (with
Borders). — The jet passes tangentially upon the vane. Fig.
IMPULSE OF JET.
801
626. B is the stationary nozzle from which a jet of water of
cross-section F (area) and velocity = c impinges tangentially
upon the vane, which has
plane borders, parallel to
paper, to prevent the lat-
eral eicape of the jet.
The curve of the vane is
not circular necessarily.
The vane being smooth,
the velocity of the water
in its curved path remains
= c at ail points a.ong
the curve. Conceive the
curve divided into a great
num.ber of small lengths each =■ ds, and subtending some
angle = dulse on thvi vane is a force
P" = VX^+T^ = 9ll ^§!(r_ cos a), . . (3)
and makes such an angle a', Fig. ^^7^, with the direction £A,
that
tan or = -TT^T =
sm a
X 1 — COS (
«•••*•
(4)
Fig. 627.
That is, tana'=cot. ia, or a' =90° -Jo:; and the force
P" bisects the angle between the original (BA), and the final
direction of the jet. For example, if a = 70°, a' = 55°; while
if a = 180°, Fig. 628, we have a'=0° and hence P" is parallel
to BA, its value being (see eq. (3)),
P"=2Qr-.
9
IMPULSE OF JET OTf VANE.
803
565. Impulse of a Jet or a Fixed Solid of E,evolutioE whose
Axis is Parallel to the Jet.- — If the
curved vane, with borders, of the pre-
ceding paragraph be replaced by a b
solid of revolution. Fig. 629, with its iT^ ^S
axis in line of the jet, the resultant
pressure of the jet upon it will simply
be the sum of the X-components (i.e., ,.„.„,,.,„^
= to JdA) 01 the pressures on all ele- Fia. eao,
ments of the surface at a given instant ; i.e.,
X^P"^ Qr -(1 - cos a);
g
(5)
while the components 1 to X, all directed toward the axis of
the solid, neutralize each other. For a fixed plate^ then, Fig.
630, at right angles to the jet, we have for the force, or " im-
pulse" (with a = 90°),
P'' =
-Qr^-^o
'^G=±^r = 2F^r' ' (6)
The experiments of Bidone, made in
1838, confirm the truth of eq. (6) quite
closely, as do also those of two students of
the University of Pennsylvania at Phila-
delphia (see Jour, of the Frank. Inst, for Oct. '87, p. 258).
We may apply eq. (6) to the
theory of Pitoes Tube (§ 539),
Fig. 631, assuming the current
to act like a jet, with a = 90°.
The water in the tube is at rest,
and its section at A. (of area=i^)
may be treated as a flat vertical
plate receiving not only the
hydrostatic pressure Fxy^ due
to the depth x below the sur- fig. esi.
face, but a continuous impulse P" = F&y -~- g [see eq. (6)].*
* This implies that the sectional area F ot the "equivalent isolated jet"
is equal to that of tlie extremity of tube and that a is 90°, an assumption
which, though simple, is largely conjectural.
804 MECHANICS OF ENGINEERING.
Hence, we write, for the equilibrium of the "plate" A,
Fxr + '-=Fxr + Fhy; i.e., h' = {2.0)^. .
(7)
But the assumed size of the "equivalent isolated jet" and
of the angle a (see foot-note, p. 803) are both probably much
too large; so that, for the factor 2.0 of eq. (7), we substi-
tute some smaller number, or coefficient, k; and hence write
. 2g
(7a)
as a theoretical relation holding good for the Pitot Tube.
The value of k can only be determined by experiment.
Pitot found k = 1.5 when the point of the tube was made
flaring hke a funnel; while Darcy, desiring that the end
of the tube should occasion but
little disturbance in the current
itself, made the extremity small and
t conically convergent.* The latter found
k practically = 1.00.
For other practical details see p. 750.
If the solid of revolution is made cup-
shaped, as in Fiii;. 632, we have (as in
Fig. 628) a = 180°, and therefore, from
eq. (5),
Fig. 632.
P'' = 2^r- = '^ = W-V.
(8)
Example. — Fig. 632. If c = 30 ft. per sec. and the jet
(cylindrical) has a diameter of 1 inch, the liquid being water,
so that y = 62.5 lbs. per cub. ft., we have [ft., lb., sec]
the impulse (force) = P" =
2f(^y9Q0X62.5
32:2
= 19.05 lbs.
Experiment would probably show a smaller result.
* See p. 833 for Mr. Freeman's Experiments.
IMPULSE OF JET OlS^ MOVING VANE.
806
Fig. 633.
566. Impulse of a Liquid Jet upon a Moving Vane having
Lateral Borders and Moving in the Direction of the Jet. — Fig.
633. The vane has a motion of translation (§ 108) in the
same direction as the jet. Call this the axis X. It is moving
with a velocity v away from the jet (or, if toward the jet, v
is negative). We con-
sider V constant, its ac-
celeration being prevented
bj a proper resistance
(such as a weight = G)
to balance the X-com-
ponents of the arc-pres-
sures. Before coming in
contact with the vane,
which it does tangentiallj
(to avoid sudden devia-
tion), the absolute velocity
(§ 83) of the water in the
jet = (3, while its velocity
relatively to the vane at A is = c — i; ; and it will now be
proved that the relative velocity along the vane is constant.
See Fig. 634. Let v = the velocity of the vane (of each
point of it, since its motion is one of translation), and u = the
velocity of a water particle (or small mass of water of length
= ds) relatively to the point of the vane which it is passing.
Then %o^ the absolute velocity of the small mass, is the diago-
nal formed on xi and v. Neglecting friction, the only actual
force acting on the mass is P^ the pressure of the vane against
it, and this is normal to the curve. Now an imaginary system
of forces, equivalent to this actual system of one force P^ i.e.,
capable of producing the same motion in the mass, may be
conceived of, consisting of the individual forces which would
produce, separately, the separate motions of which the actual
motion of this small mass M is compounded. These com-
ponent motions are as follows :
1. A horizontal uniform motion of constant velocity = -y ;
and
2. A motion in the arc of a circle of radius = r and with a
806 MECHAisrics or eistghstekrhstg.
velocity = u, which we shall consider variable until proved
otherwise.
Motion 1 is of such a nature as to call for no force (by New-
ton's tirst law of motion), while motion 2 could be maintained
bj a system of two forces, one normal, P^, = -, and the
other tangential, P^ = Jf — — [see eq. (5), p. Y6]. This imagi-
nary system of forces is shown at (IL), Fig. 634, and is equiv-
(II.) /
V
Fig. 634.
alent to the actual system at (I.). Therefore ^(tang. com-
pons..) in (I.) should be equal to '2 (tang, compons.) in (II.) ;
whence we have
P, = 0; i.e., J/| = 0; or | = 0; . . (1)
i.e., ti is constant along the vane and is equal to c — -y at every
point. (The weight of the mass has been neglected since the
height of the vane is small.) In Fig. 631 the symbol u\ has
been used instead of c, and the point corresponds to A in
Fig. 633.
[N.B. If the motion of the vane were rotary^ about an axis
1 to AB (or to 6'), this relative velocity would be different at
different points. See p. 59 of Hydraulic Motors. If the
radius of motion of the point A, however, is quite large com-
pared with the projection oi AD upon this radius, the relative
velocity is approximately = c — v at all parts of the vane,
and will be taken =c— ■y in treating the " Hurdy gurdy" in
§561]
WORK OF JET ON VANE. 807
By putting 2 (normal compons.) of (I.) = 2 (normal com-
pons.) in (II.) we have
P = P.; i.e., P = Jf g: = ^(^ - ^)' ; ... (2}
so that to find the sura of the JT-components of the pressures
exerted against the vane simultaneously by all the small masses
of water in contact with it at any instant, the analysis differs
from that in § 564 only in replacing the o of that article by
the {c — v) of this. Therefore
2(Xp.^».^) = p. = J:(« -«)■[!- cos „], . (3)
(where a is the angle of total deviation, relatively to vane, of
the stream leaving the vane, from its original direction), and
is seen to be proportional to the square of the relative velocity,
i^is the sectional area of jet, and y the heaviness (§ 7) of the
liquid. The J^-component (or P^ of the resultant impulse
is counteracted by the support EF^ Fig. 633. Hence, /br a
imifonn 7)iotion to he maintained, with a given velocity = v^
the weight G must be made = P^ of eq. (3). (We here
neglect friction and suppose the jet to preserve a practically
horizontal direction for an indefinite distance before meeting
the vane. If this uniform motion is to be toward the jet, v
will 1)6 negative in eq. (3), making ^^.(and .*. G) larger than
for a positive v of same numerical value.
As to the doing of woi-lc [§§ 128, etc.], or exchange of
energy, between the two bodies, jet and vane, during a uni-
form motion away from the jet, Pg, exerts 2k power of
Fv
L = P^v = —^ {g — vfv\l — cos or], . , . (4)
in which L denotes the number of units of work done per unit
of time hyPa,', i.e., t\\Q power (§ 130) exerted by /*«,.
If -y is negative, call it — v', and we have the
Power expended ) d , Fy / , /xs /n n /»
7 ^ ' ^t = Pa^v = — ^ (c + v)v\l ~ cos a\. . (op
by vane -upon jet ) * a *
MECHANICS OF EISTGINEEEING.
Of course, practically, we are more concerned with eq. (4)
than with (5). The power L in (4) is a maximum for v =^\c\
but in practice, since a single moving vane or float cannot
utilize the water of the jet as fast as it flows from the nozzle,
let us conceive of a succession of vanes coming into position
consecutively in front of the jet, all having the same velocity
-y ; then the portion of jet intercepted between two vanes is at
liberty to finish its work on the front vane, while additional
work is being done on the hinder one ; i.e., the water will be
utilized as fast as it issues from the nozzle.
With such a series of vcmes, then, we may put Q\ = J9cy
the volume of flow per unit of time from the nozzle, in place
of J^\g — v) = the volume of flow per unit of time over the
vane, in eq. (4) ; whence
Power exerted on
series of vanes
= L'=
g
[1 — cos a^G — v^v. . (6)
Making v variable, and putting dL'-^dv=^Q^ whence 0—2^=0,
we find that for v = |^c, L', the power, is a maximnm. As-
suming different values for or, we find that for a = 180°, i.e.,
by the use of a semicircular vane, or of a hemispherical cup,
Fig. 635, with a point in middle, 1 — cos or is a max., = 2 ;
whence, with v = ^c, we have, as the
maximitm power,
_Qy G^_Ml, ( «=180°, ) ,^
^max.- ^ '2"" 2 '|'y=4c; \"^^f
in which M' denotes the mass of the flow
y per unit of time from the stationary
\°J] nozzle. 'Now ■■ is the entire hinetio
2i
energy furnished per unit of time by the
jet ; hence the motor of Fig. 635 {series
Fig. 635. of cups) has a theoretical efficiency of
unity, utilizing all the kinetic energy of the water. If this is
true, the absolute velocity of the particles of liquid where they
leave the cup, or vane, should be zero, which is seen to be true,
POWER OF JETS.
SOD
AS follows : At ^ or H ., the velocity of the particles rela-
tively to the vane is — (? — -y = what it was at 4, and hence
G C
is ^ c — — = -- ; hence at H the absolute velocity \% w=^'
G C
(rel. veloc. — toward left)— (veloc. — of vane toward right) = ;
Q.E.D. For t; > or < ^g this efficiency will not be attained.
567 The California "Hurdy-gurdy;" or Pelton Wheel. —The
efficiency of unity in the series of cups just mentioned is in
practice reduced to 80 or 85 per cent from friction and lateral
escape of water. The
Pelton wheel or Cali-
fornia '' Hurdy-gur-
dy," shown (in prin-
ciple only) in Fig. 636,
is designed to utilize
the n)echanical rela-
tion just presented,
and yields results con-
firming the above the-
ory, viz., that with the
linear velocity of the
Fig. 636.
cup-centres regulated to equal — , and with a = 180°, the effi-
ciency approaches unity or 100 per cent. Each cup has a pro-
jecting sharp edge or rib along the middle, to split the jet ; se*>
Fig. 635. (See also p. TO, Hydraulic Motors.)
This wheel was invented to utilize small jets of very great
velocities (c) in regions just deserted by " hydraulic mining"
operators. Although g is great, still, by giving a large value
to r, the radius of the wheel, the making of -y = - does not
2
jQecessifcate an inconveniently great speed of rotation (i.e.,
revolutions per unit of time). The plane of the wheel may
be in any convenient position.
In the London Engineer of May '84, p. 397, is given an ac-
count of a test* made of a " Hurdy-gurdy," in which the motoi
* See p. 834 for further details of this test and a perspective view of wheel
810
MECHA]SriCS or ENGINEERING.
showed an efficiency of 8Y per cent. The diameter of the
wheel was onlj 6 ft., that of the jet 1.89 iD., and the head of
the supply reservoir 386 ft., the water being transmitted
through a pipe of 22 inches diameter and 6900 ft. in length.
107 H. P. was developed by the wheel.
Example. — If the jet in Fig. 636 has a velocity c = 60 ft.
per second, and is delivered through a 2-inch nozzle, the total
power due to the kinetic energy of the water is (ft., lb., sec.)
Q
'!..-= — .-(-?-)'x60x62.5xiX3500=4566.9
g 2 32.2 4V12/ ^
and if, by making the velocity of the cups =
ft. lbs.
p. sec,
30 ft. per
sec, 85 per cent of this power can be utilized, the iDower of
the wheel at this most advantageous velocity is
Z = .85 X 4566.9 = 3881 ft. lbs. per sec. = Y.05 horse-power
[since 3881 ^ 550 = Y.05] (§ 132). For a cup-velocity of 30
ft. per sec, if we make the radius, r, = 10 feet, the angular
velocity of the wheel will be (i9 = 'y-7-r=3,0 radians per
sec. (for radian see Example in § 428 ; for angula'" velocity,
§ 110), which nearly = tt, thus implying nearly a half-revolu-
tion per sec.
568. Oblique Impact of a Jet on a Moving Plate having
no Border. — The plate
has a motion of trans-
lation with a uniform
veloc = iJ in a direc-
tion parallel to jet,
whose velocity is ^ p.
At the filaments of
liquid are deviated, so
that in leaving the plate
their particles are aiv
'W07777mmm^77fim777/M/777mMmmm^^?;7MF/ found in the moving
Fig. 637. plane BB' of the plate
surface, but the respective absolute velocities of these particles
OBLIQUE JET AND PLATE. 811
depend on the location of the point of the plate where they
leave it, being found by forming a diagonal on the relative
veioc. G — v and the velocity v of the pJate. For example, at
B the absolute velocity of a liquid particle is
10 = BE = Vv' + (c — vf + 2v{G — v) cos a,
while at B' it is
w' = B'E' — Vv' + (c — vj — 2v{G — v) cos a ;
but evidently the component ~\ to plate (the other component
being parallel) of the absolute velocities of all particles leaving
the plate, is the same and = v sin a. The skin-friction of the
liquid on the plate being neglected, the resultant impulse of
the jet against the plate must be fiormal- to its surface, and its
amount, i^, is most readily found as follows :
Denoting by JM the mass of the liquid passing over the
plate in a short time /It, resolve the absolute velocities of all
the liquid particles, before and after deviation, into com-
ponents ~I to the plate (call this direction Y) and || to the
plate. Before meeting the plate the particles composing JM
have a velocity in the direction of Y oi Cy = c sin a ; on leav-
ing the plate a velocity in direction of Y of v sin a : they have
therefore lost an amount of velocity in direction of jT =
(g — v) sin (x in time /It ; i.e., they have suffered an average
retardation (or negative acceleration) in a I^-direction of
( neo:. aecelera- ] (g — v) sin a ,^.
A= Ition II to r f = Jt • •••(!)
Hence the resistance in direction of J^(i.e., the equal and op-
posite of P in figure) must be
£*Y = Tonsi&& X I^-accel. = — — {g— ^) sin or ; . . (2)
and therefore, since = M = -^ = mass of liquid passing
Jt g
812 MECHANICS OF ENGIJSTEEKING.
over the plate per unit of time (not that issuing from nozzle),
we have
sure on plate \ a 9
in which J^ = sectional area of jet before meeting plate.
[JST.B. Since eq. (3) can also be written I^ = Mcsma —
Mv sin a, and Mc sin a may be called the 1^-momentuni before
contact, while Ifv sin a is the 1^-momentum after contact (of
the mass passing over plate per unit of time), this method may
be said to be founded on the principle of momentuTn which is
nothing more than the relation that the accelerating force in
any direction = mass X acceleration in that direction ; e.g.^
P, = Mf, ; Py = Mpy ; see § 74.]
If we resolve P, Fig. 637, into two components, one, P', \\
to the direction of motion {v and c), and the other, P'', ~\ to
the same, we have
P' = P8ma = ^{c-v)sm''a, , ... (4)
and
Or
{Q = F{c — v) = volume passing over the plate per unit of
time.) The force P" does no work, while the former, P\
does an amount of work P'v per unit of time ; i.e., exerts a
power (one plate)
:=L = P'v = Sy(G-v)v&m^a. ... (6)
9
If, instead of a single plate, a series of plates, forming a
regular succession, is employed, then, as in a previous paragraph,
we may replace Q, = F{c — v), by Q' = Fc, obtaining as the
Power exerted hy jet ) ri Fey , % . , /m
on series of plates ) g ^ 7c/oxu «. . ^.^
IMPULSE OF JET ON OBLIQUE PLATE.
813
For V
and a = 90° we have
Z\
1 Fey c' _ 1 M'&
2 a 2 ~ 2 . 2
. (8)
Fi&. 638.
= only half the kinetic energy (per time-nnit) of the jet.
569. Rigid Plates Moving in a Fluid, Totally Submerged,
Fluid Moving against a Fixed Plate. Impulse and Resistance. —
If a thin flat rigid plate have a motion of uniform translation
with velocity = v tlirough a fluid
which completely surrounds it, Fig.
638, a resistance is encountered (which
must be overcome by an equal and op-
posite force, not shown in figure, to
preserve the uniform motion) consist-
ing of a normal component N^ 1 to
plate, and a (small) tangential com-
ponent, or skill-friction, T, i| to plate.
Unless the angle rv, between the surface of plate and the direc-
tion of motion 6* ... -y, is very small, i.e. unless the plate is
moving nearly edgewise through the fluid, N is usually much
greater than T. The skin-resistance between a solid and a fluid
has already been spoken of in § 510.
"When the plate and fluid are at rest the pressures on both
sides are normal and balance each other, being ordinary static
fluid pressures. When motion is in progress, however, the
normal pressures on the front surface are increased by the
components, normal to plate, of the centrifugal forces of the
curved filaments (such as AB) or "stream-lines," while on
the back surface, D, the fl.uid does not close in fast enough to
produce a pressure equal to that (even) of rest. In fact, if the
moticv is sufficiently rapid, and the fluid is inelastic (a liquid),
a vacuum may he maintained hehind the plate, in which case
there is evidently no pressure on that side of the plate.
Whatever pressure exists on the back acts, of course, to
diminish the resultant resistance. The water on turning the
sharp corners of the plate is broken up into eddies forming a
814
MECHANICS OF ENGINEERING.
" wake" behind. From the accompaniment of these eddies,
the resistance in this case (at least the component N normal to
plate) is said to be due to "" eddy-making f though logically
we should say, rather, that the body does not derive the assist-
ance (or negative resistance) from behind which it would ob-
tain if eddies were not formed ; i.e., if the fluid could close in
behind in smooth curved stream-lines symmetrical with those
in front.
The heat corresponding to the change of temperature pro-
duced in the portion of fluid acted on, by the skin-friction
and by the mutual friction of the particles in the eddies, is the
equivalent of the work done (or energy spent) by the motive
force in maintaining the uniform motion (§ 149). (Joule's
experiments to determine the Mechanical Equivalent of Heat
were made with paddles moving in water.)
If the fluid is sea-water^ the results of Col. Beaufoy's ex-
periments are applicable, viz.:
The resistance^ jper square foot of area, sustained hy a sub-
7)ierged plate momng normally to itself [i.e., a = 90°] in sea-
water with a velocity of v ^= V) ft. per second is 112 Tbs. He
also asserts that/br other velocities the resistance varies as the
square of the velocity. This latter fact we would be led to
suspect from the results obtained in § 568 for the impulse of
jets; also in §565 [see eq. (6)]. Also, that when the plate
moved obliquely to its normal (as in Fig. 638) the resistance
was nearly equal to ifhe resistance.^ at same velocity., when
oc = 90°) X {the sine of the angle a) ', also, that the depth of
submersion had no injltie7ice on the resistance.
Confining our attention to a plate moving nor-
mally to itself Fig. 639, let F= area of plate,
y = heaviness (§ 409) of the fluid, v = the uni-
form velocity of plate, and ^ = the acceleration
of gravity (=32.2 for the foot and second ;
= 9.81 for the metre and second). Then from
the analogy of eq. (6), § 565, where velocity c of
the jet against a stationary plate corresponds to
the velocity v of the plate in the present cas©
moving through a fluid at rest, we may write
Fig. 639.
PLATES MOVING IN FLUIDS. 815
Resistance of fluid \ _ p _ ^j^ '^'^ \v normal ) r^.
to momng jplate \ ~ , ~ ^ ^ 2g' ' ' ' ( to plate \ ' ' ' ^^
And similarly for the impulse of an indefinite stream of fluid
against a fixed jplate ( 1 to velocity of stream), v being the
velocity of the current,
Impulse of current \ _ p _ ^r-p '^^ \v normal ) .^\
upon fixed plate ) ~ ""'=' ^ ^ ' ' ' \ ^o plate \ ' ' °^^y
The 2^ is introduced simply for convenience; since, having
V given, we may easily tind v"^ -=- 2^ from a table of velocity-
heads; and also (a ground of greater importance) since the co-
efficients C and C' which depend on experiment are evidently
ahstract numbers in the present form of these equations (for
R and P are forces, and Fy'v" -^ 2^ is the weight (force) of
an ideal prism of fluid ; hence C and Q' must be abstract
numbers.)
From Col. Beaufoy's experiments (see above), we have for
sea-water [ft., lb., sec], putting R = 112 lbs., P= 1 sq. ft.,
y = Q4: lbs. per cub. ft., and -y = 10 ft. per second,
2X32.2X112 ^-^
1.0 X 64 X 10' * ■
Hence in eq. (1) for sea-water, we may put C = 1'13 (with
y = Q4: lbs. per cub. ft.).
From the experiments of Dubuat and Thibault, Weisbach
computes that for the plate of Fig. 639, moving through either
water or air, C = 1-25 for eq. (1), in which the y for air must
be computed from § 473 ; while for the impulse of water or
air or. fixed plates he obtains C,' = 1.86 for use in eq. (2). It
is hardly reasonable to suppose that C and C,' should not be
identical in value, and Prof. Unwin thinks that the difference
in the numbers just given must be due to errors of experi-
ment,* The latter value, C' = 1.86, agrees well with equation
(6) below. For great velocities C and Q' are greater for air
than for water, since air, being compressible, is of greater
heaviness in front of the plate than would be computed for
* Flamant thinks that this difference is due to the fact tliat the relative
conditions are not identical in the two cases; since when a current of liquid
impinges against a stationary plate there is much intricacy of internal motion
among the particles of fluid, to which there is nothing to correspond when
a plate is moved through stationary liquid.
816 MECHANICS OF ENGINEEEING.
tbe given temperature and barometric height for use in eqs.
(1) and (2)
The experiments of Borda in 1763 led to the formula
P = [0.0031 + O.OOOS^c'jSv' .... (3)
for the total pressure upon a plate moving through the air
in a direction "] to its own surface. P is the pressure in
pounds, G the length of the contour of the plate in feet, and jS
its surface in square, feet, while v is the velocity in miles per
hour. Adopting the same form of formula, Hagen found,
from experiments in 18Y3, the relation
P= [0.002894 + 0.00014c]xSV ... (4)
for the same case of fluid I'esistance.
Hagen's experiments were conducted with great care, but
like Borda's were made with a "whirling machine," in which
the plate was caused to revolve in a horizontal circle of only
7 or 8 feet radius at the end of a horizontal bar rotating about
a vertical axis. Hagen's plates ranged from 4 to 40 sq. in. in
area, and the velocities from 1 to 4 miles per hour.
The last result was quite closely confirmed by Mr. H. Allen
Hazen at Washington in JSTovember 1886, the experiments
being made with a whirling machine and plates of from 16 to
576 sq. in. area. (See the Americcm Journal of Science, Oct.
1887, p. 245.)
In Thibault's experiments plates of areas 1.16 and 1.531 sq,
ft. were exposed to direct wind-pressure, giving the formula
P = 0.00475.^^= » • (5)
Recent experiments in France (see R. R. and Eng. Journal^
Fe^ . '87), where flat boards were hung from the side of a rail-
way train run at different velocities, gave the formula
P = 0.00535^^' (6J
The highest velocity was 44 miles per hour. The magnitude
of the area did not seemingly affect the relation given.* More
* Langley found P = 0.00327*Sv^. See also Irminger's experiments
i-^ngineering News, Feb. 1895, p. 109).
PLATES IN FLUIDS. 817
extended and elaborate experiments are needed in this field,
those involving a motion of translation being considered the
better, rather than with whirling machines, in which " centrif-
ugal action"" must have a disturbing influence.*
The notation and units for eqs. (4), (5), and (6) are the same
as those gir^n for (3).
It maj be of interest to note that if equation (3) of § 568 be
considered applicable to this case of the pressure of an un-
limited stream of fluid against a plate placed at right-angles to
the current, with F put equal to the area of the plate, we ob-
tain, after reduction to the units prescribed above for the pre-
ceding equations and putting a = 90°,
P=0.0053^'y' (T)
The value y = 0.0807 lbs. per cub. ft. has been used in the
substitution, corresponding to a temperature of freezing and
a barometric height of 30 inches. At higher temperatures,
of course, y would be less, unless with very high barometer.
568a. Example. — Supposing each blade of the paddle-wheel
of a steamer to have an area of 6 sq. ft., and that when in the
lowest position its velocity [relatively to the water, not to the
vessel] is 5 ft. per second ; what resistance is it overcoming in
salt water ?
From eq. (1) of § 569, with C = 1-13 and y = %4t lbs. per
cubic foot, we have (ft., lb., sec.)
1.13X6X64X25^
2 X 32.2
If on the average there may be considered to be three pad-
dlea always overcoming this resistance on each side of the
boat, then the work lost (work of •' slip''') in overcoming these
resistances per second (i.e., power lost) is
Z, = [6 X 169.4] lbs. X 5 ft. per sec. = 5082 ft.-lbs. per sea
or 9.24 Horse Power (since 5082 -^ 550 = 9.24).
* See Capt. Bixby's article on p. 175 et seq. of the Engineering News.
March 1895.
818
MKCHAN^ICS OF EISTGINEERHSTG.
If. further, the velocity of the boat is uniform and ■= 20 ft.
per sec, the resistance of the water to the progress of the boat
at this speed being 6 X 169.4, i.e. 1016.4 lbs., the power ex-
pended in actual propulsion is
Z, = 1016.4 X 20 = 20328 ft.-lbs. per sec.
Hence the power expended in both ways (usefully in propul-
sion, uselessly in "slip") is
L, + L, = 25410 ft.-lbs. per sec. = 46.2 H. P.
Of this, 9.24 H. P., or about 20 per cent, is lost in " slip."
570. Wind-pressure
on the surface of a
roof inclined at an
angle = a with the
horizontal, i.e., with
the direction of the
wind, is usually esti-
mated according to
the empirical formula
Fig. 640.
(Hutton's)
p=p' [sin a] [i-84cosa- 1]^
(1)
in which ^' = pressure of wind per unit area against a vertical
surface ( 1 to wind), and p = that against the inclined plane
{and normal to it) at the same velocity. For a value of
p' =: 40 lbs. per square foot (as a maximum), we have the
following values iorp, computed from (1) :
For a = 5"
10°
15°
30°
25°
30°
26.5
35°
30.1
40°
33.4
45°
36.1
50°
38.1
55°
39.6
60°
40.
p={lhs. sq. ft.) 5. 2
9.6
14
18.3
22.5
Duchemln's formula for the normal pressure per unit-area is
_p=y.4M!Lf_, (2)
■^ ^ 1 + sin' a ^ ^
WIND AND SAIL.
819
with the same notation as above. Some experimenters in
London tested this latter formula by measuring the pressure
on a metal plate supported in front of the blast-pipe of a blow-
ing engine; the results were as follows:
a =
15°
20°
60°
,0°
p by experiment = (in lbs. per sq. ft.)
1.65
2.05
3.01
3.31
By Duchemin's formula p =
1.60
2.02
3.27
3.31
The scale of the Smithsonian Institution at Washington for
the estisiation and description of the velocity and pressure of
the wind is as follows :
Grade.
Velocity iu
miles per hour.
Pressure per
sq. foot in lbs.
Name.
0.00
Calm.
1
2
0.02
Very light breeze.
2
4
0.08
Gentle breeze.
3
12
0.75
Fresh wind.
4
25
3.00
Strong wind.
5
35
6
High wind.
6
45
10
Gale.
7
60
18
Strong gale.
8
75
Violent gale.
9
90
Hurricane.
10
100
Most violent hurricane.
571. Mechanics of the Sail-boat. — The action of the wind on a
sail will be understood from the following. Let Fig. 641
represent tbe boat in horizontal projection and OS the sail, O
beino; the mast. For
simplicity we consider
the sail to be a plane
and to remain vertical.
At this instant the boat
is moving in the direc-
tion M£ of its fore-and-
aft line with a velocity
= c, the wind having a velocity of the direction and magni-
tude represented by iv (purposely taken at an angle < 90° with
the direction of motion of the boat). We are now to inquire
the nature of the action of the wind on the boat, and whether
Fig. 641.
820 MECHANICS OF ENGINEERING.
in the present position its tendency is to accelerate, or retard,
the motion of the boat. If we form a parallelogram of which
w is the diagonal and c one side, then the other side OK^ mak-
ing some angle a with BM^ will be the velocity v of the wind
relaiirely to the hoat (and sail), and upon this (and not upon w)
depends the action on the sail. The sail, being so placed that
the angle 6 is smaller than a, will experience pressure from
the wind ; that is, from the impact of the particles of air which
strike the surface and glance along it. Tnis pressure, P, is
normal to the sail (considered smooth), and evidently, for the
position of the parts in the figure, the component of P along
MB points in the same direction as c, and hence if that com-
ponent is greater than the water-resistance to the boat at this
velocity, c will be accelerated; if less, c will be retarded.
Any change in c, of course, gives a different form to the
parallelogram of velocities, and thus the relative velocity v
and the pressure P, for a given position of the sail, will both
change. [The component oi P ~\ to MB tends, of course, to
cause the boat to move laterally, but the great resistance to
such movement at even a \evy slight lateral velocity will make
the resulting motion insignificant.]
As G increases, a diminishes, for a given amount and position
of w ; and the sail must be drawn nearer to the line MB, i.e.
B must be made to decrease, to derive a wind-pressure having
a forward fore-and-aft component ; and that component be-
comes smaller and smaller. But if the craft is an ice-boat, this
small component may still be of sufficient magnitude to exceed
the resistance and continue the acceleration of c until g is
larger than w ; i.e., the boat may be caused to go as fast as, or
faster than, the wind, and still be receiving from the latter a
forward pressure which exceeds the resistance. And it is
plain that there is nothing in the geometry of the figure to
preclude such a relation (i.e., c > i«, with B <, a and > 0).
5/2. Ji^esistance of Still Water to Moving Bodies, Completely
Immersed. — This resistance depends on the shape, position, and
velocity of the moving body, and also upon the roughness of
its surface. If it is pomted at both ends (Fig. 642) with its
MOVING BODIES, IMMERSED.
821
Fig 642.
A ; so that the resistance
axis parallel to the velocity, v, of its xinifornn motion, the
streamlines on closing to-
gether smoothly at the hinder —; -V-VIll
extremity, or stern, B^ exert
normal pressures against the
surface of the portion CD...B
whose longitudinal compo-
nents approximately balance
the corresponding components
of the normal pressures on CD
R^ which must be overcome to maintain the uniform velocity
V, is mainly due to the " skin-frictioti)' alone, distributed along
the external surface of the body ; the resultant of these resist-
ances is a force R acting in the line ^^ of symmetry (sup-
posing the body symmetrical about the direction of motion).
If, however. Fig. 643, the stern, E..B..F h too bluff,
eddies are formed round the corners
^and F, and the pressure on the
surface F . . . F is much less than
in Fig. 642; i.e., the water pres-
sure from behind is less than the
backward (longitudinal) pressures
from in front, and thus the resultant
resistance B is due partly to skin-
friction and partly to "eddy-making" (§ 569).
[Note. — The diminished pressure on FF is analogous to the
loss of pressure of water (flowing in a pipe) after passing a nar-
row section the enlarsjement
from which to the original
section is sudden. E.g., Fig.
644, supposing the yelocity v
and pressure p (per unit-ai'ea)
to be the same respectively
at A and A\ in the two
pipes shown, with diameter
AZ = WK=A'Z' = W'K' ;
then the pressure at M is
equal to that at A (disregarding skin-friction), whereas that at
Fig. 643.
=^-
/^
rszjr^
^1^-
ii:
---i^
zzrz
^
-_ V
L
K
A
. c
> ^
w'
-;-:
J^^^
1
-•^^' '-"---"
"-~;
~-V-
^^iP^
r_^--r
-~
;--- -.m'-
--V
ll:
-1
L'
K'
Fig.
644.
822
MEOHAlSriCS OF ENGINEEEING.
that at M is considerably less than that at A' on account
of the head lost in the sudden enlargement. (See also Fig.
575.)]
It is therefore evident that hluffness of stern may he a large
factor in the production of resistance.
In any case experiment shows that for a given body sym-
metrical about an axis and moving through a fluid (not only
water, but any fluid) in the direction of its axis with a uni-
form velocity = -WjWe may write approximately the resistance
M = {resistance at vel. v) = Q2^y
^9
(1)
As in preceding paragraphs, -F= area of the greatest section,
"I to axis, of the external surface of body (not of the sub-
stance; i.e., the sectional area of the circumscribing cylinder
(cylinder in the most general sense) with elements parallel
to the axis of the body, y = the heaviness (§ 409) of the
fluid, and v = velocity of motion ; while C is an abstract
number dependent on experiment.
According to Weisbach, who cites different experimenters,
we can put for spheres, moving in water, C = about 0.55 ;
for cannon-balls moving in water, C = .467.
According to Robins and Hutton, for spheres in mV, we
have
For 1) in mets. )
per sec. ) ^
5
25
100
200
300
400
500] -Sc.
C = .59
.63
.67
.71
.77
88
.99
1.04
For musket-balls in the air, Piobert found
C = 0.451 (1 -\- 0.0023 X veloc. in metres per sec).
From Dubuat's experiments, for the resistance of water to
a right prism moving erudwise and of length = Z,
For {I : VF) =
1
2
3
C = 1.25
1.26
1.31
1.33
For a circular cylinder moving perpendicularly to its axis
Borda claimed that C is one-half as much as for the circum-
MOVIISTG BODIES, IMMEESED. 823
scribing right parallelepiped moving with four faces parallel
to direction of motion.
Example. — The resistance of the air at a temperature of
freezing and tension of one atmosphere to a musket-ball ^ inch
in diameter when moving with r velocity of 328 ft. per sec
1,3 thus determined by Piobert's formula, above :
^,' = 0.451(1 + .0023 X 100) ^ 0.554 ;
hence, from eq. (1),
E = 0.554 X ~ (^Sx .0807 X ^|f??= 0.1018 lbs.
4 \12/ 64.4
572a. Deviation of a Spinning Ball from a Vertical Plane m
Still Air. — It is a well-known fact in base-ball playing that if a
rapid spinning motion is given to the ball about a vertical axis
as well as a forward motion of translation, its path will not
remain in its initial vertical plane, but curve out of that plane
toward the side on which the absolute velocity of an external
point of the ball's surface is least. Thus, if the ball is thrown
from llTorth to South, with a spin of such character as to ap-
pear '' clock-wise^'' seen from above, the ball will curve toward
the West, out of the vertical plane in which it started.
This could not occur if the surface of the ball were perfectly
smooth (there being also no adhesion between that surface and
the air particles), and is due to the fact that the cushion of com-
pressed air which the ball piles up in front during its progress,
and whicn would occupy a symmetrical position with respect
to the direction of motion of the centre of the ball if there
were no motion of rotation of the kind indicated, is now piled
up somewhat on the East of the centre (in example above),
crerting cf :;stantly more obstruction on that side than on the
right ; the cause of this is that the absolute velocity of the sur-
face-points, at the same level as the centre of ball, is greatest,
and the friction greatest, at the instant when they are passing-
through their extreme Easterly positions; since then that
velocity is the sum of the linear velocity of translation and
that of rotation ; whereas, in the position diametrically oppo«
824 MECHAN^ICS OF ENGINEERING.
site, on the West side, the absolute velocity is the difference ;
hence the greater accumulation of compressed air on the left
(in the case above imagined, ball thrown from ]!^orth to South,
etc.).
573. Robinson's Cup-anemometer. — This instrument, named
after Dr. T. R. Robinson of Armagh, Ireland, consists of four
hemispherical cups set at equal intervals in a circle, all facing
in the same direction round the circle, and so mounted on a
light but rigid framework as to be capable of rotating witt
but little friction about a vertical axis. When in a current oi
air (or other fluid) the apparatus begins to rotate with an ac-
celerated velocity on account of the pressure against the open
mouth of a cup on one side being greater than the resistance
met by the back of the cup diametrically opposite. Very soon,
however, the motion becomes practically uniform, the cup-
centre having a constant linear velocity v" the ratio of which
to the velocHy, v', of the wind at the same instant must be
found in some way, in order to deduce the value of the latter
from the observed amount of the former in the practical use
of the instrument. After sixteen experiments made by Dr.
Robinson on stationary cups exposed to winds of varying in-
tensities, from a gentle breeze to a hard gale, the conclusion
was reached by him that with a given wind- velocity the total
pressure on a cup with concave surface presented to the wind
was very nearly four times as great as that exerted when the
convex side was presented, whatever the velocity (see vol.
XXII of Transao. Irish Royal Acad., Part /, p. 163).
Assuming this ratio to be exactly 4.0 and neglecting axle-
friction, we have the data for obtaining an approximate value
of m, the ratio of v' to the observed -y", when the instrument is
in use. The influence of the wind on those cups the planes of
whose mouths are for the instant !| to its direction will also be
neglected.
If, then, Fig. 645, we write the imptdse on a cup when the
iiollow is presented to the wind [§ 572, eq. (1)]
P. = ZnFr''^, (1)
EOBiisrsoisr s cup-anemometer.
825
and the resistance when the convex side is presented
P. = z.Fr^,
... (2}
•we may also put
In (1) and (2) v and v^ are relative velocities.
Regarding only the two cups A and B,
whose centres at a definite instant are mov- -
ing in lines parallel to the direction of the _«l>
wind, it is evident that the motion of the
cups does not become uniform until the rel-
ative velocity v' — v" of the wind and cup
A (retreating before the wind) has become
so small, and the relative velocity v' 4- v"'
with which B advances to meet the air-
particles has become so great, that the im-
pulse of the wind on A eqtials the resist-
ance encountered by B\ i.e., these forces,
Pj^ and Pel niust be equal, having equal
lever-arms about the axis. Hence, for uniform rotary m
i^'-v'y _^ov'+v'j.
otion,
. . (4)
i.e. [see eq. (3)],
V
-1
V
+ 1 ;
4:(m~tf^{m-^l)\ .
Solving the quadratic for 'm, we obtain
m = 3.00. .
(6)
That is, the velocity of the wind is about three times that of
the cup-centre.
574. Experiments with Robinson's Cup-anemometei. -The
ratio 3.00 just obtained is the one in common use in connec-
tion with this instrum^ent in America. Experiments by Mr.
826 MECHANICS OF EISTGINEERHSTG.
Hazen at Washington in 1886 {Am. Jour. Science, Oct. '87,
p. 248) were made on a special type devised bj Lieut. Gibbon.
The anemometer was mounted on a whirh'ng machine at the
end of a 16-ft. horizontal arm, and values for m obtained, with
velocities up to 12 miles per hour, from 2.84 to 3.06 ; average
2.94. The cups were 4 in. in diameter and the distance of their
centres from the axis 6.72 in., these dimensions being those
usually adopted in America. This instrument was nearly new
and was well lubricated.
Dr. Robinson himself made an extensive series of experi-
ments, with instruments of various sizes, of which an account
may be found in the Philos. Transac. for 1878, p. 797 (see
also the volume for 1880, p. 1055). Cups of 4 in. and also of
9 in. were employed, placed first at 24 and then at 12 in. from
the axis. The cup-centres revolved in a (moving) vertical
plane perpendicular to the horizontal arm of a whirling-
machine ; this arm, however, was only 9 ft. long. A friction-
brake was attached to the axis of the instrument for testing the
effect of increased friction on the value of m. At high speeds
of 30 to 40 miles per hour (i.e., the speed of the centre of the
instrument in its horizontal circle, representing an equal speed
of wind for an instrument in actual use with axis stationary)
the effect of friction was relatively less than at low velocities.
That is, at high speeds vdth considerable friction the value of
m was nearly the same as with little friction at low speeds.
With the large 9 in. cups at a distance of either 24 or 12 in.
from the axis the value of m at 30 miles per hour ranged
generally from 2.3 to 2.6, with little or much frict'on ; while
with the minimum friction m rose slowly to about 2.9 as the
velocity diminished to 10 miles per hour. At 5 miles per
hour with minimum friction m was 3.5 for the 24 in. instru-
ment and about 5.0 for the 12 in. The effect of considerable
friction at low speeds was to increase m, making it as high as
B or 10 in some cases. With the 4 in.-cups no value was ob-
tained for m less than 3.3. On the whole, Dr. Robinson con-
cluded i:hat m is more likely to have a constant value at all
Telocities the larger the cups, the longer the arms, and the less
the friction, of the anemometer. But few straight-line experi-
VARIOUS ANEMOMETEES. 827
ments have been made with the cup-anemometer, the most
noteworthy being mentioned on p. 308 of the Engineering
News for October 1887. The instrument was placed on the
front of the locomotive of a train running between Baltimore
and Washington on a calm day. The actual distance is 40
miles between the two cities, while from the indications of the
anemometer, assuming m = 3.00, it would have been in one trip
46 miles and in another 47. The velocity of the train was 20
miles per hour in one case and 40 in the other.
575 Other Anemometers. — Both Biram's and Castello's ane-
mometers consist of a wheel furnished with radiating vanes
set obliquely to the axis of the wheel, forming a small "wind-
mill," somewhat resembling the current-meter for water shown
in Fig. 604 ; having six or eight blades, however. The wheel
revolves with but little friction, and is held in the current of
air with its axis parallel to the direction of the latter, and very
quickly assumes a steady motion of rotation. The number of
revolutions in an observed time is read from a dial. The in-
struments must be rated by experiment, and are used chiefly
in measuring the velocity of the currents of air in the galleries
of mines, of draughts of air in flues and ventilating shafts, etc.
To quote from vol. v of the Report of the Geological Sur-
vey of Ohio, p. 370 : " Approximate measurements (of the
velocity of air) are made by miners by flashing gunpowder,
and noting with a watch the speed with which the smoke
moves along the air-way of the mine. A lighted lamp is
sometimes used, the miner moving along the air-gallery, and
keeping the light in a perfectly perpendicular position, noting
the time required to pass to a given point."
Another kind makes use of the principle of Bitot's Tube
(p. 751), and consists of a U-tube partially filled with water,
one end of the tube being vertical and open, while the other
turns horizontally, and is enlarged into a wide funnel, whose
mouth receives the impulse of the current of air ; the differ-
ence of level of the water in the two parts of the U is a meas-
ure of the velocity.
828 MECHANICS OF ENGINEERING.
576. Resistance of Ships.* — We shall first suppose the ship to
te towed at a uniform speed ; i.e., to be without means of self-
propulsion (under water). This being the case, it is found that
at moderate velocities (under six miles per hour), the ship
heing of "/tM>" form (i.e., the' hull tapering both at bow and
stern, under water) the resistance in still water is almost wholly
due to skin-friction^ "eddy-making" (see § 569) being done
away with largely by avoiding a bluff stern.
When the velocity is greater than about six miles an hour
the resistance is much larger than would be accounted for by
skin-friction alone, and is found to be connected with the sur-
face-disturbance or waves produced by the motion of the hull
in (originally) still water. The recent experiments of Mr,
Froude and his son at Torquay, England, with paodels, in a tank
300 feet long, have led to important rules (see Mr. White's
Naval Architectxtre and "Hydromechanics" in the Ency.
Britann ) of so proportioning not only the total length of a
ship of given displacement, bat the length of the entrance (for-
ward taperin.-sec.]
I X 3000 X 550 = ^ X 22 ; /. R = 56250 lbs.
Further, since M varies (roughly) as the square of the veloc-
ity, and can therefore be written H, = (Const.) X 'y^ we have
from (1)
L-= a constant X v^ ...... (2)
as a roughly approximate relation between the speed and the
power necessary to maintain it uniformly. In view of eq. (3)
involving the ouhe of the velocity as it does, we can understand
why a large increase of power is necessary to secure a propor,
tionally small increase of speed.
577. " Transporting Power," or Scouring Action, of a Current,
— The capacity or power of a current of water in an open
chanTiel to carry along with it loose particles, sand, gravel,
pebbles, etc., lying upon its bed was investigated experiment
tally by Dubuat about a century ago, though on a rather small
scale. His resulrs are as follows :
The velocity of current must be at least
0.25 ft. per sec, to transport silt ;
■ 0.50 " " " loam;
1.00 " « " sand;
2.00 " " " gravel;
3.5 " " " pebbles 1 in. in diam.;
4,0 " " " broken stone ;
5.0 '^ " " chalk, soft shale.
When a current holds "silt,'' (i.e., fine clay, sand, or mud)
in suspension, the latter may be deposited if conditions of
velocity, or of depth, change. According to Kennedy's
observations on certain canals in India, silt will be de-
posited if the velocity falls below a certain critical value,
different for different depths of stream. Some of these values,
with the corresponding depths are here given (Bellasis,
Hydraulics, p. 179) :
Ford= 12 3 4 5 6 7 8 9 10 ft.
v = M 1.3 1.7 2.04 2.35 2.64 2.92 3.18 3.43 3.67 ft./sec.
In case the particles move in filaments or stream-lines
parallel to the axis of the stream the statement is sometimes
made that the "transporting power" varies as the sixth power
■ TEAI^SPORTING POWER" OF CURRENTS.
831
Fia. 646.
of the velocity of the current, hy which is meant, more defi-
nitely, the following: Fig. Q4:Q. Conceive a row of cubes (or
other solids geometri-
cally similar to each
other) of many sizes,
all of the same heavi-
ness (§ 7), and simi-
larly situated, to be
placed on the horizon-
tal bottom of a trough
and there exposed to
a current of water,
being completely im-
mersed. Suppose the coefficient of friction between the cubes
and the trough-bottom to be the same for all. Then, as the
current is given greater and greater velocity ^), the impulse
P^ (corresponding to a particular velocity 'y,„) against some
one, m, of the cubes, will be just sufficient to move it, and at
some higher velocity v^ the impulse P^ against some larger
cube, n, will be just sufficient to move it, in turn. We are to
prove that P^^ • Pn • • '^m • '^n-
Since, when a cube barely begins to move, the impulse is
equal to the friction on its base^ and the frictions under the
cubes (when motion is impending) are proportional to their
volumes (see above), we have therefore
P
(*>
Also, the impulses on the cubes, whatever the velocity, are pro-
portional to the face areas and to the squares of the velocities
(nearly ; see § 572) ; hence
JBVom (1) and
P^
'^m ^m
Pn
<^n
(2) we have
Le., —
(2)
"4>
(8)
832
MECHAlSriCS Oh ENGINEEEING.
while from (3) and (2) we have, finally,
Thus we see m a general way why it is that if the velocity
of a stream is doubled its transporting power is increased
abaut sixty -four-fold ; i.e., it can now impel along the bottom
pebbles that are sixty-four times as heavy as the heaviest which
it could move before (of same shape and heaviness).
Though rocks are generally from two to three times as
heavy as water, their loss of weight under water causes them to
encounter less friction on tlie bottom than ii not Immersed.
578. Recent Experiments with Fire-hose, Nozzles, etc. (Ad-
dendum to § 520.) — The very full and careful investigations of
Mr. J. R. Freeman^ hydraulic engineer of Boston, Mass., in this
line (see Transac. A. S. C. E., Nov. 1889) furnish the following
results: By taking piezometer readings at the ends of a portion
of fire-hose conducting a steady flow of water, the values of loss
of pressure due to fluid frictiou per 100 feet of length could be
computed ; a careful measurement being also made of the
diameter of the hose and of the volume of water transmitted iu
an observed time. The table here given presents results applica-
ble to hose of exactly 2.5 inches diameter, for a delivery of water
at the rate of 240 gallons per minute (that is, for a velocity in
the hose of 15.68 ft. per sec). (The value of / has been com-
puted by the writer.)
Sample.
Desci-iption.
Loss of pressure,
per 100 ft. of
length, in lbs. per
sq. in.
Coefficient/.
CSee § 520.)
Velocity of
water in
hose.
L
K
I
E
C
Unlined linen hose
Woven cotton, rubber-lined,
"Mill Hose"
Kx ■' cotton, rubber-lined, hose.
Ditto, but interstices between
'■'^reads well filled
Wo^fen cotton, rubber-lined,
hose. So well filled with the
rubber that the inner surface
remained smooth under pres-
sure
33.2
25.5
19.4
16.0
14,1
0.01045
0.00802
0.00610
0.00503
0.00443
15.68
15.68
15.68
15.68
15.68
It was found that with other rates of flow the friction-head
varied nearly as the square of the velocity. The great importance
of a smooth interior o^ hose is well shown by this table.
A short section of each kind of hose was filled with liquid
plaster under jjressure. After the setting of the plaster the hose
was removed and photographs taken of the cast, thus conveying
a definite idea of the degree of roughness of interior of hose.
EXPERIMENTS WITH NOZZLES.
833
As to nozzles, it was found that the plain conical nozzle gave
the best results, jets from the ring-nozzles being slightly inferior
in range.
By means of a very delicate form of Pitot's tube measurementi
were made of the velocity in different parts of the section of jets,
aear the nozzle, with the interesting result that in "about two-
thirds the whole distance from centre to circumference the ve-
locity remains the same as at centre," and that at -J inch from
th" wall of most of the orifices the velocity was only 5fo less than
at centre of jet. With a jet from a 5-foot length of brass tubing
1-g inch in diameter and used as a nozzle the velocity fell off
rapidly for filaments further from the- centre; e.g., at half the
distance from centre to circumference the velocity was 90fo of
that at the centre, and at the outside edge QOfo. Most of the
nozzles ranged from 1 in. to 1^ in. in diameter of orifice.
By using these velocity measurements to "gauge" the flow it
was found that the relation h' = — was quite closely borne out
(within Ifc) (see eq. (7V', p. 804). The point of the Pitot tube
was conically convergent, its extremity being 0.0 17 in. in external
diameter and containing an orifice of 0.006 in. diameter. A
minute passage-way led from the orifice to a Bourdon gauge.
Based on his experiments, Mr. Freeman gives tables for the
maximum vertical height, V, and also the maximum horizontal
range, II, of "good effect ive fire-streams " delivered from smooth
conical nozzles of various sizes and with different piezometer
pressures p (in lbs. p. sq. in. above atmosphere) at the base of
play-pipe, the gauge being at same level as nozzle. (The dis-
tances reached by the extreme drops are very mucn greater with
the high pressures. V and H are in feet.)
The following is a brief synopsis of this table, d is the internal
diameter of the extremity of nozzle. The maximum horizontal
range was obtained at an angle of elevation of about 32°.
Forp =
10
20
40
60
80
100
V H
V
H
V
H
V
H
V
H
V H
d = % in.
d = % "
17 ft. 19
33
29
60
44
72
54
79
62
m 88
18 21
34
.S3
62
49
77
61
85
70
90 76
d = 1 "
18 21
35
37
64
55
79
67
89
76
96 83
d = 116 '*
1 18 22
36
38
65
59
83
73
92
81
99 89
d = m "
d = 192 "
19 22
37
40
67
63
85
76
95
85
101 93
20 23
88
42
69
66
87
79
97
88
103 96
834
MECHANICS OF ENGINEERING.
579. Addendum on the Pelton Water- wheel. — The annexed cut and
additional details of the test alluded to on p. 810 are taken from cii ••alais
of the manufacturers, of San Frauciso, Cal.
The water was measured over an iron weir J" thick and 3.042 feet long
without end contraction.
The depth was measured by a Boyden hook gauge reading to .001", and
was .4146 foot. The quantity of water discharged was found to be 2.819
cubic feet per second — Fteley's formula. The head lost by friction in pipe
was 1.8 feet, reducing the effective head to 384.7 feet.
The work done was measured by a Prony brake bearing vertically down
upon a platform scale and which showed a weight of 200 pounds upon the
scale-beam when the brake gear was suspended by a cord from a point im-
mediately above the wheel-shaft. This made a constant minus correction
of 200 pounds. The friction pulley had a face of 12", and being kept wet
by a Jet of clear cold water, it developed very little heat and ran without
much jumping. Thirteen tests were made showing very uniform results,
the first four of which were as follows :
Tests.
Weight shown by
scale
Net weight
(—200 lbs.)
Rev. wheel-shaft
per min.
1
2
3
4
665
665
660
660
G = 465
4fi5
460
460
M = 254J
255
256
256f
Totals 1022
Means 255i
Gm = 118343
118575
117760
117990
472667
118167
The arm of the Prony brake was 4.775 feet from centre of wheel-shaft to
point of contact on scale and hence described a circ?e with a circumference
of 30 feet. The work done per minute was therefore Ou{2tr) — (118167
X 30) or 3,545,000 foot-pounds, equal to 107.4 horse-power. The theoret-
ical power of the water was (2.819 x 60 x 384.7 x 62.4) or 4>0S0,35? foot-
pounds. The useful effect was therefore 87.3 per cent.
APPENDIX.
Note. — ^This appendix contains addenda and tables, and also several
pages from a former edition of the book. In these last the numbering of
the articles and of the figures has been left unchanged.
48a. Addendum to § 55. Mass. — In Phtsics, the fundamental units ar«
those of
Space, involving a unit of length (and thence of area and volume) ;
Time, " a unit of time, usually the second ;
Mass, " a unit of mass, which (by Government decree) may be the
quantity of matter in a specitied piece of platinum, or specified volume of water,
etc. (a beam-balance being used to determine equal quantities of mass) ; while
FoKCE involves a derived unit, being measured by its effect in accelerating
the velocity of a moving mass, since it is proportional both to the mass and the
acceleration. The unit force (called absolute unit) is the force necessary to pro-
duce unit acceleration in a unit of mass; so that to produce an acceleration =^ in
a mass = m requires a force = i''= nip, and the force thus obtained is in absolute
units. This is called the dynamic measure of a force.
Uxample. — In the C.G.S. system of units, required the constant force necessary
to cause a mass of 400 grams to gain 200 velocity units in 2 sec ; i.e.,^ = 100
centims. per sec, per sec. From F= mp we have
i^= 400 X 100 = 40000 abs. units of force (or dynes, in C.G.S. system).
In the ft.-lb.-sec. system the absolute unit is called apoundal.
In Mechanics of Engineering, however, it is more convenient to regard the
fundamental units to be those of
Space, as ft., metre, etc., area and volume corresponding;
Time, as seconds, hours, etc. ;
Force, as lbs., grams, kilograms, tons, etc., indicated by a spring balance;
while for
Mass we assume a derived unit, a mode of measuring it being developed as
follows :
If by experiment (block on smooth table, for instance) we find that a constant
force P(lbs., tons, kilos.) will maintain an acceleration =p in the rectilinear
motion (in line of force) of a body whose weight (by previous trial with a spring
balance) is O (lbs. , tons, or other unit) ; and if in a second experiment, by allow-
ing the force O to act on the same body in vacuo, a free vertical fall with acceliBra-
tion =g is, the result, — we find that the proportion (Newton's 2d Law) P:Q::p:g
Gr
Is verified. This may be written P= — . p, and may then be read: Force = mass
X acceleration, if we call the quotient G-s-g the Mass of the body whose weight
(by spring balance) is = G at a locality where the acceleration of gravity = g ; for
this quotient will be the same at all localities on the earth's surface.
JEzample (same as above). — If a body whose weight O = 400 grams (force) is to
have its velocity increased, in 2 sec. , from 300 centims. per sec. to 500 centims.
per sec, at a uniform rate, we must provide a constant force
= 40.77 grams; or .040 kilos.
This is called the gravitation measure of a force. Hence it is evident that to r.
duce absolute units (called dynes and poxmdals) in the C.G.S. and ft.-lb.-sec.
systems, respectively) to ordinary practical units of force (lbs., tons, kilos., etc.,
of a spring balance), we divide by the value of g proper to the system of units em-
ployed : and vice versd.
835
836 APPENDIX.
(Addendum to § 49a of page 49.) Numerical Example. — A set of light
screens is set up at intervals of 100 feet apart in the horizontal path of a
cannon-bal], with the object of determining its velocity, and also the rate of
change (or negative acceleration) of that velocity, as due to the resistance
of the air.
By electrical connection the time of passing each screen is noted, and
the intervals of time are given in this diagram for four of the screens,
A, B, G, and D.
.100'.
,0.0631 sec.
....100'....
.0.0633 sec.
.100'.
.0.0643 sec.
^ 1 ^ 3 C 3 D
From these data it is required to compute, as nearly as the circumstances
allow, the velocity and acceleration (negative) of the ball at various points
(the ball moves from left to right).
Solution. — In passing from Aio B the ball has an average velocity of
1610 ft. per second, obtained by dividing the distance of 100 feet by the
time of passage, 0.0631 second. Similarly we find the average velocity
between B and G to be 1583 ft. per second, and that between Cand I) to
be 1554 ft. per second.
As the velocity is not changing very rapidly, we may claim that the ball
actually possesses the velocity Vi = 1610 ft. per second at the point 1, mid-
way between A and B, or very near that point ; and similarly the velocity
«2 = 1583 ft. per second at point 3, midway between B and G ; and
Vs = 1554 ft. per second at point 3, midway between G and I).
Hence the total gain of velocity from 1 to 3 is 1583-1610 = — 38 ft. per
second; and the time in which this gain is made is one half of the 1631
second plus one half of the 0.0633 second, i.e., 0.0636 second. Therefore
an approximate value for the average acceleration between points 1 and 2
is found by dividing the — 38 ft. per second gain in velocity by the
time 0.0636 second occupied in acquiring the gain. This gives — 447 ft.
per second per second average acceleration for portion 1. . .3 of path, and
since screen B lies at the middle of this portion, the actual acceleration of
the ball's motion as it passes the screen B is very nearly equal to this, viz.:
— 447 ft. per second per second (or " ft. per square second ")-
By a similar process the student may compute the acceleration at screen
G. Of course the reason why these results are merely approximate is that
the spaces and times concerned, though relatively small , are not infinitesimal.
[A recent English writer calls a unit of velocity a "speed;" and a unit
of acceleration, a "Tiurr3i*"i
HYPERBOLIC SINES AND COSINES.
837
TABLE OF HYPERBOLIC SINES AND COSINES (See p. 48).
r—
u
cosh >_
sinh u
u
cosh u
sinh M
0.00
1.0000
0.05
1.0013
0.0500
2.05
3.9484
3.8196
.10
1.0050
.1002
2.10
4 . 1443
4,0219
.15
1.0112
.1506
2.15
4.3507
4 2342
.20
1.0201
.2013
2.20
4 , 5679
4,4571
.25
1.0314
.2526
2.25
4.7966
4.6912
0.30
1.0453
0.3045
2. 30
5.0372
4.9369
.35
1.0619
.3572
2 35
5.2905
5.1952
.40
1 0811
.4108
2.40
5.5569
5.4662
.45
1 . 1030
.4653
2.45
5.8373
5.7510
.50
1 . 1276
.5211
2.50
6.1323
6.0502
0.55
1.1551
0.5782
2.55
6.4426
6.3645
.60
1 . 1855
.6367
2.60
6.7690
6.6947'
.65
1.2188
.6967
2.65
7 1123
7.0417
.70
1.2552
.7586
2.70
7.4735
7.4063
.75
1.2947
.8223
2.75
7.8533
7.7894
80
1.3374
0.8881
2.80
8.2527
8 1919
.85
1.3835
0.9561
2.85
8 6728
8.6150
.90
1.4331
1.0265
2.90
9.1146
9.0596
.95
1.4862
1.0995
2.95
9 5791
9,5268
1.00
1.5431
1.1752
3.00
10.0677
10.0179
1.05
1.6038
1.2539
3.05
10.5814
10.5340
1.10
1.6685
1.3356
3.10
11 1215
11.0765
1.15
1.7374
1.4208
3.15
11.6895
11.6466
1.20
1.8107 .
1.5097
3.20
12.2866
12 2459
1.25
1.8884
1 6019
3.25
12.9146
12.8758
1.30
1.9709
1.6984
3.30
13 5748
13.5379
1.35
2.0583
1.7991
3.35
14.2689
14.2338
1.40
2.1509
1.9043
3.40
14.9987
14.9654
1.45
2.2488
2.0143
3.45
15 7661
15.7343
1.50
2.3524
2 . 1293
3.50
16 5728
16.5426
1.55
2.4619
2.2496
3.55
17.4210
17.3923
1.60
2.5775
2.3757
3.60
18 3128
18.2855
1.65
2.6995
2.5075
3.65
19 2503
19 2243
1.70
2.8283
2.6456
3.70
20.2360
20.2113
1.75
2.9642
2.7904
3.75
21 2723
21.2488
1.80
3.1075
2.9422
3.80
22.3618
22.3394
1.85
3.2583
3.1013
3.85
23.5072
23.4859
1.90
3.4177
3.2682
3.90
24.7113
24.6911
1.95
3.5855
3.4432
3.95
25.9773
25.9581
2.00
3.7622
3.6269
4.00
27.3082
27.2899
838
APPENDIX.
J.+ ^J
266. The Four x-Derivatives of the Ordinate of the Elastic Curve
— If y = func. (a?) is the equation of the elastic curve for
any portion of a loaded beam, on which portion the load
per unit of length of the beam is, w — either zero, (Fig.
234:) or = constant, (Fig. 235), or = a continuous func. {x)
(as in the last §), we may prove, as fol-
lows, that w = the cc-derivative of the
^ shear. Fig. 269. Let i^ and N' be two
— »■ consecutive cross-sections of a loaded
beam, and let the block between them,
^^ bearing its portion, wdx, of a distributed
load, be considered free. The elastic
forces consist of the two stress-couples
(tensions and compressions) and the two
shears, j/and J •\- dJ, dJheing the shear-increment conse-
quent upon X receiving its increment dx. By putting
2'(vert. components) = we have
Fig. 269.
J-\-dJ- — wdx — t^^O .'. w=
dJ
dx
Q. K D. But J" itself = d3I -^ dx, (§ 240) and
M = \d?y ~r dx^] EL By substitution, then, we have th©
following relations :
?/=func.(x)= ordinate at any point of the elastic curve (1)
dy^
dx
^=z a = slope at any point of the elastic curve
(2>
d'^y
EI ~j^~ M = ordinate (to scale) of the moment curve (3)
dx
^ d^oj .,1 T f the ordinate (to scale) ) ,.^
^^ S " * ' 1 of *1^^ «lie^^ diagram \ ' ' ^^>
do^
Elp= w =
the load per unit of length "i
of beam = ordinate (to scale) >-
of a curve of loading. )
(5)
If, then, the equation of the elastic curve (the neutral line
of the beam itself ; a reality, and not artificial like th©
FLEXURE. SPECIAL PEOBLEMS. 839
other curves spoken of) is given ; we may by successive
differentiation, for a prismatic and homogeneous beam so
that both E and / are constant, find the other four quan-
tities mentioned.
As to the convei-se process, (i.e. having given w as a
function of x, to find expressions for J, M and y as func-
tions of x) this is more difficult, since in taking the
a7-anti-derivative, an unknown constant must be added and
determined. The problem just treated in § 264, however,
offers a very simple case since to is the same function of
X, along the ivJiole beam, and there is therefore but one elas-
tic curve to be determined.
We .•. begin, numbering backward, with
XTT- d*y T f since w = ybx, see ) ,- v
'dx^ ~ ^ ( last § and Fig. 268 j * ' • V0«>
[K. B. — This derivative (dJ-i-dx) is negative since dJ&nd
dx have contrary signs.]
A (sJiear=)II /^=„ r b ^-^Const
dor 2
But writing out this equation for x=0, i.e. for the point
0, where the shear=^o> we have ^o"" -{- Const. .: Const.—
Eq, and hence write
A-gain taking the a:-anti-derivative of both sides
(Moment =) E I ^^=—yh—-\-BoX+{Const.=0) . (3a)
[At 0, a;=0 also M, .-. Const. =0]. Again,
At 0, where x=0 dy -^dx=aQ=ih.e unknown slope of the
Elastic line at 0, and hence C'=EI a^
.\EI^=-rbtA-B,^-\-EI a^ . . , (2a)
aa? 24 2
840
APPENDIX.
Passing now to y itself, and remembering that at 0, botn
y and x are zero, so that the constant, if added, would==
aero, we obtain (inserting the value of Bq from last §)
^^y^-r^^-^r^V—-^- Ela^x . (la)
the equation of the elastic curve. This, however, contains
the unknown constant «^«=the slope at 0. To determine
«o write out eq. (la) for the point B, Fig. 268, where x is
known to be equal to I, and ^ to be = zero, solve for a, ,
and insert its value both in (la) and (2a). To find the
point of max. y (i.e., of greatest deflection) in the elastic
curve, write the slope, i.e. dy -^ dx, = zero [see eq. 2a] and
solve for x ; four values will be obtained, of which the one
lying between and I is obviously the one to be taken.
This value of x substituted in (la) will give the maximum
deflection. The location of this maximum deflection is
neither at the centre of action of the load ( a?= ^ I \
nor at the section of max. moment [x =1-^^/^.)
The qualities of the left hand members of equations (1)
to (5) should be carefully noted. E. g., in the inch-pound-
second system of units we should have :
1. y (a, linear quantity) = (so many) inches.
2. dy-i-dx (an abstract number) = (so many) abstract
units.
3. Jf (a moment) = (so many) inch-pounds.
4. e7^(a shear, i.e., force) = (so many) pounds.
6. w (force per linear unit) ■■= (so many) pounds per run-
jiing inch of beam's length.
As to the quantities E, and I, individually, E is pounds
per sq. in., and /has four linear dimensions, i.e. (so irany)
bi-qnadratic inches.
FLEXURE. BODIES OF UNIFORM STRENGTH. 841
287. Cantilevers of Uniform Strength. — Beams built in at
one end, horizontally, and projecting from the wall with-
out support at the other, should have the forms given be-
low, for the given cases of loading, if all cross-sections are
to be Rectangular and the weight of beam neglected. Sides
of sections horizontal and vertical. Also, the sections are
symmetrical about the axis of the piece, h and h are the
■dimensions at the wall. 1= length. No proofs given.
Fig. 390.
Width constant.
Vertical outline
parabolic. Single
end load.
^ Fig. 290, (a). {VzvfH'AWj (1)
Height constant. -|
Single end load. I „. ^^^ ,,^ , , ^ , ,t\* .v
Horizontal outline ^'g- 290. (*> (%^)H%VPi ■ (^)
triangular. J
\30
Constant ratio of-,
height V to width u.
Both outlines cu- ^^'^- ^^^- (^>-
bic parabolas.
{}4vf={y,hf'l . (3)
X
{y,uf={y,bf^ .(3y
842 APPENDIX.
Uniform L o a d. "i
Widtli constant. _. ^^^ , . , , n , ,,x^ ,.»
Vertical outline tri- r ^'S" ^^^- <''> (>^'')=(>^*)7 • (*)
angular. J
Uniform Load."
Height constant.
Horiz. outline is
two parabolas meet-
ing at (yertex)
witli geomet. axes
II to wall.
Fig. 291,(5). 5^««=(>^% • '(5)
X
{}iu)^={%hy -1, (6)
Uniform L o a d . -|
Both outlines semi-
cub i c parabolas. J- Fig. 291, (c).
Sections similar! /t/^,\3_/t/7.as ^ /av
rectangles. J (%^) -(%K} P (6)
289. — Beams and cantilevers of circular cross-sections
may be dealt with similarly, and the proper longitudinal
outline given, to constitute them *' bodies of uniform
strength." As a consequence of tlie possession of this
property, with loading and mode of support of specified
character, the following may be stated ; that to find the
equation of safe loading any cross-section tvhatever may he
employed. This refers to tension and compression. As
regards the shearing stresses in different parts of the beam
the condition of " uniform strength " is not necessarily ob-
tained at the same time with that for normal stress in the
outer fibres.
DEFLECTION OF BEAMS OF UNIFORM
STRENGTH.
290. Case of § 283, the double wedge, but symmetrical,
1.6., li=lo=^l, Fig. 292. Here we shall find the use of the
BEAMS OF UJSriFOEM STKEXGTH.
843
Fig. 293.
EI
form — (of fclie three forms for the moment of the stress
couple, see eqs. (5), (6) and (7), §§ 229 and 231) of the most
direct service in determining the form of the elastic curve
OB, which is symmetrical, and has a common tangent at
B, with the curve BC. First to find the radius of curva-
ture, p, at any section n, we have for the free body nO,
2'(moms.„=0), whence
FI
P
( TTOTYl P'O J 7/
+ y^Fx^O ; but I ^3) g 283 j ^= ^^^ ^^^ T=%uh^
E
we have Yj, - ul?= )
[P -T and .*. p=
I
.§ (1)
from which all variables have disappeared in the right
hand member; i.e., ,o is constant, the same at all points of
the elastic curve, hence the latter is the arc of a circle,
having a horizontal tangent at Bo
To find the deflection, d, at B, consider Fig. 292, (b\
where d=KB, and the full circle ol radias BR=p is
drawn.
The triangle KOB is similar to TOB,
Sind.: KB I OB :: OB : YB
But 0B=}41, KB=d and YB=2p
Pf
'bh^E ^^
.'. d= o, > and o'. , from eq. (1), d=
2p
From eq. (4) §233 we note that for a beam of the same
material but prismatic (parallelopipedical in this case,)
having the same dimensions, b and h, at all sections as at
1 PI?
the middle, deflects an amount =jq yrj-
844
APPENDIX.
load P in the middle of the span. Hence the tapering
beam of the present § has only ^ the stiffness of the pris"
matic beam, for the same h, h, I, E, and P.
291. Case of § 281 (Parabolic Body), With l^ =lo, i.e., Symmet-
rical. — Fig. 293,(a). Bequired the equation of the neutral
Fig. S93.
line OB. For the free body nO, i'(moms.n)=0 gives us
m
-y2Px
(1/
Fig. 293, (6), shows simply the geometrical relations of the
problem, position of origin, axes, etc. OnB is the neutral
line or elastic curv^* whose equation, and greatest ordinate
d, are required. (The right hand member of eq. (1)" is made
negative because d^y-^dx^ is negative, the curve being con-
cave to the axis Xin this, the first quadrant.)
Now if the beam were prismatic, /, the " moment of in-
ertia " of the cross-section would be constant, i.e., the same
for all values of x, and we might proceed by taking the x-
anti-derivative of each member of (1)" and add a constant j
but it is variable and is
hi 3
'' {'My
hence (1)" becomes
- {%iy
(fromeq. 3, §281, putting «o= ^2)
^
x^'-
'y-
dx?
y2Px
. (ly
To put this into the form Const. X ^ =func. of (x), we need
BEAMS OF UNIFORM STRENGTH. 845
3
only divide tlirougli by a5^ ,,(and for brevity denote
1 Ehh\-^ {}4l)r by A) and obtain
^w=->^^^"* • • • • ^1)
We can now take the cc-anti-derivative of each member, and
have
^^=_^P(2^+>^)+(7 .... (2/
To determine the constant C, we utilize the fact that at B,
where x=)4lt the slope dy-i-dx is zero, since the tangent
line is there horizontal, whence from (2)'
0=—F^±+G .'.C=P^.
.\ (2/ becomes A ^ =P[-^>^?-a; ^] (2>
Q/00
Ay=Pi^l.x—%xh+lG'=0-] . . . (3)
((7'=0 since for x=0, y=0). We may now find the deflec-
tion d (Fig. 293(6)) by writing 03= i^Zand 2/ =c^, whence, after
restoring the value of the constant A,
'^=>^m? • • . w
PF
and is twice as great [being=2. — rr-l* ^s if the erirder
* 3ee § 233, putting / =^ 5A in eq. (4).
were parallelopipedical. In other words, the present girder
is only half as stiff as the prismatic one.
292. Special Problem. (I.) The symmetrical beam,in Fig.
294 is of rectangular cross-section and constant width = 5,
846
APPENDIX.
but the heiglit is constant only over the extreme quartei
spans, being =^i=^7i, i. e., half the height h at mid-span.
Thft convergence of the two truncated wedges forming the
middle quarters of the beam is such that the prolongations
Fig. 294.
of the upper and lower surfaces luovM meet over the swpports
(as should be the case to make h='2hi). Neglecting the
weight of the beam, and placing a single load in middle, it
is required to find the equation for safe loading ; also the
equations of the four elastic curves ; and finally the deflec-
tion.
The solutions of this and the following problem are left
to the student, as exercises. Of course the beam here
given is not one of uniform strength.
293. Special Problem. (II). Fig. 295. Eequired the man-
ner in which the width of the beam must vary, the height
being constant, cross-sections rectangular, weight of beam
Fig. 295.
neglected, to be a beam of uniform strength, if the load is
uniformly distributed ?
HORIZONTAL STRAIGHT GIRDERS.
847
Note. — The following five pages originally formed the concluding part of
"the chapter on " Arch-ribs" of this book; and gives a graphic treatment
of straight girders considered as a particular case of curved beams (or
arch ribs).
HORIZONTAL STRAIGHT GIRDERS.
389. Ends 5'ree to Turn. — This corresponds to an arch-
rib with hinged ends, but it must be understood that there
is no hindrance to horizontal motion. (Fig. 439.) In
Fig 439.
treating a straight beam, slightly bent under vertical forcer
4)nLy (as in this case with no horizontal constraint), as a
APPENDIX.
particular case of an arch-rib, it is evident that since the
pole distance must be zero, the special equil. polygon will
have all its segments vertical, and the corresponding forco
diagram reduces to a single vertical line (the load line).
The first and last segments must pass through A and B
(points of no moment) respectively, but being vertical will
not intersect F^ and P2 ; i.e., the remainder of the special
equilibrium polygon lies at an in^nite distance above the
span AB. Hence the actual spec, equil. pol. is useless.
Hoivever. knowing that the shear, J, and the moment
M (of stress couple) are the only quantities pertaining to
any section m (Fig. 439) which we wish to determine (since
there is no thrust along the beam), and knowing that an
imaginary force H', applied horizontally at each end of the
beam, would have no influence in determining the shear
and moment at m as due to the new system of forces, we
may therefore obtain the shears and moments graphically
from this neio system (viz.: the loads Pj, etc., the vertical
reactions Fand V,„ and the two equal and opposite i^"'s).
[Evidently, since H'' has no moment about the neutral
axis (or gravity axis here), of m, the moment at m will be
unaffected by it ; and since H" has no component ~\ to the
beam at m, the shear at m is the same in the new system
of forces, as in the old, before the introduction of the
if's.]
Hence, lay off the load-line 1 . . 2. . 3, Fig. 439, and con-
struct an equil. polyg. which shall pass through A and B
and have any convenient arbitrary Zf" (forco) as a pole
distance. This is done by first determining n on the load-
line, using the auxiliary polygon A'a'B', to a pole 0' (arbi-
trary), and drawing O'n' \\ to A'B'. Taking 0'' on a hori-
zontal through n, making 0''n'=H", we complete the
force diagram, and equil. pol. AaB. Then, z being the ver-
ticalintercept heiy^Q&a. TYi dsndi the equil. polygon, we h.ive:
Moiiient at m^=M=H"z {ox — H'z' also), and shear at m, or
/,^2 . . n', i.e., = projection of the proper ray R2, or
0" . . 2, upon the vertical thrcugh m. Similarly we ob-
tain iff and J at any other sect: r n for the given load. (See
AECH-RIBS ; SPECIAL CASE ; STEAIGHT.
849
§ § 329, 337 and 367). The moment of inertia need not be
constant in this case.
390. Straight Horizontal Prismatic Girder of Fixed Ends at Same
Level. — No horizontal constraint, hence no thrust. I con-
stant. Ends at same level, with end-tangents horizontal.
We may consider the whole beam free (cutting close to the
walls) putting in the unknown upward shears J^ and o^^
and the two stress couples of unknown moments Jfl and
M,, at these end sections. Also, as in § 388, an arbitrary
H" horizontal and in line of beam at each extremity. Now
(See Fig. 33) the couple at and the force H" are equiv-
alent to a single horizontal H" at an unknown vertical dis-
tance c below ; similarly at the right hand end. The
special polygon FG is to be determined for this new sys-
tem, since the moment and shear will be the same at any
section under this new system as under the real system.
The conditions for determining it are as follows : Since
the end-tangents are fixed, IMAs={) ,*. I^^z^s=0 and since
850 APPENDIX.
O's displacement relatively to B'a tangent is zero we liave
IMxJs^O .: IN"sx/:ls=0 .'. IxzJs=0. See §374, Hence
for Equal ^s's, l'(z)=0 and l\xz)=0. Now for any pole 0'"
draw an equil. pol. F'"G"' and in it (by § 377; see Note)
locate v"'m"' so as to make 2'(3'")=0 and l{x%"'^=^.
Draw verticals through the intersections E'" and L"\ to
determine E and L on the beam, these are the points of
inflection (i.e., of zero moment), and are points in the re-
quired special polygon FG.
Draw 0'"%'' li to v'"m"' to fix n". Take a pole 0" on
the horizontal through n'\ making \)'':af'^H" (arbitrary),
draw the force diagram 0" 1234 and a corresponding
equilibrium polygon beginning at E. It should cut L,
and will fulfil the two requirments 2'^(2)=0 and 2'^(a;2!)=0,
with reference to the axis of the beam O'B'. The moment of
the stress-couple at any section m will be M=^H"z, and the
shear J = the projection of the " proper ray " of the force
diagram 0" . . 1, 2, etc., upon the vertical (not in the trial
diagram 0'". . 1, 2, etc.). As far as the moment is concern-
ed the trial polygon F'" G'" will serve as well as the special
polygon FG ; i.e., M=H"'z'" as well as H"z, R'" being the
pole-distance of 0'" ; but for the shear we must use the
rays of the final and not the trial diagram.
The peculiarity of this treatment of straight beams,
considered as a particular case of curved beams, consists
in the substitution of an imaginary system of forces in-
volving the two equal and opposite, and arbitrary H's, for
the real system in which there is no horizontal force and
consequently no " special equilibrium polygon," and thus
determining all that is desired, i.e., the moment and shear
at any section.
In the polygon FG the student will recognize the " mo-
ment-diagram " of the problems in Chaps. Ill and IV.
He will also see why the shear is proportional to the
slope — — of the moment curve in those chapters. For
ax
example, the " slope " of the second segment of the poly-
gon FG, that segment being || to 0" 2, is
AKOH-KLBS; SPECIAL CASE; STRAIGHT. 851
tang, of angle 20''w"=2%"-=-0'V'=shear -^ H"
and similarly for any other segment ; i.e., tlie tangent ol
the inclination of the " moment curve," or line, is propor-
tional to the shear.
It is also interesting to notice with the present problem
of a straight beam, that in the conditions
I{zJs)=0 and I{zAs)x=Q,
J)r locating the polygon FG^ each ds is "| tc its s, and
jhat consequently each z^s is the area of a small vertical
strip of area between the beam and the polygon; and
[z/ls)x is the " moment" of this strip of area, about 0' the
origin of x. Hence these conditions imply ; first, that the
area EWL between the polygon and the axis of the beam
on one side is equal to that (0'-P^+i^-S'6r) on the other
side, and, secondly, that the centre of gravity of EWL lies
in the same vertical as that of O'-F-E'and LB' G combined.
Another way of stating the same thing is, that, if we join
FG, the area of the trapezoid FO'B' G is equal to that of the
figure FEWLG, and tlieir centres of gravity lie in the same
vertical. A corresponding statement may be made (if we join
F'"G'") for the trapezoid F"'v'"m"'G'"' and figure
8S2
LOGAEITHMS (BKIGGS').
N
1
2
3
4
5
6
7
8
9
Dif.
10
0000
0043
0086
0128
0170
0212
0253
0394
0334
0374
42
11
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
38
12
0792
0828
0864
0899
0934
0969
1004
1088
1073
1106
35
13
1139
1173
1206
1239
1271
1303
1335
1367
1399
1430
33
14
1461
1492
1523
1553
1584
1614
1644
1673
1703
1733
30
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
3014
38
16
2041
2068
2095
2123
2148
2175
2201
2227
3353
3279
26
17
2804
2330
2355
2380
2405
2430
2455
2480
3504
3539
25
18
2553
2577
2601
2625
2648
2672
2695
3718
3743
3765
24
19
2788
2810
2833
2856
2878
2900
2923
3945
2967
3989
22
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3301
21
21
3222
3243
3263
3284
3304
8324
3345
3365
3385
3404
20
22
3424
3444
3464
3483
3502
8522
3541
3560
3579
3598
19
23
3617
3636
3655
3674
3692
3711
3729
3747
3766
3784
19
24
3802
3820
3838
3856
3874
3892
8909
3937
3945
3963
18
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
17
26
4150
4166
4183
4200
4216
4232
4349
4365
4381
4398
16
27
4314
4330
4346
4862
4878
4393
4409
4435
4440
4456
16
28
4472
4487
4502
4518
4533
4548
4564
4579
4594
4609
15
29
4624
4639
4654
4669
4683
4698
4713
4738
4743
4757
15
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
14
31
4914
4928
4942
4955
4969
4983
4997
5011
5034
5038
14
32
5051
5065
5079
5092
5105
5119
5132
5145
5159
5173^
13
33
5185
5198
5211
5224
5237
5250
5363
5376
5289
5303
13
34
5315
5328
5340
5353
5366
5378
5391
5403
5416
5438
13
35
5441
5453
5465
5478
5490
5502
5514
5537
5589
5551
13
36
5563
5575
5587
5599
5611
5623
5635
5647
5658
5670
13
37
5682
5694
5705
5717
5729
5740
5752
5763
5775
5786
12
38
5798
5809
5821
5832
5843
5855
5866
5877
5888
5899
11
39
5911
5922
5933
5944
5955
5966
5977
5988
5999
6010
11
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
11
41
6128
6138
6149
6160
6170
6180
6191
6301
6313
6333
10
42
6232
6243
6253
6263
6274
6284
6294
6304
6314
6325
10
43
6335
6345
6355
6365
6375
6385
6395
6405
6415
6425
10
44
6435
6444
6454
6464
6474
6484
6493
6503
6513
6522
10
45
6532
6542
6551
6561
6571
6580
6590
6599
6609
6618
10
46
6628
6637
6646
6656
6665
6675
6684
6693
6703
6712
9
47
6721
6730
6739
6749
6758
6767
6776
6785
6794
6803
9
48
6812
6821
6830
6839
6848
6857
6866
6875
6884
6893
9
49
6902
6911
6920
6928
6937
6946
6955
6964
6973
6981
9
50
6990
6998
7007
7016
7024
7033
7043
7050
7059
7067
9
51
7076
7084
7093
7101
7110
7118
7136
7135
7143
7152
9
52
7160
7168
7177
7185
7193
7202
7210
7318
7336
7235
8
53
7243
7251
7259
7267
7275
7284
7393
7300
7308
7316
8
54
7324
7332
7340
7348
7356
7364
7372
7380
7388
7396
8
N
. B, — Naperian log =
= Briggs' log
X 2.303.
Base of Naperian syste
in = e
= 2.71838.
LOGARITHMS (BEIGGS').
853
N
1
2
3
4
5
6
7
8
9
Dif.
55
7404
7412
7419
7437
7435
7443
7451
7459
7466
7474
8
56
7483
7490
7497
7505
7513
7530
7538
7536
7543
7551
8
57
7559
7566
7574
7583
7589
7597
7604
7613
7619
7637
8
58
7634
7642
7649
7657
7664
7673
7679
7686
7694
7701
7
59
7709
7716
7723
7731
7738
7745
7753
7760
7767
7774
7
60
7783
7789
7796
7803
7810
7818
7835
7833
7839
7846
7
61
7853
7860
7868
7875
7883
7889
7896
7903
7910
7917
7
62
7934
7931
7938
7945
7953
7959
7966
7973
7980
7987
7
63
7993
8000
8007
8014
8031
8028
8035
8041
8048
8055
7
64
8063
8069
8075
8083
8089
8096
8103
8109
8116
8132
7
65
8139
8136
8142
8149
8156
8162
8169
8176
8183
8189
7
66
8195
8302
8309
8315
8332
8228
8335
8341
8348
8254
7
67
8361
8367
8374
8280
8287
8293
8399
8306
8313
8319
6
68
8325
8331
8338
8344
8351
8357
8363
8370
8376
8382
6
69
8388
8395
8401
8407
8414
8420
8436
8433
8439
8445
6
70
8451
8457
8463
8470
8476
8483
8488
8494
8500
8506
6
71
8513
8519
8535
8531
8537
8543
8549
8555
8561
8567
6
73
8573
8579
8585
8591
8597
8603
8609
8615
8631
8627
6
73
8633
8639
8645
8651
8657
8663
8669
8675
8681
8686
6
74
8692
8698
8704
8710
8716
8733
8737
8733
8739
8745
6
75
8751
8756
8763
8768
8774
8779
8785
8791
8797
8802
6
76
8808.
8814
8830
8825
8831
8837
8843
8848
8854
8859
6
77
8865
8871
8876
8882
8887
8893
8899
8904
8910
8915
6
78
8921
8937
8933
8938
8943
8949
8954
8960
8965
8971
6
79
8976
8983
8987
8993
8998
9004
9009
9015
9030
9035
5
80
9031
9036
9043
9047
9053
9058
9063
9069
9074
9079
5
81
9085
9090
9096
9101
9106
9113
9117
9133
9138
9133
5
83
9138
9143
9149
9154
9159
9165
9170
9175
9180
9186
5
83
9191
9196
9301
9206
9212
9317
9333
9337
9333
9338
5
84
9243
9248
9353
9358
9263
9369
9374
9379
9384
9389
5
85
9394
9399
9304
9309
9315
9330
9335
9330
9335
9340
5
86
9345
9350
9355
9360
9365
9370
9375
9380
9385
9390
5
87
9395
9400
9405
9410
9415
9430
9435
9430
9435
9440
5
88
9445
9450
9455
9460
9465
9469
9474
9479
9484
9489
5
89
9494
9499
9504
9509
9513
9518
9533
9538
9533
9538
5
90
9543
9547
9553
9557
9562
9566
9571
9576
9581
9586
5
91
9590
9595
9600
9605
9609
9614
9619
9634
9638
9633
5
92
9638
9643
9647
9653
9657
9661
9666
9671
9675
9680
5
93
9685
9689
9694
9699
9703
9708
9713
9717
9733
9727
5
94
9731
9736
9741
9745
9750
9754
9759
9763
9768
9773
5
95
9777
9783
9786
9791
9795
9800
9805
9809
9814
9818
5
96
9833
9837
9833
9836
9841
9845
9850
9854
9859
9863
4
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
4
98
9912
9917
9931
9936
9930
9934
9939
9943
9948
9953
4
99
9956
9961
9965
9969
9974
9978
9983
9987
9991
9996
4
K B.—
N'aperian log =
= Briggs' log
X 3.303.
Base of
Naperian Sys
em =
e = 2.71838.
854
APPENDIX.
Trigonometric Ratios (Natural); including ''arc" by which is meant
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Steam-boiler Explosions in Theory and in Practice 12mo, i 50
Wehrenfenning's Analysis and Softening of Boiler Feed-water (Patterson) Svo, 4 00
Weisbach's Heat, Steam, and Steam-engines. (Du Bois.) Svo, 5 00
Whitham's Steam-engine Design Svo, 5 00
Wood's Thermodynamics, Heat Motors, and Refrigerating Machines. . .Svo, 4 00
MECHANICS PURE AND APPLIED.
Church's Mechanics of Engineering Svo, 6 00
Notes and Examples in Mechanics Svo, 2 00
Dana's Text-book of Elementary Mechanics for Colleges and Schools. .i2mo, i 50
Du Bois's Elementary Principles of Mechanics :
Vol. I. Kinematics Svo, 3 50
Vol. II. Statics Svo, 4 00
Mechanics of Engineering. Vol. I Small 4to, 7 50
Vol. II Small 4to, 10 00
* Greene's Structural Mechanics Svo, 2 50
James's Kinematics of a Point and the Rational Mechanics of a Particle.
Large 12mo, 2 00
* Johnson's (W. W.) Theoretical Mechanics 12mo, 3 00
Lanza's Applied Mechanics Svo, 7 50
* Martin's Text Book on Mechanics, Vol. I, Statics 12mo, i 25
* Vol. 2, Kinematics and Kinetics . .i2mo, l 50
Maurer's Technical Mechanics Svo, 4 00
* Merriman's Elements of Mechanics 12mo, i 00
Mechanics of Materials Svo, 5 00
* Michie's Elements of Analytical Mechanics Svo, 4 00
Robinson's Principles of Mechanism Svo, 3 00
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Schwamb and Merrill's Elements of Mechanism Svo, 3 00
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Principles of Elementary Mechanics • 12mo, i 25
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Abderhalden's Physiological Chemistry in Thirty Lectures. (Hall and Defren).
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von Behring's Suppression of Tuberculosis. (Bolduan.) i2mo, i 00
* Bolduan's Immune Sera i2mo, i 50
Davenport's Statistical Methods with Special Reference to Biological Varia-
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* Fischer's Physiology of Ahmentation Large i2mo, cloth, 2 00
de Fursac's Manual of Psychiatry. (Rosanoff and Collins.) Large i2mo, 2 50
Hammarsten's Text-book on Physiological Chemistry. (Mandel.) Svo, 4 00
Jackson's Directions for Laboratory Work in Physiological Chemistry. ..Svo, i 25
Lassar-Cohn's Practical Urinary Analysis. (Lorenz.) i2mo, i 00
Mandel's Hand Book for the Bio-Chemical Laboratory , . . i2mo, I 50
* Pauli's Physical Chemistry in the Service of Medicine. (Fischer.) .... i2mo, i 25
* Pozzi-Escot's Toxins and Venoms and their Antibodies. (Cohn.) i2mo, i 00
Rostoski's Serum Diagnosis. (Bolduan.) i2mo, i 00
Ruddiman's Incompatibilities in Prescriptions Svo, 2 00
Whys in Pharmacy i2mo, i 00
Salkowski's Physiological and Pathological Chemistry. (Orndorff.) Svo, 2 50
* Satterlee's Outlines of Human Embryology i2mo, i 25
Smith's Lecture Notes on Chemistry for Dental Students ijvo, 2 SJ
16
Steel's Treatise on the Diseases of the Dog 8vo, 3 so»
* Whipple's Typhoid Fever Large i2mo, 3 00
WoodhuU's Notes on MiUtary Hygiene i6mo, i 50
* Personal Hygiene i2mo, i 00
Worcester and Atkinson's Small Hospitals Establishment and Maintenance,
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METALLURGY.
Betts's Lead Refining by Electrolysis 8vo. 4 00
Holland's Encyclopedia of Founding and Dictionary of Foundry Terms Used
in the Practice of Moulding l2mo, 3 00
Iron Founder 12mo. 2 50
Supplement 12mo, 2 50
Douglas's Untechnical Addresses on Technical Subjects i2mo, i 00
Goesel's Minerals and Metals: A Reference Book , . . . . i6mo, mor. 3 00
* Iles's Lead-smelting 12mo, 2 50
Keep's Cast Iron 8vo, 2 50
Le Chatelier's High-temperature Measurements. (Boudouard — Burgess.) 12mo, 3 00
Metcalf's SteeL A Manual for Steel-users 12mo, 2 00
Miller's Cyanide Process 12mo i 00
Minet's Production of Aluminum and its Industrial Use. (Waldo.). . . . l2mo, 2 50
Robine and Lenglen's Cyanide Industry. (Le Clerc.) 8vo, 4 00
Ruer's Elements of Metallography. (Mathewson). (In Press.)
Smith's Materials of Machines 12mo, i 00
Thurston's Materials of Engineering. In Three Parts 8vo, 8 00
part I. Non-metallic Materials of Engineering, see Civil Engineering,
page 9.
Part II. Iron and Steel 8vo, 3 50
Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their
Constituents 8vo, 2 50
Ulke's Modern Electrolytic Copper Refining 8vo, 3 00
West's American Foundry Practice i2mo, 2 50
Moulders Text Book 12mo, 2 50
Wilson's Chlorination Process 12mo, r so
Cyanide Processes 12mo, 1 50
MINERALOGY.
Barringer's Description of Minerals of Commercial Value. Oblong, morocco, 2 50
Boyd's Resources of Southwest Virginia 8vo 3 00
Boyd's Map of Southwest Virginia Pocket-book form. 2 00
* Browning's Introduction to the Rarer Elements 8vo, i 50
Brush's Manual of Determinative Mineralogy. (Penfield.) 8vo, 4 OO'
Butler's Pocket Hand-Book of Minerals 16mo, mor. 3 00
Chester's Catalogue of Minerals 8vo, paper, i 00
Cloth, I 25
Crane's Gold and Silver. (In Press.)
Dana's First Appendix to Dana's New " System of Mineralogy. ." . . Large 8vo, i 00
Manual of Mineralogy and Petrography i2mo 2 no
Minerals and How to Study Them i2mo, r 50
System of Mineralogy Large 8vo, half leather, r2 50
Text-book of Mineralogy 8vo, 4 00
Douglas's Untechnical Addresses on Technical Subjects i2mo, i 00
Eakle's Mineral Tables 8vo, i 25.
Stone and Clay Products Used in Engineering. (In Preparation).
Egleston's Catalogue of Minerals and Synonyms 8vo, 2 50
Goesel's Minerals and Metals : A Reference Book i6mo. mor. 3 00
Groth's Introduction to Chemical Crystallography f Marshall) i2mo, 1 25
17
* Iddings's Rock Minerals 8vo, 5 00
Johannsen's Determination of Rock-forming Minerals in Thin Sections 8vo, 4 00
* Martin's Laboratory Guide to Qualitative Analysis vvitii the Blowpipe, izmo, 60
Merrill's Non-metallic Minerals: Their Occurrence and Uses 8vo, 4 00
Stones for Building and Decoration 8vo, 500
* Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests.
8vo, paper, 50
Tables of Minerals, Including the Use of Minerals and Statistics of
Domestic Production 8vo, i 00
Pirsson's Rocks and Rock Minerals. (In Press.)
* Richards's Synopsis of Mineral Characters i2mo, mor, 125
* Ries's Clays: Their Occurrence, Properties, and Uses 8vo, 5 00
* Tillman's Text-book of Important Minerals and Rocks 8vo, 2 00
MINING.
* Beard's Mine Gases and Explosions Large i2mo, 3 00
Boyd's Map of Southwest Virginia Pocket-book formj 2 00
Resources of Southwest Virginia 8vo, 3 00
Crane's Gold and Silver. (In Press.)
Douglas's Untefchnical Addresses on Technical Subjects i2mo, I 00
Eissler's Modern High Explosives. '. . 8vd 4 00
Goesel's Minerals and Metals : A Reference Book. . i6mo, mor. 3 00
II Iseng's Manual of Mining 8vo, 5 00
* Iles's Lead-smelting i2mo, 2 50
Miller's Cyanide Process i2mo, i 00
O'DriscoU's Notes on the Treatment of Gold Ores Svo, 2 00
Peele's Compressed Air Plant for Mines. (In Press. )
Riemer ' s Shaft Sinking Under Difficult Conditions . ( Coming and Peele) . . . Svo , 300
Robine and Lenglen's Cyanide Industry. (Le Clerc.) Svo, 4 00
* Weaver's Military Explosives 8vo, 3 00
Wilson's Chlorinati'on Process izmo, i 50
Cyanide Processes i2mo, i 50
Hydraulic and Placer Mining. 2d edition, rewritten limo, 2 50
Treatise on Practical and Theoretical Mine Ventilation T2mo, I 25
SANITARY SCIENCE.
Association of State and National Food and Dairy Departments, Hartford Meeting,
1906 Svo, 3 00
Jamestown Meeting, 1907 Svo, 3 00
* Bashore's Outlines of Practical Sanitation 12mo, i 25
Sanitation of a Country House 12mo, 1 00
Sanitation of Recreation Camps and Parks 12mo, i 00
FolweU's Sewerage. (Designing, Construction, and Maintenance.) Svo, 3 00
Water-supply Engineering Svo, 4 00
Fowler's Sewage Works Analyses 12mo, 2 00
Fuertes's Water-filtration Works 12mo, 2 50
Water and Public Health 12mo, i 50
Gerhard's Guide to Sanitary House-inspection 16mo, i 00
* Modem Baths and Bath Houses Svo, 3 00
Sanitation of Public Buildings 12mo, i 50
Hazen's Clean Water and How to Get It Large 12mo, i 50
Filtration of Public Water-supplies Svo, 3 00
Kinnicut, Winslow and Pratt's Purification of Sewage. (In Press.)
Leach's Inspection and Analysis of Food with Special Reference to State
Control Svo, 7 00
Mason's Examination of Water. (Chemical and Bacteriological) 12mo, i 25
Water-supply. (Considered principally from a Sanitary Standpoint) . . Svo, 4 00
18
* Merriman's Elements of Sanitary Engineering 8vo, 2 00
Ogden's Sewer Design l2mo, 2 00
Parsons's Disposal of Municipal Refuse 8vo, 2 00
Prescott and Winslow's Elements of Water Bacteriology, with Special Refer-
ence to Sanitary Water Analysis 12mo, i 50
* Price's Handbook on Sanitation 12mo, i 50
Richards's Cost of Food. A Study in Dietaries 12mo, i 00
Cost of Living as Modified by Sanitary Science 12mo, i 00
Cost of Shelter 12mo, i 00
* Richards and Williams's Dietary Computer 8vo, i 50
Richards and Woodman's Air, Water, and Food from a Sanitary Stand-
point 8vo, 2 00
Rideal's Disinfection and the Preservation of Food 8vo, 4 00
Sewage and Bacterial Purification of Sewage 8vo, 4 00
Soper's Air and Ventilation of Subways. (In Press.)
Turneaure and Russell's Public Water-supplies 8vo, 5 00
Venable's Garbage Crematories in America 8vo, 2 00
Method and Devices for Bacterial Treatment of Sewage 8vo, 3 00
Ward and Whipple's Freshwater Biology. (In Press. )
Whipple's Microscopy of Drinking-water Svo, 3 So
* Typhod Fever Large 12mo, 3 00
Value of Pure Water Large l2mo, i 00
Winton's Microscopy of Vegetable Foods 8vo, 7 50
MISCELLANEOUS.
Emmons's Geological Guide-book of the Rocky Mountain Excursion of the
International Congress of Geologists Large Svo,
Ferrel's Popular Treatise on the Winds 8vo,
Fitzgerald's Boston Machinist i8mo,
Gannett's Statistical Abstract of the World 24mo,
Haines's American Railway Management 12mo,
* Hanusek's The Microscopy of Technical Products. (Winton^ Svo,
Ricketts's History of Rensselaer Polytechnic Institute 1824-1894.
Large i2mo,
Rotherham's Emphasized New Testament , Large Svo,
Standage's Decoration of Wood, Glass, Metal, etc 12mo,
Thome's Structural and Physiological Botany. (Bennett) 16mo,
Westermaier's Compendium of General Botany. (Schneider) Svo,
Winslow's Elements of Applied Microscopy 12mo,
HEBREW AND CHALDEE TEXT-BOOKS.
Green's Elementary Hebrew Grammar i2mo, 1 25
Gesenius's Hebrew and Chaldee Lexicon to the Old Testament Scriptures.
(Tregelles.) Small 4to, half morocco, 5 00
19
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50
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JUL 1 1908
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