Charles Edward Fuller.

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Font size (Alt. 176) or, if we substitute the values of /i and ft in terms of the normal
stress (equation 1) and the bending moment (equation 9),

(12)

where ci Â» the distance of the inside of the cross section from the central
axis OC.

At the cross section equation (12) reduces to

â€˘'""2A"^ (B-A) \A ri-cj'

(13)

which is the maximum tensile stress intensity in the ring; and at the cross
section C equation (12) reduces to

'-,-(^(i-,7^) <">

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428 APPLIED MECHANICS

which 18 a oompreamve stress and is the Tnaxifnum intensity of stress in the
ring.

The normal stress intensity at the outside of the ring can be readily ob-
tained by substituting Ci, the distance from the center of gravity to the
outside of the cross section, for â€” Ci in equations (12), (13) and (14).

The change in the length of the diameters 00' and CC can be approxi-
mately determined by substituting the values oi ds ^ â€” n da, y ^ ri cos a,
a? - n (1 â€” sin a) and the value of M from equation (9) in equations (8)
and (9) (Art. 178) and obtaining the displacements of 0, relative to C, in the
directions OX and OF, as follows:

te =Â» â– " ^ J ^^i ( 2 ^^^ Â« V* CMada

2

Wri' ri . 1 . , no

^^ ^ MJ^^^\ir~2^'*r^^^^ -sina)(ia
2
Wri*ra,/l,l\ ^a 1 . â€ž 1Â°

and

2

-ÂĄD-l] (â– â€˘)

Since Sz is positive the diameter 00' will be shortened and the total de-
crease will be equal to

â€˘Â«^-^e-j) <")

and, since Sy is negative, the diameter CC will be lengthened and the total
increase will be equal to

MCC')-^-^(i-l) (18)

Second Solution. â€” In the preceding solution, by use of the approximate
equations (7-9) (Art. 178), the stress due to the normal force P and the effect
of the curvature on the distribution of the stress due to bending have been
neglected. A more accurate solution can be made by use of the more exact
forms of the equations for the displacements as follows:

By substituting the values Qi = (A - B) and Qf^-niA- B) (Art.
175) in equation (18) (Art. 178), we obtain

and, 'assuming noitemperature change and that gj â€” aTm "" EAr (^^^^
nearly) for any allowable bending moment, equation (19) will reduce to

â€˘Â«-r.r-[r^(H;;)+a'^- â– â– â€˘â€˘"Â»'

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THE CIRCULAR RING 429

Substituting the values of M and P, from equations (3) and (1), in equation
(20) and observing that the change in the angle between the tangents at
O and C, due to the distortion of the ring (Fig. 243), is zero, we have

2

Reducing (21), we obtain

J^[b(^' + W\-' AWsina'jda ^0; (22)

2

and integrating,

-b(^ + w)1+aw^0',

and solving for Moj we have

Substituting the value of Afo in (3), we have

and, by substituting in (5), we obtain

Af.-^ (26)

The point N, at which the bending moment is zero, may be located by
placing (24) equal to zero and solving for a', the angle between the tangent
at N and OX, which will give

a'=sin-i^ (26)

For the greatest normal stress intensity at any cross section D, we have

. . (27)

/ = ^smÂ«+ \b-A) \A^7r=Tj

and for the cross section at the greatest intensity will be

^^2A^ (B-A) \A n-c,; ^^^

and for the cross section at C,

^ "^ xB (B - A) {a^VT^J' ^^^

which is the greatest normal stress intensity in the ring. It will be observed
that, since B > A, the stress intensities given by the last three equations
are slightly less than those given by the approidmate solution. The equations
for the stress intensity at the outside of the ring at he cross sections Z>, and
C can be obtained by substituting d for â€” Ci in equations (27), (28) and (29).

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430 APPLIED MECHANICS

The variation in the bending moment and the stress intensities at the in-
side, or intrado8f and at the outside, or extradoSj of the ring is represented by
the diagram (Fig. 244).

The variation in Af for a ring of circular cross section, of radius r Â« 1
and having the proportions n = 3 r, is shown by the curve on the left of the
central axis CC'^ which is constructed by plotting the values of M at the
different cross sections radially from the central axis of the ring as a base
line, positive values of M being measured outward and negative values in-
ward from the base line.

Fia. 244.

The variation in the stress intensity at the intrados of the ring is shown
by the curve in the upper right-hand quadrant, tensile stress intensities being
from the central axis as a base line. Similarly, the diagram in the lower right-
hand quadrant represents the variation in the stress intensity at the extrados
of the ring, tensile stresses being measured outward and compressive stress
intensities inward from the central axis.

It w^ill be observed that the greatest tensile stress intensity is located at
the intrados, at the sections and O', and that the greatest compressive
stress intensity is located at the intrados, at the sections C and C, and that
the latter is the maximum stress intensity for the entire ring.

To determine the change in the diameters 00' and CC we have, by substi-
tuting the values of Qi and Qt in (1) (Art. 178),

MA MA M .

n (AEQt - MQi) " (5 - A) ri {,EAn + Af) " (B - .1) En^ ^^^ nearly;;

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THE CIRCULAR RING 431

and, ftflBiiTning that there is no change in temperature, equation (15) (Art. 178),
for the displacement of relative to C in the direction OX, will reduce to

and equation (16) (Art. 178), for the displacement of relative to C in the
direction OY, will become

Substituting the values a; = n (1 â€” sin a), y = n cos a, (fe = â€”n da^dx^

W / A 1 \

â€” n coBada, dy - â€” n sin a (ia, P Â» ^ sin oe and M = Wn (^ â€” 5 sin a j

in the above equations and observing that when x Â» 0, a Â» ^ and when x Â» Zi,
A <â–  0, we obtain

2

2

_Wri
'(B

and

(,B-A)E\&r~V~^' ^^^^

2
2

"(B-A)jjLBt 8'^2V B /J xBE ^""^

Since Sxi is positive, the diameter 00' will be shortened and the total
decrease will be equal to

and, as 5yi is negative, the diameter CC will be lengthened and the formula
for the total increase may be written

The following approximate expressions for the change in diameters, as
accurate as the conditions of loading in ordinary cases will warrant the use of,

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432

APPLIED MECHANICS

can be obtained by combining the approxiniate values for the change due to
bending only, given by (17) and (18), with the change due to the nonnal com-
ponent P, representecl by the last term in (34) or (35) :

B {00') Â«

2Wr i
EI

=e-0-

2TFri
tAE

.,.â€ž.2El'(i-i) + S'

(36)
(37)

181. Chain Links. â€” The simplest form of a chain link is

and straight sides as indicated in
Fig. (245), the pull being applied
along the central axis CC If
the theories for determining the
stresses in curved bars and in
to, there would be a sharp change
in the distribution of the stress
at the cross sections 0, 0', etc.,
where the curved and straight
parts join. Assuming that this
change takes place in a very short
distance in the neighborhood of
these cross sections, an approxi-
mate solution for the stresses
maj'^ be obtained as follows:

Letting W = the load and Afo, JIf Â« and M represent the bending moments
at 0, C and D, as before, (Art. 180), we have

M

Jlfo + ^Wl-sina).

(1)

The change in the angle between the tangents at G and C due to the load
W will be zero and, if there is no temperature change and we assume that the
change in the angles between the tangents at and (7, for the straight por-
tion OGy is equal to

g-gr (equation 10, Art. 97),

and that the change in the angle between the tangents at and C, for the
curved portion OC, is equal to

57 (" ^' â– *" ^^^ i â– *" ^' 2) (^^^i^'^ 7, Art. 180),

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THE SPIRAL SPRING 433

we have

and solving for Af o,

Â».-^(^) Â»>

Substituting in (1) and reducing we obtain
and -at C, where a Â» 0,

"-^â– (SIS) <Â«

Substituting in the equation for the greatest normal stress intensity at any
cross section D, we have

. TTsina , W / o + 2ri . \fB n \.

(6)

for a cross section just to the left of C, the greatest normal stress intensity

will be equal to

/ - ^ ( a + 2rA (B n X

â– ^ â– "2(B-A)\a + TrJU n-cj' â€˘â€˘â€˘â€˘â€˘â€˘ ^i)

for a cross section at 0, assuming the distribution of stress to be that in the
curved bar, the greatest normal stress intensity will be equal to

W W / a + 2n ^\(B n\,

^ 2A^2(B-A)Vo + irri )\A ri-cj'

(8)

and for any cross section between and G, assuming the distribution of stress
to be that in the straight bar, the greatest normal stress intensity will be
equal to

â€˘' 2^"^ 2/ VÂ«-|-^i / 2A 2/ Ka+TrrJ' ' ^^^

The preceding equations give the stress intensities at the inside of the
link, where y - â€” Ci. The expressions for the stress intensities at the outside
of the link, at the cross sections D, C, and (7, can be obtained by substitut-
ing ct for â€”Ci in equations (6), (7), (8) and (9).

A comparison of results will show that in a link of circular cross section
the greatest intensity of the tension on the cross section at C will exceedHhat
on the cross section at G, calculated from equation (9) ; and that the greatest
intensity of the normal stress in the link will be the compressive stress at the
inside of the cross section at C.

182. The Spiral Spring. â€” The flat spiral spring of the type
used for driving clocks and other mechanisms is usually made up
of a strip of metal of rectangular cross section, fastened to a pin

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434

APPLIED MECHANICS

at the center and held by a hinge connection at the end A
(Fig. 246). Usually the hinge A is stationary and the spring is
wound up by turning the pin 0, but obviously the stress and the
distortion in the spring would be the same if the pin were iBxed
and the winding were done by pulling the end A around the
central axis.

T

4-

A

Fig. 246.

If P a the force applied at A, in the direction of the tangent to the spiral,
the stress at any section D will be the resultant of a force equal and parallel
to P, acting through the center of gravity of the section, and a couple M Â«
Py, For different sections of the spiral, the couple M will have values vary-
ing from to Pyi and the average value will be nearly equal to

M ^Pa , (1)

The force P, acting at the center of gravity of any section, may be resolved
into a normal and shearing component and the values of each of these com-
ponents will vary, through the length of the spiral, from a minimum zero to
a maximum P.

Except possibly in the portion close to the center, the radius of curvature
will be so large, compared with the depth of the cross section, that the formulas
for the straight beam will give accurate values for the stress intensity.

The greatest bending moment occurs at B and is equal to

A/o = Pj/i = 2 Po (nearly); (2)

and hence the greatest stress intensity due to bending will be equal to

(3)

where - = the section modulus of the spring, tiiis stress intensity being
c

greater than that near the center, where the ciurvature is greater but the

p
bending moment is only half as large. The stress intensity -j , due to the

normal component P at the section B, will evidently be compression but
will be so smaU as to be negligible.

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THE CARRIAGE SPRING 435

The angular displacement a, of the pomt A relative to 0, may be esti-
mated by substituting the average value of M (equation 1) in eqxiation (10)
(Art. 178) and integrating over the entire length of the spiral; this angular
displacement evidently being equal to the change in the angle between the
tangents to the curve at A and B, Hence

M /â€˘â€˘-I , Ml Pal ,..

''^ElJ.^.^^El'^Ei (^)

The resilience of the spring in terms of the above value of a will be equal to

^-"2^-2^-8^^^^ (^^

where / Â» the greatest intensity of stress, p - the radius of gyration of the
cross section and V = the volume of the spring.
If the section is rectangular

Â« = 20^ (Â«>

183. The Carriage Spring. â€” The ordinary type -of carriage
spring is made up of curved bars, or leaves, clamped together at
the center in the manner indicated in Fig. (247).

â– ^^^^^H

Fig. 247.

The flexibility of the spring is increased, without diminishing
the strength, by shortening the successive leaves as shown in the
sketch. The curvature of the leaves in the initial state, before
clamping together, varies; the curvature of the shortest leaf being
the greatest and that of the others being diminished as the length
is increased.

The object of this is to attain the condition that the pressure
between any two successive leaves of the built-up spring shall
be concentrated, as nearly as possible, at the ends of the shorter
of the two leaves.

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436 APPLIED MECHANICS

Assuming that this condition is realized, the pressure at B between the

first and second leaves of the spring, due to a load W at the center C, will be

W *

equal to â€” and the bending moment in the portion OB of the first leaf will

be uniform and equal to

W
M-Y"' (!)

where a = the horizontal distance between A and B.

Therefore, if the portion of the leaf BO is of uniform section and the end
BA is tapered, by varying the breadth alone as indicated, or in any other
way, so as to form as nearly as possible a beam of imiform strength (Art.
85) the greatest fiber stress throughout the leaf will be constant; and hence,
since the radius of curvature is large compared with the depth of a cross sec-
tion, the greatest intensity of stress will be given by the formula

w

Similarly, the pressure between the second and third leaves will be -5-

and, if the horizontal distance between B and C is equal to a, the bending
moment in the portion OC of the second leaf will be uniform and equd to
that in (1). Hence, if the cross section of the portion OC is the same as that
in the first leaf and the end CB is tapered in the same manner as the end BA^
the maximum stress intensity throughout the second leaf will be the same as
in the first leaf.

It will follow, if the successive leaves are designed with the same tapered
ends and the central portions with the same cross section, that the maximum
stress intensity throughout the spring will be the same.

The deflection of the spring may be estimated as follows : Since the straight
portions of all the leaves are assumed to be subjected to the same uniform
bending moment the change in curvature of all will be the same and, since
the initial curvature is small, will be equal in magnitude to

If the ends are tapered, as shown in the sketch, the change in burvature
of the end BA will also be equal to the value represented by (3) (see
equation 2, Case a, Art. 103).

Hence, for the magnitude of the deflection of relative to the supports
A and D, we have

"^ ^ ^EI ^ IQEI^^'^''^ ^'^^^ ^^^

The actual deflection in the spring will differ from this, owing to the effect
the friction between the leaves and the fact that the ideal conditions im-
posed in the theory are not realized in the actual spring.

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PROBLEMS

437

184. Problems â€” Curved Bars.

Problem 1.

Find the moment of resistance of each of the cross sections shown in Fig.
(248) assuming the central axis of the member in each case to be a curve
lying in the plane of symmetry AB and that the radius of curvature of the
central axis is 20" and the greatest normal stress intensity is 5000 lbs. per
sq. in

(a)

(6) (c)

Fig. 248.

(d)^'

Problem 2.

Solve Problem (35) (Art. 134), using the formula for the curved bar.

Problem 3.

Find the load that may be carried by a hook, having the cross section shown
in Fig. (249), the center of curvature of the central axis being in the line AB
produced and 3" to the right of B. Assume/ = 8000 lbs. per sq. in. and that
the line of action of the resultant load passes through the center of ciurvature.

Fig. 249.

Fig. 250.

Fig. 251.

Problem 4.

Find the greatest intensity of the tensile and compressive stress in a hook,
having the cross section shown in Fig. (250), the center of curvature of the
centraJ axis being in the axis of symmetry AB and 3" to the right of B, The
load is 20,000 lbs. and its line of action passes through the center of curvature.

Problem 6.

Find the pressure that may be exerted by a curved rocker arm, having the
cross section shown in Fig. (251), the center of curvature of the central axis

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438 APPLIED MECHANICS

being in the axiB of symmetry AB and 25'' to the right of B, The line of
action of the resultant pressure, exerted by the arm is 30" to the right of B.
The greatest allowable intensity of the normal stress is 6000 lbs. per sq. in.

Problem 6.

Find the stress intensities at the intrados and the extrados at the cross
sections and C of the circular ring (Fig. 243), due to a pull of 6000 lbs. through
the center of the ring, aasuming the cross section to be 1}" diameter and the
radius of the central axis to be equal to 3":

(a) By the first method (Art. 180);

(b) By the second method (Art. 180).

Problem 7.

Find the decrease in the diameter 00' and the increase in the diameter
CC of the ring given in Problem (6) :

(a) By the first method (Art. 180);

(b) By the second method (Art. 180).

Problem 8.

Find the allowable load W for a chain link (Fig. 245) having a cross section
i" diameter, the radius of the central axis at the ends being \\," and the
length of the straight portion U". The greatest allowable intensity of nor*
mal stress is 8000 lbs. per sq. in. Find the greatest intensity of the tensile
stress in the straight portions of the link.

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CHAPTER XII.
ARCHES AND CATENARIES.

186. The Arch. â€” An arch may be defined as a member, or
structure, whose central axis is a plane curve which is attached
through hinges, or otherwise, to fixed or unyielding supports and
is usually designed in such a manner that the bending moments
due to transverse loading are offset, as largely as possible, by the
moments of the reactions" at the supports. In very special cases
bending may be eUminated entirely, the resultant of the stress
at every cross section coinciding with the central axis. The arch
may be solidy having a cross section similar to that of a beam or
built-up girder, or, it may be a braced arch, made up of tension
and compression members Uke a simple truss. We shall consider
the methods of determining the stress in the soUd type only.

The soUd arch may be treated as a curved bar, subjected to the
action of external forces acting in the plane of curvature and, when
the external forces are known, the bending moment and the normal
force, or Oirustj acting through the center of gravity, at any cross
section can be found by the method in Art. (176). In ordinary
cases, the radius of curvature is so large, compared with the
dimensions of the cross section, that the stress intensity at any
point can be calculated with sufficient accuracy by use of the
formula for the straight bar,

/ = fÂ±^(Art.l26),. ...... (1)

rather than the more complex formulas in Art. (176).

Similarly, the displacements at any point can be accurately
determined by use of the approximate formulas (19-21) (Art. 178)
and in many cases the more approximate equations (8-10), of
the same article, are sufficiently accurate.

Three cases will be considered, involving three diflPerent ways of
supporting the arch. The determination of the stresses and dis-
placements in two of these cases is somewhat difficult, owing to the
fact that the reactions at the points of support cannot be deter-
mined from the statical conditions of equiUbrium alone.

439

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440

APPLIED MECHANICS

Case I. Three Hinged Arch. â€” In tbis case, the arch is made up of two
parts, or ribSy supported on hinges at O and E and connected with a third
hinge at C (Fig. 252). When the external loads TTi, Wi, etc., are known, the
horizontal and vertical components Hoj Voj etc., of the reactions at the hinges
can be easily computed by applying the statical conditions of equilibrium.
The equation for the bending moment at any point D, whose coordinates
with respect to horizontal and vertical axes through are (x, y), may then be
written

M ^Hoy + ZWa - FoX, (2)

where XWa Â« the sum of the moments about D of the forces acting between
D and 0, moments tending to increase the curvature being taken as positive.
Resolving the loads TTi, TTj, etc., into H and V components, /fi, Fi, JTj, Vj,
etc., the equation for the thrust at the center of gravity of the cross section
at D may be written

P = SF(cosa) +2F(sina), (3)

where XH = Ho + Hi '\- etc., S7 = 7Â© â€” Fi â€” etc., the sununation being
taken between and D, and a = the angle between the tangent at D and the
horizontal axis OX.

Similarly, the magnitude of the shearing force at D will be equal to

5 = SH(sina) - SF(cosa) (4)

Having the values of M and P, the stress intensity at any point in the cross
section D may be calculated by use of equation (1).

The foregoing solution would evidently apply equally as well if the hinges
and E were not on the same horizontal level as shown (Fig. 252), the H

FiQ. 252.

and V components in such a case being taken to represent the components
respectively parallel and perpendicular to the axis OX, through the hinges
and E.

If the arch were subjected to a distribiUed load, a solution could evidently
be made by dividing the load into small parts AW, resolving each increment
into H and V components and making the smnmations indicated in (2) and
(3) as before.

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TWO HINGED ARCH

441

Casb II. Two Hinged Arch, First Solution, â€” In this case the arch con-
sists of a single rib held by hinges at the supports and C (Fig. 253). When
the loads Wi, TTi, etc., are known, the components Vo and Ve at the hinges
and C, can be calculated by use of the statical conditions of equilibrium;
but these conditions will fail to give a solution for the values of Ho and HoÂ»
The component Ho can be determined, however, on the assumption that the
supports are rigidly fixed and hence the displacement of the hinge 0, relative
to C, is equal to zero. Equation (2) may be written in the form

M = /foy-fX, (5)

where

^' (6)

which is the part of M which can be determined from the statical conditions
of equilibrium alone.

/

FlQ. 253.

ix

â– X*.

/â€˘z-2 'tfl rz-l K

Substituting this value of M in equation (8), (Art. 178), we have, on the
basis of the above asstmiption,

-Qâ€” gr-^^ = M-o:^^H-o^'

By solving (7) the value of Ho can be obtained. Except in comparatively
simple cases, however, the solution is complicated and hence the following
Olustratfons are restricted to the arch of uniform cross section and material,
in which the central axis is the arc of a circle, the end hinges are on the same
level and the loads are vertical (Fig. 253). In such a case

ÂŁro= -

fKyda
ft/^ds

(8)

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442 APPLIED MECHANICS

Having the values of H^ and Fo, the bending moment, the thrust, and the
maximum normal stress intensity, at any croes section, can be found from
equations (2), (3) and (1) in the same manner as in the three hinged arch.

Concentrated Loads, â€” We will let I Â« the span OC^ n Â» the radius of cur-
vature of the central axis, oeo Â« the angle between OX and the tangent to
the central axis at 0, a Â» the angle between OX and the tangent to the
central axis at any point D, whose codrdinates are (x, y), and 0i, ^, etc.,
equal the angles between OX and the tangents to the central axis at the
points of intersection with the loads Wu Wtt etc., which are located at dis-
tances c{], diy etc., from the axis OF. Then I Â» 2ri8inao, x Â» n (sinoo

â€” sin a), y = fi (cos a â€” cos oo), d\ =Â» r\ (sin oeo â€” sin ^i), (x â€” d\) = n (sin Bi

â€” sin a), etc., (} â€” di) = n (sin oo H- sin ^0, ds = ^ ndafdx = â€” n cos a da
and dy = â€” n sin a da.

If the load Wi were the only load on the arch, equation (6), for values of x
from to di, would take the form

K' = â€” Vo'x = â€” Fo'fi (sin<xo â€” sina) (9)

and for values of x from di to 2,

K' = TTi (x - di) - Fo'x = TTifi (sin^i - sina) - FoVi (sinoo - sin a), (10)

the value of 7o' being

jr, Wx (I - di) PTi (sinoo + sin gi) .â€ž.

Ko =* J = TTâ€”- VAl/

I 2smoo

Substituting the values of X' in (7), observing that when x = i, a = â€” oÂ©,

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